Download Instructor Solutions Manual for Physics by Halliday, Resnick, and

Instructor Solutions Manual
for
Physics
by
Halliday, Resnick, and Krane
Paul Stanley
Beloit College
Volume 1: Chapters 1-24
A Note To The Instructor...
The solutions here are somewhat brief, as they are designed for the instructor, not for the student.
Check with the publishers before electronically posting any part of these solutions; website, ftp, or
server access must be restricted to your students.
I have been somewhat casual about subscripts whenever it is obvious that a problem is one
dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution.
Although this does not change the validity of the answer, it will sometimes obfuscate the approach
if viewed by a novice.
There are some traditional formula, such as
2
vx2 = v0x
+ 2ax x,
which are not used in the text. The worked solutions use only material from the text, so there may
be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an
easier approach existed. But if it was not in the text, I did not use it here.
I also tried to avoid reinventing the wheel. There are some exercises and problems in the text
which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer
you to the previous solution.
I adopt a different approach for rounding of significant figures than previous authors; in particular, I usually round intermediate answers. As such, some of my answers will differ from those in
the back of the book.
Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manual
with considerably more detail and, when appropriate, include discussion on any physical implications
of the answer. These student solutions carefully discuss the steps required for solving problems, point
out the relevant equation numbers, or even specify where in the text additional information can be
found. When two almost equivalent methods of solution exist, often both are presented. You are
encouraged to refer students to the Student’s Solution Manual for these exercises and problems.
However, the material from the Student’s Solution Manual must not be copied.
Paul Stanley
Beloit College
[email protected]
1
E1-1 (a) Megaphones; (b) Microphones; (c) Decacards (Deck of Cards); (d) Gigalows (Gigolos);
(e) Terabulls (Terribles); (f) Decimates; (g) Centipedes; (h) Nanonanettes (?); (i) Picoboos (Peek-aBoo); (j) Attoboys (’atta boy); (k) Two Hectowithits (To Heck With It); (l) Two Kilomockingbirds
(To Kill A Mockingbird, or Tequila Mockingbird).
E1-2 (a) $36, 000/52 week = $692/week. (b) $10, 000, 000/(20 × 12 month) = $41, 700/month. (c)
30 × 109 /8 = 3.75 × 109 .
E1-3
Multiply out the factors which make up a century.
365 days
24 hours
60 minutes
1 century = 100 years
1 year
1 day
1 hour
This gives 5.256 × 107 minutes in a century, so a microcentury is 52.56 minutes.
The percentage difference from Fermi’s approximation is (2.56 min)/(50 min) × 100% or 5.12%.
E1-4 (3000 mi)/(3 hr) = 1000 mi/timezone-hour. There are 24 time-zones, so the circumference
is approximately 24 × 1000 mi = 24, 000 miles.
E1-5 Actual number of seconds in a year is
24 hr
60 min
60 s
(365.25 days)
= 3.1558 × 107 s.
1 day
1 hr
1 min
The percentage error of the approximation is then
3.1416 × 107 s − 3.1558 × 107 s
= −0.45 %.
3.1558 × 107 s
E1-6 (a) 10−8 seconds per shake means 108 shakes per second. There are
365 days
24 hr
60 min
60 s
= 3.1536 × 107 s/year.
1 year
1 day
1 hr
1 min
This means there are more shakes in a second.
(b) Humans have existed for a fraction of
106 years/1010 years = 10−4 .
That fraction of a day is
10−4 (24 hr)
60 min
1 hr
60 s
1 min
= 8.64 s.
E1-7 We’ll assume, for convenience only, that the runner with the longer time ran exactly one
mile. Let the speed of the runner with the shorter time be given by v1 , and call the distance actually
ran by this runner d1 . Then v1 = d1 /t1 . Similarly, v2 = d2 /t2 for the other runner, and d2 = 1 mile.
We want to know when v1 > v2 . Substitute our expressions for speed, and get d1 /t1 > d2 /t2 .
Rearrange, and d1 /d2 > t1 /t2 or d1 /d2 > 0.99937. Then d1 > 0.99937 mile × (5280 feet/1 mile) or
d1 > 5276.7 feet is the condition that the first runner was indeed faster. The first track can be no
more than 3.3 feet too short to guarantee that the first runner was faster.
2
E1-8 We will wait until a day’s worth of minutes have been gained. That would be
60 min
(24 hr)
= 1440 min.
1 hr
The clock gains one minute per day, so we need to wait 1,440 days, or almost four years. Of course,
if it is an older clock with hands that only read 12 hours (instead of 24), then after only 720 days
the clock would be correct.
E1-9
First find the “logarithmic average” by
log tav
=
=
=
1
log(5 × 1017 ) + log(6 × 10−15 ) ,
2
1
log 5 × 1017 × 6 × 10−15 ,
2
√
1
log 3000 = log
3000 .
2
Solve, and tav = 54.8 seconds.
E1-10 After 20 centuries the day would have increased in length by a total of 20 × 0.001 s = 0.02 s.
The cumulative effect would by the product of the average increase and the number of days; that
average is half of the maximum, so the cumulative effect is 12 (2000)(365)(0.02 s) = 7300 s. That’s
about 2 hours.
E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
the Moon has farther to go to complete a phase. In 27.3 days the Moon may have orbited through
360◦ , but since the Earth moved through (27.3/365) × 360◦ = 27◦ the Moon needs to move 27◦
farther to catch up. That will take (27◦ /360◦ ) × 27.3 days = 2.05 days, but in that time the Earth
would have moved on yet farther, and the moon will need to catch up again. How much farther?
(2.05/365) × 360◦ = 2.02◦ which means (2.02◦ /360◦ ) × 27.3 days = 0.153 days. The total so far is
2.2 days longer; we could go farther, but at our accuracy level, it isn’t worth it.
E1-12 (1.9 m)(3.281 ft/1.000 m) = 6.2 ft, or just under 6 feet, 3 inches.
E1-13 (a) 100 meters = 328.1 feet (Appendix G), or 328.1/3 = 10.9 yards. This is 28 feet longer
than 100 yards, or (28 ft)(0.3048 m/ft) = 8.5 m. (b) A metric mile is (1500 m)(6.214×10−4 mi/m) =
0.932 mi. I’d rather run the metric mile.
E1-14 There are
300, 000 years
365.25 days
1 year
24 hr
1 day
60 min
1 hr
60 s
1 min
= 9.5 × 1012 s
that will elapse before the cesium clock is in error by 1 s. This is almost 1 part in 1013 . This kind
of accuracy with respect to 2572 miles is
1609 m
10−13 (2572 mi)
= 413 nm.
1 mi
3
E1-15 The volume of Antarctica is approximated by the area of the base time the height; the
area of the base is the area of a semicircle. Then
1 2
V = Ah =
πr h.
2
The volume is
V
=
=
1
(3.14)(2000 × 1000 m)2 (3000 m) = 1.88 × 1016 m3
2
3
100 cm
16
3
1.88 × 10 m ×
= 1.88 × 1022 cm3 .
1m
E1-16 The volume is (77×104 m2 )(26 m) = 2.00×107 m3 . This is equivalent to
(2.00×107 m3 )(10−3 km/m)3 = 0.02 km3 .
E1-17 (a) C = 2πr = 2π(6.37 × 103 km) = 4.00 × 104 km. (b) A = 4πr2 = 4π(6.37 × 103 km)2 =
5.10 × 108 km. (c) V = 34 π(6.37 × 103 km)3 = 1.08 × 1012 km3 .
E1-18 The conversions: squirrel, 19 km/hr(1000 m/km)/(3600 s/hr) = 5.3 m/s;
rabbit, 30 knots(1.688ft/s/knot)(0.3048 m/ft) = 15 m/s;
snail, 0.030 mi/hr(1609 m/mi)/(3600 s/hr) = 0.013 m/s;
spider, 1.8 ft/s(0.3048 m/ft) = 0.55 m/s;
cheetah, 1.9 km/min(1000 m/km)/(60 s/min) = 32 m/s;
human, 1000 cm/s/(100 cm/m) = 10 m/s;
fox, 1100 m/min/(60 s/min) = 18 m/s;
lion, 1900 km/day(1000 m/km)/(86, 400 s/day) = 22 m/s.
The order is snail, spider, squirrel, human, rabbit, fox, lion, cheetah.
E1-19
Then
One light-year is the distance traveled by light in one year, or (3 × 108 m/s) × (1 year).
mi
light-year
1609 m
1 hr
100 year
,
19, 200
hr (3 × 108 m/s) × (1 year)
1 mi
3600 s
1 century
which is equal to 0.00286 light-year/century.
E1-20 Start with the British units inverted,
gal
231 in3
1.639 × 10−2 L
mi
= 7.84 × 10−2 L/km.
30.0 mi
gal
1.609 km
in3
E1-21 (b) A light-year is
(3.00 × 105 km/s)
3600 s
1 hr
24 hr
1 day
(365 days) = 9.46 × 1012 km.
A parsec is
1.50 × 108 km
0◦ 00 100
360◦
2π rad
=
1.50 × 108 km
(1/3600)◦
360◦
2π rad
= 3.09 × 1013 km.
(a) (1.5 × 108 km)/(3.09 × 1013 km/pc = 4.85 × 10−6 pc. (1.5 × 108 km)/(9.46 × 1012 km/ly) =
1.59 × 10−5 ly.
4
E1-22 First find the “logarithmic average” by
log dav
=
=
=
1
log(2 × 1026 ) + log(1 × 10−15 ) ,
2
1
log 2 × 1026 × 1 × 10−15 ,
2
p
1
log 2 × 1011 = log
2 × 1011 .
2
Solve, and dav = 450 km.
E1-23
atoms.
The number of atoms is given by (1 kg)/(1.00783 × 1.661 × 10−27 kg), or 5.974 × 1026
E1-24 (a) (2 × 1.0 + 16)u(1.661 × 10−27 kg) = 3.0 × 10−26 kg.
(b) (1.4 × 1021 kg)/(3.0 × 10−26 kg) = 4.7 × 1046 molecules.
E1-25 The coffee in Paris costs $18.00 per kilogram, or
0.4536 kg
−1
$18.00 kg
= $8.16 lb−1 .
1 lb
It is cheaper to buy coffee in New York (at least according to the physics textbook, that is.)
E1-26 The room volume is (21 × 13 × 12)ft3 (0.3048 m/ft)3 = 92.8 m3 . The mass contained in the
room is
(92.8 m3 )(1.21 kg/m3 ) = 112 kg.
E1-27 One mole of sugar cubes would have a volume of NA × 1.0 cm3 , where NA is the Avogadro
√
constant. Since the volume of a cube is equal to the length cubed, V = l3 , then l = 3 NA cm
= 8.4 × 107 cm.
E1-28 The number of seconds in a week is 60 × 60 × 24 × 7 = 6.05 × 105 . The “weight” loss per
second is then
(0.23 kg)/(6.05 × 105 s) = 3.80 × 10−1 mg/s.
E1-29 The definition of the meter was wavelengths per meter; the question asks for meters per
wavelength, so we want to take the reciprocal. The definition is accurate to 9 figures, so the reciprocal
should be written as 1/1, 650, 763.73 = 6.05780211 × 10−7 m = 605.780211 nm.
E1-30 (a) 37.76 + 0.132 = 37.89. (b) 16.264 − 16.26325 = 0.001.
E1-31 The easiest approach is to first solve Darcy’s Law for K, and then substitute the known
SI units for the other quantities. Then
m3 (m)
VL
K=
has units of
AHt
(m2 ) (m) (s)
which can be simplified to m/s.
5
E1-32 The Planck length, lP , is found from
[lP ] = [ci ][Gj ][hk ],
L = (LT−1 )i (L3 T−2 M−1 )j (ML2 T−1 )k ,
= Li+3j+2k T−i−2j−k M−j+k .
Equate powers on each side,
L: 1 = i + 3j + 2k,
T: 0 = −i − 2j − k,
M: 0 = −j + k.
Then j = k, and i = −3k, and 1 = 2k; so k = 1/2, j = 1/2, and i = −3/2. Then
[lP ]
= [c−3/2 ][G1/2 ][h1/2 ],
= (3.00 × 108 m/s)−3/2 (6.67 × 10−11 m3 /s2 · kg)1/2 (6.63 × 10−34 kg · m2 /s)1/2 ,
= 4.05 × 10−35 m.
E1-33 The Planck mass, mP , is found from
[mP ] = [ci ][Gj ][hk ],
M = (LT−1 )i (L3 T−2 M−1 )j (ML2 T−1 )k ,
= Li+3j+2k T−i−2j−k M−j+k .
Equate powers on each side,
L: 0 = i + 3j + 2k,
T: 0 = −i − 2j − k,
M: 1 = −j + k.
Then k = j + 1, and i = −3j − 1, and 0 = −1 + 2k; so k = 1/2, and j = −1/2, and i = 1/2. Then
[mP ]
= [c1/2 ][G−1/2 ][h1/2 ],
= (3.00 × 108 m/s)1/2 (6.67 × 10−11 m3 /s2 · kg)−1/2 (6.63 × 10−34 kg · m2 /s)1/2 ,
= 5.46 × 10−8 kg.
P1-1 There are 24 × 60 = 1440 traditional minutes in a day. The conversion plan is then fairly
straightforward
1440 trad. min
822.8 dec. min
= 1184.8 trad. min.
1000 dec. min
This is traditional minutes since midnight, the time in traditional hours can be found by dividing
by 60 min/hr, the integer part of the quotient is the hours, while the remainder is the minutes. So
the time is 19 hours, 45 minutes, which would be 7:45 pm.
P1-2 (a) By similar triangles, the ratio of the distances is the same as the ratio of the diameters—
390:1.
(b) Volume is proportional to the radius (diameter) cubed, or 3903 = 5.93 × 107 .
(c) 0.52◦ (2π/360◦ ) = 9.1 × 10−3 rad. The diameter is then (9.1 × 10−3 rad)(3.82 × 105 km) =
3500 km.
6
P1-3 (a) The circumference of the Earth is approximately 40,000 km; 0.5 seconds of an arc is
0.5/(60 × 60 × 360) = 3.9 × 10−7 of a circumference, so the north-south error is ±(3.9 × 10−7 )(4 ×
107 m) = ±15.6 m. This is a range of 31 m.
(b) The east-west range is smaller, because the distance measured along a latitude is smaller than
the circumference by a factor of the cosine of the latitude. Then the range is 31 cos 43.6◦ = 22 m.
(c) The tanker is in Lake Ontario, some 20 km off the coast of Hamlin?
P1-4 Your position is determined by the time it takes for your longitude to rotate ”underneath”
the sun (in fact, that’s the way longitude was measured originally as in 5 hours west of the Azores...)
the rate the sun sweep over at equator is 25,000 miles/86,400 s = 0.29 miles/second. The correction
factor because of latitude is the cosine of the latitude, so the sun sweeps overhead near England at
approximately 0.19 mi/s. Consequently a 30 mile accuracy requires an error in time of no more than
(30 mi)/(0.19 mi/s) = 158 seconds.
Trip takes about 6 months, so clock accuracy needs to be within (158 s)/(180 day) = 1.2 seconds/day.
(b) Same, except 0.5 miles accuracy requires 2.6 s accuracy, so clock needs to be within 0.007
s/day!
P1-5 Let B be breaths/minute while sleeping. Each breath takes in (1.43 g/L)(0.3 L) = 0.429 g;
and lets out (1.96 g/L)(0.3 L) = 0.288 g. The net loss is 0.141 g. Multiply by the number of breaths,
(8 hr)(60 min./hr)B(0.141 g) = B(67.68 g). I’ll take a short nap, and count my breaths, then finish
the problem.
I’m back now, and I found my breaths to be 8/minute. So I lose 541 g/night, or about 1 pound.
P1-6 The mass of the water is (1000 kg/m3 )(5700 m3 ) = 5.7 × 106 kg. The rate that water leaks
drains out is
(5.7 × 106 kg)
= 132 kg/s.
(12 hr)(3600 s/hr)
P1-7 Let the radius of the grain be given by rg . Then the surface area of the grain is Ag = 4πrg2 ,
and the volume is given by Vg = (4/3)πrg3 .
If N grains of sand have a total surface area equal to that of a cube 1 m on a edge, then
N Ag = 6 m2 . The total volume Vt of this number of grains of sand is N Vg . Eliminate N from these
two expressions and get
(6 m2 )
(6 m2 )rg
Vt = N V g =
Vg =
.
Ag
3
Then Vt = (2 m2 )(50 × 10−6 m) = 1 × 10−4 m3 .
The mass of a volume Vt is given by
2600 kg
−4
3
1 × 10 m
= 0.26 kg.
1 m3
P1-8 For a cylinder V = πr2 h, and A = 2πr2 + 2πrh. We want to minimize A with respect to
changes in r, so
dA
d
V
=
2πr2 + 2πr 2 ,
dr
dr
πr
V
= 4πr − 2 2 .
r
Set this equal to zero; then V = 2πr3 . Notice that h = 2r in this expression.
7
P1-9
(a) The volume per particle is
(9.27 × 10−26 kg)/(7870 kg/m3 ) = 1.178 × 10−28 m3 .
The radius of the corresponding sphere is
r
−28 m3 )
3 3(1.178 × 10
r=
= 1.41 × 10−10 m.
4π
Double this, and the spacing is 282 pm.
(b) The volume per particle is
(3.82 × 10−26 kg)/(1013 kg/m3 ) = 3.77 × 10−29 m3 .
The radius of the corresponding sphere is
r
−29 m3 )
3 3(3.77 × 10
r=
= 2.08 × 10−10 m.
4π
Double this, and the spacing is 416 pm.
P1-10
(a) The area of the plate is (8.43 cm)(5.12 cm) = 43.2 cm2 . (b) (3.14)(3.7 cm)2 = 43 cm2 .
8
E2-1 Add the vectors as is shown in Fig. 2-4. If ~a has length a = 4 m and ~b has length b = 3 m
then the sum is given by ~s. The cosine law can be used to find the magnitude s of ~s,
s2 = a2 + b2 − 2ab cos θ,
where θ is the angle between sides a and b in the figure.
(a) (7 m)2 = (4 m)2 + (3 m)2 − 2(4 m)(3 m) cos θ, so cos θ = −1.0, and θ = 180◦ . This means
that ~a and ~b are pointing in the same direction.
(b) (1 m)2 = (4 m)2 + (3 m)2 − 2(4 m)(3 m) cos θ, so cos θ = 1.0, and θ = 0◦ . This means that
~a and ~b are pointing in the opposite direction.
(c) (5 m)2 = (4 m)2 + (3 m)2 − 2(4 m)(3 m) cos θ, so cos θ = 0, and θ = 90◦ . This means that ~a
and ~b are pointing at right angles to each other.
E2-2 (a) Consider the figures below.
(b) Net displacement is 2.4 km west, (5.2 − 3.1 = 2.1) km south. A bird would fly
p
2.42 + 2.12 km = 3.2 km.
E2-3 Consider the figure below.
a
a+b
b
-b
a-b
a
◦
◦
E2-4 (a) The components
) = −6.98ĵ.
p are (7.34) cos(252 ) = −2.27î and (7.34) sin(252
−1
(b) The magnitude is (−25)2 + (43)2 = 50; the direction is θ = tan (43/ − 25) = 120◦ . We
did need to choose the correct quadrant.
E2-5
The components are given by the trigonometry relations
O = H sin θ = (3.42 km) sin 35.0◦ = 1.96 km
and
A = H cos θ = (3.42 km) cos 35.0◦ = 2.80 km.
9
The stated angle is measured from the east-west axis, counter clockwise from east. So O is measured
against the north-south axis, with north being positive; A is measured against east-west with east
being positive.
Since her individual steps are displacement vectors which are only north-south or east-west, she
must eventually take enough north-south steps to equal 1.96 km, and enough east-west steps to
equal 2.80 km. Any individual step can only be along one or the other direction, so the minimum
total will be 4.76 km.
E2-6 Let ~rf = 124î km and ~ri = (72.6î + 31.4ĵ) km. Then the ship needs to travel
∆~r = ~rf − ~ri = (51.4î + 31.4ĵ) km.
Ship needs to travel
of north.
√
51.42 + 31.42 km = 60.2 km in a direction θ = tan−1 (31.4/51.4) = 31.4◦ west
E2-7 (a) In unit vector notation we need only add the components; ~a + ~b = (5î+3ĵ)+(−3î+2ĵ) =
(5 − 3)î + (3 + 2)ĵ = 2î + 5ĵ.
q
√
(b) If we define ~c = ~a + ~b and write the magnitude of ~c as c, then c = c2x + c2y = 22 + 52 =
5.39. The direction is given by tan θ = cy /cx which gives an angle of 68.2◦ , measured counterclockwise from the positive x-axis.
E2-8 (a) ~a + ~b = (4 − 1)î + (−3 + 1)ĵ + (1 + 4)k̂ = 3î − 2ĵ + 5k̂.
(b) ~a − ~b = (4 − −1)î + (−3 − 1)ĵ + (1 − 4)k̂ = 5î − 4ĵ − 3k̂.
(c) Rearrange, and ~c = ~b − ~a, or ~b − ~a = (−1 − 4)î + (1 − −3)ĵ + (4 − 1)k̂ = −5î + 4ĵ + 3k̂.
p
E2-9 (a) The magnitude of ~a is 4.02 + (−3.0)2 = 5.0; the direction is θ = tan−1 (−3.0/4.0)
323◦ .
√
(b) The magnitude of ~b is 6.02 + 8.03 = 10.0; the direction is θ = tan−1 (6.0/8.0) = 36.9◦ .
~
î + (−3.0 + 8.0)ĵ. The magnitude of ~a + ~b
p (c) The resultant vector is ~a + b = (4.0 + 6.0)
−1
2
2
(10.0) + (5.0) = 11.2; the direction is θ = tan (5.0/10.0) = 26.6◦ .
~
+ (−3.0 − 8.0)ĵ. The magnitude of ~a − ~b
p (d) The resultant vector is ~a − b = (4.0 − 6.0)î −1
2
2
(−2.0) + (−11.0) = 11.2; the direction is θ = tan (−11.0/ − 2.0) = 260◦ .
~
î + (8.0 − −3.0)ĵ. The magnitude of ~b − ~a
p (e) The resultant vector is b − ~a = (6.0 − 4.0)
−1
2
2
(2.0) + (11.0) = 11.2; the direction is θ = tan (11.0/2.0) = 79.7◦ .
=
is
is
is
E2-10 (a) Find components of ~a; ax = (12.7) cos(28.2◦ ) = 11.2, ay = (12.7) sin(28.2◦ ) = 6.00.
Find components of ~b; bx = (12.7) cos(133◦ ) = −8.66, by = (12.7) sin(133◦ ) = 9.29. Then
~r = ~a + ~b = (11.2 − 8.66)î + (6.00 + 9.29)ĵ = 2.54î + 15.29ĵ.
√
(b) The magnitude of ~r is 2.542 + 15.292 = 15.5.
(c) The angle is θ = tan−1 (15.29/2.54) = 80.6◦ .
E2-11 Consider the figure below.
10
E2-12 Consider the figure below.
E2-13 Our axes will be chosen so that î points toward 3 O’clock and ĵ points toward 12 O’clock.
(a)
The two relevant positions are ~ri = (11.3 cm)î and ~rf = (11.3 cm)ĵ. Then
∆~r = ~rf − ~ri
= (11.3 cm)ĵ − (11.3 cm)î.
(b)
The two relevant positions are now ~ri = (11.3 cm)ĵ and ~rf = (−11.3 cm)ĵ. Then
∆~r = ~rf − ~ri
= (11.3 cm)ĵ − (−11.3 cm)ĵ
=
(22.6 cm)ĵ.
(c)
The two relevant positions are now ~ri = (−11.3 cm)ĵ and ~rf = (−11.3 cm)ĵ. Then
∆~r = ~rf − ~ri
= (−11.3 cm)ĵ − (−11.3 cm)ĵ
=
(0 cm)ĵ.
E2-14 (a) The components of ~r1 are
r1x = (4.13 m) cos(225◦ ) = −2.92 m
and
r1y = (4.13 m) sin(225◦ ) = −2.92 m.
11
The components of ~r2 are
r1x = (5.26 m) cos(0◦ ) = 5.26 m
and
r1y = (5.26 m) sin(0◦ ) = 0 m.
The components of ~r3 are
r1x = (5.94 m) cos(64.0◦ ) = 2.60 m
and
r1y = (5.94 m) sin(64.0◦ ) = 5.34 m.
(b) The resulting displacement is
h
i
(−2.92 + 5.26 + 2.60)î + (−2.92 + 0 + 5.34)ĵ m = (4.94î + 2.42ĵ) m.
√
(c) The magnitude of the resulting displacement is 4.942 + 2.422 m = 5.5 m. The direction of
the resulting displacement is θ = tan−1 (2.42/4.94) = 26.1◦ . (d) To bring the particle back to the
starting point we need only reverse the answer to (c); the magnitude will be the same, but the angle
will be 206◦ .
E2-15 The components of the initial position are
r1x = (12, 000 ft) cos(40◦ ) = 9200 ft
and
r1y = (12, 000 ft) sin(40◦ ) = 7700 ft.
The components of the final position are
r2x = (25, 8000 ft) cos(163◦ ) = −24, 700 ft
and
r2y = (25, 800 ft) sin(163◦ ) = 7540 ft.
The displacement is
h
i
~r = ~r2 − ~r1 = (−24, 700 − 9, 200)î + (7, 540 − 9, 200)ĵ) = (−33900î − 1660ĵ) ft.
E2-16 (a) The displacement vector is ~r = (410îp
− 820ĵ) mi, where positive x is east and positive
y is north. The magnitude of the displacement is (410)2 + (−820)2 mi = 920 mi. The direction is
θ = tan−1 (−820/410) = 300◦ .
(b) The average velocity is the displacement divided by the total time, 2.25 hours. Then
~vav = (180î − 360ĵ) mi/hr.
(c) The average speed is total distance over total time, or (410 + 820)/(2.25) mi/hr = 550 mi/hr.
12
E2-17
(a) Evaluate ~r when t = 2 s.
~r =
3
4
[(2 m/s )t3 − (5 m/s)t]î + [(6 m) − (7 m/s )t4 ]ĵ
3
4
=
[(2 m/s )(2 s)3 − (5 m/s)(2 s)]î + [(6 m) − (7 m/s )(2 s)4 ]ĵ
=
[(16 m) − (10 m)]î + [(6 m) − (112 m)]ĵ
=
[(6 m)]î + [−(106 m)]ĵ.
(b) Evaluate:
~v =
d~r
dt
3
4
3
4
=
[(2 m/s )3t2 − (5 m/s)]î + [−(7 m/s )4t3 ]ĵ
=
[(6 m/s )t2 − (5 m/s)]î + [−(28 m/s )t3 ]ĵ.
Into this last expression we now evaluate ~v(t = 2 s) and get
~v
3
4
=
[(6 m/s )(2 s)2 − (5 m/s)]î + [−(28 m/s )(2 s)3 ]ĵ
=
[(24 m/s) − (5 m/s)]î + [−(224 m/s)]ĵ
=
[(19 m/s)]î + [−(224 m/s)]ĵ,
for the velocity ~v when t = 2 s.
(c) Evaluate
~a =
d~v
dt
3
4
= [(6 m/s )2t]î + [−(28 m/s )3t2 ]ĵ
=
3
4
[(12 m/s )t]î + [−(84 m/s )t2 ]ĵ.
Into this last expression we now evaluate ~a(t = 2 s) and get
~a =
=
3
4
[(12 m/s )(2 s)]î + [−(84 m/s )(2 2)2 ]ĵ
2
2
[(24 m/s )]î + [−(336 m/s )]ĵ.
E2-18 (a) Let ui point north, ĵ point east, and k̂ point up. The displacement is (8.7î + 9.7ĵ +
2.9k̂) km. The average velocity is found by dividing each term by 3.4 hr; then
~vav = (2.6î + 2.9ĵ + 0.85) km/hr.
√
The magnitude of the average velocity is 2.62 + √
2.92 + 0.852 km/hr = 4.0 km/hr.
(b) The horizontal velocity has a magnitude of 2.62 + 2.92 km/hr = 3.9 km/hr. The angle with
the horizontal is given by θ = tan−1 (0.85/3.9) = 13◦ .
E2-19 (a) The derivative of the velocity is
~a = [(6.0 m/s2 ) − (8.0 m/s3 )t]î
so the acceleration at t = 3 s is ~a = (−18.0 m/s2 )î. (b) The acceleration is zero when (6.0 m/s2 ) −
(8.0 m/s3 )t = 0, or t = 0.75 s. (c) The velocity is neverpzero; there is no way to “cancel” out the
y component. (d) The speed equals 10 m/s when 10 = vx2 + 82 , or vx = ±6.0 m/s. This happens
when (6.0 m/s2 ) − (8.0 m/s3 )t = ±6.0 m/s, or when t = 0 s.
13
E2-20 If v is constant then so is v 2 = vx2 + vy2 . Take the derivative;
d
d
vx + 2vy vy = 2(vx ax + vy ay ).
dt
dt
But if the value is constant the derivative is zero.
2vx
E2-21 Let the actual flight time, as measured by the passengers, be T . There is some time
difference between the two cities, call it ∆T = Namulevu time - Los Angeles time. The ∆T will be
positive if Namulevu is east of Los Angeles. The time in Los Angeles can then be found from the
time in Namulevu by subtracting ∆T .
The actual time of flight from Los Angeles to Namulevu is then the difference between when the
plane lands (LA times) and when the plane takes off (LA time):
T
= (18:50 − ∆T ) − (12:50)
= 6:00 − ∆T,
where we have written times in 24 hour format to avoid the AM/PM issue. The return flight time
can be found from
T
= (18:50) − (1:50 − ∆T )
= 17:00 + ∆T,
where we have again changed to LA time for the purpose of the calculation.
(b) Now we just need to solve the two equations and two unknowns.
17:00 + ∆T
2∆T
∆T
= 6:00 − ∆T
= 6:00 − 17:00
= −5:30.
Since this is a negative number, Namulevu is located west of Los Angeles.
(a) T = 6:00 − ∆T = 11 : 30, or eleven and a half hours.
(c) The distance traveled by the plane is given by d = vt = (520 mi/hr)(11.5 hr) = 5980 mi.
We’ll draw a circle around Los Angeles with a radius of 5980 mi, and then we look for where it
intersects with longitudes that would belong to a time zone ∆T away from Los Angeles. Since the
Earth rotates once every 24 hours and there are 360 longitude degrees, then each hour corresponds
to 15 longitude degrees, and then Namulevu must be located approximately 15◦ × 5.5 = 83◦ west of
Los Angeles, or at about longitude 160 east. The location on the globe is then latitude 5◦ , in the
vicinity of Vanuatu.
When this exercise was originally typeset the times for the outbound and the inbound flights were
inadvertently switched. I suppose that we could blame this on the airlines; nonetheless, when the answers
were prepared for the back of the book the reversed numbers put Namulevu east of Los Angeles. That would
put it in either the North Atlantic or Brazil.
E2-22 There is a three hour time zone difference. So the flight is seven hours long, but it takes
3 hr 51 min for the sun to travel same distance. Look for when the sunset distance has caught up
with plane:
dsunset = dplane ,
v sunset (t − 1:35) = v plane t,
(t − 1:35)/3:51 = t/7:00,
so t = 3:31 into flight.
14
E2-23 The distance is
d = vt = (112 km/hr)(1 s)/(3600 s/hr) = 31 m.
E2-24 The time taken for the ball to reach the plate is
t=
(18.4 m)
d
=
(3600 s/hr)/(1000 m/km) = 0.414 s.
v
(160 km/hr)
E2-25 Speed is distance traveled divided by time taken; this is equivalent to the inverse of the
slope of the line in Fig. 2-32. The line appears to pass through the origin and through the point
(1600 km, 80 × 106 y), so the speed is v = 1600 km/80 × 106 y= 2 × 10−5 km/y. Converting,
100 cm
1000 m
−5
v = 2 × 10 km/y
= 2 cm/y
1 km
1m
E2-26 (a) For Maurice Greene v av = (100 m)/(9.81 m) = 10.2 m/s. For Khalid Khannouchi,
(26.219 mi) 1609 m
1hr
v av =
= 5.594 m/s.
(2.0950 hr)
1 mi
3600 s
(b) If Maurice Greene ran the marathon with an average speed equal to his average sprint speed
then it would take him
(26.219 mi) 1609 m
1hr
t=
= 1.149 hr,
10.2 m/s
1 mi
3600 s
or 1 hour, 9 minutes.
E2-27 The time saved is the difference,
∆t =
(700 km)
(700 km)
−
= 1.22 hr,
(88.5 km/hr) (104.6 km/hr)
which is about 1 hour 13 minutes.
E2-28 The ground elevation will increase by 35 m in a horizontal distance of
x = (35.0 m)/ tan(4.3◦ ) = 465 m.
The plane will cover that distance in
t=
(0.465 km)
(1300 km/hr)
3600 s
1hr
= 1.3 s.
E2-29 Let v1 = 40 km/hr be the speed up the hill, t1 be the time taken, and d1 be the distance
traveled in that time. We similarly define v2 = 60 km/hr for the down hill trip, as well as t2 and d2 .
Note that d2 = d1 .
15
v1 = d1 /t1 and v2 = d2 /t2 . v av = d/t, where d total distance and t is the total time. The total
distance is d1 + d2 = 2d1 . The total time t is just the sum of t1 and t2 , so
v av
=
=
=
=
d
t
2d1
t 1 + t2
2d1
d1 /v1 + d2 /v2
2
,
1/v1 + 1/v2
Take the reciprocal of both sides to get a simpler looking expression
2
1
1
=
+ .
v av
v1
v2
Then the average speed is 48 km/hr.
E2-30 (a) Average speed is total distance divided by total time. Then
v av =
(240 ft) + (240 ft)
= 5.7 ft/s.
(240 ft)/(4.0 ft/s) + (240 ft)/(10 ft/s)
(b) Same approach, but different information given, so
v av =
(60 s)(4.0 ft/s) + (60 s)(10 ft/s)
= 7.0 ft/s.
(60 s) + (60 s)
E2-31 The distance traveled is the total area under the curve. The “curve” has four regions: (I)
a triangle from 0 to 2 s; (II) a rectangle from 2 to 10 s; (III) a trapezoid from 10 to 12 s; and (IV)
a rectangle from 12 to 16 s.
The area underneath the curve is the sum of the areas of the four regions.
d=
1
1
(2 s)(8 m/s) + (8.0 s)(8 m/s) + (2 s)(8 m/s + 4 m/s) + (4.0 s)(4 m/s) = 100 m.
2
2
E2-32 The acceleration is the slope of a velocity-time curve,
a=
(8 m/s) − (4 m/s)
= −2 m/s2 .
(10 s) − (12 s)
E2-33 The initial velocity is ~vi = (18 m/s)î, the final velocity is ~vf = (−30 m/s)î. The average
acceleration is then
~vf − ~vi
∆~v
(−30 m/s)î − (18 m/s)î
~aav =
=
=
,
∆t
∆t
2.4 s
which gives ~aav = (−20.0 m/s2 )î.
16
E2-34 Consider the figure below.
10
5
0
-5
1
2
3
4
5
-10
E2-35 (a) Up to A vx > 0 and is constant. From A to B vx is decreasing, but still positive. From
B to C vx = 0. From C to D vx < 0, but |vx | is decreasing.
(b) No. Constant acceleration would appear as (part of) a parabola; but it would be challenging
to distinguish between a parabola and an almost parabola.
E2-36 (a) Up to A vx > 0 and is decreasing. From A to B vx = 0. From B to C vx > 0 and is
increasing. From C to D vx > 0 and is constant.
(b) No. Constant acceleration would appear as (part of) a parabola; but it would be challenging
to distinguish between a parabola and an almost parabola.
E2-37
Consider the figure below.
v
x
a
v
E2-38 Consider the figure below.
17
12
12
8
8
v(cm/s)
16
x(cm)
16
4
4
0
0
−4
−4
0
1
2
3
4
5
The acceleration
is a constant
2 cm/s/s during
the entire time
interval.
6 t(s)
0
1
2
3
4
5
6 t(s)
E2-39 (a) A must have units of m/s2 . B must have units of m/s3 .
(b) The maximum positive x position occurs when vx = 0, so
vx =
dx
= 2At − 3Bt2
dt
implies vx = 0 when either t = 0 or t = 2A/3B = 2(3.0 m/s2 )/3(1.0 m/s3 ) = 2.0 s.
(c) Particle starts from rest, then travels in positive direction until t = 2 s, a distance of
x = (3.0 m/s2 )(2.0 s)2 − (1.0 m/s3 )(2.0 s)3 = 4.0 m.
Then the particle moves back to a final position of
x = (3.0 m/s2 )(4.0 s)2 − (1.0 m/s3 )(4.0 s)3 = −16.0 m.
The total path followed was 4.0 m + 4.0 m + 16.0 m = 24.0 m.
(d) The displacement is −16.0 m as was found in part (c).
(e) The velocity is vx = (6.0 m/s2 )t − (3.0 m/s3 )t2 . When t = 0, vx = 0.0 m/s. When t = 1.0 s,
vx = 3.0 m/s. When t = 2.0 s, vx = 0.0 m/s. When t = 3.0 s, vx = −9.0 m/s. When t = 4.0 s,
vx = −24.0 m/s.
(f) The acceleration is the time derivative of the velocity,
ax =
dvx
= (6.0 m/s2 ) − (6.0 m/s3 )t.
dt
When t = 0 s, ax = 6.0 m/s2 . When t = 1.0 s, ax = 0.0 m/s2 . When t = 2.0 s, ax = −6.0 m/s2 .
When t = 3.0 s, ax = −12.0 m/s2 . When t = 4.0 s, ax = −18.0 m/s2 .
(g) The distance traveled was found in part (a) to be −20 m The average speed during the time
interval is then vx,av = (−20 m)/(2.0 s) = −10 m/s.
E2-40 v0x = 0, vx = 360 km/hr = 100 m/s. Assuming constant acceleration the average velocity
will be
1
vx,av = (100 m/s + 0) = 50 m/s.
2
The time to travel the distance of the runway at this average velocity is
t = (1800 m)/(50 m/s) = 36 s.
The acceleration is
ax = 2x/t2 = 2(1800 m)/(36.0 s)2 = 2.78 m/s2 .
18
E2-41
(a) Apply Eq. 2-26,
vx = v0x + ax t,
2
(3.0 × 107 m/s) = (0) + (9.8 m/s )t,
3.1 × 106 s = t.
(b) Apply Eq. 2-28 using an initial position of x0 = 0,
1
x = x0 + v0x + ax t2 ,
2
1
2
x = (0) + (0) + (9.8 m/s )(3.1 × 106 s)2 ,
2
x = 4.7 × 1013 m.
E2-42 v0x = 0 and vx = 27.8 m/s. Then
t = (vx − v0x )/a = ((27.8 m/s) − (0)) /(50 m/s2 ) = 0.56 s.
I want that car.
E2-43 The muon will travel for t seconds before it comes to a rest, where t is given by
t = (vx − v0x )/a = (0) − (5.20 × 106 m/s) /(−1.30 × 1014 m/s2 ) = 4 × 10−8 s.
The distance traveled will be
1
1
x = ax t2 + v0x t = (−1.30 × 1014 m/s2 )(4 × 10−8 s)2 + (5.20 × 106 m/s)(4 × 10−8 s) = 0.104 m.
2
2
E2-44 The average velocity of the electron was
vx,av =
1
(1.5 × 105 m/s + 5.8 × 106 m/s) = 3.0 × 106 m/s.
2
The time to travel the distance of the runway at this average velocity is
t = (0.012 m)/(3.0 × 106 m/s) = 4.0 × 10−9 s.
The acceleration is
ax = (vx − v0x )/t = ((5.8 × 106 m/s) − (1.5 × 105 m/s))/(4.0 × 10−9 s) = 1.4 × 1015 m/s2 .
E2-45
It will be easier to solve the problem if we change the units for the initial velocity,
km 1000 m
hr
m
v0x = 1020
= 283 ,
hr
km
3600 s
s
and then applying Eq. 2-26,
vx = v0x + ax t,
(0) = (283 m/s) + ax (1.4 s),
−202 m/s
2
= ax .
The problem asks for this in terms of g, so
−202 m/s
g
2
9.8 m/s
19
2
!
= 21g.
E2-46 Change miles to feet and hours to seconds. Then vx = 81 ft/s and v0x = 125 ft/s. The
time is then
2
t = ((81 ft/s) − (125 ft/s)) /(−17 ft/s ) = 2.6 s.
E2-47 (a) The time to stop is
2
t = ((0 m/s) − (24.6 m/s)) /(−4.92 m/s ) = 5.00 s.
(b) The distance traveled is
x=
1
1
2
ax t2 + v0x t = (−4.92 m/s )(5.00 s)2 + (24.6 m/s)(5.00 s) = 62 m.
2
2
E2-48 Answer part (b) first. The average velocity of the arrow while decelerating is
vy ,av =
1
((0) + (260 ft/s)) = 130 ft/s.
2
The time for the arrow to travel 9 inches (0.75 feet) is
t = (0.75 ft)/(130 ft/s) = 5.8 × 10−3 s.
(a) The acceleration of the arrow is then
2
ay = (vy − v0y )/t = ((0) − (260 ft/s))/(5.8 × 10−3 s) = −4.5 × 104 ft/s .
E2-49 The problem will be somewhat easier if the units are consistent, so we’ll write the maximum speed as
ft
min
ft
1000
= 16.7 .
min 60 s
s
(a) We can find the time required for the acceleration from Eq. 2-26,
vx
(16.7 ft/s)
4.18 s
= v0x + ax t,
2
= (0) + (4.00 ft/s )t,
= t.
And from this and Eq 2-28 we can find the distance
1
x = x0 + v0x + ax t2 ,
2
1
2
x = (0) + (0) + (4.00 ft/s )(4.18 s)2 ,
2
x = 34.9 ft.
(b) The motion of the elevator is divided into three parts: acceleration from rest, constant speed
motion, and deceleration to a stop. The total distance is given at 624 ft and in part (a) we found
the distance covered during acceleration was 34.9 ft. By symmetry, the distance traveled during
deceleration should also be 34.9 ft. The distance traveled at constant speed is then (624 − 34.9 −
34.9) ft = 554 ft. The time required for the constant speed portion of the trip is found from Eq.
2-22, rewritten as
∆x
554 ft
∆t =
=
= 33.2 s.
v
16.7 ft/s
The total time for the trip is the sum of times for the three parts: accelerating (4.18 s), constant
speed (33.2 s), and decelerating (4.18 s). The total is 41.6 seconds.
20
E2-50 (a) The deceleration is found from
ax =
2
2
(x − v0 t) =
((34 m) − (16 m/s)(4.0 s)) = −3.75 m/s2 .
t2
(4.0 s)2
(b) The impact speed is
vx = v0x + ax t = (16 m/s) + (−3.75 m/s2 )(4.0 s) = 1.0 m/s.
E2-51 Assuming the drops fall from rest, the time to fall is
s
s
2y
2(−1700 m)
t=
=
= 19 s.
ay
(−9.8 m/s2 )
The velocity of the falling drops would be
vy = ay t = (−9.8 m/s2 )(19 s) = 190 m/s,
or about 2/3 the speed of sound.
E2-52 Solve the problem out of order.
(b) The time to fall is
s
s
2y
2(−120 m)
t=
=
= 4.9 s.
ay
(−9.8 m/s2 )
(a) The speed at which the elevator hits the ground is
vy = ay t = (−9.8 m/s2 )(4.9 s) = 48 m/s.
(d) The time to fall half-way is
t=
s
2y
=
ay
s
2(−60 m)
= 3.5 s.
(−9.8 m/s2 )
(c) The speed at the half-way point is
vy = ay t = (−9.8 m/s2 )(3.5 s) = 34 m/s.
E2-53 The initial velocity of the “dropped” wrench would be zero. I choose vertical to be along
the y axis with up as positive, which is the convention of Eq. 2-29 and Eq. 2-30. It turns out that
it is much easier to solve part (b) before solving part (a).
(b) We solve Eq. 2-29 for the time of the fall.
vy
(−24.0 m/s)
2.45 s
= v0y − gt,
2
= (0) − (9.8 m/s )t,
= t.
(a) Now we can use Eq. 2-30 to find the height from which the wrench fell.
y
(0)
0
1
= y0 + v0y t − gt2 ,
2
1
2
= y0 + (0)(2.45 s) − (9.8 m/s )(2.45 s)2 ,
2
= y0 − 29.4 m
We have set y = 0 to correspond to the final position of the wrench: on the ground. This results in
an initial position of y0 = 29.4 m; it is positive because the wrench was dropped from a point above
where it landed.
21
E2-54 (a) It is easier to solve the problem from the point of view of an object which falls from the
highest point. The time to fall from the highest point is
s
s
2y
2(−53.7 m)
t=
=
= 3.31 s.
ay
(−9.81 m/s2 )
The speed at which the object hits the ground is
vy = ay t = (−9.81 m/s2 )(3.31 s) = −32.5 m/s.
But the motion is symmetric, so the object must have been launched up with a velocity of vy =
32.5 m/s.
(b) Double the previous answer; the time of flight is 6.62 s.
E2-55 (a) The time to fall the first 50 meters is
s
s
2y
2(−50 m)
t=
=
= 3.2 s.
ay
(−9.8 m/s2 )
(b) The total time to fall 100 meters is
s
s
2y
2(−100 m)
t=
=
= 4.5 s.
ay
(−9.8 m/s2 )
The time to fall through the second 50 meters is the difference, 1.3 s.
E2-56 The rock returns to the ground with an equal, but opposite, velocity. The acceleration is
then
ay = ((−14.6 m/s) − (14.6 m/s))/(7.72 s) = 3.78 m/s2 .
That would put them on Mercury.
E2-57 (a) Solve Eq. 2-30 for the initial velocity. Let the distances be measured from the ground
so that y0 = 0.
y
(36.8 m)
36.8 m
27.4 m/s
1
= y0 + v0y t − gt2 ,
2
1
2
(0) + v0y (2.25 s) − (9.8 m/s )(2.25 s)2 ,
2
= v0y (2.25 s) − 24.8 m,
= v0y .
=
(b) Solve Eq. 2-29 for the velocity, using the result from part (a).
vy
= v0y − gt,
vy
vy
=
=
2
(27.4 m/s) − (9.8 m/s )(2.25 s),
5.4 m/s.
(c) We need to solve Eq. 2-30 to find the height to which the ball rises, but we don’t know how
long it takes to get there. So we first solve Eq. 2-29, because we do know the velocity at the highest
point (vy = 0).
vy
= v0y − gt,
2
(0) = (27.4 m/s) − (9.8 m/s )t,
2.8 s = t.
22
And then we find the height to which the object rises,
y
1
= y0 + v0y t − gt2 ,
2
y
=
y
1
2
(0) + (27.4 m/s)(2.8 s) − (9.8 m/s )(2.8 s)2 ,
2
= 38.3m.
This is the height as measured from the ground; so the ball rises 38.3 − 36.8 = 1.5 m above the point
specified in the problem.
E2-58 The time it takes for the ball to fall 2.2 m is
s
s
2y
2(−2.2 m)
t=
=
= 0.67 s.
ay
(−9.8 m/s2 )
The ball hits the ground with a velocity of
vy = ay t = (−9.8 m/s2 )(0.67 s) = −6.6 m/s.
The ball then bounces up to a height of 1.9 m. It is easier to solve the falling part of the motion,
and then apply symmetry. The time is would take to fall is
s
s
2y
2(−1.9 m)
=
= 0.62 s.
t=
ay
(−9.8 m/s2 )
The ball hits the ground with a velocity of
vy = ay t = (−9.8 m/s2 )(0.62 s) = −6.1 m/s.
But we are interested in when the ball moves up, so vy = 6.1 m/s.
The acceleration while in contact with the ground is
ay = ((6.1 m/s) − (−6.6 m/s))/(0.096 s) = 130 m/s2 .
E2-59 The position as a function of time for the first object is
1
y1 = − gt2 ,
2
The position as a function of time for the second object is
1
y2 = − g(t − 1 s)2
2
The difference,
∆y = y2 − y1 = f rac12g ((2 s)t − 1) ,
is the set equal to 10 m, so t = 1.52 s.
E2-60 Answer part (b) first.
(b) Use the quadratic equation to solve
1
(−9.81 m/s2 )t2 + (12.4 m/s)t
2
for time. Get t = −3.0 s and t = 5.53 s. Keep the positive answer.
(a) Now find final velocity from
(−81.3 m) =
vy = (−9.8 m/s2 )(5.53 s) + (12.4 m/s) = −41.8 m/s.
23
E2-61 The total time the pot is visible is 0.54 s; the pot is visible for 0.27 s on the way down.
We’ll define the initial position as the highest point and make our measurements from there. Then
y0 = 0 and v0y = 0. Define t1 to be the time at which the falling pot passes the top of the window
y1 , then t2 = t1 + 0.27 s is the time the pot passes the bottom of the window y2 = y1 − 1.1 m. We
have two equations we can write, both based on Eq. 2-30,
y1
y1
1
= y0 + v0y t1 − gt21 ,
2
1
= (0) + (0)t1 − gt21 ,
2
and
y2
y1 − 1.1 m
1
= y0 + v0y t2 − gt22 ,
2
1
= (0) + (0)t2 − g(t1 + 0.27 s)2 ,
2
Isolate y1 in this last equation and then set the two expressions equal to each other so that we can
solve for t1 ,
1
− gt21
2
1
− gt21
2
1
1.1 m − g(t1 + 0.27 s)2 ,
2
1
= 1.1 m − g(t21 + [0.54 s]t1 + 0.073 s2 ),
2
1
0 = 1.1 m − g([0.54 s]t1 + 0.073 s2 ).
2
=
This last line can be directly solved to yield t1 = 0.28 s as the time when the falling pot passes the
top of the window. Use this value in the first equation above and we can find y1 = − 12 (9.8 m/s2 )(0.28
s)2 = −0.38 m. The negative sign is because the top of the window is beneath the highest point, so
the pot must have risen to 0.38 m above the top of the window.
p
P2-1 (a) The net shift is (22 m)2 + (17 m)2 ) = 28 m.
(b) The vertical displacement is (17 m) sin(52◦ ) = 13 m.
P2-2 Wheel “rolls” through half of a turn, or πr = 1.41 m. The vertical displacement is 2r =
0.90 m. The net displacement is
p
(1.41 m)2 + (0.90 m)2 = 1.67 m0.
The angle is
θ = tan−1 (0.90 m)/(1.41 m) = 33◦ .
P2-3 We align the coordinate system so that the origin corresponds to the starting position of
the fly and that all positions inside the room are given by positive coordinates.
(a) The displacement vector can just be written,
∆~r = (10 ft)î + (12 ft)ĵ + (14 ft)k̂.
√
(b) The magnitude of the displacement vector is |∆~r| = 102 + 122 + 142 ft= 21 ft.
24
(c) The straight line distance between two points is the shortest possible distance, so the length
of the path taken by the fly must be greater than or equal to 21 ft.
(d) If the fly walks it will need to cross two faces. The shortest path will be the diagonal across
these two faces. If the lengths of sides of the room are l1 , l2 , and l3 , then the diagonal length across
two faces will be given by
q
(l1 + l2 )2 + l32 ,
where we want to choose the li from the set of 10 ft, 12 ft, and 14 ft that will minimize the length.
The minimum distance is when l1 = 10 ft, l2 = 12 ft, and l3 = 14. Then the minimal distance the
fly would walk is 26 ft.
P2-4 Choose vector ~a to lie on the x axis. Then ~a = aî and ~b = bx î + by ĵ where bx = b cos θ and
by = b sin θ. The sum then has components
rx = a + b cos θ and ry = b sin θ.
Then
r2
P2-5
2
2
= (a + b cos θ) + (b sin θ) ,
= a2 + 2ab cos θ + b2 .
(a) Average speed is total distance divided by total time. Then
v av =
(35.0 mi/hr)(t/2) + (55.0 mi/hr)(t/2)
= 45.0 mi/hr.
(t/2) + (t/2)
(b) Average speed is total distance divided by total time. Then
v av =
(d/2) + (d/2)
= 42.8 mi/hr.
(d/2)/(35.0 mi/hr) + (d/2)/(55.0 mi/hr)
(c) Average speed is total distance divided by total time. Then
v av =
P2-6
d+d
= 43.9 mi/hr
(d)/(45.0 mi/hr) + (d)/(42.8 mi/hr)
(a) We’ll do just one together. How about t = 2.0 s?
x = (3.0 m/s)(2.0 s) + (−4.0 m/s2 )(2.0 s)2 + (1.0 m/s3 )(2.0 s)3 = −2.0 m.
The rest of the values are, starting from t = 0, x = 0.0 m, 0.0 m, -2.0 m, 0.0 m, and 12.0 m.
(b) Always final minus initial. The answers are xf − xi = −2.0 m − 0.0 m = −2.0 m and xf − xi =
12.0 m − 0.0 m = 12.0 m.
(c) Always displacement divided by (change in) time.
v av =
(12.0 m) − (−2.0 m)
= 7.0 m/s,
(4.0 s) − (2.0 s)
and
v av =
(0.0 m) − (0.0 m)
= 0.0 m/s.
(3.0 s) − (0.0 s)
25
P2-7 (a) Assume the bird has no size, the trains have some separation, and the bird is just
leaving one of the trains. The bird will be able to fly from one train to the other before the two
trains collide, regardless of how close together the trains are. After doing so, the bird is now on the
other train, the trains are still separated, so once again the bird can fly between the trains before
they collide. This process can be repeated every time the bird touches one of the trains, so the bird
will make an infinite number of trips between the trains.
(b) The trains collide in the middle; therefore the trains collide after (51 km)/(34 km/hr) = 1.5
hr. The bird was flying with constant speed this entire time, so the distance flown by the bird is (58
km/hr)(1.5 hr) = 87 km.
P2-8
(a) Start with a perfect square:
(v1 − v2 )2
v12 + v22
2
(v1 + v22 )t1 t2
d21 + d22 + (v12 + v22 )t1 t2
(v12 t1 + v22 t2 )(t1 + t2 )
v12 t1 + v22 t2
d1 + d2
v1 d1 + v2 d2
d1 + d2
>
>
>
>
>
0,
2v1v2,
2v1 v2 t1 t2 ,
d21 + d22 + 2v1 v2 t1 t2 ,
(d1 + d2 )2 ,
d1 + d2
>
,
t1 + t 2
v1 t1 + v2 t2
>
t 1 + t2
Actually, it only works if d1 + d2 > 0!
(b) If v1 = v2 .
P2-9
(a) The average velocity during the time interval is v av = ∆x/∆t, or
v av =
(A + B(3 s)3 ) − (A + B(2 s)3 )
3
= (1.50 cm/s )(19 s3 )/(1 s) = 28.5 cm/s.
(3 s) − (2 s)
3
(b) v = dx/dt = 3Bt2 = 3(1.50 cm/s )(2 s)2 = 18 cm/s.
3
(c) v = dx/dt = 3Bt2 = 3(1.50 cm/s )(3 s)2 = 40.5 cm/s.
3
(d) v = dx/dt = 3Bt2 = 3(1.50 cm/s )(2.5 s)2 = 28.1 cm/s.
(e) The midway position is (xf + xi )/2, or
xmid = A + B[(3 s)3 + (2 s)3 )]/2 = A + (17.5 s3 )B.
p
3
(17.5 s3 ). The instantaneous velocity at this point is
p
3
v = dx/dt = 3Bt2 = 3(1.50 cm/s )( 3 (17.5 s3 ))2 = 30.3 cm/s.
This occurs when t =
P2-10
Consider the figure below.
x
x
1
(a)
t
x
1
t
x
1
(b)
(c)
26
t
1
(d)
t
P2-11
(a) The average velocity is displacement divided by change in time,
3
v av =
3
(2.0 m/s )(2.0 s)3 − (2.0 m/s )(1.0 s)3
14.0 m
=
= 14.0 m/s.
(2.0 s) − (1.0 s)
1.0 s
The average acceleration is the change in velocity. So we need an expression for the velocity, which
is the time derivative of the position,
v=
dx
d
3
3
= (2.0 m/s )t3 = (6.0 m/s )t2 .
dt
dt
From this we find average acceleration
3
aav =
3
(6.0 m/s )(2.0 s)2 − (6.0 m/s )(1.0 s)2
18.0 m/s
2
=
= 18.0 m/s .
(2.0 s) − (1.0 s)
1.0 s
(b) The instantaneous velocities can be found directly from v = (6.0 m/s2 )t2 , so v(2.0 s) = 24.0
m/s and v(1.0 s) = 6.0 m/s. We can get an expression for the instantaneous acceleration by taking
the time derivative of the velocity
a=
d
dv
3
3
= (6.0 m/s )t2 = (12.0 m/s )t.
dt
dt
Then the instantaneous accelerations are a(2.0 s) = 24.0 m/s2 and a(1.0 s) = 12.0 m/s2
(c) Since the motion is monotonic we expect the average quantities to be somewhere between
the instantaneous values at the endpoints of the time interval. Indeed, that is the case.
P2-12
Consider the figure below.
a(m/s^2)
v(m/s)
15
75
10
50
5
25
0
0
0
2
4
6
8 10
0
t(s)
P2-13
x(m)
2
4
6
8 10
t(s)
0
2
4
6
8 10
t(s)
Start with v f = v i + at, but v f = 0, so v i = −at, then
x=
1 2
1
1
at + v i t = at2 − at2 = − at2 ,
2
2
2
q
p
2
2
so t = −2x/a = −2(19.2 ft)/(−32 ft/s ) = 1.10 s. Then v i = −(−32 ft/s )(1.10 s) = 35.2 ft/s.
Converting,
35.2 ft/s(1/5280 mi/ft)(3600 s/h) = 24 mi/h.
27
P2-14
(b) The average speed during while traveling the 160 m is
v av = (33.0 m/s + 54.0 m/s)/2 = 43.5 m/s.
The time to travel the 160 m is t = (160 m)/(43.5 m/s) = 3.68 s.
(a) The acceleration is
a=
2x 2vi
2(160 m) 2(33.0 m/s)
−
=
−
= 5.69 m/s2 .
t2
t
(3.68 s)2
(3.68 s)
(c) The time required to get up to a speed of 33 m/s is
t = v/a = (33.0 m/s)/(5.69 m/s2 ) = 5.80 s.
(d) The distance moved from start is
d=
1 2
1
at = (5.69 m/s2 )(5.80 s)2 = 95.7 m.
2
2
P2-15 (a) The distance traveled during the reaction time happens at constant speed; treac = d/v =
(15 m)/(20 m/s) = 0.75 s.
(b) The braking distance is proportional to the speed squared (look at the numbers!) and in
this case is dbrake = v 2 /(20 m/s2 ). Then dbrake = (25 m/s)2 /(20 m/s2 ) = 31.25 m. The reaction time
distance is dreac = (25 m/s)(0.75 s) = 18.75 m. The stopping distance is then 50 m.
P2-16 (a) For the car xc = ac t2 /2. For the truck xt = v t t. Set both xi to the same value, and
then substitute the time from the truck expression:
x = ac t2 /2 = ac (x/v t )2 /2,
or
x = 2v t 2 /ac = 2(9.5 m/s)2 /(2.2 m/s) = 82 m.
(b) The speed of the car will be given by v c = ac t, or
v c = ac t = ac x/v t = (2.2 m/s)(82 m)/(9.5 m/s) = 19 m/s.
P2-17 The runner covered a distance d1 in a time interval t1 during the acceleration phase and
a distance d2 in a time interval t2 during the constant speed phase. Since the runner started from
rest we know that the constant speed is given by v = at1 , where a is the runner’s acceleration.
The distance covered during the acceleration phase is given by
d1 =
1 2
at .
2 1
The distance covered during the constant speed phase can also be found from
d2 = vt2 = at1 t2 .
We want to use these two expressions, along with d1 + d2 = 100 m and t2 = (12.2 s) − t1 , to get
100 m
= d1 + d2 =
1 2
at + at1 (12.2 s − t1 ),
2 1
1
= − at21 + a(12.2 s)t1 ,
2
2
= −(1.40 m/s )t21 + (34.2 m/s)t1 .
28
This last expression is quadratic in t1 , and is solved to give t1 = 3.40 s or t1 = 21.0 s. Since the race
only lasted 12.2 s we can ignore the second answer.
(b) The distance traveled during the acceleration phase is then
d1 =
1 2
2
at = (1.40 m/s )(3.40 s)2 = 16.2 m.
2 1
P2-18 (a) The ball will return to the ground with the same speed it was launched. Then the total
time of flight is given by
t = (v f − v i )/g = (−25 m/s − 25 m/s)/(9.8 m/s2 ) = 5.1 s.
(b) For small quantities we can think in terms of derivatives, so
δt = (δv f − δv i )/g,
or τ = 2/g.
P2-19 Use y = −gt2 /2, but only keep the absolute value. Then y50 = (9.8 m/s2 )(0.05 s)2 /2 =
1.2 cm; y100 = (9.8 m/s2 )(0.10 s)2 /2 = 4.9 cm; y150 = (9.8 m/s2 )(0.15 s)2 /2 = 11 cm; y200 =
(9.8 m/s2 )(0.20 s)2 /2 = 20 cm; y250 = (9.8 m/s2 )(0.25 s)2 /2 = 31 cm.
P2-20 The truck will move 12 m in (12 m)/(55 km/h) = 0.785 s. The apple will fall y = −gt2 /2 =
−(9.81 m/s2 )(0.785 s)2 /2 = −3.02 m.
2
P2-21 The rocket travels a distance d1 = 12 at21 = 12 (20 m/s )(60 s)2 = 36, 000 m during the accel2
eration phase; the rocket velocity at the end of the acceleration phase is v = at = (20 m/s )(60 s) =
1200 m/s. The second half of the trajectory can be found from Eqs. 2-29 and 2-30, with y0 = 36, 000
m and v0y = 1200 m/s.
(a) The highest point of the trajectory occurs when vy = 0, so
vy
(0)
122 s
= v0y − gt,
2
= (1200 m/s) − (9.8 m/s )t,
= t.
This time is used to find the height to which the rocket rises,
y
1
= y0 + v0y t − gt2 ,
2
=
1
2
(36000 m) + (1200 m/s)(122s) − (9.8 m/s )(122 s)2 = 110000 m.
2
(b) The easiest way to find the total time of flight is to solve Eq. 2-30 for the time when the
rocket has returned to the ground. Then
y
(0)
1
= y0 + v0y t − gt2 ,
2
1
2
= (36000 m) + (1200 m/s)t − (9.8 m/s )t2 .
2
This quadratic expression has two solutions for t; one is negative so we don’t need to worry about
it, the other is t = 270 s. This is the free-fall part of the problem, to find the total time we need to
add on the 60 seconds of accelerated motion. The total time is then 330 seconds.
29
P2-22
(a) The time required for the player to “fall” from the highest
point
p
p
pa distance of y = 15 cm
is 2y/g; the total time spent in the top 15 cm is twice this, or 2 2y/g = 2 2(0.15 m)/(9.81 m/s) =
0.350 s.
p (b) The time required for the player to “fall” from the highest point a distance of 76 cm is
2(0.76 m)/(9.81 m/s) = 0.394 s,p
the time required for the player to fall from the highest point a
distance of (76 − 15 = 61) cm is 2(0.61 m)/g = 0.353 s. The time required to fall the bottom 15
cm is the difference, or 0.041 s. The time spent in the bottom 15 cm is twice this, or 0.081 s.
P2-23 (a) The average speed between A and B is v av = (v + v/2)/2 = 3v/4. We can also write
v av = (3.0 m)/∆t = 3v/4. Finally, v/2 = v − g∆t. Rearranging, v/2 = g∆t. Combining all of the
above,
v
4(3.0 m)
=g
or v 2 = (8.0 m)g.
2
3v
p
Then v = (8.0 m)(9.8 m/s2 ) = 8.85 m/s.
(b) The time to the highest point above B is v/2 = gt, so the distance above B is
2
g
v
g v
v v
v2
y = − t2 + t = −
+
=
.
2
2
2 2g
2 2g
8g
Then y = (8.85 m/s)2 /(8(9.8 m/s2 )) = 1.00 m.
p
p
P2-24 (a) The time in free fall is t = −2y/g = −2(−145 m)/(9.81 m/s2 ) = 5.44 s.
(b) The speed at the bottom is v = −gt = −(9.81 m/s2 )(5.44 s) = −53.4 m/s.
(c) The time for deceleration is given by v = −25gt, or t = −(−53.4 m/s)/(25 × 9.81 /s2 ) =
0.218 s. The distance through which deceleration occurred is
y=
P2-25
25g 2
t + vt = (123 m/s2 )(0.218 s)2 + (−53.4 m/s)(0.218 s) = −5.80 m.
2
Find the time she fell from Eq. 2-30,
1
2
(0 ft) = (144 ft) + (0)t − (32 ft/s )t2 ,
2
which is a simple quadratic with solutions t = ±3.0 s. Only the positive solution is of interest. Use
this time in Eq. 2-29 to find her speed when she hit the ventilator box,
2
vy = (0) − (32 ft/s )(3.0 s) = −96 ft/s.
This becomes the initial velocity for the deceleration motion, so her average speed during deceleration
is given by Eq. 2-27,
v av,y =
1
1
(vy + v0y ) = ((0) + (−96 ft/s)) = −48 ft/s.
2
2
This average speed, used with the distance of 18 in (1.5 ft), can be used to find the time of deceleration
v av,y = ∆y/∆t,
and putting numbers into the expression gives ∆t = 0.031 s. We actually used ∆y = −1.5 ft, where
the negative sign indicated that she was still moving downward. Finally, we use this in Eq. 2-26 to
find the acceleration,
(0) = (−96 ft/s) + a(0.031 s),
which gives a = +3100 ft/s2 . In terms of g this is a = 97g, which can be found by multiplying
through by 1 = g/(32 ft/s2 ).
30
P2-26 Let the speed of the disk when it comes into contact with the ground be v1 ; then the
average speed during the deceleration process is v1 /2; so the time taken for the deceleration process
is t1 = 2d/v1 , where d = −2 mm. But d is also give by d = at21 /2 + v1 t1 , so
2
100g 2d
2d
d2
d=
+ v1
= 200g 2 + 2d,
2
v1
v1
v1
or v12 = −200gd. The negative signs are necessary!.
The disk was dropped from a height h = −y and it first came into contact with the ground when
it had a speed of v1 . Then the average speed is v1 /2, and we can repeat most of the above (except
a = −g instead of 100g), and then the time to fall is t2 = 2y/v1 ,
2
g 2y
2y
y2
y=
+ v1
= 2g 2 + 2y,
2 v1
v1
v1
or v12 = −2gy. The negative signs are necessary!.
Equating, y = 100d = 100(−2 mm) = −0.2 m, so h = 0.2 m. Note that although 100g’s sounds
like plenty, you still shouldn’t be dropping your hard disk drive!
P2-27 Measure from the feet! Jim is 2.8 cm tall in the photo, so 1 cm on the photo is 60.7
cm in real-life. Then Jim has fallen a distance y1 = −3.04 m while Clare has fallen a distance
y2 = −5.77 m. Clare jumped first, and the time she has been falling is t2 ; Jim jumped p
seconds, the
2
2
time
he
has
been
falling
is
t
=
t
−
∆t.
Then
y
=
−gt
/2
and
y
=
−gt
/2,
or
t
=
−2y2 /g =
1
2
1
2
2
1
p
p2
p
2
2
−2(−5.77 m)/(9.81 m/s ) = 1.08 s and t1 = −2y1 /g = −2(3.04 m)/(9.81 m/s ) = 0.79 s. So
Jim waited 0.29 s.
P2-28 (a) Assuming she starts from rest and has a speed of v1 when she opens her chute, then
her average speed while falling freely is v1 /2, and the time taken to fallp
y1 = −52.0 m is t1 = 2y1 /v1 .
Her speed v1 is given by v1 = −gt1 , or v12 = −2gy1 . Then v1 = − −2(9.81 m/s2 )(−52.0 m) =
−31.9pm/s. We must
p use the negative answer, because she fall down! The time in the air is then
t1 = −2y1 /g = −2(52.0 m)/(9.81 m/s2 ) = 3.26 s.
Her final speed is v2 = −2.90 m/s, so the time for the deceleration is t2 = (v2 − v1 )/a, where
a = 2.10 m/s2 . Then t2 = (−2.90 m/s − −31.9 m/s)/(2.10 m/s2 ) = 13.8 s.
Finally, the total time of flight is t = t1 + t2 = 3.26 s + 13.8 s = 17.1 s.
(b) The distance fallen during the deceleration phase is
g
(2.10 m/s2 )
y2 = − t22 + v1 t2 = −
(13.8 s)2 + (−31.9 m/s)(13.8 s) = −240 m.
2
2
The total distance fallen is y = y1 + y2 = −52.0 m − 240 m = −292 m. It is negative because she was
falling down.
P2-29 Let the speed of the bearing be v1
These speeds are related by v2 = v1 − gt12 ,
bearing is at the top of the window and at
v av = (v1 + v2 )/2 = v1 − gt12 /2. The distance
at the top of the windows and v2 at the bottom.
where t12 = 0.125 s is the time between when the
the bottom of the window. The average speed is
traveled in the time t12 is y12 = −1.20 m, so
y12 = v av t12 = v1 t12 − gt212 /2,
and expression that can be solved for v1 to yield
v1 =
y12 + gt212 /2
(−1.20 m) + (9.81 m/s2 )(0.125 s)2 /2
=
= −8.99 m/s.
t12
(0.125 s)
31
Now that we know v1 we can find the height of the building above the top of the window. The time
the object has fallen to get to the top of the window is t1 = −v1 /g = −(−8.99 m/s)/(9.81 m/s2 ) =
0.916 m.
The total time for falling is then (0.916 s) + (0.125 s) + (1.0 s) = 2.04 s. Note that we remembered
to divide the last time by two! The total distance from the top of the building to the bottom is then
y = −gt2 /2 = −(9.81 m/s2 )(2.04 s)2 /2 = 20.4 m.
P2-30
Each ball falls from a highest point through a distance of 2.0 m in
p
t = −2(2.0 m)/(9.8 m/s2 ) = 0.639 s.
The time each ball spends in the air is twice this, or 1.28 s. The frequency of tosses per ball is the
reciprocal, f = 1/T = 0.781 s−1 . There are five ball, so we multiply this by 5, but there are two
hands, so we then divide that by 2. The tosses per hand per second then requires a factor 5/2, and
the tosses per hand per minute is 60 times this, or 117.
P2-31 Assume each hand can toss n objects per second. Let τ be the amount of time that any
one object is in the air. Then 2nτ is the number of objects that are in the air at any time, where
the “2” comes from the fact that (most?) jugglers have two hands. We’ll estimate n, but τ can be
found from Eq. 2-30 for an object which falls a distance h from rest:
1
0 = h + (0)t − gt2 ,
2
p
solving, t = 2h/g. But τ is twice this, because the object had to go up before it could come down.
So the number of objects that can be juggled is
p
4n 2h/g
We estimate n = 2 tosses/second. So the maximum number of objects one could juggle to a height
h would be
p
3.6 h/meters.
P2-32 (a) We need to look up the height of the leaning
p tower to solve
p this! If the height is h = 56 m,
then the time taken to fall a distance h = −y1 is t1 = −2y1 /g = −2(−56 m)/(9.81 m/s2 ) = 3.4 s.
The second object, however, has only fallen a a time t2 = t1 − ∆t = 3.3 s, so the distance the second
object falls is y2 = −gt22 /2 = −(9.81 m/s2 )(3.3 s)2 /2 = 53.4. The difference is y1 − y2 = 2.9 m.
(b) If the vertical separation is ∆y = 0.01 m, then we can approach this problem in terms of
differentials,
δy = at δt,
so δt = (0.01 m)/[((9.81 m/s2 )(3.4 s)] = 3×10−4 s.
P2-33 Use symmetry, and focus on the path from the highest point downward. Then ∆tU = 2tU ,
where tU is the time from the highest point to the upper level. A similar expression exists for
the lower level, but replace U with L. The distance from the highest point to the upper level is
yU = −gt2U /2 = −g(∆tU /2)2 /2. The distance from the highest point to the lower level is yL =
−gt2L /2 = −g(∆tL /2)2 /2. Now H = yU − yL = −g∆t2U /8 − −g∆t2L /8, which can be written as
g=
8H
.
∆t2L − ∆t2U
32
E3-1 The Earth orbits the sun with a speed of 29.8 km/s. The distance to Pluto is 5900×106 km.
The time it would take the Earth to reach the orbit of Pluto is
t = (5900×106 km)/(29.8 km/s) = 2.0×108 s,
or 6.3 years!
E3-2 (a) a = F/m = (3.8 N)/(5.5 kg) = 0.69 m/s2 .
(b) t = v f /a = (5.2 m/s)/(0.69 m/s2 ) = 7.5 s.
(c) x = at2 /2 = (0.69 m/s2 )(7.5 s)2 /2 = 20 m.
E3-3
2-27
Assuming constant acceleration we can find the average speed during the interval from Eq.
1
1
(vx + v0x ) =
(5.8×106 m/s) + (0) = 2.9×106 m/s.
2
2
From this we can find the time spent accelerating from Eq. 2-22, since ∆x = v av,x ∆t. Putting in
the numbers ∆t = 5.17×10−9 s. The acceleration is then
v av,x =
ax =
∆vx
(5.8×106 m/s) − (0)
=
= 1.1×1015 m/s2 .
∆t
(5.17×10−9 s)
The net force on the electron is from Eq. 3-5,
X
Fx = max = (9.11×10−31 kg)(1.1×1015 m/s2 ) = 1.0×10−15 N.
E3-4 The average speed while decelerating is v av = 0.7 × 107 m/s. The time of deceleration is
t = x/v av = (1.0×10−14 m)/(0.7×107 m/s) = 1.4×10−21 s. The deceleration is a = ∆v/t = (−1.4×
107 m/s)/(1.4×10−21 s) = −1.0×1028 m/s2 . The force is F = ma = (1.67×10−27 kg)(1.0×1028 m/s2 ) =
17 N.
E3-5 The net force on the sled is 92 N−90 N= 2 N; subtract because the forces are in opposite
directions. Then
P
Fx
(2 N)
ax =
=
= 8.0×10−2 m/s2 .
m
(25 kg)
E3-6 53 km/hr is 14.7 m/s. The average speed while decelerating is v av = 7.4 m/s. The time
of deceleration is t = x/v av = (0.65 m)/(7.4 m/s) = 8.8×10−2 s. The deceleration is a = ∆v/t =
(−14.7 m/s)/(8.8×10−2 s) = −17×102 m/s2 . The force is F = ma = (39kg)(1.7×102 m/s2 ) = 6600 N.
E3-7 Vertical acceleration is a = F/m = (4.5×10−15 N)/(9.11×10−31 kg) = 4.9×1015 m/s2 . The
electron moves horizontally 33 mm in a time t = x/vx = (0.033 m)/(1.2×107 m/s) = 2.8×10−9 s.
The vertical distance deflected is y = at2 /2 = (4.9×1015 m/s2 )(2.8×10−9 s)2 /2 = 1.9×10−2 m.
E3-8 (a) a = F/m = (29 N)/(930 kg) = 3.1×10−2 m/s2 .
(b) x = at2 /2 = (3.1×10−2 m/s2 )(86400 s)2 /2 = 1.2×108 m.
(c) v = at = (3.1×10−2 m/s2 )(86400 s) = 2700 m/s.
33
P
E3-9 Write the
Fx = m1 a1x and that
P expression for the motion of the first object as
P of the
second object as Fx = m2 a2x . In both cases there is only one force, F , on the object, so Fx = F .
We will solve these for the mass as m1 = F/a1 and m2 = F/a2 . Since a1 > a2 we can conclude that
m2 > m1
(a) The acceleration of and object with mass m2 − m1 under the influence of a single force of
magnitude F would be
a=
F
1
F
=
=
,
2
m2 − m1
F/a2 − F/a1
1/(3.30 m/s ) − 1/(12.0 m/s2 )
which has a numerical value of a = 4.55 m/s2 .
(b) Similarly, the acceleration of an object of mass m2 + m1 under the influence of a force of
magnitude F would be
a=
1
1
=
,
2
1/a2 + 1/a1
1/(3.30 m/s ) + 1/(12.0 m/s2 )
which is the same as part (a) except for the sign change. Then a = 2.59 m/s2 .
E3-10 (a) The required acceleration is a = v/t = 0.1c/t. The required force is F = ma = 0.1mc/t.
Then
F = 0.1(1200×103 kg)(3.00×108 m/s)/(2.59×105 s) = 1.4×108 N,
and
F = 0.1(1200×103 kg)(3.00×108 m/s)/(5.18×106 s) = 6.9×106 N,
(b) The distance traveled during the acceleration phase is x1 = at21 /2, the time required to
travel the remaining distance is t2 = x2 /v where x2 = d − x1 . d is 5 light-months, or d = (3.00×
108 m/s)(1.30×107 s) = 3.90×1015 m. Then
t = t1 + t 2 = t 1 +
d − x1
2d − at21
2d − vt1
= t1 +
= t1 +
.
v
2v
2v
If t1 is 3 days, then
t = (2.59×105 s) +
2(3.90×1015 m) − (3.00×107 m/s)(2.59×105 s)
= 1.30×108 s = 4.12 yr,
2(3.00×107 m/s)
if t1 is 2 months, then
t = (5.18×106 s) +
E3-11
2(3.90×1015 m) − (3.00×107 m/s)(5.18×106 s)
= 1.33×108 s = 4.20 yr,
2(3.00×107 m/s)
(a) The net force on the second block is given by
X
Fx = m2 a2x = (3.8 kg)(2.6 m/s2 ) = 9.9 N.
There is only one (relevant) force on the block, the force of block 1 on block 2.
(b) There is only one (relevant) force on block 1, the force of block 2 on block 1.PBy Newton’s
third law this force has a magnitude of 9.9 N. Then Newton’s second law gives
Fx = −9.9
N= m1 a1x = (4.6 kg)a1x . So a1x = −2.2 m/s2 at the instant that a2x = 2.6 m/s2 .
E3-12 (a) W = (5.00 lb)(4.448 N/lb) = 22.2 N; m = W/g = (22.2 N)/(9.81 m/s2 ) = 2.26 kg.
(b) W = (240 lb)(4.448 N/lb) = 1070 N; m = W/g = (1070 N)/(9.81 m/s2 ) = 109 kg.
(c) W = (3600 lb)(4.448 N/lb) = 16000 N; m = W/g = (16000 N)/(9.81 m/s2 ) = 1630 kg.
34
E3-13 (a) W = (1420.00 lb)(4.448 N/lb) = 6320 N; m = W/g = (6320 N)/(9.81 m/s2 ) = 644 kg.
(b) m = 412 kg; W = mg = (412 kg)(9.81 m/s2 ) = 4040 N.
E3-14 (a) W = mg = (75.0 kg)(9.81 m/s2 ) = 736 N.
(b) W = mg = (75.0 kg)(3.72 m/s2 ) = 279 N.
(c) W = mg = (75.0 kg)(0 m/s2 ) = 0 N.
(d) The mass is 75.0 kg at all locations.
E3-15 If g = 9.81 m/s2 , then m = W/g = (26.0 N)/(9.81 m/s2 ) = 2.65 kg.
(a) Apply W = mg again, but now g = 4.60 m/s2 , so at this point W = (2.65 kg)(4.60 m/s2 ) =
12.2 N.
(b) If there is no gravitational force, there is no weight, because g = 0. There is still mass,
however, and that mass is still 2.65 kg.
E3-16 Upward force balances weight, so F = W = mg = (12000 kg)(9.81 m/s2 ) = 1.2×105 N.
E3-17 Mass is m = W/g; net force is F = ma, or F = W a/g. Then
2
2
F = (3900 lb)(13 ft/s )/(32 ft/s ) = 1600 lb.
E3-18 a = ∆v/∆t = (450 m/s)/(1.82 s) = 247 m/s2 . Net force is F = ma = (523 kg)(247 m/s2 ) =
1.29×105 N.
E3-19
P
Fx = 2(1.4×105 N) = max . Then m = 1.22×105 kg and
W = mg = (1.22×105 kg)(9.81 m/s2 ) = 1.20×106 N.
E3-20 Do part (b) first; there must be a 10 lb force to support the mass. Now do part (a), but
cover up the left hand side of both pictures. If you can’t tell which picture is which, then they must
both be 10 lb!
E3-21 (b) Average speed during deceleration is 40 km/h, or 11 m/s. The time taken to stop the
car is then t = x/v av = (61 m)/(11 m/s) = 5.6 s.
(a) The deceleration is a = ∆v/∆t = (22 m/s)/(5.6 s) = 3.9 m/s2 . The braking force is F =
ma = W a/g = (13, 000 N)(3.9 m/s2 )/(9.81 m/s2 ) = 5200 N.
(d) The deceleration is same; the time to stop the car is then ∆t = ∆v/a = (11 m/s)/(3.9 m/s2 ) =
2.8 s.
(c) The distance traveled during stopping is x = v av t = (5.6 m/s)(2.8 s) = 16 m.
E3-22 Assume acceleration of free fall is 9.81 m/s2 at the altitude of the meteor. The net force is
F net = ma = (0.25 kg)(9.2 m/s2 ) = 2.30 N. The weight is W = mg = (0.25 kg)(9.81 m/s2 ) = 2.45 N.
The retarding force is F net − W = (2.3 N) − (2.45 N) = −0.15 N.
E3-23
(a) Find the time during the “jump down” phase from Eq. 2-30.
1
2
(0 m) = (0.48 m) + (0)t − (9.8 m/s )t2 ,
2
which is a simple quadratic with solutions t = ±0.31 s. Use this time in Eq. 2-29 to find his speed
when he hit ground,
2
vy = (0) − (9.8 m/s )(0.31 s) = −3.1 m/s.
35
This becomes the initial velocity for the deceleration motion, so his average speed during deceleration
is given by Eq. 2-27,
v av,y =
1
1
(vy + v0y ) = ((0) + (−3.1 m/s)) = −1.6 m/s.
2
2
This average speed, used with the distance of -2.2 cm (-0.022 m), can be used to find the time of
deceleration
v av,y = ∆y/∆t,
and putting numbers into the expression gives ∆t = 0.014 s. Finally, we use this in Eq. 2-26 to find
the acceleration,
(0) = (−3.1 m/s) + a(0.014 s),
which gives a = 220 m/s2 .
(b) The average net force on the man is
X
Fy = may = (83 kg)(220 m/s2 ) = 1.8×104 N.
E3-24 The average speed of the salmon while decelerating is 4.6 ft/s. The time required to stop
the salmon is then t = x/v av = (0.38 ft)/(4.6 ft/s) = 8.3×10−2 s. The deceleration of the salmon
2
is a = ∆v/∆t = (9.2 ft/s)/(8.2-2s) = 110 ft/s . The force on the salmon is then F = W a/g =
2
2
(19 lb)(110 ft/s )/(32 ft/s ) = 65 lb.
E3-25 From appendix G we find 1 lb = 4.448 N; so the weight is (100 lb)(4.448 N/1 lb) = 445
N; similarly the cord will break if it pulls upward on the object with a force greater than 387 N.
The mass of the object is m = W/g = (445 N)/(9.8 m/s2 ) = 45 kg.
There are two vertical forces on the 45 kg object, an upward force fromP
the cord FOC (which has a
maximum value of 387 N) and a downward force from gravity FOG . Then Fy = FOC −FOG = (387
N) − (445 N) = −58 N. Since the net force is negative, the object must be accelerating downward
according to
X
ay =
Fy /m = (−58 N)/(45 kg) = −1.3 m/s2 .
E3-26 (a) Constant speed means no acceleration, hence no net force; this means the weight is
balanced by the force from the scale, so the scale reads 65 N.
(b) Net force on mass is F net = ma = W a/g = (65 N)(−2.4 m/s2 )/(9.81 m/s2 ) = −16 N.. Since
the weight is 65 N, the scale must be exerting a force of (−16 N) − (−65 N) = 49 N.
E3-27 The magnitude of the net force is W − R = (1600 kg)(9.81 m/s2 ) − (3700 N) = 12000 N.
The acceleration is then a = F/m = (12000 N)/(1600 kg) = 7.5 m/s2 . The time to fall is
p
p
t = 2y/a = 2(−72 m)/(−7.5 m/s2 ) = 4.4 s.
The final speed is v = at = (−7.5 m/s2 )(4.4 s) = 33 m/s. Get better brakes, eh?
E3-28 The average speed during the acceleration is 140 ft/s. The time for the plane to travel 300
ft is
t = x/v av = (300 ft)/(140 ft/s) = 2.14 s.
The acceleration is then
2
a = ∆v/∆t = (280 ft/s)/(2.14 s) = 130 ft/s .
2
2
The net force on the plane is F = ma = W a/g = (52000 lb)(130 ft/s )/(32 ft/s ) = 2.1×105 lb.
The force exerted by the catapult is then 2.1×105 lb − 2.4×104 lb = 1.86×105 lb.
36
E3-29 (a) The acceleration of a hovering rocket is 0, so the net force is zero; hence the thrust must
equal the weight. Then T = W = mg = (51000 kg)(9.81 m/s2 ) = 5.0×105 N.
(b) If the rocket accelerates upward then the net force is F = ma = (51000 kg)(18 m/s2 ) =
9.2×105 N. Now F net = T − W , so T = 9.2×105 N + 5.0×105 N = 1.42×106 N.
E3-30 (a) Net force on parachute + person system is F net = ma = (77 kg + 5.2 kg)(−2.5 s2 ) =
−210 N. The weight of the system is W = mg = (77 kg +5.2 kg)(9.81 s2 ) = 810 N. If P is the upward
force of the air on the system (parachute) then P = F net + W = (−210 N) + (810 N) = 600 N.
(b) The net force on the parachute is F net = ma = (5.2 kg)(−2.5 s2 ) = −13 N. The weight of the
parachute is W = mg = (5.2 kg)(9.81 m/s2 ) = 51 N. If D is the downward force of the person on
the parachute then D = −F net − W + P = −(−13 N) − (51 N) + 600 N = 560 N.
E3-31 (a) The total mass of the helicopter+car system is 19,500 kg; and the only other force
acting on the system is the force of gravity, which is
W = mg = (19, 500 kg)(9.8 m/s2 ) = 1.91×105 N.
P
The force of gravity is directed down, so the net force on the system
Fy = FBA − (1.91×105
P is
N). The net force can also be found from Newton’s second law:
Fy = may = (19, 500 kg)(1.4
m/s2 ) = 2.7×104 N. Equate the two expressions for the net force, FBA − (1.91×105 N) = 2.7×104
N, and solve; FBA = 2.2×105 N.
(b) Repeat the above steps except: (1) the system will consist only of the car, and (2) the upward
force on the car comes from the supporting cable
P only FCC . The weight4 of the car is W = mg = (4500
2
4
kg)(9.8
m/s
)
=
4.4×10
N.
The
net
force
is
Fy = FCC − (4.4×10 N), it can also be written as
P
Fy = may = (4500 kg)(1.4 m/s2 ) = 6300 N. Equating, FCC = 50, 000 N.
P3-1 (a) The acceleration is a = F/m = (2.7×10−5 N)/(280 kg) = 9.64×10−8 m/s2 . The displacement (from the original trajectory) is
y = at2 /2 = (9.64×10−8 m/s2 )(2.4 s)2 /2 = 2.8×10−7 m.
(b) The acceleration is a = F/m = (2.7×10−5 N)/(2.1 kg) = 1.3×10−5 m/s2 . The displacement
(from the original trajectory) is
y = at2 /2 = (1.3×10−5 m/s2 )(2.4 s)2 /2 = 3.7×10−5 m.
P3-2 (a) The acceleration of the sled is a = F/m = (5.2 N)/(8.4 kg) = 0.62 m/s2 .
(b) The acceleration of the girl is a = F/m = (5.2 N)/(40 kg) = 0.13 m/s2 .
(c) The distance traveled by girl is x1 = a1 t2 /2; the distance traveled by the sled is x2 = a2 t2 /2.
Thep
two meet when x1 + x2 = 15 m. This happens when (a1 + a2 )t2 = 30 m. They then meet when
t = (30 m)/(0.13 m/s2 + 0.62 m/s2 ) = 6.3 s. The girl moves x1 = (0.13 m/s2 )(6.3 s)2 /2 = 2.6 m.
P3-3 (a) Start with block one. It starts from rest, accelerating through a distance of 16 m in a
time of 4.2 s. Applying Eq. 2-28,
1
x = x0 + v0x t + ax t2 ,
2
1
−16 m = (0) + (0)(4.2 s) + ax (4.2 s)2 ,
2
find the acceleration to be ax = −1.8 m/s2 .
37
Now for the second block. The acceleration of the second block is identical to the first for much
the same reason that all objects fall with approximately the same acceleration.
(b) The initial and final velocities are related by a sign, then vx = −v0x and Eq. 2-26 becomes
vx
−v0x
−2v0x
= v0x + ax t,
= v0x + ax t,
2
= (−1.8 m/s )(4.2 s).
which gives an initial velocity of v0x = 3.8 m/s.
(c) Half of the time is spent coming down from the highest point, so the time to “fall” is 2.1 s.
The distance traveled is found from Eq. 2-28,
1
x = (0) + (0)(2.1 s) + (−1.8 m/s2 )(2.1 s)2 = −4.0 m.
2
P3-4 (a) The weight of the engine is W = mg = (1400 kg)(9.81 m/s2 ) = 1.37×104 N. If each bolt
supports 1/3 of this, then the force on a bolt is 4600 N.
(b) If engine accelerates up at 2.60 m/s2 , then net force on the engine is
F net = ma = (1400 kg)(2.60 m/s2 ) = 3.64×103 N.
The upward force from the bolts must then be
B = F net + W = (3.64×103 N) + (1.37×104 N) = 1.73×104 N.
The force per bolt is one third this, or 5800 N.
P3-5 (a) If craft descends with constant speed then net force is zero, so thrust balances weight.
The weight is then 3260 N.
(b) If the thrust is 2200 N the net force is 2200 N − 3260 N = −1060 N. The mass is then
m = F/a = (−1060 N)/(−0.390 m/s2 ) = 2720 kg.
(c) The acceleration due to gravity is g = W/m = (3260 N)/(2720 kg) = 1.20 m/s2 .
P3-6 The weight is originally M g. The net force is originally −M a. The upward force is then
originally B = M g − M a. The goal is for a net force of (M − m)a and a weight (M − m)g. Then
(M − m)a = B − (M − m)g = M g − M a − M g + mg = mg − M a,
or m = 2M a/(a + g).
P3-7
(a) Consider all three carts as one system. Then
X
Fx = mtotal ax ,
6.5 N
0.97 m/s
2
= (3.1 kg + 2.4 kg + 1.2 kg)ax ,
= ax .
(b) Now choose your system so that it only contains the third car. Then
X
Fx = F23 = m3 ax = (1.2 kg)(0.97 m/s2 ).
The unknown can be solved to give F23 = 1.2 N directed to the right. P
(c) Consider a system involving the second and third carts. Then
Fx = F12 , so Newton’s law
applied to the system gives
F12 = (m2 + m3 )ax = (2.4 kg + 1.2 kg)(0.97 m/s2 ) = 3.5 N.
38
P3-8 (a) F = ma = (45.2 kg + 22.8 kg + 34.3 kg)(1.32 m/s2 ) = 135 N.
(b) Consider only m3 . Then F = ma = (34.3 kg)(1.32 m/s2 ) = 45.3 N.
(c) Consider m2 and m3 . Then F = ma = (22.8 kg + 34.3 kg)(1.32 m/s2 ) = 75.4 N.
P3-9 (c) The net force on each link is the same, F net = ma = (0.100 kg)(2.50 m/s2 ) = 0.250 N.
(a) The weight of each link is W = mg = (0.100 kg)(9.81 m/s2 ) = 0.981 N. On each link (except
the top or bottom link) there is a weight, an upward force from the link above, and a downward force
from the link below. Then F net = U −D−W . Then U = F net +W +D = (0.250 N)+(0.981 N)+D =
1.231 N + D. For the bottom link D = 0. For the bottom link, U = 1.23 N. For the link above,
U = 1.23 N + 1.23 N = 2.46 N. For the link above, U = 1.23 N + 2.46 N = 3.69 N. For the link above,
U = 1.23 N + 3.69 N = 4.92 N.
(b) For the top link, the upward force is U = 1.23 N + 4.92 N = 6.15 N.
P3-10 (a) The acceleration of the two blocks is a = F/(m1 + m2 ) The net force on block 2 is from
the force of contact, and is
P = m2 a = F m2 /(m1 + m2 ) = (3.2 N)(1.2 kg)/(2.3 kg + 1.2 kg) = 1.1 N.
(b) The acceleration of the two blocks is a = F/(m1 + m2 ) The net force on block 1 is from the
force of contact, and is
P = m1 a = F m1 /(m1 + m2 ) = (3.2 N)(2.3 kg)/(2.3 kg + 1.2 kg) = 2.1 N.
Not a third law pair, eh?
P3-11 (a) Treat the system as including both the block and the rope, so that
P the mass of
the system is M + m. There is one (relevant) force which acts on the system, so
Fx = P . Then
Newton’s second law would be written as P = (M +m)ax . Solve this for ax and get ax = P/(M +m).
(b) Now consider only the block. The horizontal force doesn’t act on the block; instead, there is
the force of the rope on the block.PWe’ll assume that force has a magnitude R, and this is the only
(relevant) force on the block, so
Fx = R for the net force on the block.. In this case Newton’s
second law would be written R = M ax . Yes, ax is the same in part (a) and (b); the acceleration of
the block is the same as the acceleration of the block + rope. Substituting in the results from part
(a) we find
M
R=
P.
M +m
39
E4-1 (a) The time to pass between the plates is t = x/vx = (2.3 cm)/(9.6×108 cm/s) = 2.4×10−9 s.
2
(b) The vertical displacement of the beam is then y = ay t2 /2 == (9.4 × 1016 cm/s )(2.4 ×
−9 2
10 s) /2 = 0.27 cm.
2
(c) The velocity components are vx = 9.6 × 108 cm/s and vy = ay t = (9.4 × 1016 cm/s )(2.4 ×
−9
8
10 s) = 2.3×10 cm/s.
E4-2 ~a = ∆~v/∆t = −(6.30î − 8.42ĵ)(m/s)/(3 s) = (−2.10î + 2.81ĵ)(m/s2 ).
E4-3
(a) The velocity is given by
d~r
dt
~v
=
=
d d 2 d Aî +
Bt ĵ +
Ctk̂ ,
dt
dt
dt
(0) + 2Btĵ + C k̂.
(b) The acceleration is given by
d~v
=
dt
~a =
d d 2Btĵ +
C k̂ ,
dt
dt
(0) + 2B ĵ + (0).
(c) Nothing exciting happens in the x direction, so we will focus on the yz plane. The trajectory
in this plane is a parabola.
E4-4 (a) Maximum x is when vx = 0. Since vx = ax t + vx,0 , vx = 0 when t = −vx,0 /ax =
−(3.6 m/s)/(−1.2 m/s2 ) = 3.0 s.
(b) Since vx = 0 we have |~v| = |vy |. But vy = ay t + vy,0 = −(1.4 m/s)(3.0 s) + (0) = −4.2 m/s.
Then |~v| = 4.2 m/s.
(c) ~r = ~at2 /2 + ~v0 t, so
~r = [−(0.6 m/s2 )î − (0.7 m/s2 )ĵ](3.0 s)2 + [(3.6 m/s)î](3.0 s) = (5.4 m)î − (6.3 m)ĵ.
~ =F
~1 +F
~ 2 = (3.7 N)ĵ + (4.3 N)î. Then ~a = F/m
~
E4-5 F
= (0.71 m/s2 )ĵ + (0.83 m/s2 )î.
~
E4-6 (a) The acceleration is ~a = F/m
= (2.2 m/s2 )ĵ. The velocity after 15 seconds is ~v = ~at + v~0 ,
or
~v = [(2.2 m/s2 )ĵ](15 s) + [(42 m/s)î] = (42 m/s)î + (33 m/s)ĵ.
(b) ~r = ~at2 /2 + ~v0 t, so
~r = [(1.1 m/s2 )ĵ](15 s)2 + [(42 m/s)î](15 s) = (630 m)î + (250 m)ĵ.
E4-7 The block has a weight W = mg = (5.1 kg)(9.8 m/s2 ) = 50 N.
(a) Initially P = 12 N, so Py = (12 N) sin(25◦ ) = 5.1 N and Px = (12 N) cos(25◦ ) = 11 N. Since
the upward component is less than
P the weight, the block doesn’t leave the floor, and a normal force
willPbe present which will make
Fy = 0. There is only one contribution to the horizontal force,
so
Fx = Px . Newton’s second law then gives ax = Px /m = (11 N)/(5.1 kg) = 2.2 m/s2 .
(b) As P is increased, so is Py ; eventually Py will be large enough to overcome the weight of the
block. This happens just after Py = W = 50 N, which occurs when P = Py / sin θ = 120 N.
(c) Repeat part (a), except now P = 120 N. Then Px = 110 N, and the acceleration of the block
is ax = Px /m = 22 m/s2 .
40
E4-8 (a) The block has weight W = mg = (96.0 kg)(9.81 m/s2 ) = 942 N. Px = (450 N) cos(38◦ ) =
355 N; Py = (450 N) sin(38◦ ) = 277 N. Since Py < W the crate stays on the floor and there is a
normal force N = W − Py . The net force in the x direction is Fx = Px − (125 N) = 230 N. The
acceleration is ax = Fx /m = (230 N)/(96.0 kg) = 2.40 m/s2 .
(b) The block has mass m = W/g = (96.0 N)/(9.81 m/s2 ) = 9.79 kg. Px = (450 N) cos(38◦ ) =
355 N; Py = (450 N) sin(38◦ ) = 277 N. Since Py > W the crate lifts off of the floor! The net
force in the x direction is Fx = Px − (125 N) = 230 N. The x acceleration is ax = Fx /m =
(230 N)/(9.79 kg) = 23.5 m/s2 . The net force in the y direction is Fy = Py − W = 181 N. The y
acceleration is ay = Fy /m = (181 N)/(9.79 kg) = 18.5 m/s2 . Wow.
E4-9 Let y be perpendicular and x be parallel to the incline. Then P = 4600 N;
Px = (4600 N) cos(27◦ ) = 4100 N;
Py = (4600 N) sin(27◦ ) = 2090 N.
The weight of the car is W = mg = (1200 kg)(9.81 m/s2 ) = 11800 N;
Wx = (11800 N) sin(18◦ ) = 3650 N;
Wy = (11800 N) cos(18◦ ) = 11200 N.
Since Wy > Py the car stays on the incline. The net force in the x direction is Fx = Px −Wx = 450 N.
The acceleration in the x direction is ax = Fx /m = (450 N)/(1200 kg) = 0.375 m/s2 . The distance
traveled in 7.5 s is x = ax t2 /2 = (0.375 m/s2 )(7.5 s)2 /2 = 10.5 m.
E4-10 Constant speed means zero acceleration, so net force is zero. Let y be perpendicular and x be
parallel to the incline. The weight is W = mg = (110 kg)(9.81 m/s2 ) = 1080 N; Wx = W sin(34◦ );
Wy = W cos(34◦ ). The push F has components Fx = F cos(34◦ ) and Fy = −F sin(34◦ ). The
y components will balance after a normal force is introduced; the x components will balance if
Fx = Wx , or F = W tan(34◦ ) = (1080 N) tan(34◦ ) = 730 N.
E4-11 If the x axis is parallel to the river and the y axis is perpendicular, then ~a = 0.12î m/s2 .
The net force on the barge is
X
~ = m~a = (9500 kg)(0.12î m/s2 ) = 1100î N.
F
The force exerted on the barge by the horse has components in both the x and y direction. If
~ = P cos θî + P sin θĵ =
P = 7900 N is the magnitude of the pull and θ = 18◦ is the direction, then P
(7500î + 2400ĵ) N.
P
Let the force exerted on the barge by the water be F~w = Fw,x î + Fw,y ĵ. Then
Fx = (7500
P
P~
N) + Fw,x and
Fy = (2400 N) + Fw,y . But we already found
F, so
Fx = 1100 N = 7500 N + Fw,x ,
Fy = 0 = 2400 N + Fw,y .
Solving, Fw,x = −6400 N and Fw,y = −2400 N. The magnitude is found by Fw =
6800 N.
41
q
2 + F2
Fw,x
w,y =
E4-12 (a) Let y be perpendicular and x be parallel to the direction of motion of the plane.
Then Wx = mg sin θ; Wy = mg cos θ; m = W/g. The plane is accelerating in the x direction, so
ax = 2.62 m/s2 ; the net force is in the x direction, where Fx = max . But Fx = T − Wx , so
ax
(2.62 m/s2 )
4
◦
T = Fx + W x = W
+ W sin θ = (7.93×10 N)
+ sin(27 ) = 5.72×104 N.
g
(9.81 m/s2 )
(b) There is no motion in the y direction, so
L = Wy = (7.93×104 N) cos(27◦ ) = 7.07×104 N.
E4-13
(a) The ball rolled off horizontally so v0y = 0. Then
1
= v0y t − gt2 ,
2
1
2
(−4.23 ft) = (0)t − (32.2 ft/s )t2 ,
2
y
which can be solved to yield t = 0.514 s.
(b) The initial velocity in the x direction can be found from x = v0x t; rearranging, v0x = x/t =
(5.11 ft)/(0.514 s) = 9.94 ft/s. Since there is no y component to the velocity, then the initial speed
is v0 = 9.94 ft/s.
E4-14 The electron travels for a time t = x/vx . The electron “falls” vertically through a distance
y = −gt2 /2 in that time. Then
y=−
g
2
x
vx
2
=−
(9.81 m/s2 )
2
(1.0 m)
(3.0×107 m/s)
2
= −5.5×10−15 m.
E4-15 (a) The dart “falls” vertically through a distance y = −gt2 /2 = −(9.81 m/s2 )(0.19 s)2 /2 =
−0.18 m.
(b) The dart travels horizontally x = vx t = (10 m/s)(0.19 s) = 1.9 m.
E4-16 The initial velocity components are
vx,0 = (15 m/s) cos(20◦ ) = 14 m/s
and
vy,0 = −(15 m/s) sin(20◦ ) = −5.1 m/s.
(a) The horizontal displacement is x = vx t = (14 m/s)(2.3 s) = 32 m.
(b) The vertical displacement is
y = −gt2 /2 + vy,0 t = −(9.81 m/s2 )(2.3 s)2 /2 + (−5.1 m/s)(2.3 s) = −38 m.
E4-17
Find the time in terms of the the initial y component of the velocity:
vy = v0y − gt,
(0) = v0y − gt,
t = v0y /g.
42
Use this time to find the highest point:
y
ymax
1
= v0y t − gt2 ,
2
2
1
v0y
v0y
= v0y
− g
,
g
2
g
2
v0y
=
.
2g
2
Finally, we know the initial y component of the velocity from Eq. 2-6, so ymax = (v0 sin φ0 ) /2g.
E4-18 The horizontal displacement is x = vx t. The vertical displacement is y = −gt2 /2. Combining, y = −g(x/vx )2 /2. The edge of the nth step is located at y = −nw, x = nw, where w = 2/3 ft.
If |y| > nw when x = nw then the ball hasn’t hit the step. Solving,
g(nw/vx )2 /2 < nw,
gnw/vx2 < 2,
n
2
< 2vx2 /(gw) = 2(5.0 ft/s)2 /[(32 ft/s )(2/3 ft)] = 2.34.
Then the ball lands on the third step.
E4-19 (a) Start from the observation point 9.1 m above the ground. The ball will reach the
highest point when vy = 0, this will happen at a time t after the observation such that t = vy,0 /g =
(6.1 m/s)/(9.81 m/s2 ) = 0.62 s. The vertical displacement (from the ground) will be
y = −gt2 /2 + vy,0 t + y0 = −(9.81 m/s2 )(0.62 s)2 /2 + (6.1 m/s)(0.62 s) + (9.1 m) = 11 m.
p
(b)
The
time
for
the
ball
to
return
to
the
ground
from
the
highest
point
is
t
=
2ymax /g =
p
2
2(11 m)/(9.81 m/s ) = 1.5 s. The total time of flight is twice this, or 3.0 s. The horizontal distance
traveled is x = vx t = (7.6 m/s)(3.0 s) = 23 m.
(c) The velocity of the ball just prior to hitting the ground is
~v = ~at + ~v0 = (−9.81 m/s2 )ĵ(1.5 s) + (7.6 m/s)î = 7.6 m/sî − 15 m/suj.
√
The magnitude is 7.62 + 152 (m/s) = 17 m/s. The direction is
θ = arctan(−15/7.6) = −63◦ .
E4-20 Focus on the time it takes the ball to get to the plate, assuming it traveled in a straight
line. The ball has a “horizontal” velocity of 135 ft/s. Then t = x/vx = (60.5 ft)/(135 ft/s) = 0.448 s.
2
The ball will “fall” a vertical distance of y = −gt2 /2 = −(32 ft/s )(0.448 s)2 /2 = −3.2 ft. That’s in
the strike zone.
E4-21 Since R ∝ 1/g one can write R2 /R1 = g1 /g2 , or
g1
(9.7999 m/s2 )
∆R = R2 − R1 = R1 1 −
= (8.09 m) 1 −
= 1.06 cm.
g2
(9.8128 m/s2 )
43
E4-22 If initial position is ~r0 = 0, then final position is ~r = (13 ft)î + (3 ft)ĵ. The initial velocity
is ~v0 = v cos θî + v sin θĵ. The horizontal equation is (13 ft) = v cos θt; the vertical equation is
(3 ft) = −(g/2)t2 + v sin θt. Rearrange the vertical equation and then divide by the horizontal
equation to get
3 ft + (g/2)t2
= tan θ,
(13 ft)
or
t2 = [(13 ft) tan(55◦ ) − (3 ft)][2/(32 m/s2 )] = 0.973 s2 ,
or t = 0.986 s. Then v = (13 ft)/(cos(55◦ )(0.986 s)) = 23 ft/s.
E4-23 vx = x/t = (150 ft)/(4.50 s) = 33.3 ft/s. The time to the highest point is half the hang
2
time, or 2.25 s. The vertical speed when the ball hits the ground is vy = −gt = −(32 ft/s )(2.25 s) =
72.0 ft/s. Then the initial vertical velocity is 72.0 ft/s. The magnitude of the initial velocity is
√
722 + 332 (ft/s) = 79 ft/s. The direction is
θ = arctan(72/33) = 65◦ .
E4-24 (a) The magnitude of the initial velocity of the projectile is v = 264 ft/s. The projectile
was in the air for a time t where
t=
x
x
(2300 ft)
=
=
= 9.8 s.
vx
v cos θ
(264 ft/s) cos(−27◦ )
(b) The height of the plane was −y where
−y = gt2 /2 − vy,0 t = (32 ft/s)(9.8 s)2 /2 − (264 ft/s) sin(−27◦ )(9.8 s) = 2700 ft.
E4-25 Define the point the ball leaves the racquet as ~r = 0.
(a) The initial conditions are given as v0x = 23.6 m/s and v0y = 0. The time it takes for the ball
to reach the horizontal location of the net is found from
x = v0x t,
(12 m) = (23.6 m/s)t,
0.51 s = t,
Find how far the ball has moved horizontally in this time:
1
1
2
y = v0y t − gt2 = (0)(0.51 s) − (9.8 m/s )(0.51 s)2 = −1.3 m.
2
2
Did the ball clear the net? The ball started 2.37 m above the ground and “fell” through a distance
of 1.3 m by the time it arrived at the net. So it is still 1.1 m above the ground and 0.2 m above the
net.
(b) The initial conditions are now given by v0x = (23.6 m/s)(cos[−5.0◦ ]) = 23.5 m/s and v0y =
(23.6 m/s)(sin[−5.0◦ ]) = −2.1 m/s. Now find the time to reach the net just as done in part (a):
t = x/v0x = (12.0 m)/(23.5 m/s) = 0.51 s.
Find the vertical position of the ball when it arrives at the net:
1
1
2
y = v0y t − gt2 = (−2.1 m/s)(0.51 s) − (9.8 m/s )(0.51 s)2 = −2.3 m.
2
2
Did the ball clear the net? Not this time; it started 2.37 m above the ground and then passed the
net 2.3 m lower, or only 0.07 m above the ground.
44
E4-26 The initial speed of the ball is given by v =
time of flight from the batter to the wall is
√
q
2
gR = (32 ft/s )(350 ft) = 106 ft/s. The
t = x/vx = (320 ft)/[(106 ft/s) cos(45◦ )] = 4.3 s.
The height of the ball at that time is given by y = y0 + v0y t − 12 gt2 , or
2
y = (4 ft) + (106 ft/s) sin(45◦ )(4.3 s) − (16 ft/s )(4.3 s)2 = 31 ft,
clearing the fence by 7 feet.
E4-27 The ball lands x = (358 ft) + (39 ft) cos(28◦ ) = 392 ft from the initial position. The ball
lands y = (39 ft) sin(28◦ ) − (4.60 ft) = 14 ft above the initial position. The horizontal equation
is (392 ft) = v cos θt; the vertical equation is (14 ft) = −(g/2)t2 + v sin θt. Rearrange the vertical
equation and then divide by the horizontal equation to get
14 ft + (g/2)t2
= tan θ,
(392 ft)
or
t2 = [(392 ft) tan(52◦ ) − (14 ft)][2/(32 m/s2 )] = 30.5 s2 ,
or t = 5.52 s. Then v = (392 ft)/(cos(52◦ )(5.52 s)) = 115 ft/s.
E4-28 Since ball is traveling at 45◦ when it returns to the same level from which it was thrown
for maximum range, then we can assume it actually traveled ≈ 1.6 m. farther than it would have
had it been launched from the ground level. This won’t make a big difference, but that means that
R = 60.0 m − 1.6 m = 58.4 m. If v0 is initial speed of ball thrown directly up, the ball rises to the
highest point in a time t = v0 /g, and that point is ymax = gt2 /2 = v02 /(2g) above the launch point.
But v02 = gR, so ymax = R/2 = (58.4 m)/2 = 29.2 m. To this we add the 1.60 m point of release to
get 30.8 m.
P
E4-29 The net force on the pebble is zero, so
Fy = 0. There are only two forces on the
pebble, the force of gravity W and the force of the water on the pebble FP W . These point in
opposite directions, so 0 = FP W − W . But W = mg = (0.150 kg)(9.81 m/s2 ) = 1.47 N. Since
FP W = W in this problem, the force of the water on the pebble must also be 1.47 N.
E4-30 Terminal speed is when drag force equal weight, or mg = bv T 2 . Then v T =
E4-31
p
mg/b.
Eq. 4-22 is
vy (t) = v T 1 − e−bt/m ,
where we have used Eq. 4-24 to substitute for the terminal speed. We want to solve this equation
for time when vy (t) = v T /2, so
1
−bt/m
,
2 vT = vT 1 − e
1
−bt/m
,
2 = 1−e
e−bt/m = 12
bt/m = − ln(1/2)
t= m
b ln 2
45
E4-32 The terminal speed is 7 m/s for a raindrop with r = 0.15 cm. The mass of this drop is
m = 4πρr3 /3, so
b=
4π(1.0×10−3 kg/cm3 )(0.15 cm)3 (9.81 m/s2 )
mg
=
= 2.0×10−5 kg/s.
vT
3(7 m/s)
E4-33 (a) The speed of the train is v = 9.58 m/s. The drag force on one car is f = 243(9.58) N =
2330 N. The total drag force is 23(2330 N) = 5.36×104 N. The net force exerted on the cars (treated
as a single entity) is F = ma = 23(48.6×103 kg)(0.182 m/s2 ) = 2.03×105 N. The pull of the locomotive
is then P = 2.03×105 N + 5.36×104 N = 2.57×105 N.
(b) If the locomotive is pulling the cars at constant speed up an incline then it must exert a
force on the cars equal to the sum of the drag force and the parallel component of the weight. The
drag force is the same in each case, so the parallel component of the weight is W|| = W sin θ =
2.03×105 N = ma, where a is the acceleration from part (a). Then
θ = arcsin(a/g) = arcsin[(0.182 m/s2 )/(9.81 m/s2 )] = 1.06◦ .
E4-34 (a) a = v 2 /r = (2.18×106 m/s)2 /(5.29×10−11 m) = 8.98×1022 m/s2 .
(b) F = ma = (9.11×10−31 kg)(8.98×1022 m/s2 ) = 8.18×10−8 N, toward the center.
q
√
2
E4-35 (a) v = rac = (5.2 m)(6.8)(9.8 m/s ) = 19 m/s.
(b) Use the fact that one revolution corresponds to a length of 2πr:
m
1 rev
60 s
rev
19
= 35
.
s 2π(5.2 m)
1 min
min
E4-36 (a) v = 2πr/T = 2π(15 m)/(12 s) = 7.85 m/s. Then a = v 2 /r = (7.85 m/s)2 /(15 m) =
4.11 m/s2 , directed toward center, which is down.
(b) Same arithmetic as in (a); direction is still toward center, which is now up.
(c) The magnitude of the net force in both (a) and (b) is F = ma = (75 kg)(4.11 m/s2 ) = 310 N.
The weight of the person is the same in both parts: W = mg = (75 kg)(9.81 m/s2 ) = 740 N. At
the top the net force is down, the weight is down, so the Ferris wheel is pushing up with a force of
P = W − F = (740 N) − (310 N) = 430 N. At the bottom the net force is up, the weight is down, so
the Ferris wheel is pushing up with a force of P = W + F = (740 N) + (310 N) = 1050 N.
E4-37 (a) v = 2πr/T = 2π(20×103 m)/(1.0 s) = 1.26×105 m/s.
(b) a = v 2 /r = (1.26×105 m/s)2 /(20×103 m) = 7.9×105 m/s2 .
E4-38 (a) v = 2πr/T = 2π(6.37 × 106 m)/(86400 s) = 463 m/s. a = v 2 /r = (463 m/s)2 /(6.37 ×
106 m) = 0.034 m/s2 .
(b) The net force on the object is F = ma = (25.0 kg)(0.034 m/s2 ) = 0.85 N. There are two forces
on the object: a force up from the scale (S), and the weight down, W = mg = (25.0 kg)(9.80 m/s2 ) =
245 N. Then S = F + W = 245 N + 0.85 N = 246 N.
E4-39 Let ∆x = 15 m be the length; tw = 90 s, the time to walk the stalled Escalator; ts = 60
s, the time to ride the moving Escalator; and tm , the time to walk up the moving Escalator.
The walking speed of the person relative to a fixed Escalator is vwe = ∆x/tw ; the speed of the
Escalator relative to the ground is veg = ∆x/ts ; and the speed of the walking person relative to the
46
ground on a moving Escalator is vwg = ∆x/tm . But these three speeds are related by vwg = vwe +veg .
Combine all the above:
vwg
∆x
tm
1
tm
= vwe + veg ,
∆x ∆x
=
+
,
tw
ts
1
1
=
+ .
tw
ts
Putting in the numbers, tm = 36 s.
E4-40 Let v w be the walking speed, v s be the sidewalk speed, and v m = v w + v s be Mary’s speed
while walking on the moving sidewalk. All three cover the same distance x, so vi = x/ti , where i is
one of w, s, or m. Put this into the Mary equation, and
1/tm = 1/tw + 1/ts = 1/(150 s) + 1/(70 s) = 1/48 s.
E4-41 If it takes longer to fly westward then the speed of the plane (relative to the ground)
westward must be less than the speed of the plane eastward. We conclude that the jet-stream must
be blowing east. The speed of the plane relative to the ground is v e = v p + v j when going east and
v w = v p − v j when going west. In either case the distance is the same, so x = vi ti , where i is e or
w. Since tw − te is given, we can write
tw − t e =
x
2v j
x
−
=x 2
.
vp − vj
vp + vj
vp − vj 2
Solve the quadratic if you want, but since v j v p we can neglect it in the denominator and
v j = v p 2 (0.83 h)/(2x) = (600 mi/h)2 (0.417 h)/(2700 mi) = 56 mi/hr.
E4-42 The horizontal component of the rain drop’s velocity is 28 m/s. Since vx = v sin θ, v =
(28 m/s)/ sin(64◦ ) = 31 m/s.
E4-43 (a) The position of the bolt relative to the elevator is ybe , the position of the bolt relative
to the shaft is ybs , and the position of the elevator relative to the shaft is yes . Zero all three positions
at t = 0; at this time v0,bs = v0,es = 8.0 ft/s.
The three equations describing the positions are
ybs
yes
ybe + res
1
= v0,bs t − gt2 ,
2
1
= v0,es t + at2 ,
2
= rbs ,
where a = 4.0 m/s2 is the upward acceleration of the elevator. Rearrange the last equation and
solve for ybe ; get ybe = − 12 (g + a)t2 , where advantage was taken of the fact that the initial velocities
are the same.
Then
q
p
2
2
t = −2ybe /(g + a) = −2(−9.0 ft)/(32 ft/s + 4 ft/s ) = 0.71 s
(b) Use the expression for ybs to find how the bolt moved relative to the shaft:
1
1
2
ybs = v0,bs t − gt2 = (8.0 ft)(0.71 s) − (32 ft/s )(0.71 s)2 = −2.4 ft.
2
2
47
E4-44 The speed of the plane relative to the ground is v pg = (810 km)/(1.9 h) = 426 km/h. The
velocity components of the plane relative to the air are vN = (480 km/h) cos(21◦ ) = 448 km/h and
vE = (480 km/h) sin(21◦ ) = 172 km/h. The wind must be blowing with a component of 172 km/h
to the west and a component of 448 − 426 = 22 km/h to the south.
E4-45 (a) Let î point east and ĵ point north. The velocity of the torpedo is ~v = (50 km/h)î sin θ +
(50 km/h)ĵ cos θ. The initial coordinates of the battleship are then ~r0 = (4.0 km)î sin(20◦ ) +
(4.0 km)ĵ cos(20◦ ) = (1.37 km)î + (3.76 km)ĵ. The final position of the battleship is ~r = (1.37 km +
24 km/ht)î + (3.76 km)ĵ, where t is the time of impact. The final position of the torpedo is the
same, so
[(50 km/h)î sin θ + (50 km/h)ĵ cos θ]t = (1.37 km + 24 km/ht)î + (3.76 km)ĵ,
or
[(50 km/h) sin θ]t − 24 km/ht = 1.37 km
and
[(50 km/h) cos θ]t = 3.76 km.
Dividing the top equation by the bottom and rearranging,
50 sin θ − 24 = 18.2 cos θ.
Use any trick you want to solve this. I used Maple and found θ = 46.8◦ .
(b) The time to impact is then t = 3.76 km/[(50 km/h) cos(46.8◦ )] = 0.110 h, or 6.6 minutes.
P4-1 Let ~rA be the position of particle of particle A, and ~rB be the position of particle B. The
equations for the motion of the two particles are then
~rA
= ~r0,A + ~vt,
= dĵ + vtî;
1 2
~at ,
~rB =
2
1
=
a(sin θî + cos θĵ)t2 .
2
A collision will occur if there is a time when ~rA = ~rB . Then
dĵ + vtî =
1
a(sin θî + cos θĵ)t2 ,
2
but this is really two equations: d = 21 at2 cos θ and vt = 12 at2 sin θ.
Solve the second one for t and get t = 2v/(a sin θ). Substitute that into the first equation, and
then rearrange,
d
=
d
=
sin2 θ
=
1 − cos2 θ
=
0
=
1 2
at cos θ,
2
2
1
2v
a
cos θ,
2
a sin θ
2v
cos θ,
ad
2
2v
cos θ,
ad
2v 2
cos2 θ +
cos θ − 1.
ad
48
This last expression is quadratic in cos θ. It simplifies the solution if we define b = 2v/(ad) = 2(3.0
m/s)2 /([0.4 m/s2 ][30 m]) = 1.5, then
√
−b ± b2 + 4
cos θ =
= −0.75 ± 1.25.
2
Then cos θ = 0.5 and θ = 60◦ .
P4-2 (a) The acceleration of the ball is ~a = (1.20 m/s2 )î − (9.81 m/s2 )ĵ. Since ~a is constant the
trajectory is given by ~r = ~at2 /2, since ~v0 = 0 and we choose ~r0 = 0. This is a straight line trajectory,
with a direction given by ~a. Then
θ = arctan(9.81/1.20) = 83.0◦ .
◦
and R = (39.0 m)/ tan(83.0◦ ) = 4.79 m. It will be useful to find
) = 39.3 m.
√ H = (39.0 m)/ sin(83.0
2
2
2
(b) The magnitude of the acceleration of the ball is a = 9.81 + 1.20p(m/s ) = 9.88 m/s2 . The
time for the ball to travel down the hypotenuse of the figure is then t = 2(39.3 m)/(9.88 m/s2 ) =
2.82 s.
(c) The magnitude of the speed of the ball at the bottom will then be
v = at = (9.88 m/s2 )(2.82 s) = 27.9 m/s.
~ = (61.2 kN) cos(58.0◦ )î + (61.2 kN) sin(58.0◦ )î = 32.4 kNî +
P4-3 (a) The rocket thrust is T
~ =T
~ + W,
~ or
51.9 kNĵ. The net force on the rocket is the F
~ = 32.4 kNî + 51.9 kNĵ − (3030 kg)(9.81 m/s2 )ĵ = 32.4 kNî + 22.2 kNĵ.
F
The acceleration (until rocket cut-off) is this net force divided by the mass, or
~a = 10.7m/s2 î + 7.33m/s2 ĵ.
The position at rocket cut-off is given by
~r = ~at2 /2 = (10.7m/s2 î + 7.33m/s2 ĵ)(48.0 s)2 /2,
=
1.23×104 mî + 8.44×103 mĵ.
The altitude at rocket cut-off is then 8.44 km.
(b) The velocity at rocket cut-off is
~v = ~at = (10.7m/s2 î + 7.33m/s2 ĵ)(48.0 s) = 514 m/sî + 352 m/sĵ,
this becomes the initial velocity for the “free fall” part of the journey. The rocket will hit the ground
after t seconds, where t is the solution to
0 = −(9.81 m/s2 )t2 /2 + (352 m/s)t + 8.44×103 m.
The solution is t = 90.7 s. The rocket lands a horizontal distance of x = vx t = (514 m/s)(90.7 s) =
4.66×104 m beyond the rocket cut-off; the total horizontal distance covered by the rocket is 46.6 km+
12.3 km = 58.9 km.
49
P4-4
(a) The horizontal speed of the ball is vx = 135 ft/s. It takes
t = x/vx = (30.0 ft)/(135 ft/s) = 0.222 s
to travel the 30 feet horizontally, whether the first 30 feet, the last 30 feet, or 30 feet somewhere in
the middle.
2
(b) The ball “falls” y = −gt2 /2 = −(32 ft/s )(0.222 s)2 /2 = −0.789 ft while traveling the first
30 feet.
2
(c) The ball “falls” a total of y = −gt2 /2 = −(32 ft/s )(0.444 s)2 /2 = −3.15 ft while traveling
the first 60 feet, so during the last 30 feet it must have fallen (−3.15 ft) − (−0.789 ft) = −2.36 ft.
(d) The distance fallen because of acceleration is not linear in time; the distance moved horizontally is linear in time.
P4-5
(a) The initial velocity of the ball has components
vx,0 = (25.3 m/s) cos(42.0◦ ) = 18.8 m/s
and
vy,0 = (25.3 m/s) sin(42.0◦ ) = 16.9 m/s.
The ball is in the air for t = x/vx = (21.8 m)/(18.8 m/s) = 1.16 s before it hits the wall.
(b) y = −gt2 /2 + vy,0 t = −(4.91 m/s2 )(1.16 s)2 + (16.9 m/s)(1.16 s) = 13.0 m.
(c) vx = vx,0 = 18.8 m/s. vy = −gt + vy,0 = −(9.81 m/s2 )(1.16 s) + (16.9 m/s) = 5.52 m/s.
(d) Since vy > 0 the ball is still heading up.
P4-6 (a) The initial vertical velocity is vy,0 = v0 sin φ0 . The time to the highest point is t = vy,0 /g.
The highest point is H = gt2 /2. Combining,
H = g(v0 sin φ0 /g)2 /2 = v02 sin2 φ0 /(2g).
The range is R = (v02 /g) sin 2φ0 = 2(v02 /g) sin φ0 cos φ0 . Since tan θ = H/(R/2), we have
tan θ =
v02 sin2 φ0 /g
1
2H
=
= tan φ0 .
2
R
2(v0 /g) sin φ0 cos φ0
2
(b) θ = arctan(0.5 tan 45◦ ) = 26.6◦ .
P4-7 The components of the initial velocity are given by v0x = v0 cos θ = 56 ft/s and v0y =
v0 sin θ = 106 ft/s where we used v0 = 120 ft/s and θ = 62◦ .
(a) To find h we need only find out the vertical position of the stone when t = 5.5 s.
1
1
2
y = v0y t − gt2 = (106 ft/s)(5.5 s) − (32 ft/s )(5.5 s)2 = 99 ft.
2
2
(b) Look at this as a vector problem:
~v
= ~v0 + ~at,
=
v0x î + v0y ĵ − g ĵt,
= v0x î + (v0y − gt) ĵ,
2
= (56 ft/s)î + (106 ft/s − (32 ft/s )(5.5 s) ĵ,
=
(56 ft/s)î + (−70.0 ft/s)ĵ.
50
p
The magnitude of this vector gives the speed when t = 5.5 s; v = 562 + (−70)2 ft/s= 90 ft/s.
(c) Highest point occurs when vy = 0. Solving Eq. 4-9(b) for time; vy = 0 = v0y − gt =
2
(106 ft/s) − (32 ft/s )t; t = 3.31 s. Use this time in Eq. 4-10(b),
1
1
2
y = v0y t − gt2 = (106 ft/s)(3.31 s) − (32 ft/s )(3.31 s)2 = 176 ft.
2
2
P4-8 (a) Since R = (v02 /g) sin 2φ0 , it is sufficient to prove that sin 2φ0 is the same for both
sin 2(45◦ + α) and sin 2(45◦ − α).
sin 2(45◦ ± α) = sin(90◦ ± 2α) = cos(±2α) = cos(2α).
Since the ± dropped out, the two quantities are equal.
(b) φ0 = (1/2) arcsin(Rg/v02 ) = (1/2) arcsin((20.0 m)(9.81 m/s2 )/(30.0 m/s)2 ) = 6.3◦ . The other
choice is 90◦ − 6.3◦ = 83.7◦ .
P4-9 To score the ball must pass the horizontal distance of 50 m with an altitude of no less than
3.44 m. The initial velocity components are v0x = v0 cos θ and v0y = v0 sin θ where v0 = 25 m/s,
and θ is the unknown.
The time to the goal post is t = x/v0x = x/(v0 cos θ).
The vertical motion is given by
2
1
x
1
x
y = v0y t − gt2 = (v0 sin θ)
− g
,
2
v0 cos θ
2
v0 cos θ
gx2 1
sin θ
− 2
.
= x
cos θ
2v0 cos2 θ
In this last expression y needs to be greater than 3.44 m. In this last expression use
1
1
cos2 θ
1 − cos2 θ
sin2 θ
−1+1=
−
+1=
+1=
+ 1 = tan2 θ + 1.
2
2
2
2
cos θ
cos θ cos θ
cos θ
cos2 θ
This gives for our y expression
y = x tan θ −
gx2
tan2 θ + 1 ,
2v0
which can be combined with numbers and constraints to give
2
(9.8 m/s )(50 m)2
tan2 θ + 1 ,
2(25 m/s)2
3.44 ≤ 50 tan θ − 20 tan2 θ + 1 ,
(3.44 m)
≤ (50 m) tan θ −
0 ≤
−20 tan2 θ + 50 tan θ − 23
Solve, and tan θ = 1.25 ± 0.65, so the allowed kicking angles are between θ = 31◦ and θ = 62◦ .
P4-10 (a) The height of the projectile atpthe highest
p point is H = L sin θ. The amount of time
before the projectile hits the ground is t = p
2H/g = 2L sin θ/g. The horizontal distance covered
by the projectile in this time is x = vx t = v 2L sin θ/g. The horizontal distance to the projectile
when it is at the highest point is x0 = L cos θ. The projectile lands at
p
D = x − x0 = v 2L sin θ/g − L cos θ.
(b) The projectile will pass overhead if D > 0.
51
P4-11 v 2 = vx2 + vy2 . For a projectile vx is constant, so we need only evaluate d2 (vy2 )/dt2 . The first
derivative is 2vy dvy /dt = −2vy g. The derivative of this (the second derivative) is −2g dvy /dt = 2g 2 .
P4-12
|~r| is a maximum when r2 is a maximum. r2 = x2 + y 2 , or
r2
=
(vx,0 t)2 + (−gt2 /2 + vy,0 t)2 ,
=
2
2
(v0 t cos φ0 ) + v0 t sin φ0 − gt2 /2 ,
= v02 t2 − v0 gt3 sin φ0 + g 2 t4 /4.
We want to look for the condition which will allow dr2 /dt to vanish. Since
dr2 /dt = 2v02 t − 3v0 gt2 sin φ0 + g 2 t3
we can focus on the quadratic discriminant, b2 − 4ac, which is
9v02 g 2 sin2 φ0 − 8v02 g 2 ,
a quantity which will only be greater than zero if 9 sin2 φ0 > 8. The critical angle is then
p
φc = arcsin( 8/9) = 70.5◦ .
P4-13 There is a downward force on the balloon of 10.8 kN from gravity and an upward force
of 10.3 kN from the buoyant force of the air. The resultant of these two forces is 500 N down, but
since the balloon is descending at constant speed so the net force on the balloon must be zero. This
is possible because there is a drag force on the balloon of D = bv 2 , this force is directed upward.
The magnitude must be 500 N, so the constant b is
b=
(500 N)
= 141 kg/m.
(1.88 m/s)2
If the crew drops 26.5 kg of ballast they are “lightening” the balloon by
(26.5 kg)(9.81 m/s2 ) = 260 N.
This reduced the weight, but not the buoyant force, so the drag force at constant speed will now be
500 N − 260 N = 240 N.
The new constant downward speed will be
p
p
v = D/b = (240 N)/(141 kg/m) = 1.30 m/s.
P4-14
The constant b is
b = (500 N)/(1.88 m/s) = 266 N · s/m.
The drag force after “lightening” the load will still be 240 N. The new downward speed will be
v = D/b = (240 N)/(266 N · s/m) = 0.902 m/s.
P4-15 (a) Initially v0 = 0, so D = 0, the only force is the weight, so a = −g.
p
(b) After some time the acceleration is zero, then W = D, or bv T 2 = mg, or v T = mg/b.
(c) When v = v T /2 the drag force is D = bv T 2 /4 = mg/4, so the net force is F = D − W =
−3mg/4. The acceleration is the a = −3g/4.
52
P4-16 (a) The net force on the barge is F = −D = −bv, this results in a differential equation
m dv/dt = −bv, which can be written as
dv/v
Z
dv/v
= −(b/m)dt,
Z
= −(b/m) dt,
ln(v f /v i ) = −bt/m.
Then t = (m/b) ln(v i /v f ).
(b) t = [(970 kg)/(68 N · s/m)] ln(32/8.3) = 19 s.
P4-17
(a) The acceleration is the time derivative of the velocity,
ay =
mg b
dvy
d mg =
1 − e−bt/m
=
e−bt/m ,
dt
dt b
b m
which can be simplified as ay = ge−bt/m . For large t this expression approaches 0; for small t the
exponent can be expanded to give
bt
ay ≈ g 1 −
= g − v T t,
m
where in the last line we made use of
(b) The position is the integral of
Z t
vy dt
0
Z t
dy
dt
0 dt
Z y
dy
0
Eq. 4-24.
the velocity,
Z t
mg =
1 − e−bt/m
dt,
b
0
t
mg =
t − (−m/b)e−bt/m ,
b
0
v T −vT t/g
e
−1 ,
= vT t +
g
v T −vT t/g
y = vT t +
e
−1 .
g
P4-18 (a) We have vy = v T (1 − e−bt/m ) from Eq. 4-22; this can be substituted into the last line
of the solution for P4-17 to give
vy
y95 = v T t −
.
g
We can also rearrange Eq. 4-22 to get t = −(m/b) ln(1 − vy /v T ), so
vy
2
y95 = v T /g − ln(1 − vy /v T ) −
.
vT
But vy /v T = 0.95, so
y95 = v T 2 /g (− ln(0.05) − 0.95) = v T 2 /g (ln 20 − 19/20) .
(b) y95 = (42 m/s)2 /(9.81 m/s2 )(2.05) = 370 m.
P4-19 (a) Convert units first. v = 86.1 m/s, a = 0.05(9.81 m/s2 ) = 0.491 m/s2 . The minimum
2
2
radius is r =
p(86.1 m/s)/(0.491 m/s ) = 15 km.
√v /a =
2
(b) v = ar = (0.491 m/s )(940 m) = 21.5 m/s. That’s 77 km/hr.
53
P4-20 (a) The position is given by ~r = R sin ωt î + R(1 − cos ωt)ĵ, where ω = 2π/(20 s) = 0.314 s−1
and R = 3.0 m. When t = 5.0 s ~r = (3.0 m)î + (3.0 m)ĵ; when t = 7.5 s ~r = (2.1 m)î + (5.1 m)ĵ;
when t = 10 s ~r = (6.0 m)ĵ. These vectors have magnitude 4.3 m, 5.5 m and 6.0 m, respectively. The
vectors have direction 45◦ , 68◦ and 90◦ respectively.
(b) ∆~r = (−3.0 m)î + (3.0 m)ĵ, which has magnitude 4.3 m and direction 135◦ .
(c) v av = ∆r/∆t = (4.3 m)/(5.0 s) = 0.86 m/s. The direction is the same as ∆~r.
(d) The velocity is given by ~v = Rω cos ωt î + Rω sin ωt ĵ. At t = 5.0 s ~v = (0.94 m/s)ĵ; at t = 10 s
~v = (−0.94 m/s)î.
(e) The acceleration is given by ~a = −Rω 2 sin ωt î + Rω 2 cos ωt ĵ. At t = 5.0 s ~a = (−0.30 m/s2 )î;
at t = 10 s ~a = (−0.30 m/s2 )ĵ.
P4-21 Start from where the stone lands; in order to get there the stone fell through a vertical
distance of 1.9 m while moving 11 m horizontally. Then
r
1 2
−2y
y = − gt which can be written as t =
.
2
g
Putting in the numbers, t = 0.62 s is the time of flight from the moment the string breaks. From
this time find the horizontal velocity,
vx =
x
(11 m)
=
= 18 m/s.
t
(0.62 s)
Then the centripetal acceleration is
ac =
(18 m/s)2
v2
=
= 230 m/s2 .
r
(1.4 m)
P4-22 (a) The path traced out by her feet has circumference c1 = 2πr cos 50◦ , where r is the
radius of the earth; the path traced out by her head has circumference c2 = 2π(r + h) cos 50◦ , where
h is her height. The difference is ∆c = 2πh cos 50◦ = 2π(1.6 m) cos 50◦ = 6.46 m.
(b) a = v 2 /r = (2πr/T )2 /r = 4π 2 r/T 2 . Then ∆a = 4π 2 ∆r/T 2 . Note that ∆r = h cos θ! Then
∆a = 4π 2 (1.6 m) cos 50◦ /(86400 s)2 = 5.44×10−9 m/s2 .
P4-23
(a) A cycloid looks something like this:
(b) The position of the particle is given by
~r = (R sin ωt + ωRt)î + (R cos ωt + R)ĵ.
The maximum value of y occurs whenever cos ωt = 1. The minimum value of y occurs whenever
cos ωt = −1. At either of those times sin ωt = 0.
The velocity is the derivative of the displacement vector,
~v = (Rω cos ωt + ωR)î + (−Rω sin ωt)ĵ.
When y is a maximum the velocity simplifies to
~v = (2ωR)î + (0)ĵ.
54
When y is a minimum the velocity simplifies to
~v = (0)î + (0)ĵ.
The acceleration is the derivative of the velocity vector,
~a = (−Rω 2 sin ωt)î + (−Rω 2 cos ωt)ĵ.
When y is a maximum the acceleration simplifies to
~a = (0)î + (−Rω 2 )ĵ.
When y is a minimum the acceleration simplifies to
~a = (0)î + (Rω 2 )ĵ.
P4-24 (a) The speed of the car is v c = 15.3 m/s. The snow appears to fall with an angle θ =
arctan(15.3/7.8) = 63◦ .
p
(b) The apparent speed is (15.3)2 + (7.8)2 (m/s) = 17.2 m/s.
P4-25 (a) The decimal angles are 89.994250◦ and 89.994278◦ . The earth moves in the orbit
around the sun with a speed of v = 2.98×104 m/s (Appendix C). The speed of light is then between
c = (2.98 × 104 m/s) tan(89.994250◦ ) = 2.97 × 108 m/s and c = (2.98 × 104 m/s) tan(89.994278◦ ) =
2.98×108 m/s. This method is highly sensitive to rounding. Calculating the orbital speed from the
radius and period of the Earth’s orbit will likely result in different answers!
P4-26 (a) Total distance is 2l, so t0 = 2l/v.
(b) Assume wind blows east. Time to travel out is t1 = l/(v + u), time to travel back is
t2 = l/(v − u). Total time is sum, or
tE =
l
2lv
t0
l
+
= 2
=
.
v+u v−u
v − u2
1 − u2 /v 2
If wind blows west the times reverse, but the result is otherwise the same.
(c) Assume wind blows north. The airplane will still have a speed of v relative to the wind, but
it will
√ need to fly with a heading away from east. The speed of the plane relative to the ground will
be v 2 − u2 . This will be the speed even when it flies west, so
tN = √
2l
t0
=p
.
2
−u
1 − u2 /v 2
v2
(d) If u > v the wind sweeps the plane along in one general direction only; it can never fly back.
Sort of like a black hole event horizon.
P4-27 The velocity of the police car with respect to the ground is ~vpg = −76km/hî. The velocity
of the motorist with respect the ground is ~vmg = −62 km/hĵ.
The velocity of the motorist with respect to the police car is given by solving
~vmg = ~vmp + ~vpg ,
so ~vmp = 76km/hî − 62 km/hĵ. This velocity has magnitude
p
vmp = (76km/h)2 + (−62 km/h)2 = 98 km/h.
55
The direction is
θ = arctan(−62 km/h)/(76km/h) = −39◦ ,
but that is relative to î. We want to know the direction relative to the line of sight. The line of sight
is
α = arctan(57 m)/(41 m) = −54◦
relative to î, so the answer must be 15◦ .
P4-28 (a) The velocity of the plane with respect to the air is ~vpa ; the velocity of the air with
respect to the ground is ~vag , the velocity of the plane with respect to the ground is ~vpg . Then
~vpg = ~vpa + ~vag . This can be represented by a triangle; since the sides are given we can find the
angle between ~vag and ~vpg (points north) using the cosine law
(135)2 − (135)2 − (70)2
= 75◦ .
θ = arccos
−2(135)(70)
(b) The direction of ~vpa can also be found using the cosine law,
(70)2 − (135)2 − (135)2
= 30◦ .
θ = arccos
−2(135)(135)
56
E5-1 There are three forces which act on the charged sphere— an electric force, FE , the force of
gravity, W , and the tension in the string, T . All arranged as shown in the figure on the right below.
(a) Write the vectors so that they geometrically show that the sum is zero, as in the figure
on the left below. Now W = mg = (2.8 × 10−4 kg)(9.8 m/s2 ) = 2.7 × 10−3 N. The magnitude
of the electric force can be found from the tangent relationship, so FE = W tan θ = (2.7 × 10−3
N) tan(33◦ ) = 1.8 × 10−3 N.
FE
T
θ
FE
T θ W
W
(a)
(b)
(b) The tension can be found from the cosine relation, so
T = W/ cos θ = (2.7 × 10−3 N)/ cos(33◦ ) = 3.2 × 10−3 N.
2
2
E5-2 (a) The net force on the elevator is F = ma = W a/g = (6200 lb)(3.8 ft/s )/(32 ft/s ) =
740 lb. Positive means up. There are two force on the elevator: a weight W down and a tension
from the cable T up. Then F = T − W or T = F + W = (740 lb) + (6200 lb) = 6940 lb.
(b) If the elevator acceleration is down then F = −740 lb; consequently T = F +W = (−740 lb)+
(6200 lb) = 5460 lb.
E5-3 (a) The tension T is up, the weight W is down, and the net force F is in the direction of the
acceleration (up). Then F = T − W . But F = ma and W = mg, so
m = T /(a + g) = (89 N)/[(2.4 m/s2 ) + (9.8 m/s2 )] = 7.3 kg.
(b) T = 89 N. The direction of velocity is unimportant. In both (a) and (b) the acceleration is
up.
E5-4 The average speed of the elevator during deceleration is v av = 6.0 m/s. The time to stop the
elevator is then t = (42.0 m)/(6.0 m/s) = 7.0 s. The deceleration is then a = (12.0 m/s)/(7.0 s) =
1.7 m/s2 . Since the elevator is moving downward but slowing down, then the acceleration is up,
which will be positive.
The net force on the elevator is F = ma; this is equal to the tension T minus the weight W .
Then
T = F + W = ma + mg = (1600 kg)[(1.7 m/s2 ) + (9.8 m/s2 )] = 1.8×104 N.
E5-5
(a) The magnitude of the man’s acceleration is given by
a=
m2 − m1
(110 kg) − (74 kg)
g=
g = 0.2g,
m2 + m1
(110 kg) + (74 kg)
and is directed down. The time which elapses while he falls is found by solving y = v0y t + 12 ay t2 ,
or, with numbers, (−12 m) = (0)t + 21 (−0.2g)t2 which has the solutions t = ±3.5 s. The velocity
with which he hits the ground is then v = v0y + ay t = (0) + (−0.2g)(3.5 s) = −6.9 m/s.
57
(b) Reducing the speed can be accomplished by reducing the acceleration. We can’t change Eq.
5-4 without also changing one of the assumptions that went into it. Since the man is hoping to
reduce the speed with which he hits the ground, it makes sense that he might want to climb up the
rope.
E5-6 (a) Although it might be the monkey which does the work, the upward force to lift him still
comes from the tension in the rope. The minimum tension to lift the log is T = W l = ml g. The net
force on the monkey is T − W m = mm a. The acceleration of the monkey is then
a = (ml − mm )g/mm = [(15 kg) − (11 kg)](9.8 m/s2 )/(11 kg) = 3.6 m/s2 .
(b) Atwood’s machine!
a = (ml − mm )g/(ml + mm ) = [(15 kg) − (11 kg)](9.8 m/s2 )/[(15 kg) + (11 kg)] = 1.5 m/s.
(c) Atwood’s machine!
T = 2ml mm g/(ml + mm ) = 2(15 kg)(11 kg)(9.8 m/s2 )/[(15 kg) + (11 kg)] = 120 N.
E5-7 The weight of each car has two components: a component parallel to the cables W|| = W sin θ
and a component normal to the cables W⊥ .. The normal component is “balanced” by the supporting
cable. The parallel component acts with the pull cable.
In order to accelerate a car up the incline there must be a net force up with magnitude F = ma.
Then F = T above − T below − W|| , or
∆T = ma + mg sin θ = (2800 kg)[(0.81 m/s2 ) + (9.8 m/s2 ) sin(35◦ )] = 1.8×104 N.
E5-8 The tension in the cable is T , the weight of the man + platform system is W = mg, and the
net force on the man + platform system is F = ma = W a/g = T − W . Then
2
2
T = W a/g + W = W (a/g + 1) = (180 lb + 43 lb)[(1.2 ft/s )/(32 ft/s ) + 1] = 231 lb.
E5-9
See Sample Problem 5-8. We need only apply the (unlabeled!) equation
µs = tan θ
to find the egg angle. In this case θ = tan−1 (0.04) = 2.3◦ .
E5-10 (a) The maximum force of friction is F = µs N . If the rear wheels support half of the weight
of the automobile then N = W/2 = mg/2. The maximum acceleration is then
a = F/m = µs N/m = µs g/2.
(b) a = (0.56)(9.8 m/s2 )/2 = 2.7 m/s2 .
E5-11 The maximum force of friction is F = µs N . Since there is no motion in the y direction the
magnitude of the normal force must equal the weight, N = W = mg. The maximum acceleration is
then
a = F/m = µs N/m = µs g = (0.95)(9.8 m/s2 ) = 9.3 m/s2 .
E5-12 There is no motion in the vertical direction, so N = W = mg. Then µk = F/N =
(470 N)/[(9.8 m/s2 )(79 kg)] = 0.61.
58
E5-13 A 75 kg mass has a weight of W = (75 kg)(9.8 m/s2 ) = 735 N, so the force of friction on
each end of the bar must be 368 N. Then
F ≥
(368 N)
fs
=
= 900 N.
µs
(0.41)
E5-14 (a) There is no motion in the vertical direction, so N = W = mg.
To get the box moving you must overcome static friction and push with a force of P ≥ µs N =
(0.41)(240 N) = 98 N.
(b) To keep the box moving at constant speed you must push with a force equal to the kinetic
friction, P = µk N = (0.32)(240 N) = 77 N.
(c) If you push with a force of 98 N on a box that experiences a (kinetic) friction of 77 N, then
the net force on the box is 21 N. The box will accelerate at
a = F/m = F g/W = (21 N)(9.8 m/s2 )/(240 N) = 0.86 m/s2 .
E5-15 (a) The maximum braking force is F = µs N . There is no motion in the vertical direction,
so N = W = mg. Then F = µs mg = (0.62)(1500 kg)(9.8 m/s2 ) = 9100 N.
(b) Although we still use F = µs N , N 6= W on an incline! The weight has two components;
one which is parallel to the surface and the other which is perpendicular. Since there is no motion
perpendicular to the surface we must have N = W⊥ = W cos θ. Then
F = µs mg cos θ = (0.62)(1500 kg)(9.8 m/s2 ) cos(8.6◦ ) = 9000 N.
E5-16 µs = tan θ is the condition for an object to sit without slipping on an incline. Then
θ = arctan(0.55) = 29◦ . The angle should be reduced by 13◦ .
E5-17 (a) The force of staticPfriction is less than µs N , where N is the normal force. Since the
crate isn’t moving up or down,
Fy = 0 = N − W . So in this case N = W = mg = (136 kg)(9.81
m/s2 ) = 1330 N. The force of static friction is less than or equal to (0.37)(1330 N) = 492 N; moving
the crate will require a force greater than or equal to 492 N.
(b) The second worker could lift upward with a force L, reducing the normal force, and hence
reducing the force of friction. If the first worker can move the block with a 412 N force, then
412 ≥ µs N . Solving for N , the normal force needs to be less than 1110 N. The crate doesn’t move
off the table, so then N + L = W , or L = W − N = (1330 N) − (1110 N) = 220 N.
(c) Or the second worker can help by adding a push so that the total force of both workers is
equal to 492 N. If the first worker pushes with a force of 412 N, the second would need to push with
a force of 80 N.
E5-18 The coefficient of static friction is µs = tan(28.0◦ ) = 0.532. The acceleration is a =
2(2.53 m)/(3.92 s)2 = .329 m/s2 . We will need to insert a negative sign since this is downward.
The weight has two components: a component parallel to the plane, W|| = mg sin θ; and a
component perpendicular to the plane, W⊥ = mg cos θ. There is no motion perpendicular to the
plane, so N = W⊥ . The kinetic friction is then f = µk N = µk mg cos θ. The net force parallel to
the plane is F = ma = f − W|| = µk mg cos θ − mg sin θ. Solving this for µk ,
µk
=
=
(a + g sin θ)/(g cos θ),
[(−0.329 m/s2 ) + (9.81 m/s2 ) sin(28.0◦ )]/[((9.81 m/s2 ) cos(28.0◦ )] = 0.494.
59
E5-19 The acceleration is a = −2d/t2 , where d = 203 m is the distance down the slope and t is
the time to make the run.
The weight has two components: a component parallel to the incline, W|| = mg sin θ; and a
component perpendicular to the incline, W⊥ = mg cos θ. There is no motion perpendicular tot he
plane, so N = W⊥ . The kinetic friction is then f = µk N = µk mg cos θ. The net force parallel to
the plane is F = ma = f − W|| = µk mg cos θ − mg sin θ. Solving this for µk ,
µk
=
=
(a + g sin θ)/(g cos θ),
(g sin θ − 2d/t2 )/(g cos θ).
If t = 61 s, then
µk =
(9.81 m/s2 ) sin(3.0◦ ) − 2(203 m)/(61 s)2
= 0.041;
(9.81 m/s2 ) cos(3.0◦ )
µk =
(9.81 m/s2 ) sin(3.0◦ ) − 2(203 m)/(42 s)2
= 0.029.
(9.81 m/s2 ) cos(3.0◦ )
if t = 42 s, then
E5-20 (a) If the block slides down with constant velocity then a = 0 and µk = tan θ. Not only
that, but the force of kinetic friction must be equal to the parallel component of the weight, f = W|| .
If the block is projected up the ramp then the net force is now 2W|| = 2mg sin θ. The deceleration
is a = 2g sin θ; the block will travel a time t = v0 /a before stopping, and travel a distance
d = −at2 /2 + v0 t = −a(v0 /a)2 /2 + v0 (v0 /a) = v02 /(2a) = v02 /(4g sin θ)
before stopping.
(b) Since µk < µs , the incline is not steep enough to get the block moving again once it stops.
E5-21 Let a1 be acceleration down frictionless incline of length l, and t1 the time taken. The a2
is acceleration down “rough” incline, and t2 = 2t1 is the time taken. Then
l=
1 2
1
a1 t and l = a2 (2t1 )2 .
2 1
2
Equate and find a1 /a2 = 4.
There are two force which act on the ice when it sits on the frictionless incline. The normal force
acts perpendicular to the surface, so it doesn’t contribute any components parallel to the surface.
The force of gravity has a component parallel to the surface, given by
W|| = mg sin θ,
and a component perpendicular to the surface given by
W⊥ = mg cos θ.
The acceleration down the frictionless ramp is then
a1 =
W||
= g sin θ.
m
When friction is present the force of kinetic friction is fk = µk N ; since the ice doesn’t move
perpendicular to the surface we also have N = W⊥ ; and finally the acceleration down the ramp is
a2 =
W|| − fk
= g(sin θ − µ cos θ).
m
60
Previously we found the ratio of a1 /a2 , so we now have
sin θ = 4 sin θ − 4µ cos θ,
sin 33◦ = 4 sin 33◦ − 4µ cos 33◦ ,
µ = 0.49.
E5-22 (a) The static friction between A and the table must be equal to the weight of block B to
keep A from sliding. This means mB g = µs (mA + mC )g, or mc = mB /µs − mA = (2.6 kg)/(0.18) −
(4.4 kg) = 10 kg.
(b) There is no up/down motion for block A, so f = µk N = µk mA g. The net force on the system
containing blocks A and B is F = WB − f = mB g − µk mA g; the acceleration of this system is then
a=g
mB − µk mA
(2.6 kg) − (0.15)(4.4 kg)
= (9. m/s2 )
= 2.7 m/s2 .
mA + mB
(2.6 kg) + (4.4 kg)
E5-23 There are four forces on the block— the force of gravity, W = mg; the normal force,
N ; the horizontal push, P , and the force of friction, f . Choose the coordinate system so that
components are either parallel (x-axis) to the plane or perpendicular (y-axis) to it. θ = 39◦ . Refer
to the figure below.
N
θ
f
P
W
The magnitudes of the x components of the forces are Wx = W sin θ, Px = P cos θ and f ; the
magnitudes of the y components of the forces are Wy = W cos θ, Py = P sin θ.
(a) We consider the first the case of the block moving up the ramp; then f is directed down.
Newton’s second law for each set of components then reads as
X
Fx = Px − f − Wx = P cos θ − f − W sin θ = max ,
X
Fy = N − Py − Wy = N − P sin θ − W cos θ = may
Then the second equation is easy to solve for N
2
N = P sin θ + W cos θ = (46 N) sin(39◦ ) + (4.8 kg)(9.8 m/s ) cos(39◦ ) = 66 N.
The force of friction is found from f = µk N = (0.33)(66 N) = 22 N. This is directed down the incline
while the block is moving up. We can now find the acceleration in the x direction.
max
= P cos θ − f − W sin θ,
= (46 N) cos(39◦ ) − (22 N) − (4.8 kg)(9.8 m/s2 ) sin(39◦ ) = −16 N.
So the block is slowing down, with an acceleration of magnitude 3.3 m/s2 .
(b) The block has an initial speed of v0x = 4.3 m/s; it will rise until it stops; so we can use
vy = 0 = v0y + ay t to find the time to the highest point. Then t = (vy − v0y )/ay = −(−4.3 m/s)/(3.3
m/s2 = 1.3 s. Now that we know the time we can use the other kinematic relation to find the
distance
1
1
2
y = v0y t + ay t2 = (4.3 m/s)(1.3 s) + (−3.3 m/s )(1.3 s)2 = 2.8 m
2
2
61
(c) When the block gets to the top it might slide back down. But in order to do so the frictional
force, which is now directed up the ramp, must be sufficiently small so that f + Px ≤ Wx . Solving
for f we find f ≤ Wx − Px or, using our numbers from above, f ≤ −6 N. Is this possible? No, so
the block will not slide back down the ramp, even if the ramp were frictionless, while the horizontal
force is applied.
E5-24 (a) The horizontal force needs to overcome the maximum static friction, so P ≥ µs N =
µs mg = (0.52)(12 kg)(9.8 m/s2 ) = 61 N.
(b) If the force acts upward from the horizontal then there are two components: a horizontal
component Px = P cos θ and a vertical component Py = P sin θ. The normal force is now given by
W = Py + N ; consequently the maximum force of static friction is now µs N = µs (mg − P sin θ).
The block will move only if P cos θ ≥ µs (mg − P sin θ), or
P ≥
µs mg
(0.52)(12 kg)(9.8 m/s2 )
=
= 66 N.
cos θ + µs sin θ
cos(62◦ ) + (0.52) sin(62◦ )
(c) If the force acts downward from the horizontal then θ = −62◦ , so
P ≥
(0.52)(12 kg)(9.8 m/s2 )
µs mg
=
= 5900 N.
cos θ + µs sin θ
cos(−62◦ ) + (0.52) sin(−62◦ )
E5-25 (a) If the tension acts upward from the horizontal then there are two components: a horizontal component Tx = T cos θ and a vertical component Ty = T sin θ. The normal force is now given
by W = Ty + N ; consequently the maximum force of static friction is now µs N = µs (W − T sin θ).
The crate will move only if T cos θ ≥ µs (W − T sin θ), or
P ≥
(0.52)(150 lb)
µs W
=
= 70 lb.
cos θ + µs sin θ
cos(17◦ ) + (0.52) sin(17◦ )
(b) Once the crate starts to move then the net force on the crate is F = Tx − f . The acceleration
is then
a =
=
=
g
[T cos θ − µk (W − T sin θ)],
W
2
(32 ft/s )
{(70 lb) cos(17◦ ) − (0.35)[(150 lb) − (70 lb) sin(17◦ )]},
(150 lb)
2
4.6 ft/s .
E5-26 If the tension acts upward from the horizontal then there are two components: a horizontal
component Tx = T cos θ and a vertical component Ty = T sin θ. The normal force is now given by
W = Ty + N ; consequently the maximum force of static friction is now µs N = µs (W − T sin θ). The
crate will move only if T cos θ ≥ µs (W − T sin θ), or
W ≤ T cos θ/µs + T sin θ.
We want the maximum, so we find dW/dθ,
dW/dθ = −(T /µs ) sin θ + T cos θ,
which equals zero when µs = tan θ. For this problem θ = arctan(0.35) = 19◦ , so
W ≤ (1220 N) cos(19◦ )/(0.35) + (1220 N) sin(19◦ ) = 3690 N.
62
E5-27 The three force on the know above A must add to zero. Construct a vector diagram:
~A +T
~B +T
~ d = 0, where T
~ d refers to the diagonal rope. TA and TB must be related by TA =
T
◦
TB tan θ, where θ = 41 .
TB
TA
Td
There is no up/down motion of block B, so N = WB and f = µs WB . Since block B is at rest
f = TB . Since block A is at rest WA = TA . Then
WA = WB (µs tan θ) = (712 N)(0.25) tan(41◦ ) = 155 N.
E5-28 (a) Block 2 doesn’t move up/down, so N = W2 = m2 g and the force of friction on block 2
is f = µk m2 g. Block 1 is on a frictionless incline; only the component of the weight parallel to the
surface contributes to the motion, and W|| = m1 g sin θ. There are two relevant forces on the two
mass system. The effective net force is the of magnitude W|| − f , so the acceleration is
a=g
m1 sin θ − µk m2
(4.20 kg) sin(27◦ ) − (0.47)(2.30 kg)
= (9.81 m/s2 )
= 1.25 m/s2 .
m1 + m2
(4.20 kg) + (2.30 kg)
The blocks accelerate down the ramp.
(b) The net force on block 2 is F = m2 a = T − f . The tension in the cable is then
T = m2 a + µk m2 g = (2.30 kg)[(1.25 m/s2 ) + (0.47)(9.81 m/s2 )] = 13.5 N.
E5-29 This problem is similar to Sample Problem 5-7, except now there is friction which can act
on block B. The relevant equations are now for block B
N − mB g cos θ = 0
and
T − mB g sin θ ± f = mB a,
where the sign in front of f depends on the direction in which block B is moving. If the block is
moving up the ramp then friction is directed down the ramp, and we would use the negative sign.
If the block is moving down the ramp then friction will be directed up the ramp, and then we will
use the positive sign. Finally, if the block is stationary then friction we be in such a direction as to
make a = 0.
For block A the relevant equation is
mA g − T = mA a.
Combine the first two equations with f = µN to get
T − mB g sin θ ± µmB g cos θ = mB a.
63
Take some care when interpreting friction for the static case, since the static value of µ yields the
maximum possible static friction force, which is not necessarily the actual static frictional force.
Combine this last equation with the block A equation,
mA g − mA a − mB g sin θ ± µmB g cos θ = mB a,
and then rearrange to get
a=g
mA − mB sin θ ± µmB cos θ
.
mA + mB
For convenience we will use metric units; then the masses are mA = 13.2 kg and mB = 42.6 kg. In
addition, sin 42◦ = 0.669 and cos 42◦ = 0.743.
(a) If the blocks are originally at rest then
mA − mB sin θ = (13.2 kg) − (42.6 kg)(0.669) = −15.3 kg
where the negative sign indicates that block B would slide downhill if there were no friction.
If the blocks are originally at rest we need to consider static friction, so the last term can be as
large as
µmB cos θ = (.56)(42.6 kg)(0.743) = 17.7 kg.
Since this quantity is larger than the first static friction would be large enough to stop the blocks
from accelerating if they are at rest.
(b) If block B is moving up the ramp we use the negative sign, and the acceleration is
a = (9.81 m/s2 )
(13.2 kg) − (42.6 kg)(0.669) − (.25)(42.6 kg)(0.743)
= −4.08 m/s2 .
(13.2 kg) + (42.6 kg)
where the negative sign means down the ramp. The block, originally moving up the ramp, will
slow down and stop. Once it stops the static friction takes over and the results of part (a) become
relevant.
(c) If block B is moving down the ramp we use the positive sign, and the acceleration is
a = (9.81 m/s2 )
(13.2 kg) − (42.6 kg)(0.669) + (.25)(42.6 kg)(0.743)
= −1.30 m/s2 .
(13.2 kg) + (42.6 kg)
where the negative sign means down the ramp. This means that if the block is moving down the
ramp it will continue to move down the ramp, faster and faster.
E5-30 The weight can be resolved into a component parallel to the incline, W|| = W sin θ and a
component that is perpendicular, W⊥ = W cos θ. There are two normal forces on the crate, one from
each side of the trough. By symmetry we expect them to have equal magnitudes; since they both
act perpendicular to their respective surfaces we expect them to be perpendicular to each other.
They must add to equal the perpendicular component of the weight.
√ Since they are at right angles
and equal in magnitude, this yields N 2 + N 2 = W⊥ 2 , or N = W⊥ / √2.
Each surface contributes
a frictional force f = µk N = µk W⊥ / 2; the √
total frictional force is
√
then twice this, or 2µk W⊥ . The net force on the crate is F = W sin θ − 2µk W cos θ down the
ramp. The acceleration is then
√
a = g(sin θ − 2µk cos θ).
64
E5-31 The normal force between the top slab and the bottom slab is N = W t = mt g. The force
of friction between the top and the bottom slab is f ≤ µN = µmt g. We don’t yet know if the slabs
slip relative to each other, so we don’t yet know what kind of friction to consider.
The acceleration of the top slab is
at = (110 N)/(9.7 kg) − µ(9.8 m/s2 ) = 11.3 m/s2 − µ(9.8 m/s2 ).
The acceleration of the bottom slab is
ab = µ(9.8 m/s2 )(9.7, kg)/(42 kg) = µ(2.3 m/s2 ).
Can these two be equal? Only if µ ≥ 0.93. Since the static coefficient is less than this, the block
slide. Then at = 7.6 m/s2 and ab = 0.87 m/s2 .
E5-32 (a) Convert the speed to ft/s: v = 88 ft/s. The acceleration is
2
a = v 2 /r = (88 ft/s)2 /(25 ft) = 310 ft/s .
2
2
(b) a = 310 ft/s g/(32 ft/s ) = 9.7g.
E5-33
(a) The force required to keep the car in the turn is F = mv 2 /r = W v 2 /(rg), or
F = (10700 N)(13.4 m/s)2 /[(61.0 m)(9.81 m/s2 )] = 3210 N.
(b) The coefficient of friction required is µs = F/W = (3210 N)/(10700 N) = 0.300.
E5-34 (a) The proper banking angle is given by
θ = arctan
v2
(16.7 m/s)2
= arctan
= 11◦ .
Rg
(150 m)(9.8 m/s2 )
(b) If the road is not banked then the force required to keep the car in the turn is F = mv 2 /r =
W v 2 /(Rg) and the required coefficient of friction is
µs = F/W =
v2
(16.7 m/s)2
=
= 0.19.
Rg
(150 m)(9.8 m/s2 )
E5-35 (a) This conical pendulum makes an angle θ = arcsin(0.25/1.4) = 10◦ with the vertical.
The pebble has a speed of
p
p
v = Rg tan θ = (0.25 m)(9.8 m/s2 ) tan(10◦ ) = 0.66 m/s.
(b) a = v 2 /r = (0.66 m/s)2 /(0.25 m) = 1.7 m/s2 .
(c) T = mg/cosθ = (0.053 kg)(9.8 m/s2 )/ cos(10◦ ) = 0.53 N.
E5-36 Ignoring air friction (there must be a forward component to the friction!), we have a normal force upward which is equal to the weight: N = mg = (85 kg)(9.8 m/s2 ) = 833 N. There
is a sideways component to the friction which is equal tot eh centripetal force, F = mv 2 /r =
(85 kg)(8.7 m/s)2 /(25 m) = 257 N. The magnitude of the net force of the road on the person is
p
F = (833 N)2 + (257 N)2 = 870 N,
and the direction is θ = arctan(257/833) = 17◦ off of vertical.
65
E5-37 (a) The speed is v = 2πrf = 2π(5.3×10−11 m)(6.6×1015 /s) = 2.2×106 m/s.
(b) The acceleration is a = v 2 /r = (2.2×106 m/s)2 /(5.3×10−11 m) = 9.1×1022 m/s2 .
(c) The net force is F = ma = (9.1×10−31 kg)(9.1×1022 m/s2 ) = 8.3×10−8 N.
E5-38 The basket has speed v = 2πr/t. The basket experiences a frictional force F = mv 2 /r =
m(2πr/t)2 /r = 4π 2 mr/t2 . The coefficient of static friction is µs = F/N = F/W . Combining,
µs =
4π 2 (4.6 m)
4π 2 r
=
= 0.032.
2
gt
(9.8 m/s2 )(24 s)2
E5-39 There are two forces on the hanging cylinder: the force of the cord pulling up T and the
force of gravity W = M g. The cylinder is at rest, so these two forces must balance, or T = W .
There are three forces on the disk, but only the force of the cord on the disk T is relevant here, since
there is no friction or vertical motion.
The disk undergoes circular motion, so T = mv 2 /r. We want to solve this for v and then express
the answer in terms of m, M , r, and G.
r
r
Tr
M gr
v=
=
.
m
m
E5-40 (a) The frictional force stopping the car is F = µs N = µs mg. The deceleration of the car
is then a = µs g. If the car is moving at v = 13.3 m/s then the average speed while decelerating
is half this, or v av = 6.7 m/s. The time required to stop is t = x/v av = (21 m)/(6.7 m/s) = 3.1 s.
The deceleration is a = (13.3 m/s)/(3.1 s) = 4.3 m/s2 . The coefficient of friction is µs = a/g =
(4.3 m/s2 )/(9.8 m/s2 ) = 0.44.
(b) The acceleration is the same as in part (a), so r = v 2 /a = (13.3 m/s)2 /(4.3 m/s2 ) = 41 m.
E5-41 There are three forces to consider: the normal force of the road on the car N ; the force of
gravity on the car W ; and the frictional force on the car f . The acceleration of the car in circular
motion is toward the center of the circle; this means the net force on the car is horizontal, toward
the center. We will arrange our coordinate system so that r is horizontal and z is vertical. Then the
components of the normal force are Nr = N sin θ and Nz = N cos θ; the components of the frictional
force are fr = f cos θ and fz = f sin θ.
The direction of the friction depends on the speed of the car. The figure below shows the two
force diagrams.
N
N
f
f
θ
θ
W
W
The turn is designed for 95 km/hr, at this speed a car should require no friction to stay on the
road. Using Eq. 5-17 we find that the banking angle is given by
tan θb =
v2
(26 m/s)2
=
2 = 0.33,
rg
(210 m)(9.8 m/s )
66
for a bank angle of θb = 18◦ .
(a) On the rainy day traffic is moving at 14 m/s. This is slower than the rated speed, so any
frictional force must be directed up the incline. Newton’s second law is then
X
Fr
X
Fz
mv 2
,
r
= Nz + fz − W = N cos θ + f sin θ − mg = 0.
= Nr − fr = N sin θ − f cos θ =
We can substitute f = µs N to find the minimum value of µs which will keep the cars from slipping.
There will then be two equations and two unknowns, µs and N . Solving for N ,
N (sin θ − µs cos θ) =
mv 2
and N (cos θ + µs sin θ) = mg.
r
Combining,
(sin θ − µs cos θ) mg = (cos θ + µs sin θ)
mv 2
r
Rearrange,
gr sin θ − v 2 cos θ
.
gr cos θ + v 2 sin θ
We know all the numbers. Put them in and we’ll find µs = 0.22
(b) Now the frictional force will point the other way, so Newton’s second law is now
µs =
X
Fr
X
Fz
mv 2
,
r
= Nz − fz − W = N cos θ − f sin θ − mg = 0.
= Nr + fr = N sin θ + f cos θ =
The bottom equation can be rearranged to show that
mg
N=
.
cos θ − µs sin θ
This can be combined with the top equation to give
mg
sin θ + µs cos θ
mv 2
=
.
cos θ − µs sin θ
r
We can solve this final expression for v using all our previous numbers and get v = 35 m/s. That’s
about 130 km/hr.
E5-42 (a) The net force on the person at the top of the Ferris wheel is mv 2 /r = W − N t , pointing
down. The net force on the bottom is still mv 2 /r, but this quantity now equals N b − W and is point
up. Then N b = 2W − N t = 2(150 lb) − (125 lb) = 175 lb.
(b) Doubling the speed would quadruple the net force, so the new scale reading at the top would
be (150 lb) − 4[(150 lb) − (125 lb)] = 50 lb. Wee!
E5-43 The net force on the object when it is not sliding is F = mv 2 /r; the speed of the object is
v = 2πrf (f is rotational frequency here), so F = 4π 2 mrf 2 . The coefficient of friction is then at
least µs = F/W = 4π 2 rf 2 /g. If the object stays put when the table rotates at 33 31 rev/min then
µs ≥ 4π 2 (0.13 m)(33.3/60 /s)2 /(9.8 m/s2 ) = 0.16.
If the object slips when the table rotates at 45.0 rev/min then
µs ≤ 4π 2 (0.13 m)(45.0/60 /s)2 /(9.8 m/s2 ) = 0.30.
67
E5-44 This is effectively a banked highway problem if the pilot is flying correctly.
r=
v2
(134 m/s)2
=
= 2330 m.
g tan θ
(9.8 m/s2 ) tan(38.2◦ )
E5-45 (a) Assume that frigate bird flies as well as a pilot. Then this is a banked highway problem.
The speed of the bird is given by v 2 = gr tan θ. But there is also vt = 2πr, so 2πv 2 = gvt tan θ, or
v=
(9.8 m/s2 )(13 s) tan(25◦ )
gt tan θ
=
= 9.5 m/s.
2π
2π
(b) r = vt/(2π) = (9.5 m/s)(13 s)/(2π) = 20 m.
p
E5-46 (a) The radius of the turn is r = (33 m)2 − (18 m)2 = 28 m. The speed of the plane is v =
2πrf = 2π(28 m)(4.4/60 /s) = 13 m/s. The acceleration is a = v 2 /r = (13 m/s)2 /(28 m) = 6.0 m/s2 .
(b) The tension has two components: Tx = T cos θ and Ty = T sin θ. In this case θ =
arcsin(18/33) = 33◦ . All of the centripetal force is provided for by Tx , so
T = (0.75 kg)(6.0 m/s2 )/ cos(33◦ ) = 5.4 N.
(c) The lift is balanced by the weight and Ty . The lift is then
Ty + W = (5.4 N) sin(33◦ ) + (0.75 kg)(9.8 m/s2 ) = 10 N.
E5-47 (a) The acceleration is a = v 2 /r = 4π 2 r/t2 = 4π 2 (6.37 × 106 m)/(8.64 × 104 s)2 = 3.37 ×
10−2 m/s2 . The centripetal force on the standard kilogram is F = ma = (1.00 kg)(3.37×10−2 m/s2 ) =
0.0337 N.
(b) The tension in the balance would be T = W − F = (9.80 N) − (0.0337 N) = 9.77 N.
E5-48 (a) v = 4(0.179 m/s4 )(7.18 s)3 − 2(2.08 m/s2 )(7.18 s) = 235 m/s.
(b) a = 12(0.179 m/s4 )(7.18 s)2 − 2(2.08 m/s2 ) = 107 m/s2 .
(c) F = ma = (2.17 kg)(107 m/s2 ) = 232 N.
E5-49
The force only has an x component, so we can use Eq. 5-19 to find the velocity.
Z
Z
1 t
F0 t
vx = v0x +
Fx dt = v0 +
(1 − t/T ) dt
m 0
m 0
Integrating,
vx = v0 + a0
1 2
t−
t
2T
Now put this expression into Eq. 5-20 to find the position as a function of time
Z t
Z t
1 2
x = x0 +
vx dt =
v0x + a0 t −
t
dt
2T
0
0
Integrating,
x = v0 T + a0
1 2
1 3
T −
T
2
6T
= v0 T + a0
T2
.
3
Now we can put t = T into the expression for v.
1 2
vx = v0 + a0 T −
T
= v0 + a0 T /2.
2T
68
P5-1 (a) There are two forces which accelerate block 1: the tension, T , and the parallel component
of the weight, W||,1 = m1 g sin θ1 . Assuming the block accelerates to the right,
m1 a = m1 g sin θ1 − T.
There are two forces which accelerate block 2: the tension, T , and the parallel component of the
weight, W||,2 = m2 g sin θ2 . Assuming the block 1 accelerates to the right, block 2 must also accelerate
to the right, and
m2 a = T − m2 g sin θ2 .
Add these two equations,
(m1 + m2 )a = m1 g sin θ1 − m2 g sin θ2 ,
and then rearrange:
a=
m1 g sin θ1 − m2 g sin θ2
.
m1 + m2
Or, take the two net force equations, divide each side by the mass, and set them equal to each other:
g sin θ1 − T /m1 = T /m2 − g sin θ2 .
Rearrange,
T
1
1
+
m1
m2
and then rearrange again:
T =
= g sin θ1 + g sin θ2 ,
m1 m2 g
(sin θ1 + sin θ2 ).
m1 + m2
(b) The negative sign we get in the answer means that block 1 accelerates up the ramp.
a=
(3.70 kg) sin(28◦ ) − (4.86 kg) sin(42◦ )
(9.81 m/s2 ) = −1.74 m/s2 .
(3.70 kg) + (4.86 kg)
T =
(3.70 kg)(4.86 kg)(9.81 m/s2 )
[sin(28◦ ) + sin(42◦ )] = 23.5 N.
(3.70 kg) + (4.86 kg)
(c) No acceleration happens when m2 = (3.70 kg) sin(28◦ )/ sin(42◦ ) = 2.60 kg. If m2 is more
massive than this m1 will accelerate up the plane; if m2 is less massive than this m1 will accelerate
down the plane.
P5-2 (a) Since the pulley is massless, F = 2T . The largest value of T that will allow block 2 to
remain on the floor is T ≤ W2 = m2 g. So F ≤ 2(1.9 kg)(9.8 m/s2 ) = 37 N.
(b) T = F/2 = (110 N)/2 = 55 N.
(c) The net force on block 1 is T − W1 = (55 N) − (1.2 kg)(9.8 m/s2 ) = 43 N. This will result in
an acceleration of a = (43 N)/(1.2 kg) = 36 m/s2 .
P5-3 As the string is pulled the two masses will move together so that the configuration
P will look
like the figure below. The point where the force is applied to the string is massless, so
F = 0 at
that point. We can take advantage of this fact and the figure below to find the tension in the cords,
F/2 = T cos θ. The factor of 1/2 occurs because only 1/2 of F is contained in the right triangle that
has T as the hypotenuse. From the figure we can find the x component of the force on one mass to
be Tx = T sin θ. Combining,
F sin θ
F
Tx =
= tan θ.
2 cos θ
2
69
But the tangent is equal to
tan θ =
Opposite
x
=√
2
Adjacent
L − x2
And now we have the answer in the book.
θ
T
T
F
What happens when x = L? Well, ax is infinite according to this expression. Since that could
only happen if the tension in the string were infinite, then there must be some other physics that
we had previously ignored.
P5-4 (a) The force of static friction can be as large as f ≤ µs N = (0.60)(12 lb) = 7.2 lb. That is
more than enough to hold the block up.
(b) The force of static friction is actually only large enough top
hold up the block: f = 5.0 lb.
The magnitude of the force of the wall on the block is then F bw = (5.0)2 + (12.0)2 lb = 13 lb.
P5-5 (a) The weight has two components: normal to the incline, W⊥ = mg cos θ and parallel to the
incline, W|| = mg sin θ. There is no motion perpendicular to the incline, so N = W⊥ = mg cos θ. The
force of friction on the block is then f = µN = µmg cos θ, where we use whichever µ is appropriate.
The net force on the block is F − f − W|| = F ± µmg cos θ − mg sin θ.
To hold the block in place we use µs and friction will point up the ramp so the ± is +, and
F = (7.96 kg)(9.81 m/s2 )[sin(22.0◦ ) − (0.25) cos(22.0◦ )] = 11.2 N.
(b) To find the minimum force to begin sliding the block up the ramp we still have static friction,
but now friction points down, so
F = (7.96 kg)(9.81 m/s2 )[sin(22.0◦ ) + (0.25) cos(22.0◦ )] = 47.4 N.
(c) To keep the block sliding up at constant speed we have kinetic friction, so
F = (7.96 kg)(9.81 m/s2 )[sin(22.0◦ ) + (0.15) cos(22.0◦ )] = 40.1 N.
P5-6 The sand will slide if the cone makes an angle greater than θ where µs = tan θ. But
tan θ = h/R or h = R tan θ. The volume of the cone is then
Ah/3 = πR2 h/3 = πR3 tan θ/3 = πµs R3 /3.
P5-7 There are four forces on the broom: the force of gravity W = mg; the normal force of the
floor N ; the force of friction f ; and the applied force from the person P (the book calls it F ). Then
X
Fx = Px − f = P sin θ − f = max ,
X
Fy = N − Py − W = N − P cos θ − mg = may = 0
Solve the second equation for N ,
N = P cos θ + mg.
70
(a) If the mop slides at constant speed f = µk N . Then
P sin θ − f = P sin θ − µk (P cos θ + mg) = 0.
We can solve this for P (which was called F in the book);
P =
µmg
.
sin θ − µk cos θ
This is the force required to push the broom at constant speed.
(b) Note that P becomes negative (or infinite) if sin θ ≤ µk cos θ. This occurs when tan θc ≤ µk .
If this happens the mop stops moving, to get it started again you must overcome the static friction,
but this is impossible if tan θ0 ≤ µs
P5-8 (a) The condition to slide is µs ≤ tan θ. In this case, (0.63) > tan(24◦ ) = 0.445.
(b) The normal force on the slab is N = W⊥ = mg cos θ. There are three forces parallel to the
surface: friction, f = µs N = µs mg cos θ; the parallel component of the weight, W|| = mg sin θ, and
the force F . The block will slide if these don’t balance, or
F > µs mg cos θ − mg sin θ = (1.8×107 kg)(9.8 m/s2 )[(0.63) cos(24◦ ) − sin(24◦ )] = 3.0×107 N.
P5-9 To hold up the smaller block the frictional force between the larger block and smaller block
must be as large as the weight of the smaller block. This can be written as f = mg. The normal
force of the larger block on the smaller block is N , and the frictional force is given by f ≤ µs N . So
the smaller block won’t fall if mg ≤ µs N .
There is only one horizontal force on the large block, which is the normal force of the small block
on the large block. Newton’s third law says this force has a magnitude N , so the acceleration of the
large block is N = M a.
There is only one horizontal force on the two block system, the force F . So the acceleration of
this system is given by F = (M + m)a. The two accelerations are equal, otherwise the blocks won’t
stick together. Equating, then, gives N/M = F/(M + m).
We can combine this last expression with mg ≤ µs N and get
mg ≤ µs F
or
F ≥
M
M +m
g(M + m)m
(9.81 m/s2 )(88 kg + 16 kg)(16 kg)
=
= 490 N
µs M
(0.38)(88 kg)
P5-10 The normal force on the ith block is Ni = mi g cos θ; the force of friction on the ith block
is then fi = µi mi g cos θ. The parallel component of the weight on the ith block is W||,i = mi g sin θ.
(a) The net force on the system is
X
F =
mi g(sin θ − µi cos θ).
i
Then
a =
=
(9.81 m/s2 )
(1.65 kg)(sin 29.5◦ − 0.226 cos 29.5◦ ) + (3.22 kg)(sin 29.5◦ − 0.127 cos 29.5◦ )
,
(1.65 kg) + (3.22 kg)
3.46 m/s2 .
(b) The net force on the lower mass is m2 a = W||,2 − f2 − T , so the tension is
T = (9.81 m/s2 )(3.22 kg)(sin 29.5◦ − 0.127 cos 29.5◦ ) − (3.22 kg)(3.46 m/s2 ) = 0.922 N.
71
(c) The acceleration will stay the same, since the system is still the same. Reversing the order
of the masses can only result in a reversing of the tension: it is still 0.992 N, but is now negative,
meaning compression.
P5-11 The rope wraps around the dowel and there is a contribution to the frictional force ∆f
from each small segment of the rope where it touches the dowel. There is also a normal force ∆N at
each point where the contact occurs. We can find ∆N much the same way that we solve the circular
motion problem.
In the figure on the left below we see that we can form a triangle with long side T and short side
∆N . In the figure on the right below we see a triangle with long side r and short side r∆θ. These
triangles are similar, so r∆θ/r = ∆N/T .
T
∆N
r ∆θ
T
r
r
Now ∆f = µ∆N and T (θ) + ∆f ≈ T (θ + ∆θ). Combining, and taking the limit as ∆θ → 0,
dT = df
Z
Z
1 dT
= dθ
µ T
integrating both sides of this expression,
Z
1 dT
µ T
1
T
ln T |T21
µ
T2
=
Z
dθ,
= π,
= T1 eπµ .
In this case T1 is the weight and T2 is the downward force.
P5-12 Answer this out of order!
(b) The maximum static friction between the blocks is 12.0 N; the maximum acceleration of the
top block is then a = F/m = (12.0 N)/(4.40 kg) = 2.73 m/s2 .
(a) The net force on a system of two blocks that will accelerate them at 2.73 m/s2 is F =
(4.40 kg + 5.50 kg)(2.73 m/s2 ) = 27.0 N.
(c) The coefficient of friction is µs = F/N = F/mg = (12.0 N)/[(4.40 kg)(9.81 m/s2 )] = 0.278.
P5-13 The speed is v = 23.6 m/s.
(a) The average speed while stopping is half the initial speed, or v av = 11.8 m/s. The time to stop
is t = (62 m)/(11.8 m/s) = 5.25 s. The rate of deceleration is a = (23.6 m/s)/(5.25 s) = 4.50 m/s2 .
The stopping force is F = ma; this is related to the frictional force by F = µs mg, so µs = a/g =
(4.50 m/s2 )/(9.81 m/s2 ) = 0.46.
72
(b) Turning,
a = v 2 /r = (23.6 m/s)2 /(62 m) = 8.98 m/s2 .
Then µs = a/g = (8.98 m/s2 )/(9.81 m/s2 ) = 0.92.
P5-14 (a) The net force on car as it travels at the top of a circular hill is F net = mv 2 /r = W − N ;
in this case we are told N = W/2, so F net = W/2 = (16000 N)/2 = 8000 N. When the car travels
through the bottom valley the net force at the bottom is F net = mv 2 /r = N − W . Since the
magnitude of v, r, and hence F net is the same in both cases,
N = W/2 + W = 3W/2 = 3(16000 N)/2 = 24000 N
at the bottom of the valley.
(b) You leave the hill when N = 0, or
p
√
v = rg = (250 m)(9.81 m/s2 ) = 50 m/s.
(c) At this speed F net = W , so at the bottom of the valley N = 2W = 32000 N.
P5-15 (a) v = 2πr/t = 2π(0.052 m)(3/3.3 s) = 0.30 m/s.
(b) a = v 2 /r = (0.30 m/s)2 /(0.052 m) = 1.7 m/s2 , toward center.
(c) F = ma = (0.0017 kg)(1.7 m/s2 ) = 2.9×10−3 N.
(d) If the coin can be as far away as r before slipping, then
µs = F/mg = (2πr/t)2 /(rg) = 4π 2 r/(t2 g) = 4π 2 (0.12 m)/[(3/3.3 s)2 (9.8 m/s2 )] = 0.59.
P5-16 (a) Whether you assume constant speed or constant energy, the tension in the string must
be the greatest at the bottom of the circle, so that’s where the string will break.
(b) The net force on the stone at the bottom is T − W = mv 2 /r. Then
q
p
2
v = rg[T /W − 1] = (2.9 ft)(32 ft/s )[(9.2 lb)/(0.82 lb) − 1] = 31 ft/s.
P5-17 (a) In order to keep the ball moving in a circle there must be a net centripetal force Fc
directed horizontally toward the rod. There are only three forces which act on the ball: the force
of gravity, W = mg = (1.34 kg)(9.81 m/s2 ) = 13.1 N; the tension in the top string T1 = 35.0 N, and
the tension in the bottom string, T2 .
The components of the force from the tension in the top string are
T1,x = (35.0 N) cos 30◦ = 30.3 N and T1,y = (35.0 N) sin 30◦ = 17.5 N.
The vertical components do balance, so
T1,y + T2,y = W,
or T2,y = (13.1 N) − (17.5 N) = −4.4 N. From this we can find the tension in the bottom string,
T2 = T2,y / sin(−30◦ ) = 8.8 N.
(b) The net force on the object will be the sum of the two horizontal components,
Fc = (30.3 N) + (8.8 N) cos 30◦ = 37.9 N.
(c) The speed will be found from
p
√
v =
ac r = Fc r/m,
p
=
(37.9 m)(1.70 m) sin 60◦ /(1.34 kg) = 6.45 m/s.
73
P5-18 The net force on the cube is F = mv 2 /r. The speed is 2πrω. (Note that we are using ω in
a non-standard way!) Then F = 4π 2 mrω 2 . There are three forces to consider: the normal force of
the funnel on the cube N ; the force of gravity on the cube W ; and the frictional force on the cube
f . The acceleration of the cube in circular motion is toward the center of the circle; this means the
net force on the cube is horizontal, toward the center. We will arrange our coordinate system so
that r is horizontal and z is vertical. Then the components of the normal force are Nr = N sin θ
and Nz = N cos θ; the components of the frictional force are fr = f cos θ and fz = f sin θ.
The direction of the friction depends on the speed of the cube; it will point up if ω is small and
down if ω is large.
(a) If ω is small, Newton’s second law is
X
Fr = Nr − fr = N sin θ − f cos θ = 4π 2 mrω 2 ,
X
Fz = Nz + fz − W = N cos θ + f sin θ − mg = 0.
We can substitute f = µs N . Solving for N ,
N (cos θ + µs sin θ) = mg.
Combining,
4π 2 rω 2 = g
sin θ − µs cos θ
.
cos θ + µs sin θ
Rearrange,
1
ω=
2π
s
g sin θ − µs cos θ
.
r cos θ + µs sin θ
This is the minimum value.
(b) Now the frictional force will point the other way, so Newton’s second law is now
X
Fr = Nr + fr = N sin θ + f cos θ = 4π 2 mrω 2 ,
X
Fz = Nz − fz − W = N cos θ − f sin θ − mg = 0.
We’ve swapped + and - signs, so
1
ω=
2π
s
g sin θ + µs cos θ
r cos θ − µs sin θ
is the maximum value.
P5-19 (a) The radial distance from the axis of rotation at a latitude L is R cos L. The speed
in the circle is then v = 2πR cos L/T . The net force on a hanging object is F = mv 2 /(R cos L) =
4π 2 mR cos L/T 2 . This net force is not directed toward the center of the earth, but is instead directed
toward the axis of rotation. It makes an angle L with the Earth’s vertical. The tension in the cable
must then have two components: one which is vertical (compared to the Earth) and the other which
is horizontal. If the cable makes an angle θ with the vertical, then T|| = T sin θ and T⊥ = T cos θ.
Then T|| = F|| and W − T⊥ = F⊥ . Written with a little more detail,
T sin θ = 4π 2 mR cos L sin L/T 2 ≈ T θ,
and
T cos θ = 4π 2 mR cos2 L/T 2 + mg ≈ T.
74
But 4π 2 R cos2 L/T 2 g, so it can be ignored in the last equation compared to g, and T ≈ mg.
Then from the first equation,
θ = 2π 2 R sin 2L/(gT 2 ).
(b) This is a maximum when sin 2L is a maximum, which happens when L = 45◦ . Then
θ = 2π 2 (6.37×106 m)/[(9.8 m/s2 )(86400 s)2 ] = 1.7×10−3 rad.
(c) The deflection at both the equator and the poles would be zero.
Rt
Rt
P5-20 a = (F0 /m)e−t/T . Then v = 0 a dt = (F0 T /m)e−t/T , and x = 0 v dt = (F0 T 2 /m)e−t/T .
(a) When t = T v = (F0 T /m)e−1 = 0.368(F0 T /m).
(b) When t = T x = (F0 T 2 /m)e−1 = 0.368(F0 T 2 /m).
75
E6-1 (a) v1 = (m2 /m1 )v2 = (2650 kg/816 kg)(16.0 km/h) = 52.0 km/h.
(b) v1 = (m2 /m1 )v2 = (9080 kg/816 kg)(16.0 km/h) = 178 km/h.
5
E6-2 ~pi = (2000 kg)(40 km/h)ĵ = 8.00×104 kg · km/hĵ. ~pf = (2000 kg)(50 km/h)
p î = 1.00×10 kg ·
5
4
2
km/hî. ∆~p = ~pf − ~pi = 1.00×10 kg · km/hî − 8.00×10 kg · km/hĵ. ∆p = (∆px ) + (∆py )2 =
1.28×105 kg · km/h. The direction is 38.7◦ south of east.
E6-3 The figure below shows the initial and final momentum vectors arranged to geometrically
show ~pf − ~pi = ∆~p. We can use the cosine law to find the length of ∆~p.
-pi
∆p
θ
pf
The angle α = 42◦ + 42◦ , pi = mv = (4.88 kg)(31.4 m/s) = 153 kg·m/s. Then the magnitude of
∆~p is
p
∆p = (153 kg·m/s)2 + (153 kg·m/s)2 − 2(153 kg·m/s)2 cos(84◦ ) = 205 kg·m/s,
directed up from the plate. By symmetry it must be perpendicular.
E6-4 The change in momentum is ∆p = −mv = −(2300 kg)(15 m/s) = −3.5×104 kg · m/s. The
average force is F = ∆p/∆t = (−3.5×104 kg · m/s)/(0.54 s) = −6.5×104 N.
E6-5 (a) The change in momentum is ∆p = (−mv) − mv; the average force is F = ∆p/∆t =
−2mv/∆t.
(b) F = −2(0.14 kg)(7.8 m/s)/(3.9×10−3 s) = 560 N.
E6-6 (a) J = ∆p = (0.046 kg)(52.2 m/s) − 0 = 2.4 N · s.
(b) The impulse imparted to the club is opposite that imparted to the ball.
(c) F = ∆p/∆t = (2.4 N · s)/(1.20×10−3 s) = 2000 N.
E6-7 Choose the coordinate system so that the ball is only moving along the x axis, with away
from the batter as positive. Then pf x = mv f x = (0.150 kg)(61.5 m/s) = 9.23 kg·m/s and pix =
mv ix = (0.150 kg)(−41.6 m/s) = −6.24 kg·m/s. The impulse is given by Jx = pf x − pix = 15.47
kg·m/s. We can find the average force by application of Eq. 6-7:
F av,x =
Jx
(15.47 kg · m/s)
=
= 3290 N.
∆t
(4.7 × 10−3 s)
76
E6-8 The magnitude of the impulse is J = F δt = (−984 N)(0.0270 s) = −26.6 N · s. Then pf =
pi + ∆p, so
(0.420 kg)(13.8 m/s) + (−26.6 N · s)
vf =
= −49.5 m/s.
(0.420 kg)
The ball moves backward!
E6-9 The change in momentum of the ball is ∆p = (mv) − (−mv) = 2mv = 2(0.058 kg)(32 m/s) =
3.7 kg · m/s. The impulse is the area under a force - time graph; for the trapezoid in the figure this
area is J = Fmax (2 ms + 6 ms)/2 = (4 ms)Fmax . Then Fmax = (3.7 kg · m/s)/(4 ms) = 930 N.
E6-10 The final speed of each object is given by vi = J/mi , where i refers to which object (as
opposed to “initial”). The object are going in different directions, so the relative speed will be the
sum. Then
v rel = v1 + v2 = (300 N · s)[1/(1200 kg) + 1/(1800 kg)] = 0.42 m/s.
E6-11
Use Simpson’s rule. Then the area is given by
Jx
=
=
1
h (f0 + 4f1 + 2f2 + 4f3 + ... + 4f1 3 + f1 4) ,
3
1
(0.2 ms) (200 + 4 · 800 + 2 · 1200... N)
3
which gives Jx = 4.28 kg·m/s.
Since the impulse is the change in momentum, and the ball started from rest, pf x = Jx +pix = 4.28
kg·m/s. The final velocity is then found from vx = px /m = 8.6 m/s.
E6-12 (a) The average speed during the the time the hand is in contact with the board is half of the
initial speed, or v av = 4.8 m/s. The time of contact is then t = y/v av = (0.028 m)/(4.8 m/s) = 5.8 ms.
(b) The impulse given to the board is the same as the magnitude in the change in momentum of
the hand, or J = (0.54 kg)(9.5 m/s) = 5.1 N · s. Then F av = (5.1 N · s)/(5.8 ms) = 880 N.
E6-13 ∆p = J = F ∆t = (3000 N)(65.0 s) = 1.95×105 N · s. The direction of the thrust relative to
the velocity doesn’t matter in this exercise.
E6-14 (a) p = mv = (2.14×10−3 kg)(483 m/s) = 1.03 kg · m/s.
(b) The impulse imparted to the wall in one second is ten times the above momentum, or
J = 10.3 kg · m/s. The average force is then F av = (10.3 kg · m/s)/(1.0 s) = 10.3 N.
(c) The average force for each individual particle is F av = (1.03 kg · m/s)/(1.25×10−3 s) = 830 N.
E6-15 A transverse direction means at right angles, so the thrusters have imparted a momentum
sufficient to direct the spacecraft 100+3400 = 3500 km to the side of the original path. The spacecraft
is half-way through the six-month journey, so it has three months to move the 3500 km to the side.
This corresponds to a transverse speed of v = (3500×103 m)/(90×86400 s) = 0.45 m/s. The required
time for the rocket to fire is ∆t = (5400 kg)(0.45 m/s)/(1200 N) = 2.0 s.
E6-16 Total initial momentum is zero, so
vm = −
ms
ms g
(0.158 lb)
vs = −
vs = −
(12.7 ft/s) = −1.0×10−2 ft/s.
mm
mm g
(195 lb)
77
E6-17
Conservation of momentum:
pf,m + pf,c
mm v f,m + mc v f,c
v f,c − v i,c
∆v c
= pi,m + pi,c ,
= mm v i,m + mc v i,c ,
mm v i,m − mm v f,m
=
,
mc
(75.2 kg)(2.33 m/s) − (75.2 kg)(0)
=
,
(38.6 kg)
= 4.54 m/s.
The answer is positive; the cart speed increases.
E6-18 Conservation of momentum:
pf,m + pf,c
mm (v f,c − v rel ) + mc v f,c
(mm + mc )v f,c − mm v rel
∆v c
=
=
=
=
=
pi,m + pi,c ,
(mm + mc )v i,c ,
(mm + mc )v i,c ,
mm v rel /(mm + mc ),
wv rel /(w + W ).
E6-19 Conservation of momentum. Let m refer to motor and c refer to command module:
pf,m + pf,c
mm (v f,c − v rel ) + mc v f,c
(mm + mc )v f,c − mm v rel
v f,c
= pi,m + pi,c ,
= (mm + mc )v i,c ,
= (mm + mc )v i,c ,
mm v rel + (mm + mc )v i,c
=
,
(mm + mc )
4mc (125 km/h) + (4mc + mc )(3860 km/h)
=
= 3960 km/h.
(4mc + mc )
E6-20 Conservation of momentum. The block on the left is 1, the other is 2.
m1 v 1,f + m2 v 2,f
v 1,f
= m1 v 1,i + m2 v 2,i ,
m2
= v 1,i +
(v 2,i − v 2,f ),
m1
(2.4 kg)
= (5.5 m/s) +
[(2.5 m/s) − (4.9 m/s)],
(1.6 kg)
= 1.9 m/s.
E6-21 Conservation of momentum. The block on the left is 1, the other is 2.
m1 v 1,f + m2 v 2,f
v 1,f
= m1 v 1,i + m2 v 2,i ,
m2
= v 1,i +
(v 2,i − v 2,f ),
m1
(2.4 kg)
= (5.5 m/s) +
[(−2.5 m/s) − (4.9 m/s)],
(1.6 kg)
= −5.6 m/s.
78
E6-22 Assume a completely inelastic collision. Call the Earth 1 and the meteorite 2. Then
m1 v 1,f + m2 v 2,f
v 1,f
= m1 v 1,i + m2 v 2,i ,
m2 v 2,i
=
,
m1 + m2
(5×1010 kg)(7200 m/s)
=
= 7×10−11 m/s.
(5.98×1024 kg) + (5×1010 kg)
That’s 2 mm/y!
E6-23
Conservation of momentum is used to solve the problem:
Pf
pf,bl + pf,bu
mbl v f,bl + mbu v f,bu
(715 g)v f,bl + (5.18 g)(428 m/s)
=
=
=
=
P i,
pi,bl + pi,bu ,
mbl v i,bl + mbu v i,bu ,
(715 g)(0) + (5.18 g)(672 m/s),
which has solution v f,bl = 1.77 m/s.
E6-24 The y component of the initial momentum is zero; therefore the magnitudes of the y components of the two particles must be equal after the collision. Then
mα vα sin θα
vα
= mO v O sin θO ,
mO v O sin θO
=
,
mα vα sin θα
(16 u)(1.20×105 m/s) sin(51◦ )
= 4.15×105 m/s.
=
(4.00 u) sin(64◦ )
E6-25 The total momentum is
~p =
=
(2.0 kg)[(15 m/s)î + (30 m/s)ĵ] + (3.0 kg)[(−10 m/s)î + (5 m/s)ĵ],
75 kg · m/s ĵ.
The final velocity of B is
~vB f
=
=
=
1
(~p − mA ~vAf ),
mB
1
{(75 kg · m/s ĵ) − (2.0 kg)[(−6.0 m/s)î + (30 m/s)ĵ]},
(3.0 kg)
(4.0 m/s)î + (5.0 m/s)ĵ.
E6-26 Assume electron travels in +x direction while neutrino travels in +y direction. Conservation
of momentum requires that
~p = −(1.2×10−22 kg · m/s)î − (6.4×10−23 kg · m/s)ĵ
be the momentum of the nucleus after the decay. This has a magnitude of p = 1.4×10−22 kg · m/s
and be directed 152◦ from the electron.
79
E6-27 What we know:
~p1,i
=
(1.50×105 kg)(6.20 m/s)î = 9.30×105 kg · m/s î,
~p2,i
=
(2.78×105 kg)(4.30 m/s)ĵ = 1.20×106 kg · m/s ĵ,
~p2,f
=
(2.78×105 kg)(5.10 m/s)[sin(18◦ )î + cos(18◦ )ĵ],
=
4.38×105 kg · m/s î + 1.35×106 kg · m/s ĵ.
Conservation of momentum then requires
~p1,f
=
(9.30×105 kg · m/s î) − (4.38×105 kg · m/s î)
+(1.20×106 kg · m/s ĵ) − (1.35×106 kg · m/s ĵ),
=
4.92×105 kg · m/s î − 1.50×105 kg · m/s î.
This corresponds to a velocity of
~v1,f = 3.28 m/s î − 1.00 m/s ĵ,
which has a magnitude of 3.43 m/s directed 17◦ to the right.
E6-28 v f = −2.1 m/s.
E6-29 We want to solve Eq. 6-24 for m2 given that v 1,f = 0 and v 1,i = −v 2,i . Making these
substitutions
m1 − m2
2m2
v 1,i +
(−v 1,i ),
m1 + m2
m1 + m2
0 = (m1 − m2 )v 1,i − (2m2 )v 1,i ,
3m2 = m1
(0)
=
so m2 = 100 g.
E6-30 (a) Rearrange Eq. 6-27:
m2 = m1
v 1i − v 1f
(1.24 m/s) − (0.636 m/s)
= (0.342 kg)
= 0.110 kg.
v 1i + v 1f
(1.24 m/s) + (0.636 m/s)
(b) v 2f = 2(0.342 kg)(1.24 m/s)/(0.342 kg + 0.110 kg) = 1.88 m/s.
E6-31 Rearrange Eq. 6-27:
m2 = m1
v 1i − v 1f
v 1i − v 1i /4
= (2.0 kg)
= 1.2 kg.
v 1i + v 1f
v 1i + v 1i /4
E6-32 I’ll multiply all momentum equations by g, then I can use weight directly without converting
to mass.
(a) v f = [(31.8 T)(5.20 ft/s) + (24.2 T)(2.90 ft/s)]/(31.8 T + 24.2 T) = 4.21 ft/s.
(b) Evaluate:
v 1f =
31.8 T − 24.2 T
2(24.2 T)
(5.20 ft/s) +
(2.90 ft/s) = 3.21 ft/s.
31.8 T + 24.2 T
31.8 T + 24.2 T
v 2f = −
31.8 T − 24.2 T
2(31.8 T)
(2.90 ft/s) +
(5.20 ft/s) = 5.51 ft/s.
31.8 T + 24.2 T
31.8 T + 24.2 T
80
E6-33 Let the initial momentum of the first object be ~p1,i = m~v1,i , that of the second object be
~p2,i = m~v2,i , and that of the combined final object be ~pf = 2m~vf . Then
~p1,i + ~p2,i = ~pf ,
implies that we can find a triangle with sides of length p1,i , p2,i , and pf . These lengths are
p1,i
p2,i
pf
= mv i ,
= mv i ,
= 2mv f = 2mv i /2 = mv i ,
so this is an equilateral triangle. This means the angle between the initial velocities is 120◦ .
E6-34 We need to change to the center of mass system. Since both particles have the same
mass, the conservation of momentum problem is effectively the same as a (vector) conservation of
velocity problem. Since one of the particles is originally at rest, the center of mass moves with
speed v cm = v 1i /2. In the figure below the center of mass velocities are primed; the transformation
velocity is vt .
vt
v’1f
v 1f
v’2i
v’1i
v
v’2f
2f
vt
Note that since vt = v 0 1i = v 0 2i = v 0 1f = v 0 2f the entire problem can be inscribed in a rhombus.
The diagonals of the rhombus are the directions of v 1f and v 2f ; note that the diagonals of a rhombus
are necessarily at right angles!
(a) The target proton moves off at 90◦ to the direction the incident proton moves after the
collision, or 26◦ away from the incident protons original direction.
(b) The y components of the final momenta must be equal, so v 2f sin(26◦ ) = v 1f sin(64◦ ), or v 2f =
v 1f tan(64◦ ). The x components must add to the original momentum, so (514 m/s) = v 2f cos(26◦ ) +
v 1f cos(64◦ ), or
v 1f = (514 m/s)/{tan(64◦ ) cos(26◦ ) + cos(64◦ )} = 225 m/s,
and
v 2f = (225 m/s) tan(64◦ ) = 461 m/s.
E6-35 v cm = {(3.16 kg)(15.6 m/s) + (2.84 kg)(−12.2 m/s)}/{(3.16 kg) + (2.84 kg)} = 2.44 m/s, positive means to the left.
81
P6-1 The force is the change in momentum over change in time; the momentum is the mass time
velocity, so
∆p
m∆v
m
F =
=
= ∆v
= 2uµ,
∆t
∆t
∆t
since µ is the mass per unit time.
P6-2 (a) The initial momentum is ~pi = (1420 kg)(5.28 m/s)ĵ = 7500 kg · m/s ĵ. After making the
right hand turn the final momentum is ~pf = 7500 kg · m/s î. The impulse is ~J = 7500 kg · m/s î −
7500 kg · m/s ĵ, which has magnitude J = 10600 kg · m/s.
(b) During the collision the impulse is ~J = 0−7500 kg·m/s î. The magnitude is J = 7500 kg·m/s.
(c) The average force is F = J/t = (10600 kg · m/s)/(4.60 s) = 2300 N.
(d) The average force is F = J/t = (7500 kg · m/s)/(0.350 s) = 21400 N.
P6-3
(a) Only the component of the momentum which is perpendicular to the wall changes. Then
~J = ∆~p = −2(0.325 kg)(6.22 m/s) sin(33◦ )ĵ = −2.20 kg · m/s ĵ.
~ = −~J/t = −(−2.20 kg · m/s ĵ)/(0.0104 s) = 212 N.
(b) F
P6-4 The change in momentum of one bullet is ∆p = 2mv = 2(0.0030 kg)(500 m/s) = 3.0 kg · m/s.
The average force is the total impulse in one minute divided by one minute, or
F av = 100(3.0 kg · m/s)/(60 s) = 5.0 N.
P6-5 (a) The volume of a hailstone is V = 4πr3 /3 = 4π(0.5 cm)3 /3 = 0.524 cm3 . The mass of a
hailstone is m = ρV = (9.2×10−4 kg/cm3 )(0.524 cm3 ) = 4.8×10−4 kg.
(b) The change in momentum of one hailstone when it hits the ground is
∆p = (4.8×10−4 kg)(25 m/s) = 1.2×10−2 kg · m/s.
The hailstones fall at 25 m/s, which means that in one second the hailstones in a column 25 m high
hit the ground. Over an area of 10 m × 20 m then there would be (25 m)(10 m)(20 m) = 500 m3 worth
of hailstones, or 6.00×105 hailstones per second striking the surface. Then
F av = 6.00×105 (1.2×10−2 kg · m/s)/(1 s) = 7200 N.
P6-6 Assume the links are not connected once the top link
p is released. Consider the link that
2h/g and hits the table with a speed
starts h above
the
table;
it
falls
a
distance
h
in
a
time
t
=
√
v = gt = 2hg. When the link hits the table h of the chain is already on the table, and L − h is yet
to come. The linear mass density of the chain is M/L,√
so when this link strikes the table the mass is
hitting the table at a rate dm/dt = (M/L)v = (M/L) 2hg. The average force required to stop the
falling link is then v dm/dt = (M/L)2hg = 2(M/L)hg. But the weight of the chain that is already
on the table is (M/L)hg, so the net force on the table is the sum of these two terms, or F = 3W .
P6-7 The weight of the marbles in the box after a time t is mgRt because Rt is the number of
marbles in the box.
p
The marbles fall a distance h from rest; the time required √to fall this distance is t = 2h/g,
the speed of the marbles when
√ they strike the box is v = gt = 2gh. The momentum each marble
imparts on the
box
is
then
m
2gh. If the marbles strike at a rate R then the force required to stop
√
them is Rm 2gh.
82
The reading on the scale is then
p
W = mR( 2gh + gt).
This will give a numerical result of
p
(4.60×10−3 kg)(115 s−1 )
2(9.81 m/s2 )(9.62 m) + (9.81 m/s2 )(6.50 s) = 41.0 N.
P6-8 (a) v = (108 kg)(9.74 m/s)/(108 kg + 1930 kg) = 0.516 m/s.
(b) Label the person as object 1 and the car as object 2. Then m1 v1 + m2 v2 = (108 kg)(9.74 m/s)
and v1 = v2 + 0.520 m/s. Combining,
v2 = [1050 kg · m/s − (0.520 m/s)(108 kg)]/(108 kg + 1930 kg) = 0.488 m/s.
p
P6-9 (a) It
√ takes a time t1 = 2h/g to fall h = 6.5 ft. An object will be moving at a speed
v1 = gt1 = 2hg after falling this distance. If there is an inelastic collision with the pile then the
two will move together with a speed of v2 = M v1 /(M + m) after the collision.
If the pile then stops within d = 1.5 inches, then the time of stopping is given by t2 = d/(v2 /2) =
2d/v2 .
For inelastic collisions this corresponds to an average force of
F av =
(M + m)v22
M 2 v12
(gM )2 h
(M + m)v2
=
=
=
.
t2
2d
2(M + m)d
g(M + m) d
Note that we multiply through by g to get weights. The numerical result is F av = 130 t.
(b) For an elastic collision v2 = 2M v1 /(M + m); the time of stopping is still expressed by
t2 = 2d/v2 , but we now know F av instead of d. Then
F av =
mv2
mv22
4M mv12
2(gM )(gm) h
=
=
=
.
t2
2d
(M + m)d
g(M + m) d
or
d=
2(gM )(gm) h
,
g(M + m) F av
which has a numerical result of d = 0.51 inches.
But wait! The weight, which just had an elastic collision, “bounced” off of the pile, and then hit
it again. This drives the pile deeper into the earth. The weight hits the pile a second time with a
speed of v3 = (M − m)/(M + m)v1 ; the pile will (in this second elastic collision) then have a speed of
v4 = 2M (M + m)v3 = [(M − m)/(M + m)]v2 . In other words, we have an infinite series of distances
traveled by the pile, and if α = [(M − m)/(M + m)] = 0.71, the depth driven by the pile is
df = d(1 + α2 + α4 + α6 · · ·) =
d
,
1 − α2
or d = 1.03.
P6-10
0, or
The cat jumps off of sled A; conservation of momentum requires that M vA,1 +m(vA,1 +vc ) =
vA,1 = −mvc /(m + M ) = −(3.63 kg)(3.05 m/s)/(22.7 kg + 3.63 kg) = −0.420 m/s.
The cat lands on sled B; conservation of momentum requires vB,1 = m(vA,1 + vc )/(m + M ). The
cat jumps off of sled B; conservation of momentum is now
M vB,2 + m(vB,2 − vc ) = m(vA,1 + vc ),
83
or
vB,2 = 2mvc /(m + M ) = (3.63 kg)[(−0.420 m/s) + 2(3.05 m/s)]/(22.7 kg + 3.63 kg) = 0.783 m/s.
The cat then lands on cart A; conservation of momentum requires that (M + m)vA,2 = −M vB,2 , or
vA,2 = −(22.7 kg)(0.783 m/s)/(22.7 kg + 3.63 kg) = −0.675 m/s.
P6-11 We align the coordinate system so that west is +x and south is +y. The each car
contributes the following to the initial momentum
A :
(2720 lb/g)(38.5 mi/h)î = 1.05×105 lb · mi/h/g î,
B
(3640 lb/g)(58.0 mi/h)ĵ = 2.11×105 lb · mi/h/g ĵ.
:
These become the components of the final momentum. The direction is then
θ = arctan
2.11×105 lb · mi/h/g
= 63.5◦ ,
1.05×105 lb · mi/h/g
south of west. The magnitude is the square root of the sum of the squares,
2.36×105 lb · mi/h/g,
and we divide this by the mass (6360 lb/g) to get the final speed after the collision: 37.1 mi/h.
P6-12 (a) Ball A must carry off a momentum of ~p = mB v î−mB v/2 ĵ, which would be in a direction
θ = arctan(−0.5/1) = 27◦ from the original direction of B, or 117◦ from the final direction.
(b) No.
P6-13 (a) We assume all balls have a mass m. The collision imparts a “sideways” momentum to
the cue ball of m(3.50 m/s) sin(65◦ ) = m(3.17 m/s). The other ball must have an equal, but opposite
“sideways” momentum, so −m(3.17 m/s) = m(6.75 m/s) sin θ, or θ = −28.0◦ .
(b) The final “forward” momentum is
m(3.50 m/s) cos(65◦ ) + m(6.75 m/s) cos(−28◦ ) = m(7.44 m/s),
so the initial speed of the cue ball would have been 7.44 m/s.
P6-14
Assuming M m, Eq. 6-25 becomes
v 2f = 2v 1i − v 1i = 2(13 km/s) − (−12 km/s) = 38 km/s.
P6-15
(a) We get
2(220 g)
(45.0 m/s) = 74.4 m/s.
(220 g) + (46.0 g)
(b) Doubling the mass of the clubhead we get
v 2,f =
v 2,f =
2(440 g)
(45.0 m/s) = 81.5 m/s.
(440 g) + (46.0 g)
(c) Tripling the mass of the clubhead we get
v 2,f =
2(660 g)
(45.0 m/s) = 84.1 m/s.
(660 g) + (46.0 g)
Although the heavier club helps some, the maximum speed to get out of the ball will be less than
twice the speed of the club.
84
P6-16 There will always be at least two collisions. The balls are a, b, and c from left to right.
After the first collision between a and b one has
vb,1 = v0 and va,1 = 0.
After the first collision between b and c one has
vc,1 = 2mv0 /(m + M ) and vb,2 = (m − M )v0 /(m + M ).
(a) If m ≥ M then ball b continue to move to the right (or stops) and there are no more collisions.
(b) If m < M then ball b bounces back and strikes ball a which was at rest. Then
va,2 = (m − M )v0 /(m + M ) and vb,3 = 0.
P6-17 All three balls are identical in mass and radii? Then balls 2 and 3 will move off at 30◦ to
the initial direction of the first ball. By symmetry we expect balls 2 and 3 to have the same speed.
The problem now is to define an elastic three body collision. It is no longer the case that the
balls bounce off with the same speed in the center of mass. One can’t even treat the problem as two
separate collisions, one right after the other. No amount of momentum conservation laws will help
solve the problem; we need some additional physics, but at this point in the text we don’t have it.
P6-18 The original speed is v0 in the lab frame. Let α be the angle of deflection in the cm frame
and ~v10 be the initial velocity in the cm frame. Then the velocity after the collision in the cm frame
is v10 cos α î + v10 sin α ĵ and the velocity in the lab frame is (v10 cos α + v)î + v10 sin α ĵ, where v is the
speed of the cm frame. The deflection angle in the lab frame is
θ = arctan[(v10 sin α)/(v10 cos α + v)],
but v = m1 v0 /(m1 + m2 ) and v10 = v0 − v so v10 = m2 v0 /(m1 + m2 ) and
θ = arctan[(m2 sin α)/(m2 cos α + m1 )].
(c) θ is a maximum when (cos α + m1 /m2 )/ sin α is a minimum, which happens when cos α =
−m1 /m2 if m1 ≤ m2 . Then [(m2 sin α)/(m2 cos α + m1 )] can have any value between −∞ and ∞,
so θ can be between 0 and π.
(a) If m1 > m2 then (cos α + m1 /m2 )/ sin α is a minimum when cos α = −m2 /m1 , then
q
[(m2 sin α)/(m2 cos α + m1 )] = m2 / m21 − m22 .
p
If tan θ = m2 / m21 − m22 then m1 is like a hypotenuse and m2 the opposite side. Then
q
p
cos θ = m21 − m22 /m1 = 1 − (m2 /m1 )2 .
(b) We need to change to the center of mass system. Since both particles have the same mass,
the conservation of momentum problem is effectively the same as a (vector) conservation of velocity
problem. Since one of the particles is originally at rest, the center of mass moves with speed
v cm = v 1i /2. In the figure below the center of mass velocities are primed; the transformation
velocity is vt .
85
vt
v’1f
v 1f
v’2i
v’1i
v
v’2f
2f
vt
Note that since vt = v 0 1i = v 0 2i = v 0 1f = v 0 2f the entire problem can be inscribed in a rhombus.
The diagonals of the rhombus are the directions of v 1f and v 2f ; note that the diagonals of a rhombus
are necessarily at right angles!
P6-19 (a) The speed of the bullet after leaving the first block but before entering the second can
be determined by momentum conservation.
Pf
pf,bl + pf,bu
mbl v f,bl + mbu v f,bu
(1.78kg)(1.48 m/s)+(3.54×10−3 kg)(1.48 m/s)
=
=
=
=
P i,
pi,bl + pi,bu ,
mbl v i,bl + mbu v i,bu ,
(1.78kg)(0)+(3.54×10−3 kg)v i,bu ,
which has solution v i,bl = 746 m/s.
(b) We do the same steps again, except applied to the first block,
Pf
=
pf,bl + pf,bu =
mbl v f,bl + mbu v f,bu =
(1.22kg)(0.63 m/s)+(3.54×10−3 kg)(746 m/s) =
P i,
pi,bl + pi,bu ,
mbl v i,bl + mbu v i,bu ,
(1.22kg)(0)+(3.54×10−3 kg)v i,bu ,
which has solution v i,bl = 963 m/s.
◦
P6-20 The acceleration of the block down
the ramp
). The ramp has a length of
p
p is a1 = g sin(22
◦
◦
d = h/ sin(22 ), so it takes a time t1 = 2d/a
=
2h/g/
sin(22
)
to
reach
the bottom. The speed
1
√
when it reaches the bottom is v1 = a1 t1 = 2gh. Notice that it is independent of the angle!
The collision is inelastic, so the two stick together and move with an initial speed of v2 =
m1 v1 /(m1 + m2 ). They slide a distance x before stopping; the average speed while decelerating is
v av = v2 /2, so the stopping time is t2 = 2x/v2 and the deceleration is a2 = v2 /t2 = v22 /(2x). If the
retarding force is f = (m1 + m2 )a2 , then f = µk (m1 + m2 )g. Glue it all together and
µk =
m21
h
(2.0 kg)2
(0.65 m)
=
= 0.15.
2
2
(m1 + m2 ) x
(2.0 kg + 3.5 kg) (0.57 m)
86
P6-21 (a) For an object with initial speed v and deceleration −a which travels a distance x before
2
stopping, the time
√ t to stop is t = v/a, the average speed while stopping is v/2, and d = at /2.
Combining, v = 2ax. The deceleration in this case is given by a = µk g.
Then just after the collision
p
vA = 2(0.130)(9.81 m/s2 )(8.20 m) = 4.57 m/s,
while
vB =
p
2(0.130)(9.81 m/s2 )(6.10 m) = 3.94 m/s,
(b) v0 = [(1100 kg)(4.57 m/s) + (1400kg)(3.94 m/s)]/(1400 kg) = 7.53 m/s.
87
E7-1 xcm = (7.36×1022 kg)(3.82×108 m)/(7.36×1022 kg + 5.98×1034 kg) = 4.64×106 m. This is less
than the radius of the Earth.
E7-2 If the particles are l apart then
x1 = m1 l(m1 + m2 )
is the distance from particle 1 to the center of mass and
x2 = m2 l(m1 + m2 )
is the distance from particle 2 to the center of mass. Divide the top equation by the bottom and
x1 /x2 = m1 /m2 .
E7-3
The center of mass velocity is given by Eq. 7-1,
~vcm
=
=
m1 ~v1 + m2 ~v2
,
m1 + m2
(2210 kg)(105 km/h) + (2080 kg)(43.5 km/h)
= 75.2 km/h.
(2210 kg) + (2080 kg)
E7-4 They will meet at the center of mass, so
xcm = (65 kg)(9.7 m)/(65 kg + 42 kg) = 5.9 m.
E7-5 (a) No external forces, center of mass doesn’t move.
(b) The collide at the center of mass,
xcm = (4.29 kg)(1.64 m)/(4.29 kg + 1.43 kg) = 1.23 m.
E7-6 The range of the center of mass is
R = v02 sin 2θ /g = (466 m/s)2 sin(2 × 57.4◦ )/(9.81 m/s2 ) = 2.01×104 m.
Half lands directly underneath the highest point, or 1.00×104 m. The other piece must land at x,
such that
2.01×104 m = (1.00×104 m + x)/2;
then x = 3.02×104 m.
E7-7 The center of mass of the boat + dog doesn’t move because there are no external forces on
the system. Define the coordinate system so that distances are measured from the shore, so toward
the shore is in the negative x direction. The change in position of the center of mass is given by
∆xcm =
md ∆xd + mb ∆xb
= 0,
md + mb
Both ∆xd and ∆xb are measured with respect to the shore; we are given ∆xdb = −8.50 ft, the
displacement of the dog with respect to the boat. But
∆xd = ∆xdb + ∆xb .
88
Since we want to find out about the dog, we’ll substitute for the boat’s displacement,
0=
md ∆xd + mb (∆xd − ∆xdb )
.
md + mb
Rearrange and solve for ∆xd . Use W = mg and multiply the top and bottom of the expression by
g. Then
(46.4 lb)(−8.50 ft)
mb ∆xdb g
=
= −6.90 ft.
∆xd =
md + mb g
(10.8 lb) + (46.4 lb)
The dog is now 21.4 − 6.9 = 14.5 feet from shore.
E7-8 Richard has too much time on his hands.
The center of mass of the system is xcm away from the center of the boat. Switching seats is
effectively the same thing as rotating the canoe through 180◦ , so the center of mass of the system
has moved through a distance of 2xcm = 0.412 m. Then xcm = 0.206 m. Then
xcm = (M l − ml)/(M + m + mc ) = 0.206 m,
where l = 1.47 m, M = 78.4 kg, mc = 31.6 kg, and m is Judy’s mass. Rearrange,
m=
M l − (M + mc )xcm
(78.4 kg)(1.47 m) − (78.4 kg + 31.6 kg)(0.206 m)
=
= 55.2 kg.
l + xcm
(1.47 m) + (0.206 m)
E7-9 It takes the man t = (18.2 m)/(2.08 m/s) = 8.75 s to walk to the front of the boat. During
this time the center of mass of the system has moved forward x = (4.16 m/s)(8.75 s) = 36.4 m. But
in walking forward to the front of the boat the man shifted the center of mass by a distance of
(84.4 kg)(18.2 m)/(84.4 kg + 425 kg) = 3.02 m, so the boat only traveled 36.4 m − 3.02 m = 33.4 m.
E7-10 Do each coordinate separately.
xcm =
(3 kg)(0) + (8 kg)(1 m) + (4 kg)(2 m)
= 1.07 m
(3 kg) + (8 kg) + (4 kg)
y cm =
(3 kg)(0) + (8 kg)(2 m) + (4 kg)(1 m)
= 1.33 m
(3 kg) + (8 kg) + (4 kg)
and
E7-11 The center of mass of the three hydrogen atoms will be at the center of the pyramid
base. The problem is then reduced to finding the center of mass of the nitrogen atom and the three
hydrogen atom triangle. This molecular center of mass must lie on the dotted line in Fig. 7-27.
The location of the plane of the hydrogen atoms can be found from Pythagoras theorem
p
y h = (10.14×10−11 m)2 − (9.40×10−11 m)2 = 3.8×10−11 m.
This distance can be used to find the center of mass of the molecule. From Eq. 7-2,
y cm =
(13.9mh )(0) + (3mh )(3.8×10−11 m)
mn y n + mh y h
=
= 6.75×10−12 m.
mn + mh
(13.9mh ) + (3mh )
E7-12 The velocity components of the center of mass at t = 1.42 s are v cm,x = 7.3 m/s and
v cm,y = (10.0 m/s) − (9.81 m/s)(1.42 s) = −3.93 m/s. Then the velocity components of the “other”
piece are
v2,x = [(9.6 kg)(7.3 m/s) − (6.5 kg)(11.4 m/s)]/(3.1 kg) = −1.30 m/s.
and
v2,y = [(9.6 kg)(−3.9 m/s) − (6.5 kg)(−4.6 m/s)]/(3.1 kg) = −2.4 m/s.
89
E7-13 The center of mass should lie on the perpendicular bisector of the rod of mass 3M . We
can view the system as having two parts: the heavy rod of mass 3M and the two light rods each of
mass M . The two light rods have a center of mass at the center of the square.
Both of these center of masses are located along the vertical line of symmetry for the object.
The center of mass of the heavy bar is at y h,cm = 0, while the combined center of mass of the two
light bars is at y l,cm = L/2, where down is positive. The center of mass of the system is then at
y cm =
2(L/2)
2M y l,cm + 3M y h,cm
=
= L/5.
2M + 3M
5
E7-14 The two slabs have the same volume and have mass mi = ρi V . The center of mass is
located at
3
xcm =
3
ρ1 − ρ2
(7.85 g/cm ) − (2.70 g/cm )
m1 l − m2 l
=l
= (5.5 cm)
3
3 = 2.68 cm
m1 + m2
ρ1 + ρ2
(7.85 g/cm ) + (2.70 g/cm )
from the boundary inside the iron; it is centered in the y and z directions.
E7-15 Treat the four sides of the box as one thing of mass 4m with a mass located l/2 above the
base. Then the center of mass is
z cm = (l/2)(4m)/(4m + m) = 2l/5 = 2(0.4 m)/5 = 0.16 m,
xcm = y cm = 0.2 m.
E7-16 One piece moves off with momentum m(31.4 m/s)î, another moves off with momentum
2m(31.4 m/s)ĵ. The third piece must then have momentum −m(31.4 m/s)î − 2m(31.4 m/s)ĵ and
velocity −(1/3)(31.4 m/s)î − 2/3(31.4 m/s)ĵ = −10.5 m/s î − 20.9 m/s ĵ. The magnitude of v3 is
23.4 m/s and direction 63.3◦ away from the lighter piece.
E7-17 It will take an impulse of (84.7 kg)(3.87 m/s) = 328 kg · m/s to stop the animal. This would
come from firing n bullets where n = (328 kg · m/s)/[(0.0126 kg)(975 m/s) = 27.
E7-18 Conservation of momentum for firing one cannon ball of mass m with muzzle speed vc
forward out of a cannon on a trolley of original total mass M moving forward with original speed
v0 is
M v0 = (M − m)v1 + m(vc + v1 ) = M v1 + mvc ,
where v1 is the speed of the trolley after the cannonball is fired. Then to stop the trolley we require
n cannonballs be fired so that
n = (M v0 )/(mvc ) = [(3500 kg)(45 m/s)]/[(65 kg)(625 m/s)] = 3.88,
so n = 4.
E7-19 Label the velocities of the various containers as ~vk where k is an integer between one and
twelve. The mass of each container is m. The subscript “g” refers to the goo; the subscript k refers
to the kth container.
The total momentum before the collision is given by
X
~ =
P
m~vk ,i + mg ~vg,i = 12m~vcont.,cm + mg ~vg,i .
k
90
We are told, however, that the initial velocity of the center of mass of the containers is at rest, so
~ = mg ~vg,i , and has a magnitude of 4000 kg·m/s.
the initial momentum simplifies to P
(a) Then
P
(4000 kg·m/s)
v cm =
=
= 3.2 m/s.
12m + mg
12(100.0 kg) + (50 kg)
(b) It doesn’t matter if the cord breaks, we’ll get the same answer for the motion of the center
of mass.
E7-20 (a) F = (3270 m/s)(480 kg/s) = 1.57×106 N.
(b) m = 2.55×105 kg − (480 kg/s)(250 s) = 1.35×105 kg.
(c) Eq. 7-32:
v f = (−3270 m/s) ln(1.35×105 kg/2.55×105 kg) = 2080 m/s.
E7-21 Use Eq. 7-32. The initial velocity of the rocket is 0. The mass ratio can then be found
from a minor rearrangement;
Mi
= e|vf /vrel |
Mf
The “flipping” of the left hand side of this expression is possible because the exhaust velocity is
negative with respect to the rocket. For part (a) M i /M f = e = 2.72. For part (b) M i /M f = e2 =
7.39.
E7-22 Eq. 7-32 rearranged:
Mf
= e−|∆v/vrel | = e−(22.6m/s)/(1230m/s) = 0.982.
Mi
The fraction of the initial mass which is discarded is 0.0182.
E7-23 The loaded rocket has a weight of (1.11×105 kg)(9.81 m/s2 ) = 1.09×106 N; the thrust must be
at least this large to get the rocket off the ground. Then v ≥ (1.09×106 N)/(820 kg/s) = 1.33×103 m/s
is the minimum exhaust speed.
E7-24
The acceleration down the incline is (9.8 m/s2 ) sin(26◦ ) = 4.3 m/s2 . It will take t =
p
2(93 m)/(4.3 m/s2 ) = 6.6 s. The sand doesn’t affect the problem, so long as it only “leaks” out.
E7-25 We’ll use Eq. 7-4 to solve this problem, but since we are given weights instead of mass
we’ll multiply the top and bottom by g like we did in Exercise 7-7. Then
~vcm =
m1 ~v1 + m2 ~v2 g
W1 ~v1 + W2 ~v2
=
.
m1 + m2 g
W1 + W2
Now for the numbers
v cm =
P7-1
(9.75 T)(1.36 m/s) + (0.50 T)(0)
= 1.29 m/s.
(9.75 T) + (0.50 T)
(a) The balloon moves down so that the center of mass is stationary;
0 = M v b + mv m = M v b + m(v + v b ),
or v b = −mv/(m + M ).
(b) When the man stops so does the balloon.
91
P7-2 (a) The center of mass is midway between them.
(b) Measure from the heavier mass.
xcm = (0.0560 m)(0.816 kg)/(1.700 kg) = 0.0269 m,
which is 1.12 mm closer to the heavier mass than in part (a).
(c) Think Atwood’s machine. The acceleration of the two masses is
a = 2∆m g/(m1 + m2 ) = 2(0.034 kg)g/(1.700 kg) = 0.0400g,
the heavier going down while the lighter moves up. The acceleration of the center of mass is
acm = (am1 − am2 )/(m1 + m2 ) = (0.0400g)2(0.034 kg)g/(1.700 kg) = 0.00160g.
P7-3 This is a glorified Atwood’s machine problem. The total mass on the right side is the mass
per unit length times the length, mr = λx; similarly the mass on the left is given by ml = λ(L − x).
Then
m2 − m1
λx − λ(L − x)
2x − L
a=
g=
g=
g
m2 + m1
λx + λ(L − x)
L
which solves the problem. The acceleration is in the direction of the side of length x if x > L/2.
P7-4 (a) Assume the car is massless. Then moving the cannonballs is moving the center of mass,
unless the cannonballs don’t move but instead the car does. How far? L.
(b) Once the cannonballs stop moving so does the rail car.
P7-5 By symmetry, the center of mass of the empty storage tank should be in the very center,
along the axis at a height y t,cm = H/2. We can pretend that the entire mass of the tank, mt = M ,
is located at this point.
The center of mass of the gasoline is also, by symmetry, located along the axis at half the height of
the gasoline, y g,cm = x/2. The mass, if the tank were filled to a height H, is m; assuming a uniform
density for the gasoline, the mass present when the level of gas reaches a height x is mg = mx/H.
(a) The center of mass of the entire system is at the center of the cylinder when the tank is full
and when the tank is empty. When the tank is half full the center of mass is below the center. So as
the tank changes from full to empty the center of mass drops, reaches some lowest point, and then
rises back to the center of the tank.
(b) The center of mass of the entire system is found from
y cm =
mg y g,cm + mt y t,cm
(mx/H)(x/2) + (M )(H/2)
mx2 + M H 2
=
=
.
mg + mt
(mx/H) + (M )
2mx + 2M H
Take the derivative:
m mx2 + 2xM H − M H 2
dy cm
=
dx
(mx + M H)2
Set this equal to zero to find the minimum; this means we want the numerator to vanish, or mx2 +
2xM H − M H 2 = 0. Then
√
−M + M 2 + mM
x=
H.
m
92
P7-6
The center of mass will be located along symmetry axis. Call this the x axis. Then
Z
1
xcm =
xdm,
M
Z R Z √R2 −x2
4
=
x dy dx,
πR2 0 0
Z R p
4
=
x R2 − x2 dx,
πR2 0
4
4R
=
R3 /3 =
.
2
πR
3π
P7-7
(a) The components of the shell velocity with respect to the cannon are
vx0 = (556 m/s) cos(39.0◦ ) = 432 m/s and vy0 = (556 m/s) sin(39.0◦ ) = 350 m/s.
The vertical component with respect to the ground is the same, vy = vy0 , but the horizontal component is found from conservation of momentum:
M (vx − vx0 ) + m(vx ) = 0,
so vx = (1400 kg)(432 m/s)(70.0 kg + 1400 kg) = 411 m/s. The resulting speed is v = 540 m/s.
(b) The direction is θ = arctan(350/411) = 40.4◦ .
P7-8
v = (2870 kg)(252 m/s)/(2870 kg + 917 kg) = 191 m/s.
P7-9 It takes (1.5 m/s)(20 kg) = 30 N to accelerate the luggage to the speed of the belt. The
people when taking the luggage off will (on average) also need to exert a 30 N force to remove it;
this force (because of friction) will be exerted on the belt. So the belt requires 60 N of additional
force.
P7-10
(a) The thrust must be at least equal to the weight, so
dm/dt = (5860 kg)(9.81 m/s2 )/(1170 m/s) = 49.1 kg/s.
(b) The net force on the rocket will need to be F = (5860 kg)(18.3 m/s2 ) = 107000 N. Add this
to the weight to find the thrust, so
dm/dt = [107000 N + (5860 kg)(9.81 m/s2 )]/(1170 m/s) = 141 kg/s
P7-11 Consider Eq. 7-31. We want the barges to continue at constant speed, so the left hand
side of that equation vanishes. Then
X
~ ext = −~vrel dM .
F
dt
We are told that the frictional force is independent of the weight, since the speed doesn’t change the
frictional force should be constant
P ~ and equal in magnitude to the force exerted by the engine before
the shoveling happens. Then F
ext is equal to the additional force required from the engines. We’ll
~
call it P.
The relative speed of the coal to the faster moving cart has magnitude: 21.2 − 9.65 = 11.6
km/h= 3.22 m/s. The mass flux is 15.4 kg/s, so P = (3.22 m/s)(15.4kg/s) = 49.6 N. The faster
moving cart will need to increase the engine force by 49.6 N. The slower cart won’t need to do
anything, because the coal left the slower barge with a relative speed of zero according to our
approximation.
93
P7-12 (a) Nothing is ejected from the string, so v rel = 0. Then Eq. 7-31 reduces to m dv/dt = F ext .
(b) Since F ext is from the weight of the hanging string, and the fraction that is hanging is y/L,
F ext = mgy/L. The equation of motion is then d2 y/dt2 = gy/L.
(c) Take first derivative:
√
√
y0 p
dy
= ( g/L) e g/Lt − e− g/Lt ,
dt
2
and then second derivative,
√
√
d2 y
y0 p
2
g/Lt
− g/Lt
=
(
g/L)
e
+
e
.
dt2
2
Substitute into equation of motion. It works! Note that when t = 0 we have y = y0 .
94
E8-1 An n-dimensional object can be oriented by stating the position of n different carefully
chosen points Pi inside the body. Since each point has n coordinates, one might think there are n2
coordinates required to completely specify the position of the body. But if the body is rigid then
the distances between the points are fixed. There is a distance dij for every pair of points Pi and
Pj . For each distance dij we need one fewer coordinate to specify the position of the body. There
are n(n − 1)/2 ways to connect n objects in pairs, so n2 − n(n − 1)/2 = n(n + 1)/2 is the number
of coordinates required.
E8-2 (1 rev/min)(2π rad/rev)/(60 s/min) = 0.105 rad/s.
E8-3 (a) ω = a + 3bt2 − 4ct3 .
(b) α = 6bt − 12t2 .
E8-4 (a) The radius is r = (2.3×104 ly)(3.0×108 m/s) = 6.9×1012 m · y/s. The time to make one
revolution is t = (2π6.9×1012 m · y/s)/(250×103 m/s) = 1.7×108 y.
(b) The Sun has made 4.5×109 y/1.7×108 y = 26 revolutions.
E8-5
(a) Integrate.
ωz = ω0 +
t
Z
0
4at3 − 3bt2 dt = ω0 + at4 − bt3
(b) Integrate, again.
∆θ =
Z
0
t
ωz dt =
Z
0
t
1
1
ω0 + at4 − bt3 dt = ω0 t + at5 − bt4
5
4
E8-6 (a) (1 rev/min)(2π rad/rev)/(60 s/min) = 0.105 rad/s.
(b) (1 rev/h)(2π rad/rev)/(3600 s/h) = 1.75×10−3 rad/s.
(c) (1/12 rev/h)(2π rad/rev)/(3600 s/h) = 1.45×10−3 rad/s.
E8-7 85 mi/h = 125 ft/s. The ball takes t = (60 ft)/(125 ft/s) = 0.48 s to reach the plate. It
makes (30 rev/s)(0.48 s) = 14 revolutions in that time.
p
E8-8 It takes t = 2(10 m)/(9.81 m/s2 ) = 1.43 s to fall 10 m. The average angular velocity is
then ω = (2.5)(2π rad)/(1.43 s) = 11 rad/s.
E8-9 (a) Since there are eight spokes, this means the wheel can make no more than 1/8 of a
revolution while the arrow traverses the plane of the wheel. The wheel rotates at 2.5 rev/s; it
makes one revolution every 1/2.5 = 0.4 s; so the arrow must pass through the wheel in less than
0.4/8 = 0.05 s.
The arrow is 0.24 m long, and it must move at least one arrow length in 0.05 s. The corresponding
minimum speed is (0.24 m)/(0.05 s) = 4.8 m/s.
(b) It does not matter where you aim, because the wheel is rigid. It is the angle through which
the spokes have turned, not the distance, which matters here.
95
E8-10 We look for the times when the Sun, the Earth, and the other planet are collinear in some
specified order.
Since the outer planets revolve around the Sun more slowly than Earth, after one year the Earth
has returned to the original position, but the outer planet has completed less than one revolution.
The Earth will then “catch up” with the outer planet before the planet has completed a revolution.
If θE is the angle through which Earth moved and θP is the angle through which the planet moved,
then θE = θP + 2π, since the Earth completed one more revolution than the planet.
If ωP is the angular velocity of the planet, then the angle through which it moves during the
time TS (the time for the planet to line up with the Earth). Then
θE
ωE TS
ωE
= θP + 2π,
= ωP TS + 2π,
= ωP + 2π/TS
The angular velocity of a planet is ω = 2π/T , where T is the period of revolution. Substituting this
into the last equation above yields
1/TE = 1/TP + 1/TS .
E8-11 We look for the times when the Sun, the Earth, and the other planet are collinear in some
specified order.
Since the inner planets revolve around the Sun more quickly than Earth, after one year the Earth
has returned to the original position, but the inner planet has completed more than one revolution.
The inner planet must then have “caught-up” with the Earth before the Earth has completed a
revolution. If θE is the angle through which Earth moved and θP is the angle through which the
planet moved, then θP = θE + 2π, since the inner planet completed one more revolution than the
Earth.
96
If ωP is the angular velocity of the planet, then the angle through which it moves during the
time TS (the time for the planet to line up with the Earth). Then
θP
ωP TS
ωP
= θE + 2π,
= ωE TS + 2π,
= ωE + 2π/TS
The angular velocity of a planet is ω = 2π/T , where T is the period of revolution. Substituting this
into the last equation above yields
1/TP = 1/TE + 1/TS .
2
E8-12 (a) α = (−78 rev/min)/(0.533 min) = −150 rev/min .
(b) Average angular speed while slowing down is 39 rev/min, so (39 rev/min)(0.533 min) =
21 rev.
2
E8-13 (a) α = (2880 rev/min − 1170 rev/min)/(0.210 min) = 8140 rev/min .
(b) Average angular speed while accelerating is 2030 rev/min, so (2030 rev/min)(0.210 min) =
425 rev.
E8-14 Find area under curve.
1
(5 min + 2.5 min)(3000 rev/min) = 1.13×104 rev.
2
E8-15
(a) ω0z = 25.2 rad/s; ωz = 0; t = 19.7 s; and αz and φ are unknown. From Eq. 8-6,
ωz = ω0z + αz t,
(0) = (25.2 rad/s) + αz (19.7 s),
αz
2
= −1.28 rad/s
(b) We use Eq. 8-7 to find the angle through which the wheel rotates.
1
1
2
φ = φ0 +ω0z t+ αz t2 = (0)+(25.2 rad/s)(19.7 s)+ (−1.28 rad/s )(19.7 s)2 = 248 rad.
2
2
97
(c) φ = 248 rad
1 rev
2π rad
= 39.5 rev.
2
E8-16 (a) α = (225 rev/min − 315 rev/min)/(1.00 min) = −90.0 rev/min .
2
(b) t = (0 − 225 rev/min)/(−90.0 rev/min ) = 2.50 min.
2
(c) (−90.0 rev/min )(2.50 min)2 /2 + (225 rev/min)(2.50 min) = 281 rev.
E8-17 (a) The average angular speed was (90 rev)/(15 s) = 6.0 rev/s. The angular speed at the
beginning of the interval was then 2(6.0 rev/s) − (10 rev/s) = 2.0 rev/s.
2
(b) The angular acceleration was (10 rev/s − 2.0 rev/s)/(15 s) = 0.533 rev/s . The time required
2
to get the wheel to 2.0 rev/s was t = (2.0 rev/s)/(0.533 rev/s ) = 3.8 s.
E8-18 (a) The wheel will rotate through an angle φ where
φ = (563 cm)/(8.14 cm/2) = 138 rad.
(b) t =
q
2
2(138 rad)/(1.47 rad/s ) = 13.7 s.
E8-19 (a) We are given φ = 42.3 rev= 266 rad, ω0z = 1.44 rad/s, and ωz = 0. Assuming a
uniform deceleration, the average angular velocity during the interval is
ω av,z =
1
(ω0z + ωz ) = 0.72 rad/s.
2
Then the time taken for deceleration is given by φ = ω av,z t, so t = 369 s.
(b) The angular acceleration can be found from Eq. 8-6,
ωz = ω0z + αz t,
(0) = (1.44 rad/s) + αz (369 s),
αz
2
= −3.9 × 10−3 rad/s .
(c) We’ll solve Eq. 8-7 for t,
1
φ = φ0 + ω0z t + αz t2 ,
2
1
2
(0) + (1.44 rad/s)t + (−3.9 × 10−3 rad/s )t2 ,
2
= −133 + (1.44 s−1 )t − (−1.95 × 10−3 s−2 )t2 .
(133 rad) =
0
Solving this quadratic expression yields two answers: t = 108 s and t = 630 s.
2
E8-20 The angular acceleration is α = (4.96 rad/s)/(2.33 s) = 2.13 rad/s . The angle through
2
which the wheel turned while accelerating is φ = (2.13 rad/s )(23.0 s)2 /2 = 563 rad. The angular
2
speed at this time is ω = (2.13 rad/s )(23.0 s) = 49.0 rad/s. The wheel spins through an additional
angle of (49.0 rad/s)(46 s − 23 s) = 1130 rad, for a total angle of 1690 rad.
E8-21 ω = (14.6 m/s)/(110 m) = 0.133 rad/s.
E8-22 The linear acceleration is (25 m/s − 12 m/s)/(6.2 s) = 2.1 m/s2 . The angular acceleration is
α = (2.1 m/s2 )/(0.75 m/2) = 5.6 rad/s.
98
E8-23 (a) The angular speed is given by vT = ωr. So ω = vT /r = (28, 700 km/hr)/(3220
km) = 8.91 rad/hr. That’s the same thing as 2.48×10−3 rad/s.
(b) aR = ω 2 r = (8.91 rad/h)2 (3220 km) = 256000 km/h2 , or
2
aR = 256000 km/h (1/3600 h/s)2 (1000 m/km) = 19.8 m/s2 .
(c) If the speed is constant then the tangential acceleration is zero, regardless of the shape of the
trajectory!
E8-24 The bar needs to make
(1.50 cm)(12.0 turns/cm) = 18 turns.
This will happen is (18 rev)/(237 rev/min) = 4.56 s.
E8-25 (a) The angular speed is ω = (2π rad)/(86400 s) = 7.27×10−5 rad/s.
(b) The distance from the polar axis is r = (6.37×106 m) cos(40◦ ) = 4.88×106 m. The linear speed
is then v = (7.27×10−5 rad/s)(4.88×106 m) = 355 m/s.
(c) The angular speed is the same as part (a). The distance from the polar axis is r = (6.37×
106 m) cos(0◦ ) = 6.37×106 m. The linear speed is then v = (7.27×10−5 rad/s)(6.37×106 m) = 463 m/s.
2
E8-26 (a) aT = (14.2 rad/s )(0.0283 m) = 0.402 m/s2 .
(b) Full speed is ω = 289 rad/s. aR = (289 rad/s)2 (0.0283 m) = 2360 m/s2 .
(c) It takes
2
t = (289 rad/s)/(14.2 rad/s ) = 20.4 s
to get up to full speed. Then x = (0.402 m/s2 )(20.4 s)2 /2 = 83.6 m is the distance through which a
point on the rim moves.
E8-27 (a) The pilot sees the propeller rotate, no more. So the tip of the propeller is moving with
a tangential velocity of vT = ωr = (2000 rev/min)(2π rad/rev)(1.5 m) = 18900 m/min. This is the
same thing as 315 m/s.
(b) The observer on the ground sees this tangential motion and sees the forward motion of
the plane. These two velocity components are perpendicular, so the magnitude of the sum is
p
(315 m/s)2 + (133 m/s)2 = 342 m/s.
E8-28 aT = aR when rα = rω 2 = r(αt)2 , or t =
q
2
1/(0.236 rad/s ) = 2.06 s.
E8-29 (a) aR = rω 2 = rα2 t2 .
(b) aT = rα.
p
(c) Since aR = aT tan(57.0◦ ), t = tan(57.0◦ )/α. Then
φ=
1 2
1
αt = tan(57.0◦ ) = 0.77 rad = 44.1◦ .
2
2
E8-30 (a) The tangential speed of the edge of the wheel relative axle is v = 27 m/s. ω =
(27 m/s)/(0.38 m) = 71 rad/s.
(b) The average angular speed while slowing is 71 rad/s/2, the time required to stop is then
t = (30 × 2π rad)/(71 rad/s/2) = 5.3 s. The angular acceleration is then α = (−71 rad/s)/(5.3 s) =
−13 rad/s.
(c) The car moves forward (27 m/s/2)(5.3 s) = 72 m.
99
E8-31 Yes, the speed would be wrong. The angular velocity of the small wheel would be ω = v t /rs ,
but the reported velocity would be v = ωrl = v t rl /rs . This would be in error by a fraction
∆v
(72 cm)
=
− 1 = 0.16.
vt
(62 cm)
E8-32 (a) Square both equations and then add them:
x2 + y 2 = (R cos ωt)2 + (R sin ωt)2 = R2 ,
which is the equation for a circle of radius R.
(b) vx = −Rω sin ωt = −ωy; vy = Rω cos ωt = ωx. Square and add, v = ωR. The direction is
tangent to the circle.
(b) ax = −Rω 2 cos ωt = −ω 2 x; ay = −Rω 2 sin ωt = −ω 2 y. Square and add, a = ω 2 R. The
direction is toward the center.
E8-33 (a) The object is “slowing down”, so α
~ = (−2.66 rad/s2 )k̂. We know the direction
because it is rotating about the z axis and we are given the direction of ω
~ . Then from Eq. 8-19,
~ = (14.3 rad/s)k̂ × [(1.83 m)ĵ + (1.26 m)k̂]. But only the cross term k̂ × ĵ survives, so
~v = ω
~ ×R
~v = (−26.2 m/s)î.
(b) We find the acceleration from Eq. 8-21,
~ +ω
~a = α
~ ×R
~ × ~v,
2
= (−2.66 rad/s )k̂ × [(1.83 m)ĵ + (1.26 m)k̂] + (14.3 rad/s)k̂ × (−26.2 m/s)î,
=
(4.87 m/s2 )î + (−375 m/s2 )ĵ.
~ = −2m~
E8-34 (a) F
ω × ~v = −2mωv cos θ, where θ is the latitude. Then
F = 2(12 kg)(2π rad/86400 s)(35 m/s) cos(45◦ ) = 0.043 N,
and is directed west.
(b) Reversing the velocity will reverse the direction, so east.
(c) No. The Coriolis force pushes it to the west on the way up and gives it a westerly velocity;
on the way down the Coriolis force slows down the westerly motion, but does not push it back east.
The object lands to the west of the starting point.
2
P8-1 (a) ω = (4.0 rad/s) − (6.0 rad/s )t + (3.0 rad/s)t2 . Then ω(2.0 s) = 4.0 rad/s and ω(4.0 s) =
28.0 rad/s.
2
(b) αav = (28.0 rad/s − 4.0 rad/s)/(4.0 s − 2.0 s) = 12 rad/s .
2
2
2
(c) α = −(6.0 rad/s ) + (6.0 rad/s)t. Then α(2.0 s) = 6.0 rad/s and α(4.0 s) = 18.0 rad/s .
P8-2
If the wheel really does move counterclockwise at 4.0 rev/min, then it turns through
(4.0 rev/min)/[(60 s/min)(24 frames/s)] = 2.78×10−3 rev/frame.
This means that a spoke has moved 2.78 × 10−3 rev. There are 16 spokes each located 1/16 of a
revolution around the wheel. If instead of moving counterclockwise the wheel was instead moving
clockwise so that a different spoke had moved 1/16 rev − 2.78 × 10−3 rev = 0.0597 rev, then the
same effect would be present. The wheel then would be turning clockwise with a speed of ω =
(0.0597 rev)(60 s/min)(24 frames/s) = 86 rev/min.
100
P8-3 (a) In the diagram below the Earth is shown at two locations a day apart. The Earth rotates
clockwise in this figure.
Note that the Earth rotates through 2π rad in order to be correctly oriented for a complete
sidereal day, but because the Earth has moved in the orbit it needs to go farther through an angle
θ in order to complete a solar day. By the time the Earth has gone all of the way around the sun
the total angle θ will be 2π rad, which means that there was one more sidereal day than solar day.
(b) There are (365.25 d)(24.000 h/d) = 8.7660×103 hours in a year with 265.25 solar days. But
there are 366.25 sidereal days, so each one has a length of 8.7660×103 /366.25 = 23.934 hours, or 23
hours and 56 minutes and 4 seconds.
P8-4 (a) The period is time per complete rotation, so ω = 2π/T .
(b) α = ∆ω/∆t, so
2π
2π
α =
−
/(∆t),
T0 + ∆T
T0
2π
−∆T
=
,
∆t T0 (T0 + ∆T )
2π −∆T
≈
,
∆t T02
2π
−(1.26×10−5 s)
2
= −2.30×10−9 rad/s .
=
7
(3.16×10 s)
(0.033 s)2
2
(c) t = (2π/0.033 s)/(2.30×10−9 rad/s ) = 8.28×1010 s, or 2600 years.
(d) 2π/T0 = 2π/T − αt, or
−1
2
T0 = 1/(0.033 s) − (−2.3×10−9 rad/s )(3.0×1010 s)/(2π)
= 0.024 s.
P8-5 The final angular velocity during the acceleration phase is ωz = αz t = (3.0 rad/s)(4.0
s) = 12.0 rad/s. Since both the acceleration and deceleration phases are uniform with endpoints
ωz = 0, the average angular velocity for both phases is the same, and given by half of the maximum:
ω av,z = 6.0 rad/s.
101
The angle through which the wheel turns is then
φ = ω av,z t = (6.0 rad/s)(4.1 s) = 24.6 rad.
The time is the total for both phases.
(a) The first student sees the wheel rotate through the smallest angle less than one revolution;
this student would have no idea that the disk had rotated more than once. Since the disk moved
through 3.92 revolutions, the first student will either assume the disk moved forward through 0.92
revolutions or backward through 0.08 revolutions.
(b) According to whom? We’ve already answered from the perspective of the second student.
2
2
P8-6 ω = (0.652 rad/s )t and α = (0.652 rad/s ).
2
(a) ω = (0.652 rad/s )(5.60 s) = 3.65 rad/s
(b) v T = ωr = (3.65 rad/s)(10.4 m) = 38 m/s.
2
(c) aT = αr = (0.652 rad/s )(10.4 m) = 6.78 m/s2 .
2
(d) aR = ω r = (3.65 rad/s)2 (10.4 m) = 139 m/s2 .
P8-7 (a) ω = (2π rad)/(3.16×107 s) = 1.99×10−7 rad/s.
(b) v T = ωR = (1.99×10−7 rad/s)(1.50×1011 m) = 2.99×104 m/s.
(c) aR = ω 2 R = (1.99×10−7 rad/s)2 (1.50×1011 m) = 5.94×10−3 m/s2 .
2
P8-8 (a) α = (−156 rev/min)/(2.2 × 60 min) = −1.18 rev/min .
(b) The average angular speed while slowing down is 78 rev/min, so the wheel turns through
(78 rev/min)(2.2 × 60 min) = 10300 revolutions.
2
2
(c) aT = (2π rad/rev)(−1.18 rev/min )(0.524 m) = −3.89 m/min . That’s the same as −1.08×
10−3 m/s2 .
2
(d) aR = (2π rad/rev)(72.5 rev/min)2 (0.524 m) = 1.73 × 104 m/min . That’s the same as
2
4.81m/s . This is so much larger than the aT term that the magnitude of the total linear acceleration is simply 4.81m/s2 .
P8-9 (a) There are 500 teeth (and 500 spaces between these teeth); so disk rotates 2π/500 rad
between the outgoing light pulse and the incoming light pulse. The light traveled 1000 m, so the
elapsed time is t = (1000 m)/(3×108 m/s) = 3.33×10−6 s.
Then the angular speed of the disk is ωz = φ/t = 1.26×10−2 rad)/(3.33×10−6 s) = 3800 rad/s.
(b) The linear speed of a point on the edge of the would be
vT = ωR = (3800 rad/s)(0.05 m) = 190 m/s.
P8-10 The linear acceleration of the belt is a = αA rA . The angular acceleration of C is αC =
a/rC = αA (rA /rC ). The time required for C to get up to speed is
t=
(2π rad/rev)(100 rev/min)(1/60 min/s)
2
(1.60 rad/s )(10.0/25.0)
= 16.4 s.
P8-11 (a) The final angular speed is ω o = (130 cm/s)/(5.80 cm) = 22.4 rad/s.
(b) The recording area is π(Ro 2 − Ri 2 ), the recorded track has a length l and width w, so
l=
π[(5.80 cm)2 − (2.50 cm2 )]
= 5.38×105 cm.
(1.60×10−4 cm)
(c) Playing time is t = (5.38×105 cm)/(130 cm/s) = 4140 s, or 69 minutes.
102
P8-12
The angular position is given by φ = arctan(vt/b). The derivative (Maple!) is
ω=
vb
,
b 2 + v 2 t2
and is directed up. Take the derivative again,
α=
2bv 3 t
,
(b2 + v 2 t2 )2
but is directed down.
P8-13 (a) Let the rocket sled move along the line x = b. The observer is at the origin and
sees the rocket move with a constant angular speed, so the angle made with the x axis increases
according to θ = ωt. The observer, rocket, and starting point form a right triangle; the position y
of the rocket is the opposite side of this triangle, so
tan θ = y/b implies y = b/ tan ωt.
We want to take the derivative of this with respect to time and get
v(t) = ωb/ cos2 (ωt).
(b) The speed becomes infinite (which is clearly unphysical) when t = π/2ω.
103
E9-1
~ = (5.0 N)î.
(a) First, F
~τ
=
[yFz − zFy ]î + [zFx − xFz ]ĵ + [xFy − yFx ]k̂,
=
[y(0) − (0)(0)]î + [(0)Fx − x(0)]ĵ + [x(0) − yFx ]k̂,
=
[−yFx ]k̂ = −(3.0 m)(5.0 N)k̂ = −(15.0 N · m)k̂.
~ = (5.0 N)ĵ. Ignoring all zero terms,
(b) Now F
~τ = [xFy ]k̂ = (2.0 m)(5.0 N)k̂ = (10 N · m)k̂.
~ = (−5.0 N)î.
(c) Finally, F
~τ = [−yFx ]k̂ = −(3.0 m)(−5.0 N)k̂ = (15.0 N · m)k̂.
E9-2 (a) Everything is in the plane of the page, so the net torque will either be directed normal
to the page. Let out be positive, then the net torque is τ = r1 F1 sin θ1 − r2 F2 sin θ2 .
(b) τ = (1.30 m)(4.20 N) sin(75.0◦ ) − (2.15 m)(4.90 N) sin(58.0◦ ) = −3.66N · m.
E9-4 Everything is in the plane of the page, so the net torque will either be directed normal to
the page. Let out be positive, then the net torque is
τ = (8.0 m)(10 N) sin(45◦ ) − (4.0 m)(16 N) sin(90◦ ) + (3.0 m)(19 N) sin(20◦ ) = 12N · m.
E9-5 Since ~r and ~s lie in the xy plane, then ~t = ~r × ~s must be perpendicular to that plane, and
can only point along the z axis.
The angle between ~r and ~s is 320◦ − 85◦ = 235◦ . So |~t| = rs| sin θ| = (4.5)(7.3)| sin(235◦ )| = 27.
Now for the direction of ~t. The smaller rotation to bring ~r into ~s is through a counterclockwise
rotation; the right hand rule would then show that the cross product points along the positive z
direction.
E9-6 ~a = (3.20)[cos(63.0◦ )ĵ + sin(63.0◦ )k̂] and ~b = (1.40)[cos(48.0◦ )î + sin(48.0◦ )k̂]. Then
~a × ~b =
(3.20) cos(63.0◦ )(1.40) sin(48.0◦ )î
+(3.20) sin(63.0◦ )(1.40) cos(48.0◦ )ĵ
−(3.20) cos(63.0◦ )(1.40) cos(48.0◦ )k̂
=
1.51î + 2.67ĵ − 1.36k̂.
E9-7 ~b × ~a has magnitude ab sin φ and points in the negative z direction. It is then perpendicular
to ~a, so ~c has magnitude a2 b sin φ. The direction of ~c is perpendicular to ~a but lies in the plane
containing vectors ~a and ~b. Then it makes an angle π/2 − φ with ~b.
E9-8 (a) In unit vector notation,
~c
=
[(−3)(−3) − (−2)(1)]î + [(1)(4) − (2)(−3)]ĵ + [(2)(−2) − (−3)(4)]k̂,
=
11î + 10ĵ + 8k̂.
(b) Evaluate arcsin[|~a × b|/(ab)], finding magnitudes with the Pythagoras relationship:
φ = arcsin (16.8)/[(3.74)(5.39)] = 56◦ .
104
E9-9
This exercise is a three dimensional generalization of Ex. 9-1, except nothing is zero.
~τ
=
[yFz − zFy ]î + [zFx − xFz ]ĵ + [xFy − yFx ]k̂,
=
[(−2.0 m)(4.3 N) − (1.6 m)(−2.4 N)]î + [(1.6 m)(3.5 N) − (1.5 m)(4.3 N)]ĵ
+[(1.5 m)(−2.4 N) − (−2.0 m)(3.5 N)]k̂,
=
[−4.8 N·m]î + [−0.85 N·m]ĵ + [3.4 N·m]k̂.
~ = (2.6 N)î, then ~τ = (0.85 m)(2.6 N)ĵ − (−0.36 m)(2.6 N)k̂ = 2.2 N · mĵ + 0.94 N · mk̂.
E9-10 (a) F
~
(b) F = (−2.6 N)k̂, then ~τ = (−0.36 m)(−2.6 N)î − (0.54 m)(−2.6 N)ĵ = 0.93 N · mî + 1.4 N · mĵ.
E9-11 (a) The rotational inertia about an axis through the origin is
I = mr2 = (0.025 kg)(0.74 m)2 = 1.4×10−2 kg · m2 .
(b) α = (0.74 m)(22 N) sin(50◦ )/(1.4×10−2 kg · m2 ) = 890 rad/s.
E9-12 (a) I0 = (0.052 kg)(0.27 m)2 + (0.035 kg)(0.45 m)2 + (0.024 kg)(0.65 m)2 = 2.1×10−2 kg · m2 .
(b) The center of mass is located at
xcm =
(0.052 kg)(0.27 m) + (0.035 kg)(0.45 m) + (0.024 kg)(0.65 m)
= 0.41 m.
(0.052 kg) + (0.035 kg) + (0.024 kg)
Applying the parallel axis theorem yields I cm = 2.1×10−2 kg · m2 − (0.11 kg)(0.41 m)2 = 2.5×10−3 kg ·
m2 .
E9-13 (a) Rotational inertia is additive so long as we consider the inertia about the same axis.
We can use Eq. 9-10:
X
I=
mn rn2 = (0.075 kg)(0.42 m)2 + (0.030 kg)(0.65 m)2 = 0.026 kg·m2 .
(b) No change.
E9-14 ~τ = [(0.42 m)(2.5 N) − (0.65 m)(3.6 N)]k̂ = −1.29 N · m k̂. Using the result from E9-13,
2
α
~ = (−1.29 N · m k̂)/(0.026 kg·m2 ) = 50 rad/s k̂. That’s clockwise if viewed from above.
E9-15 (a) F = mω 2 r = (110 kg)(33.5 rad/s)2 (3.90 m) = 4.81×105 N.
2
(b) The angular acceleration is α = (33.5 rad/s)/(6.70 s) = 5.00 rad/s . The rotational inertia
2
3
2
about the axis of rotation is I = (110 kg)(7.80 m) /3 = 2.23×10 kg · m . τ = Iα = (2.23×103 kg ·
2
m2 )(5.00 rad/s ) = 1.12×104 N · m.
E9-16 We can add the inertias for the three rods together,
1
I=3
M L2 = (240 kg)(5.20 m)2 = 6.49×103 kg · m2 .
3
E9-17 The diagonal distance from the axis through the center of mass and the axis through the
p
edge is h = (a/2)2 + (b/2)2 , so
1
1
1
2
2
2
2
2
I = Icm + M h =
M a + b + M (a/2) + (b/2) =
+
M a2 + b2 .
12
12 4
Simplifying, I = 13 M a2 + b2 .
105
E9-18 I = Icm + M h2 = (0.56 kg)(1.0 m)2 /12 + (0.56)(0.30 m)2 = 9.7×10−2 kg · m2 .
E9-19 For particle one I1 = mr2 = mL2 ; for particle two I2 = mr2 = m(2L)2 = 4mL2 . The
rotational inertia of the rod is I rod = 13 (2M )(2L)2 = 83 M L2 . Add the three inertias:
8
I = 5m + M L2 .
3
√
E9-20 (a) I = M R2 /2 = M (R/ 2)2 .
(b) Let I be the rotational inertia. Assuming
that k is the radius of a hoop with an equivalent
p
rotational inertia, then I = M k 2 , or k = I/M .
E9-21 Note the mistakes in the equation in the part (b) of the exercise text.
(a) mn = M/N .
(b) Each piece has a thickness t = L/N , the distance from the end to the nth piece is xn =
(n − 1/2)t = (n − 1/2)L/N . The axis of rotation is the center, so the distance from the center is
rn = xn − L/2 = nL/N − (1 + 1/2N )L.
(c) The rotational inertia is
I
=
N
X
mn rn2 ,
n=1
=
=
=
≈
N
M L2 X
(n − 1/2 − N )2 ,
N 3 n=1
N
M L2 X 2
n − (2N + 1)n + (N + 1/2)2 ,
3
N n=1
M L2 N (N + 1)(2N + 1)
N (N + 1)
2
− (2N + 1)
+ (N + 1/2) N ,
N3
6
2
M L2 2N 3
2N 3
−
+ N3 ,
N3
6
2
= M L2 /3.
E9-22 F = (46 N)(2.6 cm)/(13 cm) = 9.2 N.
E9-23 Tower topples when center of gravity is no longer above base. Assuming center of gravity
is located at the very center of the tower, then when the tower leans 7.0 m then the tower falls. This
is 2.5 m farther than the present.
(b) θ = arcsin(7.0 m/55 m) = 7.3◦ .
E9-24 If the torque from the force is sufficient to lift edge the cube then the cube will tip. The
net torque about the edge which stays in contact with the ground will be τ = F d − mgd/2 if F is
sufficiently large. Then F ≥ mg/2 is the minimum force which will cause the cube to tip.
The minimum force to get the cube to slide is F ≥ µs mg = (0.46)mg. The cube will slide first.
106
E9-25 The ladder slips if the force of static friction required to keep the ladder up exceeds µs N .
Equations 9-31 give us the normal force in terms of the masses of the ladder and the firefighter,
N = (m + M )g, and is independent of the location of the firefighter on the ladder. Also from Eq.
9-31 is the relationship between the force from the wall and the force of friction; the condition at
which slipping occurs is Fw ≥ µs (m + M )g.
Now go straight to Eq. 9-32. The a/2 in the second term is the location of the firefighter, who in
the example was halfway between the base of the ladder and the top of the ladder. In the exercise
we don’t know where the firefighter is, so we’ll replace a/2 with x. Then
−Fw h + M gx +
mga
=0
3
is an expression for rotational equilibrium. Substitute in the condition of Fw when slipping just
starts, and we get
mga
= 0.
− (µs (m + M )g) h + M gx +
3
Solve this for x,
m
ma
45 kg
(45 kg)(7.6 m)
x = µs
+1 h−
= (0.54)
+ 1 (9.3 m) −
= 6.6 m
M
3M
72 kg
3(72 kg)
This is the horizontal distance; the fraction of the total length along the ladder is then given by
x/a = (6.6 m)/(7.6 m) = 0.87. The firefighter can climb (0.87)(12 m) = 10.4 m up the ladder.
E9-26 (a) The net torque about the rear axle is (1360 kg)(9.8 m/s2 )(3.05 m−1.78 m)−F f (3.05 m) =
0, which has solution F f = 5.55×103 N. Each of the front tires support half of this, or 2.77×103 N.
(b) The net torque about the front axle is (1360 kg)(9.8 m/s2 )(1.78 m) − F f (3.05 m) = 0, which
has solution F f = 7.78×103 N. Each of the front tires support half of this, or 3.89×103 N.
E9-27 The net torque on the bridge about the end closest to the person is
(160 lb)L/4 + (600 lb)L/2 − F f L = 0,
which has a solution for the supporting force on the far end of F f = 340 lb.
The net force on the bridge is (160 lb)L/4 + (600 lb)L/2 − (340 lb) − F c = 0, so the force on the
close end of the bridge is F c = 420 lb.
E9-28 The net torque on the board about the left end is
F r (1.55 m) − (142 N)(2.24 m) − (582 N)(4.48 m) = 0,
which has a solution for the supporting force for the right pedestal of F r = 1890 N. The force on the
board from the pedestal is up, so the force on the pedestal from the board is down (compression).
The net force on the board is F l +(1890 N)−(142 N)−(582 N) = 0, so the force from the pedestal
on the left is F l = −1170 N. The negative sign means up, so the pedestal is under tension.
~ and the force of gravity W
~ act on the center of the
E9-29 We can assume that both the force F
wheel. Then the wheel will just start to lift when
~ × ~r + F
~ × ~r = 0,
W
or
W sin θ = F cos θ,
107
where θ is the angle between the vertical (pointing down) and the line between the center of the
wheel and the point of contact with the step. The use of the sine on the left is a straightforward
application of Eq. 9-2. Why the cosine on the right? Because
sin(90◦ − θ) = cos θ.
Then F = W tan θ. We can express the angle θ in terms of trig functions, h, and r. r cos θ is the
vertical distance from the center of the wheel to the top of the step, or r − h. Then
s
2
h
h
cos θ = 1 − and sin θ = 1 − 1 −
.
r
r
Finally by combining the above we get
F =W
q
2h
r
−
1−
h
r
h2
r2
√
=W
2hr − h2
.
r−h
E9-30 (a) Assume that each of the two support points for the square sign experience the same
tension, equal to half of the weight of the sign. The net torque on the rod about an axis through
the hinge is
(52.3 kg/2)(9.81 m/s2 )(0.95 m) + (52.3 kg/2)(9.81 m/s2 )(2.88 m) − (2.88 m)T sin θ = 0,
where T is the tension in the cable and θ is the angle between the cable and the rod. The angle can
be found from θ = arctan(4.12 m/2.88 m) = 55.0◦ , so T = 416 N.
(b) There are two components to the tension, one which is vertical, (416 N) sin(55.0◦ ) = 341 N,
and another which is horizontal, (416 N) cos(55.0◦ ) = 239 N. The horizontal force exerted by the wall
must then be 239 N. The net vertical force on the rod is F + (341 N) − (52.3 kg/2)(9.81 m/s2 ) = 0,
which has solution F = 172 N as the vertical upward force of the wall on the rod.
E9-31 (a) The net torque on the rod about an axis through the hinge is
τ = W (L/2) cos(54.0◦ ) − T L sin(153.0◦ ) = 0.
or T = (52.7 lb/2)(sin 54.0◦ / sin 153.0◦ ) = 47.0 lb.
(b) The vertical upward force of the wire on the rod is Ty = T cos(27.0◦ ). The vertical upward
force of the wall on the rod is Py = W − T cos(27.0◦ ), where W is the weight of the rod. Then
Py = (52.7 lb) − (47.0 lb) cos(27.0◦ ) = 10.8 lb
The horizontal force from the wall is balanced by the horizontal force from the wire. Then Px =
(47.0 lb) sin(27.0◦ ) = 21.3 lb.
E9-32 If the ladder is not slipping then the torque about an axis through the point of contact with
the ground is
τ = (W L/2) cos θ − N h/ sin θ = 0,
where N is the normal force of the edge on the ladder. Then N = W L cos θ sin θ /(2h).
N has two components; one which is vertically up, Ny = N cos θ, and another which is horizontal,
Nx = N sin θ. The horizontal force must be offset by the static friction.
The normal force on the ladder from the ground is given by
N g = W − N cos θ = W [1 − L cos2 θ sin θ /(2h)].
108
The force of static friction can be as large as f = µs N g , so
µs =
W L cos θ sin2 θ /(2h)
L cos θ sin2 θ
=
.
W [1 − L cos2 θ sin θ /(2h)]
2h − L cos2 θ sin θ
Put in the numbers and θ = 68.0◦ . Then µs = 0.407.
E9-33 Let out be positive. The net torque about the axis is then
τ = (0.118 m)(5.88 N) − (0.118 m)(4.13 m) − (0.0493 m)(2.12 N) = 0.102 N · m.
The rotational inertia of the disk is I = (1.92 kg)(0.118 m)2 /2 = 1.34 × 10−2 kg · m2 . Then α =
2
(0.102 N · m)/(1.34×10−2 kg · m2 ) = 7.61 rad/s .
2
E9-34 (a) I = τ /α = (960 N · m)/(6.23 rad/s ) = 154 kg · m2 .
(b) m = (3/2)I/r2 = (1.5)(154 kg · m2 )/(1.88 m)2 = 65.4 kg.
E9-35
(a) The angular acceleration is
α=
∆ω
6.20 rad/s
2
=
= 28.2 rad/s
∆t
0.22 s
2
(b) From Eq. 9-11, τ = Iα = (12.0 kg· m2 )(28.2 rad/s ) = 338 N·m.
2
E9-36 The angular acceleration is α = 2(π/2 rad)/(30 s)2 = 3.5×10−3 rad/s . The required force
is then
2
F = τ /r = Iα/r = (8.7×104 kg · m2 )(3.5×10−3 rad/s )/(2.4 m) = 127 N.
Don’t let the door slam...
E9-37 The torque is τ = rF , the angular acceleration is α = τ /I = rF/I. The angular velocity is
ω=
Z
0
t
α dt =
rAt2
rBt3
+
,
2I
3I
so when t = 3.60 s,
ω=
(9.88×10−2 m)(0.496 N/s)(3.60 s)2
(9.88×10−2 m)(0.305 N/s2 )(3.60 s)3
+
= 690 rad/s.
2(1.14×10−3 kg · m2 )
3(1.14×10−3 kg · m2 )
E9-38 (a) α = 2θ/t2 .
(b) a = αR = 2θR/t2 .
(c) T1 and T2 are not equal. Instead, (T1 − T2 )R = Iα. For the hanging block M g − T1 = M a.
Then
T1 = M g − 2M Rθ/t2 ,
and
T2 = M g − 2M Rθ/t2 − 2(I/R)θ/t2 .
109
E9-39
Apply a kinematic equation from chapter 2 to find the acceleration:
y
ay
1
= v0y t + ay t2 ,
2
2y
2(0.765 m)
2
=
=
= 0.0586 m/s
t2
(5.11 s)2
Closely follow the approach in Sample Problem 9-10. For the heavier block, m1 = 0.512 kg, and
Newton’s second law gives
m1 g − T1 = m1 ay ,
where ay is positive and down. For the lighter block, m2 = 0.463 kg, and Newton’s second law gives
T2 − m2 g = m2 ay ,
where ay is positive and up. We do know that T1 > T2 ; the net force on the pulley creates a torque
which results in the pulley rotating toward the heavier mass. That net force is T1 − T2 ; so the
rotational form of Newton’s second law gives
(T1 − T2 ) R = Iαz = IaT /R,
where R = 0.049 m is the radius of the pulley and aT is the tangential acceleration. But this
acceleration is equal to ay , because everything— both blocks and the pulley— are moving together.
We then have three equations and three unknowns. We’ll add the first two together,
m1 g − T1 + T2 − m2 g
T1 − T2
= m1 ay + m2 ay ,
= (g − ay )m1 − (g + ay )m2 ,
and then combine this with the third equation by substituting for T1 − T2 ,
(g − ay )m1 − (g + ay )m2
g
g
− 1 m1 −
+ 1 m2 R2
ay
ay
= Iay /R2 ,
= I.
Now for the numbers:
(9.81 m/s2 )
(9.81 m/s2 )
− 1 (0.512 kg) −
+ 1 (0.463 kg) =
(0.0586 m/s2 )
(0.0586 m/s2 )
(7.23 kg)(0.049 m)2
7.23 kg,
= 0.0174 kg·m2 .
E9-40 The wheel turns with an initial angular speed of ω0 = 88.0 rad/s. The average speed while
decelerating is ω av = ω0 /2. The wheel stops turning in a time t = φ/ω av = 2φ/ω0 . The deceleration
is then α = −ω0 /t = −ω02 /(2φ).
The rotational inertia is I = M R2 /2, so the torque required to stop the disk is τ = Iα =
−M R2 ω02 /(4φ). The force of friction on the disk is f = µN , so τ = Rf . Then
µ=
M Rω02
(1.40 kg)(0.23 m)(88.0 rad/s)2
=
= 0.272.
4N φ
4(130 N)(17.6 rad)
E9-41 (a) The automobile has an initial speed of v0 = 21.8 m/s. The angular speed is then
ω0 = (21.8 m/s)/(0.385 m) = 56.6 rad/s.
(b) The average speed while decelerating is ω av = ω0 /2. The wheel stops turning in a time
t = φ/ω av = 2φ/ω0 . The deceleration is then
α = −ω0 /t = −ω02 /(2φ) = −(56.6 rad/s)2 /[2(180 rad)] = −8.90 rad/s.
(c) The automobile traveled x = φr = (180 rad)(0.385 m) = 69.3 m.
110
E9-42 (a) The angular acceleration is derived in Sample Problem 9-13,
2
α=
g
1
(981 cm/s )
1
2
=
= 39.1 rad/s .
2
2
R0 1 + I/(M R0 )
(0.320 cm) 1 + (0.950 kg · cm )/[(0.120 kg)(0.320 cm)2 ]
2
2
The acceleration is a = αR0 = (39.1 rad/s
q )(0.320 cm) = 12.5 cm/s .
p
2
(b) Starting from rest, t = 2x/a = 2(134 cm)/(12.5 cm/s ) = 4.63 s.
2
(c) ω = αt = (39.1 rad/s )(4.63 s) = 181 rad/s. This is the same as 28.8 rev/s.
(d) The yo-yo accelerates toward the ground according to y = at2 + v0 t, where down is positive.
The time required to move to the end of the string is found from
p
p
−(1.30 m/s) + (1.30 m/s)2 + 4(0.125 m/s2 )(1.34 m)
−v0 + v02 + 4ay
t=
=
= 0.945 s
2a
2(0.125 m/s2 )
The initial rotational speed was ω0 = (1.30 m/s)/(3.2×10−3 m) = 406 rad/s. Then
2
ω = ω0 + αt = (406 rad/s) + (39.1 rad/s )(0.945 s) = 443 rad/s,
which is the same as 70.5 rev/s.
E9-43 (a) Assuming a perfect hinge at B, the only two vertical forces on the tire will be the
normal force from the belt and the force of gravity. Then N = W = mg, or N = (15.0 kg)(9.8
m/s2 ) = 147 N.
While the tire skids we have kinetic friction, so f = µk N = (0.600)(147 N) = 88.2 N. The force
of gravity and and the pull from the holding rod AB both act at the axis of rotation, so can’t
contribute to the net torque. The normal force acts at a point which is parallel to the displacement
from the axis of rotation, so it doesn’t contribute to the torque either (because the cross product
would vanish); so the only contribution to the torque is from the frictional force.
The frictional force is perpendicular to the radial vector, so the magnitude of the torque is just
τ = rf = (0.300 m)(88.2 N) = 26.5 N·m. This means the angular acceleration will be α = τ /I =
(26.5 N·m)/(0.750 kg·m2 ) = 35.3 rad/s2 .
When ωR = vT = 12.0 m/s the tire is no longer slipping. We solve for ω and get ω = 40 rad/s.
Now we solve ω = ω0 + αt for the time. The wheel started from rest, so t = 1.13 s.
(b) The length of the skid is x = vt = (12.0 m/s)(1.13 s) = 13.6 m long.
P9-1 The problem of sliding down the ramp has been solved (see Sample Problem 5-8); the
critical angle θs is given by tan θs = µs .
The problem of tipping is actually not that much harder: an object tips when the center of
gravity is no longer over the base. The important angle for tipping is shown in the figure below; we
can find that by trigonometry to be
tan θt =
O
(0.56 m)
=
= 0.67,
A
(0.56 m) + (0.28 m)
so θt = 34◦ .
111
f
N
W
θ
(a) If µs = 0.60 then θs = 31◦ and the crate slides.
(b) If µs = 0.70 then θs = 35◦ and the crate tips before sliding; it tips at 34◦ .
P9-2 (a) The total force up on the chain needs to be equal to the total force down; the force
down is W . Assuming the tension at the end points is T then T sin θ is the upward component, so
T = W/(2 sin θ).
(b) There is a horizontal component to the tension T cos θ at the wall; this must be the tension
at the horizontal point at the bottom of the cable. Then T bottom = W/(2 tan θ).
P9-3 (a) The rope exerts a force on the sphere which has horizontal T sin θ and vertical T cos θ
components, where θ = arctan(r/L). The√weight of the sphere p
is balanced by the upward force from
the rope, so T cos θ = W . But cos θ = L/ r2 + L2 , so T = W 1 + r2 /L2 .
(b) The wall pushes outward against the sphere equal to the inward push on the sphere from the
rope, or P = T sin θ = W tan θ = W r/L.
P9-4 Treat the problem as having two forces: the man at one end lifting with force F = W/3
and the two men acting together a distance x away from the first man and lifting with a force
2F = 2W/3. Then the torque about an axis through the end of the beam where the first man is
lifting is τ = 2xW/3 − W L/2, where L is the length of the beam. This expression equal zero when
x = 3L/4.
P9-5 (a) We can solve this problem with Eq. 9-32 after a few modifications. We’ll assume the
center of mass of the ladder is at the center, then the third term of Eq. 9-32 is mga/2. The cleaner
didn’t climb half-way, he climbed 3.10/5.12 = 60.5% of the way, sop
the second term of Eq. 9-32
becomes M ga(0.605). h, L, and a are related by L2 = a2 + h2 , so h = (5.12 m)2 − (2.45 m)2 = 4.5
m. Then, putting the correction into Eq. 9-32,
Fw
1h
mga i
M ga(0.605) +
,
h
2
h
1
2
=
(74.6 kg)(9.81 m/s )(2.45 m)(0.605) ,
(4.5 m)
i
2
+ (10.3 kg)(9.81 m/s )(2.45 m)/2 ,
=
=
269 N
(b) The vertical component of the force of the ground on the ground is the sum of the weight of
the window cleaner and the weight of the ladder, or 833 N.
112
The horizontal component is equal in magnitude to the force of the ladder on the window. Then
the net force of the ground on the ladder has magnitude
p
(269 N)2 + (833 N)2 = 875 N
and direction
θ = arctan(833/269) = 72◦ above the horizontal.
P9-6 (a) There are no upward forces other than the normal force on the bottom ball, so the force
exerted on the bottom ball by the container is 2W .
(c) The bottom ball must exert a force on the top ball which has a vertical component equal to the
weight of the top ball. Then W = N sin θ or the force of contact between the balls is N = W/ sin θ.
(b) The force of contact between the balls has a horizontal component P = N cos θ = W/ tan θ,
this must also be the force of the walls on the balls.
~ the normal force from the lower plane N
~ 1,
P9-7 (a) There are three forces on the ball: weight W,
~ 2 . The force from the lower plane has components
and the normal force from the upper plane N
N1,x = −N1 sin θ1 and N1,y = N1 cos θ1 . The force from the upper plane has components N2,x =
N2 sin θ2 and N2,y = −N2 cos θ2 . Then N1 sin θ1 = N2 sin θ2 and N1 cos θ1 = W + N2 cos θ2 .
Solving for N2 by dividing one expression by the other,
W
cos θ2
cos θ1
=
+
,
sin θ1
N2 sin θ2
sin θ2
or
N2
=
=
=
−1
W
cos θ2
cos θ1
−
,
sin θ2 sin θ1
sin θ2
W cos θ1 sin θ2 − cos θ2 sin θ1
,
sin θ2
sin θ1 sin θ2
W sin θ1
.
sin(θ2 − θ1 )
Then solve for N1 ,
N1 =
W sin θ2
.
sin(θ2 − θ1 )
(b) Friction changes everything.
P9-8
(a) The net torque about a line through A is
τ = W x − T L sin θ = 0,
so T = W x/(L sin θ).
(b) The horizontal force on the pin is equal in magnitude to the horizontal component of the
tension: T cos θ = W x/(L tan θ). The vertical component balances the weight: W − W x/L.
(c) x = (520 N)(2.75 m) sin(32.0◦ )/(315 N) = 2.41 m.
P9-9 (a) As long as the center of gravity of an object (even if combined) is above the base, then
the object will not tip.
Stack the bricks from the top down. The center of gravity of the top brick is L/2 from the edge
of the top brick. This top brick can be offset nor more than L/2 from the one beneath. The center
of gravity of the top two bricks is located at
xcm = [(L/2) + (L)]/2 = 3L/4.
113
These top two bricks can be offset no more than L/4 from the brick beneath. The center of gravity
of the top three bricks is located at
xcm = [(L/2) + 2(L)]/3 = 5L/6.
These top three bricks can be offset no more than L/6 from the brick beneath. The total offset is
then L/2 + L/4 + L/6 = 11L/12.
(b) Actually, we never need to know the location of the center of gravity; we now realize that
each brick is located with an offset L/(2n) with the brick beneath, where n is the number of the
brick counting from the top. The series is then of the form
(L/2)[(1/1) + (1/2) + (1/3) + (1/4) + · · ·],
a series (harmonic, for those of you who care) which does not converge.
(c) The center of gravity would be half way between the ends of the two extreme bricks. This
would be at N L/n; the pile will topple when this value exceeds L, or when N = n.
R
R
P9-10 (a)
a planar object
which lies in the x − y plane, Ix = x2 dm and Iy = y 2 dm. Then
R For
R
Ix + Iy = (x2 + y 2 )dm = r2 dm. But this is the rotational inertia about the z axis, sent r is the
distance from the z axis.
(b) Since the rotational inertia about one diameter (Ix ) should be the same as the rotational
inertia about any other (Iy ) then Ix = Iy and Ix = Iz /2 = M R2 /4.
P9-11 Problem 9-10 says that Ix + Iy = Iz for any thin, flat object which lies only in the
x − y plane.It doesn’t matter in which direction the x and y axes are chosen, so long as they are
perpendicular. We can then orient our square as in either of the pictures below:
y
y
x
x
By symmetry Ix = Iy in either picture. Consequently, Ix = Iy = Iz /2 for either picture. It is
the same square, so Iz is the same for both picture. Then Ix is also the same for both orientations.
P9-12 Let M0 be the mass of the plate before the holes are cut out. Then M1 = M0 (a/L)2 is the
mass of the part cut out of each hole and M = M0 − 9M1 is the mass of the plate. The rotational
inertia (about an axis perpendicular to the plane through the center of the square) for the large
uncut square is M0 L2 /6 and for each smaller cut out is M1 a2 /6.
From the large uncut square’s inertia we need to remove M1 a2 /6 √
for the center cut-out, M1 a2 /6+
2
2
M1 (L/3) for each of the four edge cut-outs, and M1 a /6 + M1 ( 2L/3)2 for each of the corner
sections.
114
Then
I
=
=
M0 L2
M1 a2
M1 L2
2M1 L2
−9
−4
−4
,
6
6
9
9
M0 L2
M0 a4
M0 a2
−3
−
4
.
6
2L2
3
R
P9-13 (a) From Eq. 9-15, I = r2 dm about some axis of rotation when r is measured from that
p
axis. If we consider the x axis as our axis of rotation, then r = y 2 + z 2 , since the distance to the
x axis depends only on the y and z coordinates. We have similar equations for the y and z axes, so
Z
Ix =
y 2 + z 2 dm,
Z
Iy =
x2 + z 2 dm,
Z
Iz =
x2 + y 2 dm.
These three equations can be added together to give
Z
Ix + Iy + Iz = 2
x2 + y 2 + z 2 dm,
so if we now define r to be measured from the origin (which is not the definition used above), then
we end up with the answer in the text.
(b) The right hand side of the equation is integrated over the entire body, regardless of how
the axes are defined. So the integral should be the same, no matter how the coordinate system is
rotated.
R
P9-14 R(a) Since the shell is spherically symmetric Ix = Iy = Iz , so Ix = (2/3) r2 dm =
(2R2 /3) dm = 2M R2 /3.
(b) Since the solid ball is spherically symmetric Ix = Iy = Iz , so
Ix =
P9-15
2
3
Z
r2
3M r2 dr
2
= M R2 .
R3
5
(a) A simple ratio will suffice:
dm
M
2M r
=
or dm =
dr.
2
2πr dr
πR
R2
(b) dI = r2 dm = (2M r3 /R2 )dr.
RR
(c) I = 0 (2M r3 /R2 )dr = M R2 /2.
P9-16
(a) Another simple ratio will suffice:
dm
M
3M (R2 − z 2 )
=
or dm =
dz.
2
3
πr dz
(4/3)πR
4R3
(b) dI = r2 dm/2 = [3M (R2 − z 2 )2 /8R3 ]dz.
115
(c) There are a few steps to do here:
I
R
3M (R2 − z 2 )2
dz,
8R3
−R
Z
3M R 4
(R − 2R2 z 2 + z 4 )dz,
4R3 0
3M 5
2
(R − 2R5 /3 + R5 /5) = M R2 .
3
4R
5
Z
=
=
=
P
P9-17 The rotational acceleration will be given by αz = τ /I.
The torque about the pivot comes from the force of gravity on eachP
block. This forces will both
originally be at right angles to the pivot arm, so the net torque will be
τ = mgL2 − mgL1 , where
clockwise as seen on the page is positive.
P
The rotational inertia about the pivot is given by I =
mn rn2 = m(L22 + L21 ). So we can now
find the rotational acceleration,
P
τ
mgL2 − mgL1
L2 − L1
2
α=
=
=g 2
= 8.66 rad/s .
2
2
I
m(L2 + L1 )
L2 + L21
The linear acceleration is the tangential acceleration, aT = αR. For the left block, aT = 1.73 m/s2 ;
for the right block aT = 6.93 m/s2 .
P9-18 (a) The force of friction on the hub is µk M g. The torque is τ = µk M ga. The angular
acceleration has magnitude α = τ /I = µk ga/k 2 . The time it takes to stop the wheel will be
t = ω0 /α = ω0 k 2 /(µk ga).
(b) The average rotational speed while slowing is ω0 /2. The angle through which it turns while
slowing is ω0 t/2 radians, or ω0 t/(4π) = ω02 k 2 /(4πµk ga)
P9-19
(a) Consider a differential ring of the disk. The torque on the ring because of friction is
µk M g
2µk M gr2
2πr
dr
=
dr.
πR2
R2
dτ = r dF = r
The net torque is then
τ=
Z
dτ =
Z
0
R
2µk M gr2
2
dr = µk M gR.
R2
3
(b) The rotational acceleration has magnitude α = τ /I = 34 µk g/R. Then it will take a time
t = ω0 /α =
3Rω0
4µk g
to stop.
P9-20 We need only show that the two objects have the same acceleration.
Consider first the hoop. There is a force W|| = W sin θ = mg sin θ pulling it down the ramp
and a frictional force f pulling it up the ramp. The frictional force is just large enough to cause
a torque that will allow the hoop to roll without slipping. This means a = αR; consequently,
f R = αI = aI/R. In this case I = mR2 .
The acceleration down the plane is
ma = mg sin θ − f = mg sin θ − maI/R2 = mg sin θ − ma.
116
Then a = g sin θ /2. The mass and radius are irrelevant!
For a block sliding with friction there are also two forces: W|| = W sin θ = mg sin θ and f =
µk mg cos θ. Then the acceleration down the plane will be given by
a = g sin θ − µk g cos θ,
which will be equal to that of the hoop if
µk =
sin θ − sin θ/2
1
= tan θ.
cos θ
2
P9-21 This problem is equivalent to Sample Problem 9-11, except that we have a sphere instead
of a cylinder. We’ll have the same two equations for Newton’s second law,
M g sin θ − f = M acm and N − M g cos θ = 0.
Newton’s second law for rotation will look like
−f R = I cm α.
The conditions for accelerating without slipping are acm = αR, rearrange the rotational equation to
get
I cm α
I cm (−acm )
f =−
=−
,
R
R2
and then
I cm (acm )
M g sin θ −
= M acm ,
R2
and solve for acm . For fun, let’s write the rotational inertia as I = βM R2 , where β = 2/5 for the
sphere. Then, upon some mild rearranging, we get
acm = g
sin θ
1+β
For the sphere, acm = 5/7g sin θ.
(a) If acm = 0.133g, then sin θ = 7/5(0.133) = 0.186, and θ = 10.7◦ .
(b) A frictionless block has no rotational properties; in this case β = 0! Then acm = g sin θ =
0.186g.
P9-22 (a) There are three forces on the cylinder: gravity W and the tension from each cable T .
The downward acceleration of the cylinder is then given by ma = W − 2T .
The ropes unwind according to α = a/R, but α = τ /I and I = mR2 /2. Then
a = τ R/I = (2T R)R/(mR2 /2) = 4T /m.
Combining the above, 4T = W − 2T , or T = W/6.
(b) a = 4(mg/6)/m = 2g/3.
P9-23
The force of friction required to keep the cylinder rolling is given by
f=
1
M g sin θ;
3
the normal force is given to be N = M g cos θ; so the coefficient of static friction is given by
µs ≥
f
1
= tan θ.
N
3
117
P9-24 a = F/M , since F is the net force on the disk. The torque about the center of mass is F R,
so the disk has an angular acceleration of
α=
FR
FR
2F
=
=
.
I
M R2 /2
MR
P9-25 This problem is equivalent to Sample Problem 9-11, except that we have an unknown rolling
object. We’ll have the same two equations for Newton’s second law,
M g sin θ − f = M acm and N − M g cos θ = 0.
Newton’s second law for rotation will look like
−f R = I cm α.
The conditions for accelerating without slipping are acm = αR, rearrange the rotational equation to
get
I cm α
I cm (−acm )
f =−
=−
,
R
R2
and then
I cm (acm )
= M acm ,
M g sin θ −
R2
and solve for acm . Write the rotational inertia as I = βM R2 , where β = 2/5 for a sphere, β = 1/2
for a cylinder, and β = 1 for a hoop. Then, upon some mild rearranging, we get
acm = g
sin θ
1+β
Note that a is largest when β is smallest; consequently the cylinder wins. Neither M nor R entered
into the final equation.
118
E10-1 l = rp = mvr = (13.7×10−3 kg)(380 m/s)(0.12 m) = 0.62 kg · m2 /s.
E10-2 (a) ~l = m~r × ~v, or
~l = m(yvz − zvy )î + m(zvx − xvz )ĵ + m(xvy − yvx )k̂.
(b) If ~v and ~r exist only in the xy plane then z = vz = 0, so only the uk term survives.
E10-3
If the angular momentum ~l is constant in time, then d~l/dt = 0. Trying this on Eq. 10-1,
d~l
dt
d
(~r × ~p) ,
dt
d
(~r × m~v) ,
=
dt
d~r
d~v
= m × ~v + m~r ×
,
dt
dt
= m~v × ~v + m~r × ~a.
=
Now the cross product of a vector with itself is zero, so the first term vanishes. But in the exercise
we are told the particle has constant velocity, so ~a = 0, and consequently the second term vanishes.
Hence, ~l is constant for a single particle if ~v is constant.
P
E10-4 (a) L =
li ; li = ri mi vi . Putting the numbers in for each planet and then summing (I
won’t bore you with the arithmetic details) yields L = 3.15×1043 kg · m2 /s.
(b) Jupiter has l = 1.94×1043 kg · m2 /s, which is 61.6% of the total.
E10-5 l = mvr = m(2πr/T )r = 2π(84.3 kg)(6.37×106 m)2 /(86400 s) = 2.49×1011 kg · m2 /s.
E10-6 (a) Substitute and expand:
X
~ =
L
(~rcm + ~r0i ) × (mi ~vcm + ~p0i ),
X
=
(mi~rcm × ~vcm + ~rcm × ~p0i + mi~r0i × ~vcm + ~r0i × ~p0i ),
X
X
X
~p0i ) + (
~r0i × ~p0i .
= M~rcm × ~vcm + ~rcm × (
mi~r0i ) × ~vcm +
P 0
P
~pi = 0 and
(b) But
mi~r0i = 0, because these two quantities are in the center of momentum
and center of mass. Then
X
~ = M~rcm × ~vcm +
~ 0 + M~rcm × ~vcm .
~r0i × ~p0i = L
L
E10-7 (a) Substitute and expand:
~p0i = mi
d~r0i
d~ri
d~rcm
= mi
− mi
= ~pi − mi ~vcm .
dt
dt
dt
(b) Substitute and expand:
X d~r0
X
X
~0
dL
d~p0i
d~p0i
i
~r0i ×
~r0i ×
=
× ~p0i +
=
.
dt
dt
dt
dt
The first term vanished because ~vi0 is parallel to ~p0i .
119
(c) Substitute and expand:
~0
dL
dt
d(~pi − mi ~vcm )
,
dt
X
~r0i × (mi~ai − mi~ac m),
=
X
X
~r0i × mi~ai +
=
mi~r0i × ~ac m)
X
=
~r0i ×
The second term vanishes because of the definition of the center of mass. Then
X
~0
dL
~ i,
~r0i × F
=
dt
~ i is the net force on the ith particle. The force F
~ i may include both internal and external
where F
components. If there is an internal component, say between the ith and jth particles, then the
torques from these two third law components will cancel out. Consequently,
X
~0
dL
=
~τi = ~τ ext .
dt
E10-8 (a) Integrate.
Z
~τ dt =
Z
~
dL
dt =
dt
Z
~ = ∆L.
~
dL
(b) If I is fixed, ∆L = I∆ω. Not only that,
Z
Z
Z
τ dt = F r dt = r F dt = rF av ∆t,
where we use the definition of average that depends on time.
E10-9 (a) ~τ ∆t = ∆~l. The disk starts from rest, so ∆~l = ~l − ~l0 = ~l. We need only concern
ourselves with the magnitudes, so
l = ∆l = τ ∆t = (15.8 N·m)(0.033 s) = 0.521 kg·m2 /s.
(b) ω = l/I = (0.521 kg·m2 /s)/(1.22×10−3 kg·m2 ) = 427 rad/s.
E10-10 (a) Let v0 be the initial speed; the average speed while slowing to a stop is v0 /2; the time
required to stop is t = 2x/v0 ; the acceleration is a = −v0 /t = −v02 /(2x). Then
a = −(43.3 m/s)2 /[2(225 m/s)] = −4.17 m/s2 .
2
(b) α = a/r = (−4.17 m/s2 )/(0.247 m) = −16.9 rad/s .
2
(c) τ = Iα = (0.155 kg · m2 )(−16.9 rad/s ) = −2.62 N · m.
E10-11 Let ~ri = ~z + ~r0i . From the figure, ~p1 = −~p2 and ~r01 = −~r02 . Then
~ =
L
=
=
=
~l1 + ~l2 = ~r1 × ~p1 + ~r2 × ~p2 ,
(~r1 − ~r2 ) × ~p1 ,
(~r01 − ~r02 ) × ~p1 ,
2~r01 × ~p1 .
~ must be along the z axis.
Since ~r01 and ~p1 both lie in the xy plane then L
120
E10-12 Expand:
X
X
~li =
~ri × ~pi ,
X
X
mi~ri × (~
ω × ~ri )
=
mi~ri × ~vi =
X
=
mi [(~ri · ~ri )~
ω − (~ri · ω
~ )~r)i],
X
2
2
=
mi [ri ω
~ − (zi ω)k̂ − (zi xi ω)î − (zi yi ω)ĵ],
~ =
L
but if the body is symmetric about the z axis then the last two terms vanish, leaving
X
X
~ =
L
mi [ri2 ω
~ − (zi2 ω)k̂] =
mi (x2i + yi2 )~
ω = I~
ω.
E10-13 An impulse of 12.8 N·s will change the linear momentum by 12.8 N·s; the stick starts
from rest, so the final momentum must be 12.8 N·s. Since p = mv, we then can find v = p/m = (12.8
N·s)/(4.42 kg) = 2.90 m/s.
R
~
Impulse is a vector, given by Fdt.
We can take the cross product of the impulse with the
displacement vector ~r (measured from the axis of rotation to the point where the force is applied)
and get
Z
Z
~
~
~r × Fdt ≈ ~r × Fdt,
The two sides of the above expression are only equal if ~r has a constant magnitude and direction. This
won’t be true, but if the force is of sufficiently short duration then it hopefully won’t change much.
The right hand side is an integral over a torque, and will equal the change in angular momentum of
the stick.
~ = rF , and the “torque
The exercise states that the force is perpendicular to the stick, then |~r × F|
impulse” is then (0.464 m)(12.8 N·s) = 5.94 kg·m/s. This “torque impulse” is equal to the change
in the angular momentum, but the stick started from rest, so the final angular momentum of the
stick is 5.94 kg·m/s.
But how fast is it rotating? We can use Fig. 9-15 to find the rotational inertia about the center
1
1
M L2 = 12
(4.42 kg)(1.23 m)2 = 0.557 kg·m2 . The angular velocity of the stick is
of the stick: I = 12
ω = l/I = (5.94 kg·m/s)/(0.557 kg·m2 ) = 10.7 rad/s.
E10-14 The point of rotation is the point of contact with the plane; the torque about that point
is τ = rmg sin θ. The angular momentum is Iω, so τ = Iα. In this case I = mr2 /2 + mr2 , the
second term from the parallel axis theorem. Then
a = rα = rτ /I = mr2 g sin θ/(3mr2 /2) =
E10-15 From Exercise 8 we can immediately write
I1 (ω1 − ω0 )/r1 = I2 (ω2 − 0)/r2 ,
but we also have r1 ω1 = −r2 ω2 . Then
ω2 = −
r1 r2 I1 ω0
.
r12 I2 − r22 I1
121
2
g sin θ.
3
E10-16 (a) ∆ω/ω = (1/T1 − 1/T2 )/(1/T1 ) = −(T2 − T1 )/T2 = −∆T /T , which in this case is
−(6.0×10−3 s)(8.64×104 s) = −6.9×10−8 .
(b) Assuming conservation of angular momentum, ∆I/I = −∆ω/ω. Then the fractional change
would be 6.9×10−8 .
E10-17
The rotational inertia of a solid sphere is I = 25 M R2 ; so as the sun collapses
~i
L
I iω
~i
2
M Ri 2 ω
~i
5
2
Ri ω
~i
~ f,
= L
= If ω
~ f,
2
=
M Rf 2 ω
~ f,
5
2
= Rf ω
~f.
The angular frequency is inversely proportional to the period of rotation, so
Tf = Ti
Rf 2
= (3.6×104 min)
Ri 2
(6.37×106 m)
(6.96×108 m)
2
= 3.0 min.
E10-18 The final angular velocity of the train with respect to the tracks is ω tt = Rv. The
conservation of angular momentum implies
0 = M R2 ω + mR2 (ω tt + ω),
or
ω=
−mv
.
(m + M )R
E10-19 This is much like a center of mass problem.
0 = I p φp + I m (φmp + φp ),
or
φmp = −
(12.6 kg · m2 )(25◦ )
(I p + I m )φp
≈−
= 1.28×105◦ .
Im
(2.47×10−3 kg · m2 )
That’s 354 rotations!
E10-20 ω f = (I i /I f )ω i = [(6.13 kg · m2 )/(1.97 kg · m2 )](1.22 rev/s) = 3.80 rev/s.
E10-21 We have two disks which are originally not in contact which then come into contact;
there are no external torques. We can write
~l1,i + ~l2,i
I1 ω
~ 1,i + I2 ω
~ 2,i
= ~l1,f + ~l2,f ,
= I1 ω
~ 1,f + I2 ω
~ 2,f .
The final angular velocities of the two disks will be equal, so the above equation can be simplified
and rearranged to yield
ωf =
I1
(1.27 kg · m2 )
ω 1,i =
(824 rev/min) = 171 rev/min
I1 + I2
(1.27 kg · m2 ) + (4.85 kg · m2 )
E10-22 l⊥ = l cos θ = mvr cos θ = mvh.
122
E10-23 (a) ω f = (I1 /I2 )ω i , I1 = (3.66 kg)(0.363 m)2 = 0.482 kg · m2 . Then
ω f = [(0.482 kg · m2 )/(2.88 kg · m2 )](57.7 rad/s) = 9.66 rad/s,
with the same rotational sense as the original wheel.
(b) Same answer, since friction is an internal force internal here.
E10-24 (a) Assume the merry-go-round is a disk. Then conservation of angular momentum yields
1
( mm R2 + mg R2 )ω + (mr R2 )(v/R) = 0,
2
or
ω=−
(1.13 kg)(7.82 m/s)/(3.72 m)
= −5.12×10−3 rad/s.
(827 kg)/2 + (50.6 kg)
(b) v = ωR = (−5.12×10−3 rad/s)(3.72 m) = −1.90×10−2 m/s.
E10-25 Conservation of angular momentum:
(mm k 2 + mg R2 )ω = mg R2 (v/R),
so
(44.3 kg)(2.92 m/s)/(1.22 m)
= 0.496 rad/s.
(176 kg)(0.916 m)2 + (44.3 kg)(1.22 m)2
ω=
E10-26 Use Eq. 10-22:
ωP =
M gr
(0.492 kg)(9.81 m/s2 )(3.88×10−2 m)
=
= 2.04 rad/s = 0.324 rev/s.
Iω
(5.12×10−4 kg · m2 )(2π28.6 rad/s)
E10-27 The relevant precession expression is Eq. 10-22.
The rotational inertia will be a sum of the contributions from both the disk and the axle, but
the radius of the axle is probably very small compared to the disk, probably as small as 0.5 cm.
Since I is proportional to the radius squared, we expect contributions from the axle to be less than
(1/100)2 of the value for the disk. For the disk only we use
I=
1
1
M R2 = (1.14 kg)(0.487 m)2 = 0.135 kg · m2 .
2
2
Now for ω,
ω = 975 rev/min
2π rad
1 rev
1 min
60 s
= 102 rad/s.
Then L = Iω = 13.8 kg·m2 /s.
Back to Eq. 10-22,
2
ωp =
M gr
(1.27 kg)(9.81 m/s )(0.0610 m)
=
= 0.0551 rad/s.
L
13.8 kg · m2 /s
The time for one precession is
t=
1rev
2π rad
=
= 114 s.
ωp
(0.0551 rad/s)
123
P10-1 Positive z is out of the page.
(a) ~l = rmv sin θk̂ = (2.91 m)(2.13 kg)(4.18 m) sin(147◦ )k̂ = 14.1 kg · m2 /sk̂.
(b) ~τ = rF sin θk̂ = (2.91 m)(1.88 N) sin(26◦ )k̂ = 2.40 N · mk̂.
P10-2 Regardless of where the origin is located one can orient the coordinate system so that the
two paths lie in the xy plane and are both parallel to the y axis. The one of the particles travels
along the path x = vt, y = a, z = b; the momentum of this particle is ~p1 = mv î. The other
particle will then travel along a path x = c − vt, y = a + d, z = b; the momentum of this particle is
~p2 = −mv î. The angular momentum of the first particle is
~l1 = mvbĵ − mvak̂,
while that of the second is
~l2 = −mvbĵ + mv(a + d)k̂,
so the total is ~l1 + ~l2 = mvdk̂.
P10-3 Assume that the cue stick strikes the ball horizontally with a force of constant magnitude
F for a time ∆t. Then the magnitude of the change in linear momentum of the ball is given by
F ∆t = ∆p = p, since the initial momentum is zero.
If the force is applied a distance x above the center of the ball, then the magnitude of the torque
about a horizontal axis through the center of the ball is τ = xF . The change in angular momentum
of the ball is given by τ ∆t = ∆l = l, since initially the ball is not rotating.
For the ball to roll without slipping we need v = ωR. We can start with this:
v
p
m
F ∆t
m
F
m
= ωR,
lR
=
,
I
τ ∆tR
=
,
I
xF R
=
.
I
Then x = I/mR is the condition for rolling without sliding from the start. For a solid sphere,
I = 25 mR2 , so x = 25 R.
P10-4 The change in momentum of the block is M (v2 − v1 ), this is equal to the magnitude the
impulse delivered to the cylinder. According to E10-8 we can write M (v2 − v1 )R = Iω f . But in the
end the box isn’t slipping, so ωf = v2 /R. Then
M v2 − M v1 = (I/R2 )v2 ,
or
v2 = v1 /(1 + I/M R2 ).
P10-5 Assume that the cue stick strikes the ball horizontally with a force of constant magnitude
F for a time ∆t. Then the magnitude of the change in linear momentum of the ball is given by
F ∆t = ∆p = p, since the initial momentum is zero. Consequently, F ∆t = mv0 .
If the force is applied a distance h above the center of the ball, then the magnitude of the torque
about a horizontal axis through the center of the ball is τ = hF . The change in angular momentum
of the ball is given by τ ∆t = ∆l = l0 , since initially the ball is not rotating. Consequently, the initial
angular momentum of the ball is l0 = hmv0 = Iω0 .
124
The ball originally slips while moving, but eventually it rolls. When it has begun to roll without
slipping we have v = Rω. Applying the results from E10-8,
m(v − v0 )R + I(ω − ω0 ) = 0,
or
2
v
m(v − v0 )R + mR2 − hmv0 = 0,
5
R
then, if v = 9v0 /7,
h=
P10-6
9
2
9
4
−1 R+ R
= R.
7
5
7
5
(a) Refer to the previous answer. We now want v = ω = 0, so
2
v
m(v − v0 )R + mR2 − hmv0 = 0,
5
R
becomes
−v0 R − hv0 = 0,
or h = −R. That’ll scratch the felt.
(b) Assuming only a horizontal force then
v=
(h + R)v0
,
R(1 + 2/5)
which can only be negative if h < −R, which means hitting below the ball. Can’t happen. If instead
we allow for a downward component, then we can increase the “reverse English” as much as we want
without increasing the initial forward velocity, and as such it would be possible to get the ball to
move backwards.
P10-7 We assume the bowling ball is solid, so the rotational inertia will be I = (2/5)M R2 (see
Figure 9-15).
The normal force on the bowling ball will be N = M g, where M is the mass of the bowling ball.
The kinetic friction on the bowling ball is Ff = µk N = µk M g. The magnitude of the net torque on
the bowling ball while skidding is then τ = µk M gR.
Originally the angular momentum of the ball is zero; the final angular momentum will have
magnitude l = Iω = Iv/R, where v is the final translational speed of the ball.
(a) The time requires for the ball to stop skidding is the time required to change the angular
momentum to l, so
∆l
(2/5)M R2 v/R
2v
∆t =
=
=
.
τ
µk M gR
5µk g
Since we don’t know v, we can’t solve this for ∆t. But the same time through which the angular
momentum of the ball is increasing the linear momentum of the ball is decreasing, so we also have
∆t =
∆p
M v − M v0
v0 − v
=
=
.
−Ff
−µk M g
µk g
Combining,
∆t
=
=
v0 − v
,
µk g
v0 − 5µk g∆t/2
,
µk g
125
2µk g∆t
=
∆t
=
=
2v0 − 5µk g∆t,
2v0
,
7µk g
2(8.50 m/s)
= 1.18 s.
7(0.210)(9.81 m/s2 )
(d) Use the expression for angular momentum and torque,
v = 5µk g∆t/2 = 5(0.210)(9.81 m/s2 )(1.18 s)/2 = 6.08 m/s.
(b) The acceleration of the ball is F/M = −µg. The distance traveled is then given by
1 2
at + v0 t,
2
1
= − (0.210)(9.81 m/s2 )(1.18 s)2 + (8.50 m/s)(1.18 s) = 8.6 m,
2
x =
(c) The angular acceleration is τ /I = 5µk g/(2R). Then
θ
=
=
1 2
αt + ω0 t,
2
5(0.210)(9.81 m/s2 )
(1.18 s)2 = 32.6 rad = 5.19 revolutions.
4(0.11 m)
P10-8 (a) l = Iω0 = (1/2)M R2 ω0 .
(b) The initial speed is v0 = Rω0 . The chip decelerates in a time t = v0 /g, and during this time
the chip travels with an average speed of v0 /2 through a distance of
y = v av t =
v0 v0
R2 ω 2
=
.
2 g
2g
(c) Loosing the chip won’t change the angular velocity of the wheel.
P10-9
Since L = Iω = 2πI/T and L is constant, then I ∝ T . But I ∝ R2 , so R2 ∝ T and
∆T
2R∆R
2∆R
=
=
.
2
T
R
R
Then
∆T = (86400 s)
P10-10
2(30 m)
≈ 0.8 s.
(6.37×106 m)
Originally the rotational inertia was
2
8π
M R2 =
ρ0 R 5 .
5
15
The average density can be found from Appendix C. Now the rotational inertia is
Ii =
8π
8π
(ρ1 − ρ2 )R15 +
ρ2 R5 ,
15
15
where ρ1 is the density of the core, R1 is the radius of the core, and ρ2 is the density of the mantle.
Since the angular momentum is constant we have ∆T /T = ∆I/I. Then
If =
∆T
ρ1 − ρ2 R15
ρ2
10.3 − 4.50 35705
4.50
=
+
−1=
+
− 1 = −0.127,
5
T
ρ0 R
ρ0
5.52
63705
5.52
so the day is getting longer.
126
P10-11 The cockroach initially has an angular speed of ω c,i = −v/r. The rotational inertia of
the cockroach about the axis of the turntable is I c = mR2 . Then conservation of angular momentum
gives
lc,i + ls,i
I c ω c,i + I s ω s,i
−mR2 v/r + Iω
ωf
= lc,f + ls,f ,
= I c ω c,f + I s ω s,f ,
= (mR2 + I)ω f ,
Iω − mvR
=
.
I + mR2
P10-12 (a) The skaters move in a circle of radius R = (2.92 m)/2 = 1.46 m centered midway
between the skaters. The angular velocity of the system will be ω i = v/R = (1.38 m/s)/(1.46 m) =
0.945 rad/s.
(b) Moving closer will decrease the rotational inertia, so
ωf =
2M Ri 2
(1.46 m)2
ωi =
(0.945 rad/s) = 9.12 rad/s.
2
2M Rf
(0.470 m)2
127
E11-1 (a) Apply Eq. 11-2, W = F s cos φ = (190 N)(3.3 m) cos(22◦ ) = 580 J.
(b) The force of gravity is perpendicular to the displacement of the crate, so there is no work
done by the force of gravity.
(c) The normal force is perpendicular to the displacement of the crate, so there is no work done
by the normal force.
E11-2 (a) The force required is F = ma = (106 kg)(1.97 m/s2 ) = 209 N. The object moves with
an average velocity v av = v0 /2 in a time t = v0 /a through a distance x = v av t = v02 /(2a). So
x = (51.3 m/s)2 /[2(1.97 m/s2 )] = 668 m.
The work done is W = F x = (−209 N)(668 m) = 1.40×105 J.
(b) The force required is
F = ma = (106 kg)(4.82 m/s2 ) = 511 N.
x = (51.3 m/s)2 /[2(4.82 m/s2 )] = 273 m. The work done is W = F x = (−511 N)(273 m) = 1.40×105 J.
E11-3 (a) W = F x = (120 N)(3.6 m) = 430 J.
(b) W = F x cos θ = mgx cos θ = (25 kg)(9.8 m/s2 )(3.6 m) cos(117◦ ) = 400 N.
(c) W = F x cos θ, but θ = 90◦ , so W = 0.
~ this force has components Px = P cos θ and Py = P sin θ,
E11-4 The worker pushes with a force P;
◦
where θ = −32.0 . The normal force of the ground on the crate is N = mg − Py , so the force of
friction is f = µk N = µk (mg − Py ). The crate moves at constant speed, so Px = f . Then
P cos θ
P
= µk (mg − P sin θ),
µk mg
=
.
cos θ + µk sin θ
The work done on the crate is
W
µk xmg
,
1 + µk tan θ
(0.21)(31.3 ft)(58.7 lb)
= 444 ft · lb.
1 + (0.21) tan(−32.0◦ )
~ · ~x = P x cos θ =
= P
=
~ has
E11-5 The components of the weight are W|| = mg sin θ and W⊥ = mg cos θ. The push P
components P|| = P cos θ and P⊥ = P sin θ.
The normal force on the trunk is N = W⊥ + P ⊥ so the force of friction is f = µk (mg cos θ +
P sin θ). The push required to move the trunk up at constant speed is then found by noting that
P|| = W|| + f .
Then
mg(tan θ + µk )
P =
.
1 − µk tan θ
(a) The work done by the applied force is
W = P x cos θ =
(52.3 kg)(9.81 m/s2 )[sin(28.0◦ ) + (0.19) cos(28.0◦ )](5.95 m)
= 2160 J.
1 − (0.19) tan(28.0◦ )
(b) The work done by the force of gravity is
W = mgx cos(θ + 90◦ ) = (52.3 kg)(9.81 m/s2 )(5.95 m) cos(118◦ ) = −1430 J.
128
E11-6 θ = arcsin(0.902 m/1.62 m) = 33.8◦ .
The components of the weight are W|| = mg sin θ and W⊥ = mg cos θ.
The normal force on the ice is N = W⊥ so the force of friction is f = µk mg cos θ. The push
required to allow the ice to slide down at constant speed is then found by noting that P = W|| − f .
Then P = mg(sin θ − µk cos θ).
(a) P = (47.2 kg)(9.81 m/s2 )[sin(33.8◦ ) − (0.110) cos(33.8◦ )] = 215 N.
(b) The work done by the applied force is W = P x = (215 N)(−1.62 m) = −348 J.
(c) The work done by the force of gravity is
W = mgx cos(90◦ − θ) = (47.2 kg)(9.81 m/s2 )(1.62 m) cos(56.2◦ ) = 417 J.
E11-7
Equation 11-5 describes how to find the dot product of two vectors from components,
~a · ~b = ax bx + ay by + az bz ,
= (3)(2) + (3)(1) + (3)(3) = 18.
Equation 11-3 can be used to find the angle between the vectors,
p
a =
(3)2 + (3)2 + (3)2 = 5.19,
p
b =
(2)2 + (1)2 + (3)2 = 3.74.
Now use Eq. 11-3,
cos φ =
~a · ~b
(18)
=
= 0.927,
ab
(5.19)(3.74)
and then φ = 22.0◦ .
E11-8 ~a · ~b = (12)(5.8) cos(55◦ ) = 40.
E11-9 ~r · ~s = (4.5)(7.3) cos(320◦ − 85◦ ) = −19.
E11-10 (a) Add the components individually:
~r = (5 + 2 + 4)î + (4 − 2 + 3)ĵ + (−6 − 3 + 2)k̂ = 11î + 5ĵ − 7k̂.
√
112 + 52 + 72 ) = 120◦ .
(b) θ = arccos(−7/ √
~
(c) θ = arccos(~a · b/ ab), or
(5)(−2) + (4)(2) + (−6)(3)
θ=p
= 124◦ .
(52 + 42 + 62 )(22 + 22 + 32 )
E11-11 There are two forces on the woman, the force of gravity directed down and the normal
force of the floor directed up. These will be effectively equal, so N = W = mg. Consequently, the
57 kg woman must exert a force of F = (57kg)(9.8m/s2 ) = 560N to propel herself up the stairs.
From the reference frame of the woman the stairs are moving down, and she is exerting a force
down, so the work done by the woman is given by
W = F s = (560 N)(4.5 m) = 2500 J,
this work is positive because the force is in the same direction as the displacement.
The average power supplied by the woman is given by Eq. 11-7,
P = W/t = (2500 J)/(3.5 s) = 710 W.
129
E11-12 P = W/t = mgy/t = (100 × 667 N)(152 m)/(55.0 s) = 1.84×105 W.
E11-13 P = F v = (110 N)(0.22 m/s) = 24 W.
E11-14 F = P/v, but the units are awkward.
F =
E11-15
25 hp.
(4800 hp) 1 knot
1 ft/s 745.7 W
= 9.0×104 N.
(77 knots) 1.688 ft/s 0.3048 m/s 1 hp
P = F v = (720 N)(26 m/s) = 19000 W; in horsepower, P = 19000 W(1/745.7 hp/W) =
E11-16 Change to metric units! Then P = 4920 W, and the flow rate is Q = 13.9 L/s. The
density of water is approximately 1.00 kg/L , so the mass flow rate is R = 13.9 kg/s.
y=
P
(4920 kg)
=
= 36.1 m,
gR
(9.81 m/s2 )(13.9 kg/s)
which is the same as approximately 120 feet.
E11-17
(a) Start by converting kilowatt-hours to Joules:
1 kW · h = (1000 W)(3600 s) = 3.6×106 J.
The car gets 30 mi/gal, and one gallon of gas produces 140 MJ of energy. The gas required to
produce 3.6×106 J is
1 gal
6
= 0.026 gal.
3.6×10 J
140×106 J
The distance traveled on this much gasoline is
30 mi
0.026 gal
= 0.78 mi.
1 gal
(b) At 55 mi/h, it will take
0.78 mi
1 hr
55 mi
= 0.014 h = 51 s.
The rate of energy expenditure is then (3.6×106 J)/(51 s) = 71000 W.
E11-18 The linear speed is v = 2π(0.207 m)(2.53 rev/s) = 3.29 m/s. The frictional force is f =
µk N = (0.32)(180 N) = 57.6 N. The power developed is P = F v = (57.6 N)(3.29 m) = 190 W.
E11-19 The net force required will be (1380 kg − 1220 kg)(9.81 m/s2 ) = 1570 N. The work is
W = F y, the power output is P = W/t = (1570 N)(54.5 m)/(43.0 s) = 1990 W, or P = 2.67 hp.
E11-20 (a) The momentum change of the ejected material in one second is
∆p = (70.2 kg)(497 m/s − 184 m/s) + (2.92 kg)(497 m/s) = 2.34×104 kg · m/s.
The thrust is then F = ∆p/∆t = 2.34×104 N.
(b) The power is P = F v = (2.34×104 N)(184 m/s) = 4.31×106 W. That’s 5780 hp.
130
E11-21
The acceleration on the object as a function of position is given by
20 m/s2
x,
8m
The work done on the object is given by Eq. 11-14,
Z 8
Z 8
20 m/s2
W =
Fx dx =
(10 kg)
x dx = 800 J.
8m
0
0
a=
E11-22 Work is area between the curve and the line F = 0. Then
1
1
W = (10 N)(2 s) + (10 N)(2 s) + (−5 N)(2 s) = 25 J.
2
2
E11-23
(a) For a spring, F = −kx, and ∆F = −k∆x.
k=−
∆F
(−240 N) − (−110 N)
=−
= 6500 N/m.
∆x
(0.060 m) − (0.040 m)
With no force on the spring,
∆x = −
(0) − (−110 N)
∆F
=−
= −0.017 m.
k
(6500 N/m)
This is the amount less than the 40 mm mark, so the position of the spring with no force on it is 23
mm.
(b) ∆x = −10 mm compared to the 100 N picture, so
∆F = −k∆x = −(6500 N/m)(−0.010 m) = 65 N.
The weight of the last object is 110 N − 65 N = 45 N.
E11-24 (a) W = 21 k(xf 2 − xi 2 ) = 12 (1500 N/m)(7.60×10−3 m)2 = 4.33×10−2 J.
(b) W = 21 (1500 N/m)[(1.52×10−2 m)2 − (7.60×10−3 m)2 = 1.30×10−1 J.
E11-25 Start with Eq. 11-20, and let Fx = 0 while Fy = −mg:
Z f
Z f
W =
(Fx dx + Fy dy) = −mg
dy = −mgh.
i
i
E11-26 (a) F0 = mv02 /r0 = (0.675 kg)(10.0 m/s)2 /(0.500 m) = 135 N.
(b) Angular momentum is conserved, so v = v0 (r0 /r). The force then varies as F = mv 2 /r =
2 2 3
mv0 r0 /r = F0 (r0 /r)3 . The work done is
Z
3
~ = (−135 N)(0.500 m) (0.300 m)−2 − (0.500 m)−2 = 60.0 J.
~ · dr
W = F
−2
E11-27
The kinetic energy of the electron is
1.60 × 10−19 J
4.2 eV
= 6.7 × 10−19 J.
1 eV
Then
v=
r
2K
=
m
s
2(6.7×10−19 J)
= 1.2×106 m/s.
(9.1×10−31 kg)
131
E11-28 (a) K = 21 (110 kg)(8.1 m/s)2 = 3600 J.
(b) K = 12 (4.2×10−3 kg)(950 m/s)2 = 1900 J.
(c) m = 91, 400 tons(907.2 kg/ton) = 8.29×107 kg;
v = 32.0 knots(1.688 ft/s/knot)(0.3048 m/ft) = 16.5 m/s.
K=
1
(8.29×107 kg)(16.5 m/s)2 = 1.13×1010 J.
2
E11-29 (b) ∆K = W = F x = (1.67×10−27 kg)(3.60×1015 m/s2 )(0.0350 m) = 2.10×10−13 J. That’s
2.10×10−13 J/(1.60×10−19 J/eV) = 1.31×106 eV.
(a) K f = 2.10×10−13 J + 12 (1.67×10−27 kg)(2.40×107 m/s2 ) = 6.91×10−13 J. Then
p
p
v f = 2K/m = 2(6.91×10−13 J)/(1.67×10−27 kg) = 2.88×107 m/s.
E11-30 Work is negative if kinetic energy is decreasing. This happens only in region CD. The
work is zero in region BC. Otherwise it is positive.
E11-31
(a) Find the velocity of the particle by taking the time derivative of the position:
v=
dx
= (3.0 m/s) − (8.0 m/s2 )t + (3.0 m/s3 )t2 .
dt
Find v at two times: t = 0 and t = 4 s.
v(0) = (3.0 m/s) − (8.0 m/s2 )(0) + (3.0 m/s3 )(0)2 = 3.0 m/s,
v(4) = (3.0 m/s) − (8.0 m/s2 )(4.0 s) + (3.0 m/s3 )(4.0 s)2 = 19.0 m/s
The initial kinetic energy is K i = 12 (2.80 kg)(3.0 m/s)2 = 13 J, while the final kinetic energy is
K f = 12 (2.80 kg)(19.0 m/s)2 = 505 J.
The work done by the force is given by Eq. 11-24,
W = K f − K i = 505 J − 13 J = 492 J.
(b) This question is asking for the instantaneous power when t = 3.0 s. P = F v, so first find a;
a=
dv
= −(8.0 m/s2 ) + (6.0 m/s3 )t.
dt
Then the power is given by P = mav, and when t = 3 s this gives
P = mav = (2.80 kg)(10 m/s2 )(6 m/s) = 168 W.
E11-32 W = ∆K = −K i . Then
1
W = − (5.98×1024 kg)(29.8×103 m/s)2 = 2.66×1033 J.
2
E11-33 (a) K = 12 (1600 kg)(20 m/s)2 = 3.2×105 J.
(b) P = W/t = (3.2×105 J)/(33 s) = 9.7×103 W.
(c) P = F v = mav = (1600 kg)(20 m/s/33.0 s)(20 m/s) = 1.9×104 W.
E11-34 (a) I = 1.40×104 u · pm2 (1.66×10−27 kg/mboxu) = 2.32×10−47 kg · m2 .
(b) K = 21 Iω 2 = 12 (2.32×10−47 kg · m2 )(4.30×1012 rad/s)2 = 2.14×10−22 J. That’s 1.34 meV.
132
E11-35 The translational kinetic energy is K t =
2
1
2
2 Iω = 3 K t . Then
ω=
r
2m
v=
3I
s
1
2
2 mv ,
the rotational kinetic energy is K r =
2(5.30×10−26 kg)
(500 m/s) = 6.75×1012 rad/s.
3(1.94×10−46 kg · m2 )
E11-36 K r = 12 Iω 2 = 14 (512 kg)(0.976 m)2 (624 rad/s)2 = 4.75×107 J.
(b) t = W/P = (4.75×107 J)/(8130 W) = 5840 s, or 97.4 minutes.
E11-37
From Eq. 11-29, K i = 21 Iω i 2 . The object is a hoop, so I = M R2 . Then
Ki =
1
1
M R2 ω 2 = (31.4 kg)(1.21 m)2 (29.6 rad/s)2 = 2.01 × 104 J.
2
2
Finally, the average power required to stop the wheel is
P =
W
Kf − Ki
(0) − (2.01 × 104 J)
=
=
= −1360 W.
t
t
(14.8 s)
E11-38 The wheels are connected by a belt, so rA ωA = rB ωB , or ωA = 3ωB .
(a) If lA = lB then
IA
lA /ωA
ωB
1
=
=
= .
IB
lB /ωB
ωA
3
(b) If instead KA = KB then
IA
2KA /ωA 2
ωB 2
1
=
=
= .
2
2
IB
2KB /ωB
ωA
9
E11-39 (a) ω = 2π/T , so
K=
1 2
4π 2 (5.98×1024 kg)(6.37×106 m)2
Iω =
= 2.57×1029 J
2
5
(86, 400 s)2
(b) t = (2.57×1029 J)/(6.17×1012 W) = 4.17×1016 s, or 1.3 billion years.
E11-40 (a) The velocities relative to the center of mass are m1 v1 = m2 v2 ; combine with v1 + v2 =
910.0 m/s and get
(290.0 kg)v1 = (150.0 kg)(910 m/s − v1 ),
or
v1 = (150 kg)(910 m/s)/(290 kg + 150 kg) = 310 m/s
and v2 = 600 m/s. The rocket case was sent back, so v c = 7600 m/s − 310 m/s = 7290 m/s. The
payload capsule was sent forward, so v p = 7600 m/s + 600 m/s = 8200 m/s.
(b) Before,
1
K i = (290 kg + 150 kg)(7600 m/s)2 = 1.271×1010 J.
2
After,
1
1
K f = (290 kg)(7290 m/s)2 + (150 kg)(8200 m/s)2 = 1.275×1010 J.
2
2
The “extra” energy came from the spring.
133
E11-41 Let the mass of the freight car be M and the initial speed be v i . Let the mass of the
caboose be m and the final speed of the coupled cars be v f . The caboose is originally at rest, so the
expression of momentum conservation is
M v i = M v f + mv f = (M + m)v f
The decrease in kinetic energy is given by
Ki − Kf
1
1
1
2
2
2
M vi −
M v f + mv f ,
2
2
2
1
M v i 2 − (M + m)v f 2
2
=
=
What we really want is (K i − K f )/K i , so
Ki − Kf
Ki
M v i 2 − (M + m)v f 2
,
M vi 2
2
M + m vf
= 1−
,
M
vi
2
M +m
M
= 1−
,
M
M +m
=
where in the last line we substituted from the momentum conservation expression.
Then
Ki − Kf
M
Mg
=1−
=1−
.
Ki
M +m
M g + mg
The left hand side is 27%. We want to solve this for mg, the weight of the caboose. Upon rearranging,
mg =
Mg
(35.0 ton)
− Mg =
− (35.0 ton) = 12.9 ton.
1 − 0.27
(0.73)
E11-42 Since the body splits into two parts with equal mass then the velocity gained by one is
identical to the velocity “lost” by the other. The initial kinetic energy is
1
K i = (8.0 kg)(2.0 m/s)2 = 16 J.
2
The final kinetic energy is 16 J greater than this, so
1
1
K f = 32 J = (4.0 kg)(2.0 m/s + v)2 + (4.0 kg)(2.0 m/s − v)2 ,
2
2
1
2
2
=
(8.0 kg)[(2.0 m/s) + v ],
2
so 16.0 J = (4.0 kg)v 2 . Then v = 2.0 m/s; one chunk comes to a rest while the other moves off at a
speed of 4.0 m/s.
E11-43 The initial velocity of the neutron is v0 î, the final velocity is v1 ĵ. Bypmomentum conservation the final momentum of the deuteron is mn (v0 î − v1 ĵ). Then md v2 = mn v02 + v12 .
There is also conservation of kinetic energy:
1
1
1
mn v02 = mn v12 + md v22 .
2
2
2
Rounding the numbers slightly we have md = 2mn , then 4v22 = v02 + v12 is the momentum expression
and v02 = v12 + 2v22 is the energy expression. Combining,
2v02 = 2v12 + (v02 + v12 ),
or v12 = v02 /3. So the neutron is left with 1/3 of its original kinetic energy.
134
E11-44 (a) The third particle must have a momentum
~p3
= −(16.7×10−27 kg)(6.22×106 m/s)î + (8.35×10−27 kg)(7.85×106 m/s)ĵ
=
(−1.04î + 0.655ĵ)×10−19 kg · m/s.
(b) The kinetic energy can also be written as K =
kinetic energy appearing in this process is
K
=
1
2
2 mv
=
1
2
2 m(p/m)
= p2 /2m. Then the
1
1
(16.7×10−27 kg)(6.22×106 m/s)2 + (8.35×10−27 kg)(7.85×106 m/s)2
2
2
1
−19
(1.23×10 kg · m/s)2 = 1.23×10−12 J.
+
2(11.7×10−27 kg)
This is the same as 7.66 MeV.
P11-1
Change your units! Then
F =
W
(4.5 eV)(1.6×10−19 J/eV)
=
= 2.1×10−10 N.
s
(3.4×10−9 m)
P11-2 (a) If the acceleration is −g/4 the the net force on the block is −M g/4, so the tension in
the cord must be T = 3M g/4.
~ · ~s = (3M g/4)(−d) = −(3/4)M gd.
(a) The work done by the cord is W = F
~
(b) The work done by gravity is W = F · ~s = (−M g)(−d) = M gd.
P11-3 (a) There are four cords which are attached to the bottom load L. Each supports a tension
F , so to lift the load 4F = (840 lb) + (20.0 lb), or F = 215 lb.
~ · ~s = (840 lb)(12.0 ft) = 10100 ft · lb.
(b) The work done against gravity is W = F
(c) To lift the load 12 feet each segment of the cord must be shortened by 12 ft; there are four
segments, so the end of the cord must be pulled through a distance of 48.0 ft.
~ · ~s = (215 lb)(48.0 ft) = 10300 ft · lb.
(d) The work done by the applied force is W = F
P11-4 The incline has a height h where h = W/mg = (680 J)/[(75 kg)(9.81 m/s2 )]. The work
required to lift the block is the same regardless of the path, so the length of the incline l is l =
W/F = (680 J)/(320 N). The angle of the incline is then
θ = arcsin
h
F
(320 N)
= arcsin
= arcsin
= 25.8◦ .
l
mg
(75 kg)(9.81 m/s2 )
P11-5 (a) In 12 minutes the horse moves x = (5280 ft/mi)(6.20 mi/h)(0.200 h) = 6550 ft. The
~ · ~s = (42.0 lb)(6550 ft) cos(27.0◦ ) = 2.45×105 ft · lb.
work done by the horse in that time is W = F
(b) The power output of the horse is
P =
(2.45×105 ft · lb)
1 hp
= 340 ft · lb/s
= 0.618 hp.
(720 s)
550 ft · lb/s
P11-6 In this problem θ = arctan(28.2/39.4) = 35.5◦ .
The weight of the block has components W|| = mg sin θ and W⊥ = mg cos θ. The force of friction
on the block is f = µk N = µk mg cos θ. The tension in the rope must then be
T = mg(sin θ + µk cos θ)
in order to move the block up the ramp. The power supplied by the winch will be P = T v, so
P = (1380 kg)(9.81 m/s2 )[sin(35.5◦ ) + (0.41) cos(35.5◦ )](1.34 m/s) = 1.66×104 W.
135
P11-7 If the power is constant then the force on the car is given by F = P/v. But the force is
related to the acceleration by F = ma and to the speed by F = m dv
dt for motion in one dimension.
Then
P
,
v
P
=
,
v
P
=
,
v
P
=
,
Zv
F
=
dv
dt
dx dv
m
dt dx
dv
mv
dx
Z v
mv 2 dv
m
x
=
0
P dx,
0
1
mv 3
3
= P x.
We can rearrange this final expression to get v as a function of x, v = (3xP/m)1/3 .
P11-8 (a) If the drag is D = bv 2 , then the force required to move the plane forward at constant
speed is F = D = bv 2 , so the power required is P = F v = bv 3 .
(b) P ∝ v 3 , so if the speed increases to 125% then P increases by a factor of 1.253 = 1.953, or
increases by 95.3%.
P11-9
(a) P = mgh/t, but m/t is the persons per minute times the average mass, so
P = (100 people/min)(75.0kg)(9.81 m/s2 )(8.20 m) = 1.01×104 W.
(b) In 9.50 s the Escalator has moved (0.620 m/s)(9.50 s) = 5.89 m; so the Escalator has “lifted”
the man through a distance of (5.89 m)(8.20 m/13.3 m) = 3.63 m. The man did the rest himself.
The work done by the Escalator is then W = (83.5 kg)(9.81 m/s2 )(3.63 m) = 2970 J.
(c) Yes, because the point of contact is moving in a direction with at least some non-zero component of the force. The power is
P = (83.5 m/s2 )(9.81 m/s2 )(0.620 m/s)(8.20 m/13.3 m) = 313 W.
(d) If there is a force of contact between the man and the Escalator then the Escalator is doing
work on the man.
p
P11-10 (a) dP/dv = ab − 3av 2 , so P max occurs when 3v 2 = b, or v = b/3.
(b) F = P/v, so dF/dv = −2v, which means F is a maximum when v = 0.
(c) No; P = 0, but F = ab.
P11-11
(b) Integrate,
W =
Z
0
3x0
~ · d~s = F0
F
x0
Z
3x0
(x − x0 )dx = F0 x0
0
or W = 3F0 x0 /2.
136
9
−3 ,
2
P11-12
(a) Simpson’s rule gives
W =
1
[(10 N) + 4(2.4 N) + (0.8 N)] (1.0 m) = 6.8 J.
3
R
R
(b) W = F ds = (A/x2 )dx = −A/x, evaluating this between the limits of integration gives
W = (9 N · m2 )(1/1 m − 1/3 m) = 6 J.
P11-13
The work required to stretch the spring from xi to xf is given by
Z xf
k
k
W =
kx3 dx = xf 4 − xi 4 .
4
4
xi
The problem gives
k
k 4 k 4
(l) − (0) = l4 .
4
4
4
We then want to find the work required to stretch from x = l to x = 2l, so
W0 =
Wl→2l
k
k
(2l)4 − (l)4 ,
4
4
k 4 k 4
= 16 l − l ,
4
4
k 4
= 15 l = 15W0 .
4
=
p
P11-14 (a) The spring extension is δl = l02 + x2 − l0 . The force from one spring has magnitude
kδl, but only the x component contributes to the problem, so
q
x
F = 2k
l02 + x2 − l0 p 2
l0 + x2
is the force required to move the point.
Rx
The work required is the integral, W = 0 F dx, which is
q
W = kx2 − 2kl0 l02 + x2 + 2kl02
Note that it does reduce to the expected behavior for x l0 .
(b) Binomial expansion of square root gives
q
1 x2
1 x4
2
2
l0 + x = l0 1 +
−
··· ,
2 l02
8 l04
so the first term in the above expansion cancels with the last term in W ; the second term cancels
with the first term in W , leaving
1 x4
W = k 2.
4 l0
P11-15 Number the springs clockwise from the top of the picture. Then the four forces on each
spring are
p
F1 = k(l0 − x2 + (l0 − y)2 ),
p
F2 = k(l0 − (l0 − x)2 + y 2 ),
p
F3 = k(l0 − x2 + (l0 + y)2 ),
p
F4 = k(l0 − (l0 + x)2 + y 2 ).
137
The directions are much harder to work out, but for small x and y we can assume that
p
F~1 = k(l0 − x2 + (l0 − y)2 )ĵ,
p
F~2 = k(l0 − (l0 − x)2 + y 2 )î,
p
F~3 = k(l0 − x2 + (l0 + y)2 )ĵ,
p
F~4 = k(l0 − (l0 + x)2 + y 2 )î.
Then
W =
Z
~ · d~s =
F
Z
(F1 + F3 )dy +
Z
(F2 + F4 )dx,
Since x and y are small, expand the force(s) in a binomial expansion:
∂F1 ∂F1 x+
y = ky;
F1 (x, y) ≈ F1 (0, 0) +
∂x x,y=0
∂y x,y=0
there will be similar expression for the other four forces. Then
Z
Z
W = 2ky dy + 2kx dx = k(x2 + y 2 ) = kd2 .
P11-16
(a) K i = 12 (1100 kg)(12.8 m/s)2 = 9.0×104 J. Removing 51 kJ leaves 39 kJ behind, so
p
p
v f = 2K f /m = 2(3.9×104 J)/(1100 kg) = 8.4 m/s,
or 30 km/h.
(b) 39 kJ, as was found above.
P11-17 Let M be the mass of the man and m be the mass of the boy. Let vM be the original
speed of the man and vm be the original speed of the boy. Then
1
1 1
2
2
M vM =
mv
2
2 2 m
and
1
1
2
M (vM + 1.0 m/s)2 = mvm
.
2
2
Combine these two expressions and solve for vM ,
1
1 1
2
M vM
=
M (vM + 1.0 m/s)2 ,
2
2 2
1
2
vM
=
(vM + 1.0 m/s)2 ,
2
2
0 = −vM
+ (2.0 m/s)vM + (1.0 m/s)2 .
The last line can be solved as a quadratic, and vM = (1.0 m/s) ± (1.41 m/s). Now we use the very
first equation to find the speed of the boy,
1
1 1
2
2
M vM =
mv
,
2
2 2 m
1 2
2
vM
=
v ,
4 m
2vM = vm .
138
P11-18 (a) The work done by gravity on the projectile as it is raised up to 140 m is W =
−mgy = −(0.550 kg)(9.81 m/s2 )(140 m) = −755 J. Then the kinetic energy at the highest point is
1550 J − 755 J = 795 J. Since
p the projectile must be moving horizontally at the highest point, the
horizontal velocity is vx = 2(795 J)/(0.550 kg) = 53.8 m/s.
p
(b)
pThe magnitude of the velocity at launch is v = 2(1550 J)/(0.550 kg) = 75.1 m/s. Then
vy = (75.1 m/s)2 − (53.8 m/s)2 = 52.4 m/s.
(c) The kinetic energy at that point is 21 (0.550 kg)[(65.0 m/s)2 + (53.8 m/s)2 ] = 1960 J. Since it
has extra kinetic energy, it must be below the launch point, and gravity has done 410 J of work on
it. That means it is y = (410 J)/[(0.550 kg)(9.81 m/s2 )] = 76.0 m below the launch point.
P11-19 (a) K = 21 mv 2 = 12 (8.38 × 1011 kg)(3.0 × 104 m/s)2 = 3.77 × 1020 J. In terms of TNT,
K = 9.0×104 megatons.
√
(b) The diameter will be 3 8.98×104 = 45 km.
P11-20 (a) W g = −(0.263 kg)(9.81 m/s2 )(−0.118 m) = 0.304 J.
(b) W s = − 21 (252 N/m)(−0.118 m)2 = −1.75 J.
(c) The kinetic
energy just before hitting the block would be 1.75 J − 0.304 J = 1.45 J. The speed
p
is then v = 2(1.45 J)/(0.263 kg) = 3.32 m/s.
(d) Doubling the speed quadruples the initial kinetic energy to 5.78 J. The compression will then
be given by
1
−5.78 J = − (252 N/m)y 2 − (0.263 kg)(9.81 m/s2 )y,
2
with solution y = 0.225 m.
P11-21
(a) We can solve this with a trick of integration.
Z x
W =
F dx,
0
Z x
Z t
dt
dx
=
max dx = max
dt
dt
0
0 dt
Z t
Z t
= max
vx dt = max
at dt,
0
=
0
1
ma2x t2 .
2
Basically, we changed the variable of integration from x to t, and then used the fact the the acceleration was constant so vx = v0x + ax t. The object started at rest so v0x = 0, and we are given in
the problem that v f = atf . Combining,
1
1
W = ma2x t2 = m
2
2
vf
tf
2
(b) Instantaneous power will be the derivative of this, so
dW
P =
=m
dt
139
vf
tf
2
t.
t2 .
2
P11-22 (a) α = (−39.0 rev/s)(2π rad/rev)/(32.0 s) = −7.66 rad/s .
(b) The total rotational inertia of the system about the axis of rotation is
I = (6.40 kg)(1.20 m)2 /12 + 2(1.06 kg)(1.20 m/2)2 = 1.53 kg · m2 .
2
The torque is then τ = (1.53 kg · m2 )(7.66 rad/s ) = 11.7 N · m.
(c) K = 21 (1.53 kg · m2 )(245 rad/s)2 = 4.59×104 J.
(d) θ = ω av t = (39.0 rev/s/2)(32.0 s) = 624 rev.
(e) Only the loss in kinetic energy is independent of the behavior of the frictional torque.
P11-23 The wheel turn with angular speed ω = v/r, where r is the radius of the wheel and v the
speed of the car. The total rotational kinetic energy in the four wheels is then
h i
1
v 2
1
= (11.3 kg)v 2 .
Kr = 4 Iω 2 = 2 (11.3 kg)r2
2
2
r
The translational kinetic energy is K t = 12 (1040 kg)v 2 , so the fraction of the total which is due to
the rotation of the wheels is
11.3
= 0.0213 or 2.13%.
520 + 11.3
P11-24 (a) Conservation of angular momentum: ω f = (6.13 kg · m2 /1.97 kg · m2 )(1.22 rev/s) =
3.80 rev/s.
(b) K r ∝ Iω 2 ∝ l2 /I, so
K f /K i = I i /I f = (6.13 kg · m2 )/(1.97 kg · m2 ) = 3.11.
P11-25 We did the first part of the solution in Ex. 10-21. The initial kinetic energy is (again,
ignoring the shaft),
1
~ 1,i 2 ,
K i = I1 ω
2
since the second wheel was originally at rest. The final kinetic energy is
Kf =
1
(I1 + I2 )~
ωf 2 ,
2
since the two wheels moved as one. Then
Ki − Kf
Ki
=
=
=
1
~ 1,i 2
2 I1 ω
− 12 (I1 + I2 )~
ωf 2
,
1
~ 1,i 2
2 I1 ω
(I1 + I2 )~
ωf 2
,
I1 ω
~ 1,i 2
I1
1−
,
I1 + I2
1−
where in the last line we substituted from the results of Ex. 10-21.
Using the numbers from Ex. 10-21,
Ki − Kf
(1.27 kg·m2 )
=1−
= 79.2%.
Ki
(1.27 kg·m2 ) + (4.85 kg·m2 )
140
P11-26
See the solution to P10-11.
Ki =
while
Kf =
I 2 m 2
ω + v
2
2
1
(I + mR2 )ω f 2
2
according to P10-11,
ωf =
Then
Kf =
Iω − mvR
.
I + mR2
1 (Iω − mvR)2
.
2 I + mR2
Finally,
∆K
1 (Iω − mvR)2 − (I + mR2 )(Iω 2 + mv 2 )
,
2
I + mR2
1 ImR2 ω 2 + 2mvRIω + Imv 2
= −
,
2
I + mR2
Im (Rω + v)2
= −
.
2 I + mR2
=
P11-27 See the solution to P10-12.
(a) K i = 12 I i ω i 2 , so
1
2(51.2 kg)(1.46 m)2 (0.945 rad/s)2 = 97.5 J.
2
1
(b) K f = 2 2(51.2 kg)(0.470 m)2 (9.12 rad/s)2 = 941 J. The energy comes from the work they
do while pulling themselves closer together.
Ki =
P11-28
K = 12 mv 2 =
1 2
2m p
=
1
~
2m p
Kf
=
=
∆K
=
· ~p. Then
1
(~pi + ∆~p) · (~pi + ∆~p),
2m
1
pi 2 + 2~pi · ∆~p + (∆p)2 ,
2m
1
2~pi · ∆~p + (∆p)2 .
2m
In all three cases ∆p = (3000 N)(65.0 s) = 1.95×105 N · s and pi = (2500 kg)(300 m/s) = 7.50×105 kg ·
m/s.
(a) If the thrust is backward (pushing rocket forward),
∆K =
+2(7.50×105 kg · m/s)(1.95×105 N · s) + (1.95×105 N · s)2
= +6.61×107 J.
2(2500 kg)
(b) If the thrust is forward,
∆K =
−2(7.50×105 kg · m/s)(1.95×105 N · s) + (1.95×105 N · s)2
= −5.09×107 J.
2(2500 kg)
(c) If the thrust is sideways the first term vanishes,
∆K =
+(1.95×105 N · s)2
= 7.61×106 J.
2(2500 kg)
141
P11-29
There’s nothing to integrate here! Start with the work-energy theorem
W
1
1
mv f 2 − mv i 2 ,
2
2
1
m vf 2 − vi 2 ,
2
1
m (v f − v i ) (v f + v i ) ,
2
= Kf − Ki =
=
=
where in the last line we factored the difference of two squares. Continuing,
W
=
=
1
(mv f − mv i ) (v f + v i ) ,
2
1
(∆p) (v f + v i ) ,
2
but ∆p = J, the impulse. That finishes this problem.
P11-30 Let M be the mass of the helicopter. It will take a force M g to keep the helicopter
airborne. This force comes from pushing the air down at a rate ∆m/∆t with a speed of v; so
M g = v∆m/∆t. The blades sweep out a cylinder of cross sectional area A, so ∆m/∆t
= ρAv.
p
The force is then M g = ρAv 2 ; the speed that the air must be pushed down is v = M g/ρA. The
minimum power is then
s
s
Mg
(1820 kg)3 (9.81 m/s2 )3
P = Fv = Mg
=
= 2.49×105 W.
ρA
(1.23 kg/m3 )π(4.88 m)2
P11-31 (a) Inelastic collision, so v f = mv i /(m + M ).
(b) K = 12 mv 2 = p2 /2m, so
∆K
1/m − 1/(m + M )
M
=
=
.
Ki
1/m
m+M
P11-32
Inelastic collision, so
vf =
(1.88 kg)(10.3 m/s) + (4.92 kg)(3.27 m/s)
= 5.21 m/s.
(1.88 kg) + (4.92 kg)
The loss in kinetic energy is
∆K =
(1.88 kg)(10.3 m/s)2
(4.92 kg)(3.27 m/s)2
(1.88 kg + 4.92 kg)(5.21 m/s)2
+
−
= 33.7 J.
2
2
2
This change is because of work done on the spring, so
p
x = 2(33.7 J)/(1120 N/m) = 0.245 m
P11-33 ~pf ,B = ~pi,A + ~pi,B − ~pf ,A , so
~pf ,B
=
[(2.0 kg)(15 m/s) + (3.0 kg)(−10 m/s) − (2.0 kg)(−6.0 m/s)]î
+[(2.0 kg)(30 m/s) + (3.0 kg)(5.0 m/s) − (2.0 kg)(30 m/s)]ĵ,
=
(12 kg · m/s)î + (15 kg · m/s)ĵ.
142
Then ~vf ,B = (4.0 m/s)î + (5.0 m/s)ĵ. Since K =
∆K
m 2
2 (vx
+ vy2 ), the change in kinetic energy is
(2.0 kg)[(−6.0 m/s)2 + (30 m/s)2 − (15 m/s)2 − (30 m/s)2 ]
2
(3.0 kg)[(4.0 m/s)2 + (5 m/s)2 − (−10 m/s)2 − (5.0 m/s)2 ]
+
2
= −315 J.
=
P11-34 For the observer on the train the acceleration of the particle is a, the distance traveled is
xt = 21 at2 , so the work done as measured by the train is W t = maxt = 12 a2 t2 . The final speed of
the particle as measured by the train is v t = at, so the kinetic energy as measured by the train is
K = 12 mv 2 = 12 m(at)2 . The particle started from rest, so ∆K t = W t .
For the observer on the ground the acceleration of the particle is a, the distance traveled is
xg = 12 at2 + ut, so the work done as measured by the ground is W g = maxg = 12 a2 t2 + maut.
The final speed of the particle as measured by the ground is v g = at + u, so the kinetic energy as
measured by the ground is
Kg =
1
1
1
1
mv 2 = m(at + u)2 = a2 t2 + maut + mu2 .
2
2
2
2
But the initial kinetic energy as measured by the ground is 21 mu2 , so W g = ∆K g .
P11-35 (a) K i = 12 m1 v 1,i 2 .
(b) After collision v f = m1 v 1,i /(m1 + m2 ), so
2
1
m1 v 1,i
1
m1
2
K f = (m1 + m2 )
= m1 v 1,i
.
2
m1 + m2
2
m1 + m2
(c) The fraction lost was
m1
m2
=
.
m1 + m2
m1 + m2
= v f . The initial kinetic energy of the system is
1−
(d) Note that v cm
Ki =
1
1
m1 v 0 1,i 2 + m2 v 0 2,i 2 .
2
2
The final kinetic energy is zero (they stick together!), so the fraction lost is 1. The amount lost,
however, is the same.
P11-36 Only consider the first two collisions, the one between m and m0 , and then the one between
m0 and M .
Momentum conservation applied to the first collision means the speed of m0 will be between
v 0 = mv0 /(m + m0 ) (completely inelastic)and v 0 = 2mv0 /(m + m0 ) (completely elastic). Momentum
conservation applied to the second collision means the speed of M will be between V = m0 v 0 /(m0 +M )
and V = 2m0 v 0 /(m0 + M ). The largest kinetic energy for M will occur when it is moving the fastest,
so
2mv0
2m0 v 0
4m0 mv0
v0 =
and V = 0
=
.
0
m+m
m +M
(m + m0 )(m0 + M )
We want to maximize V as a function of m0 , so take the derivative:
dV
4mv0 (mM − m02 )
=
.
0
dm
(m0 + M )2 (m + m0 )2
This vanishes when m0 =
√
mM .
143
E12-1 (a) Integrate.
U (x) = −
x
Z
G
∞
m1 m2
m1 m2
dx = −G
.
2
x
x
(b) W = U (x) − U (x + d), so
W = Gm1 m2
1
1
−
x x+d
= Gm1 m2
d
.
x(x + d)
E12-2 If d << x then x(x + d) ≈ x2 , so
W ≈G
E12-3
m1 m2
d.
x2
Start with Eq. 12-6.
U (x) − U (x0 )
= −
Finishing the integration,
x
x
Z x0
Fx (x)dx,
−αxe−βx
x0
x
−α −βx2 e
.
2β
x0
= −
=
Z
2
dx,
2
α −βx20
e
− e−βx .
2β
If we choose x0 = ∞ and U (x0 ) = 0 we would be left with
α −βx2
U (x) = −
e
.
2β
U (x) = U (x0 ) +
E12-4 ∆K = −∆U so ∆K = mg∆y. The power output is then
P = (58%)
(1000 kg/m3 )(73, 800 m3 )
(9.81 m/s2 )(96.3 m) = 6.74×108 W.
(60 s)
E12-5 ∆U = −∆K, so 21 kx2 = 12 mv 2 . Then
k=
mv 2
(2.38 kg)(10.0×103 /s)
=
= 1.10×108 N/m.
2
x
(1.47 m)2
Wow.
E12-6 ∆U g + ∆U s = 0, since K = 0 when the man jumps and when the man stops. Then
∆U s = −mg∆y = (220 lb)(40.4 ft) = 8900 ft · lb.
E12-7
Apply Eq. 12-15,
K f + U f = K i + U i,
1
1
2
mv f + mgy f =
mv i 2 + mgy i ,
2
2
1 2
1 2
v f + g(−r) =
(0) + g(0).
2
2
Rearranging,
vf =
p
p
−2g(−r) = −2(9.81 m/s2 )(−0.236 m) = 2.15 m/s.
144
E12-8 (a) K = 21 mv 2 = 12 (2.40 kg)(150 m/s)2 = 2.70×104 J.
(b) Assuming that the ground is zero, U = mgy = (2.40 kg)(9.81 m/s2 )(125 m) = 2.94×103 J.
(c) K f = K i + U i since U f = 0. Then
s
(2.70×104 J) + (2.94×103 J)
vf = 2
= 158 m/s.
(2.40 kg)
Only (a) and (b) depend on the mass.
E12-9 (a) Since ∆y = 0, then ∆U = 0 and ∆K = 0. Consequently, at B, v = v0 .
(b) At C KC = KA + UA − UC , or
or
(c) At D KD
1
1
h
mvC 2 = mv0 2 + mgh − mg ,
2
2
2
r
h p
vB = v0 2 + 2g = v0 2 + gh.
2
= KA + UA − UD , or
1
1
mvD 2 = mv0 2 + mgh − mg(0),
2
2
p
vB = v0 2 + 2gh.
or
E12-10 From the slope of the graph, k = (0.4 N)/(0.04 m) = 10 N/m.
(a) ∆K = −∆U , so 12 mv f 2 = 12 kxi 2 , or
s
(10 N/m)
vf =
(0.0550 m) = 2.82 m/s.
(0.00380 kg)
(b) ∆K = −∆U , so 21 mv f 2 = 12 k(xi 2 − xf 2 ), or
s
(10 N/m)
vf =
[(0.0550 m)2 − (0.0150 m)2 ] = 2.71 m/s.
(0.00380 kg)
E12-11
(a) The force constant of the spring is
k = F/x = mg/x = (7.94 kg)(9.81 m/s2 )/(0.102 m) = 764 N/m.
(b) The potential energy stored in the spring is given by Eq. 12-8,
U=
1 2
1
kx = (764 N/m)(0.286 m + 0.102 m)2 = 57.5 J.
2
2
(c) Conservation of energy,
Kf + U f
1
1
2
mv f + mgy f + kxf 2
2
2
1 2
1
(0) + mgh + k(0)2
2
2
= K i + U i,
1
1
=
mv i 2 + mgy i + kxi 2 ,
2
2
1 2
1
=
(0) + mg(0) + kxi 2 .
2
2
Rearranging,
h=
k
(764 N/m)
(0.388 m)2 = 0.738 m.
xi 2 =
2mg
2(7.94 kg)(9.81 m/s2 )
145
E12-12 The annual mass of water is m = (1000 kg/m3 )(8 × 1012 m2 )(0.75 m). The change in
potential energy each year is then ∆U = −mgy, where y = −500 m. The power available is then
P =
(0.75 m)
1
(1000 kg/m3 )(8×1012 m2 )
(500 m) = 3.2×107 W.
3
(3.15×107 m)
E12-13 (a) From kinematics, v = −gt, so K = 12 mg 2 t2 and U = U0 − K = mgh − 12 mg 2 t2 .
(b) U = mgy so K = U0 − U = mg(h − y).
E12-14 The potential energy is the same in both cases. Consequently, mg E ∆y E = mg M ∆y M , and
then
y M = (2.05 m − 1.10 m)(9.81 m/s2 )/(1.67 m/s2 ) + 1.10 m = 6.68 m.
E12-15
The working is identical to Ex. 12-11,
Kf + U f
1
1
mv f 2 + mgy f + kxf 2
2
2
1 2
1
(0) + mgh + k(0)2
2
2
so
h=
= K i + U i,
1
1
=
mv i 2 + mgy i + kxi 2 ,
2
2
1 2
1
=
(0) + mg(0) + kxi 2 ,
2
2
k
(2080 N/m)
xi 2 =
(0.187 m)2 = 1.92 m.
2mg
2(1.93 kg)(9.81 m/s2 )
The distance up the incline is given by a trig relation,
d = h/ sin θ = (1.92 m)/ sin(27◦ ) = 4.23 m.
E12-16 The vertical position of the pendulum is y = −l cos θ, where θ is measured from the
downward vertical and l is the length of the string. The total mechanical energy of the pendulum is
E=
1
mv b 2
2
if we set U = 0 at the bottom of the path and v b is the speed at the bottom. In this case
U = mg(l + y).
(a) K = E − U = 12 mv b 2 − mgl(1 − cos θ). Then
v=
p
(8.12 m/s)2 − 2(9.81 m/s2 )(3.82 kg)(1 − cos 58.0◦ ) = 5.54 m/s.
(b) U = E − K, but at highest point K = 0. Then
1
(8.12 m/s)2
θ = arccos 1 −
= 83.1◦ .
2 (9.81 m/s2 )(3.82 kg)
(c) E = 12 (1.33 kg)(8.12 m/s)2 = 43.8 J.
E12-17 The equilibrium position is when F = ky = mg. Then ∆U g = −mgy and ∆U s =
1
1
2 (ky)y = 2 mgy. So 2∆U s = −∆U g .
146
E12-18 Let the spring get compressed a distance x. If the object fell from a height h = 0.436 m,
then conservation of energy gives 12 kx2 = mg(x + h). Solving for x,
mg
x=
±
k
r
mg 2
k
+2
mg
h
k
only the positive answer is of interest, so
s
2
(2.14 kg)(9.81 m/s2 )
(2.14 kg)(9.81 m/s2 )
(2.14 kg)(9.81 m/s2 )
x=
±
+2
(0.436 m) = 0.111 m.
(1860 N/m)
(1860 N/m)
(1860 N/m)
E12-19 The horizontal distance traveled by the marble is R = vtf , where tf is the time of flight
and v is the speed of the marble when it leaves the gun. We find that speed using energy conservation
principles applied to the spring just before it is released and just after the marble leaves the gun.
Ki + U i
1
0 + kx2
2
= Kf + U f ,
1
=
mv 2 + 0.
2
K i = 0 because the marble isn’t moving originally, and U f = 0 because the spring is no longer
compressed. Substituting R into this,
1 2
1
kx = m
2
2
R
tf
2
.
We have two values for the compression, x1 and x2 , and two ranges, R1 and R2 . We can put both
pairs into the above equation and get two expressions; if we divide one expression by the other we
get
2 2
x2
R2
=
.
x1
R1
We can easily take the square root of both sides, then
x2
R2
=
.
x1
R1
R1 was Bobby’s try, and was equal to 2.20 − 0.27 = 1.93 m. x1 = 1.1 cm was his compression. If
Rhoda wants to score, she wants R2 = 2.2 m, then
x2 =
2.2 m
1.1 cm = 1.25 cm.
1.93 m
E12-20 Conservation of energy— U1 + K1 = U2 + K2 — but U1 = mgh, K1 = 0, and U2 = 0, so
K2 = 21 mv 2 = mgh at the bottom of the swing.
The net force on Tarzan at the bottom of the swing is F = mv 2 /r, but this net force is equal to
the tension T minus the weight W = mg. Then 2mgh/r = T − mg. Rearranging,
2(8.5 ft)
T = (180 lb)
+ 1 = 241 lb.
(50 ft)
This isn’t enough to break the vine, but it is close.
147
E12-21 Let point 1 be the start position of the first mass, point 2 be the collision point, and point
1
2
3 be the highest point in the swing after the collision. Then U√
1 = K2 , or 2 m1 v1 = m1 gd, where v1
is the speed of m1 just before it collides with m2 . Then v1 = 2gd.
After the collision the speed of both objects is, by momentum conservation, v2 = m1 v1 /(m1 +m2 ).
Then, by energy conservation, U3 = K20 , or 12 (m1 + m2 )v2 2 = (m1 + m2 )gy, where y is the height
to which the combined masses rise.
Combining,
2
m1 2 v1 2
m1
v2 2
y=
=
=
d.
2g
2(m1 + m2 )2 g
m1 + m2
E12-22 ∆K = −∆U , so
1
1
mv 2 + Iω 2 = mgh,
2
2
where I = 12 M R2 and ω = v/R. Combining,
1
1
mv 2 + M v 2 = mgh,
2
4
so
v=
r
4mgh
=
2m + M
s
4(0.0487 kg)(9.81 m/s2 )(0.540 m)
= 1.45 m/s.
2(0.0487 kg) + (0.396 kg)
E12-23 There are three contributions to the kinetic energy: rotational kinetic energy of the shell
(K s ), rotational kinetic energy of the pulley (K p ), and translational kinetic energy of the block
(K b ). The conservation of energy statement is then
K s,i + K p,i + K b,i + U i
= K s,f + K p,f + K b,f + U f ,
1
1
1
I s ω s 2 + I p ω p 2 + mv b 2 + mgy.
(0) + (0) + (0) + (0) =
2
2
2
Finally, y = −h and
ωs R = ωp r = vb .
Combine all of this together, and our energy conservation statement will look like this:
1 2
v b 2 1 v b 2 1
0=
M R2
+ Ip
+ mv b 2 − mgh
2 3
R
2
r
2
which can be fairly easily rearranged into
vb 2 =
2mgh
.
2M/3 + I P /r2 + m
E12-24 The angular speed of the flywheel and the speed of the car are related by
k=
ω
(1490 rad/s)
=
= 62.1 rad/m.
v
(24.0 m/s)
The height of the slope is h = (1500 m) sin(5.00◦ ) = 131 m. The rotational inertia of the flywheel is
I=
1 (194 N)
(0.54 m)2 = 2.88 kg · m2 .
2 (9.81 m/s2 )
148
(a) Energy is conserved as the car moves down the slope: U i = K f , or
mgh =
or
v=
r
2mgh
=
m + Ik 2
s
1
1
1
1
mv 2 + Iω 2 = mv 2 + Ik 2 v 2 ,
2
2
2
2
2(822 kg)(9.81 m/s2 )(131 m)
= 13.3 m/s,
(822 kg) + (2.88 kg · m2 )(62.1 rad/m)2
or 47.9 m/s.
(b) The average speed down the slope is 13.3 m/s/2 = 6.65 m/s. The time to get to the bottom
is t = (1500 m)/(6.65 m/s) = 226 s. The angular acceleration of the disk is
α=
ω
(13.3 m/s)(62.1 rad/m)
2
=
= 3.65 rad/s .
t
(226 s)
(c) P = τ ω = Iαω, so
2
P = (2.88 kg · m2 )(3.65 rad/s )(13.3 m/s)(62.1 rad/m) = 8680 W.
E12-25 (a) For the solid sphere I =
energy means K i = U f . Then
2
2
5 mr ;
1
1
mv 2 +
2
2
or
h=
if it rolls without slipping ω = v/r; conservation of
2 2 v 2
mr
= mgh.
5
r
(5.18 m/s)2
(5.18 m/s)2
+
= 1.91 m.
2(9.81 m/s2 ) 5(9.81 m/s2 )
The distance up the incline is (1.91 m)/ sin(34.0◦ ) = 3.42 m.
(b) The sphere will travel a distance of 3.42 m with an average speed of 5.18 m/2, so t =
(3.42 m)/(2.59 m/s) = 1.32 s. But wait, it goes up then comes back down, so double this time to get
2.64 s.
(c) The total distance is 6.84 m, so the number of rotations is (6.84 m)/(0.0472 m)/(2π) = 23.1.
E12-26 Conservation of energy means 12 mv 2 + 12 Iω 2 = mgh. But ω = v/r and we are told
h = 3v 2 /4g, so
1
1 v2
3v 2
mv 2 + I 2 = mg
,
2
2 r
4g
or
3
1
1
2
I = 2r
m − m = mr2 ,
4
2
2
which could be a solid disk or cylinder.
E12-27 We assume the cannon ball is solid, so the rotational inertia will be I = (2/5)M R2
The normal force on the cannon ball will be N = M g, where M is the mass of the bowling ball.
The kinetic friction on the cannon ball is Ff = µk N = µk M g. The magnitude of the net torque on
the bowling ball while skidding is then τ = µk M gR.
Originally the angular momentum of the cannon ball is zero; the final angular momentum will
have magnitude l = Iω = Iv/R, where v is the final translational speed of the ball.
The time requires for the cannon ball to stop skidding is the time required to change the angular
momentum to l, so
∆l
(2/5)M R2 v/R
2v
∆t =
=
=
.
τ
µk M gR
5µk g
149
Since we don’t know ∆t, we can’t solve this for v. But the same time through which the angular
momentum of the ball is increasing the linear momentum of the ball is decreasing, so we also have
∆t =
∆p
M v − M v0
v0 − v
=
=
.
−Ff
−µk M g
µk g
Combining,
v0 − v
,
µk g
= 5(v0 − v),
= 5v0 /7
=
2
4
1
3
K(x) (J)
F(x) (N)
2v
5µk g
2v
v
0
-1
-2
2
1
0
0
1
2
3
4
5
6
x(m)
0
1
2
3
4
5
6
x (m)
E12-28
E12-29 (a) F = −∆U/∆x = −[(−17 J) − (−3 J)]/[(4 m) − (1 m)] = 4.7 N.
(b) The total energy is 21 (2.0 kg)(−2.0 m/s)2 + (−7 J), or −3 J. The particle is constrained to
move between x = 1 m and x = 14 m.
p
(c) When x = 7 m K = (−3 J) − (−17 J) = 14 J. The speed is v = 2(14 J)/(2.0 kg) = 3.7 m/s.
E12-30 Energy is conserved, so
1
1
mv 2 = mv 2 + mgy,
2 0
2
q
v = v02 − 2gy,
or
which depends only on y.
E12-31
(a) We can find Fx and Fy from the appropriate derivatives of the potential,
Fx
Fy
∂U
= −kx,
∂x
∂U
= −
= −ky.
∂y
= −
The force at point (x, y) is then
~ = Fx î + Fy ĵ = −kxî − ky ĵ.
F
(b) Since the force vector points directly toward the origin there is no angular component, and
Fθ = 0. Then Fr = −kr where r is the distance from the origin.
(c) A spring which is attached to a point; the spring is free to rotate, perhaps?
150
E12-32 (a) By symmetry we expect Fx , Fy , and Fz to all have the same form.
Fx = −
∂U
−kx
= 2
,
∂x
(x + y 2 + z 2 )3/2
with similar expressions for Fy and Fz . Then
~ =
F
(x2
−k
(xî + y ĵ + z k̂).
+ y 2 + z 2 )3/2
(b) In spherical polar coordinates r2 = x2 + y 2 + z 2 . Then U = −k/r and
Fr = −
∂U
k
= − 2.
∂r
r
E12-33 We’ll just do the paths, showing only non-zero terms.
Rb
Path 1: W = 0 (−k2 a) dy) = −k2 ab.
Ra
Path 2: W = 0 (−k1 b) dx) = −k1 ab.
Rd
Path 3: W = (cos φ sin φ) 0 (−k1 − k2 )r dr = −(k1 + k2 )ab/2.
These three are only equal if k1 = k2 .
P12-1
(a) We need to integrate an expression like
Z z
k
k
−
dz =
.
2
(z
+
l)
z
+
l
∞
The second half is dealt with in a similar manner, yielding
k
k
−
.
z+l z−l
(b) If z l then we can expand the denominators, then
U (z) =
k
k
−
,
z+l z−l
kl
k
kl
k
− 2 −
+ 2 ,
≈
z
z
z
z
2kl
= − 2.
z
U (z) =
P12-2 The ball
p
√ just reaches the top, so K2 = 0.
2(mgL)/m = 2gL.
Then K1 = U2 − U1 = mgL, so v1 =
P12-3 Measure distances along the incline by x, where x = 0 is measured from the maximally
compressed spring. The vertical position of the mass is given by x sin θ. For the spring k =
(268 N)/(0.0233 m) = 1.15×104 N/m. The total energy of the system is
1
(1.15×104 N/m)(0.0548 m)2 = 17.3 J.
2
(a) The block needs to have moved a vertical distance x sin(32.0◦ ), where
17.3 J = (3.18 kg)(9.81 m/s2 )x sin(32.0◦ ),
or x = 1.05 m.
(b) When the block hits the top of the spring the gravitational potential energy has changed by
∆U = −(3.18 kg)(9.81 m/s2 )(1.05 m − 0.0548 m) sin(32.0◦ ) = 16.5, J;
p
hence the speed is v = 2(16.5 J)/(3.18 kg) = 3.22 m/s.
151
P12-4
The potential energy associated with the hanging part is
0
Z 0
M g 2 M gL
M
U=
gy dy =
y =−
,
2L
32
−L/4 L
−L/4
so the work required is W = M gL/32.
P12-5
(a) Considering points P and Q we have
KP + UP
= KQ + UQ ,
1
(0) + mg(5R) =
mv 2 + mg(R),
2
1
mv 2 ,
4mgR =
2
p
8gR = v.
There are two forces on the block, the normal force from the track,
N=
mv 2
m(8gR)
=
= 8mg,
R
R
and the force of gravity W = mg. They are orthogonal so
p
√
F net = (8mg)2 + (mg)2 = 65 mg
and the angle from the horizontal by
tan θ =
−mg
1
=− ,
8mg
8
or θ = 7.13◦ below the horizontal.
(b) If the block barely makes it over the top of the track then the speed at the top of the loop
(point S, perhaps?) is just fast enough so that the centripetal force is equal in magnitude to the
weight,
mv S 2 /R = mg.
Assume the block was released from point T . The energy conservation problem is then
KT + UT
(0) + mgyT
yT
= KS + US ,
1
=
mv 2 + mgyS ,
2 S
1
=
(R) + m(2R),
2
= 5R/2.
P12-6 The wedge slides to the left, the block to the right. Conservation of momentum requires
M v w + mv b,x = 0. The block is constrained to move on the surface of the wedge, so
tan α =
v b,y
,
v b,x − v w
or
v b,y = v b,x tan α(1 + m/M ).
152
Conservation of energy requires
1
1
mv b 2 + M v w 2 = mgh.
2
2
Combining,
2
1 m
1
m v b,x 2 + v b,y 2 + M
v b,x
2
2
M
m
2
2
tan α(1 + m/M ) + 1 +
v b,x 2
M
sin2 α(M + m)2 + M 2 cos2 α + mM cos2 α v b,x 2
M 2 + mM + mM sin2 α + m2 sin2 α v b,x 2
(M + m)(M + m sin2 α) v b,x 2
or
v w = −m cos α
=
2gh,
=
2M 2 gh cos2 α,
=
2M 2 gh cos2 α,
= 2M 2 gh cos2 α,
s
2gh
.
(M + m)(M + m sin2 α)
s
2gh
.
(M + m)(M + m sin2 α)
v b,x = M cos α
Then
= mgh,
R
P12-7 U (x) = − Fx dx = −Ax2 /2 − Bx3 /3.
(a) U = −(−3.00 N/m)(2.26 m)2 /2 − (−5.00 N/m2 )(2.26 m)3 /3 = 26.9 J.
(b) There are two points to consider:
U1
U2
K1
= −(−3.00 N/m)(4.91 m)2 /2 − (−5.00 N/m2 )(4.91 m)3 /3 = 233 J,
= −(−3.00 N/m)(1.77 m)2 /2 − (−5.00 N/m2 )(1.77 m)3 /3 = 13.9 J,
1
(1.18kg)(4.13 m/s)2 = 10.1 J.
=
2
Then
v2 =
s
2(10.1 J + 233 J − 13.9 J)
= 19.7 m/s.
(1.18 kg)
P12-8
p Assume that U0 = K0 = 0. Then conservation of energy requires K = −U ; consequently,
v = 2g(−y).
p
(a) v = p2(9.81 m/s2 )(1.20 m) = 4.85 m/s.
(b) v = 2(9.81 m/s2 )(1.20 m − 0.45 m − 0.45 m) = 2.43 m/s.
P12-9 Assume that U0 = K0 = 0. Then conservation of energy requires K = −U ; consequently,
p
v = 2g(−y). If the ball barely swings around the top of the peg then the speed at the top of the
loop is just fast enough so that the centripetal force is equal in magnitude to the weight,
mv 2 /R = mg.
The energy conservation problem is then
mv 2 = 2mg(L − 2(L − d)) = 2mg(2d − L)
mg(L − d) = 2mg(2d − L),
d = 3L/5.
153
P12-10
The speed at the top and the speed at the bottom are related by
1
1
mv b 2 = mv t 2 + 2mgR.
2
2
The magnitude of the net force is F = mv 2 /R, the tension at the top is
T t = mv t 2 /R − mg,
while tension at the bottom is
T b = mv b 2 /R + mg,
The difference is
∆T = 2mg + m(v b 2 − v t 2 )/R = 2mg + 4mg = 6mg.
P12-11 Let the angle θ be measured from the horizontal to the point on the hemisphere where
the boy is located. There are then two components to the force of gravity— a component tangent
to the hemisphere, W|| = mg cos θ, and a component directed radially toward the center of the
hemisphere, W⊥ = mg sin θ.
While the boy is in contact with the hemisphere the motion is circular so
mv 2 /R = W⊥ − N.
When the boy leaves the surface we have mv 2 /R = W⊥ , or mv 2 = mgR sin θ. Now for energy
conservation,
K +U
1
2
mv + mgy
2
1
gR sin θ + mgy
2
1
y+y
2
y
= K0 + U0 ,
1
=
m(0)2 + mgR,
2
= mgR,
= R,
= 2R/3.
P12-12 (a) To be in contact at the top requires mv t 2 /R = mg. The speed at the bottom would
be given by energy conservation
1
1
mv b 2 = mv t 2 + 2mgR,
2
2
√
so v b = 5gR is the speed at the bottom that will allow the object to make it around the circle
without loosing contact.
(b) The particle will lose contact with the track if mv 2 /R ≤ mg sin θ. Energy conservation gives
1
1
mv02 = mv 2 + mgR(1 + sin θ)
2
2
for points above the half-way point. Then the condition for “sticking” to the track is
1 2
v − 2g(1 + sin θ) ≤ g sin θ,
R 0
or, if v0 = 0.775v m ,
5(0.775)2 − 2 ≤ 3 sin θ,
or θ = arcsin(1/3).
154
P12-13
The rotational inertia is
I=
1
4
M L2 + M L2 = M L2 .
3
3
Conservation of energy is
1 2
Iω = 3M g(L/2),
2
so ω =
p
9g/(4L).
P12-14 The rotational speed of the sphere is ω = v/r; the rotational kinetic energy is K r =
1
1
2
2
2 Iω = 5 mv .
(a) For the marble to stay on the track mv 2 /R = mg at the top of the track. Then the marble
needs to be released from a point
mgh =
1
1
mv 2 + mv 2 + 2mgR,
2
5
or h = R/2 + R/5 + 2R = 2.7R.
(b) Energy conservation gives
6mgR =
1
1
mv 2 + mv 2 + mgR,
2
5
or mv 2 /R = 50mg/7. This corresponds to the horizontal force acting on the marble.
P12-15
1
2
2 mv0
+ mgy = 0, where y is the distance beneath the rim, or y = −r cos θ0 . Then
v0 =
p
p
−2gy = 2gr cos θ0 .
P12-16 (a) For E1 the atoms will eventually move apart completely.
(b) For E2 the moving atom will bounce back and forth between a closest point and a farthest
point.
(c) U ≈ −1.2×10−19 J.
(d) K = E1 − U ≈ 2.2×10−19 J.
(e) Find the slope of the curve, so
F ≈−
(−1×10−19 J) − (−2×10−19 J)
= −1×10−9 N,
(0.3×10−9 m) − (0.2×10−9 m)
which would point toward the larger mass.
P12-17 The function needs to fall off at infinity in both directions; an exponential envelope
would work, but it will need to have an −x2 term to force the potential to zero on both sides. So we
propose something of the form
2
U (x) = P (x)e−βx
where P (x) is a polynomial in x and β is a positive constant.
We proposed the polynomial because we need a symmetric function which has two zeroes. A
quadratic of the form αx2 −U0 would work, it has two zeroes, a minimum at x = 0, and is symmetric.
So our trial function is
2
U (x) = αx2 − U0 e−βx .
155
This function should have three extrema. Take the derivative, and then we’ll set it equal to zero,
2
2
dU
= 2αxe−βx − 2 αx2 − U0 βxe−βx .
dx
Setting this equal to zero leaves two possibilities,
2
2α − 2 αx − U0
x = 0,
β = 0.
The first equation is trivial, the second is easily rearranged to give
s
α + βU0
x=±
βα
These are the points ±x1 . We can, if we wanted, try to find α and β from the picture, but you
might notice we have one equation, U (x1 ) = U1 and two unknowns. It really isn’t very illuminating
to take this problem much farther, but we could.
(b) The force is the derivative of the potential; this expression was found above.
(c) As long as the energy is less than the two peaks, then the motion would be oscillatory, trapped
in the well.
P12-18
(a) F = −∂U/∂r, or
F = −U0
r0
1
+
r2
r
e−r/r0 .
(b) Evaluate the force at the four points:
F (r0 ) = −2(U0 /r0 )e−1 ,
F (2r0 ) = −(3/4)(U0 /r0 )e−2 ,
F (4r0 ) = −(5/16)(U0 /r0 )e−4 ,
F (10r0 ) = −(11/100)(U0 /r0 )e−10 .
The ratios are then
F (2r0 )/F (r0 ) = (3/8)e−1 = 0.14,
F (4r0 )/F (r0 ) = (5/32)e−3 = 7.8×10−3 ,
F (10r0 )/F (r0 ) = (11/200)e−9 = 6.8×10−6 .
156
E13-1 If the projectile had not experienced air drag it would have risen to a height y2 , but
because of air drag 68 kJ of mechanical energy was dissipated so it only rose to a height y1 . In
either case the initial velocity, and hence initial kinetic energy, was the same; and the velocity at
the highest point was zero. Then W = ∆U , so the potential energy would have been 68 kJ greater,
and
∆y = ∆U/mg = (68×103 J)/(9.4 kg)(9.81 m/s2 ) = 740 m
is how much higher it would have gone without air friction.
E13-2 (a) The road incline is θ = arctan(0.08) = 4.57◦ . The frictional forces are the same; the car
is now moving with a vertical upward speed of (15 m/s) sin(4.57◦ ) = 1.20 m/s. The additional power
required to drive up the hill is then ∆P = mgvy = (1700 kg)(9.81 m/s2 )(1.20 m/s) = 20000 W. The
total power required is 36000 W.
(b) The car will “coast” if the power generated by rolling downhill is equal to 16000 W, or
vy = (16000 W)/[(1700 kg)(9.81 m/s2 )] = 0.959 m/s,
down. Then the incline is
θ = arcsin(0.959 m/s/15 m/s) = 3.67◦ .
This corresponds to a downward grade of tan(3.67◦ ) = 6.4 %.
E13-3 Apply energy conservation:
1
1
mv 2 + mgy + ky 2 = 0,
2
2
so
v=
p
−2(9.81 m/s2 )(−0.084 m) − (262 N/m)(−0.084 m)2 /(1.25 kg) = 0.41 m/s.
E13-4 The car climbs a vertical distance of (225 m) sin(10◦ ) = 39.1 m in coming to a stop. The
change in energy of the car is then
∆E = −
1 (16400 N)
(31.4 m/s)2 + (16400 N)(39.1 m) = −1.83×105 J.
2 (9.81 m/s2 )
E13-5 (a) Applying conservation of energy to the points where the ball was dropped and where
it entered the oil,
1
1
mv f 2 + mgy f =
mv i 2 + mgy i ,
2
2
1 2
1 2
v f + g(0) =
(0) + gy i ,
2
2
p
vf =
2gy i ,
p
=
2(9.81 m/s2 )(0.76 m) = 3.9 m/s.
(b) The change in internal energy of the ball + oil can be found by considering the points where
the ball was released and where the ball reached the bottom of the container.
∆E
= K f + U f − K i − U i,
1
1
=
mv f 2 + mgy f − m(0)2 − mgy i ,
2
2
1
=
(12.2×10−3 kg)(1.48m/s)2 − (12.2×10−3 kg)(9.81m/s2 )(−0.55m − 0.76m),
2
= −0.143 J
157
E13-6 (a) U i = (25.3 kg)(9.81 m/s2 )(12.2 m) = 3030 J.
(b) K f = 12 (25.3 kg)(5.56 m/s)2 = 391 J.
(c) ∆E int = 3030 J − 391 J = 2640 J.
E13-7 (a) At atmospheric entry the kinetic energy is
K=
1
(7.9×104 kg)(8.0×103 m/s)2 = 2.5×1012 J.
2
The gravitational potential energy is
U = (7.9×104 kg)(9.8 m/s2 )(1.6×105 m) = 1.2×1011 J.
The total energy is 2.6×1012 J.
(b) At touch down the kinetic energy is
K=
1
(7.9×104 kg)(9.8×101 m/s)2 = 3.8×108 J.
2
E13-8 ∆E/∆t = (68 kg)(9.8 m/s2 )(59 m/s) = 39000 J/s.
E13-9 Let m be the mass of the water under consideration. Then the percentage of the potential
energy “lost” which appears as kinetic energy is
Kf − Ki
.
Ui − Uf
Then
Kf − Ki
Ui − Uf
1
m v f 2 − v i 2 / (mgy i − mgy f ) ,
2
vf 2 − vi 2
=
,
−2g∆y
(13 m/s)2 − (3.2 m/s)2
=
,
−2(9.81 m/s2 )(−15 m)
= 54 %.
=
The rest of the energy would have been converted to sound and thermal energy.
E13-10 The change in energy is
∆E =
1
(524 kg)(62.6 m/s)2 − (524 kg)(9.81 m/s2 )(292 m) = 4.74×105 J.
2
E13-11 U f = K i − (34.6 J). Then
h=
1 (7.81 m/s)2
(34.6 J)
−
= 2.28 m;
2
2 (9.81 m/s ) (4.26 kg)(9.81 m/s2 )
which means the distance along the incline is (2.28 m)/ sin(33.0◦ ) = 4.19 m.
158
E13-12 (a) K f = U i − U f , so
p
v f = 2(9.81 m/s2 )[(862 m) − (741 m)] = 48.7 m/s.
That’s a quick 175 km/h; but the speed at the bottom of the valley is 40% of the speed of sound!
(b) ∆E = U f − U i , so
∆E = (54.4 kg)(9.81 m/s2 )[(862 m) − (741 m)] = −6.46×104 J;
which means the internal energy of the snow and skis increased by 6.46×104 J.
E13-13
ball, so
The final potential energy is 15% less than the initial kinetic plus potential energy of the
0.85(K i + U i ) = U f .
But U i = U f , so K i = 0.15U f /0.85, and then
r
p
0.15
vi =
2gh = 2(0.176)(9.81 m/s2 )(12.4 m) = 6.54 m/s.
0.85
E13-14 Focus on the potential energy. After the nth bounce the ball will have a potential energy
at the top of the bounce of Un = 0.9Un−1 . Since U ∝ h, one can write hn = (0.9)n h0 . Solving for n,
n = log(hn /h0 )/ log(0.9) = log(3 ft/6 ft)/ log(0.9) = 6.58,
which must be rounded up to 7.
E13-15 Let m be the mass of the ball and M be the mass of the block.
The
p kinetic energy of the ball just before colliding with the block is given by K1 = U0 , so
v1 = 2(9.81 m/s2 )(0.687 m) = 3.67 m/s.
Momentum is conserved, so if v2 and v3 are velocities of the ball and block after the collision
then mv1 = mv2 + M v3 . Kinetic energy is not conserved, instead
1 1
1
1
2
mv1 = mv22 + M v32 .
2 2
2
2
Combine the energy and momentum expressions to eliminate v3 :
m
2
mv12 = 2mv22 + 2M
(v1 − v2 ) ,
M
M v12 = 2M v22 + 2mv12 − 4mv1 v2 + 2mv22 ,
which can be formed into a quadratic. The solution for v2 is
p
2m ± 2(M 2 − mM )
v1 = (0.600 ± 1.95) m/s.
v2 =
2(M + m)
The corresponding solutions for v3 are then found from the momentum expression to be v3 =
0.981 m/s and v3 = 0.219. Since it is unlikely that the ball passed through the block we can toss
out the second set of answers.
E13-16 E f = K f + U f = 3mgh, or
p
v f = 2(9.81 m/s2 )2(0.18 m) = 2.66 m/s.
159
E13-17 We can find the kinetic energy of the center of mass of the woman when her feet leave
the ground by considering energy conservation and her highest point. Then
1
mv i 2 + mgy i
2
1
mv i
2
1
mv f 2 + mgy f ,
2
=
= mg∆y,
(55.0 kg)(9.81 m/s2 )(1.20 m − 0.90 m) = 162 J.
=
(a) During the jumping phase her potential energy changed by
∆U = mg∆y = (55.0 kg)(9.81 m/s2 )(0.50 m) = 270 J
while she was moving up. Then
F ext =
∆K + ∆U
(162 J) + (270 J)
=
= 864 N.
∆s
(0.5 m)
(b) Her fastest speed was when her feet left the ground,
v=
2K
2(162 J)
=
= 2.42 m/s.
m
(55.0 kg)
E13-18 (b) The ice skater needs to lose 12 (116 kg)(3.24 m/s)2 = 609 J of internal energy.
(a) The average force exerted on the rail is F = (609 J)/(0.340 m) = 1790 N.
E13-19 12.6 km/h is equal to 3.50 m/s; the initial kinetic energy of the car is
1
(2340 kg)(3.50 m/s)2 = 1.43×104 J.
2
(a) The force exerted on the car is F = (1.43×104 J)/(0.64 m) = 2.24×104 N.
(b) The change increase in internal energy of the car is
∆E int = (2.24×104 N)(0.640 m − 0.083 m) = 1.25×104 J.
E13-20 Note that vn2 = vn0 2 − 2~vn0 · ~vcm + v cm 2 . Then
K
=
X1
n
=
X1
n
mn vn0 2 − 2mn ~vn0 · ~vcm + mn v cm 2 ,
2
2
mn vn0 2
= K int −
−
X
mn ~vn0
!
· ~vcm +
n
X
mn ~vn0
!
X1
n
2
mn
!
v cm 2 ,
· ~vcm + K cm .
n
The middle term vanishes because of the definition of velocities relative to the center of mass.
160
E13-21 Momentum conservation requires mv0 = mv+M V, where the sign indicates the direction.
We are assuming one dimensional collisions. Energy conservation requires
1
1
1
mv 2 = mv 2 + M V 2 + E.
2 0
2
2
Combining,
1
mv 2
2 0
M v02
1
1 m
m 2
mv 2 + M
v0 −
v + E,
2
2
M
M
2
= M v 2 + m (v0 − v) + 2(M/m)E.
=
Arrange this as a quadratic in v,
(M + m) v 2 − (2mv0 ) v + 2(M/m)E + mv02 − M v02 = 0.
This will only have real solutions if the discriminant (b2 − 4ac) is greater than or equal to zero. Then
2
(2mv0 ) ≥ 4 (M + m) 2(M/m)E + mv02 − M v02
is the condition for the minimum v0 . Solving the equality condition,
4m2 v02 = 4(M + m) 2(M/m)E + (m − M )v02 ,
or M 2 v02 = 2(M + m)(M/m)E. One last rearrangement, and v0 =
p
2(M + m)E/(mM ).
P13-1 (a) The initial kinetic energy will equal the potential energy at the highest point plus the
amount of energy which is dissipated because of air drag.
mgh + f h
=
h
=
1
mv 2 ,
2 0
v02
v02
=
.
2(g + f /m)
2g(1 + f /w)
(b) The final kinetic energy when the stone lands will be equal to the initial kinetic energy minus
twice the energy dissipated on the way up, so
1
mv 2
2
=
=
=
v2
=
v
=
1
mv 2 − 2f h,
2 0
1
v02
mv02 − 2f
,
2
2g(1 + f /w)
m
f
−
v02 ,
2
g(1 + f /w)
2f
1−
v02 ,
w+f
1/2
w−f
v0 .
w+f
P13-2 The object starts with U = (0.234 kg)(9.81 m/s2 )(1.05 m) = 2.41 J. It will move back and
forth across the flat portion (2.41 J)/(0.688 J) = 3.50 times, which means it will come to a rest at
the center of the flat part while attempting one last right to left journey.
161
P13-3
(a) The work done on the block block because of friction is
(0.210)(2.41 kg)(9.81 m/s2 )(1.83 m) = 9.09 J.
The energy dissipated because of friction is (9.09 J)/0.83 = 11.0 J.
The speed of the block just after the bullet comes to a rest is
p
p
v = 2K/m = 2(1.10 J)/(2.41 kg) = 3.02 m/s.
(b) The initial speed of the bullet is
v0 =
M +m
(2.41 kg) + (0.00454 kg)
v=
(3.02 m/s) = 1.60×103 m/s.
m
(0.00454 kg)
P13-4 The energy stored in the spring after compression is 12 (193 N/m)(0.0416 m)2 = 0.167 J.
Since 117 mJ was dissipated by friction, the kinetic energy of the block before colliding with the
spring was 0.284 J. The speed of the block was then
p
v = 2(0.284 J)/(1.34 kg) = 0.651 m/s.
P13-5
(a) Using Newton’s second law, F = ma, so for circular motion around the proton
mv 2
e2
= F = k 2.
r
r
The kinetic energy of the electron in an orbit is then
K=
1
1 e2
mv 2 = k .
2
2 r
The change in kinetic energy is
1
∆K = ke2
2
1
1
−
r2
r1
.
(b) The potential energy difference is
Z r2 2
1
1
ke
2
dr
=
−ke
−
.
∆U = −
r2
r2
r1
r1
(c) The total energy change is
1
∆E = ∆K + ∆U = − ke2
2
1
1
−
r2
r1
.
P13-6 (a) The initial energy of the system is (4000 lb)(12ft) = 48, 000 ft · lb. The safety device
removes (1000 lb)(12ft) = 12, 000 ft · lb before the elevator hits the spring, so the elevator has a
kinetic energy of 36, 000 ft · lb when it hits the spring. The speed of the elevator when it hits the
spring is
s
2
v=
2(36, 000 ft · lb)(32.0 ft/s )
= 24.0 ft/s.
(4000 lb)
(b) Assuming the safety clamp remains in effect while the elevator is in contact with the spring
then the distance compressed will be found from
36, 000 ft · lb =
1
(10, 000 lb/ft)y 2 − (4000 lb)y + (1000 lb)y.
2
162
This is a quadratic expression in y which can be simplified to look like
5y 2 − 3y − 36 = 0,
which has solutions y = (0.3 ± 2.7) ft. Only y = 3.00 ft makes sense here.
(c) The distance through which the elevator will bounce back up is found from
33, 000 ft = (4000 lb)y − (1000 lb)y,
where y is measured from the most compressed point of the spring. Then y = 11 ft, or the elevator
bounces back up 8 feet.
(d) The elevator will bounce until it has traveled a total distance so that the safety device
dissipates all of the original energy, or 48 ft.
P13-7
The net force on the top block while it is being pulled is
11.0 N − Ff = 11.0 N − (0.35)(2.5 kg)(9.81 m/s2 ) = 2.42 N.
Thispmeans it is accelerating at (2.42 N)/(2.5 kg) = 0.968 m/s2 . That acceleration will last a time
t = 2(0.30 m)/(0.968 m/s2 ) = 0.787 s. The speed of the top block after the force stops pulling is
then (0.968 m/s2 )(0.787 s) = 0.762 m/s. The force on the bottom block is Ff , so the acceleration of
the bottom block is
(0.35)(2.5 kg)(9.81 m/s2 )/(10.0 kg) = 0.858 m/s2 ,
and the speed after the force stops pulling on the top block is (0.858 m/s2 )(0.787 s) = 0.675 m/s.
(a) W = F s = (11.0 N)(0.30 m) = 3.3 J of energy were delivered to the system, but after the
force stops pulling only
1
1
(2.5 kg)(0.762 m/s)2 + (10.0 kg)(0.675 m/s)2 = 3.004 J
2
2
were present as kinetic energy. So 0.296 J is “missing” and would be now present as internal energy.
(b) The impulse received by the two block system is then J = (11.0 N)(0.787 s) = 8.66 N·s. This
impulse causes a change in momentum, so the speed of the two block system after the external
force stops pulling and both blocks move as one is (8.66 N·s)(12.5 kg) = 0.693 m/s. The final kinetic
energy is
1
(12.5 kg)(0.693 m/s)2 = 3.002 J;
2
this means that 0.002 J are dissipated.
P13-8
Hmm.
163
E14-1 F S /F E = M S rE 2 /M E rS 2 , since everything else cancels out in the expression. Then
FS
(1.99×1030 kg)(3.84×108 m)2
=
= 2.18
FE
(5.98×1024 )(1.50×1011 m)2
E14-2 Consider the force from the Sun and the force from the Earth. F S /F E = M S rE 2 /M E rS 2 ,
since everything else cancels out in the expression. We want the ratio to be one; we are also
constrained because rE + rS = R is the distance from the Sun to the Earth. Then
2
M E (R − rE )
R − rE
rE
= M S rE 2 ,
r
MS
=
rE ,
ME
=
11
(1.50×10 m)/ 1 +
s
(1.99×1030 kg)
(5.98×1024 )
!
= 2.6×108 m.
E14-3 The masses of each object are m1 = 20.0 kg and m2 = 7.0 kg; the distance between the
centers of the two objects is 15 + 3 = 18 m.
The magnitude of the force from Newton’s law of gravitation is then
F =
(6.67×10−11 N · m2 /kg2 )(20.0 kg)(7.0 kg)
Gm1 m2
=
= 2.9×10−11 N.
r2
(18 m)2
E14-4 (a) The magnitude of the force from Newton’s law of gravitation is
F =
(6.67×10−11 N · m2 /kg2 )(12.7 kg)(9.85×10−3 kg)
Gm1 m2
=
= 7.15×10−10 N.
r2
(0.108 m)2
(b) The torque is τ = 2(0.262 m)(7.15×10−10 N) = 3.75×10−10 N · m.
E14-5
The force of gravity on an object near the surface of the earth is given by
F =
GM m
,
(re + y)2
where M is the mass of the Earth, m is the mass of the object, re is the radius of the Earth, and
y is the height above the surface of the Earth. Expand the expression since y re . We’ll use a
Taylor expansion, where F (re + y) ≈ F (re ) + y∂F/∂re ;
F ≈
GM m
GM m
− 2y
2
re
re3
Since we are interested in the difference between the force at the top and the bottom, we really want
∆F = 2y
GM m
y GM m
y
=2
= 2 W,
re3
re re2
re
where in the last part we substituted for the weight, which is the same as the force of gravity,
W =
GM m
.
re2
Finally,
∆F = 2(411 m)/(6.37×106 m)(120 lb) = 0.015 lb.
164
E14-6 g ∝ 1/r2 , so g1 /g2 = r22 /r12 . Then
p
r2 = (9.81 m/s2 )/(7.35 m/s2 )(6.37×106 m) = 7.36×106 m.
That’s 990 kilometers above the surface of the Earth.
E14-7 (a) a = GM/r2 , or
a=
(b) v =
√
2ax =
(6.67×10−11 N · m2 /kg2 )(1.99×1030 kg)
= 1.33×1012 m/s2 .
(10.0×103 m)2
p
2(1.33×1012 m/s2 )(1.2 m/s) = 1.79×106 m/s.
E14-8 (a) g0 = GM/r2 , or
g0 =
(6.67×10−11 N · m2 /kg2 )(7.36×1022 kg)
= 1.62 m/s2 .
(1.74×106 m)2
(b) W m = W e (g m /g e ) so
W m = (100 N)(1.62 m/s2 /9.81 m/s2 ) = 16.5 N.
(c) Invert g = GM/r2 ;
q
p
r = GM/g = (6.67×10−11 N · m2 /kg2 )(5.98×1024 kg)/(1.62 m/s2 ) = 1.57×107 m.
That’s 2.46 Earth radii, or 1.46 Earth radii above the surface of the Earth.
E14-9
The object fell through y = −10.0 m; the time required to fall would then be
t=
p
p
−2y/g = −2(−10.0 m)/(9.81 m/s2 ) = 1.43 s.
We are interested in the error, that means taking the total derivative of y = − 12 gt2 . and getting
1
δy = − δg t2 − gt δt.
2
δy = 0 so − 12 δg t = gδ t, which can be rearranged as
δt = −
δg t
2g
The percentage error in t needs to be δt/t = 0.1 %/2 = 0.05 %. The absolute error is then δt =
(0.05 %)(1.43 s) = 0.7 ms.
E14-10 Treat mass which is inside a spherical shell as being located at the center of that shell.
Ignore any contributions from shells farther away from the center than the point in question.
(a) F = G(M1 + M2 )m/a2 .
(b) F = G(M1 )m/b2 .
(c) F = 0.
165
E14-11
For a sphere of uniform density and radius R > r,
M (r)
=
4
3
3 πr
M
,
4
3
3 πR
where M is the total mass.
The force of gravity on the object of mass m is then
F =
GM m r3
GM mr
=
.
2
3
r
R
R3
g is the free-fall acceleration of the object, and is the gravitational force divided by the mass, so
g=
GM r
GM r
GM R − D
= 2
= 2
.
R3
R R
R
R
Since R is the distance from the center to the surface, and D is the distance of the object beneath
the surface, then r = R − D is the distance from the center to the object. The first fraction is the
free-fall acceleration on the surface, so
GM R − D
R−D
D
g= 2
= gs
= gs 1 −
R
R
R
R
E14-12 The work required to move the object is GM S m/r, where r is the gravitational radius.
But if this equals mc2 we can write
mc2
r
= GM S m/r,
= GM S /c2 .
For the sun, r = (6.67 × 10−11 N · m2 /kg2 )(1.99 × 1030 kg)/(3.00 × 108 m/s)2 = 1.47 × 103 m. That’s
2.1×10−6 RS .
E14-13 The distance from the center is
r = (80000)(3.00×108 m/s)(3.16×107 s) = 7.6×1020 m.
The mass of the galaxy is
M = (1.4×1011 )(1.99×1030 kg) = 2.8×1041 kg.
The escape velocity is
v=
p
2GM/r =
q
2(6.67×10−11 N · m2 /kg2 )(2.8×1041 kg)/(7.6×1020 m) = 2.2×105 m/s.
E14-14 Staying in a circular orbit requires the centripetal force be equal to the gravitational force,
so
mv orb 2 /r = GM m/r2 ,
or mv orb 2 = GM m/r. But −GM m/r is the gravitational potential energy; to escape one requires a
kinetic energy
mv esc 2 /2 = GM m/r = mv orb 2 ,
√
which has solution v esc = 2v orb .
166
E14-15
(a) Near the surface of the Earth the total energy is
E =K +U =
2 GM m
1 p
E
m 2 gRE −
2
RE
but
g=
GM
,
RE 2
so the total energy is
E
GM E m
2mgRE −
,
R
E
GM
GM E m
= 2m
RE −
,
RE 2
RE
GM E m
=
RE
=
This is a positive number, so the rocket will escape.
(b) Far from earth there is no gravitational potential energy, so
1
GM E m
GM E
mv 2 =
=
mRE = gmRE,
2
RE
RE 2
with solution v =
√
2gRE .
E14-16 The rotational acceleration of the sun is related to the galactic acceleration of free fall by
4π 2 mr/T 2 = GN m2 /r2 ,
where N is the number of “sun” sized stars of mass m, r is the size of the galaxy, T is period of
revolution of the sun. Then
N=
E14-17
4π 2 r3
4π 2 (2.2×1020 m)3
=
= 5.1×1010 .
GmT 2
(6.67×10−11 N·m2 /kg2 )(2.0×1030 kg)(7.9×1015 s)2
Energy conservation is K i + U i = K f + U f , but at the highest point K f = 0, so
Uf
GM E m
−
R
1
R
1
R
R
= K i + U i,
1
GM E m
=
mv02 −
,
2
RE
1
1
=
−
v2 ,
RE
2GM E 0
1
(9.42×103 m/s)2 )
,
=
−
(6.37×106 m) 2(6.67×10−11 N·m2 /kg2 )(5.98×1024 kg)
=
2.19×107 m.
The distance above the Earth’s surface is 2.19×107 m − 6.37×106 m = 1.55×106 m.
167
E14-18 (a) Free-fall acceleration is g = GM/r2 . Escape speed is v =
p
2(1.30 m/s2 )(1.569×106 m) = 2.02×103 m/s.
(b) U f = K i + U i . But U/m = −g0 r02 /r, so
p
√
2GM/r. Then v = 2gr =
1
(1.01×103 m/s)2
1
1
=
−
=
.
6
2
6
2
rf
(1.569×10 m) 2(1.30 m/s )(1.569×10 m)
2.09×106 m
That’s 523 km above the surface.
(c) K f = U i − U f . But U/m = −g0 r02 /r, so
p
v = 2(1.30 m/s2 )(1.569×106 m)2 [1/(1.569×106 m) − 1/(2.569×106 m)] = 1260 m/s.
(d) M = gr2 /G, or
M = (1.30 m/s2 )(1.569×106 m)2 /(6.67×10−11 N·m2 /kg2 ) = 4.8×1022 kg.
E14-19 (a) Apply ∆K = −∆U . Then mv 2 = Gm2 (1/r2 − 1/r1 ), so
s
1
1
−
= 3.34×107 m/s.
v = (6.67×10−11 N·m2 /kg2 )(1.56×1030 kg)
(4.67×104 m) (9.34×104 m)
(b) Apply ∆K = −∆U . Then mv 2 = Gm2 (1/r2 − 1/r1 ), so
s
1
1
2
−11
2
30
v = (6.67×10 N·m /kg )(1.56×10 kg)
−
= 5.49×107 m/s.
(1.26×104 m) (9.34×104 m)
E14-20 Call the particles 1 and 2. Then conservation of momentum requires the particle to have
the same momentum of the same magnitude, p = mv1 = M v2 . The momentum of the particles is
given by
1 2
1 2
p +
p
2m
2M
m+M 2
p
mM
p
=
GM m
,
d
=
2GM m/d,
p
= mM 2G/d(m + M ).
Then v rel = |v1 | + |v2 | is equal to
v rel
1
1
+
,
m M
p
m+M
= mM 2G/d(m + M )
,
mM
p
=
2G(m + M )/d
p
= mM 2G/d(m + M )
E14-21 The maximum speed is mv 2 = Gm2 /d, or v =
p
Gm/d.
E14-22 T12 /r13 = T22 /r23 , or
T2 = T1 (r2 /r1 )3/2 = (1.00 y)(1.52)3/2 = 1.87 y.
168
E14-23
for M —
We can use Eq. 14-23 to find the mass of Mars; all we need to do is rearrange to solve
M=
4π 2 r3
4π 2 (9.4×106 m)3
=
= 6.5×1023 kg.
GT 2
(6.67×10−11 N·m2 /kg2 )(2.75×104 s)2
E14-24 Use GM/r2 = 4π 2 r/T 2 , so M = 4π 2 r3 /GT 2 , and
M=
4π 2 (3.82×108 m)3
= 5.93×1024 kg.
(6.67×10−11 N·m2 /kg2 )(27.3 × 86400 s)2
E14-25 T12 /r13 = T22 /r23 , or
T2 = T1 (r2 /r1 )3/2 = (1.00 month)(1/2)3/2 = 0.354 month.
E14-26 Geosynchronous orbit was found in Sample Problem 14-8 to be 4.22×107 m. The latitude
is given by
θ = arccos(6.37×106 m/4.22×107 m) = 81.3◦ .
E14-27 (b) Make the assumption that the altitude of the satellite is so low that the radius of
the orbit is effectively the radius of the moon. Then
2
4π
T2 =
r3 ,
GM
4π 2
=
(1.74×106 m)3 = 4.24×107 s2 .
(6.67×10−11 N·m2 /kg2 )(7.36×1022 kg)
So T = 6.5×103 s.
(a) The speed of the satellite is the circumference divided by the period, or
v=
2πr
2π(1.74×106 m)
=
= 1.68×103 m/s.
T
(6.5×103 s)
E14-28 The total energy is −GM m/2a. Then
so
1
GM m
GM m
mv 2 −
=−
,
2
r
2a
2 1
2
v = GM
−
.
r
a
E14-29 ra = a(1 + e), so from Ex. 14-28,
s
va =
GM
2
1
−
;
a(1 + e) a
2
1
−
;
a(1 − e) a
rp = a(1 − e), so from Ex. 14-28,
vp =
s
GM
Dividing one expression by the other,
s
s
2/(1 − e) − 1
2/0.12 − 1
vp = va
= (3.72 km/s)
= 58.3 km/s.
2/(1 + e) − 1
2/1.88 − 1
169
E14-30 (a) Convert.
G=
6.67×10−11
m3
kg · s2
1.99×1030 kg
MS
3.156×107 s
y
2 AU
1.496×1011 m
3
,
which is G = 39.49 AU3 /M S 2 · y2 .
(b) Here is a hint: 4π 2 = 39.48. Kepler’s law then looks like
M S 2 · y2 r3
2
T =
.
M
AU3
E14-31 Kepler’s third law states T 2 ∝ r3 , where r is the mean distance from the Sun and T is
the period of revolution. Newton was in a position to find the acceleration of the Moon toward the
Earth by assuming the Moon moved in a circular orbit, since ac = v 2 /r = 4π 2 r/T 2 . But this means
that, because of Kepler’s law, ac ∝ r/T 2 ∝ 1/r2 .
E14-32 (a) The force of attraction between the two bodies is
F =
GM m
.
(r + R)2
The centripetal acceleration for the body of mass m is
rω 2
=
ω2
=
T2
=
GM
,
(r + R)2
GM
,
r3 (1 + R/r)2
4π 2 3
r (1 + R/r)2 .
GM
(b) Note that r = M d/(m + M ) and R = md/(m + M ). Then R/r = m/M , so the correction is
(1 + 5.94×1024 /1.99×1030 )2 = 1.000006 for the Earth/Sun system and 1.025 for the Earth/Moon
system.
E14-33 (a) Use the results of Exercise 14-32. The center of mass is located a distance r =
2md/(m + 2m) = 2d/3 from the star of mass m and a distance R = d/3 from the star of mass 2m.
The period of revolution is then given by
4π 2
T =
G(2m)
2
2
d
3
3 d/3
1+
2d/3
2
=
4π 2 3
d .
3Gm
(b) Use Lm = mr2 ω, then
Lm
mr2
m(2d/3)2
=
=
= 2.
LM
M R2
(2m)(d/3)2
(c) Use K = Iω 2 /2 = mr2 ω 2 /2. Then
Km
mr2
m(2d/3)2
=
=
= 2.
2
KM
MR
(2m)(d/3)2
170
E14-34 Since we don’t know which direction the orbit will be, we will assume that the satellite
on the surface of the Earth starts with zero kinetic energy. Then E i = U i .
∆U = U f − U i to get the satellite up to the specified altitude. ∆K = K f = −U f /2. We want
to know if ∆U − ∆K is positive (more energy to get it up) or negative (more energy to put it in
orbit). Then we are interested in
3
1
−
.
∆U − ∆K = 3U f /2 − U i = GM m
ri
2rf
The “break-even” point is when rf = 3ri /2 = 3(6400 km)/2 = 9600 km, which is 3200 km above the
Earth.
(a) More energy to put it in orbit.
(b) Same energy for both.
(c) More energy to get it up.
E14-35
(a) The approximate force of gravity on a 2000 kg pickup truck on Eros will be
F =
GM m
(6.67×10−11 N·m2 /kg2 )(5.0 × 1015 kg)(2000 kg)
=
= 13.6 N.
r2
(7000 m)2
(b) Use
v=
r
GM
=
r
s
(6.67×10−11 N·m2 /kg2 )(5.0 × 1015 kg)
= 6.9 m/s.
(7000 m)
E14-36 (a) U = −GM m/r. The variation is then
∆U
−11
=
(6.67×10
=
1.78×1032 J.
2
2
30
24
N·m /kg )(1.99×10 kg)(5.98×10 kg)
1
1
−
11
(1.47×10 m) (1.52×1011 m)
(b) ∆K + ∆U = ∆E = 0, so |∆K| = 1.78×1032 J.
(c) ∆E = 0.
(d) Since ∆l = 0 and l = mvr, we have
(1.47×1011 m)
rp
vp − va = vp 1 −
= vp 1 −
= 3.29×10−2 v p .
ra
(1.52×1011 m)
But v p ≈ v av = 2π(1.5×1011 m)/(3.16×107 s) = 2.98×104 m/s. Then ∆v = 981 m/s.
E14-37 Draw a triangle. The angle made by Chicago, Earth center, satellite is 47.5◦ . The distance
from Earth center to satellite is 4.22×107 m. The distance from Earth center to Chicago is 6.37×106 m.
Applying the cosine law we find the distance from Chicago to the satellite is
p
(4.22×107 m)2 + (6.37×106 m)2 − 2(4.22×107 m)(6.37×106 m) cos(47.5◦ ) = 3.82×107 m.
Applying the sine law we find the angle made by Earth center, Chicago, satellite to be
(4.22×107 m)
◦
arcsin
sin(47.5
)
= 126◦ .
(3.82×107 m)
That’s 36◦ above the horizontal.
171
E14-38 (a) The new orbit is an ellipse with eccentricity given by r = a0 (1 + e). Then
e = r/a0 − 1 = (6.64×106 m)/(6.52×106 m) − 1 = 0.0184.
The distance at P 0 is given by rP 0 = a0 (1 − e). The potential energy at P 0 is
UP 0 = UP
1 + 0.0184
1+e
= 2(−9.76×1010 J)
= −2.03×1011 J.
1−e
1 − 0.0184
The kinetic energy at P 0 is then
KP 0 = (−9.94×1010 J) − (−2.03×1011 J) = 1.04×1011 J.
p
That would mean v = 2(1.04×1011 J)/(3250kg) = 8000 m/s.
(b) The average speed is
2π(6.52×106 m)
v=
= 7820 m/s.
(5240 s)
E14-39 (a) The Starshine satellite was approximately 275 km above the surface of the Earth on
1 January 2000. We can find the orbital period from Eq. 14-23,
2
4π
T2 =
r3 ,
GM
4π 2
=
(6.65×106 m)3 = 2.91×107 s2 ,
(6.67×10−11 N·m2 /kg2 )(5.98×1024 kg)
so T = 5.39×103 s.
(b) Equation 14-25 gives the total energy of the system of a satellite of mass m in a circular orbit
of radius r around a stationary body of mass M m,
E=−
GM m
.
2r
We want the rate of change of this with respect to time, so
dE
GM m dr
=
dt
2r2 dt
We can estimate the value of dr/dt from the diagram. I’ll choose February 1 and December 1 as my
two reference points.
dr
∆r
(240 km) − (300 km)
≈
=
≈ −1 km/day
dt t=t0
∆t
(62 days)
The rate of energy loss is then
dE
(6.67×10−11 N·m2 /kg2 )(5.98×1024 kg)(39 kg) −1000 m
= −2.0 J/s.
=
dt
2(6.65×106 m)2
8.64×104 s
P14-1 The object on the top experiences a force down from gravity W1 and a force down from
the tension in the rope T . The object on the bottom experiences a force down from gravity W2 and
a force up from the tension in the rope.
In either case, the magnitude of Wi is
Wi =
GM m
ri2
172
where ri is the distance of the ith object from the center of the Earth. While the objects fall they
have the same acceleration, and since they have the same mass we can quickly write
GM m
GM m
+T =
− T,
2
r1
r22
or
T
=
=
=
GM m GM m
−
,
2r22
2r12
1
GM m 1
−
,
2
r12
r22
GM m r22 − r12
.
2
r12 r22
Now r1 ≈ r2 ≈ R in the denominator, but r2 = r1 + l, so r22 − r12 ≈ 2Rl in the numerator. Then
T ≈
GM ml
.
R3
P14-2 For a planet of uniform density, g = GM/r2 = G(4πρr3 /3)/r2 = 4πGρr/3. Then if ρ is
doubled while r is halved we find that g will be unchanged.
P14-3 (a) F = GM m/r2 , a = F/m = GM/r.
(b) The acceleration of the Earth toward the center of mass is aE = F/M = Gm/r2 . The
relative acceleration is then GM/r + Gm/r = G(m + M )/r. Only if M m can we assume that a
is independent of m relative to the Earth.
P14-4
(a) g = GM/r2 , δg = −(2GM/r3 )δr. In this case δr = h and M = 4πρr3 /3. Then
δW = m δg = 8πGρmh/3.
(b) ∆W/W = ∆g/g = 2h/r. Then an error of one part in a million will occur when h is one
part in two million of r, or 3.2 meters.
P14-5
(a) The magnitude of the gravitational force from the Moon on a particle at A is
FA =
GM m
,
(r − R)2
where the denominator is the distance from the center of the moon to point A.
(b) At the center of the Earth the gravitational force of the moon on a particle of mass m is
FC = GM m/r2 .
(c) Now we want to know the difference between these two expressions:
FA − FC
GM m
GM m
−
,
2
(r − R)
r2
r2
(r − R)2
= GM m
−
,
r2 (r − R)2
r2 (r − R)2
2
r − (r − R)2
= GM m
,
r2 (r − R)2
R(2r − R)
.
= GM m
r2 (r − R)2
=
173
To simplify assume R r and then substitute (r − R) ≈ r. The force difference simplifies to
FT = GM m
R(2r)
2GM mR
=
r2 (r)2
r3
(d) Repeat part (c) except we want r + R instead of r − R. Then
FA − FC
GM m
GM m
−
,
2
(r + R)
r2
(r + R)2
r2
− 2
,
= GM m
r2 (r + R)2
r (r + R)2
2
r − (r + R)2
= GM m
,
r2 (r + R)2
−R(2r + R)
= GM m
.
r2 (r + R)2
=
To simplify assume R r and then substitute (r + R) ≈ r. The force difference simplifies to
FT = GM m
−R(2r)
2GM mR
=−
r2 (r)2
r3
The negative sign indicates that this “apparent” force points away from the moon, not toward it.
(e) Consider the directions: the water is effectively attracted to the moon when closer, but
repelled when farther.
P14-6 F net = mrω s 2 , where ω s is the rotational speed of the ship. But since the ship is moving
relative to the earth with a speed v, we can write ω s = ω ± v/r, where the sign is positive if the ship
is sailing east. Then F net = mr(ω ± v/r)2 .
The scale measures a force W which is given by mg − F net , or
W = mg − mr(ω ± v/r)2 .
Note that W0 = m(g − rω 2 ). Then
W
g − r(ω ± v/r)2
,
g − rω 2
2ωv
≈ W0 1 ±
,
1 − rω 2
≈ W0 (1 ± 2vω/g).
= W0
P14-7 (a) a = GM/r2 − rω 2 . ω is the rotational speed of the Earth. Since Frank observes a = g/2
we have
g/2 = GM/r2 − rω 2 ,
r2 = (2GM − 2r3 ω 2 )/g,
p
r =
2(GM − r3 ω 2 )/g
Note that
GM = (6.67×10−11 N · m2 /kg2 )(5.98×1024 kg) = 3.99×1014 m3 /s2
while
r3 ω 2 = (6.37×106 m)3 (2π/86400 s)2 = 1.37×1012 m3 /s2 .
174
Consequently, r3 ω 2 can be treated as a perturbation of GM bear the Earth. Solving iteratively,
p
r0 =
2[(3.99×1014 m3 /s2 ) − (6.37×106 m)3 (2π/86400 s)2 ]/(9.81 m/s2 ) = 9.00×106 m,
p
r1 =
2[(3.99×1014 m3 /s2 ) − (9.00×106 m)3 (2π/86400 s)2 ]/(9.81 m/s2 ) = 8.98×106 m,
which is close enough for me. Then h = 8.98×106 m − 6.37×106 m = 2610 km.
(b) ∆E = E f − E i = U f /2 − U i . Then
1
1
2
−11
2
24
∆E = (6.67×10 N·m /kg )(5.98×10 kg)(100 kg)
−
= 4.0×109 J.
(6.37×106 m) 2(8.98×106 m)
P14-8
(a) Equate centripetal force with the force of gravity.
4π 2 mr
T2
4π 2
T2
GM m
,
r2
G(4/3)πr3 ρ
=
,
3
r r
3π
=
Gρ
=
T
(b) T =
q
3π/(6.7×10−11 N · m2 /kg2 )(3.0×103 kg/m3 ) = 6800 s.
P14-9 (a) One can find δg by pretending the Earth is not there, but the material in the hole is.
Concentrate on the vertical component of the resulting force of attraction. Then
δg =
GM d
,
r2 r
where r is the straight line distance from the prospector to the center of the hole and M is the mass
of material that would fill the hole. A few substitutions later,
δg =
4πGρR3 d
√
.
3( d2 + x2 )3
(b) Directly above the hole x = 0, so a ratio of the two readings gives
(10.0 milligals)
=
(14.0 milligals)
d2
d2 + (150 m)2
3/2
or
(0.800)(d2 + 2.25×104 m2 ) = d2 ,
which has solution d = 300 m. Then
R3 =
3(14.0×10−5 m/s2 )(300 m)2
,
4π(6.7×10−11 N · m2 /kg2 )(2800 kg/m3 )
so R = 250 m. The top of the cave is then 300 m − 250 m = 50 m beneath the surface.
(b) All of the formulae stay the same except replace ρ with the difference between rock and water.
d doesn’t change, but R will now be given by
R3 =
3(14.0×10−5 m/s2 )(300 m)2
,
4π(6.7×10−11 N · m2 /kg2 )(1800 kg/m3 )
so R = 292 m, and then the cave is located 300 m − 292 m = 7 m beneath the surface.
175
P14-10 g = GM/r2 , where M is the mass enclosed in within the sphere of radius r. Then
dg = (G/r2 )dM − 2(GM/r3 )dr, so that g is locally constant if dM/dr = 2M/r. Expanding,
4πr2 ρl
ρl
= 8πr2 ρ/3,
= 2ρ/3.
P14-11 The force of gravity on the small sphere of mass m is equal to the force of gravity from
a solid lead sphere minus the force which would have been contributed by the smaller lead sphere
which would have filled the hole. So we need to know about the size and mass of the lead which was
removed to make the hole.
The density of the lead is given by
M
ρ= 4 3
3 πR
The hole has a radius of R/2, so if the density is constant the mass of the hole will be
3
M
R
M
4
Mh = ρV = 4 3
π
=
3
2
8
3 πR
The “hole” is closer to the small sphere; the center of the hole is d − R/2 away. The force of the
whole lead sphere minus the force of the “hole” lead sphere is
G(M/8)m
GM m
−
d2
(d − R/2)2
P14-12
p
√
(a) Use v = ω R2 − r2 , where ω = GME /R3 . Then
Z T
Z 0
Z 0
dr
dr
T =
dt =
=
,
dr/dt
0
R
R v
Z 0
dr
√
=
,
2
2
R ω R −r
π
=
2ω
Knowing that
ω=
s
(6.67×10−11 N · m2 /kg2 )(5.98×1024 kg)
= 1.24×10−3 /s,
(6.37×106 m)3
we can find T = 1260 s = 21 min.
(b) Same time, 21 minutes. To do a complete journey would require four times this, or 2π/ω.
That’s 84 minutes!
(c) The answers are the same.
P14-13
(a) g = GM/r2 and M = 1.93×1024 kg + 4.01×1024 kg + 3.94×1022 kg = 5.98×1024 kg so
g = (6.67×10−11 N · m2 /kg2 )(5.98×1024 kg)/(6.37×106 m)2 = 9.83 m/s2 .
(b) Now M = 1.93×1024 kg + 4.01×1024 kg = 5.94×1024 kg so
g = (6.67×10−11 N · m2 /kg2 )(5.94×1024 kg)/(6.345×106 m)2 = 9.84 m/s2 .
(c) For a uniform body, g = 4πGρr/3 = GM r/R3 , so
g = (6.67×10−11 N · m2 /kg2 )(5.98×1024 kg)(6.345×106 m)/(6.37×106 m)3 = 9.79 m/s2 .
176
P14-14
(a) Use g = GM/r2 , then
g = (6.67×10−11 N · m2 /kg2 )(1.93×1024 kg)/(3.490×106 m)2 = 106 m/s2 .
The variation with depth is linear if core has uniform density.
(b) In the mantle we have g = G(M c + M )/r2 , where M is the amount of the mass of the mantle
which is enclosed in the sphere of radius r. The density of the core is
ρc =
3(1.93×1024 kg)
= 1.084×104 kg/m3 .
4π(3.490×106 m)3
The density of the mantle is harder to find,
ρc =
3(4.01×1024 kg)
= 4.496×103 kg/m3 .
4π[(6.345×106 m)3 − (3.490×106 m)3
We can pretend that the core is made up of a point mass at the center and the rest has a density
equal to that of the mantle. That point mass would be
Mp =
4π(3.490×106 m)3 (1.084×104 kg/m3 − 4.496×103 kg/m3 )
= 1.130×1024 kg.
3
Then
g = GM p /r2 + 4πGρm r/3.
Find dg/dr, and set equal to zero. This happens when
2M p /r3 = 4πρm /3,
or r = 4.93×106 m. Then g = 9.29 m/s2 . Since this is less than the value at the end points it must
be a minimum.
P14-15 (a) We will use part of the hint, but we will integrate instead of assuming the bit about
g av ; doing it this way will become important for later chapters. Consider a small horizontal slice of
the column of thickness dr. The weight of the material above the slice exerts a force F (r) on the
top of the slice; there is a force of gravity on the slice given by
dF =
GM (r) dm
,
r2
where M (r) is the mass contained in the sphere of radius r,
M (r) =
4 3
πr ρ.
3
Lastly, the mass of the slice dm is related to the thickness and cross sectional area by dm = ρA dr.
Then
4πGAρ2
dF =
r dr.
3
Integrate both sides of this expression. On the left the limits are 0 to F center , on the right the limits
are R to 0; we need to throw in an extra negative sign because the force increases as r decreases.
Then
2
F = πGAρ2 R2 .
3
Divide both sides by A to get the compressive stress.
177
(b) Put in the numbers!
S=
2
π(6.67×10−11 N · m2 /kg2 )(4000 kg/m3 )2 (3.0×105 m)2 = 2.0×108 N/m2 .
3
(c) Rearrange, and then put in numbers;
s
3(4.0×107 N/m2 )
R=
= 1.8×105 m.
2π(6.67×10−11 N · m2 /kg2 )(3000 kg/m3 )2
P14-16 The two mass increments each exert a vertical and a horizontal force on the particle, but
the horizontal components will cancel. The vertical component is proportional to the sine of the
angle, so that
2Gm dm y
2Gmλ dx y
dF =
=
,
r2
r
r2
r
where r2 = x2 + y 2 . We will eventually integrate from 0 to ∞, so
Z ∞
2Gmλ dx y
F =
,
r2
r
0
Z ∞
dx
= 2Gmλy
,
(x2 + y 2 )3/2
0
2Gmλ
.
=
y
P14-17 For any arbitrary point P the cross sectional area which is perpendicular to the axis
dA0 = r2 dΩ is not equal to the projection dA onto the surface of the sphere. It depends on the angle
that the axis makes with the normal, according to dA0 = cos θdA. Fortunately, the angle made at
point 1 is identical to the angle made at point 2, so we can write
dΩ1
dA1 /r12
= dΩ2 ,
= dA2 /r22
But the mass of the shell contained in dA is proportional to dA, so
r12 dm1
Gm dm1 /r12
= r22 dm2 ,
= Gm dm2 /r22 .
Consequently, theR force on an object at point P is balanced by both cones.
(b) Evaluate dΩ for the top and bottom halves of the sphere. Since every dΩ on the top is
balanced by one on the bottom, the net force is zero.
P14-18
P14-19
W
Assume that the small sphere is always between the two spheres. Then
=
∆U1 + ∆U2 ,
=
−11
(6.67×10
=
9.85×10−11 J.
1
1
N · m /kg )(0.212 kg) [(7.16 kg) − (2.53 kg)]
−
,
(0.420 m) (1.14 m)
2
2
178
P14-20 Note that 21 mv esc 2 = −U0 , where U0 is the potential energy at the burn-out height.
Energy conservation gives
K
1
mv 2
2
v
= K0 + U0 ,
1
1
=
mv02 − mv esc 2 ,
2
2
q
=
v02 − v esc 2 .
P14-21 (a) The force of one star on the other is given by F = Gm2 /d2 , where d is the distance
between the stars. The stars revolve around the center of mass, which is halfway between the stars
so r = d/2 is the radius of the orbit of the stars. If a is the centripetal acceleration of the stars, the
period of revolution is then
r
r
r
4π 2 r
4mπ 2 r
16π 2 r3
T =
=
=
.
a
F
Gm
The numerical value is
s
T =
16π 2 (1.12×1011 m)3
= 3.21×107 s = 1.02 y.
(6.67×10−11 N · m2 /kg2 )(3.22×1030 kg)
(b) The gravitational potential energy per kilogram midway between the stars is
−2
Gm
(6.67×10−11 N · m2 /kg2 )(3.22×1030 kg)
= −2
= −3.84×109 J/kg.
r
(1.12×1011 m)
An object of mass M at the center between the stars would need (3.84×109 J/kg)M kinetic energy
to escape, this corresponds to a speed of
p
p
v = 2K/M = 2(3.84×109 J/kg) = 8.76×104 m/s.
P14-22 (a) Each differential mass segment on the ring contributes the same amount to the force
on the particle,
Gm dm x
dF =
,
r2 r
where r2 = x2 + R2 . Since the differential mass segments are all equal distance, the integration is
trivial, and the net force is
GM mx
F = 2
.
(x + R2 )3/2
(b) The potential energy can be found by integrating with respect to x,
Z ∞
Z ∞
GM mx
GM m
∆U =
F dx =
dx =
.
2
2
3/2
R
(x + R )
0
0
Then
the particle of mass m will pass through the center of the ring with a speed v =
p
2GM/R.
179
p
2∆U/m =
P14-23
(a) Consider the following diagram.
R
r
θ
R
F
R
α
F
The distance r is given by the cosine law to be
r2 = R2 + R2 − 2R2 cos θ = 2R2 (1 − cos θ).
The force between two particles is then F = Gm2 /r2 . Each particle has a symmetric partner, so
only the force component directed toward the center contributes. If we call this the R component
we have
FR = F cos α = F cos(90◦ − θ/2) = F sin(θ/2).
Combining,
Gm2 sin(θ/2)
.
2R2 1 − cos θ
But each of the other particles contributes to this force, so
FR =
F net =
Gm2 X sin(θi /2)
2R2 i 1 − cos θi
When there are only 9 particles the angles are in steps of 40◦ ; the θi are then 40◦ , 80◦ , 120◦ , 160◦ ,
200◦ , 240◦ , 280◦ , and 320◦ . With a little patience you will find
X sin(θi /2)
= 6.649655,
1 − cos θi
i
using these angles. Then F net = 3.32Gm2 /R2 .
(b) The rotational period of the ring would need to be
r
r
r
4π 2 R
4mπ 2 R
16π 2 R3
T =
=
=
.
a
F
3.32Gm
P14-24 The potential energy of the system is U = −Gm2 /r. The kinetic energy is mv 2 . The total
energy is E = −Gm2 /d. Then
p
dr
= 2 Gm(1/r − 1/d),
dt
180
so the time to come together is
T =
Z
d
0
dr
p
=
2 Gm(1/r − 1/d)
r
d3
4Gm
Z
1
r
0
x
π
dx =
1−x
4
r
d3
.
Gm
P14-25 (a) E = U/2 for each satellite, so the total mechanical energy is −GM m/r.
(b) Now there is no K, so the total mechanical energy is simply U = −2GM m/r. The factor of
2 is because there are two satellites.
(c) The wreckage falls vertically onto the Earth.
P14-26
is
Let ra = a(1 + e) and rp = a(1 − e). Then ra + re = 2a and ra − rp = 2ae. So the answer
2(0.0167)(1.50×1011 m) = 5.01×109 m,
or 7.20 solar radii.
P14-27
P14-28
The net force on an orbiting star is
F = Gm
M
2
+
m4r
.
r2
This is the centripetal force, and is equal to 4π 2 mr/T 2 . Combining,
4π 2
G
= 3 (4M + m),
T2
4r
so T = 4π
p
r3 /[G(4M + m)].
P14-29
p
(a) v = GM/r, so
q
v = (6.67×10−11 N · m2 /kg2 )(5.98×1024 kg)/(7.01×106 m) = 7.54×103 m/s.
(b) T = 2π(7.01×106 m)/(7.54×103 m/s) = 5.84×103 s.
(c) Originally E0 = U/2, or
E=−
(6.67×10−11 N · m2 /kg2 )(5.98×1024 kg)(220kg)
= −6.25×109 J.
2(7.01×106 m)
After 1500 orbits the energy is now −6.25 × 109 J − (1500)(1.40 × 105 J) = −6.46 × 109 J. The new
distance from the Earth is then
r=−
(6.67×10−11 N · m2 /kg2 )(5.98×1024 kg)(220kg)
= 6.79×106 m.
2(−6.46×109 J)
The altitude is now 6790 − 6370 = 420 km.
(d) F = (1.40×105 J)/(2π7.01×106 m) = 3.2×10−3 N.
(e) No.
181
P14-30 Let the satellite S be directly overhead at some time. The magnitude of the speed is
equal to that of a geosynchronous satellite T whose orbit is not inclined, but since there are both
parallel and perpendicular components to the motion of S it will appear to move north while “losing
ground” compared to T . Eventually, though, it must pass overhead again in 12 hours. When S is as
far north as it will go (6 hours) it has a velocity which is parallel to T , but it is located in a region
where the required speed to appear fixed is slower. Hence, it will appear to be “gaining ground”
against the background stars. Consequently, the motion against the background stars appears to be
a figure 8.
P14-31
The net force of gravity on one star because of the other two is
F =
2GM 2
cos(30◦ ).
L2
The stars orbit about a point r = L/2 cos(30◦ ) from any star. The orbital speed is then found from
M v2
2GM 2
M v2
=
=
cos(30◦ ),
◦
r
L/2 cos(30 )
L2
or v =
p
GM/L.
P14-32 A parabolic path will eventually escape; this
p means that the speed of the comet at any
distance is the escape speed for that distance, or v = 2GM/r. The angular momentum is constant,
and is equal to
p
l = mv A rA = m 2GM rA .
For a parabolic path, r = 2rA /(1 + cos θ). Combining with Eq. 14-21 and the equation before that
one we get
√
dθ
2GM rA
=
(1 + cos θ)2 .
dt
4rA 2
The time required is the integral
r
T =
8rA 3
GM
Z
π/2
0
dθ
=
(1 + cos θ)2
r
8rA 3
GM
2
.
3
p
Note that rA 3 /GM is equal to 1/2π years. Then the time for the comet to move is
T =
1 √ 2
8 y = 0.300 y.
2π
3
P14-33 There are three forces on loose matter (of mass m0 ) sitting on the moon: the force of
gravity toward the moon, Fm = Gmm0 /a2 , the force of gravity toward the planet, FM = GM m0 /(r−
a)2 , and the normal force N of the moon pushing the loose matter away from the center of the moon.
The net force on this loose matter is FM + N − Fm , this value is exactly equal to the centripetal
force necessary to keep the loose matter moving in a uniform circle. The period of revolution of the
loose matter is identical to that of the moon,
p
T = 2π r3 /GM ,
but since the loose matter is actually revolving at a radial distance r − a the centripetal force is
Fc =
4π 2 m0 (r − a)
GM m0 (r − a)
=
.
T2
r3
182
Only if the normal force is zero can the loose matter can lift off, and this will happen when F c =
FM − Fm , or
M (r − a)
r3
=
=
M a2 (r − a)3
−3r2 a3 + 3ra4 − a4
M
m
− 2,
(r − a)2
a
M a2 − m(r − a)2
,
a2 (r − a)2
= M r3 a2 − mr3 (r − a)2 ,
m
=
−r5 + 2r4 a − r3 a2
M
Let r = ax, then x is dimensionless; let β = m/M , then β is dimensionless. The expression then
simplifies to
−3x2 + 3x − 1 = β(−x5 + 2x4 − x3 ).
If we assume than x is very large (r a) then only the largest term on each side survives. This means
3x2 ≈ βx5 , or x = (3/β)1/3 . In that case, r = a(3M/m)1/3 . For the Earth’s moon rc = 1.1×107 m,
which is only 4,500 km away from the surface of the Earth. It is somewhat interesting to note that
the radius r is actually independent of both a and m if the moon has a uniform density!
183
E15-1
The pressure in the syringe is
p=
(42.3 N)
= 4.29×105 Pa.
π(1.12×10−2 m/s)2
E15-2 The total mass of fluid is
m = (0.5×10−3 m3 )(2600 kg/m3 )+(0.25×10−3 m3 )(1000 kg/m3 )+(0.4×10−3 m3 )(800 kg/m3 ) = 1.87 kg.
The weight is (18.7 kg)(9.8 m/s2 ) = 18 N.
E15-3 F = A∆p, so
F = (3.43 m)(2.08 m)(1.00 atm − 0.962 atm)(1.01×105 Pa/ atm) = 2.74×104 N.
E15-4 B∆V /V = −∆p; V = L3 ; ∆V ≈ L2 ∆L/3. Then
∆p = (140×109 Pa)
(5×10−3 m)
= 2.74×109 Pa.
3(0.85 m)
E15-5 There is an inward force F1 pushing the lid closed from the pressure of the air outside the
box; there is an outward force F2 pushing the lid open from the pressure of the air inside the box.
To lift the lid we need to exert an additional outward force F3 to get a net force of zero.
The magnitude of the inward force is F1 = P out A, where A is the area of the lid and P out is
the pressure outside the box. The magnitude of the outward force F2 is F2 = P in A. We are told
F3 = 108 lb. Combining,
F2 = F1 − F3 ,
P in A = P out A − F3 ,
P in = P out − F3 /A,
2
2
so P in = (15 lb/in − (108 lb)/(12 in2 ) = 6.0 lb/in .
E15-6 h = ∆p/ρg, so
h=
(0.05 atm)(1.01×105 Pa/ atm)
= 0.52 m.
(1000 kg/m3 )(9.8 m/s2 )
E15-7 ∆p = (1060 kg/m3 )(9.81 m/s2 )(1.83 m) = 1.90×104 Pa.
E15-8 ∆p = (1024 kg/m3 )(9.81 m/s2 )(118 m) = 1.19×106 Pa. Add this to p0 ; the total pressure is
then 1.29×106 Pa.
E15-9
The pressure differential assuming we don’t have a sewage pump:
p2 − p1 = −ρg (y2 − y1 ) = (926 kg/m3 )(9.81 m/s2 )(8.16 m − 2.08 m) = 5.52×104 Pa.
We need to overcome this pressure difference with the pump.
E15-10 (a) p = (1.00 atm)e−5.00/8.55 = 0.557 atm.
(b) h = (8.55 km) ln(1.00/0.500) = 5.93 km.
184
E15-11 The mercury will rise a distance a on one side and fall a distance a on the other so that
the difference in mercury height will be 2a. Since the masses of the “excess” mercury and the water
will be proportional, we have 2aρm = dρw , so
a=
(0.112m)(1000 kg/m3 )
= 4.12×10−3 m.
2(13600 kg/m3 )
E15-12 (a) The pressure (due to the water alone) at the bottom of the pool is
3
2
P = (62.45 lb/ft )(8.0 ft) = 500 lb/ft .
The force on the bottom is
2
F = (500 lb/ft )(80 ft)(30 ft) = 1.2×106 lb.
The average pressure on the side is half the pressure on the bottom, so
2
F = (250 lb/ft )(80 ft)(8.0 ft) = 1.6×105 lb.
The average pressure on the end is half the pressure on the bottom, so
2
F = (250 lb/ft )(30 ft)(8.0 ft) = 6.0×104 lb.
(b) No, since that additional pressure acts on both sides.
E15-13 (a) Equation 15-8 can be used to find the height y2 of the atmosphere if the density is
constant. The pressure at the top of the atmosphere would be p2 = 0, and the height of the bottom
y1 would be zero. Then
y2 = (1.01×105 Pa)/ (1.21 kg/m3 )(9.81 m/s2 ) = 8.51×103 m.
(b) We have to go back to Eq. 15-7 for an atmosphere which has a density which varies linearly
with altitude. Linear variation of density means
y
ρ = ρ0 1 −
ymax
Substitute this into Eq. 15-7,
p2 − p 1
ymax
= −
Z
= −
Z0 ymax
0
= −ρ0 g
ρg dy,
ρ0 g 1 −
y
ymax
ymax
2
y
y−
,
2ymax dy,
0
= −ρgymax /2.
In this case we have ymax = 2p1 /(ρg), so the answer is twice that in part (a), or 17 km.
E15-14 ∆P = (1000 kg/m3 )(9.8 m/s2 )(112 m) = 1.1 × 106 Pa. The force required is then F =
(1.1×106 Pa)(1.22 m)(0.590 m) = 7.9×105 N.
185
E15-15 (a) Choose any infinitesimally small spherical region where equal volumes of the two fluids
are in contact. The denser fluid will have the larger mass. We can treat the system as being a sphere
of uniform mass with a hemisphere of additional mass being superimposed in the region of higher
density. The orientation of this hemisphere is the only variable when calculating the potential energy.
The center of mass of this hemisphere will be a low as possible only when the surface is horizontal.
So all two-fluid interfaces will be horizontal.
(b) If there exists a region where the interface is not horizontal then there will be two different
values for ∆p = ρgh, depending on the path taken. This means that there will be a horizontal
pressure gradient, and the fluid will flow along that gradient until the horizontal pressure gradient
is equalized.
E15-16 The mass of liquid originally in the first vessel is m1 = ρAh1 ; the center of gravity is
at h1 /2, so the potential energy of the liquid in the first vessel is originally U1 = ρgAh21 /2. A
similar expression exists for the liquid in the second vessel. Since the two vessels have the same
cross sectional area the final height in both containers will be hf = (h1 + h2 )/2. The final potential
energy of the liquid in each container will be U f = ρgA(h1 + h2 )2 /8. The work done by gravity is
then
W
= U1 + U2 − 2U f ,
ρgA 2
2h1 + 2h22 − (h21 + 2h1 h2 + h22 ) ,
=
4
ρgA
=
(h1 − h2 )2 .
4
E15-17 There are three force on the block: gravity (W = mg), a buoyant force B0 = mw g, and a
tension T0 . When the container is at rest all three forces balance, so B0 − W − T0 = 0. The tension
in this case is T0 = (mw − m)g.
When the container accelerates upward we now have B − W − T = ma. Note that neither the
tension nor the buoyant force stay the same; the buoyant force increases according to B = mw (g +a).
The new tension is then
T = mw (g + a) − mg − ma = (mw − m)(g + a) = T0 (1 + a/g).
E15-18 (a) F1 /d21 = F2 /d22 , so
F2 = (18.6 kN)(3.72 cm)2 /(51.3 cm)2 = 97.8 N.
(b) F2 h2 = F1 h1 , so
h2 = (1.65 m)(18.6 kN)/(97.8 N) = 314 m.
E15-19 (a) 35.6 kN; the boat doesn’t get heavier or lighter just because it is in different water!
(b) Yes.
(35.6×103 N)
1
1
∆V =
−
= −8.51×10−2 m3 .
(9.81 m/s2 ) (1024 kg/m3 ) (1000 kg/m3 )
E15-20 (a) ρ2 = ρ1 (V1 /V2 ) = (1000 kg/m3 )(0.646) = 646 kg/m3 .
(b) ρ2 = ρ1 (V1 /V2 ) = (646 kg/m3 )(0.918)−1 = 704 kg/m3 .
186
E15-21 The can has a volume of 1200 cm3 , so it can displace that much water. This would
provide a buoyant force of
B = ρV g = (998kg/m3 )(1200×10−6 m3 )(9.81 m/s2 ) = 11.7 N.
This force can then support a total mass of (11.7 N)/(9.81 m/s2 ) = 1.20 kg. If 130 g belong to the
can, then the can will be able to carry 1.07 kg of lead.
3
3
E15-22 ρ2 = ρ1 (V1 /V2 ) = (0.98 g/cm )(2/3)−1 = 1.47 g/cm .
E15-23 Let the object have a mass m. The buoyant force of the air on the object is then
Bo =
ρa
mg.
ρo
There is also a buoyant force on the brass, equal to
Bb =
ρa
mg.
ρb
The fractional error in the weighing is then
3
3
Bo − Bb
(0.0012 g/cm ) (0.0012 g/cm )
=
−
= 2.0×10−4
3
3
mg
(3.4 g/cm )
(8.0 g/cm )
E15-24 The volume of iron is
V i = (6130 N)/(9.81 m/s2 )(7870 kg/m3 ) = 7.94×10−2 m3 .
The buoyant force of water is 6130 N − 3970 N = 2160 N. This corresponds to a volume of
V w = (2160 N)/(9.81 m/s2 )(1000 kg/m3 ) = 2.20×10−1 m3 .
The volume of air is then 2.20×10−1 m3 − 7.94×10−2 m3 = 1.41×10−1 m3 .
E15-25 (a) The pressure on the top surface is p = p0 + ρgL/2. The downward force is
Ft
= (p0 + ρgL/2)L2 ,
= (1.01×105 Pa) + (944 kg/m3 )(9.81 m/s2 )(0.608 m)/2 (0.608 m)2 = 3.84×104 N.
(b) The pressure on the bottom surface is p = p0 + 3ρgL/2. The upward force is
Fb
= (p0 + 3ρgL/2)L2 ,
= (1.01×105 Pa) + 3(944 kg/m3 )(9.81 m/s2 )(0.608 m)/2 (0.608 m)2 = 4.05×104 N.
(c) The tension in the wire is given by T = W + F t − F b , or
T = (4450 N) + (3.84×104 N) − (4.05×104 N) = 2350 N.
(d) The buoyant force is
B = L3 ρg = (0.6083 )(944 kg/m3 )(9.81 m/s2 ) = 2080 N.
187
E15-26 The fish has the (average) density of water if
ρw =
mf
Vc+Va
Va =
mf
− V c.
ρw
or
We want the fraction V a /(V c + V a ), so
Va
Vc+Va
Vc
,
mf
=
1 − ρw
=
1 − ρw /ρc = 1 − (1.024 g/cm )/(1.08 g/cm ) = 5.19×10−2 .
3
3
E15-27 There are three force on the dirigible: gravity (W = mg g), a buoyant force B = ma g,
and a tension T . Since these forces must balance we have T = B − W . The masses are related to
the densities, so we can write
T = (ρa − ρg )V g = (1.21 kg/m3 − 0.796 kg/m3 )(1.17×106 m3 )(9.81m/s2 ) = 4.75×106 N.
E15-28 ∆m = ∆ρV , so
∆m = [(0.160 kg/m3 ) − (0.0810 kg/m3 )](5000 m3 ) = 395 kg.
E15-29 The volume of one log is π(1.05/2 ft)2 (5.80 ft) = 5.02 ft3 . The weight of the log is
3
(47.3 lb/ft )(5.02 ft3 ) = 237 lb. Each log if completely submerged will displace a weight of wa3
ter (62.4 lb/ft )(5.02 ft3 ) = 313 lb. So each log can support at most 313 lb − 237 lb = 76 lb. The
three children have a total weight of 247 lb, so that will require 3.25 logs. Round up to four.
E15-30 (a) The ice will hold up the automobile if
ρw >
Then
A=
ma + mi
ma
=
+ ρi .
Vi
At
(1120 kg)
= 44.2 m2 .
(0.305 m)[(1000 kg/m3 ) − (917 kg/m3 )]
E15-31 If there were no water vapor pressure above the barometer then the height of the water
would be y1 = p/(ρg), where p = p0 is the atmospheric pressure. If there is water vapor where there
should be a vacuum, then p is the difference, and we would have y2 = (p0 − pv )/(ρg). The relative
error is
(y1 − y2 )/y1
= [p0 /(ρg) − (p0 − pv )/(ρg)] / [p0 /(ρg)] ,
= pv /p0 = (3169 Pa)/(1.01×105 Pa) = 3.14 %.
E15-32 ρ = (1.01×105 Pa)/(9.81 m/s2 )(14 m) = 740 kg/m3 .
E15-33 h = (90)(1.01×105 Pa)/(8.60 m/s2 )(1.36×104 kg/m3 ) = 78 m.
E15-34 ∆U = 2(4.5×10−2 N/m)4π(2.1×10−2 m)2 = 5.0×10−4 J.
188
E15-35 The force required is just the surface tension times the circumference of the circular
patch. Then
F = (0./072 N/m)2π(0.12 m) = 5.43×10−2 N.
E15-36 ∆U = 2(2.5×10−2 N/m)4π(1.4×10−2 m)2 = 1.23×10−4 J.
P15-1 (a) One can replace the two hemispheres with an open flat end with two hemispheres with
a closed flat end. Then the area of relevance is the area of the flat end, or πR2 . The net force from
the pressure difference is ∆pA = ∆pπR2 ; this much force must be applied to pull the hemispheres
apart.
(b) F = π(0.9)(1.01×105 Pa)(0.305 m)2 = 2.6×104 N.
P15-2
The pressure required is 4×109 Pa. This will happen at a depth
h=
P15-3
(4×109 Pa)
= 1.3×105 m.
(9.8 m/s2 )(3100 kg/m3 )
(a) The resultant force on the wall will be
Z Z
F =
P dx dy,
Z
=
(−ρgy)W dy,
= ρgD2 W/2.
(b) The torque will is given by τ = F (D −y) (the distance is from the bottom) so if we generalize,
Z Z
τ =
P y dx dy,
Z
=
(−ρg(D − y)) yW dy,
= ρgD3 W/6.
(c) Dividing to find the location of the equivalent resultant force,
d = τ /F = (ρgD3 W/6)/(ρgD2 W/2) = D/3,
this distance being measured from the bottom.
P15-4 p = ρgy = ρg(3.6 m); the force on the bottom is F = pA = ρg(3.6 m)π(0.60 m)2 = 1.296πρg.
The volume of liquid is
V = (1.8 m) π(0.60 m) + 4.6×10−4 m2 = 2.037 m3
The weight is W = ρg(2.037 m3 ). The ratio is 2.000.
P15-5 The pressure at b is ρc (3.2×104 m) + ρm y. The pressure at a is ρc (3.8×104 m + d) + ρm (y − d).
Set these quantities equal to each other:
ρc (3.8×104 m + d) + ρm (y − d) =
ρc (6×103 m + d) =
d =
=
ρc (3.2×104 m) + ρm y,
ρm d,
ρc (6×103 m)/(ρm − ρc ),
(2900 kg/m3 )(6×103 m)/(400 kg/m3 ) = 4.35×104 m.
189
P15-6
(a) The pressure (difference) at a depth y is ∆p = ρgy. Since ρ = m/V , then
∆ρ ≈ −
∆p
m ∆V
= ρs
.
V V
B
Then
ρ ≈ ρs + ∆ρ = ρs +
ρs 2 gy
.
B
(b) ∆ρ/ρs = ρs gy/B, so
∆ρ/ρ ≈ (1000 kg/m3 )(9.8 m/s2 )(4200 m)/(2.2×109 Pa) = 1.9 %.
P15-7
(a) Use Eq. 15-10, p = (p0 /ρ0 )ρ, then Eq. 15-13 will look like
(p0 /ρ0 )ρ = (p0 /ρ0 )ρ0 e−h/a .
(b) The upward velocity of the rocket as a function of time is given by v = ar t. The height of
the rocket above the ground is given by y = 21 ar t2 . Combining,
r
2y p
v = ar
= 2yar .
ar
Put this into the expression for drag, along with the equation for density variation with altitude;
D = CAρv 2 = CAρ0 e−y/a 2yar .
Now take the derivative with respect to y,
dD/dy = (−1/a)CAρ0 e−y/a (2yar ) + CAρ0 e−y/a (2ar ).
This will vanish when y = a, regardless of the acceleration ar .
P15-8 (a) Consider a slice of cross section A a depth h beneath the surface. The net force on the
fluid above the slice will be
F net = ma = ρhAg,
Since the weight of the fluid above the slice is
W = mg = ρhAg,
then the upward force on the bottom of the fluid at the slice must be
W + F net = ρhA(g + a),
so the pressure is p = F/A = ρh(g + a).
(b) Change a to −a.
(c) The pressure is zero (ignores atmospheric contributions.)
~ =
P15-9 (a) Consider a portion of the liquid at the surface. The net force on this portion is F
~ = −mg ĵ. There must then be a buoyant
m~a = maî. The force of gravity on this portion is W
~ =F
~ −W
~ = m(aî + g ĵ). The buoyant force makes an angle
force on the portion with direction B
θ = arctan(a/g) with the vertical. The buoyant force must be perpendicular to the surface of the
fluid; there are no pressure-related forces which are parallel to the surface. Consequently, the surface
must make an angle θ = arctan(a/g) with the horizontal.
(b) It will still vary as ρgh; the derivation on page 334 is still valid for vertical displacements.
190
P15-10
dp/dr = −ρg, but now g = 4πGρr/3. Then
Z r
Z p
4
r dr,
dp = − πGρ2
3
R
0
2
p =
πGρ2 R2 − r2 .
3
P15-11 We can start with Eq. 15-11, except that we’ll write our distance in terms of r instead
if y. Into this we can substitute our expression for g,
g = g0
R2
.
r2
Substituting, then integrating,
Z
p
p0
ln
dp
p
dp
p
dp
p
p
p0
gρ0
dr,
p0
g0 ρ0 R2 dr
= −
,
p0 r 2
Z r
g0 ρ0 R2 dr
= −
,
p0 r 2
R
g0 ρ0 R2 1
1
=
−
p0
r
R
= −
If k = g0 ρ0 R2 /p0 , then
p = p0 ek(1/r−1/R) .
P15-12 (a) The net force on a small volume of the fluid is dF = rω 2 dm directed toward the
center. For radial displacements, then, dF/dr = −rω 2 dm/dr or dp/dr = −rω 2 ρ.
(b) Integrating outward,
Z r
1
p = pc +
ρω 2 r dr = pc + ρr2 ω 2 .
2
0
(c) Do part (d) first.
(d) It will still vary as ρgh; the derivation on page 334 is still valid for vertical displacements.
(c) The pressure anywhere in the liquid is then given by
1
p = p0 + ρr2 ω 2 − ρgy,
2
where p0 is the pressure on the surface, y is measured from the bottom of the paraboloid, and r is
measured from the center. The surface is defined by p = p0 , so
1 2 2
ρr ω − ρgy = 0,
2
or y = r2 ω 2 /2g.
P15-13 The total mass of the shell is m = ρw πdo 3 /3, or it wouldn’t barely float. The mass of iron
in the shell is m = ρi π(do 3 − di 3 )/3, so
ρi − ρw 3
di 3 =
do ,
ρi
so
s
(7870 kg/m3 ) − (1000 kg/m3 )
di = 3
(0.587 m) = 0.561 m.
(7870 kg/m3 )
191
P15-14 The wood will displace a volume of water equal to (3.67 kg)/(594 kg/m3 )(0.883) = 5.45×
10−3 m3 in either case. That corresponds to a mass of (1000 kg/m3 )(5.45×10−3 m3 ) = 5.45 kg that
can be supported.
(a) The mass of lead is 5.45 kg − 3.67 kg = 1.78 kg.
(b) When the lead is submerged beneath the water it displaces water, which affects the “apparent”
mass of the lead. The true weight of the lead is mg, the buoyant force is (ρw /ρl )mg, so the apparent
weight is (1 − ρw /ρl )mg. This means the apparent mass of the submerged lead is (1 − ρw /ρl )m.
This apparent mass is 1.78 kg, so the true mass is
m=
P15-15
(11400 kg/m3 )
(1.78 kg) = 1.95 kg.
(11400 kg/m3 ) − (1000 kg)
We initially have
ρo
1
=
.
4
ρmercury
When water is poured over the object the simple relation no longer works.
Once the water is over the object there are two buoyant forces: one from mercury, F1 , and one
from the water, F2 . Following a derivation which is similar to Sample Problem 15-3, we have
F1 = ρ1 V1 g and F2 = ρ2 V2 g
where ρ1 is the density of mercury, V1 the volume of the object which is in the mercury, ρ2 is the
density of water, and V2 is the volume of the object which is in the water. We also have
F1 + F2 = ρo V o g and V1 + V2 = V o
as expressions for the net force on the object (zero) and the total volume of the object. Combining
these four expressions,
ρ1 V1 + ρ2 V2 = ρo V o ,
or
ρ1 V1 + ρ2 (V o − V1 ) = ρo V o ,
(ρ1 − ρ2 ) V1 = (ρo − ρ2 ) V o ,
V1
ρo − ρ2
=
.
Vo
ρ1 − ρ2
The left hand side is the fraction that is submerged in the mercury, so we just need to substitute
our result for the density of the material from the beginning to solve the problem. The fraction
submerged after adding water is then
V1
Vo
=
=
=
ρo − ρ2
,
ρ1 − ρ2
ρ1 /4 − ρ2
,
ρ1 − ρ2
(13600 kg/m3 )/4 − (998 kg/m3 )
= 0.191.
(13600 kg/m3 ) − (998 kg/m3 )
P15-16 (a) The car floats if it displaces a mass of water equal to the mass of the car. Then
V = (1820 kg)/(1000 kg/m3 ) = 1.82 m3 .
(b) The car has a total volume of 4.87 m3 + 0.750 m3 + 0.810 m3 = 6.43 m3 . It will sink if the total
mass inside the car (car + water) is then (6.43 m3 )(1000 kg/m3 ) = 6430 kg. So the mass of the water
in the car is 6430 kg−1820 kg = 4610 kg when it sinks. That’s a volume of (4610 kg)/(1000 kg/m3 ) =
4.16 m3 .
192
P15-17 When the beaker is half filled with water it has a total mass exactly equal to the maximum
amount of water it can displace. The total mass of the beaker is the mass of the beaker plus the
mass of the water inside the beaker. Then
ρw (mg /ρg + V b ) = mg + ρw V b /2,
where mg /ρg is the volume of the glass which makes up the beaker. Rearrange,
ρg =
P15-18
(0.390 kg)
mg
=
= 2790 kg/m3 .
mg /ρw − V b /2
(0.390 kg)/(1000 kg/m3 ) − (5.00×10−4 m3 )/2
(a) If each atom is a cube then the cube has a side of length
p
l = 3 (6.64×10−27 kg)/(145 kg/m3 ) = 3.58×10−10 m.
Then the atomic surface density is l−2 = (3.58×10−10 m)−2 = 7.8×1018 /m2 .
(b) The bond surface density is twice the atomic surface density. Show this by drawing a square
array of atoms and then joining each adjacent pair with a bond. You will need twice as many bonds
as there are atoms. Then the energy per bond is
(3.5×10−4 N/m)
= 1.4×10−4 eV.
2(7.8×1018 /m2 )(1.6×10−19 J/eV)
P15-19 Pretend the bubble consists of two hemispheres. The force from surface tension holding
the hemispheres together is F = 2γL = 4πrγ. The “extra” factor of two occurs because each
hemisphere has a circumference which “touches” the boundary that is held together by the surface
tension of the liquid. The pressure difference between the inside and outside is ∆p = F/A, where A
is the area of the flat side of one of the hemispheres, so ∆p = (4πrγ)/(πr2 ) = 4γ/r.
P15-20 Use the results of Problem 15-19. To get a numerical answer you need to know the surface
tension; try γ = 2.5×10−2 N/m. The initial pressure inside the bubble is pi = p0 + 4γ/ri . The final
pressure inside the bell jar is p = pf − 4γ/rf . The initial and final pressure inside the bubble are
related by pi ri 3 = pf rf 3 . Now for numbers:
pi = (1.00×105 Pa) + 4(2.5×10−2 N/m)/(1.0×10−3 m) = 1.001×105 Pa.
pf = (1.0×10−3 m/1.0×10−2 m)3 (1.001×105 Pa) = 1.001×102 Pa.
p = (1.001×102 Pa) − 4(2.5×10−2 N/m)/(1.0×10−2 m) = 90.1Pa.
P15-21 The force on the liquid in the space between the rod and the cylinder is F = γL =
2πγ(R + r). This force can support a mass of water m = F/g. This mass has a volume V = m/ρ.
The cross sectional area is π(R2 − r2 ), so the height h to which the water rises is
h
=
=
2πγ(R + r)
2γ
=
,
ρgπ(R2 − r2 )
ρg(R − r)
2(72.8×10−3 N/m)
= 3.71×10−3 m.
(1000 kg/m3 )(9.81 m/s2 )(4.0×10−3 m)
193
P15-22
(a) Refer to Problem 15-19. The initial pressure difference is
4(2.6×10−2 N/m)/(3.20×10−2 m) = 3.25Pa.
(b) The final pressure difference is
4(2.6×10−2 N/m)/(5.80×10−2 m) = 1.79Pa.
(c) The work done against the atmosphere is p∆V , or
(1.01×105 Pa)
4π
[(5.80×10−2 m)3 − (3.20×10−2 m)3 ] = 68.7 J.
3
(d) The work done in stretching the bubble surface is γ∆A, or
(2.60×10−2 N/m)4π[(5.80×10−2 m)2 − (3.20×10−2 m)2 ] = 7.65×10−4 J.
194
E16-1 R = Av = πd2 v/4 and V = Rt, so
t=
4(1600 m3 )
= 6530 s
π(0.345 m)2 (2.62 m)
E16-2 A1 v1 = A2 v2 , A1 = πd21 /4 for the hose, and A2 = N πd22 for the sprinkler, where N = 24.
Then
(0.75 in)2
v2 =
(3.5 ft/s) = 33 ft/s.
(24)(0.050 in)2
E16-3 We’ll assume that each river has a rectangular cross section, despite what the picture
implies. The cross section area of the two streams is then
A1 = (8.2 m)(3.4 m) = 28 m2 and A2 = (6.8 m)(3.2 m) = 22 m2 .
The volume flow rate in the first stream is
R1 = A1 v1 = (28 m2 )(2.3 m/s) = 64 m3 /s,
while the volume flow rate in the second stream is
R2 = A2 v2 = (22 m2 )(2.6 m/s) = 57 m3 /s.
The amount of fluid in the stream/river system is conserved, so
R3 = R1 + R2 = (64 m3 /s) + (57 m3 /s) = 121 m3 /s.
where R3 is the volume flow rate in the river. Then
D3 = R3 /(v3 W3 ) = (121 m3 /s)/[(10.7 m)(2.9 m/s)] = 3.9 m.
E16-4 The speed of the water is originally zero so both the kinetic and potential energy is zero.
When it leaves the pipe at the top it has a kinetic energy of 12 (5.30 m/s)2 = 14.0 J/kg and a
potential energy of (9.81 m/s2 )(2.90 m) = 28.4 J/kg. The water is flowing out at a volume rate of
R = (5.30 m/s)π(9.70×10−3 m)2 = 1.57×10−3 m3 /s. The mass rate is ρR = (1000 kg/m3 )(1.57×
10−3 m3 /s) = 1.57 kg/s.
The power supplied by the pump is (42.8 J/kg)(1.57 kg/s) = 67.2W.
E16-5 There are 8500 km2 which collects an average of (0.75)(0.48 m/y), where the 0.75 reflects
the fact that 1/4 of the water evaporates, so
1y
R = 8500(103 m)2 (0.75)(0.48 m/y)
= 97 m3 /s.
365 × 24 × 60 × 60 s
Then the speed of the water in the river is
v = R/A = (97 m3 /s)/[(21 m)(4.3 m)] = 1.1 m/s.
E16-7 (a) ∆p = ρg(y1 − y2 ) + ρ(v12 − v22 )/2. Then
3
3
2
2
∆p = (62.4 lb/ft )(572 ft) + [(62.4 lb/ft )/(32 ft/s )][(1.33 ft/s)2 − (31.0 ft/s)2 ]/2 = 3.48×104 lb/ft .
(b) A2 v2 = A1 v1 , so
A2 = (7.60 ft2 )(1.33 ft/s)/(31.0 ft/s) = 0.326 ft2 .
195
E16-8 (a) A2 v2 = A1 v1 , so
v2 = (2.76 m/s)[(0.255 m)2 − (0.0480 m)2 ]/(0.255 m)2 = 2.66 m/s.
(b) ∆p = ρ(v12 − v22 )/2,
∆p = (1000 kg/m3 )[(2.66 m/s)2 − (2.76 m/s)2 ]/2 = −271 Pa
E16-9
(b) We will do part (b) first.
R = (100 m2 )(1.6 m/y)
1y
365 × 24 × 60 × 60 s
= 5.1×10−6 m3 /s.
(b) The speed of the flow R through a hole of cross sectional area a will be v = R/a. p = p0 +ρgh,
where h = 2.0 m is the depth of the hole. Bernoulli’s equation can be applied to find the speed of
the water as it travels a horizontal stream line out the hole,
1
p0 + ρv 2 = p,
2
where we drop any terms which are either zero or the same on both sides. Then
p
p
p
v = 2(p − p0 )/ρ = 2gh = 2(9.81 m/s2 )(2.0 m) = 6.3 m/s.
Finally, a = (5.1×10−6 m3 /s)/(6.3 m/s) = 8.1×10−7 m2 , or about 0.81 mm2 .
E16-10 (a) v2 = (A1 /A2 )v1 = (4.20 cm2 )(5.18 m/s)/(7.60 cm2 ) = 2.86 m/s.
(b) Use Bernoulli’s equation:
1
1
p2 + ρgy2 + ρv22 = p1 + ρgy1 + ρv12 .
2
2
Then
p2
=
1
1
2
2
2
(1.52×10 Pa) + (1000 kg/m ) (9.81 m/s )(9.66 m) + (5.18 m/s) − (2.86 m/s) ,
2
2
=
2.56×105 Pa.
5
3
E16-11 (a) The wind speed is (110 km/h)(1000 m/km)/(3600 s/h) = 30.6 m/s. The pressure
difference is then
1
∆p = (1.2 kg/m3 )(30.6 m/s)2 = 562 Pa.
2
(b) The lifting force would be F = (562 Pa)(93 m2 ) = 52000 N.
E16-12 The pressure difference is
∆p =
1
(1.23 kg/m3 )(28.0 m/s)2 = 482 N.
2
The net force is then F = (482 N)(4.26 m)(5.26 m) = 10800 N.
196
E16-13 The lower pipe has a radius r1 = 2.52 cm, and a cross sectional area of A1 = πr12 . The
speed of the fluid flow at this point is v1 . The higher pipe has a radius of r2 = 6.14 cm, a cross
sectional area of A2 = πr22 , and a fluid speed of v2 . Then
A1 v1 = A2 v2 or r12 v1 = r22 v2 .
Set y1 = 0 for the lower pipe. The problem specifies that the pressures in the two pipes are the
same, so
1
p0 + ρv12 + ρgy1
2
1 2
v
2 1
1
= p0 + ρv22 + ρgy2 ,
2
1 2
=
v + gy2 ,
2 2
We can combine the results of the equation of continuity with this and get
v12
v12
= v22 + 2gy2 ,
2
= v1 r12 /r22 + 2gy2 ,
v12
1 − r14 /r24
= 2gy2 ,
= 2gy2 / 1 − r14 /r24 .
v12
Then
v12 = 2(9.81 m/s2 )(11.5 m)/ 1 − (0.0252 m)4 /(0.0614 m)4 = 232 m2 /s2
The volume flow rate in the bottom (and top) pipe is
R = πr12 v1 = π(0.0252 m)2 (15.2 m/s) = 0.0303 m3 /s.
E16-14 (a) As instructed,
1
= p0 + ρv32 + ρgy3 ,
2
1 2
0 =
v + g(y3 − y1 ),
2 3
1
p0 + ρv12 + ρgy1
2
√
But y3 − y1 = −h, so v3 = 2gh.
(b) h above the hole. Just reverse your streamline!
(c) It won’t come out as fast and it won’t rise as high.
E16-15
Then
Sea level will be defined as y = 0, and at that point the fluid is assumed to be at rest.
1
p0 + ρv12 + ρgy1
2
1
= p0 + ρv22 + ρgy2 ,
2
1 2
0 =
v + gy2 ,
2 2
where y2 = −200 m. Then
v2 =
p
p
−2gy2 = −2(9.81 m/s2 )(−200 m) = 63 m/s.
197
E16-16 Assume streamlined flow, then
1
p1 + ρv12 + ρgy1
2
1
= p2 + ρv22 + ρgy2 ,
2
1 2
v .
(p1 − p2 )/ρ + g(y1 − y2 ) =
2 2
Then upon rearranging
p
v2 = 2 [(2.1)(1.01×105 Pa)/(1000 kg/m3 ) + (9.81 m/s2 )(53.0 m)] = 38.3 m/s.
E16-17 (a) Points 1 and 3 are both at atmospheric pressure, and both will move at the same
speed. But since they are at different heights, Bernoulli’s equation will be violated.
(b) The flow isn’t steady.
E16-18 The atmospheric pressure difference between the two sides will be ∆p = 21 ρa v 2 . The height
difference in the U-tube is given by ∆p = ρw gh. Then
h=
(1.20 kg/m3 )(15.0 m/s)2
= 1.38×10−2 m.
2(1000 kg/m3 )(9.81 m/s2 )
E16-19 (a) There are three forces on the plug. The force from the pressure of the water, F1 =
P1 A, the force from the pressure of the air, F2 = P2 A, and the force of friction, F3 . These three
forces must balance, so F3 = F1 − F2 , or F3 = P1 A − P2 A. But P1 − P2 is the pressure difference
between the surface and the water 6.15 m below the surface, so
F3
= ∆P A = −ρgyA,
= −(998 kg/m3 )(9.81 m/s2 )(−6.15 m)π(0.0215 m)2 ,
= 87.4 N
(b) To find the volume of water which flows out in three hours we need to know the volume
flow rate, and for that we need both the cross section area of the hole and the speed of the flow.
The speed of the flow can be found by an application of Bernoulli’s equation. We’ll consider the
horizontal motion only— a point just inside the hole, and a point just outside the hole. These points
are at the same level, so
1
p1 + ρv12 + ρgy1
2
p1
1
= p2 + ρv22 + ρgy2 ,
2
1
= p2 + ρv22 .
2
Combine this with the results of Pascal’s principle above, and
p
p
p
v2 = 2(p1 − p2 )/ρ = −2gy = −2(9.81 m/s2 )(−6.15 m) = 11.0 m/s.
The volume of water which flows out in three hours is
V = Rt = (11.0 m/s)π(0.0215 m)2 (3 × 3600 s) = 173 m3 .
E16-20 Apply Eq. 16-12:
p
v1 = 2(9.81 m/s2 )(0.262 m)(810 kg/m3 )/(1.03 kg/m3 ) = 63.6 m/s.
198
E16-21 We’ll assume that the central column of air down the pipe exerts minimal force on the
card when it is deflected to the sides. Then
1
p1 + ρv12 + ρgy1
2
1
= p2 + ρv22 + ρgy2 ,
2
1
= p2 + ρv22 .
2
p1
The resultant upward force on the card is the area of the card times the pressure difference, or
F = (p1 − p2 )A =
1
ρAv 2 .
2
E16-22 If the air blows uniformly over the surface of the plate then there can be no torque about
any axis through the center of mass of the plate. Since the weight also doesn’t introduce a torque,
then the hinge can’t exert a force on the plate, because any such force would produce an unbalanced
torque. Consequently mg = ∆pA. ∆p = ρv 2 /2, so
s
r
2mg
2(0.488 kg)(9.81 m/s2 )
v=
=
= 30.9 m/s.
ρA
(1.21 kg/m3 )(9.10×10−2 m)2
E16-23 Consider a streamline which passes above the wing and a streamline which passes beneath
the wing. Far from the wing the two streamlines are close together, move with zero relative velocity,
and are effectively at the same pressure. So we can pretend they are actually one streamline. Then,
since the altitude difference between the two points above and below the wing (on this new, single
streamline) is so small we can write
∆p =
1
ρ(v t 2 − v u 2 )
2
The lift force is then
L = ∆p A =
1
ρA(v t 2 − v u 2 )
2
E16-24 (a) From Exercise 16-23,
L=
1
(1.17 kg/m3 )(2)(12.5 m2 ) (49.8 m/s)2 − (38.2 m/s)2 = 1.49×104 N.
2
The mass of the plane must be m = L/g = (1.49×104 N)/(9.81 m/s2 ) = 1520 kg.
(b) The lift is directed straight up.
(c) The lift is directed 15◦ off the vertical toward the rear of the plane.
(d) The lift is directed 15◦ off the vertical toward the front of the plane.
E16-25 The larger pipe has a radius r1 = 12.7 cm, and a cross sectional area of A1 = πr12 . The
speed of the fluid flow at this point is v1 . The smaller pipe has a radius of r2 = 5.65 cm, a cross
sectional area of A2 = πr22 , and a fluid speed of v2 . Then
A1 v1 = A2 v2 or r12 v1 = r22 v2 .
Now Bernoulli’s equation. The two pipes are at the same level, so y1 = y2 . Then
1
p1 + ρv12 + ρgy1
2
1
p1 + ρv12
2
1
= p2 + ρv22 + ρgy2 ,
2
1
= p2 + ρv22 .
2
199
Combining this with the results from the equation of continuity,
1
p1 + ρv12
2
v12
v12
1−
v12
4
r1
r24
v12
1
= p2 + ρv22 ,
2
2
2
= v2 + (p2 − p1 ) ,
ρ
2
r12
2
=
v1 2
+ (p2 − p1 ) ,
r2
ρ
2
=
(p2 − p1 ) ,
ρ
2 (p2 − p1 )
=
.
ρ (1 − r14 /r24 )
It may look a mess, but we can solve it to find v1 ,
s
2(32.6×103 Pa − 57.1×103 Pa)
v1 =
= 1.41 m/s.
(998 kg/m3 )(1 − (0.127 m)4 /(0.0565 m)4 )
The volume flow rate is then
R = Av = π(0.127 m)2 (1.41 m/s) = 7.14×10−3 m3 /s.
That’s about 71 liters/second.
E16-26 The lines are parallel and equally spaced, so the velocity is everywhere the same. We can
transform to a reference frame where the liquid appears to be at rest, so the Pascal’s equation would
apply, and p + ρgy would be a constant. Hence,
1
p0 = p + ρgy + ρv 2
2
is the same for all streamlines.
E16-27 (a) The “particles” of fluid in a whirlpool would obey conservation of angular momentum,
meaning a particle off mass m would have l = mvr be constant, so the speed of the fluid as a function
of radial distance from the center would be given by k = vr, where k is some constant representing
the angular momentum per mass. Then v = k/r.
(b) Since v = k/r and v = 2πr/T , the period would be T ∝ r2 .
(c) Kepler’s third law says T ∝ r3/2 .
E16-28 Rc = 2000. Then
v<
E16-29
pipe.
Rc η
(2000)(4.0×10−3 N · s/m2 )
= 2.0 m/s.
=
ρD
(1060 kg/m3 )(3.8×10−3 m)
(a) The volume flux is given; from that we can find the average speed of the fluid in the
v=
5.35 × 10−2 L/min
2
= 4.81×10−3 L/cm · min.
π(1.88 cm)2
But 1 L is the same as 1000 cm3 and 1 min is equal to 60 seconds, so v = 8.03×10−4 m/s.
200
Reynold’s number from Eq. 16-22 is then
R=
(13600 kg/m3 )(0.0376 m)(8.03×10−4 m/s)
ρDv
=
= 265.
η
(1.55 × 10−3 N · s/m2 )
This is well below the critical value of 2000.
(b) Poiseuille’s Law, Eq. 16-20, can be used to find the pressure difference between the ends of
the pipe. But first, note that the mass flux dm/dt is equal to the volume rate times the density
when the density is constant. Then ρ dV /dt = dm/dt, and Poiseuille’s Law can be written as
δp =
P16-1
8(1.55 × 10−3 N · s/m2 )(1.26 m)
8ηL dV
=
(8.92×10−7 m3 /s) = 0.0355 Pa.
4
πR dt
π(1.88×10−2 m)4
The volume of water which needs to flow out of the bay is
V = (6100 m)(5200 m)(3 m) = 9.5×107 m3
during a 6.25 hour (22500 s) period. The average speed through the channel must be
v=
(9.5×107 m3 )
= 3.4 m/s.
(22500 s)(190 m)(6.5 m)
P16-2 (a) The speed of the fluid through either hole is v =
is Q = ρAv, so ρ1 A1 = ρ2 A2 . Then ρ1 /ρ2 = A2 /A1 = 2.
(b) R = Av, so R1 /R2 =
√ A1 /A2 = 1/2.
√
(c) If R1 = R2 then A1 2gh1 = A2 2gh2 . Then
√
2gh. The mass flux through a hole
h2 /h1 = (A1 /A2 )2 = (1/2)2 = 1/4.
So h2 = h1 /4.
√
P16-3 (a) Apply Torricelli’s law (Exercise 16-14): v = 2gh. The speed v is a horizontal velocity,
and serves as the initial horizontal velocity of the fluid “projectile” after it leaves the tank. There
is no initial vertical velocity.
This fluid “projectile” falls through a vertical distance H −h before splashing against the ground.
The equation governing the time t for it to fall is
1
−(H − h) = − gt2 ,
2
p
Solve this for the time, and t = 2(H − h)/g. The equation which governs the horizontal distance
traveled during the fall is x = vx t, but vx = v and we just found t, so
p
p
p
x = vx t = 2gh 2(H − h)/g = 2 h(H − h).
(b) How many values of h will lead to a distance of x? We need to invert the expression, and
we’ll start by squaring both sides
x2 = 4h(H − h) = 4hH − 4h2 ,
and then solving the resulting quadratic expression for h,
√
p
4H ± 16H 2 − 16x2
1
h=
=
H ± H 2 − x2 .
8
2
201
For values of x between 0 and H there are two real solutions, if x = H there is one real solution,
and if x > H there are no real solutions.
If h1 is a solution, then we can write h1 = (H + ∆)/2, where ∆ = 2h1 − H could be positive or
negative. Then h2 = (H + ∆)/2 is also a solution, and
h2 = (H + 2h1 − 2H)/2 = h1 − H
is also a solution.
(c) The farthest distance is x = H, and this happens when h = H/2, as we can see from the
previous section.
p
P16-4 (a) Apply Torricelli’s law (Exercise 16-14): v = 2g(d + h2 ), assuming that the liquid
remains in contact with the walls of the tube until it exits at the bottom.
(b) The speed of the fluid in the tube is everywhere the same. Then the pressure difference at
various points are only functions of height. The fluid exits at C, and assuming that it remains in
contact with the walls of the tube the pressure difference is given by ∆p = ρ(h1 + d + h2 ), so the
pressure at B is
p = p0 − ρ(h1 + d + h2 ).
(c) The lowest possible pressure at B is zero. Assume the flow rate is so slow that Pascal’s
principle applies. Then the maximum height is given by 0 = p0 + ρgh1 , or
h1 = (1.01×105 Pa)/[(9.81 m/s2 )(1000 kg/m3 )] = 10.3 m.
P16-5 (a) The momentum per kilogram of the fluid in the smaller pipe is v1 . The momentum per
kilogram of the fluid in the larger pipe is v2 . The change in momentum per kilogram is v2 − v1 .
There are ρa2 v2 kilograms per second flowing past any point, so the change in momentum per
second is ρa2 v2 (v2 − v1 ). The change in momentum per second is related to the net force according
to F = ∆p/∆t, so F = ρa2 v2 (v2 − v1 ). But F ≈ ∆p/a2 , so p1 − p2 ≈ ρv2 (v2 − v1 ).
(b) Applying the streamline equation,
1
1
p1 + ρgy1 + ρv12 = p2 + ρgy2 + ρv22 ,
2
2
1
2
2
ρ(v1 − v2 ) = p2 − p1
2
(c) This question asks for the loss of pressure beyond that which would occur from a gradually
widened pipe. Then we want
∆p
=
=
=
1
ρ(v12 − v22 ) − ρv2 (v1 − v2 ),
2
1
ρ(v12 − v22 ) − ρv2 v1 + ρv22 ,
2
1
1
ρ(v12 − 2v1 v2 + v22 ) = ρ(v1 − v2 )2 .
2
2
√
P16-6 The juice leaves the jug with a speed v = 2gy, where y is the height of the juice in the
jug. If A is the cross sectional area of the base of the jug and a√the cross sectional area of the hole,
then the juice flows out the
√ hole with a rate dV /dt = va = a 2gy, which means the level of jug
varies as dy/dt = −(a/A) 2gy. Rearrange and integrate,
Z
h
√
dy/ y
y
Z tp
=
2g(a/A)dt,
0
202
√
p
√
2( h − y) =
2gat/A.
s !
A 2h √
√
( h − y) = t
a
g
When y = 14h/15 we have t = 12.0 s. Then the part in the parenthesis on the left is 3.539×102 s.
The time to empty completely is then 354 seconds, or 5 minutes and 54 seconds. But we want the
remaining time, which is 12 seconds less than this.
P16-7 The greatest possible value for v will be the value over the wing which results in an air
pressure of zero. If the air at the leading edge is stagnant (not moving) and has a pressure of p0 ,
then Bernoulli’s equation gives
1
p0 = ρv 2 ,
2
p
p
or v = 2p0 /ρ = 2(1.01×105 Pa)/(1.2 kg/m3 ) = 410 m/s. This value is only slightly larger than
the speed of sound; they are related because sound waves involve the movement of air particles which
“shove” other air particles out of the way.
P16-8
Bernoulli’s equation together with continuity gives
1
p1 + ρv12
2
p1 − p2
1
= p2 + ρv22 ,
2
1
=
ρ v22 − v12 ,
2 1
A21 2
2
=
ρ
v − v1 ,
2
A22 1
v12
=
ρ A21 − A22 .
2
2A2
But p1 − p2 = (ρ0 − ρ)gh. Note that we are not assuming ρ is negligible compared to ρ0 . Combining,
s
2(ρ0 − ρ)gh
.
v1 = A2
ρ(A21 − A22 )
P16-9
(a) Bernoulli’s equation together with continuity gives
1
p1 + ρv12
2
p1
Then
v1 =
s
1
= p2 + ρv22 ,
2
1
=
ρ v22 − v12 ,
2 1
A21 2
2
=
ρ
v − v1 ,
2
A22 1
= v12 ρ (4.75)2 − 1 /2.
2(2.12)(1.01×105 Pa)
= 4.45 m/s,
(1000 kg/m3 )(21.6)
and then v2 = (4.75)(4.45 m/s) = 21.2 m/s.
(b) R = π(2.60×10−2 m)2 (4.45 m/s) = 9.45×10−3 m3 /s.
203
(a) For Fig. 16-13 the velocity is constant, or ~v = v î. d~s = îdx + ĵdy. Then
I
I
~v · d~s = v dx = 0,
P16-10
H
because dx = 0.
(b) For Fig. 16-16 the velocity is ~v = (k/r)r̂. d~s = r̂dr + θ̂r dφ. Then
I
I
~v · d~s = v dr = 0,
because
H
dr = 0.
P16-11 (a) For an element of the fluid of mass dm the net force as it moves around the circle is
dF = (v 2 /r)dm. dm/dV = ρ and dV = A dr and dF/A = dp. Then dp/dr = ρv 2 /r.
(b) From Bernoulli’s equation p + ρv 2 /2 is a constant. Then
dv
dp
+ ρv
= 0,
dr
dr
or v/r + dv/dr = 0, or d(vr) = 0. Consequently vr is a constant.
(c) The velocity is ~v = (k/r)r̂. d~s = r̂dr + θ̂r dφ. Then
I
I
~v · d~s = v dr = 0,
because
P16-12
H
dr = 0. This means the flow is irrotational.
F/A = ηv/D, so
F/A = (4.0×1019 N · s/m2 )(0.048 m/3.16×107 s)/(1.9×105 m) = 3.2×105 Pa.
H
P16-13 A flow will be irrotational if and only if ~v · d~s = 0 for all possible paths. It is fairly
easy to construct a rectangular path which is parallel to the flow on the top and bottom sides, but
perpendicular on the left and right sides. Then only the top and bottom paths contribute to the
integral. ~v is constant for either path (but not the same), so the magnitude v will come out of the
integral sign. Since the lengths of the two paths are the same but v is different the two terms don’t
cancel, so the flow is not irrotational.
P16-14 (a) The area of a cylinder is A = 2πrL. The velocity gradient if dv/dr. Then the retarding
force on the cylinder is F = −η(2πrL)dv/dr.
(b) The force pushing a cylinder through is F 0 = A∆p = πr2 ∆p.
(c) Equate, rearrange, and integrate:
πr2 ∆p
∆p
R
Z
r dr
dv
= −η(2πrL) ,
dr
Z v
= 2ηL
dv,
r
0
1
∆p (R2 − r2 )
2
Then
v=
=
2ηLv.
∆p 2
(R − r2 ).
4ηL
204
P16-15 The volume flux (called Rf to distinguish it from the radius R) through an annular ring
of radius r and width δr is
δRf = δA v = 2πr δr v,
where v is a function of r given by Eq. 16-18. The mass flux is the volume flux times the density,
so the total mass flux is
Z R
dm
δRf
= ρ
dr,
dt
δr
0
Z R
∆p 2
= ρ
2πr
(R − r2 ) dr,
4ηL
0
Z R
πρ∆p
=
(rR2 − r3 )dr,
2ηL 0
πρ∆p 4
=
(R /2 − R4 /4),
2ηL
πρ∆pR4
=
.
8ηL
P16-16 The pressure difference in the tube is ∆p = 4γ/r, where r is the (changing) radius of the
bubble. The mass flux through the tube is
4ρπR4 γ
dm
=
,
dt
8ηLr
R is the radius of the tube. dm = ρdV , and dV = 4πr2 dr. Then
Z r2
Z t 4
R γ
3
r dr =
dt,
r1
0 8ηL
r14 − r24
=
ρR4 γ
t,
2ηL
Then
t=
2(1.80×10−5 N · s/m2 )(0.112 m)
[(38.2×10−3 m)4 − (21.6×10−3 m)4 ] = 3630 s.
(0.54×10−3 m)4 (2.50×10−2 N/m)
205
E17-1 For a perfect spring |F | = k|x|. x = 0.157 m when 3.94 kg is suspended from it. There
would be two forces on the object— the force of gravity, W = mg, and the force of the spring, F .
These two force must balance, so mg = kx or
k=
mg
(3.94 kg)(9.81 m/s2 )
=
= 0.246 N/m.
x
(0.157 m)
Now that we know k, the spring constant, we can find the period of oscillations from Eq. 17-8,
s
r
m
(0.520 kg)
T = 2π
= 2π
= 0.289 s.
k
(0.246 N/m)
E17-2 (a) T = 0.484 s.
(b) f = 1/T = 1/(0.484 s) = 2.07 s−1 .
(c) ω = 2πf = 13.0 rad/s.
(d) k = mω 2 = (0.512 kg)(13.0 rad/s)2 = 86.5 N/m.
(e) v m = ωxm = (13.0 rad/s)(0.347 m) = 4.51 m/s.
(f) F m = mam = (0.512 kg)(13.0 rad/s)2 (0.347 m) = 30.0 N.
E17-3 am = (2πf )2 xm . Then
p
f = (9.81 m/s2 )/(1.20×10−6 m)/(2π) = 455 Hz.
E17-4 (a) ω = (2π)/(0.645 s) = 9.74 rad/s. k = mω 2 = (5.22 kg)(9.74 rad/s)2 = 495 N/m.
(b) xm = v m /ω = (0.153 m/s)/(9.74 rad/s) = 1.57×10−2 m.
(c) f = 1/(0.645 s) = 1.55 Hz.
E17-5 (a) The amplitude is half of the distance between the extremes of the motion, so A = (2.00
mm)/2 = 1.00 mm.
(b) The maximum blade speed is given by v m = ωxm . The blade oscillates with a frequency of
120 Hz, so ω = 2πf = 2π(120 s−1 ) = 754 rad/s, and then v m = (754 rad/s)(0.001 m) = 0.754 m/s.
(c) Similarly, am = ω 2 xm , am = (754 rad/s)2 (0.001 m) = 568 m/s2 .
2
5
E17-6 (a)p
k = mω 2 = (1460
p kg/4)(2π2.95/s) = 1.25×10 N/m
5
(b) f = k/m/2π = (1.25×10 N/m)/(1830 kg/4)/2π = 2.63/s.
E17-7 (a) x = (6.12 m) cos[(8.38 rad/s)(1.90 s) + 1.92 rad] = 3.27 m.
(b) v = −(6.12 m)(8.38/s) sin[(8.38 rad/s)(1.90 s) + 1.92 rad] = 43.4 m/s.
(c) a = −(6.12 m)(8.38/s)2 cos[(8.38 rad/s)(1.90 s) + 1.92 rad] = −229 m/s2 .
(d) f = (8.38 rad/s)/2π = 1.33/s.
(e) T = 1/f = 0.750 s.
E17-8 k = (50.0 lb)/(4.00 in) = 12.5 lb/in.
2
mg =
E17-9
(32 ft/s )(12 in/ft)(12.5 lb/in)
= 30.4 lb.
[2π(2.00/s)]2
If the drive wheel rotates at 193 rev/min then
ω = (193 rev/min)(2π rad/rev)(1/60 s/min) = 20.2 rad/s,
then v m = ωxm = (20.2 rad/s)(0.3825 m) = 7.73 m/s.
206
E17-10 k = (0.325 kg)(9.81 m/s2 )/(1.80×10−2 m) = 177 N/m..
s
r
m
(2.14 kg)
T = 2π
= 2π
= 0.691 s.
k
(177 N/m)
E17-11 For the tides ω = 2π/(12.5 h). Half the maximum occurs when cos ωt = 1/2, or ωt = π/3.
Then t = (12.5 h)/6 = 2.08 h.
E17-12 The two will separate if the (maximum) acceleration exceeds g.
(a) Since ω = 2π/T = 2π/(1.18 s) = 5.32 rad/s the maximum amplitude is
xm = (9.81 m/s2 )/(5.32 rad/s)2 = 0.347 m.
(b) In this case ω =
p
(9.81 m/s2 )/(0.0512 m) = 13.8 rad/s. Then f = (13.8 rad/s)/2π = 2.20/s.
E17-13 (a) ax /x = −ω 2 . Then
p
ω = −(−123 m/s)/(0.112 m) = 33.1 rad/s,
so f = (33.1 rad/s)/2π = 5.27/s.
(b) m = k/ω 2 = (456 N/m)/(33.1 rad/s)2 = 0.416 kg.
(c) x = xm cos ωt; v = −xm ω sin ωt. Combining,
x2 + (v/ω)2 = xm 2 cos2 ωt + xm 2 sin2 ωt = xm 2 .
Consequently,
xm =
p
(0.112 m)2 + (−13.6 m/s)2 /(33.1 rad/s)2 = 0.426 m.
E17-14 x1 = xm cos ωt, x2 = xm cos(ωt + φ). The crossing happens when x1 = xm /2, or when
ωt = π/3 (and other values!). The same constraint happens for x2 , except that it is moving in the
other direction. The closest value is ωt + φ = 2π/3, or φ = π/3.
E17-15 (a) The net force on the three cars is zero before the cable breaks. There are three forces
on the cars: the weight, W , a normal force, N , and the upward force from the cable, F . Then
F = W sin θ = 3mg sin θ.
This force is from the elastic properties of the cable, so
k=
F
3mg sin θ
=
x
x
The frequency of oscillation of the remaining two cars after the bottom car is released is
r
r
r
1
k
1
3mg sin θ
1
3g sin θ
f=
=
=
.
2π 2m
2π
2mx
2π
2x
Numerically, the frequency is
s
r
1
3g sin θ
1
3(9.81 m/s2 ) sin(26◦ )
f=
=
= 1.07 Hz.
2π
2x
2π
2(0.142 m)
(b) Each car contributes equally to the stretching of the cable, so one car causes the cable to
stretch 14.2/3 = 4.73 cm. The amplitude is then 4.73 cm.
207
E17-16 Let the height of one side over the equilibrium position be x. The net restoring force
on the liquid is 2ρAxg, where A is the cross sectional area of the tube and g is the acceleration of
free-fall. This corresponds to a spring constant of k = 2ρAg. The mass of the fluid is m = ρAL.
The period of oscillation is
s
r
m
2L
T = 2π
=π
.
k
g
E17-17 (a) There are two forces on the log. The weight, W = mg, and the buoyant force B.
We’ll assume the log is cylindrical. If x is the length of the log beneath the surface and A the
cross sectional area of the log, then V = Ax is the volume of the displaced water. Furthermore,
mw = ρw V is the mass of the displaced water and B = mw g is then the buoyant force on the log.
Combining,
B = ρw Agx,
where ρw is the density of water. This certainly looks similar to an elastic spring force law, with
k = ρw Ag. We would then expect the motion to be simple harmonic.
(b) The period of the oscillation would be
r
r
m
m
T = 2π
= 2π
,
k
ρw Ag
where m is the total mass of the log and lead. We are told the log is in equilibrium when x = L = 2.56
m. This would give us the weight of the log, since W = B is the condition for the log to float. Then
m=
B
ρw AgL
=
= ρAL.
g
g
From this we can write the period of the motion as
s
s
p
ρAL
(2.56 m)
T = 2π
= 2π L/g = 2π
= 3.21 s.
ρw Ag
(9.81 m/s2 )
E17-18 (a) k = 2(1.18 J)/(0.0984 m)2 = 244 N/m.
(b) m = 2(1.18 J)/(1.22 m/s)2 = 1.59 kg.
(c) f = [(1.22 m/s)/(0.0984 m)]/(2π) = 1.97/s.
E17-19 (a) Equate the kinetic energy of the object just after it leaves the slingshot with the
potential energy of the stretched slingshot.
k=
mv 2
(0.130 kg)(11.2×103 m/s)2
=
= 6.97×106 N/m.
x2
(1.53 m)2
(b) N = (6.97×106 N/m)(1.53 m)/(220 N) = 4.85×104 people.
E17-20 (a) E = kxm 2 /2, U = kx2 /2 = k(xm /2)2 /2 = E/4. K = E − U = 3E/4. The the energy
is 25% potential and 75% kinetic.
√
(b) If U = E/2 then kx2 /2 = kxm 2 /4, or x = xm / 2.
208
E17-21
(a) am = ω 2 xm so
ω=
r
am
=
xm
s
(7.93 × 103 m/s2 )
= 2.06 × 103 rad/s
(1.86 × 10−3 m)
The period of the motion is then
2π
= 3.05×10−3 s.
ω
(b) The maximum speed of the particle is found by
T =
v m = ωxm = (2.06 × 103 rad/s)(1.86 × 10−3 m) = 3.83 m/s.
(c) The mechanical energy is given by Eq. 17-15, except that we will focus on when vx = v m ,
because then x = 0 and
E=
1
1
mv m 2 = (12.3 kg)(3.83 m/s)2 = 90.2 J.
2
2
p
p
E17-22 (a) f = k/m/2π = (988 N/m)/(5.13 kg)/2π = 2.21/s.
(b) U i = (988 N/m)(0.535 m)2 /2 = 141 J.
2
(c) K i = (5.13
p kg)(11.2
p m/s) /2 = 322 J.
(d) xm = 2E/k = 2(322 J + 141 J)/(988 N/m) = 0.968 m.
E17-23 (a) ω =
p
(538 N/m)/(1.26 kg) = 20.7 rad/s.
p
xm = (0.263 m)2 + (3.72 m/s)2 /(20.7 rad/s)2 = 0.319 m.
(b) φ = arctan {−(−3.72 m/s)/[(20.7 rad/s)(0.263 m)]} = 34.3◦ .
E17-24 Before doing anything else apply conservation of momentum. If v0 is the speed of the
bullet just before hitting the block and v1 is the speed of the bullet/block system just after the two
begin moving as one, then v1 = mv0 /(m + M ), where m is the mass of the bullet and M is the mass
of the block.
p
For this system ω = k/(m + M ).
(a) The total energy of the oscillation is 12 (m + M )v12 , so the amplitude is
xm =
r
r
m+M
v1 =
k
m + M mv0
= mv0
k
m+M
s
1
.
k(m + M )
The numerical value is
xm = (0.050 kg)(150 m/s)
s
1
= 0.167 m.
(500 N/m)(0.050 kg + 4.00 kg)
(b) The fraction of the energy is
(m + M )v12
m+M
=
2
mv0
m
m
m+M
2
=
m
(0.050 kg)
=
= 1.23×10−2 .
m+M
(0.050 kg + 4.00 kg)
E17-25 L = (9.82 m/s2 )(1.00 s/2π)2 = 0.249 m.
209
E17-26 T = (180 s)/(72.0). Then
g=
2π(72.0)
(180 s)
2
(1.53 m) = 9.66 m/s.
E17-27 We are interested in the value of θm which will make the second term 2% of the first
term. We want to solve
θm
1
,
0.02 = 2 sin2
2
2
which has solution
√
θm
sin
= 0.08
2
◦
or θm = 33 .
(b) How large is the third term at this angle?
32
32
4 θm
sin
=
22 42
2
22
1
θm
sin2
2
2
2
2
=
9
(0.02)2
4
or 0.0009, which is very small.
E17-28 Since T ∝
p
1/g we have
Tp = Te
s
q
g e /g p = (1.00 s)
(9.78 m/s2 )
= 0.997 s.
(9.834 m/s2 )
E17-29 Let the period of the clock in Paris be T1 . In a day of length D1 = 24 hours it will
undergo n = D/T1 oscillations. In Cayenne the period is T2 . n oscillations should occur in 24 hours,
but since the clock runs slow, D2 is 24 hours + 2.5 minutes elapse. So
T2 = D2 /n = (D2 /D1 )T1 = [(1442.5 min)/(1440.0 min)]T1 = 1.0017T1 .
p
Since the ratio of the periods is (T2 /T1 ) = (g1 /g2 ), the g2 in Cayenne is
g2 = g1 (T1 /T2 )2 = (9.81 m/s2 )/(1.0017)2 = 9.78 m/s2 .
E17-30 (a) Take the differential of
g=
2π(100)
T
2
(10 m) =
4π 2 ×105 m
,
T2
so δg = (−8π 2 ×105 m/T 3 )δT . Note that T is not the period here, it is the time for 100 oscillations!
The relative error is then
δg
δT
= −2 .
g
T
If δg/g = 0.1% then δT /T = 0.05%.
(b) For g ≈ 10 m/s2 we have
p
T ≈ 2π(100) (10 m)/(10 m/s2 ) = 628 s.
Then δT ≈ (0.0005)(987 s) ≈ 300 ms.
210
E17-31 T = 2π
p
(17.3 m)/(9.81 m/s2 ) = 8.34 s.
E17-32 The spring will extend until the force from the spring balances the weight, or when M g =
kh. The frequency of this system is then
r
r
r
k
1
M g/h
1
g
1
=
=
,
f=
2π M
2π
M
2π h
which is the frequency of a pendulum of length h. The mass of the bob is irrelevant.
E17-33
The frequency of oscillation is
1
f=
2π
r
M gd
,
I
where d is the distance from the pivot about which the hoop oscillates and the center of mass of the
hoop.
The rotational inertia I is about an axis through the pivot, so we apply the parallel axis theorem.
Then
I = M d2 + I cm = M d2 + M r2 .
But d is r, since the pivot point is on the rim of the hoop. So I = 2M d2 , and the frequency is
s
r
r
M gd
1
g
1
(9.81 m/s2 )
1
f=
=
=
= 0.436 Hz.
2π 2M d2
2π 2d
2π 2(0.653 m)
(b) Note the above expression looks like the simple pendulum equation if we replace 2d with l.
Then the equivalent length of the simple pendulum is 2(0.653 m) = 1.31 m.
E17-34 Apply Eq. 17-21:
I=
T 2κ
(48.7 s/20.0)2 (0.513 N · m)
=
= 7.70 ×10−2 kg · m2 .
2
4π
4π 2
E17-35 κ = (0.192 N · m)/(0.850 rad) = 0.226 N · m. I = 52 (95.2 kg)(0.148 m)2 = 0.834 kg · m2 .
Then
p
p
T = 2π I/κ = 2π (0.834 kg · m2 )/(0.226 N · m) = 12.1 s.
E17-36 x is d in Eq. 17-29. Since the hole is drilled off center we apply the principle axis theorem
to find the rotational inertia:
1
I=
M L2 + M x2 .
12
Then
1
M L2 + M x2
12
1
(1.00 m)2 + x2
12
(8.33×10−2 m2 ) − (1.55 m)x + x2
T 2 M gx
,
4π 2
2
(2.50 s) (9.81 m/s2 )
=
x,
4π 2
= 0.
=
This has solutions x = 1.49 m and x = 0.0557 m. Use the latter.
211
E17-37 For a stick of length L which can pivot about the end, I = 13 M L2 . The center of mass
of such a stick is located d = L/2 away from the end.
The frequency of oscillation of such a stick is
r
M gd
1
f =
,
2π
I
s
1
M g (L/2)
f =
,
1
2
2π
3ML
r
3g
1
.
f =
2π 2L
p
This means that f is proportional to 1/L, regardless
p of the mass or density of the stick. The ratio
of the frequency of two such sticks is then f2 /f1 = L1 /L2 , which in our case gives
p
p
f2 = f1 L2 /L1 = f1 (L1 )/(2L1 /3) = 1.22f1 .
E17-38 The rotational inertia of the pipe section about the cylindrical axis is
I cm =
M
M 2
r + r22 =
(0.102 m)2 + (0.1084 m)2 = (1.11×10−2 m2 )M
2 1
2
(a) The total rotational inertia about the pivot axis is
I = 2I cm + M (0.102 m)2 + M (0.3188 m)2 = (0.134 m2 )M.
The period of oscillation is
T = 2π
s
(0.134 m2 )M
= 1.60 s
M (9.81 m/s2 )(0.2104 m)
(b) The rotational inertia of the pipe section about a diameter is
I cm =
M
M 2
r1 + r22 =
(0.102 m)2 + (0.1084 m)2 = (5.54×10−3 m2 )M
4
4
The total rotational inertia about the pivot axis is now
I = M (1.11×10−2 m2 ) + M (0.102 m)2 + M (5.54×10−3 m2 ) + M (0.3188 m)2 = (0.129 m2 )M
The period of oscillation is
T = 2π
s
(0.129 m2 )M
= 1.57 s.
M (9.81 m/s2 )(0.2104 m)
The percentage difference with part (a) is (0.03 s)/(1.60 s) = 1.9%.
E17-39
E17-40
E17-41 (a) Since effectively x = y, the path is a diagonal line.
(b) The path will be an ellipse which is symmetric about the line x = y.
(c) Since cos(ωt + 90◦ ) = − sin(ωt), the path is a circle.
212
E17-42 (a)
(b) Take two time derivatives and multiply by m,
~ = −mAω 2 î cos ωt + 9ĵ cos 3ωt .
F
R
~ · d~r, so
(c) U = − F
U=
1
mA2 ω 2 cos2 ωt + 9 cos2 3ωt .
2
K=
1
mA2 ω 2 sin2 ωt + 9 sin2 3ωt ;
2
(d) K = 12 mv 2 , so
And then E = K + U = 5mA2 ω 2 .
(e) Yes; the period is 2π/ω.
E17-43 The ω which describes the angular velocity in uniform circular motion is effectively the
samep
ω which describes the angular frequency of the corresponding simple harmonic motion. Since
ω = k/m, we can find the effective force constant k from knowledge of the Moon’s mass and the
period of revolution.
The moon orbits with a period of T , so
ω=
2π
2π
=
= 2.66×10−6 rad/s.
T
(27.3 × 24 × 3600 s)
This can be used to find the value of the effective force constant k from
k = mω 2 = (7.36×1022 kg)(2.66×10−6 rad/s)2 = 5.21×1011 N/m.
E17-44 (a) We want to know when e−bt/2m = 1/3, or
t=
2m
2(1.52 kg)
ln 3 =
ln 3 = 14.7 s
b
(0.227 kg/s)
(b) The (angular) frequency is
s
2
(8.13 N/m)
(0.227 kg/s)
0
ω =
−
= 2.31 rad/s.
(1.52 kg)
2(1.52 kg)
The number of oscillations is then
(14.7 s)(2.31 rad/s)/2π = 5.40
E17-45
The first derivative of Eq. 17-39 is
dx
dt
= xm (−b/2m)e−bt/2m cos(ω 0 t + φ) + xm e−bt/2m (−ω 0 ) sin(ω 0 t + φ),
= −xm e−bt/2m ((b/2m) cos(ω 0 t + φ) + ω 0 sin(ω 0 t + φ))
The second derivative is quite a bit messier;
d2
dx2
= −xm (−b/2m)e−bt/2m ((b/2m) cos(ω 0 t + φ) + ω 0 sin(ω 0 t + φ))
−xm e−bt/2m (b/2m)(−ω 0 ) sin(ω 0 t + φ) + (ω 0 )2 cos(ω 0 t + φ) ,
= xm e−bt/2m (ω 0 b/m) sin(ω 0 t + φ) + (b2 /4m2 − ω 02 ) cos(ω 0 t + φ) .
213
Substitute these three expressions into Eq. 17-38. There are, however, some fairly obvious simplifications. Every one of the terms above has a factor of xm , and every term above has a factor of
e−bt/2m , so simultaneously with the substitution we will cancel out those factors. Then Eq. 17-38
becomes
m (ω 0 b/m) sin(ω 0 t + φ) + (b2 /4m2 − ω 02 ) cos(ω 0 t + φ)
−b [(b/2m) cos(ω 0 t + φ) + ω 0 sin(ω 0 t + φ)] + k cos(ω 0 t + φ) = 0
Now we collect terms with cosine and terms with sine,
(ω 0 b − ω 0 b) sin(ω 0 t + φ) + mb2 /4m2 − ω 02 − b2 /2m + k cos(ω 0 t + φ) = 0.
The coefficient for the sine term is identically zero; furthermore, because the cosine term must then
vanish regardless of the value of t, the coefficient for the sine term must also vanish. Then
mb2 /4m2 − mω 02 − b2 /2m + k = 0,
or
k
b2
−
.
m 4m2
If this condition is met, then Eq. 17-39 is indeed a solution of Eq. 17-38.
ω 02 =
E17-46 (a) Four complete cycles requires a time t4 = 8π/ω 0 . The amplitude decays to 3/4 the
original value in this time, so 0.75 = e−bt4 /2m , or
ln(4/3) =
8πb
.
2mω 0
It is probably reasonable at this time to assume that b/2m is small compared to ω so that ω 0 ≈ ω.
We’ll do it the hard way anyway. Then
k
−
m
8π
ln(4/3)
2 b
2m
2
ω 02
=
2
2 2
8π
b
=
,
ln(4/3)
2m
2
b
= (7630)
2m
b
2m
k
m
,
Numerically, then,
b=
s
4(1.91 kg)(12.6 N/m)
= 0.112 kg/s.
(7630)
(b)
E17-47 (a) Use Eqs. 17-43 and 17-44. At resonance ω 00 = ω, so
√
G = b2 ω 2 = bω,
and then xm = F m /bω.
(b) v m = ωxm = F m /b.
214
E17-48 We need the first two derivatives of
x=
Fm
cos(ω 00 t − β)
G
The derivatives are easy enough to find,
dx
Fm
=
(−ω 00 ) sin(ω 00 t − β),
dt
G
and
F m 00 2
d2 x
=−
(ω ) cos(ω 00 t − β),
dt2
G
We’ll substitute this into Eq. 17-42,
F m 00 2
00
,m −
(ω ) cos(ω t − β)
G
Fm
Fm
00
00
+b
(−ω ) sin(ω t − β) + k
cos(ω 00 t − β) = F m cos ω 00 t.
G
G
Then we’ll cancel out as much as we can and collect the sine and cosine terms,
k − m(ω 00 )2 cos(ω 00 t − β) − (bω 00 ) sin(ω 00 t − β) = G cos ω 00 t.
We can write the left hand side of this equation in the form
A cos α1 cos α2 − A sin α1 sin α2 ,
if we let α2 = ω 00 t − β and choose A and α1 correctly. The best choice is
A cos α1
A sin α1
= k − m(ω 00 )2 ,
= bω 00 ,
and then taking advantage of the fact that sin2 + cos2 = 1,
2
A2 = k − m(ω 00 )2 + (bω 00 )2 ,
which looks like Eq. 17-44! But then we can apply the cosine angle addition formula, and
A cos(α1 + ω 00 t − β) = G cos ω 00 t.
This expression needs to be true for all time. This means that A = G and α1 = β.
E17-49
The derivatives are easy enough to find,
dx
Fm
=
(−ω 000 ) sin(ω 000 t − β),
dt
G
and
d2 x
F m 000 2
=−
(ω ) cos(ω 000 t − β),
dt2
G
We’ll substitute this into Eq. 17-42,
F m 000 2
,m −
(ω ) cos(ω 000 t − β)
G
Fm
Fm
+b
(−ω 000 ) sin(ω 000 t − β) + k
cos(ω 000 t − β) = F m cos ω 00 t.
G
G
215
Then we’ll cancel out as much as we can and collect the sine and cosine terms,
k − m(ω 000 )2 cos(ω 000 t − β) − (bω 000 ) sin(ω 000 t − β) = G cos ω 00 t.
We can write the left hand side of this equation in the form
A cos α1 cos α2 − A sin α1 sin α2 ,
if we let α2 = ω 000 t − β and choose A and α1 correctly. The best choice is
= k − m(ω 000 )2 ,
= bω 000 ,
A cos α1
A sin α1
and then taking advantage of the fact that sin2 + cos2 = 1,
A2 = k − m(ω 000 )2
2
+ (bω 000 )2 ,
which looks like Eq. 17-44! But then we can apply the cosine angle addition formula, and
A cos(α1 + ω 000 t − β) = G cos ω 00 t.
This expression needs to be true for all time. This means that A = G and α1 + ω 000 t − β = ω 00 t and
α1 = β and ω 000 = ω 00 .
E17-50 Actually, Eq. 17-39 is not a solution to Eq. 17-42 by itself, this is a wording mistake in
the exercise. Instead, Eq. 17-39 can be added to any solution of Eq. 17-42 and the result will still
be a solution.
Let xn be any solution to Eq. 17-42 (such as Eq. 17-43.) Let xh be given by Eq. 17-39. Then
x = xn + xh .
Take the first two time derivatives of this expression.
dx
dt
d2 x
dt2
=
=
dxn
dxh
+
,
dt
dt
2
2
d xn
d xh
+
2
dt
dt2
Substitute these three expressions into Eq. 17-42.
2
d xn
d2 xh
dxn
dxh
m
+
+
b
+
+ k (xn + xh ) = F m cos ω 00 t.
dt2
dt2
dt
dt
Rearrange and regroup.
d2 xn
dxn
d2 xh
dxh
m 2 +b
+ kxn + m 2 + b
+ kxh = F m cos ω 00 t.
dt
dt
dt
dt
Consider the second term on the left. The parenthetical expression is just Eq. 17-38, the damped
harmonic oscillator equation. It is given in the text (and proved in Ex. 17-45) the xh is a solution,
so this term is identically zero. What remains is Eq. 17-42; and we took as a given that xn was a
solution.
(b) The “add-on” solution of xh represents the transient motion that will die away with time.
216
E17-51
The time between “bumps” is the solution to
vt = x,
t
=
(13 ft)
(10 mi/hr)
1 mi
5280 ft
3600 s
1 hr
= 0.886 s
The angular frequency is
2π
= 7.09 rad/s
T
This is the driving frequency, and the problem states that at this frequency the up-down bounce
oscillation is at a maximum. This occurs when the driving frequency is approximately equal to the
natural frequency of oscillation. The force constant for the car is k, and this is related to the natural
angular frequency by
W 2
k = mω 2 =
ω ,
g
ω=
where W = (2200 + 4 × 180) lb= 2920 lb is the weight of the car and occupants. Then
k=
(2920 lb)
2
(32 ft/s )
(7.09 rad/s)2 = 4590 lb/ft
When the four people get out of the car there is less downward force on the car springs. The
important relationship is
∆F = k∆x.
In this case ∆F = 720 lb, the weight of the four people who got out of the car. ∆x is the distance
the car will rise when the people get out. So
∆x =
∆F
(720 lb)
=
= 0.157 ft ≈ 2 in.
k
4590 lb/ft
E17-52 The derivative is easy enough to find,
dx
Fm
=
(−ω 00 ) sin(ω 00 t − β),
dt
G
The velocity amplitude is
vm
F m 00
ω ,
G
=
Fm
=
1
ω 00
p
=
m2 (ω 002 − ω 2 )2 + b2 ω 002
Fm
p
.
00
(mω − k/ω 00 )2 + b2
,
Note that this is exactly a maximum when ω 00 = ω.
E17-53 The reduced mass is
m = (1.13 kg)(3.24 kg)/(1.12 kg + 3.24 kg) = 0.840 kg.
The period of oscillation is
T = 2π
p
(0.840 kg)/(252 N/m) = 0.363 s
217
E17-54
E17-55 Start by multiplying the kinetic energy expression by (m1 + m2 )/(m1 + m2 ).
K
=
=
(m1 + m2 )
m1 v12 + m2 v22 ,
2(m1 + m2 )
1
m21 v12 + m1 m2 (v12 + v22 ) + m22 v22 ,
2(m1 + m2 )
and then add 2m1 m2 v1 v2 − 2m1 m2 v1 v2 ,
K
=
=
1
m21 v12 + 2m1 m2 v1 v2 + m22 v22 + m1 m2 (v12 + v22 − 2v1 v2 ) ,
2(m1 + m2 )
1
(m1 v1 + m2 v2 )2 + m1 m2 (v1 − v2 )2 .
2(m1 + m2 )
But m1 v2 + m2 v2 = 0 by conservation of momentum, so
K
=
=
(m1 m2 )
2
(v1 − v2 ) ,
2(m1 + m2 )
m
(v1 − v2 )2 .
2
P17-1 The mass of one silver atom is (0.108 kg)/(6.02×1023 ) = 1.79×10−25 kg. The effective spring
constant is
k = (1.79×10−25 kg)4π 2 (10.0×1012 /s)2 = 7.07×102 N/m.
P17-2 (a) Rearrange Eq. 17-8 except replace m with the total mass, or m+M . Then (M +m)/k =
T 2 /(4π 2 ), or
M = (k/4π 2 )T 2 − m.
(b) When M = 0 we have
m = [(605.6 N/m)/(4π 2 )](0.90149 s)2 = 12.467 kg.
(c) M = [(605.6 N/m)/(4π 2 )](2.08832 s)2 − (12.467 kg) = 54.432 kg.
P17-3
The maximum static friction is Ff ≤ µs N . Then
Ff = µs N = µs W = µs mg
is the maximum available force to accelerate the upper block. So the maximum acceleration is
am =
Ff
= µs g
m
The maximum possible amplitude of the oscillation is then given by
xm =
am
µs g
=
,
2
ω
k/(m + M )
where in the last part we substituted the total mass of the two blocks because both blocks are
oscillating. Now we put in numbers, and find
xm =
(0.42)(1.22 kg + 8.73 kg)(9.81 m/s2 )
= 0.119 m.
(344 N/m)
218
P17-4 (a) Equilibrium occurs when F = 0, or b/r3 = a/r2 . This happens when r = b/a.
(b) dF/dr = 2a/r3 − 3b/r4 . At r = b/a this becomes
dF/dr = 2a4 /b3 − 3a4 /b3 = −a4 /b3 ,
4 3
which corresponds
√ constant of a /b .
p to a force
2
3
(c) T = 2π m/k = 2π mb /a , where m is the reduced mass.
P17-5 Each spring helps to restore the block. The net force on the block is then of magnitude
F1 + F2 = k1 x + k2 x = (k1 + k2 )x = kx. We can then write the frequency as
r
r
k
1
k1 + k2
1
=
.
f=
2π m
2π
m
With a little algebra,
f
1
=
2π
r
k1 + k2
,
m
1 k1
1 k2
+ 2 ,
4π 2 m
4π m
=
=
r
q
f12 + f22 .
P17-6 The tension in the two spring is the same, so k1 x1 = k2 x2 , where xi is the extension of the
ith spring. The total extension is x1 + x2 , so the effective spring constant of the combination is
F
F
1
1
k1 k2
=
=
=
=
.
x
x1 + x2
x1 /F + x2 /F
1/k1 + 1/k2
k1 + k2
The period is then
1
T =
2π
r
k
1
=
m
2π
s
k1 k2
(k1 + k2 )m
With a little algebra,
f
=
1
2π
s
k1 k2
,
(k1 + k2 )m
=
1
2π
s
1
,
m/k1 + m/k2
=
1
2π
s
1
,
1/ω12 + 1/ω22
=
1
2π
s
ω12 ω22
,
ω12 + ω22
=
f1 f2
p
f12 + f22
219
.
P17-7 (a) When a spring is stretched the tension is the same everywhere in the spring. The
stretching, however, is distributed over the entire length of the spring, so that the relative amount
of stretch is proportional to the length of the spring under consideration. Half a spring, half the
extension. But k = −F/x, so half the extension means twice the spring constant.
In short, cutting the spring in half will create two stiffer springs with twice the spring constant,
so k = 7.20 N/cm for each spring.
(b) The two spring halves now support a mass M . We can view this as each spring is holding
one-half of the total mass, so in effect
s
1
k
f=
2π M/2
or, solving for M ,
M=
2(720 N/m)
2k
=
= 4.43 kg.
4π 2 f 2
4π 2 (2.87 s−1 )2
P17-8 Treat the spring a being composed of N massless springlets each with a point mass ms /N
at the end. The spring constant for each springlet will be kN . An expression for the conservation
of energy is then
N
N
m 2 ms X 2 N k X 2
v +
vn +
xn = E.
2
2N 1
2 1
Since the spring stretches proportionally along the length then we conclude that each springlet
compresses the same amount, and then xn = A/N sin ωt could describe the change in length of each
springlet. The energy conservation expression becomes
N
m 2 ms X 2 k 2 2
v +
vn + A sin ωt = E.
2
2N 1
2
v = Aω cos ωt. The hard part to sort out is the vn , since the displacements for all springlets to one
side of the nth must be added to get the net displacement. Then
vn = n
A
ω cos ωt,
N
and the energy expression becomes
N
m
ms X 2
+
n
2
2N 3 1
!
A2 ω 2 cos2 ωt +
k 2 2
A sin ωt = E.
2
Replace the sum with an integral, then
1
N3
Z
N
n2 dn =
0
1
,
3
and the energy expression becomes
1
ms 2 2
k
m+
A ω cos2 ωt + A2 sin2 ωt = E.
2
3
2
This will only be constant if
or T = 2π
ms ω2 = m +
/k,
3
p
(m + ms /3)/k.
220
P17-9
(a) Apply conservation of energy. When x = xm v = 0, so
1
kxm 2
2
xm 2
xm
1
1
m[v(0)]2 + k[x(0)]2 ,
2
2
m
=
[v(0)]2 + [x(0)]2 ,
k
p
=
[v(0)/ω]2 + [x(0)]2 .
=
(b) When t = 0 x(0) = xm cos φ and v(0) = −ωxm sin φ, so
sin φ
v(0)
=−
= tan φ.
ωx(0)
cos φ
P17-10
P17-11
Conservation of momentum for the bullet block collision gives mv = (m + M )v f or
vf =
m
v.
m+M
This v f will be equal to the maximum oscillation speed v m . The angular frequency for the oscillation
is given by
r
k
ω=
.
m+M
Then the amplitude for the oscillation is
r
vm
m
m+M
mv
xm =
=v
=p
.
ω
m+M
k
k(m + M )
P17-12 (a) W = F s , or mg = kx, so x = mg/k.
(b) F = ma, but F = W − F s = mg − kx, and since ma = m d2 x/dt2 ,
m
d2 x
+ kx = mg.
dt2
The solution can be verified by direct substitution.
(c) Just look at the answer!
(d) dE/dt is
mv
dv
dx
dx
+ kx
− mg
= 0,
dt
dt
dt
dv
m
+ kx = mg.
dt
P17-13 The initial energy stored in the spring is kxm 2 /2. When the cylinder passes through the
equilibrium point it has a translational velocity v m and a rotational velocity ω r = v m /R, where R
is the radius of the cylinder. The total kinetic energy at the equilibrium point is
1
1
1
1
2
2
mv m + Iω r =
m + m vm 2 .
2
2
2
2
Then the kinetic energy is 2/3 translational and 1/3 rotational. The total energy of the system is
E=
1
(294 N/ m)(0.239 m)2 = 8.40 J.
2
221
(a) K t = (2/3)(8.40 J) = 5.60 J.
(b) K r = (1/3)(8.40 J) = 2.80 J.
(c) The energy expression is
1
2
3m
2
1
v 2 + kx2 = E,
2
which leads to a standard expression for the period with 3M/2 replacing m. Then T = 2π
p
3M/2k.
P17-14 (a) Integrate the potential energy expression over one complete period and then divide by
the time for one period:
ω
2π
Z
2π/ω
0
1 2
kx dt
2
=
=
=
Z
kω 2π/ω
xm 2 cos2 ωt dt,
4π 0
kω 2 π
xm ,
4π
ω
1
2
kxm .
4
This is half the total energy; since the average total energy is E, then the average kinetic energy
must be the other half of the average total energy, or (1/4)kxm 2 .
(b) Integrate over half a cycle and divide by twice the amplitude.
Z xm
1
1 2
1 1
kx dx =
kxm 3 ,
2xm −xm 2
2xm 3
1
=
kxm 2 .
6
This is one-third the total energy. The average kinetic energy must be two-thirds the total energy,
or (1/3)kxm 2 .
P17-15
The rotational inertia is
I=
1
M R 2 + M d2 = M
2
1
(0.144 m)2 + (0.102 m)2
2
= (2.08×10−2 m2 )M.
The period of oscillation is
T = 2π
s
I
= 2π
M gd
s
(2.08×10−2 m2 )M
= 0.906 s.
M (9.81 m/s2 )(0.102 m)
P17-16
(a) The rotational inertia of the pendulum about the pivot is
1
1
(0.488 kg)
(0.103 m)2 + (0.103 m + 0.524 m)2 + (0.272 kg)(0.524 m)2 = 0.219 kg · m2 .
2
3
(b) The center of mass location is
d=
(0.524 m)(0.272 kg)/2 + (0.103 m + 0.524 m)(0.488 kg)
= 0.496 m.
(0.272 kg) + (0.488 kg)
(c) The period of oscillation is
p
T = 2π (0.291 kg · m2 )/(0.272 kg + 0.488 kg)(9.81 m/s2 )(0.496 m) = 1.76 s.
222
P17-17 (a) The rotational inertia of a stick about an axis through a point which is a distance d
from the center of mass is given by the parallel axis theorem,
I = I cm + md2 =
1
mL2 + md2 .
12
The period of oscillation is given by Eq. 17-28,
s
s
I
L2 + 12d2
T = 2π
= 2π
mgd
12gd
(b) We want to find the minimum period, so we need to take the derivative of T with respect to
d. It’ll look weird, but
dT
12d2 − L2
= πp
.
dd
12gd3 (L2 + 12d2 )
√
This will vanish when 12d2 = L2 , or when d = L/ 12.
P17-18
is
The energy stored in the spring is given by kx2 /2, the kinetic energy of the rotating wheel
v 2
1
(M R2 )
,
2
r
where v is the tangential velocity of the point of attachment of the spring to the wheel. If x =
xm sin ωt, then v = xm ω cos ωt, and the energy will only be constant if
ω2 =
k r2
.
M R2
P17-19 The method of solution is identical to the approach for the simple pendulum on page 381
except replace the tension with the normal force of the bowl on the particle. The effective pendulum
with have a length R.
P17-20 Let x be the distance from the center of mass to the first pivot point. Then the period is
given by
s
I + M x2
T = 2π
.
M gx
Solve this for x by expressing the above equation as a quadratic:
M gT 2
x = I + M x2 ,
4π 2
or
2
Mx −
M gT 2
4π 2
x+I =0
There are two solutions. One corresponds to the first location, the other the second location. Adding
the two solutions together will yield L; in this case the discriminant of the quadratic will drop out,
leaving
M gT 2
gT 2
L = x1 + x2 =
=
.
2
M 4π
4π 2
Then g = 4π 2 L/T 2 .
223
P17-21
In this problem
I = (2.50 kg)
(0.210 m)2
+ (0.760 m + 0.210 m)2
2
= 2.41 kg/ · m2 .
The center of mass
p is at the center of the disk.
(a) T = 2π (2.41 kg/ · m2 )/(2.50 kg)(9.81 m/s2 )(0.760 m + 0.210 m) = 2.00 s.
(b) Replace M gd with M gd + κ and 2.00 s with 1.50 s. Then
κ=
4π 2 (2.41 kg/ · m2 )
− (2.50 kg)(9.81 m/s2 )(0.760 m + 0.210 m) = 18.5 N · m/rad.
(1.50 s)2
P17-22 The net force on the bob is toward the center of the circle, and has magnitude F net =
mv 2 /R. This net force comes from the horizontal component of the tension. There is also a vertical
component of the tension of magnitude mg. The tension then has magnitude
p
p
T = (mg)2 + (mv 2 /R)2 = m g 2 + v 4 /R2 .
It is this tension
pwhich is important in finding the restoring force in Eq. 17-22; in effect we want to
replace g with g 2 + v 4 /R2 in Eq. 17-24. The frequency will then be
s
L
1
p
.
f=
2
2π
g + v 4 /R2
P17-23 (a) Consider an object of mass m at a point P on the axis of the ring. It experiences a
gravitational force of attraction to all points on the ring; by symmetry, however, the net force is not
directed toward the circumference of the ring, but instead along the axis of the ring. There is then
a factor of cos θ which will be thrown in to the mix.
√
The distance from P to any point on the ring is r = R2 + z 2 , and θ is the angle between the
axis on the line which connects P and any point on the circumference. Consequently,
cos θ = z/r,
and then the net force on the star of mass m at P is
F =
GM mz
GM mz
GM m
cos θ =
=
.
r2
r3
(R2 + z 2 )3/2
(b) If z R we can apply the binomial expansion to the denominator, and
2
2 −3/2
(R + z )
=R
−3
1+
z 2 −3/2
R
≈R
−3
3 z 2
1−
.
2 R
Keeping terms only linear in z we have
F =
GM m
z,
R3
which corresponds to a spring constant k = GM m/R3 . The frequency of oscillation is then
p
p
f = k/m/(2π) = GM/R3 /(2π).
(c) Using some numbers from the Milky Way galaxy,
q
f = (7×10−11 N · m2 /kg2 )(2×1043 kg)/(6×1019 m)3 /(2π) = 1×10−14 Hz.
224
P17-24 (a) The acceleration of the center of mass (point C) is a = F/M . The torque about an
axis through the center of mass is tau = F R/2, since O is R/2 away from the center of mass. The
angular acceleration of the disk is then
α = τ /I = (F R/2)/(M R2 /2) = F/(M R).
Note that the angular acceleration will tend to rotate the disk anti-clockwise. The tangential component to the angular acceleration at P is aT = −αR = −F/M ; this is exactly the opposite of the
linear acceleration, so P will not (initially) accelerate.
(b) There is no net force at P .
P17-25
The value for k is closest to
k ≈ (2000 kg/4)(9.81 m/s2 )/(0.10 m) = 4.9×104 N/m.
One complete oscillation requires a time t1 = 2π/ω 0 . The amplitude decays to 1/2 the original
value in this time, so 0.5 = e−bt1 /2m , or
ln(2) =
2πb
.
2mω 0
It is not reasonable at this time to assume that b/2m is small compared to ω so that ω 0 ≈ ω. Then
k
−
m
2π
ln(2)
2 b
2m
2
ω 02
=
2
2 2
2π
b
=
,
ln(2)
2m
2
b
= (81.2)
2m
b
2m
k
m
,
Then the value for b is
b=
P17-26
s
4(2000 kg/4)(4.9×104 N/m)
= 1100 kg/s.
(81.2)
a = d2 x/dt2 = −Aω 2 cos ωt. Substituting into the non-linear equation,
−mAω 2 cos ωt + kA3 cos3 ωt = F cos ω d t.
Now let ω d = 3ω. Then
kA3 cos3 ωt − mAω 2 cos ωt = F cos 3ωt
Expand the right hand side as cos 3ωt = 4 cos3 ωt − 3 cos ωt, then
kA3 cos3 ωt − mAω 2 cos ωt = F (4 cos3 ωt − 3 cos ωt)
This will only work if 4F = kA3 and 3F = mAω 2 . Dividing one condition by the other means
4mAω 2 = kA3 , so A ∝ ω and then F ∝ ω 3 ∝ ω d 3 .
225
P17-27
(a) Divide the top and the bottom by m2 . Then
m1 m2
m1
=
,
m1 + m2
(m1 /m2 ) + 1
and in the limit as m2 → ∞ the value of (m1 /m2 ) → 0, so
lim
m2 →∞
m1 m2
m1
= lim
= m1 .
m
→∞
m1 + m2
(m1 /m2 ) + 1
2
(b) m is called the reduced mass because it is always less than either m1 or m2 . Think about it
in terms of
m2
m1
m=
=
.
(m1 /m2 ) + 1
(m2 /m1 ) + 1
Since mass is always positive, the denominator ispalways greater than or equal to 1. Equality only
occurs if one of the masses is infinite. Now ω = k/m, and since m is always less than m1 , so the
existence of a finite wall will cause ω to be larger, and the period to be smaller.
p
(c) If the bodies have equal mass then m = m1 /2. This corresponds to a value of ω = 2k/m1 .
In effect, the spring constant is doubled, which is what happens if a spring is cut in half.
226
E18-1 (a) f = v/λ = (243 m/s)/(0.0327 m) = 7.43×103 Hz.
(b) T = 1/f = 1.35×10−4 s.
E18-2 (a) f = (12)/(30 s) = 0.40 Hz.
(b) v = (15 m)/(5.0 s) = 3.0 m/s.
(c) λ = v/f = (3.0 m/s)/(0.40 Hz) = 7.5 m.
E18-3 (a) The time for a particular point to move from maximum displacement to zero displacement is one-quarter of a period; the point must then go to maximum negative displacement,
zero displacement, and finally maximum positive displacement to complete a cycle. So the period is
4(178 ms) = 712 ms.
(b) The frequency is f = 1/T = 1/(712×10−3 s) = 1.40 Hz.
(c) The wave-speed is v = f λ = (1.40 Hz)(1.38 m) = 1.93 m/s.
E18-4 Use Eq. 18-9, except let f = 1/T :
x
y = (0.0213 m) sin 2π
− (385 Hz)t = (0.0213 m) sin [(55.1 rad/m)x − (2420 rad/s)t] .
(0.114 m)
E18-5 The dimensions for tension are [F] = [M][L]/[T]2 where M stands for mass, L for length, T
for time, and F stands for force. The dimensions for linear mass density are [M]/[L]. The dimensions
for velocity are [L]/[T].
Inserting this into the expression v = F a /µb ,
[L]
=
[T]
[M][L]
[T]2
a b
[M]
/
,
[L]
[L]
[M]a [L]a [L]b
=
,
[T]
[T]2a [M]b
[L]
[M]a−b [L]a+b
=
[T]
[T]2a
There are three equations here. One for time, −1 = −2a; one for length, 1 = a + b; and one for
mass, 0 = a − b. We need to satisfy all three equations. The first is fairly quick; a = 1/2. Either of
the other equations can be used to show that b = 1/2.
E18-6 (a) y m = 2.30 mm.
(b) f = (588 rad/s)/(2π rad) = 93.6 Hz.
(c) v = (588 rad/s)/(1822 rad/m) = 0.323 m/s.
(d) λ = (2π rad)/(1822 rad/m) = 3.45 mm.
(e) uy = y m ω = (2.30 mm)(588 rad/s) = 1.35 m/s.
E18-7 (a) y m = 0.060 m.
(b) λ = (2π rad)/(2.0π rad/m) = 1.0 m.
(c) f = (4.0π rad/s)/(2π rad) = 2.0 Hz.
(d) v = (4.0π rad/s)/(2.0π rad/m) = 2.0 m/s.
(e) Since the second term is positive the wave is moving in the −x direction.
(f) uy = y m ω = (0.060 m)(4.0π rad/s) = 0.75 m/s.
E18-8 v =
p
p
F/µ = (487 N)/[(0.0625 kg)/(2.15 m)] = 129 m/s.
227
E18-9
We’ll first find the linear mass density by rearranging Eq. 18-19,
µ=
F
v2
Since this is the same string, we expect that changing the tension will not significantly change the
linear mass density. Then for the two different instances,
F2
F1
= 2
v12
v2
We want to know the new tension, so
F2 = F1
v22
(180 m/s)2
= (123 N)
= 135 N
2
v1
(172 m/s)2
E18-10 First v = (317 rad/s)/(23.8 rad/m) = 13.32 m/s. Then
µ = F/v 2 = (16.3 N)/(13.32 m/s)2 = 0.0919 kg/m.
E18-11 (a) ym = 0.05 m.
(b) λ = p
(0.55 m) −
p(0.15 m) = 0.40 m.
(c) v = F/µ = (3.6 N)/(0.025 kg/m) = 12 m/s.
(d) T = 1/f = λ/v = (0.40 m)/(12 m/s) = 3.33×10−2 s.
(e) uy = y m ω = 2πy m /T = 2π(0.05 m)/(3.33×10−2 s) = 9.4 m/s.
(f) k = (2π rad)/(0.40 m) = 5.0π rad/m; ω = kv = (5.0π rad/m)(12 m/s) = 60π rad/s. The
phase angle can be found from
(0.04 m) = (0.05 m) sin(φ),
or φ = 0.93 rad. Then
y = (0.05 m) sin[(5.0π rad/m)x + (60π rad/s)t + (0.93 rad)].
E18-12 (a) The tensions in the two strings are equal, so F = (0.511 kg)(9.81 m/s2 )/2 = 2.506 N.
The wave speed in string 1 is
p
p
v = F/µ = (2.506 N)/(3.31×10−3 kg/m) = 27.5 m/s,
while the wave speed in string 2 is
p
p
v = F/µ = (2.506 N)/(4.87×10−3 kg/m) = 22.7 m/s.
p
p
(b) We have F1 /µ1 = F2 /µ2 , or F1 /µ1 = F2 /µ2 . But Fi = Mi g, so M1 /µ1 = M2 /µ2 . Using
M = M1 + M2 ,
M1
µ1
M1
M1
+
µ1
µ2
M1
=
=
=
=
=
M − M1
,
µ2
M
,
µ2
M
1
1
+
,
µ2
µ1
µ2
(0.511 kg)
1
1
+
,
(4.87×10−3 kg/m)
(3.31×10−3 kg/m) (4.87×10−3 kg/m)
0.207 kg
and M2 = (0.511 kg) − (0.207 kg) = 0.304 kg.
228
E18-13
18-19,
We need to know the wave speed before we do anything else. This is found from Eq.
v=
s
F
=
µ
s
F
=
m/L
s
(248 N)
= 162 m/s.
(0.0978 kg)/(10.3 m)
The two pulses travel in opposite directions on the wire; one travels as distance x1 in a time t, the
other travels a distance x2 in a time t + 29.6 ms, and since the pulses meet, we have x1 + x2 = 10.3
m.
Our equations are then x1 = vt = (162 m/s)t, and x2 = v(t+29.6 ms) = (162 m/s)(t+29.6 ms) =
(162 m/s)t + 4.80 m. We can add these two expressions together to solve for the time t at which the
pulses meet,
10.3 m = x1 + x2 = (162 m/s)t + (162 m/s)t + 4.80 m = (324 m/s)t + 4.80 m.
which has solution t = 0.0170 s. The two pulses meet at x1 = (162 m/s)(0.0170 s) = 2.75 m, or
x2 = 7.55 m.
E18-14 (a) ∂y/∂r = (A/r)k cos(kr − ωt) − (A/r2 ) sin(kr − ωt). Multiply this by r2 , and then find
∂ 2 ∂y
r
= Ak cos(kr − ωt) − Ak 2 r sin(kr − ωt) − Ak cos(kr − ωt).
∂r ∂r
Simplify, and then divide by r2 to get
1 ∂ 2 ∂y
r
= −(Ak 2 /r) sin(kr − ωt).
r2 ∂r ∂r
Now find ∂ 2 y/∂t2 = −Aω 2 sin(kr − ωt). But since 1/v 2 = k 2 /ω 2 , the two sides are equal.
(b) [length]2 .
E18-15
The liner mass density is µ = (0.263 kg)/(2.72 m) = 9.669×10−2 kg/m. The wave speed is
p
v = (36.1 N)/(9.669×10−2 kg/m) = 19.32 m/s.
P av = 21 µω 2 y m 2 v, so
ω=
s
2(85.5 W)
(9.669×10−2 kg/m)(7.70×10−3 m)2 (19.32 m/s)
= 1243 rad/s.
Then f = (1243 rad/s)/2π = 199 Hz.
E18-16 (a) If the medium absorbs no energy then the power flow through any closed surface
which contains the source must be constant. Since for a cylindrical surface the area grows as r, then
intensity must fall of as 1/r.
√
(b) Intensity is proportional to the amplitude squared, so the amplitude must fall off as 1/ r.
E18-17 The intensity is the average power per unit area (Eq. 18-33); as you get farther from the
source the intensity falls off because the perpendicular area increases. At some distance r from the
source the total possible area is the area of a spherical shell of radius r, so intensity as a function of
distance from the source would be
P av
I=
4πr2
229
We are given two intensities: I1 = 1.13 W/m2 at a distance r1 ; I2 = 2.41 W/m2 at a distance
r2 = r1 − 5.30 m. Since the average power of the source is the same in both cases we can equate
these two values as
4πr12 I1
4πr12 I1
=
=
4πr22 I2 ,
4π(r1 − d)2 I2 ,
where d = 5.30 m, and then solve for r1 . Doing this we find a quadratic expression which is
(r2 − 2dr1 + d2 )I2 ,
1
I1
0 =
1−
r12 − 2dr1 + d2 ,
I2
(1.13 W/m2 )
r2 − 2(5.30 m)r1 + (5.30 m)2 ,
0 =
1−
(2.41 W/m2 ) 1
r12 I1
=
0
=
(0.531)r12 − (10.6 m)r1 + (28.1 m2 ).
The solutions to this are r1 = 16.8 m and r1 = 3.15 m; but since the person walked 5.3 m toward
the lamp we will assume they started at least that far away. Then the power output from the light
is
P = 4πr12 I1 = 4π(16.8 m)2 (1.13 W/m2 ) = 4.01×103 W.
E18-18 Energy density is energy per volume, or u = U/V . A wave front of cross sectional area A
sweeps out a volume of V = Al when it travels a distance l. The wave front travels that distance
l in a time t = l/v. The energy flow per time is the power, or P = U/t. Combine this with the
definition of intensity, I = P/A, and
I=
E18-19
P
U
uV
uAl
=
=
=
= uv.
A
At
At
At
Refer to Eq. 18-40, where the amplitude of the combined wave is
2y m cos(∆φ/2),
where y m is the amplitude of the combining waves. Then
cos(∆φ/2) = (1.65y m )/(2y m ) = 0.825,
which has solution ∆φ = 68.8◦ .
E18-20 Consider only the point x = 0. The displacement y at that point is given by
y = y m1 sin(ωt) + y m2 sin(ωt + π/2) = y m1 sin(ωt) + y m2 cos(ωt).
This can be written as
y = y m (A1 sin ωt + A2 cos ωt),
where Ai = y mi /y m . But if y m is judiciously chosen, A1 = cos β and A2 = sin β, so that
y = y m sin(ωt + β).
Since we then require A21 + A22 = 1, we must have
p
y m = (3.20 cm)2 + (4.19 cm)2 = 5.27 cm.
230
E18-21 The easiest approach is to use a phasor representation of the waves.
Write the phasor components as
x1
y1
x2
y2
=
=
=
=
ym1 cos φ1 ,
ym1 sin φ1 ,
ym2 cos φ2 ,
ym2 sin φ2 ,
and then use the cosine law to find the magnitude of the resultant.
The phase angle can be found from the arcsine of the opposite over the hypotenuse.
E18-22 (a) The diagrams for all times except t = 15 ms should show two distinct pulses, first
moving closer together, then moving farther apart. The pulses don not flip over when passing each
other. The t = 15 ms diagram, however, should simply be a flat line.
E18-23 Use a program such as Maple or Mathematica to plot this.
E18-24 Use a program such as Maple or Mathematica to plot this.
E18-25 (a) The wave speed can be found from Eq. 18-19; we need to know the linear mass
density, which is µ = m/L = (0.122 kg)/(8.36 m) = 0.0146 kg/m. The wave speed is then given by
s
s
F
(96.7 N)
=
= 81.4 m/s.
v=
µ
(0.0146 kg/m)
(b) The longest possible standing wave will be twice the length of the string; so λ = 2L = 16.7 m.
(c) The frequency of the wave is found from Eq. 18-13, v = f λ.
f=
(81.4 m/s)
v
=
= 4.87 Hz
λ
(16.7 m)
p
E18-26 (a) v = (152 N)/(7.16×10−3 kg/m) = 146 m/s.
(b) λ = (2/3)(0.894 m) = 0.596 m.
(c) f = v/λ = (146 m/s)/(0.596 m) = 245 Hz.
E18-27 (a) y = −3.9 cm.
(b) y = (0.15 m) sin[(0.79 rad/m)x + (13 rad/s)t].
(c) y = 2(0.15 m) sin[(0.79 rad/m)(2.3 m)] cos[(13 rad/s)(0.16 s)] = −0.14 m.
E18-28 (a) The amplitude is half of 0.520 cm, or 2.60 mm. The speed is
v = (137 rad/s)/(1.14 rad/cm) = 1.20 m/s.
(b) The nodes are (πrad)/(1.14 rad/cm) = 2.76 cm apart.
(c) The velocity of a particle on the string at position x and time t is the derivative of the wave
equation with respect to time, or
uy = −(0.520 cm)(137 rad/s) sin[(1.14 rad/cm)(1.47 cm)] sin[(137 rad/s)(1.36 s)] = −0.582 m/s.
231
E18-29 (a) We are given the wave frequency and the wave-speed, the wavelength is found from
Eq. 18-13,
(388 m/s)
v
= 0.624 m
λ= =
f
(622 Hz)
The standing wave has four loops, so from Eq. 18-45
L=n
(0.624 m)
λ
= (4)
= 1.25 m
2
2
is the length of the string.
(b) We can just write it down,
y = (1.90 mm) sin[(2π/0.624 m)x] cos[(2π622 s−1 )t].
E18-30 (a) fn = nv/2L = (1)(250 m/s)/2(0.150 m) = 833 Hz.
(b) λ = v/f = (348 m/s)/(833 Hz) = 0.418 m.
E18-31 v =
p
F/µ =
p
p
F L/m. Then fn = nv/2L = n F/4mL, so
p
f1 = (1) (236 N)/4(0.107 kg)(9.88 m)7.47 Hz,
and f2 = 2f1 = 14.9 Hz while f3 = 3f1 = 22.4 Hz.
p
p
p
E18-32 (a) v = F/µ = F L/m = (122 N)(1.48 m)/(8.62×10−3 kg) = 145 m/s.
(b) λ1 = 2(1.48 m) = 2.96 m; λ2 = 1.48 m.
(c) f1 = (145 m/s)/(2.96 m) = 49.0 Hz; f2 = (145 m/s)/(1.48 m) = 98.0 Hz.
E18-33 Although the tied end of the string forces it to be a node, the fact that the other end
is loose means that it should be an anti-node. The discussion of Section 18-10 indicated that the
spacing between nodes is always λ/2. Since anti-nodes occur between nodes, we can expect that the
distance between a node and the nearest anti-node is λ/4.
The longest possible wavelength will have one node at the tied end, an anti-node at the loose
end, and no other nodes or anti-nodes. In this case λ/4 = 120 cm, or λ = 480 cm.
The next longest wavelength will have a node somewhere in the middle region of the string. But
this means that there must be an anti-node between this new node and the node at the tied end
of the string. Moving from left to right, we then have an anti-node at the loose end, a node, and
anti-node, and finally a node at the tied end. There are four points, each separated by λ/4, so the
wavelength would be given by 3λ/4 = 120 cm, or λ = 160 cm.
To progress to the next wavelength we will add another node, and another anti-node. This will
add another two lengths of λ/4 that need to be fit onto the string; hence 5λ/4 = 120 cm, or λ = 100
cm.
In the figure below we have sketched the first three standing waves.
232
E18-34 (a) Note that fn = nf1 . Then fn+1 − fn = f1 . Since there is no resonant frequency
between these two then they must differ by 1, and consequently f1 = (420 Hz) − (315 Hz) = 105 Hz.
(b) v = f λ = (105 Hz)[2(0.756 m)] = 159 m/s.
P18-1
(a) λ = v/f and k = 360◦ /λ. Then
x = (55◦ )λ/(360◦ ) = 55(353 m/s)/360(493 Hz) = 0.109 m.
(b) ω = 360◦ f , so
φ = ωt = (360◦ )(493 Hz)(1.12×10−3 s) = 199◦ .
P18-2
ω = (2π rad)(548 Hz) = 3440 rad/s; λ = v/f and then
k = (2π rad)/[(326 m/s)/(548 Hz)] = 10.6 rad/m.
Finally, y = (1.12×10−2 m) sin[(10.6 rad/m)x + (3440 rad/s)t].
P18-3 (a) This problem really isn’t as bad as it might look. The tensile stress S is tension per
unit cross sectional area, so
F
S=
or F = SA.
A
We already know that linear mass density is µ = m/L, where L is the length of the wire. Substituting
into Eq. 18-19,
s
s
s
F
SA
S
v=
=
.
µ
m/L m/(AL)
But AL is the volume of the wire, so the denominator is just the mass density ρ.
(b) The maximum speed of the transverse wave will be
s
s
S
(720 × 106 Pa)
v=
=
= 300 m/s.
3
ρ
(7800 kg/m )
P18-4 (a) f = ω/2π = (4.08 rad/s)/(2π rad) = 0.649 Hz.
(b) λ = v/f = (0.826 m/s)/(0.649 Hz) = 1.27 m.
(c) k = (2π rad)/(1.27 m) = 4.95 rad/m, so
y = (5.12 cm) sin[(4.95 rad/mx − (4.08 rad/s)t + φ],
where φ is determined by (4.95 rad/m)(9.60×10−2 m) + φ = (1.16 rad), or φ = 0.685 rad.
(d) F = µv 2 = (0.386 kg/m)(0.826 m/s)2 = 0.263 m/s.
P18-5
We want to show that dy/dx = uy /v. The easy way, although not mathematically rigorous:
dy
dy dt
dy dt
1
uy
=
=
= uy =
.
dx
dx dt
dt dx
v
x
P18-6 The maximum value for uy occurs when the cosine function in Eq. 18-14 returns unity.
Consequently, um /y m = ω.
233
P18-7
(a) The linear mass density changes as the rubber band is stretched! In this case,
m
µ=
.
L + ∆L
The tension in the rubber band is given by F = k∆L. Substituting this into Eq. 18-19,
s
r
F
k∆L(L + ∆L)
v=
=
.
µ
m
(b) We want to know the time it will take to travel the length of the rubber band, so
L + ∆L
L + ∆L
or t =
.
t
v
Into this we will substitute our expression for wave speed
r
r
m
m(L + ∆L)
=
t = (L + ∆L)
k∆L(L + ∆L)
k∆L
v=
We have to possibilities to consider: either ∆L L or ∆L L. In either case we are only
interested in the part of the expression with L + ∆L; whichever term is much larger than the other
will be the only significant part.
Then if ∆L L we get L + ∆L ≈ L and
r
r
m(L + ∆L)
mL
≈
,
t=
k∆L
k∆L
√
so that t is proportional to 1/ ∆L.
But if ∆L L we get L + ∆L ≈ ∆L and
r
r
r
m(L + ∆L)
m∆L
m
t=
≈
=
,
k∆L
k∆L
k
so that t is constant.
P18-8 (a) The tension in the rope at some point is a function of the weight of the cable beneath
it. If the bottom of the rope is y = 0, then
p the weightpbeneath some point y√is W = y(m/L)g. The
speed of the wave at that point is v = T /(m/L) = y(M/L)g/(m/L) = gy.
√
(b) dy/dt = gy, so
dt
t
dy
√ ,
gy
Z L
p
dy
=
√ = 2 L/g.
gy
0
=
(c) No.
P18-9
(a) M =
R
µ dx, so
M=
Z
L
kx dx =
0
1 2
kL .
2
2
Then k = 2M/L
. p
p
(b) v = F/µ = F/kx, then
p
dt =
kx/F dx,
Z Lp
p
√
2p
t =
2M/F L2 x dx =
2M/F L2 L3/2 = 8M L/9F .
3
0
234
P18-10 Take a cue from pressure and surface tension. In the rotating non-inertial reference frame
for which the hoop appears to be at rest there is an effective force per unit length acting to push on
each part of the loop directly away from the center. This force per unit length has magnitude
∆F
v2
v2
= (∆m/∆L) = µ .
∆L
r
r
There must be a tension T in the string to hold the loop together. Imagine the loop to be replaced
with two semicircular loops. Each semicircular loop has a diameter part; the force tending to pull
off the diameter section is (∆F/∆L)2r = 2µv 2 . There are two connections to the diameter section,
2
so the tension in the string must
p be half the force on the diameter section, or T = µv .
The wave speed is v w = T /µ = v.
Note that the wave on the string can travel in either direction relative to an inertial observer.
One wave will appear to be fixed in space; the other will move around the string with twice the
speed of the string.
P18-11 If we assume that Handel wanted his violins to play in tune with the other instruments
then all we need to do is find an instrument from Handel’s time that will accurately keep pitch over
a period of several hundred years. Most instruments won’t keep pitch for even a few days because of
temperature and humidity changes; some (like the piccolo?) can’t even play in tune for more than
a few notes! But if someone found a tuning fork...
Since the length of the string doesn’t change, and we are√using a string with the same mass
density, the only choice is to change the tension. But f ∝ v ∝ T , so the percentage change in the
tension of the string is
Tf − Ti
f f 2 − f i2
(440 Hz)2 − (422.5 Hz)2
=
=
= 8.46 %.
2
Ti
fi
(422.5 Hz)2
P18-12
P18-13 (a) The point sources emit spherical waves; the solution to the appropriate wave equation
is found in Ex. 18-14:
A
yi = sin(kri − ωt).
ri
If ri is sufficiently large compared to A, and r1 ≈ r2 , then let r1 = r − δr and r2 = r − δr;
A
A
2A
+
≈
,
r1
r2
r
with an error of order (δr/r)2 . So ignore it.
Then
y1 + y 2
≈
=
ym
=
A
[sin(kr1 − ωt) + sin(kr2 − ωt)] ,
r
2A
k
sin(kr − ωt) cos (r1 − r2 ),
r
2
2A
k
cos (r1 − r2 ).
r
2
(b) A maximum (minimum) occurs when the operand of the cosine, k(r1 − r2 )/2 is an integer
multiple of π (a half odd-integer multiple of π)
235
P18-14 The direct wave
√ travels a distance d from S to D. The wave which reflects off the original
layer travels a distance d2 + 4H 2 betweenpS and D. The wave which reflects off the layer after
it has risen a distance h travels a distance d2 + 4(H + h)2 . Waves will interfere constructively if
there is a difference of an integer number of wavelengths between the two path lengths. In other
words originally we have
p
d2 + 4H 2 − d = nλ,
and later we have destructive interference so
p
d2 + 4(H + h)2 − d = (n + 1/2)λ.
We don’t know n, but we can subtract the top equation from the bottom and get
p
p
d2 + 4(H + h)2 − d2 + 4H 2 = λ/2
P18-15
The wavelength is
λ = v/f = (3.00×108 m/s)/(13.0×106 Hz) = 23.1 m.
The direct wave√travels a distance d from S to D. The wave which reflects off the original layer
2
travels a distance d2 +
p4H between S and D. The wave which reflects off the layer one minute
2
later travels a distance d + 4(H + h)2 . Waves will interfere constructively if there is a difference
of an integer number of wavelengths between the two path lengths. In other words originally we
have
p
d2 + 4H 2 − d = n1 λ,
and then one minute later we have
p
d2 + 4(H + h)2 − d = n2 λ.
We don’t know either n1 or n2 , but we do know the difference is 6, so we can subtract the top
equation from the bottom and get
p
p
d2 + 4(H + h)2 − d2 + 4H 2 = 6λ
We could use that expression as written, do some really obnoxious algebra, and then get the
answer. But we don’t want to; we want to take advantage of the fact that h is small compared to d
and H. Then the first term can be written as
p
p
d2 + 4(H + h)2 =
d2 + 4H 2 + 8Hh + 4h2 ,
p
≈
d2 + 4H 2 + 8Hh,
r
p
8H
≈
d2 + 4H 2 1 + 2
h,
d + 4H 2
p
1
8H
≈
d2 + 4H 2 1 +
h
.
2 d2 + 4H 2
Between the second and the third lines we factored out d2 + 4H 2 ; that last line is from the binomial
expansion theorem. We put this into the previous expression, and
p
p
d2 + 4(H + h)2 − d2 + 4H 2 = 6λ,
p
p
4H
d2 + 4H 2 1 + 2
h
− d2 + 4H 2 = 6λ,
d + 4H 2
4H
√
h = 6λ.
2
d + 4H 2
236
Now what were we doing? We were trying to find the speed at which the layer is moving. We know
H, d, and λ; we can then find h,
h=
6(23.1 m) p
(230×103 m)2 + 4(510×103 m)2 = 71.0 m.
4(510×103 m)
The layer is then moving at v = (71.0 m)/(60 s) = 1.18 m/s.
P18-16
The equation of the standing wave is
y = 2y m sin kx cos ωt.
The transverse speed of a point on the string is the derivative of this, or
uy = −2y m ω sin kx sin ωt,
this has a maximum value when ωt − π/2 is a integer multiple of π. The maximum value is
um = 2y m ω sin kx.
Each mass element on the string dm then has a maximum kinetic energy
dK m = (dm/2)um 2 = y m 2 ω 2 sin2 kx dm.
Using dm = µdx, and integrating over one loop from kx = 0 to kx = π, we get
K m = y m 2 ω 2 µ/2k = 2π 2 y m 2 f µv.
P18-17 (a) For 100% reflection the amplitudes of the incident and reflected wave are equal, or
Ai = Ar , which puts a zero in the denominator of the equation for SWR. If there is no reflection,
Ar = 0 leaving the expression for SWR to reduce to Ai /Ai = 1.
(b) Pr /Pi = A2r /A2i . Do the algebra:
Ai + Ar
Ai − Ar
Ai + Ar
Ar (SWR + 1)
Ar /Ai
=
SWR,
= SWR(Ai − Ar ),
= Ai (SWR − 1),
= (SWR − 1)/(SWR + 1).
Square this, and multiply by 100.
P18-18 Measure with a ruler; I get 2Amax = 1.1 cm and 2Amin = 0.5 cm.
(a) SWR = (1.1/0.5) = 2.2
(b) (2.2 − 1)2 /(2.2 + 1)2 = 0.14 %.
P18-19
(a) Call the three waves
yi
yt
yr
= A sin k1 (x − v1 t),
= B sin k2 (x − v2 t),
= C sin k1 (x + v1 t),
where the subscripts i, t, and r refer to the incident, transmitted, and reflected waves respectively.
237
Apply the principle of superposition. Just to the left of the knot the wave has amplitude y i + y r
while just to the right of the knot the wave has amplitude y t . These two amplitudes must line up
at the knot for all times t, or the knot will come undone. Remember the knot is at x = 0, so
yi + yr = yt ,
A sin k1 (−v1 t) + C sin k1 (+v1 t) = B sin k2 (−v2 t),
−A sin k1 v1 t + C sin k1 v1 t = −B sin k2 v2 t
We know that k1 v1 = k2 v2 = ω, so the three sin functions are all equivalent, and can be canceled.
This leaves A = B + C.
(b) We need to match more than the displacement, we need to match the slope just on either
side of the knot. In that case we need to take the derivative of
yi + yr = yt
with respect to x, and then set x = 0. First we take the derivative,
d
d
(y i + y r ) =
(y t ) ,
dx
dx
k1 A cos k1 (x − v1 t) + k1 C cos k1 (x + v1 t) = k2 B cos k2 (x − v2 t),
and then we set x = 0 and simplify,
k1 A cos k1 (−v1 t) + k1 C cos k1 (+v1 t) = k2 B cos k2 (−v2 t),
k1 A cos k1 v1 t + k1 C cos k1 v1 t = k2 B cos k2 v2 t.
This last expression simplifies like the one in part (a) to give
k1 (A + C) = k2 B
We can combine this with A = B + C to solve for C,
k1 (A + C) = k2 (A − C),
C(k1 + k2 ) = A(k2 − k1 ),
k2 − k1
.
C = A
k1 + k2
If k2 < k1 C will be negative; this means the reflected wave will be inverted.
P18-20
P18-21
Find the wavelength from
λ = 2(0.924 m)/4 = 0.462 m.
Find the wavespeed from
v = f λ = (60.0 Hz)(0.462 m) = 27.7 m/s.
Find the tension from
F = µv 2 = (0.0442 kg)(27.7 m/s)2 /(0.924 m) = 36.7 N.
238
P18-22 (a) The frequency of vibration f is the same for both the aluminum and steel wires; they
don not, however, need to vibrate in the same mode. The speed of waves in the aluminum is v1 ,
that in the steel is v2 . The aluminum vibrates in a mode given by n1 = 2L1 f /v1 , the steel vibrates
in a mode given by n2 = 2L2 f /v2 . Both n1 and n2 need be integers, so the ratio must be a rational
fraction. Note that the ratio is independent of f , so that L1 and L2 must be chosen correctly for
this problem to work at all!
This ratio is
s
r
n2
L2 µ2
(0.866 m) (7800 kg/m3 )
5
=
=
= 2.50 ≈
n1
L1 µ1
(0.600 m) (2600 kg/m3 )
2
Note that since the wires have the same tension and the same cross sectional area it is acceptable
to use the volume density instead of the linear density in the problem.
The smallest integer solution is then n1 = 2 and n2 = 5. The frequency of vibration is then
s
s
n1 v
n1
T
(2)
(10.0 kg)(9.81 m/s2 )
f=
=
=
= 323 Hz.
2L1
2L1 ρ1 A
2(0.600 m) (2600 kg/m3 )(1.00×10−6 m2 )
(b) There are three nodes in the aluminum and six in the steel. But one of those nodes is shared,
and two are on the ends of the wire. The answer is then six.
239
E19-1 (a) v = f λ = (25 Hz)(0.24 m) = 6.0 m/s.
(b) k = (2π rad)/(0.24 m) = 26 rad/m; ω = (2π rad)(25 Hz) = 160 rad/s. The wave equation is
s = (3.0×10−3 m) sin[(26 rad/m)x + (160 rad/s)t]
E19-2 (a) [∆P ]m = 1.48 Pa.
(b) f = (334π rad/s)/(2π rad) = 167 Hz.
(c) λ = (2π rad)/(1.07π rad/m) = 1.87 m.
(d) v = (167 Hz)(1.87 m) = 312 m/s.
E19-3 (a) The wavelength is given by λ = v/f = (343 m/s)/(4.50×106 Hz = 7.62×10−5 m.
(b) The wavelength is given by λ = v/f = (1500 m/s)/(4.50×106 Hz = 3.33×10−4 m.
E19-4 Note: There is a typo; the mean free path should have been measured in “µm” instead of
“pm”.
λmin = 1.0×10−6 m; fmax = (343 m/s)/(1.0×10−6 m) = 3.4×108 Hz.
E19-5 (a) λ = (240 m/s)/(4.2×109 Hz) = 5.7×10−8 m.
E19-6 (a) The speed of sound is
v = (331 m/s)(6.21×10−4 mi/m) = 0.206 mi/s.
In five seconds the sound travels (0.206 mi/s)(5.0 s) = 1.03 mi, which is 3% too large.
(b) Count seconds and divide by 3.
E19-7 Marching at 120 paces per minute means that you move a foot every half a second. The
soldiers in the back are moving the wrong foot, which means they are moving the correct foot half
a second later than they should. If the speed of sound is 343 m/s, then the column of soldiers must
be (343 m/s)(0.5 s) = 172 m long.
E19-8 It takes (300 m)/(343 m/s) = 0.87 s for the concert goer to hear the music after it has passed
the microphone. It takes (5.0×106 m)/(3.0×108 m/s) = 0.017 s for the radio listener to hear the music
after it has passed the microphone. The radio listener hears the music first, 0.85 s before the concert
goer.
E19-9 x/v P = tP and x/v S = tS ; subtracting and rearranging,
x = ∆t/[1/v S − 1/v P ] = (180 s)/[1/(4.5 km/s) − 1/(8.2 km/s)] = 1800 km.
E19-10 Use Eq. 19-8, sm = [∆p]m /kB, and Eq. 19-14, v =
p
B/ρ0 . Then
[∆p]m = kBsm = kv 2 ρ0 sm = 2πf vρ0 sm .
Insert into Eq. 19-18, and
I = 2π 2 ρvf 2 sm 2 .
E19-11 If the source emits equally in all directions the intensity at a distance r is given by the
average power divided by the surface area of a sphere of radius r centered on the source.
The power output of the source can then be found from
P = IA = I(4πr2 ) = (197×10−6 W/m2 )4π(42.5 m)2 = 4.47 W.
240
E19-12 Use the results of Exercise 19-10.
s
(1.13×10−6 W/m2 )
= 3.75×10−8 m.
sm =
2π 2 (1.21 kg/m3 )(343 m/s)(313 Hz)2
E19-13 U = IAt = (1.60×10−2 W/m2 )(4.70×10−4 m2 )(3600 s) = 2.71×10−2 W.
E19-14 Invert Eq. 19-21:
I1 /I2 = 10(1.00dB)/10 = 1.26.
E19-15
(a) Relative sound level is given by Eq. 19-21,
SL1 − SL2 = 10 log
I1
I1
or
= 10(SL1 −SL2 )/10 ,
I2
I2
so if ∆SL = 30 then I1 /I2 = 1030/10 = 1000.
(b) Intensity is proportional to pressure amplitude squared according to Eq. 19-19; so
p
√
∆pm,1 /∆pm,2 = I1 /I2 = 1000 = 32.
E19-16 We know where her ears hurt, so we know the intensity at that point. The power output
is then
P = 4π(1.3 m)2 (1.0 W/m2 ) = 21 W.
This is less than the advertised power.
E19-17 Use the results of Exercise 18-18, I = uv. The intensity is
I = (5200 W)/4π(4820 m)2 = 1.78×10−5 W/m2 ,
so the energy density is
u = I/v = (1.78×10−5 W/m2 )/(343 m/s) = 5.19×10−8 J/m3 .
E19-18 I2 = 2I1 , since I ∝ 1/r2 then r12 = 2r22 . Then
√
D =
2(D − 51.4 m),
√
√
D( 2 − 1) =
2(51.4 m),
D = 176 m.
E19-19
The sound level is given by Eq. 19-20,
SL = 10 log
I
I0
where I0 is the threshold intensity of 10−12 W/m2 . Intensity is given by Eq. 19-19,
I=
(∆pm )2
2ρv
If we assume the maximum possible pressure amplitude is equal to one atmosphere, then
I=
(∆pm )2
(1.01×105 Pa)2
=
= 1.22×107 W/m2 .
2ρv
2(1.21 kg/m3 )(343 m/s)
241
The sound level would then be
SL = 10 log
I
1.22×107 W/m2
= 10 log
= 191 dB
2
I0
(10−12 W/m )
E19-20 Let one person speak with an intensity I1 . N people would have an intensity N I1 . The
ratio is N , so by inverting Eq. 19-21,
N = 10(15dB)/10 = 31.6,
so 32 people would be required.
E19-21 Let one leaf rustle with an intensity I1 . N leaves would have an intensity N I1 . The ratio
is N , so by Eq. 19-21,
SLN = (8.4 dB) + 10 log(2.71×105 ) = 63 dB.
E19-22 Ignoring the finite time means that we can assume the sound waves travels vertically,
which considerably simplifies the algebra.
The intensity ratio can be found by inverting Eq. 19-21,
I1 /I2 = 10(30dB)/10 = 1000.
But intensity is proportional to the inverse distance squared, so I1 /I2 = (r2 /r1 )2 , or
p
r2 = (115 m) (1000) = 3640 m.
E19-23 A minimum will be heard at the detector if the path length difference between the
straight path and the path through the curved tube is half of a wavelength. Both paths involve a
straight section from the source to the start of the curved tube, and then from the end of the curved
tube to the detector. Since it is the path difference that matters, we’ll only focus on the part of
the path between the start of the curved tube and the end of the curved tube. The length of the
straight path is one diameter, or 2r. The length of the curved tube is half a circumference, or πr.
The difference is (π − 2)r. This difference is equal to half a wavelength, so
(π − 2)r
r
= λ/2,
λ
(42.0 cm)
=
= 18.4 cm.
=
2π − 4
2π − 4
E19-24 The path length difference here is
p
(3.75 m)2 + (2.12 m)2 − (3.75 m) = 0.5578 m.
(a) A minimum will occur if this is equal to a half integer number of wavelengths, or (n−1/2)λ =
0.5578 m. This will occur when
f = (n − 1/2)
(343 m/s)
= (n − 1/2)(615 Hz).
(0.5578 m)
(b) A maximum will occur if this is equal to an integer number of wavelengths, or nλ = 0.5578 m.
This will occur when
(343 m/s)
f =n
= n(615 Hz).
(0.5578 m)
242
E19-25 The path length difference here is
p
p
(24.4 m + 6.10 m)2 + (15.2 m)2 − (24.4 m)2 + (15.2 m)2 = 5.33 m.
A maximum will occur if this is equal to an integer number of wavelengths, or nλ = 5.33 m. This
will occur when
f = n(343 m/s)/(5.33 m) = n(64.4 Hz)
The two lowest frequencies are then 64.4 Hz and 129 Hz.
E19-26 The wavelength is λ = (343 m/s)/(300 Hz) = 1.143 m. This means that the sound maxima
will be half of this, or 0.572 m apart. Directly in the center the path length difference is zero, but
since the waves are out of phase, this will be a minimum. The maxima should be located on either
side of this, a distance (0.572 m)/2 = 0.286 m from the center. There will then be maxima located
each 0.572 m farther along.
E19-27 (a) f1 = v/2L and f2 = v/2(L − ∆L). Then
f1
L − ∆L
∆L
1
=
=
=1−
,
r
f2
L
L
or ∆L = L(1 − 1/r).
(b) The answers are ∆L = (0.80 m)(1 − 5/6) = 0.133 m; ∆L = (0.80 m)(1 − 4/5) = 0.160 m;
∆L = (0.80 m)(1 − 3/4) = 0.200 m; and ∆L = (0.80 m)(1 − 2/3) = 0.267 m.
E19-28 The wavelength is twice the distance between the nodes in this case, so λ = 7.68 cm. The
frequency is
f = (1520 m/s)/(7.68×10−2 m) = 1.98×104 Hz.
E19-29 The well is a tube open at one end and closed at the other; Eq. 19-28 describes the
allowed frequencies of the resonant modes. The lowest frequency is when n = 1, so f1 = v/4L. We
know f1 ; to find the depth of the well, L, we need to know the speed of sound.
We should use the information provided, instead of looking up the speed of sound, because maybe
the well is filled with some kind of strange gas.
Then, from Eq. 19-14,
s
s
B
(1.41 × 105 Pa)
=
= 341 m/s.
v=
ρ
(1.21 kg/m3 )
The depth of the well is then
L = v/(4f1 ) = (341 m/s)/[4(7.20 Hz)] = 11.8 m.
E19-30 (a) The resonant frequencies of the pipe are given by fn = nv/2L, or
fn = n(343 m/s)/2(0.457 m) = n(375 Hz).
The lowest frequency in the specified range is f3 = 1130 Hz; the other allowed frequencies in the
specified range are f4 = 1500 Hz, and f5 = 1880 Hz.
243
E19-31 The maximum reflected frequencies will be the ones that undergo constructive interference, which means the path length difference will be an integer multiple of a wavelength. A
wavefront will strike a terrace wall and part will reflect, the other part will travel on to the next
terrace, and then reflect. Since part of the wave had to travel to the next terrace and back, the path
length difference will be 2 × 0.914 m = 1.83 m.
If the speed of sound is v = 343 m/s, the lowest frequency wave which undergoes constructive
interference will be
(343 m/s)
v
= 187 Hz
f= =
λ
(1.83 m)
Any integer multiple of this frequency will also undergo constructive interference, and will also be
heard. The ear and brain, however, will most likely interpret the complex mix of frequencies as a
single tone of frequency 187 Hz.
E19-32 Assume there is no frequency between these two that is amplified. Then one of these
frequencies is fn = nv/2L, and the other is fn+1 = (n + 1)v/2L. Subtracting the larger from the
smaller, ∆f = v/2L, or
L = v/2∆f = (343 m/s)/2(138 Hz − 135 Hz) = 57.2 m.
E19-33 (a) v = 2Lf = 2(0.22 m)(920 Hz) = 405 m/s.
(b) F = v 2 µ = (405 m/s)2 (820×10−6 kg)/(0.220 m) = 611 N.
E19-34 f ∝ v, and v ∝
√
F , so f ∝
√
F . Doubling f requires F increase by a factor of 4.
E19-35 The speed of a wave on the string is the same, regardless of where you put your finger,
so f λ is a constant. The string will vibrate (mostly) in the lowest harmonic, so that λ = 2L, where
L is the length of the part of the string that is allowed to vibrate. Then
f2 λ2
2f2 L2
L2
= f1 λ1 ,
= 2f1 L1 ,
f1
(440 Hz)
= L1
= (30 cm)
= 25 cm.
f2
(528 Hz)
So you need to place your finger 5 cm from the end.
E19-36 The open organ pipe has a length
Lo = v/2f1 = (343 m/s)/2(291 Hz) = 0.589 m.
The second harmonic of the open pipe has frequency 2f1 ; this is the first overtone of the closed pipe,
so the closed pipe has a length
Lc = (3)v/4(2f1 ) = (3)(343 m/s)/4(2)(291 Hz) = 0.442 m.
E19-37 The unknown frequency is either 3 Hz higher or lower than the standard fork. A small
piece of wax placed
on the fork of this unknown frequency tuning fork will result in a lower frequency
p
because f ∝ k/m. If the beat frequency decreases then the two tuning forks are getting closer
in frequency, so the frequency of the first tuning fork must be above the frequency of the standard
fork. Hence, 387 Hz.
244
E19-38 If the string is too taut then the frequency is too high, or f = (440 + 4)Hz. Then
T = 1/f = 1/(444 Hz) = 2.25×10−3 s.
E19-39 One of the tuning forks need to have a frequency 1 Hz different from another. Assume
then one is at 501 Hz. The next fork can be played against the first or the second, so it could have
a frequency of 503 Hz to pick up the 2 and 3 Hz beats. The next one needs to pick up the 5, 7, and
8 Hz beats, and 508 Hz will do the trick. There are other choices.
E19-40 f = v/λ = (5.5 m/s)/(2.3 m) = 2.39 Hz. Then
f 0 = f (v + vO )/v = (2.39 Hz)(5.5 m/s + 3.3 m/s)/(5.5 m/s) = 3.8 Hz.
E19-41
We’ll use Eq. 19-44, since both the observer and the source are in motion. Then
f0 = f
v ± vO
(343 m/s) + (246 m/s)
= (15.8 kHz)
= 17.4 kHzr
v ∓ vS
(343 m/s) + (193 m/s)
E19-42 Solve Eq. 19-44 for vS ;
vS = (v + vO )f /f 0 − v = (343 m/s + 2.63 m/s)(1602 Hz)/(1590 Hz) − (343 m/s) = 5.24 m/s.
E19-43 vS = (14.7 Rad/s)(0.712 m) = 10.5 m/s.
(a) The low frequency heard is
f 0 = (538 Hz)(343 m/s)/(343 m/s + 10.5 m/s) = 522 Hz.
(a) The high frequency heard is
f 0 = (538 Hz)(343 m/s)/(343 m/s − 10.5 m/s) = 555 Hz.
E19-44 Solve Eq. 19-44 for vS ;
vS = v − vf /f 0 = (343 m/s) − (343 m/s)(440 Hz)/(444 Hz) = 3.1 m/s.
E19-45
E19-46 The approaching car “hears”
f0 = f
v + vO
(343 m/s) + (44.7 m/s)
= (148 Hz)
= 167 Hz
v − vS
(343 m/s) − (0)
This sound is reflected back at the same frequency, so the police car “hears”
f0 = f
v + vO
(343 m/s) + (0)
= (167 Hz)
= 192 Hz
v − vS
(343 m/s) − (44.7 m/s)
E19-47 The departing intruder “hears”
f0 = f
v − vO
(343 m/s) − (0.95 m/s)
= (28.3 kHz)
= 28.22 kHz
v + vS
(343 m/s) − (0)
This sound is reflected back at the same frequency, so the alarm “hears”
f0 = f
v − vO
(343 m/s) + (0)
= (28.22 kHz)
= 28.14 kHz
v + vS
(343 m/s) + (0.95 m/s)
The beat frequency is 28.3 kHz − 28.14 kHz = 160 Hz.
245
E19-48 (a) f 0 = (1000 Hz)(330 m/s)(330 m/s + 10.0 m/s) = 971 Hz.
(b) f 0 = (1000 Hz)(330 m/s)(330 m/s − 10.0 m/s) = 1030 Hz.
(c) 1030 Hz − 971 Hz = 59 Hz.
E19-49
(a) The frequency “heard” by the wall is
f0 = f
v + vO
(343 m/s) + (0)
= (438 Hz)
= 464 Hz
v − vS
(343 m/s) − (19.3 m/s)
(b) The wall then reflects a frequency of 464 Hz back to the trumpet player. Sticking with Eq.
19-44, the source is now at rest while the observer moving,
f0 = f
(343 m/s) + (19.3 m/s)
v + vO
= (464 Hz)
= 490 Hz
v − vS
(343 m/s) − (0)
E19-50 The body part “hears”
v + vb
.
v
This sound is reflected back to the detector which then “hears”
v
v + vb
f 00 = f 0
=f
.
v − vb
v − vb
f0 = f
Rearranging,
vb /v =
f 00 − f
1 ∆f
≈
,
f 00 + f
2 f
so v ≈ 2(1×10−3 m/s)/(1.3×10−6 ) ≈ 1500 m/s.
E19-51 The wall “hears”
f0 = f
v + vO
(343 m/s) + (0)
= (39.2 kHz)
= 40.21 kHz
v − vS
(343 m/s) − (8.58 m/s)
This sound is reflected back at the same frequency, so the bat “hears”
f0 = f
P19-1
v + vO
(343 m/s) + (8.58 m/s)
= (40.21 kHz)
= 41.2 kHz.
v − vS
(343 m/s) − (0)
(a) tair = L/v air and tm = L/v, so the difference is
∆t = L(1/v air − 1/v)
(b) Rearrange the above result, and
L = (0.120 s)/[1/(343 m/s) − 1/(6420 m/s)] = 43.5 m.
P19-2 The stone falls for a time t1 where y = gt21 /2 is the depth of the well. Note y is positive in
this equation. The sound travels back in a time t2 where v = y/t2 is the speed of sound in the well.
t1 + t2 = 3.00 s, so
2y = g(3.00 s − t2 )2 = g[(9.00 s2 ) − (6.00 s)y/v + y 2 /v 2 ],
or, using g = 9.81 m/s2 and v = 343 m/s,
y 2 − (2.555×105 m)y + (1.039×107 m2 ) = 0,
which has a positive solution y = 40.7 m.
246
P19-3
(a) The intensity at 28.5 m is found from the 1/r2 dependence;
I2 = I1 (r1 /r2 )2 = (962 µW/m2 )(6.11 m/28.5 m)2 = 44.2 µW/m2 .
(c) We’ll do this part first. The pressure amplitude is found from Eq. 19-19,
p
p
∆pm = 2ρvI = 2(1.21 kg/m3 )(343 m/s)(962×10−6 W/m2 ) = 0.894 Pa.
(b) The displacement amplitude is found from Eq. 19-8,
sm = ∆pm /(kB),
where k = 2πf /v is the wave number. From Eq. 19-14 w know that B = ρv 2 , so
sm =
P19-4
∆pm
(0.894 Pa)
=
= 1.64×10−7 m.
2πf ρv
2π(2090 Hz)(1.21 kg/m3 )(343 m/s)
√
(a) If the intensities are equal, then ∆pm ∝ ρv, so
s
[∆pm ]water
(998 kg/m3 )(1482 m/s)
=
= 59.9.
[∆pm ]air
(1.2 kg/m3 )(343 m/s)
(b) If the pressure amplitudes are equal, then I ∝∝ 1/ρv, so
I water
(1.2 kg/m3 )(343 m/s)
=
= 2.78×10−4 .
I air
(998 kg/m3 )(1482 m/s)
P19-5 The energy is dissipated on a cylindrical surface which grows in area as r, so the intensity
√ is
proportional to 1/r. The amplitude is proportional to the square root of the intensity, so sm ∝ 1/ r.
P19-6 (a) The first position corresponds to maximum destructive interference, so the waves are half
a wavelength out of phase; the second position corresponds to maximum constructive interference, so
the waves are in phase. Shifting the tube has in effect added half a wavelength to the path through
B. But each segment is added, so
λ = (2)(2)(1.65 cm) = 6.60 cm,
and f = (343 m/s)/(6.60 cm) = 5200 Hz.
(b) I min ∝ (s1 −s2 )2 , I max ∝ (s1 +s2 )2 , then dividing one expression by the other and rearranging
we find
√
√
√
√
s1
I max + I min
90 + 10
√
√ =2
=√
=√
s2
I max − I min
90 − 10
P19-7 (a) I = P/4πr2 = (31.6 W)/4π(194 m)2 = 6.68×10−5 W/m2 .
(b) P = IA = (6.68×10−5 W/m2 )(75.2×10−6 m2 ) = 5.02×10−9 W.
(c) U = P t = (5.02×10−9 W)(25.0 min)(60.0 s/min) = 7.53 µJ.
P19-8 Note that the reverberation time is logarithmically related to the intensity, but linearly
related to the sound level. As such, the reverberation time is the amount of time for the sound level
to decrease by
∆SL = 10 log(10−6 ) = 60 dB.
Then
t = (87 dB)(2.6 s)/(60 dB) = 3.8 s
247
P19-9 What the device is doing is taking all of the energy which strikes a large surface area and
concentrating it into a small surface area. It doesn’t succeed; only 12% of the energy is concentrated.
We can think, however, in terms of power: 12% of the average power which strikes the parabolic
reflector is transmitted into the tube.
If the sound intensity on the reflector is I1 , then the average power is P1 = I1 A1 = I1 πr12 , where
r1 is the radius of the reflector. The average power in the tube will be P2 = 0.12P1 , so the intensity
in the tube will be
P2
0.12I1 πr12
r12
=
=
0.12I
I2 =
1
A2
πr22
r22
2
Since the lowest audible sound has an intensity of I0 = 10−12 W/m , we can set I2 = I0 as the
condition for “hearing” the whisperer through the apparatus. The minimum sound intensity at the
parabolic reflector is
I0 r22
I1 =
.
0.12 r12
Now for the whisperers. Intensity falls off as 1/d2 , where d is the distance from the source. We
are told that when d = 1.0 m the sound level is 20 dB; this sound level has an intensity of
I = I0 1020/10 = 100I0
Then at a distance d from the source the intensity must be
I1 = 100I0
(1 m)2
.
d2
This would be the intensity “picked-up” by the parabolic reflector. Combining this with the condition
for being able to hear the whisperers through the apparatus, we have
I0 r22
(1 m)2
=
100I
0
0.12 r12
d2
or, upon some rearranging,
√
√
r1
(0.50 m)
d = ( 12 m) = ( 12 m)
= 346 m.
r2
(0.005 m)
P19-10 (a) A displacement node; at the center the particles have nowhere to go.
(b) This
p systems acts like a pipe which is closed at one end.
(c) v B/ρ, so
p
T = 4(0.009)(6.96×108 m) (1.0×1010 kg/m3 )/(1.33×1022 Pa) = 22 s.
P19-11 The cork filing collect at pressure antinodes when standing waves are present, and the
antinodes are each half a wavelength apart. Then v = f λ = f (2d).
P19-12 (a) f = v/4L = (343 m/s)/4(1.18 m) = 72.7 Hz.
(b) F = µv 2 = µf 2 λ2 , or
F = (9.57×10−3 kg/0.332 m)(72.7 Hz)2 [2(0.332 m)]2 = 67.1 N.
248
P19-13 In this problem the string is observed to resonate at 880 Hz and then again at 1320 Hz,
so the two corresponding values of n must differ by 1. We can then write two equations
(880 Hz) =
nv
(n + 1)v
and (1320 Hz) =
2L
2L
and solve these for v. It is somewhat easier to first solve for n. Rearranging both equations, we get
v
(1320 Hz)
v
(880 Hz)
=
and
=
.
n
2L
n+1
2L
Combining these two equations we get
(1320 Hz)
,
n+1
= n(1320 Hz),
(880 Hz)
n =
= 2.
(1320 Hz) − (880 Hz)
(880 Hz)
n
(n + 1)(880 Hz)
=
Now that we know n we can find v,
v = 2(0.300 m)
(880 Hz)
= 264 m/s
2
And, finally, we are in a position to find the tension, since
F = µv 2 = (0.652×10−3 kg/m)(264 m/s)2 = 45.4 N.
P19-14 (a) There are five choices for the first fork, and four for the second. That gives 20 pairs.
But order doesn’t matter, so we need divide that by two to get a maximum of 10 possible beat
frequencies.
(b) If the forks are ordered to have equal differences (say, 400 Hz, 410 Hz, 420 Hz, 430 Hz, and
440 Hz) then there will actually be only 4 beat frequencies.
P19-15
v = (2.25×108 m/s)/ sin(58.0◦ ) = 2.65×108 m/s.
P19-16
(a) f1 = (442 Hz)(343 m/s)/(343 m/s − 31.3 m/s) = 486 Hz, while
f2 = (442 Hz)(343 m/s)/(343 m/s + 31.3 m/s) = 405 Hz,
so ∆f = 81 Hz.
(b) f1 = (442 Hz)(343 m/s − 31.3 m/s)/(343 m/s) = 402 Hz, while
f2 = (442 Hz)(343 m/s + 31.3 m/s)/(343 m/s) = 482 Hz,
so ∆f = 80 Hz.
P19-17 The sonic boom that you hear is not from the sound given off by the plane when it is
overhead, it is from the sound given off before the plane was overhead. So this problem isn’t as
simple as distance equals velocity × time. It is very useful to sketch a picture.
249
x2
H
x1
S
θ
O
We can find the angle θ from the figure, we’ll get Eq. 19-45, so
sin θ =
v
(330 m/s)
=
= 0.833 or θ = 56.4◦
vs
(396 m/s)
Note that vs is the speed of the source, not the speed of sound!
Unfortunately t = 12 s is not the time between when the sonic boom leaves the plane and when
it arrives at the observer. It is the time between when the plane is overhead and when the sonic
boom arrives at the observer. That’s why there are so many marks and variables on the figure. x1
is the distance from where the sonic boom which is heard by the observer is emitted to the point
directly overhead; x2 is the distance from the point which is directly overhead to the point where
the plane is when the sonic boom is heard by the observer. We do have x2 = vs (12.0 s). This length
forms one side of a right triangle HSO, the opposite side of this triangle is the side HO, which is
the height of the plane above the ground, so
h = x2 tan θ = (343 m/s)(12.0 s) tan(56.4◦ ) = 7150 m.
P19-18
(a) The target “hears”
v+V
.
v
This sound is reflected back to the detector which then “hears”
f 0 = fs
fr = f 0
v
v+V
= fs
.
v−V
v−V
(b) Rearranging,
V /v =
fr − fs
1 fr − fs
≈
,
fr + fs
2 fs
where we have assumed that the source frequency and the reflected frequency are almost identical,
so that when added fr + fs ≈ 2fs .
P19-19
(a) We apply Eq. 19-44
f0 = f
v + vO
(5470 km/h) + (94.6 km/h)
= (1030 Hz)
= 1050 Hz
v − vS
(5470 km/h) − (20.2 km/h)
250
(b) The reflected signal has a frequency equal to that of the signal received by the second sub
originally. Applying Eq. 19-44 again,
f0 = f
P19-20
(5470 km/h) + (20.2 km/h)
v + vO
= (1050 Hz)
= 1070 Hz
v − vS
(5470 km/h) − (94.6 km/h)
In this case vS = 75.2 km/h − 30.5 km/h = 12.4 m/s. Then
f 0 = (989 Hz)(1482 m/s)(1482 m/s − 12.4 m/s) = 997 Hz.
P19-21 There is no relative motion between the source and observer, so there is no frequency shift
regardless of the wind direction.
P19-22
(a) vS = 34.2 m/s and vO = 34.2 m/s, so
f 0 = (525 Hz)(343 m/s + 34.2 m/s)/(343 m/s − 34.2 m/s) = 641 Hz.
(b) vS = 34.2 m/s + 15.3 m/s = 49.5 m/s and vO = 34.2 m/s − 15.3 m/s = 18.9 m/s, so
f 0 = (525 Hz)(343 m/s + 18.9 m/s)/(343 m/s − 49.5 m/s) = 647 Hz.
(c) vS = 34.2 m/s − 15.3 m/s = 18.9 m/s and vO = 34.2 m/s + 15.3 m/s = 49.5 m/s, so
f 0 = (525 Hz)(343 m/s + 49.5 m/s)/(343 m/s − 18.9 m/s) = 636 Hz.
251
E20-1 (a) t = x/v = (0.20 m)/(0.941)(3.00×108 m/s) = 7.1×10−10 s.
(b) y = −gt2 /2 = −(9.81 m/s2 )(7.1×10−10 s)2 /2 = 2.5×10−18 m.
E20-2 L = L0
E20-3
p
p
1 − u2 /c2 = (2.86 m) 1 − (0.999987)2 = 1.46 cm.
L = L0
p
p
1 − u2 /c2 = (1.68 m) 1 − (0.632)2 = 1.30 m.
p
E20-4 Solve ∆t = ∆t0 / 1 − u2 /c2 for u:
s
s
2
2
∆t0
(2.20 µs)
8
u=c 1−
= (3.00×10 m/s) 1 −
= 2.97×108 m/s.
∆t
(16.0 µs)
E20-5
We can apply ∆x = v∆t to find the time the particle existed before it decayed. Then
∆t =
(1.05 × 10−3 m)
x
= 3.53 × 10−12 s.
v (0.992)(3.00 × 108 m/s)
The proper lifetime of the particle is
p
p
∆t0 = ∆t 1 − u2 /c2 = (3.53 × 10−12 s) 1 − (0.992)2 = 4.46 × 10−13 s.
E20-6 Apply Eq. 20-12:
v=
(0.43c) + (0.587c)
= 0.812c.
1 + (0.43c)(0.587c)/c2
p
p
E20-7 (a) L = L0 1 − u2 /c2 = (130 m) 1 − (0.740)2 = 87.4 m
(b) ∆t = L/v = (87.4 m)/(0.740)(3.00×108 m/s) = 3.94×10−7 s.
p
p
E20-8 ∆t = ∆t0 / 1 − u2 /c2 = (26 ns) 1 − (0.99)2 = 184 ns. Then
L = v∆t = (0.99)(3.00×108 m/s)(184×10−9 s) = 55 m.
E20-9 (a) v g = 2v = (7.91 + 7.91) km/s = 15.82 km/s.
(b) A relativistic treatment yields v r = 2v/(1 + v 2 /c2 ). The fractional error is
vg
v2
v2
(7.91×103 m/s)2
−1= 1+ 2 −1= 2 =
= 6.95×10−10 .
vr
c
c
(3.00×108 m/s)2
p
2
E20-10 Invert
p Eq. 20-15 to get β = 1 − 1/γ .
(a) β = p1 − 1/(1.01)2 = 0.140.
(b) β = p 1 − 1/(10.0)2 = 0.995.
(c) β = p1 − 1/(100)2 = 0.99995.
(d) β = 1 − 1/(1000)2 = 0.9999995.
252
E20-11 The distance traveled by the particle is (6.0 y)c; the time required for the particle to
travel this distance is 8.0 years. Then the speed of the particle is
v=
(6.0 y)c
3
∆x
=
= c.
∆t
(8.0 y)
4
The speed parameter β is given by
β=
3
c
3
v
= 4 = .
c
c
4
p
E20-12 γ = 1/ 1 − (0.950)2 = 3.20. Then
x0
t0
= (3.20)[(1.00×105 m) − (0.950)(3.00×108 m/s)(2.00×10−4 s)] = 1.38×105 m,
= (3.20)[(2.00×10−4 s) − (1.00×105 m)(0.950)/(3.00×108 m/s)] = −3.73×10−4 s.
p
E20-13 (a) γ = 1/ 1 − (0.380)2 = 1.081. Then
x0
t0
= (1.081)[(3.20×108 m) − (0.380)(3.00×108 m/s)(2.50 s)] = 3.78×107 m,
= (1.081)[(2.50 s) − (3.20×108 m)(0.380)/(3.00×108 m/s)] = 2.26 s.
p
(b) γ = 1/ 1 − (0.380)2 = 1.081. Then
x0
t0
=
=
(1.081)[(3.20×108 m) − (−0.380)(3.00×108 m/s)(2.50 s)] = 6.54×108 m,
(1.081)[(2.50 s) − (3.20×108 m)(−0.380)/(3.00×108 m/s)] = 3.14 s.
E20-14
p
E20-15 (a) vx0 = (−u)/(1 − 0) and vy0 = c 1 − u2 /c2 .
(b) (vx0 )2 + (vy0 )2 = u2 + c2 − u2 = c2 .
E20-16 v 0 = (0.787c + 0.612c)/[1 + (0.787)(0.612)] = 0.944c.
E20-17 (a) The first part is easy; we appear to be moving away from A at the same speed as A
appears to be moving away from us: 0.347c.
(b) Using the velocity transformation formula, Eq. 20-18,
vx0 =
vx − u
(0.347c) − (−0.347c)
=
= 0.619c.
1 − uvx /c2
1 − (−0.347c)(0.347c)/c2
The negative sign reflects the fact that these two velocities are in opposite directions.
E20-18 v 0 = (0.788c − 0.413c)/[1 + (0.788)(−0.413)] = 0.556c.
p
E20-19 (a) γ = 1/ 1 − (0.8)2 = 5/3.
vx0
=
vy0
=
vx
3(0.8c)
12
=
=
c,
2
γ(1 − uvy /c )
5[1 − (0)]
25
vy − u
(0) − (0.8c)
4
=
= − c.
2
1 − uvy /c
1 − (0)
5
p
Then v 0 = c (−4/5)2 + (12/25)2 = 0.933c directed θ = arctan(−12/20) = 31◦ East of South.
253
p
(b) γ = 1/ 1 − (0.8)2 = 5/3.
vx0
=
vy0
=
vx − u
(0) − (−0.8c)
4
=
= + c,
1 − uvx /c2
1 − (0)
5
3(0.8c)
12
vy
=
=
c.
γ(1 − uvx /c2 )
5[1 − (0)]
25
p
Then v 0 = c (4/5)2 + (12/25)2 = 0.933c directed θ = arctan(20/12) = 59◦ West of North.
E20-20 This exercise should occur in Section 20-9.
8
(a) v = 2π(6.37×106 m)c/(1
ps)(3.00×10 m/s) = 0.133c.
(b) K = (γ − 1)mc2 = (1/ 1 − (0.133)2 − 1)(511 keV) = 4.58 keV.
(c) K c = mv 2 /2 = mc2 (v 2 /c2 )/2 = (511 keV)(0.133)2 /2 = 4.52 keV. The percent error is
(4.52 − 4.58)/(4.58) = −1.31%.
E20-21 ∆L = L0 − L0 so
∆L = 2(6.370×106 m)(1 −
p
1 − (29.8×103 m/s)2 /(3.00×108 m/s)2 ) = 6.29×10−2 m.
E20-22 (a) ∆L/L0 = 1 − L0 /L0 so
p
∆L = (1 − 1 − (522 m/s)2 /(3.00×108 m/s)2 ) = 1.51×10−12 .
(b) We want to solve ∆t − ∆t0 = 1 µs, or
p
1 µs = ∆t(1 − 1/ 1 − (522 m/s)2 /(3.00×108 m/s)2 ),
which has solution ∆t = 6.61×105 s. That’s 7.64 days.
E20-23
The length of the ship as measured in the “certain” reference frame is
L = L0
p
p
1 − v 2 /c2 = (358 m) 1 − (0.728)2 = 245 m.
In a time ∆t the ship will move a distance x1 = v1 ∆t while the micrometeorite will move a distance
x2 = v2 ∆t; since they are moving toward each other then the micrometeorite will pass then ship
when x1 + x2 = L. Then
∆t = L/(v1 + v2 ) = (245 m)/[(0.728 + 0.817)(3.00×108 m/s)] = 5.29×10−7 s.
This answer is the time measured in the “certain” reference frame. We can use Eq. 20-21 to find
the time as measured on the ship,
∆t =
∆t0 + u∆x0 /c2
(5.29×10−7 s) + (0.728c)(116 m)/c2
p
p
=
= 1.23×10−6 s.
2
2
1 − u /c
1 − (0.728)2
p
E20-24 (a) γ = 1/ 1 − (0.622)2 = 1.28.
(b) ∆t = (183 m)/(0.622)(3.00×108 m/s) = 9.81×10−7 s. On the clock, however,
∆t0 = ∆t/γ = (9.81×10−7 s)/(1.28) = 7.66×10−7 s.
254
E20-25 (a) ∆t = (26.0 ly)/(0.988)(1.00 ly/y) = 26.3 y.
(b) The signal takes
p 26 years to return, so 26 + 26.3 = 52.3 years.
(c) ∆t0 = (26.3 y) 1 − (0.988)2 = 4.06 y.
E20-26 (a) γ = (1000 y)(1 y) = 1000;
p
v = c 1 − 1/γ 2 ≈ c(1 − 1/2γ 2 ) = 0.9999995c
(b) No.
E20-27 (5.61×1029 MeV/c2 )c/(3.00×108 m/s) = 1.87×1021 MeV/c.
E20-28 p2 = m2 c2 = m2 v 2 /(1 − v 2 /c2 ), so 2v 2 /c2 = 1, or v =
√
2c.
E20-29 The magnitude of the momentum of a relativistic particle in terms of the magnitude of
the velocity is given by Eq. 20-23,
mv
p= p
.
1 − v 2 /c2
The speed parameter, β, is what we are looking for, so we need to rearrange the above expression
for the quantity v/c.
mv/c
p
,
1 − v 2 /c2
mβ
= p
,
1 − β2
p
1 − β2
=
,
β
p
=
1/β 2 − 1.
p/c =
p
c
mc
p
mc
p
Rearranging,
p
mc2
=
1/β 2 − 1,
pc
2 2
mc
1
=
− 1,
pc
β2
s
2
mc2
1
+1 =
,
pc
β
pc
p
= β
m2 c4 + p2 c2
(a) For the electron,
(12.5 MeV/c)c
β=p
= 0.999.
(0.511 MeV/c2 )2 c4 + (12.5 MeV/c)2 c2
(b) For the proton,
(12.5 MeV/c)c
β=p
(938 MeV/c2 )2 c4 + (12.5 MeV/c)2 c2
255
= 0.0133.
p
E20-30 K = mc2 (γ − 1), so γ = 1 + K/mc2 . β = 1 − 1/γ 2 .
(a) γ = 1 + (1.0 keV)/(511 keV) = 1.00196. β = 0.0625c.
(b) γ = 1 + (1.0 MeV)/(0.511 MeV) = 2.96. β = 0.941c.
(c) γ = 1 + (1.0 GeV)/(0.511 MeV) = 1960. β = 0.99999987c.
E20-31
The kinetic energy is given by Eq. 20-27,
mc2
K=p
1 − v 2 /c2
− mc2 .
We rearrange this to solve for β = v/c,
β=
s
1−
mc2
K + mc2
2
.
It is actually much easier to find γ, since
1
γ=p
1 − v 2 /c2
so K = γmc2 − mc2 implies
γ=
,
K + mc2
mc2
(a) For the electron,
β=
s
(0.511 MeV/c2 )c2
(10 MeV) + (0.511 MeV/c2 )c2
2
1−
γ=
(10 MeV) + (0.511 MeV/c2 )c2
= 20.6.
(0.511 MeV/c2 )c2
and
= 0.9988,
(b) For the proton,
β=
s
2
γ=
(10 MeV) + (938 MeV/c2 )c2
= 1.01.
(938 MeV/c2 )c2
and
(b) For the alpha particle,
s
β=
(938 MeV/c2 )c2
(10 MeV) + (938 MeV/c2 )c2
1−
1−
and
γ=
4(938 MeV/c2 )c2
(10 MeV) + 4(938 MeV/c2 )c2
= 0.0145,
2
= 0.73,
(10 MeV) + 4(938 MeV/c2 )c2
= 1.0027.
4(938 MeV/c2 )c2
p
E20-32 γ = 1/ 1 − (0.99)2 = 7.089.
(a) E = γmc2 = (7.089)(938.3 MeV) = 6650 MeV. K = E − mc2 = 5710 MeV. p = mvγ =
(938.3 MeV/c2 )(0.99c)(7.089) = 6580 MeV/c.
(b) E = γmc2 = (7.089)(0.511 MeV) = 3.62 MeV. K = E − mc2 = 3.11 MeV. p = mvγ =
(0.511 MeV/c2 )(0.99c)(7.089) = 3.59 MeV/c.
256
E20-33 ∆m/∆t = (1.2×1041 W)/(3.0×108 m/s)2 = 1.33×1024 kg/s, which is
∆m
(1.33×1024 kg/s)(3.16×107 s/y)
=
= 21.1
∆t
(1.99×1030 kg/sun)
E20-34 (a) If K = E − mc2 = 2mc2 , then E = 3mc2 , so γ = 3, and
p
p
v = c 1 − 1/γ 2 = c 1 − 1/(3)2 = 0.943c.
(b) If E = 2mc2 , then γ = 2, and
p
p
v = c 1 − 1/γ 2 = c 1 − 1/(2)2 = 0.866c.
E20-35
(a) The kinetic energy is given by Eq. 20-27,
−1/2
mc2
K=p
− mc2 = mc2 1 − β 2
−1 .
1 − v 2 /c2
We want to expand the 1 − β 2 part for small β,
1 − β2
−1/2
1
3
= 1 + β2 + β4 + · · ·
2
8
Inserting this into the kinetic energy expression,
K=
1 2 2 3 2 4
mc β + mc β + · · ·
2
8
But β = v/c, so
1
3 v4
mv 2 + m 2 + · · ·
2
8 c
(b) We want to know when the error because of neglecting the second (and higher) terms is 1%;
K=
or
3 v4
1
3 v 2
0.01 = ( m 2 )/( mv 2 ) =
.
8 c
2
4 c
p
This will happen when v/c = (0.01)4/3) = 0.115.
E20-36 Kc = (1000 kg)(20 m/s)2 /2 = 2.0×105 J. The relativistic calculation is slightly harder:
p
Kr = (1000 kg)(3×108 m/s)2 (1/ 1 − (20 m/s)2 /(3×108 m/s)2 − 1),
1
3
2
4
8
2
≈ (1000 kg) (20 m/s) + (20 m/s) /(3×10 m/s) + ... ,
2
8
=
E20-37
2.0×105 J + 6.7×10−10 J.
Start with Eq. 20-34 in the form
E 2 = (pc)2 + (mc2 )2
The rest energy is mc2 , and if the total energy is three times this then E = 3mc2 , so
(3mc2 )2 = (pc)2 + (mc2 )2 ,
8(mc2 )2 = (pc)2 ,
√
8 mc = p.
257
E20-38 The initial kinetic energy is
1
m
2
Ki =
2v
1 + v 2 /c2
=
2mv 2
.
(1 + v 2 /c2 )2
The final kinetic energy is
2
1 p
K f = 2 m v 2 − v 2 /c2 = mv 2 (2 − v 2 /c2 ).
2
E20-39 This exercise is much more involved than the previous one!
The initial kinetic energy is
Ki
=
mc2
r
1−
=
=
=
2v
1+v 2 /c2
2
− mc2 ,
/c2
mc2 (1 + v 2 /c2 )
p
(1 + v 2 /c2 )2 − 4v 2 /c2
− mc2 ,
m(c2 + v 2 ) m(c2 − v 2 )
−
,
1 − v 2 /c2
1 − v 2 /c2
2mv 2
.
1 − v 2 /c2
The final kinetic energy is
Kf
=
=
mc2
− 2mc2 ,
p
2
1 − v 2 − v 2 /c2 /c2
2r
mc2
2p
− 2mc2 ,
2
2
2
2
1 − (v /c )(2 − v /c )
mc2
− 2mc2 ,
1 − v 2 /c2
mc2
m(c2 − v 2 )
= 2
−
2
,
1 − v 2 /c2
1 − v 2 /c2
2mv 2
=
.
1 − v 2 /c2
=
2
E20-40 For a particle with mass, γ = K/mc2 + 1. For the electron, γ = (0.40)/(0.511) + 1 = 1.78.
For the proton, γ = (10)/(938) + 1 = 1.066.
p
For the photon, pc = E. For a particle with mass, pc = (K + mc2 )2 − m2 c4 . For the electron,
p
pc = [(0.40 MeV) + (0.511 MeV)]2 − (0.511 MeV)2 = 0.754 MeV.
For the proton,
pc =
p
[(10 MeV) + (938 MeV)]2 − (938 MeV)2 = 137 MeV.
(a) Only photons move at the speed of light, so it is moving the fastest.
(b) The proton, since it has smallest value for γ.
(c) The proton has the greatest momentum.
(d) The photon has the least.
258
E20-41
Work is change in energy, so
p
p
W = mc2 / 1 − (v f /c)2 − mc2 / 1 − (v i /c)2 .
(a) Plug in the numbers,
p
p
W = (0.511 MeV)(1/ 1 − (0.19)2 − 1/ 1 − (0.18)2 ) = 0.996 keV.
(b) Plug in the numbers,
p
p
W = (0.511 MeV)(1/ 1 − (0.99)2 − 1/ 1 − (0.98)2 ) = 1.05 MeV.
E20-42 E = 2γm0 c2 = mc2 , so
p
m = 2γm0 = 2(1.30 mg)/ 1 − (0.580)2 = 3.19 mg.
E20-43 (a) Energy conservation requires E k = 2Eπ , or mk c2 = 2γmπ c2 . Then
γ = (498 MeV)/2(140 MeV) = 1.78
p
This corresponds to a speed of v = c 1 − 1/(1.78)2 = 0.827c. p
(b) γ = (498 MeV + 325 MeV)/(498 MeV) = 1.65, so v = c 1 − 1/(1.65)2 = 0.795c.
(c) The lab frame velocities are then
v10 =
(0.795) + (−0.827)
c = −0.0934c,
1 + (0.795)(−0.827)
and
v20 =
(0.795) + (0.827)
c = 0.979c,
1 + (0.795)(0.827)
The corresponding kinetic energies are
p
K1 = (140 MeV)(1/ 1 − (−0.0934)2 − 1) = 0.614 MeV
and
p
K1 = (140 MeV)(1/ 1 − (0.979)2 − 1) = 548 MeV
E20-44
P20-1 (a) γ = 2, so v =
(b) γ = 2.
P20-2
p
1 − 1/(2)2 = 0.866c.
(a) Classically, v 0 = (0.620c) + (0.470c) = 1.09c. Relativistically,
v0 =
(0.620c) + (0.470c)
= 0.844c.
1 + (0.620)(0.470)
(b) Classically, v 0 = (0.620c) + (−0.470c) = 0.150c. Relativistically,
v0 =
(0.620c) + (−0.470c)
= 0.211c.
1 + (0.620)(−0.470)
259
P20-3
p
(a) γ = 1/ 1 − (0.247)2 = 1.032. Use the equations from Table 20-2.
∆t = (1.032)[(0) − (0.247)(30.4×103 m)/(3.00×108 m/s)] = −2.58×10−5 s.
(b) The red flash appears to go first.
P20-4 Once
p again, the “pico” should have been a µ.
γ = 1/ 1 − (0.60)2 = 1.25. Use the equations from Table 20-2.
∆t = (1.25)[(4.0×10−6 s) − (0.60)(3.0×103 m)/(3.00×108 m/s)] = −2.5×10−6 s.
P20-5 We can choose our coordinate system so that u is directed along the x axis without any
loss of generality. Then, according to Table 20-2,
∆x0
∆y 0
∆z 0
c∆t0
=
=
=
=
γ(∆x − u∆t),
∆y,
∆z,
γ(c∆t − u∆x/c).
Square these expressions,
(∆x0 )2
(∆y 0 )2
(∆z 0 )2
c2 (∆t0 )2
= γ 2 (∆x − u∆t)2 = γ 2 (∆x)2 − 2u(∆x)(∆t) + (∆t)2 ,
= (∆y)2 ,
= (∆z)2 ,
= γ 2 (c∆t − u∆x/c)2 = γ 2 c2 (∆t)2 − 2u(∆t)(∆x) + u2 (∆x)2 /c2 .
We’ll add the first three equations and then subtract the fourth. The left hand side is the equal to
(∆x0 )2 + (∆y 0 )2 + (∆z 0 )2 − c2 (∆t0 )2 ,
while the right hand side will equal
γ 2 (∆x)2 + u2 (∆t)2 − c2 (∆t)2 − u2 /c2 (∆x)2 + (∆y)2 + (∆z)2 ,
which can be rearranged as
γ 2 1 − u2 /c2 (∆x)2 + γ 2 u2 − c2 (∆t)2 + (∆y)2 + (∆z)2 ,
γ 2 1 − u2 /c2 (∆x)2 + (∆y)2 + (∆z)2 − c2 γ 2 1 − u2 /c2 (∆t)2 .
But
γ2 =
1
,
1 − u2 /c2
so the previous expression will simplify to
(∆x)2 + (∆y)2 + (∆z)2 − c2 (∆t)2 .
P20-6 (a) vx = [(0.780c) + (0.240c)]/[1 + (0.240)(0.780)] = 0.859c.
(b) vx = [(0) + (0.240c)]/[1 + (0)] = 0.240c, while
p
vy = (0.780c) 1 − (0.240)2 /[1 + (0)] = 0.757c.
p
Then v = (0.240c)2 + (0.757c)2 = 0.794c.
(b) vx0 = [(0) − (0.240c)]/[1 + (0)] = −0.240c, while
p
vy0 = (0.780c) 1 − (0.240)2 /[1 + (0)] = 0.757c.
p
Then v 0 = (−0.240c)2 + (0.757c)2 = 0.794c.
260
P20-7 If we look back at the boost equation we might notice that it looks very similar to the
rule for the tangent of the sum of two angles. It is exactly the same as the rule for the hyperbolic
tangent,
tanh α1 + tanh α2
tanh(α1 + α2 ) =
.
1 + tanh α1 tanh α2
This means that each boost of β = 0.5 is the same as a “hyperbolic” rotation of αr where tanh αr =
0.5. We need only add these rotations together until we get to αf , where tanh αf = 0.999.
αf = 3.800, and αR = 0.5493. We can fit (3.800)/(0.5493) = 6.92 boosts, but we need an integral
number, so there are seven boosts required. The final speed after these seven boosts will be 0.9991c.
P20-8
(a) If ∆x0 = 0, then ∆x = u∆t, or
u = (730 m)/(4.96×10−6 s) = 1.472×108 m/s = 0.491c.
p
(b) γ = 1/ 1 − (0.491)2 = 1.148,
∆t0 = (1.148)[(4.96×10−6 s) − (0.491)(730 m)/(3×108 )] = 4.32×10−6 s.
P20-9
Since the maximum value for u is c, then the minimum ∆t is
∆t ≥ (730 m)/(3.00×108 m/s) = 2.43×10−6 s.
P20-10 (a) Yes.
(b) The speed will be very close to the speed of light, consequently γ ≈ (23, 000)/(30) = 766.7.
Then
p
v = 1 − 1/γ 2 ≈ 1 − 1/2γ 2 = 1 − 1/2(766.7)2 = 0.99999915c.
p
P20-11 (a) ∆t0 = (5.00 µs) 1 − (0.6)2 = 4.00 µs.
(b) Note: it takes time for the reading on the S 0 clock to be seen by the S clock. In this case,
∆t1 + ∆t2 = 5.00 µs, where ∆t1 = x/u and ∆t2 = x/c. Solving for ∆t1 ,
∆t1 =
(5.00 µs)/(0.6c)
= 3.125 s,
1/(0.6c) + 1/c
and
∆t01 = (3.125 µs)
p
1 − (0.6)2 = 2.50 µs.
P20-12 The only change in the components of ∆r occur parallel to the boost. Then we can choose
the boost to be parallel to ∆r and then
∆r0 = γ[∆r − u(0)] = γ∆r ≥ ∆r,
since γ ≥ 1.
P20-13
(a) Start with Eq. 20-34,
E 2 = (pc)2 + (mc2 )2 ,
and substitute into this E = K + mc2 ,
K 2 + 2Kmc2 + (mc2 )2 = (pc)2 + (mc2 )2 .
261
We can rearrange this, and then
K 2 + 2Kmc2
= (pc)2 ,
(pc)2 − K 2
m =
2Kc2
(b) As v/c → 0 we have K → 12 mv 2 and p → mv, the classical limits. Then the above expression
becomes
m2 v 2 c2 − 14 m2 v 4
,
mv 2 c2
v 2 c2 − 14 v 4
,
= m
2 2
v c 2
1v
= m 1−
4 c2
m =
But v/c → 0, so this expression reduces to m = m in the classical limit, which is a good thing.
(c) We get
(121 MeV)2 − (55.0 MeV)2
m=
= 1.06 MeV/c2 ,
2(55.0 MeV)c2
which is (1.06 MeV/c2 )/(0.511 MeV/c2 ) = 207me . A muon.
P20-14 Since E mc2 the particle is ultra-relativistic and v ≈ c. γ = (135)/(0.1396) = 967.
Then the particle has a lab-life of ∆t0 = (967)(35.0×10−9 s) = 3.385×10−5 s. The distance traveled
is
x = (3.00×108 m/s)(3.385×10−5 s) = 1.016×104 m,
so the pion decays 110 km above the Earth.
P20-15 (a) A completely inelastic collision means the two particles, each of mass m1 , stick
together after the collision, in effect becoming a new particle of mass m2 . We’ll use the subscript 1
for moving particle of mass m1 , the subscript 0 for the particle which is originally at rest, and the
subscript 2 for the new particle after the collision. We need to conserve momentum,
p 1 + p0 = p 2 ,
γ1 m1 u1 + (0) = γ2 m2 u2 ,
and we need to conserve total energy,
E1 + E0
γ1 m1 c + m1 c2
2
= E2 ,
= γ2 m2 c2 ,
Divide the momentum equation by the energy equation and then
γ1 u1
= u2 .
γ1 + 1
p
But u1 = c 1 − 1/γ12 , so
u2
= c
γ1
p
1 − 1/γ12
,
γ1 + 1
p
γ12 − 1
= c
,
γ1 + 1
262
p
(γ1 + 1)(γ1 − 1)
,
= c
γ1 + 1
r
γ1 − 1
= c
.
γ1 + 1
(b) Using the momentum equation,
m2
γ1 u1
,
γ2 u2
p
cγ1 1 − 1/γ12
p
m1
,
u2 / 1 − (u2 /c)2
p
γ2 − 1
m1 p 1
,
1/ (c/u2 )2 − 1
p
γ12 − 1
m1 p
,
1/ (γ1 + 1)/(γ1 − 1) − 1
p
(γ1 + 1)(γ1 − 1)
m1 p
,
(γ1 − 1)/2
p
m1 2(γ1 + 1).
= m1
=
=
=
=
=
R
R
R
R
P20-16 (a) K = W = F dx = (dp/dt)dx = (dx/dt)dp = v dp.
(b) dp = mγ dv + mv(dγ/dv)dv. Now use Maple or Mathematica to save time, and get
dp =
m dv
mv 2 dv
+ 2
.
2
2
1/2
(1 − v /c )
c (1 − v 2 /c2 )3/2
Now integrate:
K
=
=
P20-17
m
mv 2
v
+
(1 − v 2 /c2 )1/2
c2 (1 − v 2 /c2 )3/2
2
mv
p
.
1 − v 2 /c2
Z
dv,
(a) Since E = K + mc2 , then
E new = 2E = 2mc2 + 2K = 2mc2 (1 + K/mc2 ).
(b) E new = 2(0.938 GeV) + 2(100 GeV) = 202 GeV.
(c) K = (100 GeV)/2 − (0.938 GeV) = 49.1 GeV.
P20-18 (a) Assume only one particle is formed. That particle can later decay, but it sets the
standard on energy and momentum conservation. The momentum of this one particle must equal
that of the incident proton, or
p2 c2 = [(mc2 + K)2 − m2 c4 ].
The initial energy was K + 2mc2 , so the mass of the “one” particle is given by
M 2 c4 = [(K + 2mc2 )2 − p2 c2 ] = 2Kmc2 + 4m2 c4 .
This is a measure of the available energy; the remaining energy is required to conserve momentum.
Then
√
p
E new = M 2 c4 = 2mc2 1 + K/2mc2 .
263
P20-19 The initial momentum is mγ i v i . The final momentum is (M − m)γ f v f . Manipulating the
momentum conservation equation,
= (M − m)γ f v f ,
p
1 − βf 2
1
=
,
mγ i β i
(M − m)β f
M −m
1
=
−1 ,
mγ i β i
βf 2
M −m
1
+1 =
,
mγ i β i
βf 2
mγ i v i
264
E21-1 (a) We’ll assume that the new temperature scale is related to the Celsius scale by a linear
transformation; then TS = mTC + b, where m and b are constants to be determined, TS is the
temperature measurement in the “new” scale, and TC is the temperature measurement in Celsius
degrees.
One of our known points is absolute zero;
TS = mTC + b,
(0) = m(−273.15◦ C) + b.
We have two other points, the melting and boiling points for water,
(TS )bp
(TS )mp
= m(100◦ C) + b,
= m(0◦ C) + b;
we can subtract the top equation from the bottom equation to get
(TS )bp − (Ts )S mp = 100 C◦ m.
We are told this is 180 S◦ , so m = 1.8 S◦ /C◦ . Put this into the first equation and then find b,
b = 273.15◦ Cm = 491.67◦ S. The conversion is then
TS = (1.8 S◦ /C◦ )TC + (491.67◦ S).
(b) The melting point for water is 491.67◦ S; the boiling point for water is 180 S◦ above this, or
671.67◦ S.
E21-2 T F = 9(−273.15 deg C)/5 + 32◦ F = −459.67◦ F.
E21-3 (a) We’ll assume that the new temperature scale is related to the Celsius scale by a linear
transformation; then TS = mTC + b, where m and b are constants to be determined, TS is the
temperature measurement in the “new” scale, and TC is the temperature measurement in Celsius
degrees.
One of our known points is absolute zero;
TS = mTC + b,
(0) = m(−273.15◦ C) + b.
We have two other points, the melting and boiling points for water,
(TS )bp
(TS )mp
= m(100◦ C) + b,
= m(0◦ C) + b;
we can subtract the top equation from the bottom equation to get
(TS )bp − (Ts )S mp = 100 C◦ m.
We are told this is 100 Q◦ , so m = 1.0 Q◦ /C◦ . Put this into the first equation and then find b,
b = 273.15◦ C = 273.15◦ Q. The conversion is then
TS = TC + (273.15◦ S).
(b) The melting point for water is 273.15◦ Q; the boiling point for water is 100 Q◦ above this, or
373.15◦ Q.
(c) Kelvin Scale.
265
E21-4 (a) T = (9/5)(6000 K − 273.15) + 32 = 10000◦ F.
(b) T = (5/9)(98.6◦ F − 32) = 37.0◦ C.
(c) T = (5/9)(−70◦ F − 32) = −57◦ C.
(d) T = (9/5)(−183◦ C) + 32 = −297◦ F.
(e) It depends on what you think is hot. My mom thinks 79◦ F is too warm; that’s T =
(5/9)(79◦ F − 32) = 26◦ C.
E21-5 T = (9/5)(310 K − 273.15) + 32 = 98.3◦ F, which is fine.
E21-6 (a) T = 2(5/9)(T − 32), so −T /10 = −32, or T = 320◦ F.
(b) 2T = (5/9)(T − 32), so 13T /5 = −32, or T = −12.3◦ F.
E21-7 If the temperature (in Kelvin) is directly proportional to the resistance then T = kR,
where k is a constant of proportionality. We are given one point, T = 273.16 K when R = 90.35 Ω,
but that is okay; we only have one unknown, k. Then (273.16 K) = k(90.35 Ω) or k = 3.023 K/Ω.
If the resistance is measured to be R = 96.28 Ω, we have a temperature of
T = kR = (3.023 K/Ω)(96.28 Ω) = 291.1 K.
E21-8 T = (510◦ C)/(0.028 V)V , so T = (1.82×104◦ C/V)(0.0102 V) = 186◦ C.
E21-9 We must first find the equation which relates gain to temperature, and then find the gain
at the specified temperature. If we let G be the gain we can write this linear relationship as
G = mT + b,
where m and b are constants to be determined. We have two known points:
(30.0)
(35.2)
= m(20.0◦ C) + b,
= m(55.0◦ C) + b.
If we subtract the top equation from the bottom we get 5.2 = m(35.0◦ C), or m = 1.49 C−1 . Put
this into either of the first two equations and
(30.0) = (0.149 C−1 )(20.0◦ C) + b,
which has a solution b = 27.0
Now to find the gain when T = 28.0 ◦ C:
G = mT + b = (0.149 C−1 )(28.0◦ C) + (27.0) = 31.2
E21-10 p/ptr = (373.15 K)/(273.16 K) = 1.366.
E21-11 100 cm Hg is 1000 torr. P He = (100 cm Hg)(373 K)/(273.16 K) = 136.550 cm Hg. Nitrogen records a temperature which is 0.2 K higher, so P N = (100 cm Hg)(373.2 K)/(273.16 K) =
136.623 cm Hg. The difference is 0.073 cm Hg.
E21-12 ∆L = (23×10−6 /C◦ )(33 m)(15C◦ ) = 1.1×10−2 m.
E21-13 ∆L = (3.2×10−6 /C◦ )(200 in)(60C◦ ) = 3.8×10−2 in.
266
E21-14 L0 = (2.725cm)[1 + (23×10−6 /C◦ )(128C◦ )] = 2.733 cm.
E21-15 We want to focus on the temperature change, not the absolute temperature. In this
case, ∆T = T f − T i = (42◦ C) − (−5.0◦ C) = 47 C◦ .
Then
∆L = (11 × 10−6 C−1 )(12.0 m)(47 C◦ ) = 6.2 × 10−3 m.
E21-16 ∆A = 2αA∆T , so
∆A = 2(9×10−6 /C◦ )(2.0 m)(3.0 m)(30C◦ ) = 3.2×10−3 m2 .
E21-17 (a) We’ll apply Eq. 21-10. The surface area of a cube is six times the area of one face,
which is the edge length squared. So A = 6(0.332 m)2 = 0.661 m2 . The temperature change is
∆T = (75.0◦ C) − (20.0◦ C) = 55.0 C◦ . Then the increase in surface area is
∆A = 2αA∆T = 2(19 × 10−6 C−1 )(0.661 m2 )(55.0 C◦ ) = 1.38 × 10−3 m2
(b) We’ll now apply Eq. 21-11. The volume of the cube is the edge length cubed, so
V = (0.332 m)3 = 0.0366 m3 .
and then from Eq. 21-11,
∆V = 2αV ∆T = 3(19 × 10−6 C−1 )(0.0366 m3 )(55.0 C◦ ) = 1.15 × 10−4 m3 ,
is the change in volume of the cube.
E21-18 V 0 = V (1 + 3α∆T ), so
V 0 = (530 cm3 )[1 + 3(29×10−6 /C◦ )(−172 C◦ )] = 522 cm3 .
E21-19 (a) The slope is approximately 1.6×10−4 /C◦ .
(b) The slope is zero.
E21-20 ∆r = (β/3)r∆T , so
∆r = [(3.2×10−5 /K)/3](6.37×106 m)(2700 K) = 1.8×105 m.
E21-21 We’ll assume that the steel ruler measures length correctly at room temperature. Then
the 20.05 cm measurement of the rod is correct. But both the rod and the ruler will expand in the
oven, so the 20.11 cm measurement of the rod is not the actual length of the rod in the oven. What
is the actual length of the rod in the oven? We can only answer that after figuring out how the 20.11
cm mark on the ruler moves when the ruler expands.
Let L = 20.11 cm correspond to the ruler mark at room temperature. Then
∆L = αsteel L∆T = (11 × 10−6 C−1 )(20.11 cm)(250 C◦ ) = 5.5 × 10−2 cm
is the shift in position of the mark as the ruler is raised to the higher temperature. Then the change
in length of the rod is not (20.11 cm) − (20.05 cm) = 0.06 cm, because the 20.11 cm mark is shifted
out. We need to add 0.055 cm to this; the rod changed length by 0.115 cm.
The coefficient of thermal expansion for the rod is
α=
(0.115 cm)
∆L
=
= 23 × 10−6 C−1 .
L∆T
(20.05 cm)(250 C◦ )
267
E21-22 A = ab, A0 = (a + ∆a)(b + ∆b) = ab + a∆b + b∆a + ∆a∆b, so
∆A = a∆b + b∆a + ∆a∆b,
= A(∆b/b + ∆a/a + ∆a∆b/ab),
≈ A(α∆T + α∆T ),
= 2αA∆T.
E21-23 Solve this problem by assuming the solid is in the form of a cube.
If the length of one side of a cube is originally L0 , then the volume is originally V0 = L30 . After
heating, the volume of the cube will be V = L3 , where L = L0 + ∆L.
Then
V
=
=
=
=
L3 ,
(L0 + ∆L)3 ,
(L0 + αL0 ∆T )3 ,
L30 (1 + α∆T )3 .
As long as the quantity α∆T is much less than one we can expand the last line in a binomial
expansion as
V ≈ V0 (1 + 3α∆T + · · ·),
so the change in volume is ∆V ≈ 3αV0 ∆T .
E21-24 (a) ∆A/A = 2(0.18%) = (0.36%).
(b) ∆L/L = 0.18%.
(c) ∆V /V = 3(0.18%) = (0.54%).
(d) Zero.
(e) α = (0.0018)/(100 C◦ ) = 1.8×10−5 /C◦ .
E21-25 ρ0 − ρ = m/V 0 − m/V = m/(V + ∆V ) − m/V ≈ −m∆V /V 2 . Then
∆ρ = −(m/V )(∆V /V ) = −ρβ∆T.
E21-26 Use the results of Exercise 21-25.
(a) ∆V /V = 3∆L/L = 3(0.092%) = 0.276%. The change in density is
∆ρ/ρ = −∆V /V = −(0.276%) = −0.28
(b) α = β/3 = (0.28%)/3(40 C◦ ) = 2.3×10−5 /C◦ . Must be aluminum.
E21-27
The diameter of the rod as a function of temperature is
ds = ds,0 (1 + αs ∆T ),
The diameter of the ring as a function of temperature is
db = db,0 (1 + αb ∆T ).
268
We are interested in the temperature when the diameters are equal,
ds,0 (1 + αs ∆T ) = db,0 (1 + αb ∆T ),
αs ds,0 ∆T − αb db,0 ∆T = db,0 − ds,0 ,
db,0 − ds,0
∆T =
,
αs ds,0 − αb db,0
(2.992 cm) − (3.000 cm)
∆T =
,
−6
◦
(11×10 /C )(3.000 cm) − (19×10−6 /C◦ )(2.992 cm)
= 335 C◦ .
The final temperature is then T f = (25◦ ) + 335 C◦ = 360◦ .
E21-28 (a) ∆L = ∆L1 + ∆L2 = (L1 α1 + L2 α2 )∆T . The effective value for α is then
α=
∆L
α1 L1 + α2 L2
=
.
L∆T
L
(b) Since L2 = L − L1 we can write
α1 L1 + α2 (L − L1 )
L1
= αL,
α − α2
= L
,
α1 − α2
(13×10−6 ) − (11×10−6 )
= (0.524 m)
= 0.131 m.
(19×10−6 ) − (11×10−6 )
The brass length is then 13.1 cm and the steel is 39.3 cm.
E21-29 At 100◦ C the glass and mercury each have a volume V0 . After cooling, the difference in
volume changes is given by
∆V = V0 (3αg − β m )∆T.
Since m = ρV , the mass of mercury that needs to be added can be found by multiplying though by
the density of mercury. Then
∆m = (0.891 kg)[3(9.0×10−6 /C◦ ) − (1.8×10−4 /C◦ )](−135C◦ ) = 0.0184 kg.
This is the additional amount required, so the total is now 909 g.
R
E21-30 (a) The rotational inertia is given by I = r2 dm; changing the temperature requires
r → r0 = r + ∆r = r(1 + α∆T ). Then
Z
Z
I 0 = (1 + α∆T )2 r2 dm ≈ (1 + 2α∆T ) r2 dm,
so ∆I = 2αI∆T .
(b) Since L = Iω, then 0 = ω∆I + I∆ω. Rearranging, ∆ω/ω = −∆I/I = −2α∆T . Then
∆ω = −2(19×10−6 /C◦ )(230 rev/s)(170 C◦ ) = −1.5 rev/s.
269
E21-31 This problem is related to objects which expand when heated, but we never actually need
to calculate any temperature changes. We will, however, be interested in the change in rotational
inertia. Rotational inertia is directly proportional to the square of the (appropriate) linear dimension,
so
I f /I i = (rf /ri )2 .
(a) If the bearings are frictionless then there are no external torques, so the angular momentum
is constant.
(b) If the angular momentum is constant, then
Li
I i ωi
= Lf ,
= I f ωf .
We are interested in the percent change in the angular velocity, which is
2
2
ωf
Ii
ri
1
ωf − ωi
=
−1=
−1=
−1=
− 1 = −0.36%.
ωi
ωi
If
rf
1.0018
(c) The rotational kinetic energy is proportional to Iω 2 = (Iω)ω = Lω, but L is constant, so
Kf − Ki
ωf − ωi
=
= −0.36%.
Ki
ωi
E21-32 (a) The period of a physical pendulum is given by Eq. 17-28. There are two variables
in the equation that depend on length. I, which is proportional to a length squared, and d, which
is proportional √
to a length. This means that the period have an overall dependence on length
proportional to r. Taking the derivative,
∆P ≈ dP =
1P
1
dr ≈ P α∆T.
2r
2
(b) ∆P/P = (0.7×10−6 C◦ )(10C◦ )/2 = 3.5×10−6 . After 30 days the clock will be slow by
∆t = (30 × 24 × 60 × 60 s)(3.5×10−6 ) = 9.07 s.
E21-33 Refer to the Exercise 21-32.
∆P = (3600 s)(19×10−6 C◦ )(−20C◦ )/2 = 0.68 s.
E21-34 At 22◦ C the aluminum cup and glycerin each have a volume V0 . After heating, the
difference in volume changes is given by
∆V = V0 (3αa − β g )∆T.
The amount that spills out is then
∆V = (110 cm3 )[3(23×10−6 /C◦ ) − (5.1×10−4 /C◦ )](6C◦ ) = −0.29 cm3 .
E21-35 At 20.0◦ C the glass tube is filled with liquid to a volume V0 . After heating, the difference
in volume changes is given by
∆V = V0 (3αg − β l )∆T.
The cross sectional area of the tube changes according to
∆A = A0 2αg ∆T.
270
Consequently, the height of the liquid changes according to
∆V
∆V /V0
= (h0 + ∆h)(A0 + ∆A) − h0 A,
≈ h0 ∆A + A0 ∆h,
= ∆A/A0 + ∆h/h0 .
Then
∆h = (1.28 m/2)[(1.1×10−5 /C◦ ) − (4.2×10−5 /C◦ )](13 C◦ ) = 2.6×10−4 m.
E21-36 (a) β = (dV /dT )/V . If pV = nRT , then p dV = nR dT , so
β = (nR/p)/V = nR/pV = 1/T.
(b) Kelvins.
(c) β ≈ 1/(300/K) = 3.3×10−3 /K.
E21-37 (a) V = (1 mol)(8.31 J/mol · K)(273 K)/(1.01×105 Pa) = 2.25×10−2 m3 .
(b) (6.02×1023 mol−1 )/(2.25×104 /cm3 ) = 2.68×1019 .
E21-38 n/V = p/kT , so
3
n/V = (1.01×10−13 Pa)/(1.38×10−23 J/K)(295 K) = 25 part/cm .
E21-39
(a) Using Eq. 21-17,
n=
pV
(108×103 Pa)(2.47 m3 )
=
= 113 mol.
RT
(8.31 J/mol·K)([12 + 273] K)
(b) Use the same expression again,
V =
nRT
(113 mol)(8.31 J/mol·K)([31 + 273] K)
=
= 0.903 m3 .
p
(316×103 Pa)
E21-40 (a) n = pV /RT = (1.01×105 Pa)(1.13×10−3 m3 )/(8.31 J/mol·K)(315 K) = 4.36×10−2 mol.
(b) T f = T i pf V f /pi V i , so
Tf =
(315 K)(1.06×105 Pa)(1.530×10−3 m3 )
= 448 K.
(1.01×105 Pa)(1.130×10−3 m3 )
2
2
E21-41 pi = (14.7 + 24.2) lb/in = 38.9 lb/in . pf = pi T f V i /T i V f , so
2
pf =
(38.9 lb/in )(299K)(988 in3 )
2
= 41.7 lb/in .
(270 K)(1020 in3 )
2
2
The gauge pressure is then (41.7 − 14.7) lb/in = 27.0 lb/in .
E21-42 Since p = F/A and F = mg, a reasonable estimate for the mass of the atmosphere is
m = pA/g = (1.01×105 Pa)4π(6.37×106 m)2 /(9.81 m/s2 ) = 5.25×1018 kg.
271
E21-43 p = p0 + ρgh, where h is the depth. Then P f = 1.01×105 Pa and
pi = (1.01×105 Pa) + (998 kg/m3 )(9.81 m/s2 )(41.5 m) = 5.07×105 Pa.
V f = V i pi T f /pf T i , so
Vf =
(19.4 cm3 )(5.07×105 Pa.)(296 K)
= 104 cm3 .
(1.01×105 Pa)(277 K)
E21-44 The new pressure in the pipe is
pf = pi V i /V f = (1.01×105 Pa)(2) = 2.02×105 Pa.
The water pressure at some depth y is given by p = p0 + ρgy, so
y=
(2.02×105 Pa) − (1.01×105 Pa)
= 10.3 m.
(998 kg/m3 )(9.81 m/s2 )
Then the water/air interface inside the tube is at a depth of 10.3 m; so h = (10.3 m) + (25.0 m)/2 =
22.8 m.
P21-1 (a) The dimensions of A must be [time]−1 , as can be seen with a quick inspection of the
equation. We would expect that A would depend on the surface area at the very least; however,
that means that it must also depend on some other factor to fix the dimensionality of A.
(b) Rearrange and integrate,
Z T
Z t
d∆T
= −
A dt,
∆T0 ∆T
0
ln(∆T /∆T0 ) = −At,
∆T = ∆T0 e−At .
P21-2
First find A.
ln(∆T0 /∆T )
ln[(29 C◦ )/(25 C◦ )]
=
= 3.30×10−3 /min.
t
(45 min)
A=
Then find time to new temperature difference.
t=
ln(∆T0 /∆T )
ln[(29 C◦ )/(21 C◦ )]
=
= 97.8min
t
(3.30×10−3 /min)
This happens 97.8 − 45 = 53 minutes later.
P21-3 If we neglect the expansion of the tube then we can assume the cross sectional area of the
tube is constant. Since V = Ah, we can assume that ∆V = A∆h. Then since ∆V = βV0 ∆T , we
can write ∆h = βh0 ∆T .
P21-4 For either container we can write pi Vi = ni RTi . We are told that Vi and ni are constants.
Then ∆p = AT1 − BT2 , where A and B are constants. When T1 = T2 ∆p = 0, so A = B. When
T1 = T tr and T2 = T b we have
(120 mm Hg) = A(373 K − 273.16 K),
so A = 1.202 mm Hg/K. Then
T =
(90 mm Hg) + (1.202 mm Hg/K)(273.16 K)
= 348 K.
(1.202 mm Hg/K)
Actually, we could have assumed A was negative, and then the answer would be 198 K.
272
P21-5
Start with a differential form for Eq. 21-8, dL/dT = αL0 , rearrange, and integrate:
Z L
Z T
dL =
αL0 dT,
L0
T0
L − L0
= L0
T
Z
α dT,
T0
L = L0
1+
Z
T
α dT
!
.
T0
P21-6
∆L = αL∆T , so
1 ∆L
(96×10−9 m/s)
∆T
=
=
= 0.23◦ C/s.
∆t
αL ∆t
(23×10−6 /C◦ )(1.8×10−2 m)
P21-7
(a) Consider the work that was done for Ex. 21-27. The length of rod a is
La = La,0 (1 + αa ∆T ),
while the length of rod b is
Lb = Lb,0 (1 + αb ∆T ).
The difference is
La − Lb
= La,0 (1 + αa ∆T ) − Lb,0 (1 + αb ∆T ),
= La,0 − Lb,0 + (La,0 αa − Lb,0 αb )∆T,
which will be a constant is La,0 αa = Lb,0 αb or
Li,0 ∝ 1/αi .
(b) We want La,0 − Lb,0 = 0.30 m so
k/αa − k/αb = 0.30 m,
where k is a constant of proportionality;
k = (0.30 m)/ 1/(11×10−6 /C◦ ) − 1/(19×10−6 /C◦ ) = 7.84×10−6 m/C◦ .
The two lengths are
La = (7.84×10−6 m/C◦ )/(11×10−6 /C◦ ) = 0.713 m
for steel and
Lb = (7.84×10−6 m/C◦ )/(19×10−6 /C◦ ) = 0.413 m
for brass.
P21-8 The fractional increase in length of the bar is ∆L/L0 = α∆T. The right triangle on the left
has base L0 /2, height x, and hypotenuse (L0 + ∆L)/2. Then
r
q
1
L
∆L
0
2
x=
(L0 + ∆L)2 − L0 =
2
.
2
2
L0
With numbers,
x=
(3.77 m) p
2(25×10−6 /C◦ )(32 C◦ ) = 7.54×10−2 m.
2
273
P21-9 We want to evaluate V = V0 (1 +
graph looks like a triangle, so the result is
R
β dT ); the integral is the area under the graph; the
V = V0 [1 + (16 C◦ )(0.0002/C◦ )/2] = (1.0016)V0 .
The density is then
ρ = ρ0 (V0 /V ) = (1000 kg/m3 )/(1.0016) = 0.9984 kg/m3 .
P21-10 At 0.00◦ C the glass bulb is filled with mercury to a volume V0 . After heating, the difference
in volume changes is given by
∆V = V0 (β − 3α)∆T.
Since T0 = 0.0◦ C, then ∆T = T , if it is measured in ◦ C. The amount of mercury in the capillary is
∆V , and since the cross sectional area is fixed at A, then the length is L = ∆V /A, or
L=
P21-11
V
(β − 3α)∆T.
A
Let a, b, and c correspond to aluminum, steel, and invar, respectively. Then
cos C =
a2 + b2 − c2
.
2ab
We can replace a with a0 (1 + αa ∆T ), and write similar expressions for b and c. Since a0 = b0 = c0 ,
this can be simplified to
cos C =
(1 + αa ∆T )2 + (1 + αb ∆T )2 − (1 + αc ∆T )2
.
2(1 + αa ∆T )(1 + αb ∆T )
Expand this as a Taylor series in terms of ∆T , and we find
cos C ≈
1 1
+ (αa + αb − 2αc ) ∆T.
2 2
Now solve:
∆T =
2 cos(59.95◦ ) − 1
= 46.4◦ C.
(23×10−6 /C◦ ) + (11×10−6 /C◦ ) − 2(0.7×10−6 /C◦ )
The final temperature is then 66.4◦ C.
P21-12 The bottom of the iron bar moves downward according to ∆L = αL∆T . The center of
mass of the iron bar is located in the center; it moves downward half the distance. The mercury
expands in the glass upwards; subtracting off the distance the iron moves we get
∆h = βh∆T − ∆L = (βh − αL)∆T.
The center of mass in the mercury is located in the center. If the center of mass of the system is to
remain constant we require
mi ∆L/2 = mm (∆h − ∆L)/2;
or, since ρ = mV = mAy,
ρi αL = ρm (βh − 2αL).
Solving for h,
h=
(12×10−6 /C◦ )(1.00 m)[(7.87×103 kg/m3 ) + 2(13.6×103 kg/m3 )]
= 0.17 m.
(13.6×103 kg/m3 )(18×10−5 /C◦ )
274
P21-13 The volume of the block which is beneath the surface of the mercury displaces a mass
of mercury equal to the mass of the block. The mass of the block is independent of the temperature
but the volume of the displaced mercury changes according to
V m = V m,0 (1 + β m ∆T ).
This volume is equal to the depth which the block sinks times the cross sectional area of the block
(which does change with temperature). Then
hs hb 2 = hs,0 hb,0 2 (1 + β m ∆T ),
where hs is the depth to which the block sinks and hb,0 = 20 cm is the length of the side of the
block. But
hb = hb,0 (1 + αb ∆T ),
so
hs = hs,0
1 + β m ∆T
.
(1 + αb ∆T )2
Since the changes are small we can expand the right hand side using the binomial expansion; keeping
terms only in ∆T we get
hs ≈ hs,0 (1 + (β m − 2αb )∆T ),
which means the block will sink a distance hs − hs,0 given by
hs,0 (β m − 2αb )∆T = hs,0 (1.8×10−4 /C◦ ) − 2(23×10−6 /C◦ ) (50 C◦ ) = (6.7×10−3 )hs,0 .
In order to finish we need to know how much of the block was submerged in the first place. Since
the fraction submerged is equal to the ratio of the densities, we have
hs,0 /hb,0 = ρb /ρm = (2.7×103 kg/m3 )/(1.36×104 kg/m3 ),
so hs,0 = 3.97 cm, and the change in depth is 0.27 mm.
P21-14 The area of glass expands according to ∆Ag = 2αg Ag ∆T . The are of Dumet wire expands
according to
∆Ac + ∆Ai = 2(αc Ac + αi Ai )∆T.
We need these to be equal, so
αg Ag = αc Ac + αi Ai ,
αg rg 2 = αc (rc 2 − ri 2 ) + αi ri 2 ,
αg (rc 2 + ri 2 ) = αc (rc 2 − ri 2 ) + αi ri 2 ,
ri 2
rc 2
=
αc − αg
.
αc − αi
P21-15
P21-16
V2 = V1 (p1 /p2 )(T1 /T2 ), so
V2 = (3.47 m3 )[(76 cm Hg)/(36 cm Hg)][(225 K)/(295 K)] = 5.59 m3 .
275
P21-17 Call the containers one and two so that V1 = 1.22 L and V2 = 3.18 L. Then the initial
number of moles in the two containers are
n1,i =
pi V 1
pi V 2
and n2,i =
.
RT i
RT i
The total is
n = pi (V1 + V2 )/(RT i ).
Later the temperatures are changed and then the number of moles of gas in each container is
n1,f =
pf V 2
pf V 1
and n2,f =
.
RT 1,f
RT 2,f
The total is still n, so
pf
R
V1
V2
+
T 1,f
T 2,f
=
pi (V1 + V2 )
.
RT i
We can solve this for the final pressure, so long as we remember to convert all temperatures to
Kelvins,
−1
V2
pi (V1 + V2 ) V1
pf =
+
,
Ti
T 1,f
T 2,f
or
−1
(1.44 atm)(1.22L + 3.18 L) (1.22 L) (3.18 L)
pf =
+
= 1.74 atm.
(289 K)
(289 K)
(381 K)
P21-18 Originally nA = pA VA /RTA and nB = pB VB /RTB ; VB = 4VA . Label the final state of
A as C and the final state of B as D. After mixing, nC = pC VA /RTA and nD = pD VB /RTB , but
PC = PD and nA + nB = nC + nD . Then
pA /TA + 4pB /TB = pC (1/TA + 4/TB ),
or
pC =
(5×105 Pa)/(300 K) + 4(1×105 Pa)/(400 K)
= 2.00×105 Pa.
1/(300 K) + 4/(400 K)
P21-19 If the temperature is uniform then all that is necessary is to substitute p0 = nRT /V and
p = nRT /V ; cancel RT from both sides, and then equate n/V with nV .
P21-20 Use the results of Problem 15-19. The initial pressure inside the bubble is pi = p0 + 4γ/ri .
The final pressure inside the bell jar is zero, sopf = 4γ/rf . The initial and final pressure inside the
bubble are related by pi ri 3 = pf rf 3 = 4γrf 2 . Now for numbers:
pi = (1.01×105 Pa) + 4(2.5×10−2 N/m)/(2.0×10−3 m) = 1.0105×105 Pa.
and
rf =
s
(1.0105×105 Pa)(2.0×10−3 m)3
= 8.99×10−2 m.
4(2.5×10−2 N/m)
P21-21
P21-22
276
E22-1 (a) n = (2.56 g)/(197 g/mol) = 1.30×10−2 mol.
(b) N = (6.02×1023 mol−1 )(1.30×10−2 mol) = 7.83×1021 .
E22-2 (a) N = pV /kT = (1.01×105 Pa)(1.00 m3 )/(1.38×10−23 J/K)(293 K) = 2.50×1025 .
(b) n = (2.50×1025 )/(6.02×1023 mol−1 ) = 41.5 mol. Then
m = (41.5 mol)[75%(28 g/mol) + 25%(32 g/mol)] = 1.20 kg.
E22-3
(a) We first need to calculate the molar mass of ammonia. This is
M = M (N) + 3M (H) = (14.0 g/mol) + 3(1.01 g/mol) = 17.0 g/mol
The number of moles of nitrogen present is
n = m/M r = (315 g)/(17.0 g/mol) = 18.5 mol.
The volume of the tank is
V = nRT /p = (18.5 mol)(8.31 J/mol · K)(350 K)/(1.35×106 Pa) = 3.99×10−2 m3 .
(b) After the tank is checked the number of moles of gas in the tank is
n = pV /(RT ) = (8.68×105 Pa)(3.99×10−2 m3 )/[(8.31 J/mol · K)(295 K)] = 14.1 mol.
In that case, 4.4 mol must have escaped; that corresponds to a mass of
m = nM r = (4.4 mol)(17.0 g/mol) = 74.8 g.
E22-4 (a) The volume per particle is V /N = kT /P , so
V /N = (1.38×10−23 J/K)(285 K)/(1.01×105 Pa) = 3.89×10−26 m3 .
The edge length is the cube root of this, or 3.39×10−9 m. The ratio is 11.3.
(b) The volume per particle is V /NA , so
V /NA = (18×10−6 m3 )/(6.02×1023 ) = 2.99×10−29 m3 .
The edge length is the cube root of this, or 3.10×10−10 m. The ratio is 1.03.
E22-5 The volume per particle is V /N = kT /P , so
V /N = (1.38×10−23 J/K)(308 K)/(1.22)(1.01×105 Pa) = 3.45×10−26 m3 .
The fraction actually occupied by the particle is
4π(0.710×10−10 m)3 /3)
= 4.34×10−5 .
(3.45×10−26 m3 )
E22-6 The component of the momentum normal to the wall is
py = (3.3×10−27 kg)(1.0×103 m/s) cos(55◦ ) = 1.89×10−24 kg · m/s.
The pressure exerted on the wall is
p=
F
(1.6×1023 /s)2(1.89×10−24 kg · m/s)
=
= 3.0×103 Pa.
A
(2.0×10−4 m2 )
277
E22-7
(a) From Eq. 22-9,
v rms =
Then
p = 1.23 × 10−3 atm
r
3p
.
ρ
1.01 × 105 Pa
1 atm
and
−5
ρ = 1.32 × 10
g/cm
3
1kg
1000 g
100 cm
1m
3
= 124 Pa
= 1.32 × 10−2 kg/m3 .
Finally,
v rms =
s
3(1240 Pa)
= 531 m/s.
(1.32 × 10−2 kg/m3 )
(b) The molar density of the gas is just n/V ; but this can be found quickly from the ideal gas
law as
n
p
(1240 Pa)
=
=
= 4.71 × 10−1 mol/m3 .
V
RT
(8.31 J/mol · K)(317 K)
(c) We were given the density, which is mass per volume, so we could find the molar mass from
ρ
(1.32 × 10−2 kg/m3 )
=
= 28.0 g/mol.
n/V
(4.71 × 10−1 mol/m3 )
But what gas is it? It could contain any atom lighter than silicon; trial and error is the way to go.
Some of my guesses include C2 H4 (ethene), CO (carbon monoxide), and N2 . There’s no way to tell
which is correct at this point, in fact, the gas could be a mixture of all three.
E22-8 The density is ρ = m/V = nMr /V , or
ρ = (0.350 mol)(0.0280 kg/mol)/π(0.125 m/2)2 (0.560 m) = 1.43 kg/m3 .
The rms speed is
v rms =
s
3(2.05)(1.01×105 Pa)
= 659 m/s.
(1.43 kg/m3 )
E22-9 (a) N/V = p/kT = (1.01×105 Pa)/(1.38×10−23 J/K)(273 K) = 2.68×1025 /m3 .
(b) Note that Eq. 22-11 is wrong; for the explanation
read the
√
√ last two paragraphs in the first
column on page 502. We need an extra factor of 2, so πd2 = V / 2N λ, so
q √
d = 1/ 2π(2.68×1025 /m3 )(285×10−9 m) = 1.72×10−10 m.
√
E22-10 (a) λ = V / 2N πd2 , so
λ= √
1
2(1.0×106 /m3 )π(2.0×10−10 m)2
(b) Particles effectively follow ballistic trajectories.
278
= 5.6×1012 m.
E22-11 We have v = f λ, where λ is the wavelength (which we will set equal to the mean free
path), and v is the speed of sound. The mean free path is, from Eq. 22-13,
λ= √
so
√
f=
2πd2 pv
=
kT
√
kT
2πd2 p
2π(315×10−12 m)2 (1.02 × 1.01×105 Pa)(343 m/s)
= 3.88×109 Hz.
(1.38×10−23 J/K)(291 K)
E22-12 (a) p = (1.10×10−6 mm Hg)(133 Pa/mm Hg) = 1.46×10−4 Pa. The particle density is
N/V = (1.46×10−4 Pa)/(1.38×10−23 J/K)(295 K) = 3.59×1016 /m3 .
(b) The mean free path is
p
λ = 1/ (2)(3.59×1016 /m3 )π(2.20×10−10 m)2 = 130 m.
E22-13 Note that v av ∝
√
√
T , while λ ∝ T . Then the collision rate is proportional to 1/ T . Then
T = (300 K)
(5.1×109 /s)2
= 216 K.
(6.0×109 /s)2
E22-14 (a) vp
av = (65 km/s)/(10) = 6.5 km/s.
(b) v rms = (505 km/s)/(10) = 7.1 km/s.
E22-15
(a) The average is
4(200 s) + 2(500 m/s) + 4(600 m/s)
= 420 m/s.
4+2+4
The mean-square value is
4(200 s)2 + 2(500 m/s)2 + 4(600 m/s)2
= 2.1 × 105 m2 /s2 .
4+2+4
The root-mean-square value is the square root of this, or 458 m/s.
(b) I’ll be lazy. Nine particles are not moving, and the tenth has a speed of 10 m/s. Then the
average speed is 1 m/s, and the root-mean-square speed is 3.16 m/s. Look, v rms is larger than v av !
(c) Can v rms =v av ? Assume that the speeds are not all the same. Transform to a frame of
reference where v av = 0, then some of the individual speeds must be greater than zero, and some
will be less than zero. Squaring these speeds will result in positive, non-zero, numbers; the mean
square will necessarily be greater than zero, so v rms > 0.
Only if all of the particles have the same speed will v rms =v av .
E22-16 Use Eq. 22-20:
v rms =
s
3(1.38×10−23 J/K)(329 K)
= 694 m/s.
(2.33×10−26 kg + 3 × 1.67×10−27 kg)
E22-17 Use Eq. 22-20:
v rms =
s
3(1.38×10−23 J/K)(2.7 K)
= 180 m/s.
(2 × 1.67×10−27 kg)
279
2
E22-18 Eq. 22-14 is in the form N = Av 2 e−Bv . Taking the derivative,
2
2
dN
= 2Ave−Bv − 2ABv 3 e−Bv ,
dv
and setting this equal to zero,
v 2 = 1/B = 2kT /m.
E22-19
We want to integrate
v av
=
=
=
Z
1 ∞
N (v)v dv,
N 0
Z ∞
m 3/2
2
1
4πN
v 2 e−mv /2kT v dv,
N 0
2πkT
m 3/2 Z ∞
2
4π
v 2 e−mv /2kT v dv.
2πkT
0
The easiest way to attack this is first with a change √
of variables— let x = mv 2 /2kT , then kT dx =
mv dv. The limits of integration don’t change, since ∞ = ∞. Then
m 3/2 Z ∞ 2kT
kT
v av = 4π
xe−x
dx,
2πkT
m
m
0
1/2 Z ∞
2kT
= 2
xe−αx dx
πm
0
The factor of α that was introduced in the last line is a Feynman trick; we’ll set it equal to one when
we are finished, so it won’t change the result.
Feynman’s trick looks like
Z
Z
Z
∂ −αx
d
−αx
e
dx =
e
dx = (−x)e−αx dx.
dα
∂α
Applying this to our original problem,
1/2 Z ∞
2kT
v av = 2
xe−αx dx,
πm
0
1/2 Z ∞
d
2kT
= − 2
e−αx dx,
dα
πm
0
∞ 1/2
2kT
d
−1 −αx = −2
,
e
πm
dα
α
0
1/2
2kT
d
1
= −2
,
πm
dα α
1/2
−1
2kT
= −2
.
πm
α2
We promised, however, that we would set α = 1 in the end, so this last line is
1/2
2kT
v av = 2
,
πm
r
8kT
=
.
πm
280
E22-20 We want to integrate
(v 2 )av
=
=
=
Z
1 ∞
N (v)v 2 dv,
N 0
Z
m 3/2
2
1 ∞
4πN
v 2 e−mv /2kT v 2 dv,
N 0
2πkT
m 3/2 Z ∞
2
4π
v 2 e−mv /2kT v 2 dv.
2πkT
0
The easiest way to attack this is first with a change of variables— let x2 = mv 2 /2kT , then
p
2kT /mdx = dv. The limits of integration don’t change. Then
2
(v )av
m 3/2 Z ∞ 2kT 5/2
2
= 4π
x4 e−x dx,
2πkT
m
0
Z ∞
8kT
4 −x2
= √
x e
dx
πm 0
Look up the integral; although you can solve it by first applying a Feynman trick (see solution to
Exercise√22-21) and then squaring the integral and changing to polar coordinates. I looked it up. I
found 3 π/8, so
8kT √
(v 2 )av = √
3 π/8 = 3kT /m.
πm
E22-21 Apply Eq. 22-20:
p
v rms = 3(1.38×10−23 J/K)(287 K)/(5.2×10−17 kg) = 1.5×10−2 m/s.
E22-22 Since v rms ∝
p
T /m, we have
T = (299 K)(4/2) = 598 K,
or 325◦ C.
E22-23 (a) The escape speed is found on page 310; v = 11.2×103 m/s. For hydrogen,
T = (2)(1.67×10−27 kg)(11.2×103 m/s)2 /3(1.38×10−23 J/K) = 1.0×104 K.
For oxygen,
T = (32)(1.67×10−27 kg)(11.2×103 m/s)2 /3(1.38×10−23 J/K) = 1.6×105 K.
(b) The escape speed is found on page 310; v = 2.38×103 m/s. For hydrogen,
T = (2)(1.67×10−27 kg)(2.38×103 m/s)2 /3(1.38×10−23 J/K) = 460K.
For oxygen,
T = (32)(1.67×10−27 kg)(2.38×103 m/s)2 /3(1.38×10−23 J/K) = 7300K.
(c) There should be more oxygen than hydrogen.
E22-24 (a) vq
av = (70 km/s)/(22) = 3.18 km/s.
(b) v rms = (250 km2 /s2 )/(22) = 3.37 km/s.
(c) 3.0 km/s.
281
E22-25
According to the equation directly beneath Fig. 22-8,
ω = vφ/L = (212 m/s)(0.0841 rad)/(0.204 m) = 87.3 rad/s.
E22-26 If v p = v rms then 2T2 = 3T1 , or T2 /T1 = 3/2.
E22-27 Read the last paragraph on the first column of page 505. The distribution of speeds is
proportional to
2
2
v 3 e−mv /2kT = v 3 e−Bv ,
taking the derivative dN/dv and setting equal to zero yields
2
2
dN
= 3v 2 e−Bv − 2Bv 4 e−Bv ,
dv
and setting this equal to zero,
v 2 = 3/2B = 3kT /m.
p
E22-28 (a) v = 3(8.31 J/mol · K)(4220 K)/(0.07261 kg/mol) = 1200 m/s.
(b) Half of the sum of the diameters, or 273 pm.
(c) The mean free path of the germanium in the argon is
√
λ = 1/ 2(4.13×1025 /m3 )π(273×10−12 m)2 = 7.31×10−8 m.
The collision rate is
(1200 m/s)/(7.31×10−8 m) = 1.64×1010 /s.
E22-29
The fraction of particles that interests us is
Z 0.03kT
1
2
√
E 1/2 e−E/kT dE.
π (kT )3/2 0.01kT
Change variables according to E/kT = x, so that dE = kT dx. The integral is then
Z 0.03
2
√
x1/2 e−x dx.
π 0.01
Since the value of x is so small compared to 1 throughout the range of integration, we can expand
according to
e−x ≈ 1 − x for x 1.
The integral then simplifies to
0.03
Z 0.03
2
2 2 3/2 2 5/2
√
x1/2 (1 − x) dx = √
x − x
= 3.09×10−3 .
5
π 0.01
π 3
0.01
E22-30 Write N (E) = N (E p + ). Then
N (E p + ) ≈ N (E p ) + dN (E) + ...
dE E p
But the very definition of E p implies that the first derivative is zero. Then the fraction of [particles
with energies in the range E p ± 0.02kT is
2
1
√
(kT /2)1/2 e−1/2 (0.02kT ),
π (kT )3/2
p
or 0.04 1/2eπ = 9.68×10−3 .
282
E22-31 The volume correction is on page 508; we need first to find d. If we assume that the
particles in water are arranged in a cubic lattice (a bad guess, but we’ll use it anyway), then 18
grams of water has a volume of 18×10−6 m3 , and
d3 =
(18×10−6 m3 )
= 3.0×10−29 m3
(6.02×1023 )
is the volume allocated to each water molecule. In this case d = 3.1×10−10 m. Then
b=
1
4
(6.02×1023 )( π(3.1×10−10 m)3 ) = 3.8 m3 /mol.
2
3
E22-32 d3 = 3b/2πNA , or
d=
s
3
3(32×10−6 m3 /mol)
= 2.9×10−10 m.
2π(6.02×1023 /mol)
E22-33 a has units of energy volume per square mole, which is the same as energy per mole times
volume per mole.
P22-1
Solve (1 − x)(1.429) + x(1.250) = 1.293 for x. The result is x = 0.7598.
P22-2
P22-3 The only thing that matters is the total number of moles of gas (2.5) and the number of
moles of the second gas (0.5). Since 1/5 of the total number of moles of gas is associated with the
second gas, then 1/5 of the total pressure is associated with the second gas.
P22-4
Use Eq. 22-11 with the appropriate
λ= √
P22-5
√
2 inserted.
(1.0×10−3 m3 )
= 6.4×10−2 m.
2(35)π(1.0×10−2 m)2
(a) Since λ ∝ 1/d2 , we have
da
=
dn
r
λn
=
λa
s
(27.5×10−8 m)
= 1.67.
(9.90×10−8 m)
(b) Since λ ∝ 1/p, we have
λ2 = λ1
p1
(75.0 cm Hg)
= (9.90×10−8 m)
= 49.5×10−8 m.
p2
(15.0 cm Hg)
(c) Since λ ∝ T , we have
λ 2 = λ1
T2
(233 K)
= (9.90×10−8 m)
= 7.87×10−8 m.
T1
(293 K)
283
P22-6
We can assume the molecule will collide with something. Then
Z ∞
1=
Ae−cr dr = A/c,
0
so A = c. If the molecule has a mean free path of λ, then
Z ∞
λ=
rce−cr dr = 1/c,
0
so A = c = 1/λ.
P22-7 What is important here is the temperature; since the temperatures are the same then the
average kinetic energies per particle are the same. Then
1
1
m1 (v rms,1 )2 = m2 (v rms,2 )2 .
2
2
We are given in the problem that v av,2 = 2v rms,1 . According to Eqs. 22-18 and 22-20 we have
r
r r
r
3RT
3π 8RT
3π
v rms =
=
=
v av .
M
8
πM
8
Combining this with the kinetic energy expression above,
m1
=
m2
v rms,2
v rms,1
2
=
r
2
3π
8
!2
= 4.71.
P22-8 (a) Assume that the speeds are not all the same. Transform to a frame of reference where
v av = 0, then some of the individual speeds must be greater than zero, and some will be less
than zero. Squaring these speeds will result in positive, non-zero, numbers; the mean square will
necessarily be greater than zero, so v rms > 0.
(b) Only if all of the particles have the same speed will v rms =v av .
P22-9
(a) We need to first find the number of particles by integrating
Z ∞
N =
N (v) dv,
Z0 v0
Z ∞
Z v0
C
2
=
Cv dv +
(0) dv = C
v 2 dv = v03 .
3
0
v0
0
Invert, then C = 3N/v03 .
(b) The average velocity is found from
v av
1
=
N
Z
∞
N (v)v dv.
0
Using our result from above,
v av
=
=
Z 1 v0 3N 2
v v dv,
N 0
v03
Z v0
3
3 v04
3
3
v
dv
=
= v0 .
3
3
v0 0
v0 4
4
284
As expected, the average speed is less than the maximum speed. We can make a prediction about
the root mean square speed; it will be larger than the average speed (see Exercise 22-15 above) but
smaller than the maximum speed.
(c) The root-mean-square velocity is found from
Z
1 ∞
2
v rms =
N (v)v 2 dv.
N 0
Using our results from above,
v 2rms
=
=
Z 1 v0 3N 2 2
v
v dv,
N 0
v03
Z v0
3 v5
3
3
v 4 dv = 3 0 = v02 .
3
v0 0
v0 5
5
Then, taking the square root,
v 2rms
Is
=
r
3
v0
5
p
3/5 > 3/4? It had better be.
P22-10
P22-11
P22-12
P22-13
P22-14
P22-15 The mass of air displaced by 2180 m3 is m = (1.22 kg/m3 )(2180 m3 ) = 2660 kg. The mass
of the balloon and basket is 249 kg and we want to lift 272 kg; this leaves a remainder of 2140 kg for
the mass of the air inside the balloon. This corresponds to (2140 kg)/(0.0289 kg/mol) = 7.4×104 mol.
The temperature of the gas inside the balloon is then
T = (pV )/(nR) = [(1.01×105 Pa)(2180 m3 )]/[(7.4×104 mol)(8.31 J/mol · K) = 358 K.
That’s 85◦ C.
P22-16
P22-17
285
E23-1
We apply Eq. 23-1,
∆T
∆x
The rate at which heat flows out is given as a power per area (mW/m2 ), so the quantity given is
really H/A. Then the temperature difference is
H = kA
∆T =
(33, 000 m)
H ∆x
= (0.054 W/m2 )
= 710 K
A k
(2.5 W/m · K)
The heat flow is out, so that the temperature is higher at the base of the crust. The temperature
there is then
710 + 10 = 720 ◦ C.
E23-2 We apply Eq. 23-1,
H = kA
∆T
(44 C◦ )
= (0.74 W/m · K)(6.2 m)(3.8 m)
= 2400 W.
∆x
(0.32 m)
E23-3 (a) ∆T /∆x = (136 C◦ )/(0.249 m) = 546 C◦ /m.
(b) H = kA∆T /∆x = (401 W/m · K)(1.80 m2 )(546 C◦ /m) = 3.94×105 W.
(c) T H = (−12◦ C + 136 C◦ ) = 124◦ C. Then
T = (124◦ C) − (546 C◦ /m)(0.11 m) = 63.9◦ C.
E23-4 (a) H = (0.040 W/m · K)(1.8 m2 )(32 C◦ )/(0.012 m) = 190 W.
(b) Since k has increased by a factor of (0.60)/(0.04) = 15 then H should also increase by a
factor of 15.
E23-5 There are three possible arrangements: a sheet of type 1 with a sheet of type 1; a sheet
of type 2 with a sheet of type 2; and a sheet of type 1 with a sheet of type 2. We can look back on
Sample Problem 23-1 to see how to start the problem; the heat flow will be
H12 =
A∆T
(L/k1 ) + (L/k2 )
for substances of different types; and
H11 =
A∆T /L
1 A∆T k1
=
(L/k1 ) + (L/k1 )
2
L
for a double layer if substance 1. There is a similar expression for a double layer of substance 2.
For configuration (a) we then have
H11 + H22 =
1 A∆T k1
1 A∆T k2
A∆T
+
=
(k1 + k2 ),
2
L
2
L
2L
while for configuration (b) we have
H12 + H21 = 2
A∆T
2A∆T
−1
=
((1/k1 ) + (1/k2 )) .
(L/k1 ) + (L/k2 )
L
We want to compare these, so expanding the relevant part of the second configuration
−1
((1/k1 ) + (1/k2 ))
−1
= ((k1 + k2 )/(k2 k2 ))
286
=
k1 k2
.
k1 + k2
Then which is larger
(k1 + k2 )/2 or
2k1 k2
?
k1 + k2
If k1 k2 then the expression become
k1 /2 and 2k2 ,
so the first expression is larger, and therefore configuration (b) has the lower heat flow. Notice that
we get the same result if k1 k2 !
E23-6 There’s a typo in the exercise.
H = A∆T /R; since the heat flows through one slab and then through the other, we can write
(T1 − Tx )/R1 = (Tx − T2 )/R2 . Rearranging,
Tx = (T1 R2 + T2 R1 )/(R1 + R2 ).
E23-7 Use the results of Exercise 23-6. At the interface between ice and water Tx = 0◦ C. Then
R1 T2 + R2 T1 = 0, or k1 T1 /L1 + k2 T2 /L2 = 0. Not only that, L1 + L2 = L, so
k1 T1 L2 + (L − L2 )k2 T2 = 0,
so
L2 =
(1.42 m)(1.67 W/m · K)(−5.20◦ C)
= 1.15 m.
(1.67 W/m · K)(−5.20◦ C) − (0.502 W/m · K)(3.98◦ C)
E23-8 ∆T is the same in both cases. So is k. The top configuration has H t = kA∆T /(2L). The
bottom configuration has H b = k(2A)∆T /L. The ratio of H b /H t = 4, so heat flows through the
bottom configuration at 4 times the rate of the top. For the top configuration H t = (10 J)/(2 min) =
5 J/min. Then H b = 20 J/min. It will take
t = (30 J)/(20 J/min) = 1.5 min.
E23-9 (a) This exercise has a distraction: it asks about the heat flow through the window, but
what you need to find first is the heat flow through the air near the window. We are given the
temperature gradient both inside and outside the window. Inside,
∆T
(20◦ C) − (5◦ C)
=
= 190 C◦ /m;
∆x
(0.08 m)
a similar expression exists for outside.
From Eq. 23-1 we find the heat flow through the air;
∆T
= (0.026 W/m · K)(0.6 m)2 (190 C◦ /m) = 1.8 W.
∆x
The value that we arrived at is the rate that heat flows through the air across an area the size of
the window on either side of the window. This heat flow had to occur through the window as well,
so
H = 1.8 W
H = kA
answers the window question.
(b) Now that we know the rate that heat flows through the window, we are in a position to find
the temperature difference across the window. Rearranging Eq. 32-1,
∆T =
H∆x
(1.8 W)(0.005 m)
=
= 0.025 C◦ ,
kA
(1.0 W/m · K)(0.6 m)2
so we were well justified in our approximation that the temperature drop across the glass is very
small.
287
E23-10 (a) W = +214 J, done on means positive.
(b) Q = −293 J, extracted from means negative.
(c) ∆E int = Q + W = (−293 J) + (+214 J) = −79.0 J.
E23-11 (a) ∆E int along any path between these two points is
∆E int = Q + W = (50 J) + (−20 J) = 30 J.
Then along ibf W = (30 J) − (36 J) = −6 J.
(b) Q = (−30 J) − (+13 J) = −43 J.
(c) E int,f = E int,i + ∆E int = (10 J) + (30 J) = 40 J.
(d) ∆E intib = (22 J) − (10 J) = 12 J; while ∆E intbf = (40 J) − (22 J) = 18 J. There is no work
done on the path bf , so
Qbf = ∆E intbf − Wbf = (18 J) − (0) = 18 J,
and Qib = Qibf − Qbf = (36 J) − (18 J) = 18 J.
E23-12 Q = mL = (0.10)(2.1×108 kg)(333×103 J/kg) = 7.0×1012 J.
E23-13 We don’t need to know the outside temperature because the amount of heat energy
required is explicitly stated: 5.22 GJ. We just need to know how much water is required to transfer
this amount of heat energy. Use Eq. 23-11, and then
m=
Q
(5.22 × 109 J)
=
= 4.45 × 104 kg.
c∆T
(4190 J/kg · K)(50.0◦ C − 22.0◦ C)
This is the mass of the water, we want to know the volume, so we’ll use the density, and then
V =
(4.45 × 104 kg)
m
=
= 44.5 m3 .
ρ
(998 kg/m3 )
E23-14 The heat energy required is Q = mc∆T . The time required is t = Q/P . Then
t=
(0.136 kg)(4190 J/kg · K)(100◦ C − 23.5◦ C)
= 198 s.
(220 W)
E23-15 Q = mL, so m = (50.4 × 103 J)/(333 × 103 J/kg) = 0.151 kg is the mount which freezes.
Then (0.258 kg) − (0.151 kg) = 0.107 kg is the amount which does not freeze.
E23-16 (a) W = mg∆y; if |Q| = |W |, then
∆T =
mg∆y
(9.81 m/s2 )(49.4 m)
=
= 0.112 C◦ .
mc
(4190 J/kg · K)
E23-17 There are three “things” in this problem: the copper bowl (b), the water (w), and the
copper cylinder (c). The total internal energy changes must add up to zero, so
∆E int,b + ∆E int,w + ∆E int,c = 0.
As in Sample Problem 23-3, no work is done on any object, so
Qb + Qw + Qc = 0.
288
The heat transfers for these three objects are
Qb
Qw
Qc
= mb cb (T f,b − T i,b ),
= mw cw (T f,w − T i,w ) + Lv m2 ,
= mc cc (T f,c − T i,c ).
For the most part, this looks exactly like the presentation in Sample Problem 23-3; but there is an
extra term in the second line. This term reflects the extra heat required to vaporize m2 = 4.70 g of
water at 100◦ C into steam 100◦ C.
Some of the initial temperatures are specified in the exercise: T i,b = T i,w = 21.0◦ C and T f,b =
T f,w = T f,c = 100◦ C.
(a) The heat transferred to the water, then, is
Qw
(0.223 kg)(4190 J/kg·K) ((100◦ C) − (21.0◦ C)) ,
+(2.26×106 J/kg)(4.70×10−3 kg),
= 8.44 × 104 J.
=
This answer differs from the back of the book. I think that they (or was it me) used the latent heat
of fusion when they should have used the latent heat of vaporization!
(b) The heat transfered to the bowl, then, is
Qw = (0.146 kg)(387 J/kg · K) ((100◦ C) − (21.0◦ C)) = 4.46 × 103 J.
(c) The heat transfered from the cylinder was transfered into the water and bowl, so
Qc = −Qb − Qw = −(4.46 × 103 J) − (8.44 × 104 J) = −8.89 × 104 J.
The initial temperature of the cylinder is then given by
T i,c = T f,c −
(−8.89 × 104 J)
Qc
= (100◦ C) −
= 832◦ C.
mc cc
(0.314 kg)(387 J/kg · K)
E23-18 The temperature of the silver must be raised to the melting point and then the heated
silver needs to be melted. The heat required is
Q = mL + mc∆T = (0.130 kg)[(105×103 J/kg) + (236 J/kg · K)(1235 K − 289 K)] = 4.27×104 J.
E23-19 (a) Use Q = mc∆T , m = ρV , and t = Q/P . Then
t
=
=
[ma ca + ρw V w cw ]∆T
,
P
[(0.56 kg)(900 J/kg·K) + (998 kg/m3 )(0.64×10−3 m3 )(4190 J/kg·K)](100◦ C − 12◦ C)
= 117 s.
(2400 W)
(b) Use Q = mL, m = ρV , and t = Q/P . Then
t=
ρw V w Lw
(998 kg/m3 )(0.640×10−3 m3 )(2256×103 J/kg)
=
= 600 s
P
(2400 W)
is the additional time required.
289
E23-20 The heat given off by the steam will be
Qs = ms Lv + ms cw (50 C◦ ).
The hear taken in by the ice will be
Qi = mi Lf + mi cw (50 C◦ ).
Equating,
ms
Lf + cw (50 C◦ )
,
Lv + cw (50 C◦ )
(333×103 J/kg) + (4190 J/kg·K)(50 C◦ )
(0.150 kg)
= 0.033 kg.
(2256×103 J/kg) + (4190 J/kg·K)(50 C◦ )
= mrmi
=
E23-21 The linear dimensions of the ring and sphere change with the temperature change according to
∆dr
∆ds
= αr dr (T f,r − T i,r ),
= αs ds (T f,s − T i,s ).
When the ring and sphere are at the same (final) temperature the ring and the sphere have the
same diameter. This means that
dr + ∆dr = ds + ∆ds
when T f,s = T f,r . We’ll solve these expansion equations first, and then go back to the heat equations.
dr + ∆dr = ds + ∆ds ,
dr (1 + αr (T f,r − T i,r )) = ds (1 + αs (T f,s − T i,s )) ,
which can be rearranged to give
αr dr T f,r − αs ds T f,s = ds (1 − αs T i,s ) − dr (1 − αr T i,r ) ,
but since the final temperatures are the same,
Tf =
ds (1 − αs T i,s ) − dr (1 − αr T i,r )
α r dr − α s ds
Putting in the numbers,
Tf =
(2.54533cm)[1−(23×10−6 /C◦ )(100◦ C)]−(2.54000cm)[1−(17×10−6 /C◦ )(0◦ C)]
,
(2.54000cm)(17×10−6 /C◦ )−(2.54533cm)(23×10−6 /C◦ )
= 34.1◦ C.
No work is done, so we only have the issue of heat flow, then
Qr + Qs = 0.
Where “r” refers to the copper ring and “s” refers to the aluminum sphere. The heat equations are
Qr
Qs
= mr cr (T f − T i,r ),
= ms cs (T f − T i,s ).
Equating and rearranging,
ms =
or
ms =
mr cr (T i,r − T f )
cs (T f − T i,s )
(21.6 g)(387 J/kg·K)(0◦ C − 34.1◦ C)
= 4.81 g.
(900 J/kg·K)(34.1◦ C − 100◦ C)
290
E23-22 The problem is compounded because we don’t know if the final state is only water, only
ice, or a mixture of the two.
Consider first the water. Cooling it to 0◦ C would require the removal of
Qw = (0.200 kg)(4190 J/kg · K)(0◦ C − 25◦ C) = −2.095×104 J.
Consider now the ice. Warming the ice to would require the addition of
Qi = (0.100 kg)(2220 J/kg · K)(0◦ C + 15◦ C) = 3.33×103 J.
The heat absorbed by the warming ice isn’t enough to cool the water to freezing. However, the ice
can melt; and if it does it will require the addition of
Qim = (0.100 kg)(333×103 J/kg) = 3.33×104 J.
This is far more than will be liberated by the cooling water, so the final temperature is 0◦ C, and
consists of a mixture of ice and water.
(b) Consider now the ice. Warming the ice to would require the addition of
Qi = (0.050 kg)(2220 J/kg · K)(0◦ C + 15◦ C) = 1.665×103 J.
The heat absorbed by the warming ice isn’t enough to cool the water to freezing. However, the ice
can melt; and if it does it will require the addition of
Qim = (0.050 kg)(333×103 J/kg) = 1.665×104 J.
This is still not enough to cool the water to freezing. Hence, we need to solve
Qi + Qim + mi cw (T − 0◦ C) + mw cw (T − 25◦ C) = 0,
which has solution
(4190 J/kg · K)(0.200 kg)(25◦ C) − (1.665×103 J) + (1.665×104 J)
= 2.5◦ C.
T =
(4190 J/kg · K)(0.250 kg)
E23-23 (a) c = (320 J)/(0.0371 kg)(42.0◦ C − 26.1◦ C) = 542 J/kg · K.
(b) n = m/M = (37.1 g)/(51.4 g/mol) = 0.722 mol.
(c) c = (542 J/kg · K)(51.4×10−3 kg/mol) = 27.9 J/mol · K.
E23-24 (1) W = −p∆V = (15 Pa)(4 m3 ) = −60 J for the horizontal path; no work is done during
the vertical path; the net work done on the gas is −60 J.
(2) It is easiest to consider work as the (negative of) the area under the curve; then W =
−(15 Pa + 5 Pa)(4 m3 )/2 = −40 J.
(3) No work is done during the vertical path; W = −p∆V = (5 Pa)(4 m3 ) = −20 J for the
horizontal path; the net work done on the gas is −20 J.
E23-25
Net work done on the gas is given by Eq. 23-15,
Z
W = − p dV.
But integrals are just the area under the curve; and that’s the easy way to solve this problem. In the
case of closed paths, it becomes the area inside the curve, with a clockwise sense giving a positive
value for the integral.
The magnitude of the area is the same for either path, since it is a rectangle divided in half by
a square. The area of the rectangle is
(15×103 Pa)(6 m3 ) = 90×103 J,
so the area of path 1 (counterclockwise) is -45 kJ; this means the work done on the gas is -(-45 kJ)
or 45 kJ. The work done on the gas for path 2 is the negative of this because the path is clockwise.
291
E23-26 During the isothermal expansion,
W1 = −nRT ln
V2
p1
= −p1 V1 ln .
V1
p2
During cooling at constant pressure,
W2 = −p2 ∆V = −p2 (V1 − V2 ) = −p2 V1 (1 − p1 /p2 ) = V1 (p1 − p2 ).
The work done is the sum, or
−(204×103 Pa)(0.142 m3 ) ln
(204×103 Pa)
+ (0.142 m3 )(103 Pa) = −5.74×103 J.
(101×103 Pa)
E23-27 During the isothermal expansion,
W = −nRT ln
V2
V2
= −p1 V1 ln ,
V1
V1
so
W = −(1.32)(1.01×105 Pa)(0.0224 m3 ) ln
(0.0153 m3 )
= 1.14×103 J.
(0.0224 m3 )
E23-28 (a) pV γ is a constant, so
p2 = p1 (V1 /V2 )γ = (1.00 atm)[(1 l)/(0.5 l)]1.32 = 2.50 atm;
T2 = T1 (p2 /p1 )(V2 /V1 ), so
T2 = (273 K)
(2.50 atm) (0.5 l)
= 341 K.
(1.00 atm) (1 l)
(b) V3 = V2 (p2 /p1 )(T3 /T2 ), so
V3 = (0.5 l)
(273 K)
= 0.40 l.
(341 K)
(c) During the adiabatic process,
W12 =
(1.01×105 Pa/atm)(1×10−3 m3 /l)
[(2.5 atm)(0.5 l) − (1.0 atm)(1 l)] = 78.9 J.
(1.32) − 1
During the cooling process,
W23 = −p∆V = −(1.01×105 Pa/atm)(2.50 atm)(1×10−3 m3 /l)[(0.4 l) − (0.5 l)] = 25.2 J.
The net work done is W123 = 78.9 J + 25.2 J = 104.1 J.
E23-29
(a) According to Eq. 23-20,
pf =
pi V i γ
(1.17 atm)(4.33 L)( 1.40)
= 8.39 atm.
=
γ
Vf
(1.06 L)( 1.40)
(b) The final temperature can be found from the ideal gas law,
Tf = Ti
pf V f
(8.39 atm)(1.06 L)
= (310 K)
= 544 K.
pi V i
(1.17 atm)(4.33 L)
(c) The work done (for an adiabatic process) is given by Eq. 23-22,
1
W =
(8.39 × 1.01×105 Pa)(1.06×10−3 m3 )
(1.40) − 1
−(1.17 × 1.01×105 Pa)(4.33×10−3 m3 ) ,
= 966 J.
292
E23-30 Air is mostly diatomic (N2 and O2 ), so use γ = 1.4.
(a) pV γ is a constant, so
p
p
V2 = V1 γ p1 /p2 = V1 1.4 (1.0 atm)/(2.3 atm) = 0.552V1 .
T2 = T1 (p2 /p1 )(V2 /V1 ), so
T2 = (291 K)
(2.3 atm) (0.552V1 )
= 369 K,
(1.0 atm)
V1
or 96◦ C.
(b) The work required for delivering 1 liter of compressed air is
W12 =
(1.01×105 Pa/atm)(1×10−3 m3 /l)
[(2.3 atm)(1.0 l) − (1.0 atm)(1.0 l/0.552)] = 123 J.
(1.40) − 1
The number of liters per second that can be delivered is then
∆V /∆t = (230 W)/(123 J/l) = 1.87 l.
E23-31 E int,rot = nRT = (1 mol)(8.31 J/mol · K)(298 K) = 2480 J.
E23-32 E int,rot = 23 nRT = (1.5)(1 mol)(8.31 J/mol · K)(523 K) = 6520 J.
E23-33
(a) Invert Eq. 32-20,
γ=
ln(p1 /p2 )
ln(122 kPa/1450 kPa)
=
= 1.20.
ln(V2 /V1 )
ln(1.36 m3 /10.7 m3 )
(b) The final temperature is found from the ideal gas law,
Tf = Ti
(1450×103 Pa)(1.36 m3 )
pf V f
= (250 K)
= 378 K,
pi V i
(122×103 Pa)(10.7 m3 )
which is the same as 105◦ C.
(c) Ideal gas law, again:
n = [pV ]/[RT ] = [(1450×103 Pa)(1.36 m3 )]/[(8.31 J/mol · K)(378 K)] = 628 mol.
(d) From Eq. 23-24,
E int =
3
3
nRT = (628 mol)(8.31 J/mol · K)(250 K) = 1.96×106 J
2
2
before the compression and
E int =
3
3
nRT = (628 mol)(8.31 J/mol · K)(378 K) = 2.96×106 J
2
2
after the compression.
(e) The ratio of the rms speeds will be proportional to the square root of the ratio of the internal
energies,
p
(1.96×106 J)/(2.96×106 J) = 0.813;
we can do this because the number of particles is the same before and after, hence the ratio of the
energies per particle is the same as the ratio of the total energies.
293
E23-34 We can assume neon is an ideal gas. Then ∆T = 2∆E int /3nR, or
∆T =
2(1.34×1012 eV)(1.6×10−19 J/eV)
= 1.43×10−7 J.
3(0.120 mol)(8.31 J/mol · K)
E23-35 At constant pressure, doubling the volume is the same as doubling the temperature. Then
7
Q = nCp ∆T = (1.35 mol) (8.31 J/mol · K)(568 K − 284 K) = 1.12×104 J.
2
E23-36 (a) n = m/M = (12 g)/(28 g/mol) = 0.429 mol.
(b) This is a constant volume process, so
5
Q = nCV ∆T = (0.429 mol) (8.31 J/mol · K)(125◦ C − 25◦ C) = 891J.
2
E23-37
(a) From Eq. 23-37,
Q = ncp ∆T = (4.34 mol)(29.1 J/mol · K)(62.4 K) = 7880 J.
(b) From Eq. 23-28,
E int =
5
5
nR∆T = (4.34 mol)(8.31 J/mol · K)(62.4 K) = 5630 J.
2
2
(c) From Eq. 23-23,
K trans =
3
5
nR∆T = (4.34 mol)(8.31 J/mol · K)(62.4 K) = 3380 J.
2
2
E23-38 cV = 32 (8.31 J/mol · K)/(4.00 g/mol) = 3120 J/kg · K.
E23-39 Each species will experience the same temperature change, so
Q = Q1 + Q2 + Q3 ,
= n1 C1 ∆T + n2 C2 ∆T + n3 C3 ∆T,
Dividing this by n = n1 + n2 + n3 and ∆T will return the specific heat capacity of the mixture, so
C=
n1 C1 + n2 C2 + n3 C3
.
n1 + n2 + n3
E23-40 WAB = 0, since it is a constant volume process, consequently, W = WAB +WABC = −15 J.
But around a closed path Q = −W , so Q = 15 J. Then
QCA = Q − QAB − QBC = (15 J) − (20 J) − (0 J) = −5 J.
Note that this heat is removed from the system!
294
E23-41
According to Eq. 23-25 (which is specific to ideal gases),
∆E int =
3
nR∆T,
2
and for an isothermal process ∆T = 0, so for an ideal gas ∆E int = 0. Consequently, Q + W = 0 for
an ideal gas which undergoes an isothermal process.
But we know W for an isotherm, Eq. 23-18 shows
W = −nRT ln
Vf
Vi
Then finally
Q = −W = nRT ln
Vf
Vi
E23-42 Q is greatest for constant pressure processes and least for adiabatic. W is greatest (in
magnitude, it is negative for increasing volume processes) for constant pressure processes and least
for adiabatic. ∆E int is greatest for constant pressure (for which it is positive), and least for adiabatic
(for which is is negative).
E23-43 (a) For a monatomic gas, γ = 1.667. Fast process are often adiabatic, so
T2 = T1 (V1 /V2 )γ−1 = (292 K)[(1)(1/10)]1.667−1 = 1360 K.
(b) For a diatomic gas, γ = 1.4. Fast process are often adiabatic, so
T2 = T1 (V1 /V2 )γ−1 = (292 K)[(1)(1/10)]1.4−1 = 733 K.
E23-44 This problem cannot be solved without making some assumptions about the type of process occurring on the two curved portions.
E23-45 If the pressure and volume are both doubled along a straight line then the process can
be described by
p1
p=
V
V1
The final point involves the doubling of both the pressure and the volume, so according to the ideal
gas law, pV = nRT , the final temperature T2 will be four times the initial temperature T1 .
Now for the exercises.
(a) The work done on the gas is
W =−
Z
2
p dV = −
1
Z
1
2
p1
p1
V dV = −
V1
V1
V22
V2
− 1
2
2
We want to express our answer in terms of T1 . First we take advantage of the fact that V2 = 2V1 ,
then
p1 4V12
V12
3
3
W =−
−
= − p1 V1 = − nRT1
V1
2
2
2
2
(b) The nice thing about ∆E int is that it is path independent, we care only of the initial and
final points. From Eq. 23-25,
∆E int =
3
3
9
nR∆T = nR (T2 − T1 ) = nRT1
2
2
2
295
(c) Finally we are in a position to find Q by applying the first law,
3
9
nRT1 + nRT1 = 6nRT1 .
2
2
(d) If we define specific heat as heat divided by temperature change, then
Q = ∆E int − W =
c=
Q
6RT1
=
= 2R.
n∆T
4T1 − T1
E23-46 The work done is the area enclosed by the path. If the pressure is measured in units of
10MPa, then the shape is a semi-circle, and the area is
W = (π/2)(1.5)2 (10MPa)(1×10−3 m3 ) = 3.53×104 J.
The heat is given by Q = −W = −3.53×104 J.
E23-47 (a) Internal energy changes according to ∆E int = Q + W , so
∆E int = (20.9 J) − (1.01×105 Pa)(113×10−6 m3 − 63×10−6 m3 ) = 15.9 J.
(b) T1 = p1 V1 /nR and T2 = p2 V2 /nR, but p is constant, so ∆T = p∆V /nR. Then
CP =
Q
QR
(20.9 J)(8.31 J/mol · K)
=
=
= 34.4 J/mol · K.
n∆T
p∆V
(1.01×105 Pa)(113×10−6 m3 − 63×10−6 m3 )
(c) CV = CP − R = (34.4 J/mol · K) − (8.31 J/mol · K) = 26.1 J/mol · K.
E23-48 Constant Volume
(a) Q = 3(3.15 mol)(8.31 J/mol · K)(52.0 K) = 4080 J.
(b) W = 0.
(c) ∆Ermint = 3(3.15 mol)(8.31 J/mol · K)(52.0 K) = 4080 J.
Constant Pressure
(a) Q = 4(3.15 mol)(8.31 J/mol · K)(52.0 K) = 5450 J.
(b) W = −p∆V = −nR∆T = −(3.15 mol)(8.31 J/mol · K)(52.0 K) = −1360 J.
(c) ∆Ermint = 3(3.15 mol)(8.31 J/mol · K)(52.0 K) = 4080 J.
Adiabatic
(a) Q = 0.
(b) W = (pf V f − pi V i )/(γ − 1) = nR∆T /(γ − 1) = 3(3.15 mol)(8.31 J/mol · K)(52.0 K) = 4080 J.
(c) ∆Ermint = 3(3.15 mol)(8.31 J/mol · K)(52.0 K) = 4080 J.
P23-1
(a) The temperature difference is
(5 C◦ /9 F◦ )(72◦ F − −20◦ F) = 51.1 C◦ .
The rate of heat loss is
H = (1.0 W/m · K)(1.4 m2 )(51.1 C◦ )/(3.0×10−3 m) = 2.4×104 W.
(b) Start by finding the R values.
Rg = (3.0×10−3 m)/(1.0 W/m · K) = 3.0×10−3 m2 · K/W,
Ra = (7.5×10−2 m)/(0.026W/m · K) = 2.88m2 · K/W.
Then use Eq. 23-5,
H=
(1.4 m2 )(51.1 C◦ )
= 25 W.
· K/W) + (2.88m2 · K/W)
2(3.0×10−3 m2
Get double pane windows!
296
P23-2 (a) H = (428 W/m · K)(4.76×10−4 m2 )(100 C◦ )/(1.17 m) = 17.4 W.
(b) ∆m/∆t = H/L = (17.4 W)/(333×103 J/kg) = 5.23×10−5 kg/s, which is the same as 188 g/h.
P23-3
Follow the example in Sample Problem 23-2. We start with Eq. 23-1:
H
H
Z
r2
dr
2
4πr
r
1
1
1
H
−
4pi r1
r2
r2 − r1
H
r1 r2
H
H
= kA
dT
,
dr
dT
= k(4πr2 ) ,
dr
Z T2
=
kdT,
T1
= k(T1 − T2 ),
= 4πk(T1 − T2 ),
=
4πk(T1 − T2 )r1 r2
.
r2 − r1
P23-4 (a) H = (54×10−3 W/m2 )4π(6.37×106 m)2 = 2.8×1013 W.
(b) Using the results of Problem 23-3,
∆T =
(2.8×1013 W)(6.37×106 m − 3.47×106 m)
= 7.0×104 C◦ .
4π(4.2 W/m · K)(6.37×106 m)(3.47×106 m)
Since T2 = 0◦ C, we expect T1 = 7.0×104 C◦ .
P23-5 Since H = −kA dT /dx, then H dx = −aT dT . H is a constant, so integrate both side
according to
Z
Z
H dx = − aT dT,
1
HL = −a (T22 − T12 ),
2
aA 2
H =
(T − T22 ).
2L 1
P23-6 Assume the water is all at 0◦ C. The heat flow through the ice is then H = kA∆T /x; the
rate of ice formation is ∆m/∆t = H/L. But ∆m = ρA∆x, so
∆x
∆t
(1.7 W/m · K)(10 C◦ )
= 1.11×10−6 m/s.
(920 kg/m3 )(333×103 J/kg)(0.05 m)
That’s the same as 0.40 cm/h.
P23-7
(a) Start with the heat equation:
Qt + Qi + Qw = 0,
297
=
H
k∆T
=
,
ρAL
ρLx
where Qt is the heat from the tea, Qi is the heat from the ice when it melts, and Qw is the heat
from the water (which used to be ice). Then
mt ct (T f − T t,i ) + mi Lf + mw cw (T f − T w,i ) = 0,
which, since we have assumed all of the ice melts and the masses are all equal, can be solved for Tf
as
Tf
ct T t,i + cw T w,i − Lf
,
ct + cw
(4190J/kg · K)(90◦ C) + (4190J/kg · K)(0◦ C) − (333×103 J/kg)
=
,
(4190J/kg · K) + (4190J/kg · K)
= 5.3◦ C.
=
(b) Once again, assume all of the ice melted. Then we can do the same steps, and we get
Tf
ct T t,i + cw T w,i − Lf
,
ct + cw
(4190J/kg · K)(70◦ C) + (4190J/kg · K)(0◦ C) − (333×103 J/kg)
=
,
(4190J/kg · K) + (4190J/kg · K)
= −4.7◦ C.
=
So we must have guessed wrong when we assumed that all of the ice melted. The heat equation
then simplifies to
mt ct (T f − T t,i ) + mi Lf = 0,
and then
mi
=
=
=
P23-8
mt ct (T t,i − T f )
,
Lf
(0.520 kg)(4190J/kg · K)(90◦ C − 0◦ )
,
(333×103 J/kg)
0.458 kg.
c = Q/m∆T = H/(∆m/∆t)∆T . But ∆m/∆t = ρ∆V /∆t. Combining,
c=
P23-9
H
(250 W)
=
= 2.4×103 J/kg · K.
−6
3
(∆V /∆t)ρ∆T
(8.2×10 m /s)(0.85×103 kg/m3 )(15 C◦ )
(a) n = NA /M , so
=
(b) E av = 32 kT , so
P23-10
(2256×103 J/kg)
= 6.75×10−20 J.
(6.02×1023 /mol)/(0.018 kg/mol)
2(6.75×10−20 J)
= 10.7.
=
E av
3(1.38×10−23 J/K)(305 K)
Qw + Qt = 0, so
C t ∆T t + mw cw (T f − T i ) = 0,
or
Ti =
(0.3 kg)(4190 J/kg · m)(44.4◦ C) + (46.1 J/K)(44.4◦ C − 15.0◦ C)
= 45.5◦ C.
(0.3 kg)(4190 J/kg · m)
298
P23-11 We can use Eq. 23-10, but we will need to approximate c first. If we assume that the
line is straight then we use c = mT + b. I approximate m from
m=
(14 J/mol · K) − (3 J/mol · K)
= 3.67×10−2 J/mol.
(500 K) − (200 K)
Then I find b from those same data points,
b = (3 J/mol · K) − (3.67×10−2 J/mol)(200 K) = −4.34 J/mol · K.
Then from Eq. 23-10,
Q = n
= n
Z
Tf
c dT,
Ti
Z Tf
(mT + b) dT,
Ti
= n
hm
T 2 + bT
2
m
iT f
,
Ti
(T f 2 − T i 2 ) + b(T f − T i ) ,
2
(3.67×10−2 J/mol)
= (0.45mol)
((500 K)2 − (200 K)2 )
2
+ (−4.34 J/mol · K)(500 K − 200 K)) ,
= 1.15×103 J.
= n
P23-12
δQ = nCδT , so
Z
Q = n C dT,
90 K
= n (0.318 J/mol · K2 )T 2 /2 − (0.00109 J/mol · K3 )T 3 /3 − (0.628 J/mol · K)T 50 K ,
= n(645.8 J/mol).
Finally,
Q = (645.8 J/mol)(316 g)/(107.87 g/mol) = 189 J.
P23-13
T V γ−1 is a constant, so
T2 = (292 K)(1/1.28)(1.40)−1 = 265 K
P23-14
R
W = − p dV , so
W
nRT
an2
− 2 dV,
V − nb
V
f
an2 = − nRT ln(V − nb) −
,
V i
V f − nb
1
1
2
= −nRT ln
− an
−
.
V i − nb
Vf
Vi
= −
Z 299
P23-15 When the tube is horizontal there are two regions filled with gas, one at p1,i , V 1,i ;
other at p2,i , V 2,i . Originally p1,i = p2,i = 1.01×105 Pa and V 1,i = V 2,i = (0.45 m)A, where A is
cross sectional area of the tube.
When the tube is held so that region 1 is on top then the mercury has three forces on it:
force of gravity, mg; the force from the gas above pushing down p2,f A; and the force from the
below pushing up p1,f A. The balanced force expression is
the
the
the
gas
p1,f A = p2,f A + mg.
If we write m = ρlm A where lm = 0.10 m, then
p1,f = p2,f + ρglm .
Finally, since the tube has uniform cross section, we can write V = Al everywhere.
(a) For an isothermal process pi li = pf lf , where we have used V = Al, and then
p1,i
l1,i
l2,i
− p2,i
= ρglm .
l1,f
l2,f
But we can factor out p1,i = p2,i and l1,i = l2,i , and we can apply l1,f + l2,f = 0.90 m. Then
1
1
ρglm
−
=
.
l1,f
0.90 m − l1,f
pi l i
Put in some numbers and rearrange,
0.90 m − 2l1,f = (0.294 m−1 )l1,f (0.90 m − l1,f ),
which can be written as an ordinary quadratic,
(0.294 m−1 )l1,f 2 − (2.265)l1,f + (0.90 m) = 0
The solutions are l1,f = 7.284 m and 0.421 m. Only one of these solutions is reasonable, so the
mercury shifted down 0.450 − 0.421 = 0.029 m.
(b) The math is a wee bit uglier here, but we can start with pi li γ = pf lf γ , and this means that
everywhere we had a l1,f in the previous derivation we need to replace it with l1,f γ . Then we have
1
l1,f
γ
−
1
ρglm
=
.
γ
(0.90 m − l1,f )
pi l i γ
This can be written as
(0.90 m − l1,f )γ − l1,f γ − = (0.404 m−γ )l1,f γ (0.90 m − l(0.1,f )γ ,
which looks nasty to me! I’ll use Maple to get the answer, and find l1,f = 0.429, so the mercury
shifted down 0.450 − 0.429 = 0.021 m.
Which is more likely? Turn the tube fast, and the adiabatic approximation works. Eventually
the system will return to room temperature, and then the isothermal approximation is valid.
P23-16
Internal energy for an ideal diatomic gas can be written as
E int =
5
5
nRT = pV,
2
2
simply by applying the ideal gas law. The room, however, has a fixed pressure and volume, so the
internal energy is independent of the temperature. As such, any energy supplied by the furnace
leaves the room, either as heat or as expanding gas doing work on the outside.
300
P23-17
The speed of sound in the iodine gas is
v = f λ = (1000 Hz)(2 × 0.0677 m) = 135 m/s.
Then
γ=
M v2
n(0.127 kg/mol)(135 m/s)2
=
= n(0.696).
RT
(8.31 J/mol · K)(400 K)
Since γ is greater than one, n ≥ 2. If n = 2 then γ = 1.39, which is consistent; if n = 3 then
γ = 2.08, which is not consistent.
Consequently, iodine gas is diatomic.
P23-18
W = −Q = mL = (333×103 J/kg)(0.122 kg) = 4.06×104 J.
P23-19 (a)
Process AB
Q = 32 (1.0 mol)(8.31 J/mol · K)(300 K) = 3740 J.
W = 0.
∆E int = Q + W = 3740 J.
Process BC
Q = 0.
W = (pf V f − pi V i )/(γ − 1) = nR∆T /(γ − 1) = (1.0 mol)(8.31 J/mol · K)(−145 K)/(1.67 − 1) =
−1800 J
∆E int = Q + W = −1800 J.
Process AB
Q = 52 (1.0 mol)(8.31 J/mol · K)(−155 K) = −3220 J.
W = −p∆V = −nR∆T = −(1.0 mol)(8.31 J/mol · K)(−155 K) = 1290 J.
∆E int = Q + W = 1930 J.
Cycle
Q = 520 J; W = −510 J (rounding error!); ∆E int = 10 J (rounding error!)
P23-20
P23-21
is
pf = (16.0 atm)(50/250)1.40 = 1.68 atm. The work done by the gas during the expansion
W
=
=
=
[pi V i − pf V f ]/(γ − 1),
(16.0 atm)(50×10−6 m3 ) − (1.68 atm)(250×10−6 m3 )
(1.01×105 Pa/atm),
(1.40) − 1
96.0 J.
This process happens 4000 times per minute, but the actual time to complete the process is half of
the cycle, or 1/8000 of a minute. Then P = (96 J)(8000)/(60 s) = 12.8×103 W.
301
E24-1 For isothermal processes the entropy expression is almost trivial, ∆S = Q/T, where if Q
is positive (heat flow into system) the entropy increases.
Then Q = T ∆S = (405 K)(46.2 J/K) = 1.87×104 J.
E24-2 Entropy is a state variable and is path independent, so
(a) ∆Sab,2 = ∆Sab,1 = +0.60 J/K,
(b) ∆Sba,2 = −∆Sab,2 = −0.60 J/K,
E24-3 (a) Heat only enters along the top path, so
Qin = T ∆S = (400 K)(0.6 J/K − 0.1 J/K) = 200 J.
(b) Heat leaves only bottom path, so
Qout = T ∆S = (250 K)(0.1 J/K − 0.6 J/K) = −125 J.
Since Q + W = 0 for a cyclic path,
W = −Q = −[(200 J) + (−125 J)] = −75 J.
E24-4 (a) The work done for isothermal expansion is given by Eq. 23-18,
W = −(4.00 mol)(8.31 J/mol · K)(410 K) ln
3.45V1
= −1.69×104 J.
V1
(b) For isothermal process, Q = −W , then
∆S = Q/T = (1.69×104 J)/(410 K) = 41.2 J/K.
(c) Entropy change is zero for reversible adiabatic processes.
E24-5
(a) We want to find the heat absorbed, so
Q = mc∆T = (1.22 kg)(387 J/mol · K) ((105◦ C) − (25.0◦ C)) = 3.77×104 J.
(b) We want to find the entropy change, so, according to Eq. 24-1,
∆S
=
Z
Tf
Ti
Tf
dQ
,
T
Z
mc dT
,
T
Ti
Tf
= mc ln .
Ti
=
The entropy change of the copper block is then
∆S = mc ln
Tf
(378 K)
= (1.22 kg)(387 J/mol · K) ln
= 112 J/K.
Ti
(298 K)
E24-6 ∆S = Q/T = mL/T , so
∆S = (0.001 kg)(−333×103 J/kg)/(263 K) = −1.27J/K.
302
E24-7 Use the first equation on page 551.
n=
∆S
(24 J/K)
=
= 3.00 mol.
R ln(V f /V i )
(8.31 J/mol · K) ln(3.4/1.3)
E24-8 ∆S = Q/T c − Q/T h .
(a) ∆S = (260 J)(1/100 K − 1/400 K) = 1.95 J/K.
(b) ∆S = (260 J)(1/200 K − 1/400 K) = 0.65 J/K.
(c) ∆S = (260 J)(1/300 K − 1/400 K) = 0.217 J/K.
(d) ∆S = (260 J)(1/360 K − 1/400 K) = 0.0722 J/K.
E24-9 (a) If the rod is in a steady state we wouldn’t expect the entropy of the rod to change.
Heat energy is flowing out of the hot reservoir into the rod, but this process happens at a fixed
temperature, so the entropy change in the hot reservoir is
∆S H =
(−1200 J)
QH
=
= −2.98 J/K.
TH
(403 K)
The heat energy flows into the cold reservoir, so
∆S C =
QH
(1200 J)
=
= 4.04 J/K.
TH
(297 K)
The total change in entropy of the system is the sum of these two terms
∆S = ∆S H + ∆S C = 1.06 J/K.
(b) Since the rod is in a steady state, nothing is changing, not even the entropy.
E24-10 (a) Qc + Ql = 0, so
mc cc (T − T c ) + ml cl (T − T l ) = 0,
which can be solved for T to give
T =
(0.05 kg)(387 J/kg · K)(400 K) + (0.10 kg)(129 J/kg · K)(200 K)
= 320 K.
(0.05 kg)(387 J/kg · K) + (0.10 kg)(129 J/kg · K)
(b) Zero.
(c) ∆S = mc ln T f /T i , so
∆S = (0.05 kg)(387 J/kg · K) ln
(320 K)
(320 K)
+ (0.10 kg)(129 J/kg · K) ln
= 1.75 J/K.
(400 K)
(200 K)
E24-11 The total mass of ice and water is 2.04 kg. If eventually the ice and water have the same
mass, then the final state will have 1.02 kg of each. This means that 1.78 kg − 1.02 kg = 0.76 kg of
water changed into ice.
(a) The change of water at 0◦ C to ice at 0◦ C is isothermal, so the entropy change is
∆S =
Q
−mL
(0.76 kg)(333×103 J/kg)
=
=
= −927 J/K.
T
T
(273 K)
(b) The entropy change is now +927 J/K.
303
E24-12 (a) Qa + Qw = 0, so
ma ca (T − T a ) + mw cw (T − T w ) = 0,
which can be solved for T to give
T =
(0.196 kg)(900 J/kg · K)(380 K) + (0.0523 kg)(4190 J/kg · K)(292 K)
= 331 K.
(0.196 kg)(900 J/kg · K) + (0.0523 kg)(4190 J/kg · K)
That’s the same as 58◦ C.
(b) ∆S = mc ln T f /T i , so
∆S a = (0.196 kg)(900 J/kg · K) ln
(331 K)
= −24.4 J/K.
(380 K)
(c) For the water,
(0.0523 kg)(4190 J/kg · K) ln
(331 K)
= 27.5 J/K.
(292 K)
(d) ∆S = (27.5 J/K) + (−24.4 J/K) = 3.1 J/K.
E24-13 (a) e = 1 − (36.2 J/52.4 J) = 0.309.
(b) W = Qh − Qc = (52.4 J) − (36.2 J) = 16.2 J.
E24-14 (a) Qh = (8.18 kJ)/(0.25) = 32.7 kJ, Qc = Qh − W = (32.7 kJ) − (8.18 kJ) = 24.5 kJ.
(b) Qh = (8.18 kJ)/(0.31) = 26.4 kJ, Qc = Qh − W = (26.4 kJ) − (8.18 kJ) = 18.2 kJ.
E24-15
One hour’s worth of coal, when burned, will provide energy equal to
(382×103 kg)(28.0×106 J/kg) = 1.07×1013 J.
In this hour, however, the plant only generates
(755×106 W)(3600 s) = 2.72×1012 J.
The efficiency is then
e = (2.72×1012 J)/(1.07×1013 J) = 25.4%.
E24-16 We use the convention that all quantities are positive, regardless of direction. W A = 5W B ;
Qi,A = 3Qi,B ; and Qo,A = 2Qo,B . But W A = Qi,A − Qo,A , so
5W B = 3Qi,B − 2Qo,B ,
or, applying W B = Qi,B − Qo,B ,
5W B
3W B
W B /Qi,B
Then
eA =
= 3Qi,B − 2(Qi,B − W B ),
= Qi,B ,
= 1/3 = eB .
WA
5W B
51
5
=
=
= .
Qi,A
3Qi,B
33
9
304
E24-17 (a) During an isothermal process W = −Q = −2090 J. The negative indicates that the
gas did work on the environment.
(b) The efficiency is e = 1 − (297 K)/(412 K) = 0.279. Then
Qo = Qi (1 − e) = (2090 J)[1 − (0.279)] = 1510 J.
Since this is rejected heat it should actually be negative.
(c) During an isothermal process W = −Q = 1510 J. Positive indicates that the gas did work on
the environment.
E24-18 1 − e = T c /T h , or T c = T h (1 − e). The difference is
∆T = T h − T c = T h e,
so T h = (75 C◦ )/(0.22) = 341 K, and
T c = (341 K)[(1 − (0.22)] = 266 K.
E24-19 The BC and DA processes are both adiabatic; so if we could find an expression for work
done during an adiabatic process we might be almost done. But what is an adiabatic process? It is
a process for which Q = 0, so according to the first law
∆E int = W.
But for an ideal gas
∆E int = nCV ∆T,
as was pointed out in Table 23-5. So we have
|W | = nCV |∆T |
and since the adiabatic paths BC and DA operate between the same two isotherms, we can conclude
that the magnitude of the work is the same for both paths.
E24-20 (a) To save typing, assume that all quantities are positive. Then
e1 = 1 − T2 /T1 ,
W1 = e1 Q1 , and Q2 = Q1 − W1 . Not only that, but
e2 = 1 − T3 /T2 ,
and W2 = e2 Q2 . Combining,
e=
W1 + W2
e1 Q1 + e2 (Q1 − W1 )
=
= e1 + e2 (1 − e1 ) = e1 + e2 − e1 e2 ,
Q1
Q1
or
e=1−
T2
T3
T2
T3
T3
T3
+1−
−1+
+
−
=1−− .
T1
T2
T1
T2
T1
T1
(b) e = 1 − (311 K)/(742 K) = 0.581.
E24-21 (a) p2 = (16.0 atm)(1/5.6)(1.33) = 1.62 atm.
(b) T2 = T1 (1/5.6)(1.33)−1 = (0.567)T1 , so
e = 1 − (0.567) = 0.433.
305
E24-22 (a) The area of the cycle is ∆V ∆p = p0 V0 , so the work done by the gas is
W = (1.01×105 Pa)(0.0225 m3 ) = 2270 J.
(b) Let the temperature at a be Ta . Then
Tb = Ta (Vb /Va )(pb /pa ) = 2Ta .
Let the temperature at c be Tc . Then
Tc = Ta (Vc /Va )(pc /pa ) = 4Ta .
Consequently, ∆Tab = Ta and ∆Tbc = 2Ta . Putting this information into the constant volume and
constant pressure heat expressions,
Qab =
3
3
3
nR∆Tab = nRTa = pa Va ,
2
2
2
Qbc =
5
5
nR∆Tbc = nR2Ta = 5pa Va ,
2
2
and
so that Qac =
13
2 p0 V 0 ,
or
Qac =
13
(1.01×105 Pa)(0.0225 m3 ) = 1.48×104 J.
2
(c) e = (2270 J)/(14800 J) = 0.153.
(d) ec = 1 − (Ta /4Ta ) = 0.75.
E24-23
According to Eq. 24-15,
K=
TL
(261 K)
=
= 6.87
T H − TL
(299 K) − (261 K)
Now we solve the question out of order.
(b) The work required to run the freezer is
|W | = |QL |/K = (185 kJ)/(5.70) = 32.5 kJ.
(a) The freezer will discharge heat into the room equal to
|QL | + |W | = (185 kJ) + (32.5 kJ) = 218 kJ.
E24-24 (a) K = |QL |/|W | = (568 J)/(153 J) = 3.71.
(b) |QH | = |QL | + |W | = (568 J) + (153 J) = 721 J.
E24-25 K = TL /(TH − TL ); |W | = |QL |/K = |QL |(TH /TL − 1).
(a) |W | = (10.0 J)(300 K/280K − 1) = 0.714 J.
(b) |W | = (10.0 J)(300 K/200K − 1) = 5.00 J.
(c) |W | = (10.0 J)(300 K/100K − 1) = 20.0 J.
(d) |W | = (10.0 J)(300 K/50K − 1) = 50.0 J.
E24-26 K = TL /(TH − TL ); |W | = |QL |/K = |QL |(TH /TL − 1). Then
|QH | = |QL | + |W | = |QL |(TH /TL ) = (0.150 J)(296 K/4.0 K) = 11 J.
306
E24-27 We will start with the assumption that the air conditioner is a Carnot refrigerator.
K = TL /(TH − TL ); |W | = |QL |/K = |QL |(TH /TL − 1). For fun, I’ll convert temperature to the
absolute Rankine scale! Then
|QL | = (1.0 J)/(555◦ R/530◦ R − 1) = 21 J.
E24-28 The best coefficient of performance is
K c = (276 K)/(308 K − 276 K) = 8.62.
The inventor claims they have a machine with
K = (20 kW − 1.9kW)/(1.9 kW) = 9.53.
Can’t be done.
E24-29 (a) e = 1 − (258 K/322 K) = 0.199. |W | = (568 J)(0.199) = 113 J.
(b) K = (258 K)/(322 K − 258 K) = 4.03. |W | = (1230 J)/(4.03) = 305 J.
E24-30 The temperatures are distractors!
|W | = |QH | − |QL | = |QH | − K|W |,
so
|W | = |QH |/(1 + K) = (7.6 MJ)/(1 + 3.8) = 1.58 MJ.
Then P = (1.58 MJ)/(3600 s) = 440 W.
E24-31 K = (260 K)/(298 K − 260 K) = 6.8.
E24-32 K = (0.85)(270K)/(299K−270K) = 7.91. In 15 minutes the motor can do (210 W)(900 s) =
1.89×105 J of work. Then
|QL | = K|W | = (7.91)(1.89×105 J) = 1.50×106 J.
E24-33
The Carnot engine has an efficiency
=1−
T2
|W |
=
.
T1
|Q1 |
The Carnot refrigerator has a coefficient of performance
K=
T4
|Q4 |
=
.
T3 − T4
|W |
Lastly, |Q4 | = |Q3 | − |W |. We just need to combine these three expressions into one. Starting with
the first, and solving for |W |,
T1 − T2
|W | = |Q1 |
.
T1
Then we combine the last two expressions, and
T4
|Q3 | − |W |
|Q3 |
=
=
− 1.
T3 − T4
|W |
|W |
307
Finally, combine them all,
T4
|Q3 | T1
=
− 1.
T3 − T4
|Q1 | T1 − T2
Now, we rearrange,
|Q3 |
|Q1 |
T4
T1 − T 2
=
+1
,
T 3 − T4
T1
T3
T 1 − T2
=
,
T 3 − T4
T1
= (1 − T2 /T1 )/(1 − T4 /T3 ).
E24-34 (a) Integrate:
ln N ! ≈
Z
N
ln x dx = N ln N − N + 1 ≈ N ln N − N.
1
(b) 91, 752, and about 615,000. You will need to use the Stirling approximation extended to a
double inequality to do the last two:
√
√
2πnn+1/2 e−n+1/(12n+1) < n! < 2πnn+1/2 e−n+1/(12n) .
E24-35 (a) For this problem we don’t care how the particles are arranged inside a section, we
only care how they are divided up between the two sides.
Consequently, there is only one way to arrange the particles: you put them all on one side, and
you have no other choices. So the multiplicity in this case is one, or w1 = 1.
(b) Once the particles are allowed to mix we have more work in computing the multiplicity.
Using Eq. 24-19, we have
N!
N!
w2 =
=
2
(N/2)!(N/2)!
((N/2)!)
(c) The entropy of a state of multiplicity w is given by Eq. 24-20,
S = k ln w
For part (a), with a multiplicity of 1, S1 = 0. Now for part (b),
!
N!
S2 = k ln
= k ln N ! − 2k ln(N/2)!
2
((N/2)!)
and we need to expand each of those terms with Stirling’s approximation.
Combining,
S2
= = k (N ln N − N ) − 2k ((N/2) ln(N/2) − (N/2)) ,
= kN lnN − kN − kN ln N + kN ln 2 + kN,
= kN ln 2
Finally, ∆S = S2 − S1 = kN ln 2.
(d) The answer should be the same; it is a free expansion problem in both cases!
308
P24-1
We want to evaluate
∆S
Tf
=
Z
=
Z
=
Z
=
nA
T f 3 − T i3 .
3
Ti
Tf
Ti
Tf
nCV dT
,
T
nAT 3 dT
,
T
nAT 2 dT ,
Ti
Into this last expression, which is true for many substances at sufficiently low temperatures, we
substitute the given numbers.
∆S =
(4.8 mol)(3.15×10−5 J/mol · K4 )
(10 K)3 − (5.0 K)3 = 4.41×10−2 J/K.
3
P24-2
P24-3 (a) Work is only done along path ab, where Wab = −p∆V = −3p0 ∆V0 . So Wabc = −3p0 V0 .
(b) ∆E R bc = 23 nR∆Tbc , with a little algebra,
∆E intbc =
3
3
3
(nRTc − nRTb ) = (pc Vc − pb Vb ) = (8 − 4)p0 V0 = 6p0 V0 .
2
2
2
∆Sbc = 32 nR ln(Tc /Tb ), with a little algebra,
∆Sbc =
3
3
nR ln(pc /pb ) = nR ln 2.
2
2
(c) Both are zero for a cyclic process.
P24-4
(a) For an isothermal process,
p2 = p1 (V1 /V2 ) = p1 /3.
For an adiabatic process,
p3 = p1 (V1 /V2 )γ = p1 (1/3)1.4 = 0.215p1 .
For a constant volume process,
T3 = T2 (p3 /p2 ) = T1 (0.215/0.333) = 0.646T1 .
(b) The easiest ones first: ∆E i nt12 = 0, W23 = 0, Q31 = 0, ∆S31 = 0. The next easier ones:
∆E i nt23 =
5
5
nR∆T23 = nR(0.646T1 − T1 ) = −0.885p1 V1 ,
2
2
Q23 = ∆E int23 − W23 = −0.885p1 V1 ,
∆E int31 = −∆E int23 − ∆E i nt12 = 0.885p1 V1 ,
W31 = ∆E int31 − Q31 = 0.885p1 V1 .
309
Finally, some harder ones:
W12 = −nRT1 ln(V2 /V1 ) = −p1 V1 ln(3) = −1.10p1 V1 ,
Q12 = ∆E int12 − W12 = 1.10p1 V1 .
And now, the hardest:
∆S12 = Q12 /T1 = 1.10nR,
∆S23 = −∆S12 − ∆S31 = −1.10nR.
P24-5 Note that TA = TB = TC /4 = TD .
Process I: ABC
(a) QAB = −WAB = nRT0 ln(VB /VA ) = p0 V0 ln 2. QBC = 32 nR(TC − TB ) = 32 (pC VC − pB VB ) =
3
2 (4p0 V0 − p0 V0 ) = 4.5p0 V0 .
(b) WAB = −nRT0 ln(VB /VA ) = −p0 V0 ln 2. WBC = 0.
(c) E int = 32 nR(TC − TA ) = 32 (pC VC − pA VA ) = 23 (4p0 V0 − p0 V0 ) = 4.5p0 V0 .
(d) ∆SAB = nR ln(VB /VA ) = nR ln 2; ∆SBC = 32 nR ln(TC /TB ) = 32 nR ln 4 = 3nR ln 2. Then
∆SAC = 4nR.
Process II: ADC
(a) QAD = −WAD = nRT0 ln(VD /VA ) = −p0 V0 ln 2. QDC = 52 nR(TC − TD ) = 52 (pC VC −
pD VD ) = 52 (4p0 V0 − p0 V0 ) = 10p0 V0 .
(b) WAB = −nRT0 ln(VD /VA ) = p0 V0 ln 2. WDC = −p∆V = −p0 (2V0 − V0 /2) = − 23 p0 V0 .
(c) E int = 23 nR(TC − TA ) = 32 (pC VC − pA VA ) = 32 (4p0 V0 − p0 V0 ) = 4.5p0 V0 .
(d) ∆SAD = nR ln(VD /VA ) = −nR ln 2; ∆SDC = 52 nR ln(TC /TD ) = 25 nR ln 4 = 5nR ln 2. Then
∆SAC = 4nR.
P24-6
The heat required to melt the ice is
Q = m(cw ∆T23 + L + ci ∆T12 ),
= (0.0126 kg)[(4190 J/kg · K)(15 C◦ ) + (333×103 J/kg) + (2220 J/kg · K)(10 C◦ )],
= 5270 J.
The change in entropy of the ice is
∆S i
= m[cw ln(T3 /T2 ) + L/T2 + ci ln(T2 /T1 )],
= (0.0126kg)[(4190J/kg·K) ln(288/273) + (333×103 J/kg)/(273K),
+(2220J/kg·K) ln(273/263)],
= 19.24 J/K
The change in entropy of the lake is ∆S l = (−5270 J)/(288 K) = 18.29 J/K. The change in entropy
of the system is 0.95 J/kg.
P24-7 (a) This is a problem where the total internal energy of the two objects doesn’t change,
but since no work is done during the process, we can start with the simpler expression Q1 + Q2 = 0.
The heat transfers by the two objects are
Q1
Q2
= m1 c1 (T 1 − T 1,i ),
= m2 c2 (T 2 − T 2,i ).
Note that we don’t call the final temperature T f here, because we are not assuming that the two
objects are at equilibrium.
310
We combine these three equations,
m2 c2 (T 2 − T 2,i ) = −m1 c1 (T 1 − T 1,i ),
m2 c2 T2 = m2 c2 T 2,i + m1 c1 (T 1,i − T 1 ),
m1 c1
T2 = T 2,i +
(T 1,i − T 1 )
m2 c2
As object 1 “cools down”, object 2 “heats up”, as expected.
(b) The entropy change of one object is given by
∆S =
Z
Tf
Ti
Tf
mc dT
= mc ln ,
T
Ti
and the total entropy change for the system will be the sum of the changes for each object, so
∆S = m1 c1 ln
T1
T2
+ m2 c2 ln
.
T i,1
T i,2
Into the this last equation we need to substitute the expression for T2 in as a function of T1 . There’s
no new physics in doing this, just a mess of algebra.
(c) We want to evaluate d(∆S)/dT1 . To save on algebra we will work with the last expression,
remembering that T2 is a function, not a variable. Then
m1 c1
m2 c2 dT2
d(∆S)
=
+
.
dT1
T1
T2 dT1
We’ve saved on algebra, but now we need to evaluate dT2 /dT1 . Starting with the results from part
(a),
dT2
d
m1 c1
=
T 2,i +
(T 1,i − T 1 ) ,
dT1
dT1
m2 c2
m1 c1
= −
.
m2 c2
Now we collect the two results and write
d(∆S)
dT1
m1 c1
m2 c2
m1 c1
+
−
,
T1
T2
m2 c2
1
1
= m1 c1
−
.
T1
T2
=
We could consider writing T2 out in all of its glory, but what would it gain us? Nothing. There is
actually considerably more physics in the expression as written, because...
(d) ...we get a maximum for ∆S when d(∆S)/dT1 = 0, and this can only occur when T1 = T2
according to the expression.
P24-8 Tb = (10.4 × 1.01×105 Pa)(1.22 m3 )/(2 mol)(8.31 J/mol · K) = 7.71×104 K. Maybe not so
realistic? Ta can be found after finding
pc = pb (Vb /Vc )γ = (10.4 × 1.01×105 Pa)(1.22/9.13)1.67 = 3.64×104 Pa,
Then
Ta = Tb (pa /pb ) = (7.71×104 K)(3.64×104 /1.05×106 ) = 2.67×103 K.
311
Similarly,
Tc = Ta (Vc /Va ) = (2.67×103 K)(9.13/1.22) = 2.00×104 K.
(a) Heat is added during process ab only;
Qab =
3
3
nR(Tb − Ta ) = (2 mol)(8.31 J/mol · K)(7.71×104 K − 2.67×103 K) = 1.85×106 J.
2
2
(b) Heat is removed during process ca only;
Qca =
5
5
nR(Ta − Tc ) = (2 mol)(8.31 J/mol · K)(2.67×103 K − 2.00×104 K) = −0.721×106 J.
2
2
(c) W = |Qab | − |Qca | = (1.85×106 J) − (0.721×106 J) = 1.13×106 J.
(d) e = W/Qab = (1.13×106 )/(1.85×106 ) = 0.611.
P24-9 The pV diagram for this process is Figure 23-21, except the cycle goes clockwise.
(a) Heat is input during the constant volume heating and the isothermal expansion. During
heating,
3
3
Q1 = nR∆T = (1 mol)(8.31 J/mol · K)(600 K − 300 K) = 3740 J;
2
2
During isothermal expansion,
Q2 = −W2 = nRT ln(V f /V i ) = (1 mol)(8.31 J/mol · K)(600 K) ln(2) = 3460 J;
so Qin = 7200 J.
(b) Work is only done during the second and third processes; we’ve already solved the second,
W2 = −3460 J;
W3 = −p∆V = pa Vc − pa Va = nR(Tc − Ta ) = (1 mol)(8.31 J/mol · K)(600 K − 300 K) = 2490 J;
So W = −970 J.
(c) e = |W |/|Qin | = (970 J)/(7200 J) = 0.13.
P24-10
(a) Tb = Ta (pb /pa ) = 3Ta ;
Tc = Tb (Vb /Vc )γ−1 = 3Ta (1/4)0.4 = 1.72Ta ;
pc = pb (Vb /Vc )γ = 3pa (1/4)1.4 = 0.430pa ;
Td = Ta (Va /Vd )γ−1 = Ta (1/4)0.4 = 0.574Ta ;
pd = pa (Va /Vd )γ = pa (1/4)1.4 = 0.144pa .
(b) Heat in occurs during process ab, so Qi =
process cd, so Qo = 52 nR∆Tcd = 2.87nRTa . Then
5
2 nR∆Tab
= 5nRTa ; Heat out occurs during
e = 1 − (2.87nRTa /5nRTa ) = 0.426.
P24-11 (c) (VB /VA ) = (pA /pB ) = (0/0.5) = 2. The work done on the gas during the isothermal
compression is
W = −nRT ln(VB /VA ) = −(1 mol)(8.31 J/mol · K)(300 K) ln(2) = −1730 J.
Since ∆E int = 0 along an isotherm Qh = 1730 J.
The cycle has an efficiency of e = 1 − (100/300) = 2/3. Then for the cycle,
W = eQh = (2/3)(1730 J) = 1150 J.
312
Instructor Solutions Manual
for
Physics
by
Halliday, Resnick, and Krane
Paul Stanley
Beloit College
Volume 2
A Note To The Instructor...
The solutions here are somewhat brief, as they are designed for the instructor, not for the student.
Check with the publishers before electronically posting any part of these solutions; website, ftp, or
server access must be restricted to your students.
I have been somewhat casual about subscripts whenever it is obvious that a problem is one
dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution.
Although this does not change the validity of the answer, it will sometimes obfuscate the approach
if viewed by a novice.
There are some traditional formula, such as
2
vx2 = v0x
+ 2ax x,
which are not used in the text. The worked solutions use only material from the text, so there may
be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an
easier approach existed. But if it was not in the text, I did not use it here.
I also tried to avoid reinventing the wheel. There are some exercises and problems in the text
which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer
you to the previous solution.
I adopt a different approach for rounding of significant figures than previous authors; in particular, I usually round intermediate answers. As such, some of my answers will differ from those in
the back of the book.
Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manual
with considerably more detail and, when appropriate, include discussion on any physical implications
of the answer. These student solutions carefully discuss the steps required for solving problems, point
out the relevant equation numbers, or even specify where in the text additional information can be
found. When two almost equivalent methods of solution exist, often both are presented. You are
encouraged to refer students to the Student’s Solution Manual for these exercises and problems.
However, the material from the Student’s Solution Manual must not be copied.
Paul Stanley
Beloit College
[email protected]
1
E25-1
The charge transferred is
Q = (2.5 × 104 C/s)(20 × 10−6 s) = 5.0 × 10−1 C.
E25-2 Use Eq. 25-4:
s
r=
(8.99×109 N·m2 /C2 )(26.3×10−6 C)(47.1×10−6 C)
= 1.40 m
(5.66 N)
E25-3 Use Eq. 25-4:
F =
(8.99×109 N·m2 /C2 )(3.12×10−6 C)(1.48×10−6 C)
= 2.74 N.
(0.123 m)2
E25-4 (a) The forces are equal, so m1 a1 = m2 a2 , or
m2 = (6.31×10−7 kg)(7.22 m/s2 )/(9.16 m/s2 ) = 4.97×10−7 kg.
(b) Use Eq. 25-4:
q=
E25-5
s
(6.31×10−7 kg)(7.22 m/s2 )(3.20×10−3 m)2
= 7.20×10−11 C
(8.99×109 N·m2 /C2 )
(a) Use Eq. 25-4,
F =
1 q1 q2
1
(21.3 µC)(21.3 µC)
=
= 1.77 N
2
−12
2
2
4π0 r12
4π(8.85×10
C /N · m )
(1.52 m)2
(b) In part (a) we found F12 ; to solve part (b) we need to first find F13 . Since q3 = q2 and
r13 = r12 , we can immediately conclude that F13 = F12 .
We must assess the direction of the force of q3 on q1 ; it will be directed along the line which
connects the two charges, and will be directed away from q3 . The diagram below shows the directions.
F
θ
F 12
F
23
23
F net
F 12
From this diagram we want to find the magnitude of the net force on q1 . The cosine law is
appropriate here:
F net 2
F net
2
2
= F12
+ F13
− 2F12 F13 cos θ,
2
= (1.77 N) + (1.77 N)2 − 2(1.77 N)(1.77 N) cos(120◦ ),
= 9.40 N2 ,
= 3.07 N.
2
E25-6 Originally F0 = CQ20 = 0.088 N, where C is a constant. When sphere 3 touches 1 the
charge on both becomes Q0 /2. When sphere 3 the touches sphere 2 the charge on each becomes
(Q0 + Q0 /2)/2 = 3Q0 /4. The force between sphere 1 and 2 is then
F = C(Q0 /2)(3Q0 /4) = (3/8)CQ20 = (3/8)F0 = 0.033 N.
~ 31 and F
~ 32 . These forces are given by the vector form of Coulomb’s
E25-7 The forces on q3 are F
Law, Eq. 25-5,
~ 31
F
=
~ 32
F
=
1 q3 q1
1 q3 q1
2 r̂31 = 4π (2d)2 r̂31 ,
4π0 r31
0
1 q3 q2
1 q3 q2
2 r̂32 = 4π (d)2 r̂32 .
4π0 r32
0
These two forces are the only forces which act on q3 , so in order to have q3 in equilibrium the forces
must be equal in magnitude, but opposite in direction. In short,
~ 31
F
1 q3 q1
r̂31
4π0 (2d)2
q1
r̂31
4
~ 32 ,
= −F
1 q3 q2
= −
r̂32 ,
4π0 (d)2
q2
= − r̂32 .
1
Note that r̂31 and r̂32 both point in the same direction and are both of unit length. We then get
q1 = −4q2 .
E25-8 The horizontal and vertical contributions from the upper left charge and lower right charge
are straightforward to find.
√ The contributions from the upper left charge require√slightly more work.
The diagonal distance is 2a; the components will be weighted by cos 45◦ = 2/2. The diagonal
charge will contribute
√
√
1 (q)(2q) 2
2 q2
√
Fx =
î =
î,
4π0 ( 2a)2 2
8π0 a2
√
√
1 (q)(2q) 2
2 q2
√
Fy =
ĵ =
ĵ.
4π0 ( 2a)2 2
8π0 a2
(a) The horizontal component of the net force is then
√
1 (2q)(2q)
2 q2
Fx =
î
+
î,
4π0
a2
8π0 a2
√
4 + 2/2 q 2
=
î,
4π0 a2
=
(4.707)(8.99×109 N · m2 /C2 )(1.13×10−6 C)2 /(0.152 m)2 î = 2.34 N î.
(b) The vertical component of the net force is then
√
1 (q)(2q)
2 q2
ĵ +
ĵ,
Fy = −
2
4π0 a
8π0 a2
√
−2 + 2/2 q 2
=
ĵ,
8π0
a2
=
(−1.293)(8.99×109 N · m2 /C2 )(1.13×10−6 C)2 /(0.152 m)2 ĵ = −0.642 N ĵ.
3
E25-9 The magnitude of the force on the negative charge from each positive charge is
F = (8.99×109 N · m2 /C2 )(4.18×10−6 C)(6.36×10−6 C)/(0.13 m)2 = 14.1 N.
The force from each positive charge is directed along the side of the triangle; but from symmetry
only the component along the bisector is of interest. This means that we need to weight the above
answer by a factor of 2 cos(30◦ ) = 1.73. The net force is then 24.5 N.
E25-10 Let the charge on one sphere be q, then the charge on the other sphere is Q = (52.6 ×
10−6 C) − q. Then
1 qQ
= F,
4π0 r2
(8.99×109 N·m2 /C2 )q(52.6×10−6 C − q) = (1.19 N)(1.94 m)2 .
Solve this quadratic expression for q and get answers q1 = 4.02×10−5 C and q2 = 1.24×10−6 N.
E25-11 This problem is similar to Ex. 25-7. There are some additional issues, however. It is
easy enough to write expressions for the forces on the third charge
~ 31
F
=
~ 32
F
=
1 q3 q1
2 r̂31 ,
4π0 r31
1 q3 q2
2 r̂32 .
4π0 r32
Then
~ 31
F
1 q3 q1
2 r̂31
4π0 r31
q1
2 r̂31
r31
~ 32 ,
= −F
1 q3 q2
= −
2 r̂32 ,
4π0 r32
q2
= − 2 r̂32 .
r32
The only way to satisfy the vector nature of the above expression is to have r̂31 = ±r̂32 ; this means
that q3 must be collinear with q1 and q2 . q3 could be between q1 and q2 , or it could be on either
side. Let’s resolve this issue now by putting the values for q1 and q2 into the expression:
(1.07 µC)
r̂31
2
r31
2
r32
r̂31
= −
(−3.28 µC)
r̂32 ,
2
r32
2
= (3.07)r31
r̂32 .
Since squared quantities are positive, we can only get this to work if r̂31 = r̂32 , so q3 is not between
q1 and q2 . We are then left with
2
2
r32
= (3.07)r31
,
so that q3 is closer to q1 than it is to q2 . Then r32 = r31 + r12 = r31 + 0.618 m, and if we take the
square root of both sides of the above expression,
p
r31 + (0.618 m) =
(3.07)r31 ,
p
(0.618 m) =
(3.07)r31 − r31 ,
(0.618 m) = 0.752r31 ,
0.822 m = r31
4
E25-12 The magnitude of the magnetic force between any two charges is kq 2 /a2 , where a =
0.153 m. The force between each charge is directed along the side of the triangle; but from symmetry
only the component along the bisector is of interest. This means that we need to weight the above
answer by a factor of 2 cos(30◦ ) = 1.73. The net force on any charge is then 1.73kq 2 /a2 .
The length of the angle bisector, d, is given by d = a cos(30◦ ).
The distance from any charge to the center of the equilateral triangle is x, given by x2 =
(a/2)2 + (d − x)2 . Then
x = a2 /8d + d/2 = 0.644a.
The angle between the strings and the plane of the charges is θ, given by
sin θ = x/(1.17 m) = (0.644)(0.153 m)/(1.17 m) = 0.0842,
or θ = 4.83◦ .
The force of gravity on each ball is directed vertically and the electric force is directed horizontally.
The two must then be related by
tan θ = F E /F G ,
so
1.73(8.99×109 N · m2 /C2 )q 2 /(0.153 m)2 = (0.0133 kg)(9.81 m/s2 ) tan(4.83◦ ),
or q = 1.29×10−7 C.
E25-13 On any corner charge there are seven forces; one from each of the other seven charges.
The net force will be the sum. Since all eight charges are the same all of the forces will be repulsive.
We need to sketch a diagram to show how the charges are labeled.
2
1
4
6
7
3
8
5
The magnitude of the force of charge 2 on charge 1 is
F12 =
1 q2
2 ,
4π0 r12
where r12 = a, the length of a side. Since both charges are the same we wrote q 2 . By symmetry we
expect that the magnitudes of F12 , F13 , and F14 will all be the same and they will all be at right
angles to each other directed along the edges of the cube. Written in terms of vectors the forces
5
would be
~ 12
F
=
~ 13
F
=
~ 14
F
=
1 q2
î,
4π0 a2
1 q2
ĵ,
4π0 a2
1 q2
k̂.
4π0 a2
The force from charge 5 is
1 q2
2 ,
4π0 r15
and is directed along the side diagonal away from charge 5. The distance r15 is also the side diagonal
distance, and can be found from
2
r15
= a2 + a2 = 2a2 ,
F15 =
then
1 q2
.
4π0 2a2
By symmetry we expect that the magnitudes of F15 , F16 , and F17 will all be the same and they will
all be directed along the diagonals of the faces of the cube. In terms of components we would have
√ 1 q2 √
~ 15 =
ĵ/ 2 + k̂/ 2 ,
F
2
4π0 2a
√ 1 q2 √
~ 16 =
F
î/
2
+
k̂/
2 ,
4π0 2a2
√ 1 q2 √
~ 17 =
F
î/
2
+
ĵ/
2 .
4π0 2a2
F15 =
The last force is the force from charge 8 on charge 1, and is given by
F18 =
1 q2
2 ,
4π0 r18
and is directed along the cube diagonal away from charge 8. The distance r18 is also the cube
diagonal distance, and can be found from
2
r18
= a2 + a2 + a2 = 3a2 ,
then in term of components
~ 18 =
F
√
√ 1 q2 √
î/
3
+
ĵ/
3
+
k̂/
3 .
4π0 3a2
We can add the components together. By symmetry we expect the same answer for each components, so we’ll just do one. How about î. This component has contributions from charge 2, 6, 7,
and 8:
1
1 q2 1
2
,
+ √ + √
4π0 a2 1 2 2 3 3
or
1 q2
(1.90)
4π0 a2
√
The three components add according to Pythagoras to pick up a final factor of 3, so
F net = (0.262)
6
q2
.
0 a2
E25-14 (a) Yes. Changing the sign of y will change the sign of Fy ; since this is equivalent to
putting the charge q0 on the “other” side, we would expect the force to also push in the “other”
direction.
(b) The equation should look Eq. 25-15, except all y’s should be replaced by x’s. Then
Fx =
1
q q
p 0
.
4π0 x x2 + L2 /4
(c) Setting the particle a distance d away should give a force with the same magnitude as
F =
1
q q
p 0
.
4π0 d d2 + L2 /4
This force is directed along the p
45◦ line, so Fx = F cos 45◦ and Fy = F sin 45◦ .
(d) Let the distance be d = x2 + y 2 , and then use the fact that Fx /F = cos θ = x/d. Then
Fx = F
1
x q0 q
x
=
.
2
2
d
4π0 (x + y + L2 /4)3/2
Fy = F
1
y q0 q
y
=
.
2
2
d
4π0 (x + y + L2 /4)3/2
and
E25-15 (a) The equation is valid for both positive and negative z, so in vector form it would read
~ = Fz k̂ =
F
1
q0 q z
k̂.
4π0 (z 2 + R2 )3/2
(b) The equation is not valid for both positive and negative√z. Reversing the sign of z should
reverse the sign of Fz , and one way to fix this is to write 1 = z/ z 2 . Then
~ = Fz k̂ = 1 2q0 qz √1 − √1
F
k̂.
4π0 R2
z2
z2
E25-16 Divide the rod into small differential lengths dr, each with charge dQ = (Q/L)dr. Each
differential length contributes a differential force
dF =
1 q dQ
1 qQ
=
dr.
2
4π0 r
4π0 r2 L
Integrate:
F
=
=
Z
Z
x+L
1 qQ
dr,
2
4π
0 r L
x
1 qQ 1
1
−
4π0 L x x + L
dF =
E25-17 You must solve Ex. 16 before solving this problem! q0 refers to the charge that had been
called q in that problem. In either case the distance from q0 will be the same regardless of the sign
of q; if q = Q then q will be on the right, while if q = −Q then q will be on the left.
Setting the forces equal to each other one gets
1 qQ 1
1
1 qQ
−
=
,
4π0 L x x + L
4π0 r2
or
r=
p
x(x + L).
7
E25-18 You must solve Ex. 16 and Ex. 17 before solving this problem.
If all charges are positive then moving q0 off axis will result in a net force away from the axis.
That’s unstable.
If q = −Q then both q and Q are on the same side of q0 . Moving q0 closer to q will result in the
attractive force growing faster than the repulsive force, so q0 will move away from equilibrium.
E25-19 We can start with the work that was done for us on Page 577, except since we are
concerned with sin θ = z/r we would have
dFx = dF sin θ =
1
q0 λ dz
z
p
.
4π0 (y 2 + z 2 ) y 2 + z 2
We will need to take into consideration that λ changes sign for the two halves of the rod. Then
!
Z L/2
Z 0
q0 λ
−z dz
+z dz
Fx =
+
,
2
2 3/2
4π0
(y 2 + z 2 )3/2
−L/2 (y + z )
0
Z L/2
q0 λ
z dz
=
,
2π0 0
(y 2 + z 2 )3/2
L/2
q0 λ
−1
p
=
,
2π0
y 2 + z 2 0
!
q0 λ 1
1
=
−p
.
2π0 y
y 2 + (L/2)2
E25-20 Use Eq. 25-15 to find the magnitude of the force from any one rod, but write it as
F =
1
qQ
p
,
4π0 r r2 + L2 /4
where r2 = z 2 + L2 /4. The component of this along the z axis is Fz = F z/r. Since there are 4 rods,
we have
1
qQz
1
qQz
p
p
F =
,=
,
2
2
2
2
2
π0 r r + L /4
π0 (z + L /4) z 2 + L2 /2
Equating the electric force with the force of gravity and solving for Q,
p
π0 mg 2
Q=
(z + L2 /4) z 2 + L2 /2;
qz
putting in the numbers,
p
π(8.85×10−12 C2 /N·m2 )(3.46×10−7 kg)(9.8m/s2 )
((0.214m)2+(0.25m)2 /4) (0.214m)2 +(0.25m)2 /2
−12
(2.45×10 C)(0.214 m)
so Q = 3.07×10−6 C.
E25-21 In each case we conserve charge by making sure that the total number of protons is the
same on both sides of the expression. We also need to conserve the number of neutrons.
(a) Hydrogen has one proton, Beryllium has four, so X must have five protons. Then X must be
Boron, B.
(b) Carbon has six protons, Hydrogen has one, so X must have seven. Then X is Nitrogen, N.
(c) Nitrogen has seven protons, Hydrogen has one, but Helium has two, so X has 7 + 1 − 2 = 6
protons. This means X is Carbon, C.
8
E25-22 (a) Use Eq. 25-4:
(8.99×109 N·m2 /C2 )(2)(90)(1.60×10−19 C)2
= 290 N.
(12×10−15 m)2
F =
(b) a = (290 N)/(4)(1.66×10−27 kg) = 4.4×1028 m/s2 .
E25-23 Use Eq. 25-4:
(8.99×109 N·m2 /C2 )(1.60×10−19 C)2
= 2.89×10−9 N.
(282×10−12 m)2
F =
E25-24 (a) Use Eq. 25-4:
s
q=
(3.7×10−9 N)(5.0×10−10 m)2
= 3.20×10−19 C.
(8.99×109 N·m2 /C2 )
(b) N = (3.20×10−19 C)/(1.60×10−19 C) = 2.
E25-25
Use Eq. 25-4,
F =
( 13 1.6 × 10−19 C)( 13 1.6 × 10−19 C)
1 q1 q2
=
= 3.8 N.
2
4π0 r12
4π(8.85 × 10−12 C2 /N · m2 )(2.6 × 10−15 m)2
E25-26 (a) N = (1.15×10−7 C)/(1.60×10−19 C) = 7.19×1011 .
(b) The penny has enough electrons to make a total charge of −1.37×105 C. The fraction is then
(1.15×10−7 C)/(1.37×105 C) = 8.40×10−13 .
E25-27 Equate the magnitudes of the forces:
1 q2
= mg,
4π0 r2
so
r=
s
(8.99×109 N·m2 /C2 )(1.60×10−19 C)2
= 5.07 m
(9.11×10−31 kg)(9.81 m/s2 )
E25-28 Q = (75.0 kg)(−1.60×10−19 C)/(9.11×10−31 kg) = −1.3×1013 C.
3
E25-29 The mass of water is (250 cm3 )(1.00 g/cm ) = 250 g. The number of moles of water is
(250 g)/(18.0 g/mol) = 13.9 mol. The number of water molecules is (13.9 mol)(6.02×1023 mol−1 ) =
8.37×1024 . Each molecule has ten protons, so the total positive charge is
Q = (8.37×1024 )(10)(1.60×10−19 C) = 1.34×107 C.
E25-30 The total positive charge in 0.250 kg of water is 1.34×107 C. Mary’s imbalance is then
q1 = (52.0)(4)(1.34×107 C)(0.0001) = 2.79×105 C,
while John’s imbalance is
q2 = (90.7)(4)(1.34×107 C)(0.0001) = 4.86×105 C,
The electrostatic force of attraction is then
5
5
1 q1 q2
9
2
2 (2.79×10 )(4.86×10 )
F =
=
(8.99×10
N
·
m
/C
)
= 1.6×1018 N.
4π0 r2
(28.0 m)2
9
E25-31
(a) The gravitational force of attraction between the Moon and the Earth is
FG =
GM E M M
,
R2
where R is the distance between them. If both the Earth and the moon are provided a charge q,
then the electrostatic repulsion would be
FE =
1 q2
.
4π0 R2
Setting these two expression equal to each other,
q2
= GM E M M ,
4π0
which has solution
p
q =
4π0 GM E M M ,
q
=
4π(8.85×10−12 C2/Nm2 )(6.67×10−11 Nm2/kg2 )(5.98×1024 kg)(7.36×1022 kg),
=
5.71 × 1013 C.
(b) We need
(5.71 × 1013 C)/(1.60 × 10−19 C) = 3.57 × 1032
protons on each body. The mass of protons needed is then
(3.57 × 1032 )(1.67 × 10−27 kg) = 5.97 × 1065 kg.
Ignoring the mass of the electron (why not?) we can assume that hydrogen is all protons, so we
need that much hydrogen.
P25-1
them is
Assume that the spheres initially have charges q1 and q2 . The force of attraction between
F1 =
1 q1 q2
= −0.108 N,
2
4π0 r12
where r12 = 0.500 m. The net charge is q1 + q2 , and after the conducting wire is connected each
sphere will get half of the total. The spheres will have the same charge, and repel with a force of
F2 =
1 12 (q1 + q2 ) 12 (q1 + q2 )
= 0.0360 N.
2
4π0
r12
Since we know the separation of the spheres we can find q1 + q2 quickly,
q
2 (0.0360 N) = 2.00 µC
q1 + q2 = 2 4π0 r12
We’ll put this back into the first expression and solve for q2 .
−0.108 N
=
1 (2.00 µC − q2 )q2
,
2
4π0
r12
−3.00 × 10−12 C2 = (2.00 µC − q2 )q2 ,
0 = −q22 + (2.00 µC)q2 + (1.73 µC)2 .
The solution is q2 = 3.0 µC or q2 = −1.0 µC. Then q1 = −1.0 µC or q1 = 3.0 µC.
10
P25-2 The electrostatic force on Q from each q has magnitude qQ/4π0 a2 , where a is the length
of the√side of the square. The magnitude of the vertical (horizontal) component of the force of Q on
Q is 2Q2 /16π0 a2 .
(a) In order to have a zero net force on Q the magnitudes of the two contributions must balance,
so
√ 2
2Q
qQ
=
,
2
16π0 a
4π0 a2
√
or q = 2Q/4. The charges must actually have opposite charge.
(b) No.
P25-3 (a) The third charge, q3 , will be between the first two. The net force on the third charge
will be zero if
1 q q3
1 4q q3
=
2
2 ,
4π0 r31
4π0 r32
which will occur if
1
2
=
r31
r32
The total distance is L, so r31 + r32 = L, or r31 = L/3 and r32 = 2L/3.
Now that we have found the position of the third charge we need to find the magnitude. The
second and third charges both exert a force on the first charge; we want this net force on the first
charge to be zero, so
1 q q3
1 q 4q
2 = 4π r 2 ,
4π0 r13
0 12
or
q3
4q
= 2,
(L/3)2
L
which has solution q3 = −4q/9. The negative sign is because the force between the first and second
charge must be in the opposite direction to the force between the first and third charge.
(b) Consider what happens to the net force on the middle charge if is is displaced a small distance
z. If the charge 3 is moved toward charge 1 then the force of attraction with charge 1 will increase.
But moving charge 3 closer to charge 1 means moving charge 3 away from charge 2, so the force of
attraction between charge 3 and charge 2 will decrease. So charge 3 experiences more attraction to
ward the charge that it moves toward, and less attraction to the charge it moves away from. Sounds
unstable to me.
P25-4
(a) The electrostatic force on the charge on the right has magnitude
F =
q2
,
4π0 x2
The weight of the ball is W = mg, and the two forces are related by
F/W = tan θ ≈ sin θ = x/2L.
Combining, 2Lq 2 = 4π0 mgx3 , so
x=
q2 L
2π0
1/3
.
(b) Rearrange and solve for q,
s
2π(8.85×10−12 C2 /N · m2 )(0.0112 kg)(9.81 m/s2 )(4.70×10−2 m)3
q=
= 2.28×10−8 C.
(1.22 m)
11
P25-5 (a) Originally the balls would not repel, so they would move together and touch; after
touching the balls would “split” the charge ending up with q/2 each. They would then repel again.
(b) The new equilibrium separation is
x0 =
P25-6
(q/2)2 L
2π0 mg
1/3
=
1/3
1
x = 2.96 cm.
4
Take the time derivative of the expression in Problem 25-4. Then
dx
2 x dq
2 (4.70×10−2 m)
=
=
(−1.20×10−9 C/s) = 1.65×10−3 m/s.
dt
3 q dt
3 (2.28×10−8 C)
P25-7
The force between the two charges is
F =
1 (Q − q)q
.
2
4π0
r12
We want to maximize this force with respect to variation in q, this means finding dF/dq and setting
it equal to 0. Then
d
1 (Q − q)q
1 Q − 2q
dF
=
=
.
2
2
dq
dq 4π0
r12
4π0 r12
This will vanish if Q − 2q = 0, or q = 12 Q.
P25-8
Displace the charge q a distance y. The net restoring force on q will be approximately
F ≈2
qQ
1
y
qQ 16
=
y.
4π0 (d/2)2 (d/2)
4π0 d3
Since F/y is effectively a force constant, the period of oscillation is
T = 2π
r
m
=
k
0 mπ 3 d3
qQ
1/2
.
P25-9 Displace the charge q a distance x toward one of the positive charges Q. The net restoring
force on q will be
qQ
1
1
F =
−
,
4π0 (d/2 − x)2
(d/2 + x)2
qQ 32
≈
x.
4π0 d3
Since F/x is effectively a force constant, the period of oscillation is
T = 2π
r
m
=
k
12
0 mπ 3 d3
2qQ
1/2
.
P25-10 (a) Zero, by symmetry.
(b) Removing a positive Cesium ion is equivalent to adding a singly charged negative ion at that
same location. The net force is then
F = e2 /4π0 r2 ,
where r is the distance between the Chloride ion and the newly placed negative ion, or
p
r = 3(0.20×10−9 m)2
The force is then
F =
(1.6×10−19 C)2
= 1.92×10−9 N.
4π(8.85×10−12 C2 /N · m2 )3(0.20×10−9 m)2
P25-11 We can pretend that this problem is in a single plane containing all three charges. The
magnitude of the force on the test charge q0 from the charge q on the left is
Fl =
q q0
1
.
4π0 (a2 + R2 )
A force of identical magnitude exists from the charge on the right. we need to add these two forces
as vectors. Only the components along R will survive, and each force will contribute an amount
F l sin θ = F l √
R
,
+ a2
R2
so the net force on the test particle will be
q q0
R
2
√
.
4π0 (a2 + R2 ) R2 + a2
We want to find the maximum value as a function of R. This means take the derivative, and set it
equal to zero. The derivative is
2q q0
1
3R2
−
,
4π0 (a2 + R2 )3/2
(a2 + R2 )5/2
which will vanish when
a2 + R2 = 3R2 ,
√
a simple quadratic equation with solutions R = ±a/ 2.
13
E26-1 E = F/q = ma/q. Then
E = (9.11×10−31 kg)(1.84×109 m/s2 )/(1.60×10−19 C) = 1.05×10−2 N/C.
E26-2 The answers to (a) and (b) are the same!
F = Eq = (3.0×106 N/C)(1.60×10−19 C) = 4.8×10−13 N.
E26-3
F = W , or Eq = mg, so
E=
mg
(6.64 × 10−27 kg)(9.81 m/s2 )
=
= 2.03 × 10−7 N/C.
q
2(1.60 × 10−19 C)
The alpha particle has a positive charge, this means that it will experience an electric force which
is in the same direction as the electric field. Since the gravitational force is down, the electric force,
and consequently the electric field, must be directed up.
E26-4 (a) E = F/q = (3.0×10−6 N)/(2.0×10−9 C) = 1.5×103 N/C.
(b) F = Eq = (1.5×103 N/C)(1.60×10−19 C) = 2.4×10−16 N.
(c) F = mg = (1.67×10−27 kg)(9.81 m/s2 ) = 1.6×10−26 N.
(d) (2.4×10−16 N)/(1.6×10−26 N) = 1.5×1010 .
E26-5 Rearrange E = q/4π0 r2 ,
q = 4π(8.85×10−12 C2 /N · m2 )(0.750 m)2 (2.30 N/C) = 1.44×10−10 C.
E26-6 p = qd = (1.60×10−19 C)(4.30×10−9 ) = 6.88×10−28 C · m.
E26-7
Use Eq. 26-12 for points along the perpendicular bisector. Then
E=
−29
1 p
C · m)
9
2
2 (3.56 × 10
=
(8.99
×
10
N
·
m
/C
)
= 1.95 × 104 N/C.
3
−9
4π0 x
(25.4 × 10 m)3
E26-8 If the charges on the line x = a where +q and −q instead of +2q and −2q then at the
center of the square E = 0 by symmetry. This simplifies the problem into finding E for a charge +q
at (a, 0) and −q at (a, a). This is a dipole, and the field is given by Eq. 26-11. For this exercise we
have x = a/2 and d = a, so
1
qa
E=
,
4π0 [2(a/2)2 ]3/2
or, putting in the numbers, E = 1.11×105 N/C.
E26-9 The charges at 1 and 7 are opposite and can be effectively replaced with a single charge of
−6q at 7. The same is true for 2 and 8, 3 and 9, on up to 6 and 12. By symmetry we expect the
field to point along a line so that three charges are above and three below. That would mean 9:30.
E26-10 If both charges are positive then Eq. 26-10 would read E = 2E+ sin θ, and Eq. 26-11
would look like
1
q
x
p
,
E = 2
2
2
2
4π0 x + (d/2)
x + (d/2)2
1 q x
√
≈ 2
4π0 x2 x2
when x d. This can be simplified to E = 2q/4π0 x2 .
14
E26-11 Treat the two charges on the left as one dipole and treat the two charges on the right as
a second dipole. Point P is on the perpendicular bisector of both dipoles, so we can use Eq. 26-12
to find the two fields.
For the dipole on the left p = 2aq and the electric field due to this dipole at P has magnitude
El =
2aq
1
4π0 (x + a)3
and is directed up.
For the dipole on the right p = 2aq and the electric field due to this dipole at P has magnitude
Er =
1
2aq
4π0 (x − a)3
and is directed down.
The net electric field at P is the sum of these two fields, but since the two component fields point
in opposite directions we must actually subtract these values,
E
= Er − El,
2aq
1
1
=
−
,
4π0 (x − a)3
(x + a)3
aq 1
1
1
=
−
.
2π0 x3 (1 − a/x)3
(1 + a/x)3
We can use the binomial expansion on the terms containing 1 ± a/x,
E
≈
=
=
aq 1
((1 + 3a/x) − (1 − 3a/x)) ,
2π0 x3
aq 1
(6a/x) ,
2π0 x3
3(2qa2 )
.
2π0 x4
E26-12 Do a series expansion on the part in the parentheses
1 R2
R2
1
=
.
1− p
≈1− 1−
2 z2
2z 2
1 + R2 /z 2
Substitute this in,
Ez ≈
σ R2 π
Q
=
.
2
20 2z π
4π0 z 2
E26-13 At the surface z = 0 and Ez = σ/20 . Half of this value occurs when z is given by
1
z
,
=1− √
2
z 2 + R2
√
which can be written as z 2 + R2 = (2z)2 . Solve this, and z = R/ 3.
E26-14 Look at Eq. 26-18. The electric field will be a maximum when z/(z 2 + R2 )3/2 is a
maximum. Take the derivative of this with respect to z, and get
1
3
2z 2
z 2 + R2 − 3z 2
−
=
.
2 (z 2 + R2 )5/2
(z 2 + R2 )3/2
(z 2 + R2 )5/2
√
This will vanish when the numerator vanishes, or when z = R/ 2.
15
E26-15 (a) The electric field strength just above the center surface of a charged disk is given by
Eq. 26-19, but with z = 0,
σ
E=
20
The surface charge density is σ = q/A = q/(πR2 ). Combining,
q = 20 πR2 E = 2(8.85 × 10−12 C2 /N · m2 )π(2.5 × 10−2 m)2 (3 × 106 N/C) = 1.04 × 10−7 C.
Notice we used an electric field strength of E = 3 × 106 N/C, which is the field at air breaks down
and sparks happen.
(b) We want to find out how many atoms are on the surface; if a is the cross sectional area of
one atom, and N the number of atoms, then A = N a is the surface area of the disk. The number
of atoms is
π(0.0250 m)2
A
=
= 1.31 × 1017
N=
a
(0.015 × 10−18 m2 )
(c) The total charge on the disk is 1.04 × 10−7 C, this corresponds to
(1.04 × 10−7 C)/(1.6 × 10−19 C) = 6.5 × 1011
electrons. (We are ignoring the sign of the charge here.) If each surface atom can have at most one
excess electron, then the fraction of atoms which are charged is
(6.5 × 1011 )/(1.31 × 1017 ) = 4.96 × 10−6 ,
which isn’t very many.
E26-16 Imagine switching the positive and negative charges. The electric field would also need
to switch directions. By symmetry, then, the electric field can only point vertically down. Keeping
only that component,
Z π/2
1 λdθ
E = 2
sin θ,
2
4π
0 r
0
2 λ
.
=
4π0 r2
But λ = q/(π/2), so E = q/π 2 0 r2 .
E26-17
We want to fit the data to Eq. 26-19,
σ
z
.
Ez =
1− √
20
z 2 + R2
There are only two variables, R and q, with q = σπR2 .
We can find σ very easily if we assume that the measurements have no error because then at the
surface (where z = 0), the expression for the electric field simplifies to
E=
σ
.
20
Then σ = 20 E = 2(8.854 × 10−12 C2 /N · m2 )(2.043 × 107 N/C) = 3.618 × 10−4 C/m2 .
Finding the radius will take a little more work. We can choose one point, and make that the
reference point, and then solve for R. Starting with
σ
z
1− √
,
Ez =
20
z 2 + R2
16
and then rearranging,
20 Ez
σ
20 Ez
σ
1
p
1 + (R/z)2
1 + (R/z)2
R
z
z
,
z 2 + R2
1
= 1− p
,
1 + (R/z)2
20 Ez
= 1−
,
σ
1
=
2,
(1 − 20 Ez /σ)
s
1
=
2 − 1.
(1 − 20 Ez /σ)
=
1− √
Using z = 0.03 m and Ez = 1.187 × 107 N/C, along with our value of σ = 3.618 × 10−4 C/m2 , we
find
s
R
1
=
2 − 1,
−12
2
2
z
(1 − 2(8.854×10 C /Nm )(1.187×107 N/C)/(3.618×10−4 C/m2 ))
R
=
2.167(0.03 m) = 0.065 m.
(b) And now find the charge from the charge density and the radius,
q = πR2 σ = π(0.065 m)2 (3.618 × 10−4 C/m2 ) = 4.80 µC.
E26-18 (a) λ = −q/L.
(b) Integrate:
E
=
=
=
Z
L+a
1
λ dxx2 ,
4π
0
a
1
1
λ
−
,
4π0 a L + a
q
1
,
4π0 a(L + a)
since λ = q/L.
(c) If a L then L can be replaced with 0 in the above expression.
E26-19 A sketch of the field looks like this.
17
E26-20 (a) F = Eq = (40 N/C)(1.60×10−19 C) = 6.4×10−18 N
(b) Lines are twice as far apart, so the field is half as large, or E = 20N/C.
E26-21 Consider a view of the disk on edge.
E26-22 A sketch of the field looks like this.
18
E26-23 To the right.
E26-24 (a) The electric field is zero nearer to the smaller charge; since the charges have opposite
signs it must be to the right of the +2q charge. Equating the magnitudes of the two fields,
2q
5q
=
,
4π0 x2
4π0 (x + a)2
or
√
5x =
which has solution
√
2(x + a),
√
x= √
2a
√ = 2.72a.
5− 2
E26-25 This can be done quickly with a spreadsheet.
E
x
d
E26-26 (a) At point A,
E=
1
4π0
−
q
−2q
−
d2
(2d)2
19
=
1 −q
,
4π0 2d2
~ is directed to the left.
where the negative sign indicates that E
At point B,
1
q
−2q
1 6q
E=
−
=
,
2
2
4π0 (d/2)
(d/2)
4π0 d2
~ is directed to the right.
where the positive sign indicates that E
At point C,
q
−2q
1 −7q
1
+ 2
=
,
E=
2
4π0 (2d)
d
4π0 4d2
~ is directed to the left.
where the negative sign indicates that E
E26-27 (a) The electric field does (negative) work on the electron. The magnitude of this work
is W = F d, where F = Eq is the magnitude of the electric force on the electron and d is the distance
through which the electron moves. Combining,
~ · ~d = q E
~ · ~d,
W =F
which gives the work done by the electric field on the electron. The electron originally possessed a
kinetic energy of K = 21 mv 2 , since we want to bring the electron to a rest, the work done must be
~ and ~d are pointing in the same direction,
negative. The charge q of the electron is negative, so E
~
~
and E · d = Ed.
By the work energy theorem,
1
W = ∆K = 0 − mv 2 .
2
We put all of this together and find d,
d=
−mv 2
−(9.11×10−31 kg)(4.86 × 106 m/s)2
W
=
=
= 0.0653 m.
qE
2qE
2(−1.60×10−19 C)(1030 N/C)
(b) Eq = ma gives the magnitude of the acceleration, and v f = v i + at gives the time. But
v f = 0. Combining these expressions,
t=−
mv i
(9.11×10−31 kg)(4.86 × 106 m/s)
=−
= 2.69×10−8 s.
Eq
(1030 N/C)(−1.60×10−19 C)
(c) We will apply the work energy theorem again, except now we don’t assume the final kinetic
energy is zero. Instead,
W = ∆K = K f − K i ,
and dividing through by the initial kinetic energy to get the fraction lost,
W
Kf − Ki
=
= fractional change of kinetic energy.
Ki
Ki
But K i = 12 mv 2 , and W = qEd, so the fractional change is
W
(−1.60×10−19 C)(1030 N/C)(7.88×10−3 m)
qEd
=
= −12.1%.
= 1
1
2
−31 kg)(4.86 × 106 m/s)2
Ki
2 mv
2 (9.11×10
4
−19
−27
E26-28 (a)
kg) = 2.07×1012 m/s2 .
p = (2.16×10 N/C)(1.60×10 C)/(1.67×10
√ a = Eq/m
5
(b) v = 2ax = 2(2.07×1012 m/s2 )(1.22×10−2 m) = 2.25×10 m/s.
20
E26-29 (a) E = 2q/4π0 r2 , or
E=
(1.88×10−7 C)
= 5.85×105 N/C.
2π(8.85×10−12 C2 /N · m2 )(0.152 m/2)2
(b) F = Eq = (5.85×105 N/C)(1.60×10−19 C) = 9.36×10−14 N.
E26-30 (a) The average speed between the plates is (1.95×10−2 m)/(14.7×10−9 s) = 1.33×106 m/s.
The speed with which the electron hits the plate is twice this, or 2.65×106 m/s.
(b) The acceleration is a = (2.65×106 m/s)/(14.7×10−9 s) = 1.80×1014 m/s2 . The electric field
then has magnitude E = ma/q, or
E = (9.11×10−31 kg)(1.80×1014 m/s2 )/(1.60×10−19 C) = 1.03×103 N/C.
E26-31 The drop is balanced if the electric force is equal to the force of gravity, or Eq = mg.
The mass of the drop is given in terms of the density by
4
m = ρV = ρ πr3 .
3
Combining,
q=
mg
4πρr3 g
4π(851 kg/m3 )(1.64×10−6 m)3 (9.81 m/s2 )
=
=
= 8.11×10−19 C.
E
3E
3(1.92×105 N/C)
We want the charge in terms of e, so we divide, and get
q
(8.11×10−19 C)
=
= 5.07 ≈ 5.
e
(1.60×10−19 C)
E26-32 (b) F = (8.99×109 N · m2 /C2 )(2.16×10−6 C)(85.3×10−9 C)/(0.117m)2 = 0.121 N.
(a) E2 = F/q1 = (0.121 N)/(2.16×10−6 C) = 5.60×104 N/C.
E1 = F/q2 = (0.121 N)/(85.3×10−9 C) = 1.42×106 N/C.
E26-33 If each value of q measured by Millikan was a multiple of e, then the difference between
any two values of q must also be a multiple of q. The smallest difference would be the smallest
multiple, and this multiple might be unity. The differences are 1.641, 1.63, 1.60, 1.63, 3.30, 3.35,
3.18, 3.24, all times 10−19 C. This is a pretty clear indication that the fundamental charge is on the
order of 1.6 × 10−19 C. If so, the likely number of fundamental charges on each of the drops is shown
below in a table arranged like the one in the book:
4
5
7
8
10
11
12
14
16
The total number of charges is 87, while the total charge is 142.69 × 10−19 C, so the average charge
per quanta is 1.64 × 10−19 C.
21
E26-34 Because of the electric field the acceleration toward the ground of a charged particle is
not g, but g ± Eq/m, where the sign depends on the direction of the electric field.
(a) If the lower plate is positively charged then a = g − Eq/m. Replace g in the pendulum period
expression by this, and then
s
L
T = 2π
.
g − Eq/m
(b) If the lower plate is negatively charged then a = g + Eq/m. Replace g in the pendulum
period expression by this, and then
s
L
T = 2π
.
g + Eq/m
E26-35 The ink drop travels an additional time t0 = d/vx , where d is the additional horizontal
distance between the plates and the paper. During this time it travels an additional vertical distance
y 0 = vy t0 , where vy = at = 2y/t = 2yvx /L. Combining,
y0 =
2yvx t0
2yd
2(6.4×10−4 m)(6.8×10−3 m)
=
=
= 5.44×10−4 m,
L
L
(1.6×10−2 m)
so the total deflection is y + y 0 = 1.18×10−3 m.
E26-36 (a) p = (1.48×10−9 C)(6.23×10−6 m) = 9.22×10−15 C · m.
(b) ∆U = 2pE = 2(9.22×10−15 C · m)(1100 N/C) = 2.03×10−11 J.
~ For this dipole p = qd = 2ed
E26-37 Use τ = pE sin θ, where θ is the angle between ~p and E.
−19
−9
−28
or p = 2(1.6 × 10
C)(0.78 × 10 m) = 2.5 × 10
C · m. For all three cases
pE = (2.5 × 10−28 C · m)(3.4 × 106 N/C) = 8.5 × 10−22 N · m.
The only thing we care about is the angle.
(a) For the parallel case θ = 0, so sin θ = 0, and τ = 0.
(b) For the perpendicular case θ = 90◦ , so sin θ = 1, and τ = 8.5 × 10−22 N · m..
(c) For the anti-parallel case θ = 180◦ , so sin θ = 0, and τ = 0.
E26-38 (c) Equal and opposite, or 5.22×10−16 N.
(d) Use Eq. 26-12 and F = Eq. Then
p
=
=
=
E26-39
4π0 x3 F
,
q
4π(8.85×10−12 C2 /N · m2 )(0.285m)3 (5.22×10−16 N)
,
(3.16×10−6 C)
4.25×10−22 C · m.
The point-like nucleus contributes an electric field
E+ =
1 Ze
,
4π0 r2
while the uniform sphere of negatively charged electron cloud of radius R contributes an electric
field given by Eq. 26-24,
1 −Zer
E− =
.
4π0 R3
22
The net electric field is just the sum,
Ze
E=
4π0
1
r
− 3
r2
R
E26-40 The shell theorem first described for gravitation in chapter 14 is applicable here since both
electric forces and gravitational forces fall off as 1/r2 . The net positive charge inside the sphere of
radius d/2 is given by Q = 2e(d/2)3 /R3 = ed3 /4R3 .
The net force on either electron will be zero when
eQ
4e2 d3
e2 d
e2
=
=
=
,
d2
(d/2)2
d2 4R3
R3
which has solution d = R.
P26-1 (a) Let the positive charge be located closer to the point in question, then the electric
field from the positive charge is
q
1
E+ =
4π0 (x − d/2)2
and is directed away from the dipole.
The negative charge is located farther from the point in question, so
E− =
1
q
4π0 (x + d/2)2
and is directed toward the dipole.
The net electric field is the sum of these two fields, but since the two component fields point in
opposite direction we must actually subtract these values,
E
= E + − E− ,
1
q
1
q
−
,
=
4π0 (z − d/2)2
4π0 (z + d/2)2
1
1
1 q
−
=
4π0 z 2 (1 − d/2z)2
(1 + d/2z)2
We can use the binomial expansion on the terms containing 1 ± d/2z,
E
≈
=
1 q
((1 + d/z) − (1 − d/z)) ,
4π0 z 2
1 qd
2π0 z 3
(b) The electric field is directed away from the positive charge when you are closer to the positive
charge; the electric field is directed toward the negative charge when you are closer to the negative
charge. In short, along the axis the electric field is directed in the same direction as the dipole
moment.
P26-2
The key to this problem will be the expansion of
1
1
≈ 2
(x2 + (z ± d/2)2 )3/2
(x + z 2 )3/2
23
3 zd
1∓
2 x2 + z 2
.
for d √
x2 + z 2 . Far from the charges the electric field of the positive charge has magnitude
E+ =
1
q
,
4π0 x2 + (z − d/2)2
the components of this are
Ex,+
=
Ez,+
=
q
x
1
p
,
2
2
2
4π0 x + z
x + (z − d/2)2
q
(z − d/2)
1
p
.
4π0 x2 + z 2 x2 + (z − d/2)2
Expand both according to the first sentence, then
Ex,+
≈
Ez,+
=
1
xq
1+
4π0 (x2 + z 2 )3/2
1 (z − d/2)q
1+
4π0 (x2 + z 2 )3/2
3 zd
,
2 x2 + z 2
3 zd
.
2 x2 + z 2
Similar expression exist for the negative charge, except we must replace q with −q and the + in the
parentheses with a −, and z − d/2 with z + d/2 in the Ez expression. All that is left is to add the
expressions. Then
1
xq
3 zd
1
−xq
3 zd
Ex =
1
+
+
1
−
,
4π0 (x2 + z 2 )3/2
2 x2 + z 2
4π0 (x2 + z 2 )3/2
2 x2 + z 2
3xqzd
1
=
,
4π0 (x2 + z 2 )5/2
3 zd
3 zd
1 (z − d/2)q
1 −(z + d/2)q
Ez =
1
+
+
1
−
,
4π0 (x2 + z 2 )3/2
2 x2 + z 2
4π0 (x2 + z 2 )3/2
2 x2 + z 2
1
3z 2 dq
1
dq
=
−
,
2
4π0 (x + z 2 )5/2
4π0 (x2 + z 2 )3/2
1 (2z 2 − x2 )dq
=
.
4π0 (x2 + z 2 )5/2
√
P26-3 (a) Each point on the ring is a distance z 2 + R2 from the point on the axis in question.
Since all points are equal distant and subtend the same angle from the axis then the top half of the
ring contributes
q1
z
√
E1z =
,
2
2
2
4π0 (x + R ) z + R2
while the bottom half contributes a similar expression. Add, and
Ez =
q1 + q2
z
q
z
=
,
4π0 (z 2 + R2 )3/2
4π0 (z 2 + R2 )3/2
which is identical to Eq. 26-18.
(b) The perpendicular component would be zero if q1 = q2 . It isn’t, so it must be the difference
q1 − q2 which is of interest. Assume this charge difference is evenly distributed on the top half of
the ring. If it is a positive difference, then E⊥ must point down. We are only interested then in the
vertical component as we integrate around the top half of the ring. Then
Z π
1 (q1 − q2 )/π
E⊥ =
cos θ dθ,
2
2
4π
0 z +R
0
q1 − q2
1
=
.
2
2
2π 0 z + R2
24
P26-4 Use the approximation 1/(z ± d)2 ≈ (1/z 2 )(1 ∓ 2d/z + 3d2 /z 2 ).
Add the contributions:
q
2q
q
1
E =
−
+
,
4π0 (z + d)2
z2
(z − d)2
q
2d 3d2
2d 3d2
≈
1
−
+
−
2
+
1
+
+
,
4π0 z 2
z
z2
z
z2
q 6d2
3Q
=
=
,
4π0 z 2 z 2
4π0 z 4
where Q = 2qd2 .
P26-5 A monopole field falls off as 1/r2 . A dipole field falls off as 1/r3 , and consists of two
oppositely charge monopoles close together. A quadrupole field (see Exercise 11 above or read
Problem 4) falls off as 1/r4 and (can) consist of two otherwise identical dipoles arranged with antiparallel dipole moments. Just taking a leap of faith it seems as if we can construct a 1/r6 field
behavior by extending the reasoning.
First we need an octopole which is constructed from a quadrupole. We want to keep things as
simple as possible, so the construction steps are
1. The monopole is a charge +q at x = 0.
2. The dipole is a charge +q at x = 0 and a charge −q at x = a. We’ll call this a dipole at
x = a/2
3. The quadrupole is the dipole at x = a/2, and a second dipole pointing the other way at
x = −a/2. The charges are then −q at x = −a, +2q at x = 0, and −q at x = a.
4. The octopole will be two stacked, offset quadrupoles. There will be −q at x = −a, +3q at
x = 0, −3q at x = a, and +q at x = 2a.
5. Finally, our distribution with a far field behavior of 1/r6 . There will be +q at x = 2a, −4q at
x = −a, +6q at x = 0, −4q at x = a, and +q at x = 2a.
~ is simply half of Eq. 26-17. The horizontal component is
P26-6 The vertical component of E
given by a variation of the work required to derive Eq. 26-16,
dEz = dE sin θ =
1
λ dz
z
p
,
4π0 y 2 + z 2 y 2 + z 2
which integrates to zero if the limits are −∞ to +∞, but in this case,
Z ∞
λ 1
Ez =
dEz =
.
4π
0 z
0
~ makes an angle of 45◦ .
Since the vertical and horizontal components are equal then E
P26-7 (a) Swap all positive and negative charges in the problem and the electric field must reverse
direction. But this is the same as flipping the problem over; consequently, the electric field must
point parallel to the rod. This only holds true at point P , because point P doesn’t move when you
flip the rod.
25
(b) We are only interested in the vertical component of the field as contributed from each point
on the rod. We can integrate only half of the rod and double the answer, so we want to evaluate
Z L/2
1
λ dz
z
p
Ez = 2
,
2 + z2
2
4π
y
0
y + z2
0
p
(L/2)2 + y 2 − y
2λ
p
=
.
4π0 y (L/2)2 + y 2
(c) The previous expression is exact. If y L, then the expression simplifies with a Taylor
expansion to
λ L2
,
Ez =
4π0 y 3
which looks similar to a dipole.
P26-8
Evaluate
E=
R
Z
0
1
z dq
,
4π0 (z 2 + r2 )3/2
where r is the radius of the ring, z the distance to the plane of the ring, and dq the differential
charge on the ring. But r2 + z 2 = R2 , and dq = σ(2πr dr), where σ = q/2πR2 . Then
√
Z R
q
R2 − r2 r dr
E =
,
R5
0 4π0
q
1
=
.
4π0 3R2
P26-9 The key statement is the second italicized paragraph on page 595; the number of field
lines through a unit cross-sectional area is proportional to the electric field strength. If the exponent
is n, then the electric field strength a distance r from a point charge is
E=
kq
,
rn
and the total cross sectional area at a distance r is the area of a spherical shell, 4πr2 . Then the
number of field lines through the shell is proportional to
EA =
kq
4πr2 = 4πkqr2−n .
rn
Note that the number of field lines varies with r if n 6= 2. This means that as we go farther from
the point charge we need more and more field lines (or fewer and fewer). But the field lines can only
start on charges, and we don’t have any except for the point charge. We have a problem; we really
do need n = 2 if we want workable field lines.
P26-10 The distance traveled by the electron will be d1 = a1 t2 /2; the distance traveled by the
proton will be d2 = a2 t2 /2. a1 and a2 are related by m1 a1 = m2 a2 , since the electric force is the
same (same charge magnitude). Then d1 + d2 = (a1 + a2 )t2 /2 is the 5.00 cm distance. Divide by
the proton distance, and then
d1 + d2
a1 + a2
m2
=
=
+ 1.
d2
a2
m1
Then
d2 = (5.00×10−2 m)/(1.67×10−27 /9.11×10−31 + 1) = 2.73×10−5 m.
26
P26-11 This is merely a fancy projectile motion problem. vx = v0 cos θ while vy,0 = v0 sin θ. The
x and y positions are x = vx t and
y=
1 2
ax2
at + vy,0 t = 2
+ x tan θ.
2
2v0 cos2 θ
The acceleration of the electron is vertically down and has a magnitude of
a=
F
Eq
(1870 N/C)(1.6×10−19 C)
=
=
= 3.284×1014 m/s2 .
m
m
(9.11×10−31 kg)
We want to find out how the vertical velocity of the electron at the location of the top plate. If
we get an imaginary answer, then the electron doesn’t get as high as the top plate.
q
vy =
vy,0 2 + 2a∆y,
p
=
(5.83×106 m/s)2 sin(39◦ )2 + 2(−3.284×1014 m/s2 )(1.97×10−2 m),
= 7.226×105 m/s.
This is a real answer, so this means the electron either hits the top plate, or it misses both plates.
The time taken to reach the height of the top plate is
t=
∆vy
(7.226×105 m/s) − (5.83×106 m/s) sin(39◦ )
=
= 8.972×10−9 s.
a
(−3.284×1014 m/s2 )
In this time the electron has moved a horizontal distance of
x = (5.83×106 m/s) cos(39◦ )(8.972×10−9 s) = 4.065×10−2 m.
This is clearly on the upper plate.
P26-12
Near the center of the ring z R, so a Taylor expansion yields
E=
λ z
.
20 R2
The force on the electron is F = Ee, so the effective “spring” constant is k = eλ/20 R2 . This means
r
r
r
k
eλ
eq
ω=
=
=
.
2
m
20 mR
4π0 mR3
P26-13
U = −pE cos θ, so the work required to flip the dipole is
W = −pE [cos(θ0 + π) − cos θ0 ] = 2pE cos θ0 .
P26-14 If the torque on a system
pis given by |τ | = κθ, where κ is a constant, then the frequency
of oscillation of the system is f = κ/I/2π. In this case τ = pE sin θ ≈ pEθ, so
p
f = pE/I/2π.
27
P26-15 Use the a variation of the exact result from Problem 26-1. The two charge are positive,
but since we will eventually focus on the area between the charges we must subtract the two field
contributions, since they point in opposite directions. Then
1
1
q
Ez =
−
4π0 (z − a/2)2
(z + a/2)2
and then take the derivative,
dEz
q
=−
dz
2π0
1
1
−
(z − a/2)3
(z + a/2)3
.
Applying the binomial expansion for points z a,
8q 1
1
1
dEz
= −
−
,
dz
2π0 a3 (2z/a − 1)3
(2z/a + 1)3
8q 1
≈ −
(−(1 + 6z/a) − (1 − 6z/a)) ,
2π0 a3
8q 1
=
.
π0 a3
There were some fancy sign flips in the second line, so review those steps carefully!
(b) The electrostatic force on a dipole is the difference in the magnitudes of the electrostatic
forces on the two charges that make up the dipole. Near the center of the above charge arrangement
the electric field behaves as
dEz Ez ≈ Ez (0) +
z + higher ordered terms.
dz z=0
The net force on a dipole is
dEz dEz F+ − F− = q(E+ − E− ) = q Ez (0) +
z+ − Ez (0) −
z−
dz z=0
dz z=0
where the “+” and “-” subscripts refer to the locations of the positive and negative charges. This
last line can be simplified to yield
dEz dEz q
(z
−
z
)
=
qd
.
+
−
dz z=0
dz z=0
28
E27-1 ΦE = (1800 N/C)(3.2×10−3 m)2 cos(145◦ ) = −7.8×10−3 N · m2 /C.
~ = (1.4 m)2 ĵ.
E27-2 The right face has an area element given by A
2
~
~
(a) ΦE = A · E = (2.0 m )ĵ · (6 N/C)î = 0.
(b) ΦE = (2.0 m2 )ĵ · (−2 N/C)ĵ = −4N · m2 /C.
(c) ΦE = (2.0 m2 )ĵ · [(−3 N/C)î + (4 N/C)k̂] = 0.
~ · A,
~ where A
~ has six
(d) In each case the field is uniform so we can simply evaluate ΦE = E
parts, one for every face. The faces, however, have the same size but are organized in pairs with
opposite directions. These will cancel, so the total flux is zero in all three cases.
E27-3
(a) The flat base is easy enough, since according to Eq. 27-7,
Z
~ · dA.
~
ΦE = E
~ is parallel to the
There are two important facts to consider in order to integrate this expression. E
~ points outward on the flat base. E
~ points inward while dA
~ is uniform, so
axis of the hemisphere, E
~ is anti-parallel to dA,
~ E
~ · dA
~ = −E dA, then
it is everywhere the same on the flat base. Since E
Z
Z
~ · dA
~ = − E dA.
ΦE = E
~ is uniform we can simplify this as
Since E
Z
Z
ΦE = − E dA = −E dA = −EA = −πR2 E.
The last steps are just substituting the area of a circle for the flat side of the hemisphere.
(b) We must first sort out the dot product
E
dA
R
θ
φ
~ · dA
~ = cos θE dA, so
We can simplify the vector part of the problem with E
Z
Z
~ · dA
~ = cos θE dA
ΦE = E
~ is uniform, so we can take it out of the integral,
Once again, E
Z
Z
ΦE = cos θE dA = E cos θ dA
Finally, dA = (R dθ)(R sin θ dφ) on the surface of a sphere centered on R = 0.
29
We’ll integrate φ around the axis, from 0 to 2π. We’ll integrate θ from the axis to the equator,
from 0 to π/2. Then
ΦE = E
Z
cos θ dA = E
2π
Z
Z
0
π/2
R2 cos θ sin θ dθ dφ.
0
Pulling out the constants, doing the φ integration, and then writing 2 cos θ sin θ as sin(2θ),
ΦE = 2πR2 E
Z
π/2
cos θ sin θ dθ = πR2 E
0
Z
π/2
sin(2θ) dθ,
0
Change variables and let β = 2θ, then we have
Z π
1
ΦE = πR2 E
sin β dβ = πR2 E.
2
0
E27-4 Through S1 , ΦE = q/0 . Through S2 , ΦE = −q/0 . Through S3 , ΦE = q/0 . Through S4 ,
ΦE = 0. Through S5 , ΦE = q/0 .
E27-5
By Eq. 27-8,
ΦE =
q
(1.84 µC)
=
= 2.08×105 N · m2 /C.
0
(8.85×10−12 C2 /N · m2 )
E27-6 The total flux through the sphere is
ΦE = (−1 + 2 − 3 + 4 − 5 + 6)(×103 N · m2 /C) = 3×103 N · m2 /C.
The charge inside the die is (8.85×10−12 C2 /N · m2 )(3×103 N · m2 /C) = 2.66×10−8 C.
E27-7 The total flux through a cube would be q/0 . Since the charge is in the center of the cube
we expect that the flux through any side would be the same, or 1/6 of the total flux. Hence the flux
through the square surface is q/60 .
E27-8 If the electric field is uniform then there are no free charges near (or inside) the net. The
flux through the netting must be equal to, but opposite in sign, from the flux through the opening.
The flux through the opening is Eπa2 , so the flux through the netting is −Eπa2 .
E27-9 There is no flux through the sides of the cube. The flux through the top of the cube is
(−58 N/C)(100 m)2 = −5.8×105 N · m2 /C. The flux through the bottom of the cube is
(110 N/C)(100 m)2 = 1.1×106 N · m2 /C.
The total flux is the sum, so the charge contained in the cube is
q = (8.85×10−12 C2 /N · m2 )(5.2×105 N · m2 /C) = 4.60×10−6 C.
E27-10 (a) There is only a flux through the right and left faces. Through the right face
ΦR = (2.0 m2 )ĵ · (3 N/C · m)(1.4 m)ĵ = 8.4 N · m2 /C.
The flux through the left face is zero because y = 0.
30
E27-11 There are eight cubes which can be “wrapped” around the charge. Each cube has three
external faces that are indistinguishable for a total of twenty-four faces, each with the same flux
ΦE . The total flux is q/0 , so the flux through one face is ΦE = q/240 . Note that this is the flux
through faces opposite the charge; for faces which touch the charge the electric field is parallel to
the surface, so the flux would be zero.
E27-12 Use Eq. 27-11,
λ = 2π0 rE = 2π(8.85×10−12 C2 /N · m2 )(1.96 m)(4.52×104 N/C) = 4.93×10−6 C/m.
E27-13 (a) q = σA = (2.0×10−6 C/m2 )π(0.12 m)(0.42 m) = 3.17×10−7 C.
(b) The charge density will be the same! q = σA = (2.0 × 10−6 C/m2 )π(0.08 m)(0.28 m) =
1.41×10−7 C.
E27-14 The electric field from the sheet on the left is of magnitude E l = σ/20 , and points directly
away from the sheet. The magnitude of the electric field from the sheet on the right is the same,
but it points directly away from the sheet on the right.
(a) To the left of the sheets the two fields add since they point in the same direction. This means
~ = −(σ/0 )î.
that the electric field is E
~ = 0.
(b) Between the sheets the two electric fields cancel, so E
(c) To the right of the sheets the two fields add since they point in the same direction. This
~ = (σ/0 )î.
means that the electric field is E
E27-15 The electric field from the plate on the left is of magnitude E l = σ/20 , and points directly
toward the plate. The magnitude of the electric field from the plate on the right is the same, but it
points directly away from the plate on the right.
(a) To the left of the plates the two fields cancel since they point in the opposite directions. This
~ = 0.
means that the electric field is E
(b) Between the plates the two electric fields add since they point in the same direction. This
~ = −(σ/0 )î.
means that the electric field is E
(c) To the right of the plates the two fields cancel since they point in the opposite directions.
~ = 0.
This means that the electric field is E
E27-16 The magnitude of the electric field is E = mg/q. The surface charge density on the plates
is σ = 0 E = 0 mg/q, or
σ=
(8.85×10−12 C2 /N · m2 )(9.11×10−31 kg)(9.81 m/s2 )
= 4.94×10−22 C/m2 .
(1.60×10−19 C)
E27-17 We don’t really need to write an integral, we just need the charge per unit length in the
cylinder to be equal to zero. This means that the positive charge in cylinder must be +3.60nC/m.
This positive charge is uniformly distributed in a circle of radius R = 1.50 cm, so
ρ=
3.60nC/m
3.60nC/m
=
= 5.09µC/m3 .
πR2
π(0.0150 m)2
31
E27-18 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell.
~ field will be perpendicular to the surface, so Gauss’ law will simplify to
The E
I
I
I
~ · dA
~ = E dA = E dA = 4πr2 E.
q enc /0 = E
(a) For point P1 the charge enclosed is q enc = 1.26×10−7 C, so
E=
(1.26×10−7 C)
= 3.38×106 N/C.
· m2 )(1.83×10−2 m)2
4π(8.85×10−12 C2 /N
(b) Inside a conductor E = 0.
E27-19 The proton orbits with a speed v, so the centripetal force on the proton is FC = mv 2 /r.
This centripetal force is from the electrostatic attraction with the sphere; so long as the proton is
outside the sphere the electric field is equivalent to that of a point charge Q (Eq. 27-15),
E=
1 Q
.
4π0 r2
If q is the charge on the proton we can write F = Eq, or
1 Q
mv 2
=q
r
4π0 r2
Solving for Q,
Q =
=
4π0 mv 2 r
,
q
4π(8.85×10−12 C2 /N · m2 )(1.67×10−27 kg)(294×103 m/s)2 (0.0113 m)
,
(1.60×10−19 C)
= −1.13×10−9 C.
E27-20 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell.
~ field will be perpendicular to the surface, so Gauss’ law will simplify to
The E
I
I
I
~ · dA
~ = E dA = E dA = 4πr2 E.
q enc /0 = E
(a) At r = 0.120 m q enc = 4.06×10−8 C. Then
E=
(4.06×10−8 C)
= 2.54×104 N/C.
· m2 )(1.20×10−1 m)2
4π(8.85×10−12 C2 /N
(b) At r = 0.220 m q enc = 5.99×10−8 C. Then
E=
(5.99×10−8 C)
= 1.11×104 N/C.
· m2 )(2.20×10−1 m)2
4π(8.85×10−12 C2 /N
(c) At r = 0.0818 m q enc = 0 C. Then E = 0.
32
E27-21 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical
~ field will be perpendicular to the curved surface and parallel to the end surfaces, so
shell. The E
Gauss’ law will simplify to
I
Z
Z
~ · dA
~ = E dA = E dA = 2πrLE,
q enc /0 = E
where L is the length of the cylinder. Note that σ = q/2πrL represents a surface charge density.
(a) r = 0.0410 m is between the two cylinders. Then
E=
(24.1×10−6 C/m2 )(0.0322 m)
= 2.14×106 N/C.
(8.85×10−12 C2 /N · m2 )(0.0410 m)
It points outward.
(b) r = 0.0820 m is outside the two cylinders. Then
E=
(24.1×10−6 C/m2 )(0.0322 m) + (−18.0×10−6 C/m2 )(0.0618 m)
= −4.64×105 N/C.
(8.85×10−12 C2 /N · m2 )(0.0820 m)
The negative sign is because it is pointing inward.
E27-22 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical
~ field will be perpendicular to the curved surface and parallel to the end surfaces, so
shell. The E
Gauss’ law will simplify to
I
Z
Z
~
~
q enc /0 = E · dA = E dA = E dA = 2πrLE,
where L is the length of the cylinder. The charge enclosed is
Z
q enc = ρdV = ρπL r2 − R2
The electric field is given by
E=
ρ r 2 − R2
ρπL r2 − R2
=
.
2π0 rL
20 r
At the surface,
Es =
ρ (2R)2 − R2
3ρR
=
.
20 2R
40
Solve for r when E is half of this:
3R
r 2 − R2
=
,
8
2r
3rR = 4r2 − 4R2 ,
0 = 4r2 − 3rR − 4R2 .
The solution is r = 1.443R. That’s (2R − 1.443R) = 0.557R beneath the surface.
E27-23 The electric field must do work on the electron to stop it. The electric field is given by
E = σ/20 . The work done is W = F d = Eqd. d is the distance in question, so
d=
20 K
2(8.85×10−12 C2 /N · m2 )(1.15×105 eV)
=
= 0.979 m
σq
(2.08×10−6 C/m2 )e
33
E27-24 Let the spherical Gaussian surface have a radius of R and be centered on the origin.
Choose the orientation of the axis so that the infinite line of charge is along the z axis. The electric
field is then directed radially outward from the z axis with magnitude E = λ/2π0 ρ, where ρ is the
perpendicular distance from the z axis. Now we want to evaluate
I
~ · dA,
~
ΦE = E
over the surface of the sphere. In spherical coordinates, dA = R2 sin θ dθ dφ, ρ = R sin θ, and
~ · dA
~ = EA sin θ. Then
E
I
2λR
λ
sin θR dθ dφ =
.
ΦE =
2π0
0
E27-25 (a) The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical
~ field will be perpendicular to the curved surface and parallel to the end surfaces, so
shell. The E
Gauss’ law will simplify to
Z
Z
I
~ · dA
~ = E dA = E dA = 2πrLE,
q enc /0 = E
where L is the length of the cylinder. Now for the q enc part. If the (uniform) volume charge density
is ρ, then the charge enclosed in the Gaussian cylinder is
Z
Z
q enc = ρdV = ρ dV = ρV = πr2 Lρ.
Combining, πr2 Lρ/0 = E2πrL or E = ρr/20 .
(b) Outside the charged cylinder the charge enclosed in the Gaussian surface is just the charge
in the cylinder. Then
Z
Z
q enc =
dV = ρV = πR2 Lρ.
ρdV = ρ
and
πR2 Lρ/0 = E2πrL,
and then finally
E=
R2 ρ
.
20 r
E27-26 (a) q = 4π(1.22 m)2 (8.13×10−6 C/m2 ) = 1.52×10−4 C.
(b) ΦE = q/0 = (1.52×10−4 C)/(8.85×10−12 C2 /N · m2 ) = 1.72×107 N · m2 /C.
(c) E = σ/0 = (8.13×10−6 C/m2 )/(8.85×10−12 C2 /N · m2 ) = 9.19×105 N/C
E27-27 (a) σ = (2.4×10−6 C)/4π(0.65 m)2 = 4.52×10−7 C/m2 .
(b) E = σ/0 = (4.52×10−7 C/m2 )/(8.85×10−12 C2 /N · m2 ) = 5.11×104 N/C.
E27-28 E = σ/0 = q/4πr2 0 .
E27-29
(a) The near field is given by Eq. 27-12, E = σ/20 , so
E≈
(6.0×10−6 C)/(8.0×10−2 m)2
= 5.3×107 N/C.
2(8.85×10−12 C2 /N · m2 )
(b) Very far from any object a point charge approximation is valid. Then
E=
1 q
1
(6.0×10−6 C)
=
= 60N/C.
4π0 r2
4π(8.85×10−12 C2 /N · m2 ) (30 m)2
34
P27-1
For a spherically symmetric mass distribution choose a spherical Gaussian shell. Then
I
I
I
~ = g dA = g dA = 4πr2 g.
~g · dA
Then
Φg
gr2
=
= −m,
4πG
G
or
Gm
.
r2
The negative sign indicates the direction; ~g point toward the mass center.
g=−
P27-2 (a) The flux through all surfaces except the right and left faces will be zero. Through the
left face,
√
Φl = −Ey A = −b aa2 .
Through the right face,
√
Φr = Ey A = b 2aa2 .
The net flux is then
√
√
Φ = ba5/2 ( 2 − 1) = (8830 N/C · m1/2 )(0.130 m)5/2 ( 2 − 1) = 22.3 N · m2 /C.
(b) The charge enclosed is q = (8.85×10−12 C2 /N · m2 )(22.3 N · m2 /C) = 1.97×10−10 C.
P27-3 The net force on the small sphere is zero; this force is the vector sum of the force of gravity
W , the electric force FE , and the tension T .
T
θ
FE
W
These forces are related by Eq = mg tan θ. We also have E = σ/20 , so
σ
=
=
=
20 mg tan θ
,
q
2(8.85×10−12 C2 /N · m2 )(1.12×10−6 kg)(9.81 m/s2 ) tan(27.4◦ )
,
(19.7×10−9 C)
5.11×10−9 C/m2 .
35
P27-4 The materials are conducting, so all charge will reside on the surfaces. The electric field
inside any conductor is zero. The problem has spherical symmetry, so use a Gaussian surface which
~ field will be perpendicular to the surface, so Gauss’ law will simplify to
is a spherical shell. The E
I
I
I
~
~
q enc /0 = E · dA = E dA = E dA = 4πr2 E.
Consequently, E = q enc /4π0 r2 .
(a) Within the sphere E = 0.
(b) Between the sphere and the shell q enc = q. Then E = q/4π0 r2 .
(c) Within the shell E = 0.
(d) Outside the shell q enc = +q − q = 0. Then E = 0.
(e) Since E = 0 inside the shell, q enc = 0, this requires that −q reside on the inside surface. This
is no charge on the outside surface.
P27-5 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical
~ field will be perpendicular to the curved surface and parallel to the end surfaces, so
shell. The E
Gauss’ law will simplify to
I
Z
Z
~ · dA
~ = E dA = E dA = 2πrLE,
q enc /0 = E
where L is the length of the cylinder. Consequently, E = q enc /2π0 rL.
(a) Outside the conducting shell q enc = +q − 2q = −q. Then E = −q/2π0 rL. The negative sign
indicates that the field is pointing inward toward the axis of the cylinder.
(b) Since E = 0 inside the conducting shell, q enc = 0, which means a charge of −q is on the
inside surface of the shell. The remaining −q must reside on the outside surface of the shell.
(c) In the region between the cylinders q enc = +q. Then E = +q/2π0 rL. The positive sign
indicates that the field is pointing outward from the axis of the cylinder.
P27-6
Subtract Eq. 26-19 from Eq. 26-20. Then
z
σ
√
E=
.
20 z 2 + R2
P27-7 This problem is closely related to Ex. 27-25, except for the part concerning q enc . We’ll
set up the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered on
the axis of the physical cylinder. For Gaussian surfaces of radius r < R, there is no charge enclosed
while for Gaussian surfaces of radius r > R, q enc = λl.
We’ve already worked out the integral
Z
~ · dA
~ = 2πrlE,
E
tube
for the cylinder, and then from Gauss’ law,
Z
q enc = 0
~ · dA
~ = 2π0 rlE.
E
tube
(a) When r < R there is no enclosed charge, so the left hand vanishes and consequently E = 0
inside the physical cylinder.
(b) When r > R there is a charge λl enclosed, so
E=
λ
.
2π0 r
36
P27-8 This problem is closely related to Ex. 27-25, except for the part concerning q enc . We’ll set
up the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered on the
axis of the physical cylinders. For Gaussian surfaces of radius r < a, there is no charge enclosed
while for Gaussian surfaces of radius b > r > a, q enc = λl.
We’ve already worked out the integral
Z
~ · dA
~ = 2πrlE,
E
tube
for the cylinder, and then from Gauss’ law,
Z
q enc = 0
~ · dA
~ = 2π0 rlE.
E
tube
(a) When r < a there is no enclosed charge, so the left hand vanishes and consequently E = 0
inside the inner cylinder.
(b) When b > r > a there is a charge λl enclosed, so
E=
λ
.
2π0 r
P27-9 Uniform circular orbits require a constant net force towards the center, so F = Eq =
λq/2π0 r. The speed of the positron is given by F = mv 2 /r; the kinetic energy is K = mv 2 /2 =
F r/2. Combining,
K
=
=
=
P27-10
λq
,
4π0
(30×10−9 C/m)(1.6×10−19 C)
,
4π((8.85 × 10−12 C2 /N · m2 )
4.31×10−17 J = 270 eV.
λ = 2π0 rE, so
q = 2π(8.85×10−12 C2 /N · m2 )(0.014 m)(0.16 m)(2.9×104 N/C) = 3.6×10−9 C.
P27-11 (a) Put a spherical Gaussian surface inside the shell centered on the point charge. Gauss’
law states
I
~ · dA
~ = q enc .
E
0
Since there is spherical symmetry the electric field is normal to the spherical Gaussian surface, and
~ · dA
~ = E dA, while since
it is everywhere the same on this surface. The dot product simplifies to E
E is a constant on the surface we can pull it out of the integral, and we end up with
I
q
E dA = ,
0
H
where q is the point charge in the center. Now dA = 4πr2 , where r is the radius of the Gaussian
surface, so
q
E=
.
4π0 r2
(b) Repeat the above steps, except put the Gaussian surface outside the conducting shell. Keep
it centered on the charge. Two things are different from the above derivation: (1) r is bigger, and
37
(2) there is an uncharged spherical conducting shell inside the Gaussian surface. Neither change will
affect the surface integral or q enc , so the electric field outside the shell is still
E=
q
,
4π0 r2
(c) This is a subtle question. With all the symmetry here it appears as if the shell has no effect;
the field just looks like a point charge field. If, however, the charge were moved off center the field
inside the shell would become distorted, and we wouldn’t be able to use Gauss’ law to find it. So
the shell does make a difference.
Outside the shell, however, we can’t tell what is going on inside the shell. So the electric field
outside the shell looks like a point charge field originating from the center of the shell regardless of
where inside the shell the point charge is placed!
(d) Yes, q induces surface charges on the shell. There will be a charge −q on the inside surface
and a charge q on the outside surface.
(e) Yes, as there is an electric field from the shell, isn’t there?
(f) No, as the electric field from the outside charge won’t make it through a conducting shell.
The conductor acts as a shield.
(g) No, this is not a contradiction, because the outside charge never experienced any electrostatic
attraction or repulsion from the inside charge. The force is between the shell and the outside charge.
P27-12
Then
The repulsive electrostatic forces must exactly balance the attractive gravitational forces.
√
or m = q/ 4π0 G. Numerically,
1 q2
m2
=
G
,
4π0 r2
r2
(1.60×10−19 C)
= 1.86×10−9 kg.
m= q
2
4π(8.85×10−12 C2 /N · m2 )(6.67×10−11 N · m2 /kg )
P27-13 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell.
~ field will be perpendicular to the surface, so Gauss’ law will simplify to
The E
I
I
I
~
~
q enc /0 = E · dA = E dA = E dA = 4πr2 E.
Consequently, E =
/4π0 r2 .
R q enc
2
q enc = q + 4π ρ r dr, or
q enc = q + 4π
Z
r
Ar dr = q + 2πA(r2 − a2 ).
a
The electric field will be constant if q enc behaves as r2 , which requires q = 2πAa2 , or A = q/2πa2 .
P27-14 (a) The problem has spherical symmetry, so use a Gaussian surface which is a spherical
~ field will be perpendicular to the surface, so Gauss’ law will simplify to
shell. The E
I
I
I
~
~
q enc /0 = E · dA = E dA = E dA = 4πr2 E.
Consequently,R E = q enc /4π0 r2 .
q enc = 4π ρ r2 dr = 4πρr3 /3, so
E = ρr/30
38
~ = ρ~r/30 .
and is directed radially out from the center. Then E
~
~ −E
~ b , where E
~ is the field from part (a) and
(b) The electric field in the hole is given by Eh = E
~ b is the field that would be produced by the matter that would have been in the hole had the hole
E
not been there. Then
~ b = ρ~b/30 ,
E
where ~b is a vector pointing from the center of the hole. Then
~
~ h = ρ~r − ρb = ρ (~r − ~b).
E
30
30
30
~ h = ρ~a/30 .
But ~r − ~b = ~a, so E
P27-15 If a point is an equilibrium point then the electric field at that point should be zero.
If it is a stable point then moving the test charge (assumed positive) a small distance from the
equilibrium point should result in a restoring force directed back toward the equilibrium point. In
other words, there will be a point where the electric field is zero, and around this point there will be
an electric field pointing inward. Applying Gauss’ law to a small surface surrounding our point P ,
we have a net inward flux, so there must be a negative charge inside the surface. But there should
be nothing inside the surface except an empty point P , so we have a contradiction.
P27-16 (a) Follow the example on Page 618. By symmetry E = 0 along the median plane. The
charge enclosed between the median plane and a surface a distance x from the plane is q = ρAx.
Then
E = ρAx/0 A = ρA/0 .
(b) Outside the slab the charge enclosed between the median plane and a surface a distance x
from the plane is is q = ρAd/2, regardless of x. The
E = ρAd/2/0 A = ρd/20 .
P27-17
(a) The total charge is the volume integral over the whole sphere,
Z
Q = ρ dV.
This is actually a three dimensional integral, and dV = A dr, where A = 4πr2 . Then
Z
Q =
ρ dV,
Z R
ρS r =
4πr2 dr,
R
0
4πρS 1 4
=
R ,
R 4
= πρS R3 .
(b) Put a spherical Gaussian surface inside the sphere centered on the center. We can use Gauss’
law here because there is spherical symmetry in the entire problem, both inside and outside the
Gaussian surface. Gauss’ law states
I
~ · dA
~ = q enc .
E
0
39
Since there is spherical symmetry the electric field is normal to the spherical Gaussian surface, and
~ · dA
~ = E dA, while since
it is everywhere the same on this surface. The dot product simplifies to E
E is a constant on the surface we can pull it out of the integral, and we end up with
I
q enc
E dA =
,
0
H
Now dA = 4πr2 , where r is the radius of the Gaussian surface, so
E=
q enc
.
4π0 r2
We aren’t done yet, because the charge enclosed depends on the radius of the Gaussian surface. We
need to do part (a) again, except this time we don’t want to do the whole volume of the sphere, we
only want to go out as far as the Gaussian surface. Then
Z
q enc =
ρ dV,
Z r
ρS r =
4πr2 dr,
R
0
4πρS 1 4
=
r ,
R 4
r4
= πρS .
R
Combine these last two results and
E
=
=
=
πρS r4
,
4π0 r2 R
πρS r2
,
4π0 R
Q r2
.
4π0 R4
In the last line we used the results of part (a) to eliminate ρS from the expression.
P27-18 (a) Inside the conductor E = 0, so a Gaussian surface which is embedded in the conductor
but containing the hole must have a net enclosed charge of zero. The cavity wall must then have a
charge of −3.0 µC.
(b) The net charge on the conductor is +10.0 µC; the charge on the outer surface must then be
+13.0 µC.
P27-19 (a) Inside the shell E = 0, so the net charge inside a Gaussian surface embedded in the
shell must be zero, so the inside surface has a charge −Q.
(b) Still −Q; the outside has nothing to do with the inside.
(c) −(Q + q); see reason (a).
(d) Yes.
40
Throughout this chapter we will use the convention that V (∞) = 0 unless explicitly stated
otherwise. Then the potential in the vicinity of a point charge will be given by Eq. 28-18,
V = q/4π0 r.
E28-1 (a) Let U12 be the potential energy of the interaction between the two “up” quarks. Then
U12 = (8.99×109 N · m2 /C2 )
(2/3)2 e(1.60×10−19 C)
= 4.84×105 eV.
(1.32×10−15 m)
(b) Let U13 be the potential energy of the interaction between an “up” quark and a “down”
quark. Then
U13 = (8.99×109 N · m2 /C2 )
(−1/3)(2/3)e(1.60×10−19 C)
= −2.42×105 eV
(1.32×10−15 m)
Note that U13 = U23 . The total electric potential energy is the sum of these three terms, or zero.
E28-2 There are six interaction terms, one for every charge pair. Number the charges clockwise
from the upper left hand corner. Then
U12
U23
U34
U41
U13
U24
=
=
=
=
−q 2 /4π0 a,
−q 2 /4π0 a,
−q 2 /4π0 a,
−q 2 /4π0 a,
√
= (−q)2 /4π0 ( 2a),
√
= q 2 /4π0 ( 2a).
Add these terms and get
2
q2
q2
U = √ −4
= −0.206
4π0 a
0 a
2
The amount of work required is W = U .
E28-3 (a) We build the electron one part at a time; each part has a charge q = e/3. Moving the
first part from infinity to the location where we want to construct the electron is easy and takes no
work at all. Moving the second part in requires work to change the potential energy to
U12 =
1 q1 q2
,
4π0 r
which is basically Eq. 28-7. The separation r = 2.82 × 10−15 m.
Bringing in the third part requires work against the force of repulsion between the third charge
and both of the other two charges. Potential energy then exists in the form U13 and U23 , where all
three charges are the same, and all three separations are the same. Then U12 = U13 = U12 , so the
total potential energy of the system is
U =3
1 (e/3)2
3
(1.60×10−19 C/3)2
=
= 2.72×10−14 J
4π0 r
4π(8.85×10−12 C2 /N · m2 ) (2.82×10−15 m)
(b) Dividing our answer by the speed of light squared to find the mass,
m=
2.72 × 10−14 J
= 3.02 × 10−31 kg.
(3.00 × 108 m/s)2
41
E28-4 There are three interaction terms, one for every charge pair. Number the charges from the
left; let a = 0.146 m. Then
U12
=
U13
=
U23
=
(25.5×10−9 C)(17.2×10−9 C)
,
4π0 a
(25.5×10−9 C)(−19.2×10−9 C)
,
4π0 (a + x)
(17.2×10−9 C)(−19.2×10−9 C)
.
4π0 x
Add these and set it equal to zero. Then
(25.5)(17.2)
(25.5)(19.2) (17.2)(19.2)
=
+
,
a
a+x
x
which has solution x = 1.405a = 0.205 m.
E28-5 The volume of the nuclear material is 4πa3 /3, where a = 8.0×10−15 m.
√ Upon dividing in
half each part will have a radius r where 4πr3 /3 = 4πa3 /6. Consequently, r = a/ 3 2 = 6.35×10−15 m.
Each fragment will have a charge of +46e.
(a) The force of repulsion is
F =
(46)2 (1.60×10−19 C)2
= 3000 N
4π(8.85×10−12 C2 /N · m2 )[2(6.35×10−15 m)]2
(b) The potential energy is
U=
(46)2 e(1.60×10−19 C)
= 2.4×108 eV
4π(8.85×10−12 C2 /N · m2 )2(6.35×10−15 m)
E28-6 This is a work/kinetic energy problem:
v0 =
s
1
2
2 mv0
= q∆V . Then
2(1.60×10−19 C)(10.3×103 V)
= 6.0×107 m/s.
(9.11×10−31 kg)
E28-7 (a) The energy released is equal to the charges times the potential through which the
charge was moved. Then
∆U = q∆V = (30 C)(1.0 × 109 V) = 3.0 × 1010 J.
(b) Although the problem mentions acceleration, we want to focus on energy. The energy will
change the kinetic energy of the car from 0 to K f = 3.0 × 1010 J. The speed of the car is then
s
r
2K
2(3.0 × 1010 J)
=
= 7100 m/s.
v=
m
(1200 kg)
(c) The energy required to melt ice is given by Q = mL, where L is the latent heat of fusion.
Then
Q
(3.0 × 1010 J)
m=
=
= 90, 100kg.
L
(3.33×105 J/kg)
42
E28-8 (a) ∆U = (1.60×10−19 C)(1.23×109 V) = 1.97×10−10 J.
(b) ∆U = e(1.23×109 V) = 1.23×109 eV.
E28-9 This is an energy conservation problem: 12 mv 2 = q∆V ; ∆V = q/4π0 (1/r1 − 1/r2 ). Combining,
s
q2
1
1
v =
−
,
2π0 m r1
r2
s
(3.1×10−6 C)2
1
1
−
,
=
2π(8.85×10−12 C2 /N · m2 )(18×10−6 kg) (0.90×10−3 m) (2.5×10−3 m)
=
2600 m/s.
E28-10 This is an energy conservation problem:
1
q2
1
m(2v)2 −
= mv 2 .
2
4π0 r
2
Rearrange,
r
=
q2
,
6π0 mv 2
=
(1.60×10−19 C)2
= 1.42×10−9 m.
6π(8.85×10−12 C2 /N · m2 )(9.11×10−31 kg)(3.44×105 m/s)2 )
E28-11 (a) V = (1.60×10−19 C)/4π(8.85×10−12 C2 /N · m2 )(5.29×10−11 m) = 27.2 V.
(b) U = qV = (−e)(27.2 V) = −27.2 eV.
(c) For uniform circular orbits F = mv 2 /r; the force is electrical, or F = e2 /4π0 r2 . Kinetic
energy is K = mv 2 /2 = F r/2, so
K=
e2
(1.60×10−19 C)
=
= 13.6 eV.
−12
8π0 r
8π(8.85×10 C2 /N · m2 )(5.29×10−11 m)
(d) The ionization energy is −(K + U ), or
E ion = −[(13.6 eV) + (−27.2 eV)] = 13.6 eV.
E28-12 (a) The electric potential at A is
1
q1
q2
(−5.0×10−6 C) (2.0×10−6 C)
VA =
+
= (8.99×109 N · m2 /C)
+
= 6.0×104 V.
4π0 r1
r2
(0.15 m)
(0.05 m)
The electric potential at B is
1
q1
q2
(−5.0×10−6 C) (2.0×10−6 C)
9
2
= (8.99×10 N · m /C)
VB =
+
+
= −7.8×105 V.
4π0 r2
r1
(0.05 m)
(0.15 m)
(b) W = q∆V = (3.0×10−6 C)(6.0×104 V − −7.8×105 V) = 2.5 J.
(c) Since work is positive then external work is converted to electrostatic potential energy.
43
E28-13
(a) The magnitude of the electric field would be found from
E=
F
(3.90 × 10−15 N)
=
= 2.44 × 104 N/C.
q
(1.60 × 10−19 C)
(b) The potential difference between the plates is found by evaluating Eq. 28-15,
∆V = −
Z
b
~ · d~s.
E
a
The electric field between two parallel plates is uniform and perpendicular to the plates. Then
~ · d~s = E ds along this path, and since E is uniform,
E
∆V = −
Z
b
~ · d~s = −
E
a
b
Z
E ds = −E
Z
a
b
ds = E∆x,
a
where ∆x is the separation between the plates. Finally, ∆V = (2.44 × 104 N/C)(0.120 m) = 2930 V.
E28-14 ∆V = E∆x, so
∆x =
20
2(8.85×10−12 C2 /N · m2 )
∆V =
(48 V) = 7.1×10−3 m
σ
(0.12×10−6 C/m2 )
E28-15 The electric field around an infinitely long straight wire is given by E = λ/2π0 r. The
potential difference between the inner wire and the outer cylinder is given by
∆V = −
Z
b
(λ/2π0 r) dr = (λ/2π0 ) ln(a/b).
a
The electric field near the surface of the wire is then given by
E=
λ
∆V
(−855 V)
= 1.32×108 V/m.
=
=
2π0 a
a ln(a/b)
(6.70×10−7 m) ln(6.70×10−7 m/1.05×10−2 m)
The electric field near the surface of the cylinder is then given by
E=
∆V
(−855 V)
λ
=
=
= 8.43×103 V/m.
−2
2π0 a
a ln(a/b)
(1.05×10 m) ln(6.70×10−7 m/1.05×10−2 m)
E28-16 ∆V = E∆x = (1.92×105 N/C)(1.50×10−2 m) = 2.88×103 V.
E28-17 (a) This is an energy conservation problem:
K=
1 (2)(79)e2
(2)(79)e(1.60×10−19 C)
= (8.99×109 N · m2 /C)
= 3.2×107 eV
4π0
r
(7.0×10−15 m)
(b) The alpha particles used by Rutherford never came close to hitting the gold nuclei.
E28-18 This is an energy conservation problem: mv 2 /2 = eq/4π0 r, or
s
(1.60×10−19 C)(1.76×10−15 C)
v=
= 2.13×104 m/s
−12
2π(8.85×10 C2 /N · m2 )(1.22×10−2 m)(9.11×10−31 kg)
44
E28-19
(a) We evaluate VA and VB individually, and then find the difference.
VA =
1 q
1
(1.16µC)
=
= 5060 V,
−12
2
2
4π0 r
4π(8.85 × 10
C /N · m ) (2.06 m)
VB =
1 q
1
(1.16µC)
=
= 8910 V,
4π0 r
4π(8.85 × 10−12 C2 /N · m2 ) (1.17 m)
and
The difference is then VA − VB = −3850 V.
(b) The answer is the same, since when concerning ourselves with electric potential we only care
about distances, and not directions.
E28-20 The number of “excess” electrons on each grain is
n=
4π0 rV
4π(8.85×10−12 C2 /N · m)(1.0×10−6 m)(−400 V)
=
= 2.8×105
e
(−1.60×10−19 C)
E28-21 The excess charge on the shuttle is
q = 4π0 rV = 4π(8.85×10−12 C2 /N · m)(10 m)(−1.0 V) = −1.1×10−9 C
E28-22 q = 1.37×105 C, so
V = (8.99×109 N · m2 /C2 )
E28-23
(1.37×105 C)
= 1.93×108 V.
(6.37×106 m)
The ratio of the electric potential to the electric field strength is
1 q
1 q
V
=
/
= r.
E
4π0 r
4π0 r2
In this problem r is the radius of the Earth, so at the surface of the Earth the potential is
V = Er = (100 V/m)(6.38×106 m) = 6.38×108 V.
E28-24 Use Eq. 28-22:
V = (8.99×109 N · m2 /C2 )
(1.47)(3.34×10−30 C · m)
= 1.63×10−5 V.
(52.0×10−9 m)2
E28-25 (a) When finding VA we need to consider the contribution from both the positive and
the negative charge, so
1
−q
VA =
qa +
4π0
a+d
There will be a similar expression for VB ,
1
VB =
4π0
q
−qa +
a+d
45
.
Now to evaluate the difference.
V A − VB
=
=
=
=
−q
1
q
1
qa +
−
−qa +
,
4π0
a+d
4π0
a+d
q
1
1
−
,
2π0 a a + d
q
a+d
a
−
,
2π0 a(a + d) a(a + d)
q
d
.
2π0 a(a + d)
(b) Does it do what we expect when d = 0? I expect it the difference to go to zero as the
two points A and B get closer together. The numerator will go to zero as d gets smaller. The
denominator, however, stays finite, which is a good thing. So yes, Va − VB → 0 as d → 0.
E28-26 (a) Since both charges are positive the electric potential from both charges will be positive.
There will be no finite points where V = 0, since two positives can’t add to zero.
~ will
(b) Between the charges the electric field from each charge√ points toward the√other, so E
vanish when q/x2 = 2q/(d − x)2 . This happens when d − x = 2x, or x = d/(1 + 2).
√
E28-27 The distance from C to either charge is 2d/2 = 1.39×10−2 m.
(a) V at C is
2(2.13×10−6 C)
V = (8.99×109 N · m2 /C2 )
= 2.76×106 V
(1.39×10−2 m)
(b) W = qδV = (1.91×10−6 C)(2.76×106 V) = 5.27 J.
(c) Don’t forget about the potential energy of the original two charges!
U0 = (8.99×109 N · m2 /C2 )
(2.13×10−6 C)2
= 2.08 J
(1.96×10−2 m)
Add this to the answer from part (b) to get 7.35 J.
E28-28 The potential is given by Eq. 28-32;
p
R2 + z 2 − z
R2 + z 2
3R/4
E28-29
at the surface V s = σR/20 , half of this occurs when
= R/2,
= R2 /4 + Rz + z 2 ,
= z.
We can find the linear charge density by dividing the charge by the circumference,
Q
,
2πR
where Q refers to the charge on the ring. The work done to move a charge q from a point x to the
origin will be given by
λ=
W
W
= q∆V,
= q (V (0) − V (x)) ,
1
Q
1
Q
√
√
,
−
= q
4π0 R2 + x2
4π0 R2
qQ
1
1
.
=
−√
4π0 R
R2 + x2
46
Putting in the numbers,
(−5.93×10−12 C)(−9.12×10−9 C)
4π(8.85×10−12 C2 /N · m2 )
1
1
−p
1.48m
(1.48m)2 + (3.07m)2
!
= 1.86×10−10 J.
E28-30 (a) The electric field strength is greatest where the gradient of V is greatest. That is
between d and e.
(b) The least absolute value occurs where the gradient is zero, which is between b and c and
again between e and f .
E28-31 The potential on the positive plate is 2(5.52 V) = 11.0 V; the electric field between the
plates is E = (11.0 V)/(1.48×10−2 m) = 743 V/m.
E28-32 Take the derivative: E = −∂V /∂z.
E28-33
field,
The radial potential gradient is just the magnitude of the radial component of the electric
Er = −
∂V
∂r
Then
∂V
∂r
= −
=
1 q
,
4π0 r2
79(1.60 × 10−19 C)
,
4π(8.85 × 10−12 C2 /N · m2 ) (7.0 × 10−15 m)2
1
= −2.32×1021 V/m.
E28-34 Evaluate ∂V /∂r, and
Ze
E=−
4π0
−1
r
+2 3
2
r
2R
.
E28-35 Ex = −∂V /∂x = −2(1530 V/m2 )x. At the point in question, E = −2(1530 V/m2 )(1.28×
10−2 m) = 39.2 V/m.
E28-36 Draw the wires so that they are perpendicular to the plane of the page; they will then
“come out of” the page. The equipotential surfaces are then lines where they intersect the page,
and they look like
47
E28-37 (a) |VB − VA | = |W/q| = |(3.94 × 10−19 J)/(1.60 × 10−19 C)| = 2.46 V. The electric field
did work on the electron, so the electron was moving from a region of low potential to a region of
high potential; or VB > VA . Consequently, VB − VA = 2.46 V.
(b) VC is at the same potential as VB (both points are on the same equipotential line), so
VC − VA = VB − VA = 2.46 V.
(c) VC is at the same potential as VB (both points are on the same equipotential line), so
VC − VB = 0 V.
E28-38 (a) For point charges r = q/4π0 V , so
r = (8.99×109 N · m2 /C2 )(1.5×10−8 C)/(30 V) = 4.5 m
(b) No, since V ∝ 1/r.
E28-39 The dotted lines are equipotential lines, the solid arrows are electric field lines. Note that
there are twice as many electric field lines from the larger charge!
48
E28-40 The dotted lines are equipotential lines, the solid arrows are electric field lines.
E28-41 This can easily be done with a spreadsheet. The following is a sketch; the electric field is
the bold curve, the potential is the thin curve.
49
sphere radius
r
E28-42 Originally V = q/4π0 r, where r is the radius of the smaller sphere.
(a) Connecting the spheres will bring them to the same potential, or V1 = V2 .
(b) q1 + q2 = q; V1 = q1 /4π0 r and V2 = q2 /4π0 2r; combining all of the above q2 = 2q1 and
q1 = q/3 and q2 = 2q/3.
E28-43 (a) q = 4πR2 σ, so V = q/4π0 R = σR/0 , or
V = (−1.60×10−19 C/m2 )(6.37×106 m)/(8.85×10−12 C2 /N · m2 ) = 0.115 V
(b) Pretend the Earth is a conductor, then E = σ/epsilon0 , so
E = (−1.60×10−19 C/m2 )/(8.85×10−12 C2 /N · m2 ) = 1.81×10−8 V/m.
E28-44 V = q/4π0 R, so
V = (8.99×109 N · m2 /C2 )(15×10−9 C)/(0.16 m) = 850 V.
E28-45 (a) q = 4π0 RV = 4π(8.85×10−12 C2 /N · m2 )(0.152 m)(215 V) = 3.63×10−9 C
(b) σ = q/4πR2 = (3.63×10−9 C)/4π(0.152 m)2 = 1.25×10−8 C/m2 .
E28-46 The dotted lines are equipotential lines, the solid arrows are electric field lines.
50
E28-47 (a) The total charge (Q = 57.2nC) will be divided up between the two spheres so that
they are at the same potential. If q1 is the charge on one sphere, then q2 = Q − q1 is the charge on
the other. Consequently
V1
1 q1
4π0 r1
q 1 r2
q1
= V2 ,
1 Q − q1
=
,
4π0 r2
= (Q − q1 )r1 ,
Qr2
=
.
r2 + r1
Putting in the numbers, we find
q1 =
Qr1
(57.2 nC)(12.2 cm)
=
= 38.6 nC,
r2 + r1
(5.88 cm) + (12.2 cm)
and q2 = Q − q1 = (57.2 nC) − (38.6 nC) = 18.6 nC.
(b) The potential on each sphere should be the same, so we only need to solve one. Then
1 q1
1
(38.6 nC)
=
= 2850 V.
−12
2
2
4π0 r1
4π(8.85 × 10
C /N · m ) (12.2 cm)
E28-48 (a) V = (8.99×109 N · m2 /C2 )(31.5×10−9 C)/(0.162 m) = 1.75×103 V.
(b) V = q/4π0 r, so r = q/4π0 V , and then
r = (8.99×109 N · m2 /C2 )(31.5×10−9 C)/(1.20×103 V) = 0.236 m.
That is (0.236 m) − (0.162 m) = 0.074 m above the surface.
51
E28-49
(a) Apply the point charge formula, but solve for the charge. Then
1 q
4π0 r
q
q
= V,
=
=
4π0 rV,
4π(8.85 × 10−12 C2 /N · m2 )(1 m)(106 V) = 0.11 mC.
Now that’s a fairly small charge. But if the radius were decreased by a factor of 100, so would
the charge (1.10 µC). Consequently, smaller metal balls can be raised to higher potentials with less
charge.
(b) The electric field near the surface of the ball is a function of the surface charge density,
E = σ/0 . But surface charge density depends on the area, and varies as r−2 . For a given potential,
the electric field near the surface would then be given by
E=
σ
q
V
=
= .
2
0
4π0 r
r
Note that the electric field grows as the ball gets smaller. This means that the break down field is
more likely to be exceeded with a low voltage small ball; you’ll get sparking.
E28-50 A “Volt” is a Joule per Coulomb. The power required by the drive belt is the product
(3.41×106 V)(2.83×10−3 C/s) = 9650 W.
P28-1 (a) According to Newtonian mechanics we want K = 12 mv 2 to be equal to W = q∆V
which means
mv 2
(0.511 MeV)
∆V =
=
= 256 kV.
2q
2e
mc2 is the rest mass energy of an electron.
(b) Let’s do some rearranging first.
2
K
= mc
K
mc2
"
1
p
−1 ,
1 − β2
1
p
− 1,
1 − β2
K
1
,
+1 = p
mc2
1 − β2
p
1
=
1 − β2,
K
mc2 + 1
1
2
2 = 1 − β ,
K
mc2 + 1
=
and finally,
β=
s
1−
1
K
mc2
2
+1
Putting in the numbers,
v
u
u1 −
t
(256
(511
1
= 0.746,
keV) + 1 2
keV)
so v = 0.746c.
52
#
P28-2 (a) The potential of the hollow sphere is V = q/4π0 r. The work required to increase the
charge by an amount dq is dW = V /, dq. Integrating,
Z e
e2
q
W =
dq =
.
8π0 r
0 4π0 r
This corresponds to an electric potential energy of
W =
e(1.60×10−19 C)
= 2.55×105 eV = 4.08×10−14 J.
· m2 )(2.82×10−15 m)
8π(8.85×10−12 C2 /N
(b) This would be a mass of m = (4.08×10−14 J)/(3.00×108 m/s)2 = 4.53×10−31 kg.
P28-3
The negative charge is held in orbit by electrostatic attraction, or
mv 2
qQ
=
.
r
4π0 r2
The kinetic energy of the charge is
K=
1
qQ
mv 2 =
.
2
8π0 r
The electrostatic potential energy is
U =−
qQ
,
4π0 r
E=−
qQ
.
8π0 r
so the total energy is
The work required to change orbit is then
qQ
W =
8π0
P28-4
1
1
−
r1
r2
.
R
(a) V = − E dr, so
V =−
Z
0
r
qr
qr2
dr = −
.
3
4π0 R
8π0 R3
(b) ∆V = q/8π0 R.
(c) If instead of V = 0 at r = 0 as was done in part (a) we take V = 0 at r = ∞, then
V = q/4π0 R on the surface of the sphere. The new expression for the potential inside the sphere
will look like V = V 0 + Vs , where V 0 is the answer from part (a) and Vs is a constant so that the
surface potential is correct. Then
Vs =
and then
V =−
q
qR2
3qR2
+
=
,
4π0 R 8π0 R3
8π0 R3
qr2
3qR2
q(3R2 − r2 )
+
=
.
3
3
8π0 R
8π0 R
8π0 R3
53
P28-5 The total electric potential energy of the system is the sum of the three interaction pairs.
One of these pairs does not change during the process, so it can be ignored when finding the change
in potential energy. The change in electrical potential energy is then
q2
q2
q2
1
1
∆U = 2
−2
=
−
.
4π0 rf
4π0 ri
2π0 rf
ri
In this case ri = 1.72 m, while rf = 0.86 m. The change in potential energy is then
1
1
∆U = 2(8.99×109 N · m2 /C2 )(0.122 C)2
−
= 1.56×108 J
(0.86 m) (1.72 m)
The time required is
t = (1.56×108 )/(831 W) = 1.87×105 s = 2.17 days.
P28-6
(a) Apply conservation of energy:
K=
qQ
qQ
, or d =
,
4π0 d
4π0 K
where d is the distance of closest approach.
(b) Apply conservation of energy:
1
qQ
+ mv 2 ,
4π0 (2d) 2
p
so, combining with the results in part (a), v = K/m.
K=
P28-7
(a) First apply Eq. 28-18, but solve for r. Then
r=
(32.0 × 10−12 C)
q
=
= 562 µm.
4π0 V
4π(8.85 × 10−12 C2 /N · m2 )(512 V)
(b) If two such drops join together the charge doubles,
√ and the volume of water doubles, but the
radius of the new drop only increases by a factor of 3 2 = 1.26 because volume is proportional to
the radius cubed.
The potential on the surface of the new drop will be
V new
=
=
=
1 q new
,
4π0 rnew
1 2q old
√
,
4π0 3 2 rold
1 q old
(2)2/3
= (2)2/3 V old .
4π0 rold
The new potential is 813 V.
P28-8 (a) The work done is W = −F z = −Eqz = −qσz/20 .
(b) Since W = q∆V , ∆V = −σz/20 , so
V = V0 − (σ/20 )z.
54
P28-9
(a) The potential at any point will be the sum of the contribution from each charge,
V =
1 q1
1 q2
+
,
4π0 r1
4π0 r2
where r1 is the distance the point in question from q1 and r2 is the distance the point in question
from q2 . Pick a point, call it (x, y). Since q1 is at the origin,
p
r1 = x2 + y 2 .
Since q2 is at (d, 0), where d = 9.60 nm,
r2 =
p
(x − d)2 + y 2 .
Define the “Stanley Number” as S = 4π0 V . Equipotential surfaces are also equi-Stanley surfaces.
In particular, when V = 0, so does S. We can then write the potential expression in a sightly
simplified form
q1
q2
S=
+ .
r1
r2
If S = 0 we can rearrange and square this expression.
q1
r1
r12
q12
x2 + y 2
q12
= −
=
=
q2
,
r2
r22
,
q22
(x − d)2 + y 2
,
q22
Let α = q2 /q1 , then we can write
α2 x2 + y 2
α2 x2 + α2 y 2
(α − 1)x + 2xd + (α2 − 1)y 2
2
2
=
(x − d)2 + y 2 ,
= x2 − 2xd + d2 + y 2 ,
= d2 .
We complete the square for the (α2 − 1)x2 + 2xd term by adding d2 /(α2 − 1) to both sides of the
equation. Then
"
#
2
d
1
2
2
2
(α − 1) x + 2
+y =d 1+ 2
.
α −1
α −1
The center of the circle is at
−
α2
d
(9.60 nm)
=
= −5.4 nm.
−1
(−10/6)2 − 1
(b) The radius of the circle is
v u
u
1 + α21−1
t
d2
,
α2 − 1
which can be simplified to
d
α
|(−10/6)|
= (9.6 nm)
= 9.00 nm.
α2 − 1
(−10/6)2 − 1
(c) No.
55
P28-10 An annulus is composed of differential rings of varying radii r and width dr; the charge
on any ring is the product of the area of the ring, dA = 2πr dr, and the surface charge density, or
dq = σ dA =
k
2πk
2πr dr = 2 dr.
3
r
r
The potential at the center can be found by adding up the contributions from each ring. Since we
are at the center, the contributions will each be dV = dq/4π0 r. Then
Z b
k
1
1
k b2 − a2
k dr
V =
=
−
.
=
.
3
40 a2
b2
40 b2 a2
a 20 r
The total charge on the annulus is
Z b
2πk
1 1
b−a
Q=
dr
=
2πk
−
= 2πk
.
2
r
a
b
ba
a
Combining,
V =
Q a+b
.
8π0 ab
P28-11
Add the three contributions, and then do a series expansion for d r.
q
−1
1
1
V =
+ +
,
4π0 r + d r
r−d
q
−1
1
=
+1+
,
4π0 r 1 + d/r
1 − d/r
d
d
q
≈
−1 + + 1 + 1 +
,
4π0 r
r
r
q
2d
≈
1+
.
4π0 r
r
P28-12
(a) Add the contributions from each differential charge: dq = λ dy. Then
Z y+L
λ
λ
y+L
V =
dy =
ln
.
4π0 y
4π0
y
y
(b) Take the derivative:
Ey = −
∂V
λ
y −L
λ
L
=−
=
.
∂y
4π0 y + L y 2
4π0 y(y + L)
(c) By symmetry it must be zero, since the system is invariant under rotations about the axis
of the rod. Note that we can’t determine E⊥ from derivatives because we don’t have the functional
form of V for points off-axis!
P28-13 (a) We follow the work done in Section 28-6 for a uniform line of charge, starting with
Eq. 28-26,
dV
=
dV
=
1
λ dx
p
,
4π0 x2 + y 2
Z L
1
kx dx
p
,
4π0 0
x2 + y 2
56
=
=
L
k p 2
x + y2 ,
4π0
0
k p 2
L + y2 − y .
4π0
(b) The y component of the electric field can be found from
Ey = −
∂V
,
∂y
which (using a computer-aided math program) is
k
Ey =
4π0
1− p
y
!
L2 + y 2
.
(c) We could find Ex if we knew the x variation of V . But we don’t; we only found the values of
V along a fixed value of x.
(d) We want to find y such that the ratio
k
k p 2
L + y2 − y /
(L)
4π0
4π0
p
is one-half. Simplifying, L2 + y 2 − y = L/2, which can be written as
L2 + y 2 = L2 /4 + Ly + y 2 ,
or 3L2 /4 = Ly, with solution y = 3L/4.
P28-14 The spheres are small compared to the separation distance. Assuming only one sphere at
a potential of 1500 V, the charge would be
q = 4π0 rV = 4π(8.85×10−12 C2 /N · m)(0.150 m)(1500 V) = 2.50×10−8 C.
The potential from the sphere at a distance of 10.0 m would be
V = (1500 V)
(0.150 m)
= 22.5 V.
(10.0 m)
This is small compared to 1500 V, so we will treat it as a perturbation. This means that we can
assume that the spheres have charges of
q = 4π0 rV = 4π(8.85×10−12 C2 /N · m)(0.150 m)(1500 V + 22.5 V) = 2.54×10−8 C.
P28-15 Calculating the fraction of excess electrons is the same as calculating the fraction of
excess charge, so we’ll skip counting the electrons. This problem is effectively the same as Exercise
28-47; we have a total charge that is divided between two unequal size spheres which are at the same
potential on the surface. Using the result from that exercise we have
q1 =
Qr1
,
r2 + r1
where Q = −6.2 nC is the total charge available, and q1 is the charge left on the sphere. r1 is the
radius of the small ball, r2 is the radius of Earth. Since the fraction of charge remaining is q1 /Q,
we can write
q1
r1
r1
=
≈
= 2.0 × 10−8 .
Q
r 2 + r1
r2
57
P28-16
The positive charge on the sphere would be
q = 4π0 rV = 4π(8.85×10−12 C2 /N · m2 )(1.08×10−2 m)(1000 V) = 1.20×10−9 C.
The number of decays required to build up this charge is
n = 2(1.20×10−9 C)/(1.60×10−19 C) = 1.50×1010 .
The extra factor of two is because only half of the decays result in an increase in charge. The time
required is
t = (1.50×1010 )/(3.70×108 s−1 ) = 40.6 s.
P28-17 (a) None.
(b) None.
(c) None.
(d) None.
(e) No.
P28-18 (a) Outside of an isolated charged spherical object E = q/4π0 r2 and V = q/4π0 r.
Then E = V /r. Consequently, the sphere must have a radius larger than r = (9.15×106 V)/(100×
106 V/m) = 9.15×10−2 m.
(b) The power required is (320×10−6 C/s)(9.15×106 V) = 2930 W.
(c) σwv = (320×10−6 C/s), so
σ=
(320×10−6 C/s)
= 2.00×10−5 C/m2 .
(0.485 m)(33.0 m/s)
58
E29-1 (a) The charge which flows through a cross sectional surface area in a time t is given by
q = it, where i is the current. For this exercise we have
q = (4.82 A)(4.60 × 60 s) = 1330 C
as the charge which passes through a cross section of this resistor.
(b) The number of electrons is given by (1330 C)/(1.60 × 10−19 C) = 8.31 × 1021 electrons.
E29-2 Q/t = (200×10−6 A/s)(60s/min)/(1.60×10−19 C) = 7.5×1016 electrons per minute.
E29-3 (a) j = nqv = (2.10×1014 /m3 )2(1.60×10−19 C)(1.40×105 m/s) = 9.41 A/m2 . Since the ions
have positive charge then the current density is in the same direction as the velocity.
(b) We need an area to calculate the current.
E29-4 (a) j = i/A = (123×10−12 A)/π(1.23×10−3 m)2 = 2.59×10−5 A/m2 .
(b) v d = j/ne = (2.59×10−5 A/m2 )/(8.49×1028 /m3 )(1.60×10−19 C) = 1.91×10−15 m/s.
E29-5
The current rating of a fuse of cross sectional area A would be
imax = (440 A/cm2 )A,
and if the fuse wire is cylindrical A = πd2 /4. Then
s
4 (0.552 A)
d=
= 4.00×10−2 cm.
π (440 A/m2 )
E29-6 Current density is current divided by cross section of wire, so the graph would look like:
I (A/mil^2 x10^−3)
4
3
2
1
50
100
59
150
200
d(mils)
E29-7 The current is in the direction of the motion of the positive charges. The magnitude of the
current is
i = (3.1×1018 /s + 1.1×1018 /s)(1.60×10−19 C) = 0.672 A.
E29-8 (a) The total current is
i = (3.50×1015 /s + 2.25×1015 /s)(1.60×10−19 C) = 9.20×10−4 A.
(b) The current density is
j = (9.20×10−4 A)/π(0.165×10−3 m)2 = 1.08×104 A/m2 .
E29-9 (a) j = (8.70×106 /m3 )(1.60×10−19 C)(470×103 m/s) = 6.54×10−7 A/m2 .
(b) i = (6.54×10−7 A/m2 )π(6.37×106 m)2 = 8.34×107 A.
E29-10 i = σwv, so
σ = (95.0×10−6 A)/(0.520 m)(28.0 m/s) = 6.52×10−6 C/m2 .
E29-11
The drift velocity is given by Eq. 29-6,
vd =
j
i
(115 A)
=
=
= 2.71×10−4 m/s.
ne
Ane
(31.2×10−6 m2 )(8.49×1028 /m3 )(1.60×10−19 C)
The time it takes for the electrons to get to the starter motor is
t=
x
(0.855 m)
=
= 3.26×103 s.
v
(2.71×10−4 m/s)
That’s about 54 minutes.
E29-12 ∆V = iR = (50×10−3 A)(1800 Ω) = 90 V.
E29-13
The resistance of an object with constant cross section is given by Eq. 29-13,
R=ρ
L
(11, 000 m)
= (3.0 × 10−7 Ω · m)
= 0.59 Ω.
A
(0.0056 m2 )
E29-14 The slope is approximately [(8.2 − 1.7)/1000]µΩ · cm/◦ C, so
α=
1
6.5×10−3 µΩ · cm/◦ C ≈ 4×10−3 /C◦
1.7µΩ · cm
E29-15 (a) i = ∆V /R = (23 V)/(15×10−3 Ω) = 1500 A.
(b) j = i/A = (1500 A)/π(3.0×10−3 m)2 = 5.3×107 A/m2 .
(c) ρ = RA/L = (15×10−3 Ω)π(3.0×10−3 m)2 /(4.0 m) = 1.1×10−7 Ω · m. The material is possibly
platinum.
E29-16 Use the equation from Exercise 29-17. ∆R = 8 Ω; then
∆T = (8 Ω)/(50 Ω)(4.3×10−3 /C◦ ) = 37 C◦ .
The final temperature is then 57◦ C.
60
E29-17
Start with Eq. 29-16,
ρ − ρ0 = ρ0 αav (T − T0 ),
and multiply through by L/A,
L
L
(ρ − ρ0 ) = ρ0 αav (T − T0 ),
A
A
to get
R − R0 = R0 αav (T − T0 ).
E29-18 The wire has a length L = (250)2π(0.122 m) = 192 m. The diameter is 0.129 inches; the
cross sectional area is then
A = π(0.129 × 0.0254 m)2 /4 = 8.43×10−6 m2 .
The resistance is
R = ρL/A = (1.69×10−8 Ω · m)(192 m)/(8.43×10−6 m2 ) = 0.385 Ω.
E29-19
If the length of each conductor is L and has resistivity ρ, then
RA = ρ
and
RB = ρ
L
4L
=ρ
πD2 /4
πD2
4L
L
=ρ
.
(π4D2 /4 − πD2 /4)
3πD2
The ratio of the resistances is then
RA
= 3.
RB
E29-20 R = R, so ρ1 L1 /π(d1 /2)2 = ρ2 L2 /π(d2 /2)2 . Simplifying, ρ1 /d21 = ρ2 /d22 . Then
p
d2 = (1.19×10−3 m) (9.68×10−8 Ω · m)/(1.69×10−8 Ω · m) = 2.85×10−3 m.
E29-21 (a) (750×10−3 A)/(125) = 6.00×10−3 A.
(b) ∆V = iR = (6.00×10−3 A)(2.65×10−6 Ω) = 1.59×10−8 V.
(c) R = ∆V /i = (1.59×10−8 V)/(750×10−3 A) = 2.12×10−8 Ω.
E29-22 Since ∆V = iR, then if ∆V and i are the same, then R must be the same.
(a) Since R = R, ρ1 L1 /πr12 = ρ2 L2 /πr22 , or ρ1 /r12 = ρ2 /r22 . Then
p
riron /rcopper = (9.68×10−8 Ω · m)(1.69×10−8 Ω · m) = 2.39.
(b) Start with the definition of current density:
j=
i
∆V
∆V
=
=
.
A
RA
ρL
Since ∆V and L is the same, but ρ is different, then the current densities will be different.
61
~ If the wire is long and thin, then the
E29-23 Conductivity is given by Eq. 29-8, ~j = σ E.
magnitude of the electric field in the wire will be given by
E ≈ ∆V /L = (115 V)/(9.66 m) = 11.9 V/m.
We can now find the conductivity,
σ=
(1.42×104 A/m2 )
j
=
= 1.19×103 (Ω · m)−1 .
E
(11.9 V/m)
E29-24 (a) v d = j/en = σE/en. Then
v d = (2.70×10−14 /Ω · m)(120 V/m)/(1.60×10−19 C)(620×106 /m3 + 550×106 /m3 ) = 1.73×10−2 m/s.
(b) j = σE = (2.70×10−14 /Ω · m)(120 V/m) = 3.24×10−14 A/m2 .
E29-25 (a) R/L = ρ/A, so j = i/A = (R/L)i/ρ. For copper,
j = (0.152×10−3 Ω/m)(62.3 A)/(1.69×10−8 Ω · m) = 5.60×105 A/m2 ;
for aluminum,
j = (0.152×10−3 Ω/m)(62.3 A)/(2.75×10−8 Ω · m) = 3.44×105 A/m2 .
(b) A = ρL/R; if δ is density, then m = δlA = lδρ/(R/L). For copper,
m = (1.0 m)(8960 kg/m3 )(1.69×10−8 Ω · m)/(0.152×10−3 Ω/m) = 0.996 kg;
for aluminum,
m = (1.0 m)(2700 kg/m3 )(2.75×10−8 Ω · m)/(0.152×10−3 Ω/m) = 0.488 kg.
E29-26 The resistance for potential differences less than 1.5 V are beyond the scale.
R (Kilo−ohms)
10
8
6
4
2
1
2
62
3
4
V(Volts)
E29-27
(a) The resistance is defined as
R=
∆V
(3.55 × 106 V/A2 )i2
=
= (3.55 × 106 V/A2 )i.
i
i
When i = 2.40 mA the resistance would be
R = (3.55 × 106 V/A2 )(2.40 × 10−3 A) = 8.52 kΩ.
(b) Invert the above expression, and
i = R/(3.55 × 106 V/A2 ) = (16.0 Ω)/(3.55 × 106 V/A2 ) = 4.51 µA.
E29-28 First, n = 3(6.02×1023 )(2700 kg/m3 )(27.0×10−3 kg) = 1.81×1029 /m3 . Then
τ=
m
(9.11×10−31 kg)
=
= 7.15×10−15 s.
2
29
3
ne ρ
(1.81×10 /m )(1.60×10−19 C)2 (2.75×10−8 Ω · m)
E29-29 (a) E = E0 /κe = q/4π0 κe R2 , so
E=
(1.00×10−6 C)
=
4π(8.85×10−12 C2 /N · m2 )(4.7)(0.10 m)2
(b) E = E0 = q/4π0 R2 , so
E=
(1.00×10−6 C)
=
4π(8.85×10−12 C2 /N · m2 )(0.10 m)2
(c) σ ind = 0 (E0 − E) = q(1 − 1/κe )/4πR2 . Then
(1.00×10−6 C)
1
1−
= 6.23×10−6 C/m2 .
σ ind =
4π(0.10 m)2
(4.7)
E29-30 Midway between the charges E = q/π0 d, so
q = π(8.85×10−12 C2 /N · m2 )(0.10 m)(3×106 V/m) = 8.3×10−6 C.
E29-31 (a) At the surface of a conductor of radius R with charge Q the magnitude of the electric
field is given by
1
E=
QR2 ,
4π0
while the potential (assuming V = 0 at infinity) is given by
V =
1
QR.
4π0
The ratio is V /E = R.
The potential on the sphere that would result in “sparking” is
V = ER = (3×106 N/C)R.
(b) It is “easier” to get a spark off of a sphere with a smaller radius, because any potential on
the sphere will result in a larger electric field.
(c) The points of a lighting rod are like small hemispheres; the electric field will be large near
these points so that this will be the likely place for sparks to form and lightning bolts to strike.
63
P29-1 If there is more current flowing into the sphere than is flowing out then there must be a
change in the net charge on the sphere. The net current is the difference, or 2 µA. The potential on
the surface of the sphere will be given by the point-charge expression,
V =
1 q
,
4π0 r
and the charge will be related to the current by q = it. Combining,
V =
or
t=
1 it
,
4π0 r
4π0 V r
4π(8.85 × 10−12 C2 /N · m2 )(980 V)(0.13 m)
=
= 7.1 ms.
i
(2 µA)
P29-2 The net current density is in the direction of the positive charges, which is to the east. There
are two electrons for every alpha particle, and each alpha particle has a charge equal in magnitude
to two electrons. The current density is then
j
P29-3
= q e ne v e + qα + nα vα ,
= (−1.6×10−19 C)(5.6×1021 /m3 )(−88 m/s) + (3.2×10−19 C)(2.8×1021 /m3 )(25 m/s),
= 1.0×105 C/m2 .
(a) The resistance of the segment of the wire is
R = ρL/A = (1.69×10−8 Ω · m)(4.0×10−2 m)/π(2.6×10−3 m)2 = 3.18×10−5 Ω.
The potential difference across the segment is
∆V = iR = (12 A)(3.18×10−5 Ω) = 3.8×10−4 V.
(b) The tail is negative.
(c) The drift speed is v = j/en = i/Aen, so
v = (12 A)/π(2.6×10−3 m)2 (1.6×10−19 C)(8.49×1028 /m3 ) = 4.16×10−5 m/s.
The electrons will move 1 cm in (1.0×10−2 m)/(4.16×10−5 m/s) = 240 s.
P29-4 (a) N = it/q = (250×10−9 A)(2.9 s)/(3.2×10−19 C) = p
2.27×1012 .
(b) The speed of the particles in the beam is given by v = 2K/m, so
p
v = 2(22.4 MeV)/4(932 MeV/c2 ) = 0.110c.
It takes (0.180 m)/(0.110)(3.00×108 m/s) = 5.45×10−9 s for the beam to travel 18.0 cm. The number
of charges is then
N = it/q = (250×10−9 A)(5.45×10−9 s)/(3.2×10−19 C) = 4260.
(c) W = q∆V , so ∆V = (22.4 MeV)/2e = 11.2 MV.
64
P29-5
(a) The time it takes to complete one turn is t = (250 m)/c. The total charge is
q = it = (30.0 A)(950 m)/(3.00×108 m/s) = 9.50×10−5 C.
(b) The number of charges is N = q/e, the total energy absorbed by the block is then
∆U = (28.0×109 eV)(9.50×10−5 C)/e = 2.66×106 J.
This will raise the temperature of the block by
∆T = ∆U/mC = (2.66×106 J)/(43.5 kg)(385J/kgC◦ ) = 159 C◦ .
P29-6
(a) i =
R
R
j dA = 2π jr dr;
Z
i = 2π −0R j0 (1 − r/R)r dr = 2πj0 (R2 /2 − R3 /3R) = πj0 R2 /6.
(b) Integrate, again:
i = 2π
P29-7
Z
−0R j0 (r/R)r dr = 2πj0 (R3 /3R) = πj0 R2 /3.
(a) Solve 2ρ0 = ρ0 [1 + α(T − 20◦ C)], or
T = 20◦ C + 1/(4.3×10−3 /C◦ ) = 250◦ C.
(b) Yes, ignoring changes in the physical dimensions of the resistor.
P29-8
The resistance when on is (2.90 V)/(0.310 A) = 9.35 Ω. The temperature is given by
T = 20◦ C + (9.35 Ω − 1.12 Ω)/(1.12 Ω)(4.5×10−3 /◦ C) = 1650◦ C.
P29-9 Originally we have a resistance R1 made out of a wire of length l1 and cross sectional area
A1 . The volume of this wire is V1 = A1 l1 . When the wire is drawn out to the new length we have
l2 = 3l1 , but the volume of the wire should be constant so
A2 l2 = A1 l1 ,
A2 (3l1 ) = A1 l1 ,
A2 = A1 /3.
The original resistance is
R1 = ρ
The new resistance is
R2 = ρ
l1
.
A1
l2
3l1
=ρ
= 9R1 ,
A2
A1 /3
or R2 = 54 Ω.
P29-10 (a) i = (35.8 V)/(935 Ω) = 3.83×10−2 A.
(b) j = i/A = (3.83×10−2 A)/(3.50×10−4 m2 ) = 109 A/m2 .
(c) v = (109 A/m2 )/(1.6×10−19 C)(5.33×1022 /m3 ) = 1.28×10−2 m/s.
(d) E = (35.8 V)/(0.158 m) = 227 V/m.
65
P29-11 (a) ρ = (1.09×10−3 Ω)π(5.5×10−3 m)2 /4(1.6 m) = 1.62×10−8 Ω · m. This is possibly silver.
(b) R = (1.62×10−8 Ω · m)(1.35×10−3 m)4/π(2.14×10−2 m)2 = 6.08×10−8 Ω.
P29-12 (a) ∆L/L = 1.7×10−5 for a temperature change of 1.0 C◦ . Area changes are twice this,
or ∆A/A = 3.4×10−5 .
Take the differential of RA = ρL: R dA+A dR = ρ dL+L dρ, or dR = ρ dL/A+L dρ/A−R dA/A.
For finite changes this can be written as
∆R
∆L ∆ρ ∆A
=
+
−
.
R
L
ρ
A
∆ρ/ρ = 4.3×10−3 . Since this term is so much larger than the other two it is the only significant
effect.
P29-13
We will use the results of Exercise 29-17,
R − R0 = R0 αav (T − T0 ).
To save on subscripts we will drop the “av” notation, and just specify whether it is carbon “c” or
iron “i”.
The disks will be effectively in series, so we will add the resistances to get the total. Looking
only at one disk pair, we have
Rc + Ri
= R0,c (αc (T − T0 ) + 1) + R0,i (αi (T − T0 ) + 1) ,
= R0,c + R0,i + (R0,c αc + R0,i αi ) (T − T0 ).
This last equation will only be constant if the coefficient for the term (T − T0 ) vanishes. Then
R0,c αc + R0,i αi = 0,
but R = ρL/A, and the disks have the same cross sectional area, so
Lc ρc αc + Li ρi αi = 0,
or
Lc
ρi α i
(9.68×10−8 Ω · m)(6.5×10−3 /C◦ )
=−
=−
= 0.036.
Li
ρc α c
(3500×10−8 Ω · m)(−0.50×10−3 /C◦ )
P29-14 The current entering the cone is i. The current density as a function of distance x from
the left end is then
i
j=
.
π[a + x(b − a)/L]2
The electric field is given by E = ρj. The potential difference between the ends is then
∆V =
Z
L
E dx =
0
Z
0
L
iρ
iρL
dx =
2
π[a + x(b − a)/L]
πab
The resistance is R = ∆V /i = ρL/πab.
66
P29-15
The current is found from Eq. 29-5,
Z
~
i = ~j · dA,
where the region of integration is over a spherical shell concentric with the two conducting shells
but between them. The current density is given by Eq. 29-10,
~j = E/ρ,
~
and we will have an electric field which is perpendicular to the spherical shell. Consequently,
Z
Z
1
~ · dA
~ = 1 E dA
i=
E
ρ
ρ
By symmetry we expect the electric field to have the same magnitude anywhere on a spherical shell
which is concentric with the two conducting shells, so we can bring it out of the integral sign, and
then
Z
1
4πr2 E
i = E dA =
,
ρ
ρ
where E is the magnitude of the electric field on the shell, which has radius r such that b > r > a.
The above expression can be inverted to give the electric field as a function of radial distance,
since the current is a constant in the above expression. Then E = iρ/4πr2 The potential is given by
Z a
~ · d~s,
∆V = −
E
b
we will integrate along a radial line, which is parallel to the electric field, so
Z a
∆V = −
E dr,
Zb a
iρ
dr,
= −
2
4πr
b
Z
iρ a dr
= −
,
4π b r
iρ 1 1
=
−
.
4π a b
We divide this expression by the current to get the resistance. Then
ρ 1 1
R=
−
4π a b
P29-16 Since τ =√λ/v d , √ρ ∝ v d . For an ideal gas the kinetic energy is proportional to the
temperature, so ρ ∝ K ∝ T .
67
E30-1
We apply Eq. 30-1,
q = C∆V = (50 × 10−12 F)(0.15 V) = 7.5 × 10−12 C;
E30-2 (a) C = ∆V /q = (73.0×10−12 C)/(19.2 V) = 3.80×10−12 F.
(b) The capacitance doesn’t change!
(c) ∆V = q/C = (210×10−12 C)/(3.80×10−12 F) = 55.3 V.
E30-3 q = C∆V = (26.0×10−6 F)(125 V) = 3.25×10−3 C.
E30-4 (a) C = 0 A/d = (8.85×10−12 F/m)π(8.22×10−2 m)2 /(1.31×10−3 m) = 1.43×10−10 F.
(b) q = C∆V = (1.43×10−10 F)(116 V) = 1.66×10−8 C.
E30-5
Eq. 30-11 gives the capacitance of a cylinder,
C = 2π0
L
(0.0238 m)
= 2π(8.85×10−12 F/m)
= 5.46×10−13 F.
ln(b/a)
ln((9.15mm)/(0.81mm))
E30-6 (a) A = Cd/0 = (9.70×10−12 F)(1.20×10−3 m)/(8.85×10−12 F/m) = 1.32×10−3 m2 .
(b) C = C0 d0 /d = (9.70×10−12 F)(1.20×10−3 m)/(1.10×10−3 m) = 1.06×10−11 F.
(c) ∆V = q0 /C = [∆V ]0 C0 /C = [∆V ]0 d/d0 . Using this formula, the new potential difference
would be [∆V ]0 = (13.0 V)(1.10×10−3 m)/(1.20×10−3 m) = 11.9 V. The potential energy has changed
by (11.9 V) − (30.0 V) = −1.1 V.
E30-7 (a) From Eq. 30-8,
C = 4π(8.85×10−12 F/m)
(0.040 m)(0.038 m)
= 8.45×10−11 F.
(0.040 m) − (0.038 m)
(b) A = Cd/0 = (8.45×10−11 F)(2.00×10−3 m)/(8.85×10−12 F/m) = 1.91×10−2 m2 .
E30-8 Let a = b + d, where d is the small separation between the shells. Then
C
(b + d)b
ab
= 4π0
,
a−b
d
b2
≈ 4π0 = 0 A/d.
d
=
4π0
E30-9 The potential difference across each capacitor in parallel is the same; it is equal to 110 V.
The charge on each of the capacitors is then
q = C∆V = (1.00 × 10−6 F)(110 V) = 1.10 × 10−4 C.
If there are N capacitors, then the total charge will be N q, and we want this total charge to be
1.00 C. Then
(1.00 C)
(1.00 C)
N=
=
= 9090.
q
(1.10 × 10−4 C)
68
E30-10 First find the equivalent capacitance of the parallel part:
C eq = C1 + C2 = (10.3×10−6 F) + (4.80×10−6 F) = 15.1×10−6 F.
Then find the equivalent capacitance of the series part:
1
1
1
=
+
= 3.23×105 F−1 .
−6
C eq
(15.1×10 F) (3.90×10−6 F)
Then the equivalent capacitance of the entire arrangement is 3.10×10−6 F.
E30-11 First find the equivalent capacitance of the series part:
1
1
1
=
+
= 3.05×105 F−1 .
C eq
(10.3×10−6 F) (4.80×10−6 F)
The equivalent capacitance is 3.28×10−6 F. Then find the equivalent capacitance of the parallel part:
C eq = C1 + C2 = (3.28×10−6 F) + (3.90×10−6 F) = 7.18×10−6 F.
This is the equivalent capacitance for the entire arrangement.
E30-12 For one capacitor q = C∆V = (25.0×10−6 F)(4200 V) = 0.105 C. There are three capacitors, so the total charge to pass through the ammeter is 0.315 C.
E30-13
(a) The equivalent capacitance is given by Eq. 30-21,
1
1
1
1
1
5
=
+
=
+
=
C eq
C1
C2
(4.0µF) (6.0µF)
(12.0µF)
or C eq = 2.40µF.
(b) The charge on the equivalent capacitor is q = C∆V = (2.40µF)(200 V) = 0.480 mC. For
series capacitors, the charge on the equivalent capacitor is the same as the charge on each of the
capacitors. This statement is wrong in the Student Solutions!
(c) The potential difference across the equivalent capacitor is not the same as the potential
difference across each of the individual capacitors. We need to apply q = C∆V to each capacitor
using the charge from part (b). Then for the 4.0µF capacitor,
∆V =
q
(0.480 mC)
=
= 120 V;
C
(4.0µF)
and for the 6.0µF capacitor,
q
(0.480 mC)
=
= 80 V.
C
(6.0µF)
Note that the sum of the potential differences across each of the capacitors is equal to the potential
difference across the equivalent capacitor.
∆V =
E30-14 (a) The equivalent capacitance is
C eq = C1 + C2 = (4.0µF) + (6.0µF) = (10.0µF).
(c) For parallel capacitors, the potential difference across the equivalent capacitor is the same as
the potential difference across either of the capacitors.
(b) For the 4.0µF capacitor,
q = C∆V = (4.0µF)(200 V) = 8.0×10−4 C;
and for the 6.0µF capacitor,
q = C∆V = (6.0µF)(200 V) = 12.0×10−4 C.
69
E30-15 (a) C eq = C + C + C = 3C;
deq =
0 A
d
0 A
=
= .
C eq
3C
3
(b) 1/C eq = 1/C + 1/C + 1/C = 3/C;
deq =
0 A
0 A
=
= 3d.
C eq
C/3
E30-16 (a) The maximum potential across any individual capacitor is 200 V; so there must be
at least (1000 V)/(200 V) = 5 series capacitors in any parallel branch. This branch would have an
equivalent capacitance of C eq = C/5 = (2.0×10−6 F)/5 = 0.40×10−6 F.
(b) For parallel branches we add, which means we need (1.2×10−6 F)/(0.40×10−6 F) = 3 parallel
branches of the combination found in part (a).
E30-17 Look back at the solution to Ex. 30-10. If C3 breaks down electrically then the circuit is
effectively two capacitors in parallel.
(b) ∆V = 115 V after the breakdown.
(a) q1 = (10.3×10−6 F)(115 V) = 1.18×10−3 C.
E30-18 The 108µF capacitor originally has a charge of q = (108×10−6 F)(52.4 V) = 5.66×10−3 C.
After it is connected to the second capacitor the 108µF capacitor has a charge of q = (108 ×
10−6 F)(35.8 V) = 3.87×10−3 C. The difference in charge must reside on the second capacitor, so the
capacitance is C = (1.79×10−3 C)/(35.8 V) = 5.00×10−5 F.
E30-19 Consider any junction other than A or B. Call this junction point 0; label the four nearest
junctions to this as points 1, 2, 3, and 4. The charge on the capacitor that links point 0 to point 1 is
q1 = C∆V01 , where ∆V01 is the potential difference across the capacitor, so ∆V01 = V0 − V1 , where
V0 is the potential at the junction 0, and V1 is the potential at the junction 1. Similar expressions
exist for the other three capacitors.
For the junction 0 the net charge must be zero; there is no way for charge to cross the plates of
the capacitors. Then q1 + q2 + q3 + q4 = 0, and this means
C∆V01 + C∆V02 + C∆V03 + C∆V04 = 0
or
∆V01 + ∆V02 + ∆V03 + ∆V04 = 0.
Let ∆V0i = V0 − Vi , and then rearrange,
4V0 = V1 + V2 + V3 + V4 ,
or
V0 =
1
(V1 + V2 + V3 + V4 ) .
4
E30-20 U = uV = 0 E 2 V /2, where V is the volume. Then
U=
1
(8.85×10−12 F/m)(150 V/m)2 (2.0 m3 ) = 1.99×10−7 J.
2
70
E30-21 The total capacitance is (2100)(5.0×10−6 F) = 1.05×10−2 F. The total energy stored is
U=
1
1
C(∆V )2 = (1.05×10−2 F)(55×103 V)2 = 1.59×107 J.
2
2
The cost is
(1.59×107 J)
$0.03
3600×103 J
= $0.133.
E30-22 (a) U = 21 C(∆V )2 = 12 (0.061 F)(1.0×104 V)2 = 3.05×106 J.
(b) (3.05×106 J)/(3600×103 J/kW · h) = 0.847kW · h.
E30-23
(a) The capacitance of an air filled parallel-plate capacitor is given by Eq. 30-5,
C=
0 A
(8.85×10−12 F/m)(42.0 × 10−4 m2 )
=
= 2.86×10−11 F.
d
(1.30 × 10−3 m)
(b) The magnitude of the charge on each plate is given by
q = C∆V = (2.86×10−11 F)(625 V) = 1.79×10−8 C.
(c) The stored energy in a capacitor is given by Eq. 30-25, regardless of the type or shape of the
capacitor, so
1
1
U = C(∆V )2 = (2.86×10−11 F)(625 V)2 = 5.59 µJ.
2
2
(d) Assuming a parallel plate arrangement with no fringing effects, the magnitude of the electric
field between the plates is given by Ed = ∆V , where d is the separation between the plates. Then
E = ∆V /d = (625 V)/(0.00130 m) = 4.81×105 V/m.
(e) The energy density is Eq. 30-28,
u=
1
1
0 E 2 = ((8.85×10−12 F/m))(4.81×105 V/m)2 = 1.02 J/m3 .
2
2
E30-24 The equivalent capacitance is given by
1/C eq = 1/(2.12×10−6 F) + 1/(3.88×10−6 F) = 1/(1.37×10−6 F).
The energy stored is U = 12 (1.37×10−6 F)(328 V)2 = 7.37×10−2 J.
E30-25 V /r = q/4π0 r2 = E, so that if V is the potential of the sphere then E = V /r is the
electric field on the surface. Then the energy density of the electric field near the surface is
u=
1
(8.85×10−12 F/m)
0 E 2 =
2
2
(8150 V)
(0.063 m)
2
= 7.41×10−2 J/m3 .
E30-26 The charge on C3 can be found from considering the equivalent capacitance. q3 = (3.10×
10−6 F)(112 V) = 3.47×10−4 C. The potential across C3 is given by [∆V ]3 = (3.47×10−4 C)/(3.90×
10−6 F) = 89.0 V.
The potential across the parallel segment is then (112 V) − (89.0 V) = 23.0 V. So [∆V ]1 =
[∆V ]2 = 23.0 V.
Then q1 = (10.3×10−6 F)(23.0 V) = 2.37×10−4 C and q2 = (4.80×10−6 F)(23.0 V) = 1.10×10−4 C..
71
E30-27 There is enough work on this problem without deriving once again the electric field
between charged cylinders. I will instead refer you back to Section 26-4, and state
E=
1 q
,
2π0 Lr
where q is the magnitude of the charge on a cylinder and L is the length of the cylinders.
The energy density as a function of radial distance is found from Eq. 30-28,
u=
1
q2
1
0 E 2 =
2
2
8π 0 L2 r2
The total energy stored in the electric field is given by Eq. 30-24,
U=
1 q2
q 2 ln(b/a)
=
,
2C
2 2π0 L
where we substituted into the last part Eq. 30-11, the capacitance of a cylindrical
√ capacitor.
We want to show that integrating a volume integral from r = a to r = ab over the energy
density function will yield U/2. Since we want to do this problem the hard way, we will pretend we
don’t know the answer, and integrate from r = a to r = c, and then find out what c is.
Then
Z
1
U =
u dV,
2
Z c Z 2π Z L 1
q2
r dr dφ dz,
=
8π 2 0 L2 r2
a
0
0
Z c Z 2π Z L
q2
dr
=
dφ dz,
8π 2 0 L2 a 0
r
0
Z
c
q2
dr
=
,
4π0 L a r
c
q2
=
ln .
4π0 L a
Now we equate this to the value for U that we found above, and we solve for c.
1 q 2 ln(b/a)
2 2 2π0 L
ln(b/a)
(b/a)
√
ab
q2
c
ln ,
4π0 L a
= 2 ln(c/a),
= (c/a)2 ,
=
= c.
E30-28 (a) d = 0 A/C, or
d = (8.85×10−12 F/m)(0.350 m2 )/(51.3×10−12 F) = 6.04×10−3 m.
(b) C = (5.60)(51.3×10−12 F) = 2.87×10−10 F.
E30-29 Originally, C1 = 0 A/d1 . After the changes, C2 = κ0 A/d2 . Dividing C2 by C1 yields
C2 /C1 = κd1 /d2 , so
κ = d2 C2 /d1 C1 = (2)(2.57×10−12 F)/(1.32×10−12 F) = 3.89.
72
E30-30 The required capacitance is found from U = 12 C(∆V )2 , or
C = 2(6.61×10−6 J)/(630 V)2 = 3.33×10−11 F.
The dielectric constant required is κ = (3.33×10−11 F)/(7.40×10−12 F) = 4.50. Try transformer oil.
E30-31
Capacitance with dielectric media is given by Eq. 30-31,
κe 0 A
.
d
The various sheets have different dielectric constants and different thicknesses, and we want to
maximize C, which means maximizing κe /d. For mica this ratio is 54 mm−1 , for glass this ratio is
35 mm−1 , and for paraffin this ratio is 0.20 mm−1 . Mica wins.
C=
E30-32 The minimum plate separation is given by
d = (4.13×103 V)/(18.2×106 V/m) = 2.27×10−4 m.
The minimum plate area is then
A=
dC
(2.27×10−4 m)(68.4×10−9 F)
=
= 0.627 m2 .
κ0
(2.80)(8.85×10−12 F/m)
E30-33 The capacitance of a cylindrical capacitor is given by Eq. 30-11,
C = 2π(8.85×10−12 F/m)(2.6)
1.0×103 m
= 8.63×10−8 F.
ln(0.588/0.11)
E30-34 (a) U = C 0 (∆V )2 /2, C 0 = κe 0 A/d, and ∆V /d is less than or equal to the dielectric
strength (which we will call S). Then ∆V = Sd and
U=
1
κe 0 AdS 2 ,
2
so the volume is given by
V = 2U/κe 0 S 2 .
This quantity is a minimum for mica, so
V = 2(250×103 J)/(5.4)(8.85×10−12 F/m)(160×106 V/m)2 = 0.41 m3 .
(b) κe = 2U/V 0 S 2 , so
κe = 2(250×103 J)/(0.087m3 )(8.85×10−12 F/m)(160×106 V/m)2 = 25.
E30-35
(a) The capacitance of a cylindrical capacitor is given by Eq. 30-11,
C = 2π0 κe
L
.
ln(b/a)
The factor of κe is introduced because there is now a dielectric (the Pyrex drinking glass) between
the plates. We can look back to Table 29-2 to get the dielectric properties of Pyrex. The capacitance
of our “glass” is then
C = 2π(8.85×10−12 F/m)(4.7)
(0.15 m)
= 7.3×10−10 F.
ln((3.8 cm)/(3.6 cm)
(b) The breakdown potential is (14 kV/mm)(2 mm) = 28 kV.
73
E30-36 (a) C 0 = κe C = (6.5)(13.5×10−12 F) = 8.8×10−11 F.
(b) Q = C 0 ∆V = (8.8×10−11 F)(12.5 V) = 1.1×10−9 C.
(c) E = ∆V /d, but we don’t know d.
(d) E 0 = E/κe , but we couldn’t find E.
E30-37 (a) Insert the slab so that it is a distance a above the lower plate. Then the distance
between the slab and the upper plate is d − a − b. Inserting the slab has the same effect as having
two capacitors wired in series; the separation of the bottom capacitor is a, while that of the top
capacitor is d − a − b.
The bottom capacitor has a capacitance of C1 = 0 A/a, while the top capacitor has a capacitance
of C2 = 0 A/(d − a − b). Adding these in series,
1
C eq
=
=
=
1
1
+
,
C1
C2
a
d−a−b
+
,
0 A
0 A
d−b
.
0 A
So the capacitance of the system after putting the copper slab in is C = 0 A/(d − b).
(b) The energy stored in the system before the slab is inserted is
Ui =
q2 d
q2
=
2C i
2 0 A
while the energy stored after the slab is inserted is
Uf =
q2
q2 d − b
=
2C f
2 0 A
The ratio is U i /U f = d/(d − b).
(c) Since there was more energy before the slab was inserted, then the slab must have gone in
willingly, it was pulled in!. To get the slab back out we will need to do work on the slab equal to
the energy difference.
q2 d
q2 d − b
q2 b
Ui − Uf =
−
=
.
2 0 A
2 0 A
2 0 A
E30-38 (a) Insert the slab so that it is a distance a above the lower plate. Then the distance
between the slab and the upper plate is d − a − b. Inserting the slab has the same effect as having
two capacitors wired in series; the separation of the bottom capacitor is a, while that of the top
capacitor is d − a − b.
The bottom capacitor has a capacitance of C1 = 0 A/a, while the top capacitor has a capacitance
of C2 = 0 A/(d − a − b). Adding these in series,
1
C eq
=
=
=
1
1
+
,
C1
C2
a
d−a−b
+
,
0 A
0 A
d−b
.
0 A
So the capacitance of the system after putting the copper slab in is C = 0 A/(d − b).
74
(b) The energy stored in the system before the slab is inserted is
Ui =
C i (∆V )2
(∆V )2 0 A
=
2
2
d
while the energy stored after the slab is inserted is
Uf =
C f (∆V )2
(∆V )2 0 A
=
2
2 d−b
The ratio is U i /U f = (d − b)/d.
(c) Since there was more energy after the slab was inserted, then the slab must not have gone in
willingly, it was being repelled!. To get the slab in we will need to do work on the slab equal to the
energy difference.
Uf − Ui =
(∆V )2 0 A
(∆V )2 0 A
(∆V )2 0 Ab
−
=
.
2 d−b
2
d
2 d(d − b)
E30-39 C = κe 0 A/d, so d = κe 0 A/C.
(a) E = ∆V /d = C∆V /κe 0 A, or
E=
(112×10−12 F)(55.0 V)
= 13400 V/m.
(5.4)(8.85×10−12 F/m)(96.5×10−4 m2 )
(b) Q = C∆V = (112×10−12 F)(55.0 V) = 6.16×10−9 C..
(c) Q0 = Q(1 − 1/κe ) = (6.16×10−9 C)(1 − 1/(5.4)) = 5.02×10−9 C.
E30-40 (a) E = q/κe 0 A, so
κe =
(890×10−9 C)
= 6.53
(1.40×106 V/m)(8.85×10−12 F/m)(110×10−4 m2 )
(b) q 0 = q(1 − 1/κe ) = (890×10−9 C)(1 − 1/(6.53)) = 7.54×10−7 C.
P30-1
The capacitance of the cylindrical capacitor is from Eq. 30-11,
C=
2π0 L
.
ln(b/a)
If the cylinders are very close together we can write b = a + d, where d, the separation between the
cylinders, is a small number, so
C=
2π0 L
2π0 L
=
.
ln ((a + d)/a)
ln (1 + d/a)
Expanding according to the hint,
C≈
2π0 L
2πa0 L
=
d/a
d
Now 2πa is the circumference of the cylinder, and L is the length, so 2πaL is the area of a cylindrical
plate. Hence, for small separation between the cylinders we have
C≈
0 A
,
d
which is the expression for the parallel plates.
75
P30-2
(a) C = 0 A/x; take the derivative and
dC
dT
0 dA 0 A dx
− 2
,
x dT
x dT 1 dA 1 dx
= C
−
.
A dT
x dT
=
(b) Since (1/A)dA/dT = 2αa and (1/x)dx/dT = αs , we need
αs = 2αa = 2(23×10−6 /C◦ ) = 46×10−6 /C◦ .
P30-3 Insert the slab so that it is a distance d above the lower plate. Then the distance between
the slab and the upper plate is a−b−d. Inserting the slab has the same effect as having two capacitors
wired in series; the separation of the bottom capacitor is d, while that of the top capacitor is a−b−d.
The bottom capacitor has a capacitance of C1 = 0 A/d, while the top capacitor has a capacitance
of C2 = 0 A/(a − b − d). Adding these in series,
1
C eq
=
=
=
1
1
+
,
C1
C2
d
a−b−d
+
,
0 A
0 A
a−b
.
0 A
So the capacitance of the system after putting the slab in is C = 0 A/(a − b).
P30-4 The potential difference between any two adjacent plates is ∆V . Each interior plate has a
charge q on each surface; the exterior plate (one pink, one gray) has a charge of q on the interior
surface only.
The capacitance of one pink/gray plate pair is C = 0 A/d. There are n plates, but only n − 1
plate pairs, so the total charge is (n − 1)q. This means the total capacitance is C = 0 (n − 1)A/d.
P30-5 Let ∆V0 = 96.6 V.
As far as point e is concerned point a looks like it is originally positively charged, and point d is
originally negatively charged. It is then convenient to define the charges on the capacitors in terms
of the charges on the top sides, so the original charge on C1 is q 1,i = C1 ∆V0 while the original charge
on C2 is q 2,i = −C2 ∆V0 . Note the negative sign reflecting the opposite polarity of C2 .
(a) Conservation of charge requires
q 1,i + q 2,i = q 1,f + q 2,f ,
but since q = C∆V and the two capacitors will be at the same potential after the switches are closed
we can write
C1 ∆V0 − C2 ∆V0 = C1 ∆V + C2 ∆V,
(C1 − C2 ) ∆V0 = (C1 + C2 ) ∆V,
C1 − C2
∆V0 = ∆V.
C1 + C2
With numbers,
∆V = (96.6 V)
(1.16 µF) − (3.22 µF)
= −45.4 V.
(1.16 µF) + (3.22 µF)
76
The negative sign means that the top sides of both capacitor will be negatively charged after the
switches are closed.
(b) The charge on C1 is C1 ∆V = (1.16 µF)(45.4 V) = 52.7µC.
(c) The charge on C2 is C2 ∆V = (3.22 µF)(45.4 V) = 146µC.
P30-6 C2 and C3 form an effective capacitor with equivalent capacitance Ca = C2 C3 /(C2 + C3 ).
The charge on C1 is originally q0 = C1 ∆V0 . After throwing the switch the potential across C1
is given by q1 = C1 ∆V1 . The same potential is across Ca ; q2 = q3 , so q2 = Ca ∆V1 . Charge is
conserved, so q1 + q2 = q0 . Combining some of the above,
∆V1 =
and then
q1 =
C1
q0
=
∆V0 ,
C1 + Ca
C1 + Ca
C12
C12 (C2 + C3 )
∆V0 =
∆V0 .
C1 + Ca
C1 C2 + C1 C3 + C2 C3
Similarly,
Ca C1
q2 =
∆V0 =
C1 + Ca
1
1
1
+
+
C1
C2
C3
−1
∆V0 .
q3 = q2 because they are in series.
P30-7 (a) If terminal a is more positive than terminal b then current can flow that will charge the
capacitor on the left, the current can flow through the diode on the top, and the current can charge
the capacitor on the right. Current will not flow through the diode on the left. The capacitors are
effectively in series.
Since the capacitors are identical and series capacitors have the same charge, we expect the
capacitors to have the same potential difference across them. But the total potential difference
across both capacitors is equal to 100 V, so the potential difference across either capacitor is 50 V.
The output pins are connected to the capacitor on the right, so the potential difference across
the output is 50 V.
(b) If terminal b is more positive than terminal a the current can flow through the diode on the
left. If we assume the diode is resistanceless in this configuration then the potential difference across
it will be zero. The net result is that the potential difference across the output pins is 0 V.
In real life the potential difference across the diode would not be zero, even if forward biased. It
will be somewhere around 0.5 Volts.
P30-8 Divide the strip of width a into N segments, each of width ∆x = a/N . The capacitance of
each strip is ∆C = 0 a∆x/y. If θ is small then
1
1
d
1
=
≈
≈ 1 − xθ/d).
y
d + x sin θ
d + xθ
(
Since parallel capacitances add,
C=
X
∆C =
0 a
d
Z
a
(1 − xθ/d)dx =
0
77
0 a2
d
1−
aθ
2d
.
P30-9 (a) When S2 is open the circuit acts as two parallel capacitors. The branch on the left has
an effective capacitance given by
1
1
1
1
=
+
=
,
Cl
(1.0×10−6 F) (3.0×10−6 F)
7.5×10−7 F
while the branch on the right has an effective capacitance given by
1
1
1
1
=
+
=
.
Cl
(2.0×10−6 F) (4.0×10−6 F)
1.33×10−6 F
The charge on either capacitor in the branch on the left is
q = (7.5×10−7 F)(12 V) = 9.0×10−6 C,
while the charge on either capacitor in the branch on the right is
q = (1.33×10−6 F)(12 V) = 1.6×10−5 C.
(b) After closing S2 the circuit is effectively two capacitors in series. The top part has an effective
capacitance of
C t = (1.0×10−6 F) + (2.0×10−6 F) = (3.0×10−6 F),
while the effective capacitance of the bottom part is
C b = (3.0×10−6 F) + (4.0×10−6 F) = (7.0×10−6 F).
The effective capacitance of the series combination is given by
1
1
1
1
=
+
=
.
−6
−6
C eq
(3.0×10 F) (7.0×10 F)
2.1×10−6 F
The charge on each part is q = (2.1×10−6 F)(12 V) = 2.52×10−5 C. The potential difference across
the top part is
∆V t = (2.52×10−5 C)/(3.0×10−6 F) = 8.4 V,
and then the charge on the top two capacitors is q1 = (1.0 × 10−6 F)(8.4 V) = 8.4 × 10−6 C and
q2 = (2.0×10−6 F)(8.4 V) = 1.68×10−5 C. The potential difference across the bottom part is
∆V t = (2.52×10−5 C)/(7.0×10−6 F) = 3.6 V,
and then the charge on the top two capacitors is q1 = (3.0 × 10−6 F)(3.6 V) = 1.08 × 10−5 C and
q2 = (4.0×10−6 F)(3.6 V) = 1.44×10−5 C.
P30-10 Let ∆V = ∆Vxy . By symmetry ∆V2 = 0 and ∆V1 = ∆V4 = ∆V5 = ∆V3 = ∆V /2.
Suddenly the problem is very easy. The charges on each capacitor is q1 , except for q2 = 0. Then the
equivalent capacitance of the circuit is
C eq =
q1 + q4
q
=
= C1 = 4.0×10−6 F.
∆V
2∆V1
78
P30-11
(a) The charge on the capacitor with stored energy U0 = 4.0 J is q0 , where
U0 =
q02
.
2C
When this capacitor is connected to an identical uncharged capacitor the charge is shared equally,
so that the charge on either capacitor is now q = q0 /2. The stored energy in one capacitor is then
U=
q2
q 2 /4
1
= 0 = U0 .
2C
2C
4
But there are two capacitors, so the total energy stored is 2U = U0 /2 = 2.0 J.
(b) Good question. Current had to flow through the connecting wires to get the charge from one
capacitor to the other. Originally the second capacitor was uncharged, so the potential difference
across that capacitor would have been zero, which means the potential difference across the connecting wires would have been equal to that of the first capacitor, and there would then have been
energy dissipation in the wires according to
P = i2 R.
That’s where the missing energy went.
P30-12
R = ρL/A and C = 0 A/L. Combining, R = ρ0 /C, or
R = (9.40 Ω · m)(8.85×10−12 F/m)/(110×10−12 F) = 0.756 Ω.
P30-13 (a)R u = 21 0 E 2 = e2 /32π 2 0 r4 .
(b) U = u dV where dV = 4πr2 dr. Then
U = 4pi
Z
∞
R
e2
e2 1
r2 dr =
.
2
4
32π 0 r
8π0 R
(c) R = e2 /8π0 mc2 , or
R=
P30-14
P30-15
(1.60×10−19 C)2
8π(8.85×10−12 F/m)(9.11×10−31 kg)(3.00×108 m/s)2
= 1.40×10−15 m.
U = 21 q 2 /C = q 2 x/2A0 . F = dU/dx = q 2 /2A0 .
According to Problem 14, the force on a plate of a parallel plate capacitor is
F =
q2
.
20 A
The force per unit area is then
q2
σ2
F
=
=
,
A
20 A2
20
where σ = q/A is the surface charge density. But we know that the electric field near the surface of
a conductor is given by E = σ/0 , so
F
1
= 0 E 2 .
A
2
79
P30-16 A small surface area element dA carries a charge dq = q dA/4πR2 . There are three forces
on the elements which balance, so
p(V0 /V )dA + q dq/4π0 R2 = p dA,
or
pR03 + q 2 /16π 2 0 R = pR3 .
This can be rearranged as
q 2 = 16π 2 0 pR(R3 − R03 ).
P30-17 The magnitude of the electric field in the cylindrical region is given by E = λ/2π0 r,
where λ is the linear charge density on the anode. The potential difference is given by ∆V =
(λ/2π0 ) ln(b/a), where a is the radius of the anode b the radius of the cathode. Combining, E =
∆V /r ln(b/a), this will be a maximum when r = a, so
∆V = (0.180×10−3 m) ln[(11.0×10−3 m)/(0.180×10−3 m)](2.20×106 V/m) = 1630 V.
P30-18
This is effectively two capacitors in parallel, each with an area of A/2. Then
0 A/2
0 A κe1 + κe2
0 A/2
+ κe2
=
.
C eq = κe1
d
d
d
2
P30-19 We will treat the system as two capacitors in series by pretending there is an infinitesimally thin conductor between them. The slabs are (I assume) the same thickness. The capacitance
of one of the slabs is then given by Eq. 30-31,
C1 =
κe1 0 A
,
d/2
where d/2 is the thickness of the slab. There would be a similar expression for the other slab. The
equivalent series capacitance would be given by Eq. 30-21,
1
C eq
=
=
=
C eq
=
1
1
+
,
C1
C2
d/2
d/2
+
,
κe1 0 A κe2 0 A
d κe2 + κe1
,
20 A κe1 κe2
20 A κe1 κe2
.
d κe2 + κe1
P30-20 Treat this as three capacitors. Find the equivalent capacitance of the series combination
on the right, and then add on the parallel part on the left. The right hand side is
1
d
d
2d κe2 + κe3
=
+
=
.
C eq
κe2 0 A/2 κe3 0 A/2
0 A
κe2 κe3
Add this to the left hand side, and
C eq
=
=
κe1 0 A/2 0 A
κe2 κe3
+
,
2d
2d κe2 + κe3
0 A κe1
κe2 κe3
+
.
2d
2
κe2 + κe3
80
P30-21 (a) q doesn’t change, but C 0 = C/2. Then ∆V 0 = q/C = 2∆V .
(b) U = C(∆V )2 /2 = 0 A(∆V )2 /2d. U 0 = C 0 (∆V 0 )2 /2 = 0 A(2∆V )2 /4d = 2U .
(c) W = U 0 − U = 2U − U = U = 0 A(∆V )2 /2d.
P30-22 The total energy is U = qδV /2 = (7.02×10−10 C)(52.3 V)/2 = 1.84×10−8 J.
(a) In the air gap we have
Ua =
(8.85×10−12 F/m)(6.9×103 V/m)2 (1.15×10−2 m2 )(4.6×10−3 m)
0 E02 V
=
= 1.11×10−8 J.
2
2
That is (1.11/1.85) = 60% of the total.
(b) The remaining 40% is in the slab.
P30-23 (a) C = 0 A/d = (8.85×10−12 F/m)(0.118 m2 )/(1.22×10−2 m) = 8.56×10−11 F.
(b) Use the results of Problem 30-24.
C0 =
(4.8)(8.85×10−12 F/m)(0.118 m2 )
= 1.19×10−10 F
(4.8)(1.22×10−2 m) − (4.3×10−3 m)(4.8 − 1)
(c) q = C∆V = (8.56×10−11 F)(120 V) = 1.03×10−8 C; since the battery is disconnected q 0 = q.
(d) E = q/0 A = (1.03×10−8 C)/(8.85×10−12 F/m)(0.118 m2 ) = 9860 V/m in the space between
the plates.
(e) E 0 = E/κe = (9860 V/m)/(4.8) = 2050 V/m in the dielectric.
(f) ∆V 0 = q/C 0 = (1.03×10−8 C)/(1.19×10−10 F) = 86.6 V.
(g) W = U 0 − U = q 2 (1/C − 1/C 0 )/2, or
W =
(1.03×10−8 C)2
[1/(8.56×10−11 F) − 1/(1.19×10−10 F)] = 1.73×10−7 J.
2
P30-24 The result is effectively three capacitors in series. Two are air filled with thicknesses of
x and d − b − x, the third is dielectric filled with thickness b. All have an area A. The effective
capacitance is given by
1
C
=
=
C
=
=
x
d−b−x
b
+
+
,
0 A
0 A
κ A
e 0
1
b
(d − b) +
,
0 A
κe
0 A
,
d − b + b/κe
κe 0 A
.
κe − b(κe − 1)
81
E31-1 (5.12 A)(6.00 V)(5.75 min)(60 s/min) = 1.06×104 J.
E31-2 (a) (12.0 V)(1.60×10−19 C) = 1.92×10−18 J.
(b) (1.92×10−18 J)(3.40×1018 /s) = 6.53 W.
E31-3
If the energy is delivered at a rate of 110 W, then the current through the battery is
i=
P
(110 W)
=
= 9.17 A.
∆V
(12 V)
Current is the flow of charge in some period of time, so
∆t =
∆q
(125 A · h)
=
= 13.6 h,
i
(9.2 A)
which is the same as 13 hours and 36 minutes.
E31-4 (100 W)(8 h) = 800 W · h.
(a) (800 W · h)/(2.0 W · h) = 400 batteries, at a cost of (400)($0.80) = $320.
(b) (800 W · h)($0.12×10−3 W · h) = $0.096.
E31-5 Go all of the way around the circuit. It is a simple one loop circuit, and although it does
not matter which way we go around, we will follow the direction of the larger emf. Then
(150 V) − i(2.0 Ω) − (50 V) − i(3.0 Ω) = 0,
where i is positive if it is counterclockwise. Rearranging,
100 V = i(5.0 Ω),
or i = 20 A.
Assuming the potential at P is VP = 100 V, then the potential at Q will be given by
VQ = VP − (50 V) − i(3.0 Ω) = (100 V) − (50 V) − (20 A)(3.0 Ω) = −10 V.
E31-6 (a) Req = (10 Ω) + (140 Ω) = 150 Ω. i = (12.0 V)/(150 Ω) = 0.080 A.
(b) Req = (10 Ω) + (80 Ω) = 90 Ω. i = (12.0 V)/(90 Ω) = 0.133 A.
(c) Req = (10 Ω) + (20 Ω) = 30 Ω. i = (12.0 V)/(30 Ω) = 0.400 A.
E31-7 (a) Req = (3.0 V − 2.0 V)/(0.050 A) = 20 Ω. Then R = (20 Ω) − (3.0 Ω) − (3.0 Ω) = 14 Ω.
(b) P = i∆V = i2 R = (0.050 A)2 (14 Ω) = 3.5×10−2 W.
E31-8 (5.0 A)R1 = ∆V . (4.0 A)(R1 +2.0 Ω) = ∆V . Combining, 5R1 = 4R1 +8.0 Ω, or R1 = 8.0 Ω.
E31-9 (a) (53.0 W)/(1.20 A) = 44.2 V.
(b) (1.20 A)(19.0 Ω) = 22.8 V is the potential difference across R. Then an additional potential
difference of (44.2 V) − (22.8 V) = 21.4 V must exist across C.
(c) The left side is positive; it is a reverse emf.
p
E31-10 (a) The current in the resistor is (9.88 W)/(0.108 Ω) = 9.56 A. The total resistance of
the circuit is (1.50 V)/(9.56 A) = 0.157 Ω. The internal resistance of the battery is then (0.157 Ω) −
(0.108 Ω) = 0.049 Ω.
(b) (9.88 W)/(9.56 A) = 1.03 V.
82
E31-11
We assign directions to the currents through the four resistors as shown in the figure.
1
2
a
b
3
4
Since the ammeter has no resistance the potential at a is the same as the potential at b. Consequently the potential difference (∆V b ) across both of the bottom resistors is the same, and the
potential difference (∆V t ) across the two top resistors is also the same (but different from the
bottom). We then have the following relationships:
∆V t + ∆V b
i1 + i2
∆Vj
= E,
= i3 + i4 ,
= i j Rj ,
where the j subscript in the last line refers to resistor 1, 2, 3, or 4.
For the top resistors,
∆V1 = ∆V2 implies 2i1 = i2 ;
while for the bottom resistors,
∆V3 = ∆V4 implies i3 = i4 .
Then the junction rule requires i4 = 3i1 /2, and the loop rule requires
(i1 )(2R) + (3i1 /2)(R) = E or i1 = 2E/(7R).
The current that flows through the ammeter is the difference between i2 and i4 , or 4E/(7R) −
3E/(7R) = E/(7R).
E31-12 (a) Define the current i1 as moving to the left through r1 and the current i2 as moving
to the left through r2 . i3 = i1 + i2 is moving to the right through R. Then there are two loop
equations:
E1
E2
= i1 r1 + i3 R,
= (i3 − i1 )r2 + i3 R.
Multiply the top equation by r2 and the bottom by r1 and then add:
r2 E1 + r1 E2 = i3 r1 r2 + i3 R(r1 + r2 ),
which can be rearranged as
i3 =
r2 E1 + r1 E2
.
r1 r2 + Rr1 + Rr2
(b) There is only one current, so
E1 + E2 = i(r1 + r2 + R),
or
i=
E1 + E2
.
r1 + r 2 + R
83
E31-13 (a) Assume that the current flows through each source of emf in the same direction as
the emf. The the loop rule will give us three equations
E1 − i1 R1 + i2 R2 − E2 − i1 R1
E2 − i2 R2 + i3 R1 − E3 + i3 R1
E1 − i1 R1 + i3 R1 − E3 + i3 R1 − i1 R1
= 0,
= 0,
= 0.
The junction rule (looks at point a) gives us i1 + i2 + i3 = 0. Use this to eliminate i2 from the second
loop equation,
E2 + i1 R2 + i3 R2 + 2i3 R1 − E3 = 0,
and then combine this with the the third equation to eliminate i3 ,
E1 R2 − E3 R2 + 2i3 R1 R2 + 2E2 R1 + 2i3 R1 R2 + 4i3 R12 − 2E3 R1 = 0,
or
i3 =
2E3 R1 + E3 R2 − E1 R2 − 2E2 R1
= 0.582 A.
4R1 R2 + 4R12
Then we can find i1 from
i1 =
E3 − E2 − i3 R2 − 2i3 R1
= −0.668 A,
R2
where the negative sign indicates the current is down.
Finally, we can find i2 = −(i1 + i3 ) = 0.0854 A.
(b) Start at a and go to b (final minus initial!),
+i2 R2 − E2 = −3.60 V.
E31-14 (a) The current through the circuit is i = E/(r + R). The power delivered to R is then
P = i∆V = i2 R = E 2 R/(r + R)2 . Evaluate dP/dR and set it equal to zero to find the maximum.
Then
dP
r−R
0=
= E 2R
,
dR
(r + R)3
which has the solution r = R.
(b) When r = R the power is
P = E 2R
1
E2
=
.
(R + R)2
4r
E31-15 (a) We first use P = F v to find the power output by the electric motor. Then P =
(2.0 N)(0.50 m/s) = 1.0 W.
The potential difference across the motor is ∆V m = E − ir. The power output from the motor is
the rate of energy dissipation, so P m = ∆V m i. Combining these two expressions,
= (E − ir) i,
= Ei − i2 r,
0 = −i2 r + Ei − P m ,
0 = (0.50 Ω)i2 − (2.0 V)i + (1.0 W).
Pm
Rearrange and solve for i,
i=
(2.0 V) ±
p
(2.0 V)2 − 4(0.50 Ω)(1.0 W)
,
2(0.50 Ω)
84
which has solutions i = 3.4 A and i = 0.59 A.
(b) The potential difference across the terminals of the motor is ∆V m = E − ir which if i = 3.4 A
yields ∆V m = 0.3 V, but if i = 0.59 A yields ∆V m = 1.7 V. The battery provides an emf of 2.0 V; it
isn’t possible for the potential difference across the motor to be larger than this, but both solutions
seem to satisfy this constraint, so we will move to the next part and see what happens.
(c) So what is the significance of the two possible solutions? It is a consequence of the fact that
power is related to the current squared, and with any quadratics we expect two solutions. Both
are possible, but it might be that only one is stable, or even that neither is stable, and a small
perturbation to the friction involved in turning the motor will cause the system to break down. We
will learn in a later chapter that the effective resistance of an electric motor depends on the speed
at which it is spinning, and although that won’t affect the problem here as worded, it will affect the
physical problem that provided the numbers in this problem!
E31-16 req = 4r = 4(18 Ω) = 72 Ω. The current is i = (27 V)/(72 Ω) = 0.375 A.
E31-17 In parallel connections of two resistors the effective resistance is less than the smaller
resistance but larger than half the smaller resistance. In series connections of two resistors the
effective resistance is greater than the larger resistance but less than twice the larger resistance.
Since the effective resistance of the parallel combination is less than either single resistance
and the effective resistance of the series combinations is larger than either single resistance we can
conclude that 3.0 Ω must have been the parallel combination and 16 Ω must have been the series
combination.
The resistors are then 4.0 Ω and 12 Ω resistors.
E31-18 Points B and C are effectively the same point!
(a) The three resistors are in parallel. Then req = R/3.
(b) See (a).
(c) 0, since there is no resistance between B and C.
E31-19 Focus on the loop through the battery, the 3.0 Ω, and the 5.0 Ω resistors. The loop rule
yields
(12.0 V) = i[(3.0 Ω) + (5.0 Ω)] = i(8.0 Ω).
The potential difference across the 5.0 Ω resistor is then
∆V = i(5.0 Ω) = (5.0 Ω)(12.0 V)/(8.0 Ω) = 7.5 V.
E31-20 Each lamp draws a current of (500 W)/(120 V) = 4.17 A. Furthermore, the fuse can
support (15 A)/(4.17 A) = 3.60 lamps. That is a maximum of 3.
E31-21 The current in the series combination is is = E/(R1 + R2 ). The power dissipated is
P s = iE = E 2 /(R1 + R2 ).
In a parallel arrangement R1 dissipates P1 = i1 E = E 2 /R1 . A similar expression exists for R2 ,
so the total power dissipated is P p = E 2 (1/R1 + 1/R2 ).
The ratio is 5, so 5 = P p /P s = (1/R1 + 1/R2 )(R1 + R2 ), or 5R1 R2 = (R1 + R2 )2 . Solving for
R2 yields 2.618R1 or 0.382R1 . Then R2 = 262 Ω or R2 = 38.2 Ω.
85
E31-22 Combining n identical resistors in series results in an equivalent resistance of req = nR.
Combining n identical resistors in parallel results in an equivalent resistance of req = R/n. If the
resistors are arranged in a square array consisting of n parallel branches of n series resistors, then
the effective resistance is R. Each will dissipate a power P , together they will dissipate n2 P .
So we want nine resistors, since four would be too small.
E31-23
parallel:
(a) Work through the circuit one step at a time. We first “add” R2 , R3 , and R4 in
1
1
1
1
1
=
+
+
=
Reff
42.0 Ω 61.6 Ω 75.0 Ω
18.7 Ω
We then “add” this resistance in series with R1 ,
Reff = (112 Ω) + (18.7 Ω) = 131 Ω.
(b) The current through the battery is i = E/R = (6.22 V)/(131 Ω) = 47.5 mA. This is also the
current through R1 , since all the current through the battery must also go through R1 .
The potential difference across R1 is ∆V1 = (47.5 mA)(112 Ω) = 5.32 V. The potential difference
across each of the three remaining resistors is 6.22 V − 5.32 V = 0.90 V.
The current through each resistor is then
i2
i3
i4
= (0.90 V)/(42.0 Ω) = 21.4 mA,
= (0.90 V)/(61.6 Ω) = 14.6 mA,
= (0.90 V)/(75.0 Ω) = 12.0 mA.
E31-24 The equivalent resistance of the parallel part is r0 = R2 R/(R2 + R). The equivalent
resistance for the circuit is r = R1 + R2 R/(R2 + R). The current through the circuit is i0 = E/r.
The potential difference across R is ∆V = E − i0 R1 , or
∆V
= E(1 − R1 /r),
R2 + R
= E 1 − R1
,
R1 R2 + R1 R + RR2
RR2
= E
.
R1 R2 + R1 R + RR2
Since P = i∆V = (∆V )2 /R,
P = E2
RR22
.
(R1 R2 + R1 R + RR2 )2
Set dP/dR = 0, the solution is R = R1 R2 /(R1 + R2 ).
E31-25 (a) First “add” the left two resistors in series; the effective resistance of that branch is
2R. Then “add” the right two resistors in series; the effective resistance of that branch is also 2R.
Now we combine the three parallel branches and find the effective resistance to be
1
1
1
1
4
=
+ +
=
,
Reff
2R R 2R
2R
or Reff = R/2.
(b) First we “add” the right two resistors in series; the effective resistance of that branch is 2R.
We then combine this branch with the resistor which connects points F and H. This is a parallel
connection, so the effective resistance is
1
1
1
3
=
+
=
,
Reff
2R R
2R
86
or 2R/3.
This value is effectively in series with the resistor which connects G and H, so the “total” is
5R/3.
Finally, we can combine this value in parallel with the resistor that directly connects F and G
according to
1
1
3
8
=
+
=
,
Reff
R 5R
5R
or Reff = 5R/8.
E31-26 The resistance of the second resistor is r2 = (2.4 V)/(0.001 A) = 2400 Ω. The potential
difference across the first resistor is (12 V) − (2.4 V) = 9.6 V. The resistance of the first resistor is
(9.6 V)/(0.001 A) = 9600 Ω.
E31-27 See Exercise 31-26. The resistance ratio is
(0.95 ± 0.1 V)
r1
=
,
r 1 + r2
(1.50 V)
or
r2
(1.50 V)
=
− 1.
r1
(0.95 ± 0.1 V)
The allowed range for the ratio r2 /r1 is between 0.5625 and 0.5957.
We can choose any standard resistors we want, and we could use any tolerance, but then we
will need to check our results. 22Ω and 39Ω would work; as would 27Ω and 47Ω. There are other
choices.
E31-28 Consider any junction other than A or B. Call this junction point 0; label the four
nearest junctions to this as points 1, 2, 3, and 4. The current through the resistor that links
point 0 to point 1 is i1 = ∆V01 /R, where ∆V01 is the potential difference across the resistor, so
∆V01 = V0 − V1 , where V0 is the potential at the junction 0, and V1 is the potential at the junction
1. Similar expressions exist for the other three resistor.
For the junction 0 the net current must be zero; there is no way for charge to accumulate on the
junction. Then i1 + i2 + i3 + i4 = 0, and this means
∆V01 /R + ∆V02 /R + ∆V03 /R + ∆V04 /R = 0
or
∆V01 + ∆V02 + ∆V03 + ∆V04 = 0.
Let ∆V0i = V0 − Vi , and then rearrange,
4V0 = V1 + V2 + V3 + V4 ,
or
V0 =
1
(V1 + V2 + V3 + V4 ) .
4
E31-29 The current through the radio is i = P/∆V = (7.5 W)/(9.0 V) = 0.83 A. The radio
was left one for 6 hours, or 2.16×104 s. The total charge to flow through the radio in that time is
(0.83 A)(2.16×104 s) = 1.8×104 C.
E31-30 The power dissipated by the headlights is (9.7 A)(12.0 V) = 116 W. The power required
by the engine is (116 W)/(0.82) = 142 W, which is equivalent to 0.190 hp.
87
E31-31 (a) P = (120 V)(120 V)/(14.0 Ω) = 1030 W.
(b) W = (1030 W)(6.42 h) = 6.61 kW · h. The cost is $0.345.
E31-32
E31-33
We want to apply either Eq. 31-21,
PR = i2 R,
or Eq. 31-22,
PR = (∆VR )2 /R,
depending on whether we are in series (the current is the same through each bulb), or in parallel
(the potential difference across each bulb is the same. The brightness of a bulb will be measured by
P , even though P is not necessarily a measure of the rate radiant energy is emitted from the bulb.
(b) If the bulbs are in parallel then PR = (∆VR )2 /R is how we want to compare the brightness.
The potential difference across each bulb is the same, so the bulb with the smaller resistance is
brighter.
(b) If the bulbs are in series then PR = i2 R is how we want to compare the brightness. Both
bulbs have the same current, so the larger value of R results in the brighter bulb.
One direct consequence of this can be tried at home. Wire up a 60 W, 120 V bulb and a 100 W,
120 V bulb in series. Which is brighter? You should observe that the 60 W bulb will be brighter.
2
E31-34 (a) j = i/A = (25 A)/π(0.05 in) = 3180 A/in = 4.93×106 A/m2 .
(b) E = ρj = (1.69×10−8 Ω · m)(4.93×106 A/m2 ) = 8.33×10−2 V/m.
(c) ∆V = Ed = (8.33×10−2 V/m)(305 m) = 25 V.
(d) P = i∆V = (25 A)(25 V) = 625 W.
E31-35 (a) The bulb is on for 744 hours. The energy consumed is (100 W)(744 h) = 74.4 kW · h,
at a cost of (74.4)(0.06) = $4.46.
(b) r = V 2 /P = (120 V)2 /(100 W) = 144 Ω.
(c) i = P/V = (100 W)/(120 V) = 0.83 A.
E31-36 P = (∆V )2 /r and r = r0 (1 + α∆T ). Then
P =
(500 W)
P0
=
= 660 W
1 + α∆T
1 + (4.0×10−4 /C◦ )(−600C◦ )
E31-37 (a) n = q/e = it/e, so
n = (485×10−3 A)(95×10−9 s)/(1.6×10−19 C) = 2.88×1011 .
(b) iav = (520/s)(485×10−3 A)(95×10−9 s) = 2.4×10−5 A.
(c) P p = ip ∆V = (485×10−3 A)(47.7×106 V) = 2.3×106 W; while P a = ia ∆V = (2.4×10−5 A)(47.7×
6
10 V) = 1.14×103 W.
E31-38 rp
= ρL/A p
= (3.5×10−5 Ω · m)(1.96×10−2 m)/π(5.12×10−3 m)2 = 8.33×10−3 Ω.
(a) i = P/r = (1.55 W)/(8.33×10−3 Ω) = 13.6 A, so
j = i/A = (13.6 A)/π(5.12×10−3 m)2 = 1.66×105 A/m2 .
p
√
(b) ∆V = P r = (1.55 W)(8.33×10−3 Ω) = 0.114 V.
88
E31-39
(a) The current through the wire is
i = P/∆V = (4800 W)/(75 V) = 64 A,
The resistance of the wire is
R = ∆V /i = (75 V)/(64 A) = 1.17 Ω.
The length of the wire is then found from
L=
RA
(1.17 Ω)(2.6×10−6 m2 )
=
= 6.1 m.
ρ
(5.0×10−7 Ωm)
One could easily wind this much nichrome to make a toaster oven. Of course allowing 64 Amps to
be drawn through household wiring will likely blow a fuse.
(b) We want to combine the above calculations into one formula, so
L=
then
L=
RA
A∆V /i
A(∆V )2
=
=
,
ρ
ρ
Pρ
(2.6×10−6 m2 )(110 V)2
= 13 m.
(4800 W)(5.0×10−7 Ωm)
Hmm. We need more wire if the potential difference is increased? Does this make sense? Yes, it
does. We need more wire because we need more resistance to decrease the current so that the same
power output occurs.
E31-40 (a) The energy required to bring the water to boiling is Q = mC∆T . The time required
is
Q
(2.1 kg)(4200 J/kg)(100◦ C − 18.5◦ C)
t=
=
= 2.22×103 s
0.77P
0.77(420 W)
(b) The additional time required to boil half of the water away is
t=
E31-41
mL/2
(2.1 kg)(2.26×106 J/kg)/2
=
= 7340 s.
0.77P
0.77(420 W)
(a) Integrate both sides of Eq. 31-26;
Z
0
q
dq
q − EC
q
ln(q − EC)|0
q − EC
ln
−EC
q − EC
−EC
q
Z
t
dt
,
RC
t
t = −
,
RC 0
t
= −
,
RC
= −
0
= e−t/RC ,
= EC 1 − e−t/RC .
That wasn’t so bad, was it?
89
(b) Rearrange Eq. 31-26 in order to get q terms on the left and t terms on the right, then
integrate;
Z q
Z t
dq
dt
= −
,
q
RC
q0
0
t
t q
ln q|q0 = −
,
RC 0
q
t
ln
= −
,
q0
RC
q
= e−t/RC ,
q0
q
= q0 e−t/RC .
That wasn’t so bad either, was it?
E31-42 (a) τC = RC = (1.42×106 Ω)(1.80×10−6 F) = 2.56 s.
(b) q0 = C∆V = (1.80×10−6 F)(11.0 V) = 1.98×10−5 C.
(c) t = −τC ln(1 − q/q0 ), so
t = −(2.56 s) ln(1 − 15.5×10−6 C/1.98×10−5 C) = 3.91 s.
E31-43 Solve n = t/τC = − ln(1 − 0.99) = 4.61.
E31-44 (a) ∆V = E(1 − e−t/τC ), so
τC = −(1.28×10−6 s)/ ln(1 − 5.00 V/13.0 V) = 2.64×10−6 s
(b) C = τC /R = (2.64×10−6 s)/(15.2×103 Ω) = 1.73×10−10 F
E31-45 (a) ∆V = Ee−t/τC , so
τC = −(10.0 s)/ ln(1.06 V/100 V) = 2.20 s
(b) ∆V = (100 V)e−17 s/2.20 s = 4.4×10−2 V.
E31-46 ∆V = Ee−t/τC and τC = RC, so
t
t
t
R=−
=−
=
.
C ln(∆V /∆V0 )
(220×10−9 F) ln(0.8 V/5 V)
4.03×10−7 F
If t is between 10.0 µs and 6.0 ms, then R is between
R = (10×10−6 s)/(4.03×10−7 F) = 24.8Ω,
and
R = (6×10−3 s)/(4.03×10−7 F) = 14.9×103 Ω.
E31-47 The charge on the capacitor needs to build up to a point where the potential across the
capacitor is VL = 72 V, and this needs to happen within 0.5 seconds. This means that we want to
solve
C∆VL = CE 1 − eT /RC
for R knowing that T = 0.5 s. This expression can be written as
R=−
T
(0.5 s)
=−
= 2.35×106 Ω.
C ln(1 − VL /E)
(0.15 µC) ln(1 − (72 V)/(95 V))
90
p
√
E31-48 (a) q0 = 2U C = 2(0.50 J)(1.0×10−6 F) = 1×10−3 C.
(b) i0 = ∆V0 /R = q0 /RC = (1×10−3 C)/(1.0×106 Ω)(1.0×10−6 F) = 1×10−3 A.
(c) ∆VC = ∆V0 e−t/τC , so
∆VC =
(1×10−3 C) −t/(1.0×106 Ω)(1.0×10−6 F)
e
= (1000 V)e−t/(1.0 s)
(1.0×10−6 F)
Note that ∆VR = ∆VC .
(d) PR = (∆VR )2 /R, so
PR = (1000 V)2 e−2t/(1.0 s) /(1×106 ω) = (1 W)e−2t/(1.0 s) .
E31-49 (a) i = dq/dt = Ee−t/τC /R, so
i=
6
−6
(4.0 V)
e−(1.0 s)/(3.0×10 Ω)(1.0×10 F) = 9.55×10−7 A.
(3.0×106 Ω)
(b) PC = i∆V = (E 2 /R)e−t/τC (1 − e−t/τC ), so
6
−6
(4.0 V)2 −(1.0 s)/(3.0×106 Ω)(1.0×10−6 F) PC =
e
1 − e−(1.0 s)/(3.0×10 Ω)(1.0×10 F) = 1.08×10−6 W.
6
(3.0×10 Ω)
(c) PR = i2 R = (E 2 /R)e−2t/τC , so
PR =
(4.0 V)2 −2(1.0 s)/(3.0×106 Ω)(1.0×10−6 F)
e
= 2.74×10−6 W.
(3.0×106 Ω)
(d) P = PR + PC , or
P = 2.74×10−6 W + 1.08×10−6 W = 3.82×10−6 W
E31-50 The rate of energy dissipation in the resistor is
PR = i2 R = (E 2 /R)e−2t/τC .
Evaluating
Z
E 2 ∞ −2t/RC
E2
PR dt =
e
dt =
C,
R 0
2
0
but that is the original energy stored in the capacitor.
Z
P31-1
∞
The terminal voltage of the battery is given by V = E − ir, so the internal resistance is
r=
E −V
(12.0 V) − (11.4 V)
=
= 0.012 Ω,
i
(50 A)
so the battery appears within specs.
The resistance of the wire is given by
R=
∆V
(3.0 V)
=
= 0.06 Ω,
i
(50 A)
so the cable appears to be bad.
What about the motor? Trying it,
R=
∆V
(11.4 V) − (3.0 V)
=
= 0.168 Ω,
i
(50 A)
so it appears to be within spec.
91
P31-2
Traversing the circuit we have
E − ir1 + E − ir2 − iR = 0,
so i = 2E/(r1 + r2 + R). The potential difference across the first battery is then
r2 − r 1 + R
2r1
∆V1 = E − ir1 = E 1 −
=E
r 1 + r2 + R
r1 + r 2 + R
This quantity will only vanish if r2 − r1 + R = 0, or r1 = R + r2 . Since r1 > r2 this is actually
possible; R = r1 − r2 .
P31-3
∆V = E − iri and i = E/(ri + R), so
∆V = E
R
,
ri + R
There are then two simultaneous equations:
(0.10 V)(500 Ω) + (0.10 V)ri = E(500 Ω)
and
(0.16 V)(1000 Ω) + (0.16 V)ri = E(1000 Ω),
with solution
(a) ri = 1.5×103 Ω and
(b) E = 0.400 V.
2
(c) The cell receives energy from the sun at a rate (2.0 mW/cm )(5.0 cm2 ) = 0.010 W. The cell
2
2
converts energy at a rate of V /R = (0.16 V) /(1000 Ω) = 0.26 %
P31-4
(a) The emf of the battery can be found from
E = iri + ∆V l = (10 A)(0.05 Ω) + (12 V) = 12.5 V
(b) Assume that resistance is not a function of temperature. The resistance of the headlights is
then
rl = (12.0 V)/(10.0 A) = 1.2 Ω.
The potential difference across the lights when the starter motor is on is
∆V l = (8.0 A)(1.2 Ω) = 9.6 V,
and this is also the potential difference across the terminals of the battery. The current through the
battery is then
E − ∆V
(12.5 V) − (9.6 V)
i=
=
= 58 A,
ri
(0.05 Ω)
so the current through the motor is 50 Amps.
P31-5
(a) The resistivities are
ρA = rA A/L = (76.2×10−6 Ω)(91.0×10−4 m2 )/(42.6 m) = 1.63×10−8 Ω · m,
and
ρB = rB A/L = (35.0×10−6 Ω)(91.0×10−4 m2 )/(42.6 m) = 7.48×10−9 Ω · m.
(b) The current is i = ∆V /(rA + rB ) = (630 V)/(111.2 µΩ) = 5.67×106 A. The current density
is then
j = (5.67×106 A)/(91.0×10−4 m2 ) = 6.23×108 A/m2 .
(c) EA = ρA j = (1.63×10−8 Ω · m)(6.23×108 A/m2 ) = 10.2 V/m and EB = ρB j = (7.48×10−9 Ω ·
m)(6.23×108 A/m2 ) = 4.66 V/m.
(d) ∆VA = EA L = (10.2 V/m)(42.6 m) = 435 V and ∆VB = EB L = (4.66 V/m)(42.6 m) = 198 V.
92
P31-6 Set up the problem with the traditional presentation of the Wheatstone bridge problem.
Then the symmetry of the problem (flip it over on the line between x and y) implies that there is no
current through r. As such, the problem is equivalent to two identical parallel branches each with
two identical series resistances.
Each branch has resistance R + R = 2R, so the overall circuit has resistance
1
1
1
1
=
+
= ,
Req
2R 2R
R
so Req = R.
P31-7
P31-8 (a) The loop through R1 is trivial: i1 = E2 /R1 = (5.0 V)/(100 Ω) = 0.05 A. The loop
through R2 is only slightly harder: i2 = (E2 + E3 − E1 )/R2 = 0.06 A.
(b) ∆Vab = E3 + E2 = (5.0 V) + (4.0 V) = 9.0 V.
P31-9 (a) The three way light-bulb has two filaments (or so we are told in the question). There
are four ways for these two filaments to be wired: either one alone, both in series, or both in
parallel. Wiring the filaments in series will have the largest total resistance, and since P = V 2 /R
this arrangement would result in the dimmest light. But we are told the light still operates at the
lowest setting, and if a filament burned out in a series arrangement the light would go out.
We then conclude that the lowest setting is one filament, the middle setting is another filament,
and the brightest setting is both filaments in parallel.
(b) The beauty of parallel settings is that then power is additive (it is also addictive, but that’s
a different field.) One filament dissipates 100 W at 120 V; the other filament (the one that burns
out) dissipates 200 W at 120 V, and both together dissipate 300 W at 120 V.
The resistance of one filament is then
R=
(120 V)2
(∆V )2
=
= 144 Ω.
P
(100 W)
The resistance of the other filament is
R=
(∆V )2
(120 V)2
=
= 72 Ω.
P
(200 W)
P31-10 We can assume that R “contains” all of the resistance of the resistor, the battery and the
ammeter, then
R = (1.50 V)/(1.0 m/A) = 1500 Ω.
For each of the following parts we apply R + r = ∆V /i, so
(a) r = (1.5 V)/(0.1 mA) − (1500 Ω) = 1.35×104 Ω,
(b) r = (1.5 V)/(0.5 mA) − (1500 Ω) = 1.5×103 Ω,
(c) r = (1.5 V)/(0.9 mA) − (1500 Ω) = 167Ω.
(d) R = (1500 Ω) − (18.5 Ω) = 1482 Ω
P31-11
(a) The effective resistance of the parallel branches on the middle and the right is
R2 R 3
.
R2 + R3
93
The effective resistance of the circuit as seen by the battery is then
R1 +
R2 R3
R1 R 2 + R1 R3 + R2 R3
=
,
R2 + R3
R 2 + R3
The current through the battery is
i=E
R 2 + R3
,
R1 R 2 + R1 R3 + R2 R3
The potential difference across R1 is then
∆V1 = E
R2 + R 3
R1 ,
R 1 R2 + R 1 R 3 + R 2 R 3
while ∆V3 = E − ∆V1 , or
∆V3 = E
R 2 R3
,
R1 R2 + R 1 R3 + R 2 R 3
so the current through the ammeter is
i3 =
or
i3 = (5.0 V)
∆V3
R2
=E
,
R3
R1 R2 + R 1 R3 + R 2 R 3
(4 Ω)
= 0.45 A.
(2 Ω)(4 Ω) + (2 Ω)(6 Ω) + (4 Ω)(6 Ω)
(b) Changing the locations of the battery and the ammeter is equivalent to swapping R1 and
R3 . But since the expression for the current doesn’t change, then the current is the same.
P31-12
∆V1 + ∆V2 = ∆VS + ∆VX ; if Va = Vb , then ∆V1 = ∆VS . Using the first expression,
ia (R1 + R2 ) = ib (RS + RX ),
using the second,
i a R1 = i b R2 .
Dividing the first by the second,
1 + R2 /R1 = 1 + RX /RS ,
or RX = RS (R2 /R1 ).
P31-13
P31-14
Lv = ∆Q/∆m and ∆Q/∆t = P = i∆V , so
Lv =
i∆V
(5.2 A)(12 V)
=
= 2.97×106 J/kg.
∆m/∆t
(21×10−6 kg/s)
P31-15 P = i2 R. W = p∆V , where V is volume. p = mg/A and V = Ay, where y is the height
of the piston. Then P = dW/dt = mgv. Combining all of this,
v=
i2 R
(0.240 A)2 (550 Ω)
=
= 0.274 m/s.
mg
(11.8 kg)(9.8 m/s2 )
94
P31-16
(a) Since q = CV , then
q = (32×10−6 F) (6 V) + (4 V/s)(0.5 s) − (2 V/s2 )(0.5 s)2 = 2.4×10−4 C.
(b) Since i = dq/dt = C dV /dt, then
i = (32×10−6 F) (4 V/s) − 2(2 V/s2 )(0.5 s) = 6.4×10−5 A.
(c) Since P = iV ,
P = [ (4 V/s) − 2(2 V/s2 )(0.5 s) (6 V) + (4 V/s)(0.5 s) − (2 V/s2 )(0.5 s)2 = 4.8×10−4 W.
P31-17
(a) We have P = 30P0 and i = 4i0 . Then
R=
P
30P0
30
=
=
R0 .
i2
(4i0 )2
16
We don’t really care what happened with the potential difference, since knowing the change in
resistance of the wire should give all the information we need.
The volume of the wire is a constant, even upon drawing the wire out, so LA = L0 A0 ; the
product of the length and the cross sectional area must be a constant.
Resistance is given by R = ρL/A, but A = L0 A0 /L, so the length of the wire is
s
s
A0 L0 R
30 A0 L0 R0
L=
=
= 1.37L0 .
ρ
16
ρ
(b) We know that A = L0 A0 /L, so
A=
P31-18
A0
L
A0 =
= 0.73A0 .
L0
1.37
(a) The capacitor charge as a function of time is given by Eq. 31-27,
q = CE 1 − e−t/RC ,
while the current through the circuit (and the resistor) is given by Eq. 31-28,
i=
E −t/RC
e
.
R
The energy supplied by the emf is
U=
Z
Ei dt = E
Z
dq = Eq;
but the energy in the capacitor is UC = q∆V /2 = Eq/2.
(b) Integrating,
Z
Z
E2
E2
Eq
UR = i2 Rdt =
e−2t/RC dt =
=
.
R
2C
2
95
P31-19
The capacitor charge as a function of time is given by Eq. 31-27,
q = CE 1 − e−t/RC ,
while the current through the circuit (and the resistor) is given by Eq. 31-28,
i=
E −t/RC
e
.
R
The energy stored in the capacitor is given by
U=
q2
,
2C
so the rate that energy is being stored in the capacitor is
PC =
dU
q dq
q
=
= i.
dt
C dt
C
The rate of energy dissipation in the resistor is
PR = i2 R,
so the time at which the rate of energy dissipation in the resistor is equal to the rate of energy
storage in the capacitor can be found by solving
PC
2
i R
iRC
ECe−t/RC
e−t/RC
t
= PR ,
q
=
i,
C
= q,
= CE 1 − e−t/RC ,
= 1/2,
= RC ln 2.
96
~ = q~v × B.
~
E32-1 Apply Eq. 32-3, F
All of the paths which involve left hand turns are positive particles (path 1); those paths which
involve right hand turns are negative particle (path 2 and path 4); and those paths which don’t turn
involve neutral particles (path 3).
E32-2 (a) The greatest magnitude of force is F = qvB = (1.6×10−19 C)(7.2×106 m/s)(83×10−3 T) =
9.6×10−14 N. The least magnitude of force is 0.
(b) The force on the electron is F = ma; the angle between the velocity and the magnetic field
is θ, given by ma = qvB sin θ. Then
(9.1×10−31 kg)(4.9×1016 m/s2 )
θ = arcsin
= 28◦ .
(1.6×10−19 C)(7.2×106 m/s)(83×10−3 T)
E32-3 (a) v = E/B = (1.5×103 V/m)/(0.44 T) = 3.4×103 m/s.
E32-4 (a) v = F/qB sin θ = (6.48×10−17 N/(1.60×10−19 C)(2.63×10−3 T) sin(23.0◦ ) = 3.94×105 m/s.
(b) K = mv 2 /2 = (938 MeV/c2 )(3.94×105 m/s)2 /2 = 809 eV.
E32-5
The magnetic force on the proton is
FB = qvB = (1.6×10−19 C)(2.8×107 m/s)(30eex−6 T) = 1.3×10−16 N.
The gravitational force on the proton is
mg = (1.7×10−27 kg)(9.8 m/s2 ) = 1.7×10−26 N.
The ratio is then 7.6×109 . If, however, you carry the number of significant digits for the intermediate
answers farther you will get the answer which is in the back of the book.
p
E32-6 The speed of the electron is given by v = 2q∆V /m, or
p
v = 2(1000 eV)/(5.1×105 eV/c2 ) = 0.063c.
The electric field between the plates is E = (100 V)/(0.020 m) = 5000 V/m. The required magnetic
field is then
B = E/v = (5000 V/m)/(0.063c) = 2.6×10−4 T.
E32-7 Both have the same velocity. Then K p /K e = mp v 2 /me v 2 = mp /me =.
p
E32-8 The speed of the ion is given by v = 2q∆V /m, or
p
v = 2(10.8 keV)/(6.01)(932 MeV/c2 ) = 1.96×10−3 c.
The required electric field is E = vB = (1.96×10−3 c)(1.22 T) = 7.17×105 V/m.
E32-9
(a) For a charged particle moving in a circle in a magnetic field we apply Eq. 32-10;
r=
mv
(9.11×10−31 kg)(0.1)(3.00×108 m/s)
=
= 3.4×10−4 m.
|q|B
(1.6×10−19 C)(0.50 T)
(b) The (non-relativistic) kinetic energy of the electron is
K=
1
1
mv 2 = (0.511 MeV)(0.10c)2 = 2.6×10−3 MeV.
2
2
97
p
p
E32-10 (a) v = 2K/m = 2(1.22 keV)/(511 keV/c2 ) = 0.0691c.
(b) B = mv/qr = (9.11×10−31 kg)(0.0691c)/(1.60×10−19 C)(0.247 m) = 4.78×10−4 T.
(c) f = qB/2πm = (1.60×10−19 C)(4.78×10−4 T)/2π(9.11×10−31 kg) = 1.33×107 Hz.
(d) T = 1/f = 1/(1.33×107 Hz) = 7.48×10−8 s.
p
p
E32-11 (a) v = 2K/m = 2(350 eV)/(511 keV/c2 ) = 0.037c.
(b) r = mv/qB = (9.11×10−31 kg)(0.037c)/(1.60×10−19 C)(0.20T) = 3.16×10−4 m.
E32-12 The frequency is f = (7.00)/(1.29×10−3 s) = 5.43×103 Hz. The mass is given by m =
qB/2πf , or
(1.60×10−19 C)(45.0×10−3 T)
m=
= 2.11×10−25 kg = 127 u.
2π(5.43×103 Hz)
E32-13
(a) Apply Eq. 32-10, but rearrange it as
v=
|q|rB
2(1.6×10−19 C)(0.045 m)(1.2 T)
=
= 2.6×106 m/s.
m
4.0(1.66×10−27 kg)
(b) The speed is equal to the circumference divided by the period, so
T =
2πr
2πm
2π4.0(1.66×10−27 kg)
=
=
= 1.1×10−7 s.
v
|q|B
2(1.6 × 10−19 C)(1.2 T)
(c) The (non-relativistic) kinetic energy is
K=
|q|2 r2 B
(2×1.6×10−19 C)2 (0.045 m)2 (1.2 T)2
=
= 2.24×10−14 J.
2m
2(4.0×1.66×10−27 kg))
To change to electron volts we need merely divide this answer by the charge on one electron, so
K=
(d) ∆V =
K
q
(2.24×10−14 J)
= 140 keV.
(1.6×10−19 C)
= (140 keV)/(2e) = 70 V.
E32-14 (a) R = mv/qB = (938 MeV/c2 )(0.100c)/e(1.40 T) = 0.223 m.
(b) f = qB/2πm = e(1.40 T)/2π(938 MeV/c2 ) = 2.13×107 Hz.
E32-15 (a) Kα /K p = (qα2 /mα )/(q p 2 /mp ) = 22 /4 = 1.
(b) K d /K p = (q d 2 /md )/(q p 2 /mp ) = 12 /2 = 1/2.
E32-16 (a) K = q∆V . Then K p = e∆Vp, K d = e∆V , and Kα√= 2e∆V .
(b) r = sqrt2mK/qB. Then rd /rp = p (2/1)(1/1)/(1/1) = 2.
√
(c) r = sqrt2mK/qB. Then rα /rp = (4/1)(2/1)/(2/1) = 2.
√
√
√
E32-17 r = 2mK/|q|B = ( m/|q|)( 2K/B). All three particles are traveling with the same
kinetic energy in the same magnetic field. The relevant factors are in front; we just need to compare
the mass and charge of each of the three particles.
√
(a) The radius of the deuteron path is 12 rp .
√
(b) The radius of the alpha particle path is 24 rp = rp .
98
E32-18 The neutron, being neutral, is unaffected by the magnetic field and moves off in a line
tangent to the original path. The proton moves at the same original speed as the deuteron and has
the same charge, but since it has half the mass it moves in a circle with half the radius.
E32-19 (a) The proton momentum would be pc = qcBR = e(3.0×108 m/s)(41×10−6 T)(6.4×106 m) =
7.9×104 MeV. Since 79000 MeV is much, much greater than 938 MeV the proton is ultra-relativistic.
Then E ≈ pc, and since γ = E/mc2 we have γ = p/mc. Inverting,
s
r
v
1
m2 c2
m2 c2
≈ 0.99993.
= 1− 2 = 1− 2 ≈1−
c
γ
p
2p2
√
E32-20 (a) Classically, R = 2mK/qB, or
p
R = 2(0.511 MeV/c2 )(10.0 MeV)/e(2.20 T) = 4.84×10−3 m.
(b) This would be an ultra-relativistic electron, so K ≈ E ≈ pc, then R = p/qB = K/qBc, or
R = (10.0 MeV)/e(2.2 T)(3.00×108 m/s) = 1.52×10−2 m.
(c) The electron is effectively traveling at the speed of light, so T = 2πR/c, or
T = 2π(1.52×10−2 m)/(3.00×108 m/s) = 3.18×10−10 s.
This result does depend on the speed!
E32-21
Use Eq. 32-10, except we rearrange for the mass,
m=
|q|rB
2(1.60×10−19 C)(4.72 m)(1.33 T)
=
= 9.43×10−27 kg
v
0.710(3.00×108 m/s)
However, if it is moving at this velocity then the “mass” which we have here is not the true mass,
but a relativistic correction. For a particle moving at 0.710c we have
1
γ=p
1−
v 2 /c2
1
=p
1 − (0.710)2
= 1.42,
so the true mass of the particle is (9.43×10−27 kg)/(1.42) = 6.64×10−27 kg. The number of nucleons
present in this particle is then (6.64×10−27 kg)/(1.67×10−27 kg) = 3.97 ≈ 4. The charge was +2,
which implies two protons, the other two nucleons would be neutrons, so this must be an alpha
particle.
E32-22 (a) Since 950 GeV is much, much greater than 938 MeV the proton is ultra-relativistic.
γ = E/mc2 , so
r
r
v
1
m2 c4
m2 c4
= 1− 2 = 1−
≈
1
−
≈ 0.9999995.
c
γ
E2
2E 2
(b) Ultra-relativistic motion requires pc ≈ E, so
B = pc/qRc = (950 GeV)/e(750 m)(3.00×108 m/s) = 4.44 T.
99
2
E32-23 First use 2πf = qB/m. The use K = q 2 B 2 R2 /2m = mR2 (2πf
√ ) /2. The number of
turns is n = K/2q∆V , on average
is located at a distance R/ 2 from the center, so the
√ the particle
√
distance traveled is x = n2πR/ 2 = n 2πR. Combining,
√ 3 3
√ 3
2π R mf 2
2π (0.53 m)3 (2 × 932×103 keV/c2 )(12×106 /s)2
x=
=
= 240 m.
q∆V
e(80 kV)
E32-24 The particle moves in a circle. x = R sin ωt and y = R cos ωt.
E32-25 We will use Eq. 32-20, E H = v d B, except we will not take the derivation through to Eq.
32-21. Instead, we will set the drift velocity equal to the speed of the strip. We will, however, set
E H = ∆V H /w. Then
v=
EH
∆V H /w
(3.9×10−6 V)/(0.88×10−2 m)
=
=
= 3.7×10−1 m/s.
B
B
(1.2×10−3 T)
E32-26 (a) v = E/B = (40×10−6 V)/(1.2×10−2 m)/(1.4 T) = 2.4×10−3 m/s.
(b) n = (3.2 A)(1.4 T)/(1.6×10−19 C)(9.5×10−6 m)(40×10−6 V) = 7.4×1028 /m3 .; Silver.
E32-27 E H = v d B and v d = j/ne. Combine and rearrange.
E32-28 (a) Use the result of the previous exercise and E c = ρj.
(b) (0.65 T)/(8.49×1028 /m3 )(1.60×10−19 C)(1.69×10−8 Ω · m) = 0.0028.
E32-29
~ is perpendicular to B
~ can use
Since L
FB = iLB.
Equating the two forces,
iLB
i
= mg,
(0.0130 kg)(9.81 m/s2 )
mg
=
=
= 0.467 A.
LB
(0.620 m)(0.440 T)
Use of an appropriate right hand rule will indicate that the current must be directed to the right
in order to have a magnetic force directed upward.
E32-30 F = iLB sin θ = (5.12 × 103 A)(100 m)(58 × 10−6 T) sin(70◦ ) = 27.9 N. The direction is
horizontally west.
E32-31 (a) We use Eq. 32-26 again, and since the (horizontal) axle is perpendicular to the
vertical component of the magnetic field,
i=
F
(10, 000 N)
=
= 3.3×108 A.
BL
(10 µT)(3.0 m)
(b) The power lost per ohm of resistance in the rails is given by
P/r = i2 = (3.3×108 A)2 = 1.1×1017 W.
(c) If such a train were to be developed the rails would melt well before the train left the station.
100
E32-32 F = idB, so a = F/m = idB/m. Since a is constant, v = at = idBt/m. The direction is
to the left.
R
R
~ is of interest. Then F = dF = i By dx, or
E32-33 Only the ĵ component of B
Z 3.2
−3
2
F = (5.0 A)(8×10 T/m )
x2 dx = 0.414 N.
1.2
The direction is −k̂.
E32-34 The magnetic force will have two components: one will lift vertically (Fy = F sin α), the
other push horizontally (Fx = F cos α). The rod will move when Fx > µ(W − Fy ). We are interested
in the minimum value for F as a function of α. This occurs when
d
µW
dF
=
= 0.
dα
dα cos α + µ sin α
This happens when µ = tan α. Then α = arctan(0.58) = 30◦ , and
F =
(0.58)(1.15 kg)(9.81 m/s2 )
= 5.66 N
cos(30◦ ) + (0.58) sin(30◦ )
is the minimum force. Then B = (5.66 N)/(53.2 A)(0.95 m) = 0.112 T.
E32-35 We choose that the field points from the shorter side to the longer side.
(a) The magnetic field is parallel to the 130 cm side so there is no magnetic force on that side.
The magnetic force on the 50 cm side has magnitude
FB = iLB sin θ,
where θ is the angle between the 50 cm side and the magnetic field. This angle is larger than 90◦ ,
but the sine can be found directly from the triangle,
sin θ =
(120 cm)
= 0.923,
(130 cm)
and then the force on the 50 cm side can be found by
FB = (4.00 A)(0.50 m)(75.0×10−3 T)
(120 cm)
= 0.138 N,
(130 cm)
and is directed out of the plane of the triangle.
The magnetic force on the 120 cm side has magnitude
FB = iLB sin θ,
where θ is the angle between the 1200 cm side and the magnetic field. This angle is larger than
180◦ , but the sine can be found directly from the triangle,
sin θ =
(−50 cm)
= −0.385,
(130 cm)
and then the force on the 50 cm side can be found by
FB = (4.00 A)(1.20 m)(75.0×10−3 T)
and is directed into the plane of the triangle.
(b) Look at the three numbers above.
101
(−50 cm)
= −0.138 N,
(130 cm)
E32-36 τ = N iAB sin θ, so
τ = (20)(0.1 A)(0.12 m)(0.05 m)(0.5 T) sin(90◦ − 33◦ ) = 5.0×10−3 N · m.
E32-37 The external magnetic field must be in the plane of the clock/wire loop. The clockwise
current produces a magnetic dipole moment directed into the plane of the clock.
(a) Since the magnetic field points along the 1 pm line and the torque is perpendicular to both
the external field and the dipole, then the torque must point along either the 4 pm or the 10 pm line.
Applying Eq. 32-35, the direction is along the 4 pm line. It will take the minute hand 20 minutes
to get there.
(b) τ = (6)(2.0 A)π(0.15 m)2 (0.07 T) = 0.059 N · m.
~ must be perpendicular to B
~ then B
~ must be along k̂. The magnitude of v is
P32-1
Since F
p
p
2
2
(40) + (35) km/s = 53.1 km/s; the magnitude of F is (−4.2)2 + (4.8)2 fN = 6.38 fN. Then
B = F/qv = (6.38×10−15 N)/(1.6×10−19 C)(53.1×103 m/s) = 0.75 T.
~ = 0.75 T k̂.
or B
~ + ~v × B).
~ For the initial velocity given,
P32-2 ~a = (q/m)(E
~ = (15.0×103 m/s)(400×10−6 T)ĵ − (12.0×103 m/s)(400×10−6 T)k̂.
~v × B
But since there is no acceleration in the ĵ or k̂ direction this must be offset by the electric field.
Consequently, two of the electric field components are Ey = −6.00 V/m and Ez = 4.80 V/m. The
third component of the electric field is the source of the acceleration, so
Ex = max /q = (9.11×10−31 kg)(2.00×1012 m/s2 )/(−1.60×10−19 C) = −11.4 V/m.
~ The electron moves horizontally, there is a
P32-3 (a) Consider first the cross product, ~v × B.
~ which is down, so the cross product results in a vector which points to the left
component of the B
of the electron’s path.
~ = q~v × B,
~ and since the electron has a negative charge
But the force on the electron is given by F
the force on the electron would be directed to the right of the electron’s path.
(b) The kinetic energy of the electrons is much less than
p the rest mass energy, so this is nonrelativistic motion. The speed of the electron is then v = 2K/m, and the magnetic force on the
electron is FB = qvB, where we are assuming sin θ = 1 because the electron moves horizontally
through a magnetic field with a vertical component. We can ignore the effect of the magnetic field’s
horizontal component because the electron is moving parallel to this component.
The acceleration of the electron because of the magnetic force is then
r
qvB
qB 2K
a =
=
,
m
m
m
s
(1.60×10−19 C)(55.0×10−6 T) 2(1.92×10−15 J)
= 6.27×1014 m/s2 .
=
(9.11×10−31 kg)
(9.11×10−31 kg)
(c) The electron travels a horizontal distance of 20.0 cm in a time of
(20.0 cm)
(20.0 cm)
t= p
=p
= 3.08×10−9 s.
−15
2K/m
2(1.92×10 J)/(9.11×10−31 kg)
In this time the electron is accelerated to the side through a distance of
d=
1
1 2
at = (6.27×1014 m/s2 )(3.08×10−9 s)2 = 2.98 mm.
2
2
102
2 2
2
2
P32-4
p (a) d needs to be larger than the turn radius, so R ≤ d; but 2mK/q B = R ≤ d , or
2
2
B ≥ 2mK/q d .
(b) Out of the page.
P32-5 Only undeflected ions emerge from the velocity selector, so v = E/B. The ions are then
deflected by B 0 with a radius of curvature of r = mv/qB; combining and rearranging, q/m =
E/rBB 0 .
P32-6 The ions are given a kinetic energy K = q∆V ; they are then deflected with a radius of
curvature given by R2 = 2mK/q 2 B 2 . But x = 2R. Combine all of the above, and m = B 2 qx2 /8∆V.
P32-7
(a) Start with the equation in Problem 6, and take the square root of both sides to get
√
m=
B2q
8∆V
12
x,
12
dx,
and then take the derivative of x with respect to m,
1 dm
√ =
2 m
B2q
8∆V
and then consider finite differences instead of differential quantities,
∆m =
mB 2 q
2∆V
12
∆x,
(b) Invert the above expression,
∆x =
2∆V
mB 2 q
12
∆m,
and then put in the given values,
∆x =
=
2(7.33×103 V)
−27
(35.0)(1.66×10 kg)(0.520 T)2 (1.60×10−19 C)
8.02 mm.
12
(2.0)(1.66×10−27 kg),
Note that we used 35.0 u for the mass; if we had used 37.0 u the result would have been closer to
the answer in the back of the book.
p
p
P32-8 (a) B = 2∆V m/qr2 = 2(0.105 MV)(238)(932 MeV/c2 )/2e(0.973 m)2 = 5.23×10−7 T.
(b) The number of atoms in a gram is 6.02×1023 /238 = 2.53×1021 . The current is then
(0.090)(2.53×1021 )(2)(1.6×10−19 C)/(3600 s) = 20.2 mA.
P32-9 (a) −q.
(b) Regardless of speed, the orbital period is T = 2πm/qB. But they collide halfway around a
complete orbit, so t = πm/qB.
P32-10
103
P32-11
(a) The period of motion can be found from the reciprocal of Eq. 32-12,
T =
2πm
2π(9.11×10−31 kg)
=
= 7.86×10−8 s.
|q|B
(1.60×10−19 C)(455×10−6 T)
(b) We need to find the velocity of the electron from the kinetic energy,
p
p
v = 2K/m = 2(22.5 eV)(1.60×10−19 J/eV)/(9.11×10−31 kg) = 2.81×106 m/s.
The velocity can written in terms of components which are parallel and perpendicular to the magnetic
field. Then
v|| = v cos θ and v⊥ = v sin θ.
The pitch is the parallel distance traveled by the electron in one revolution, so
p = v|| T = (2.81×106 m/s) cos(65.5◦ )(7.86×10−8 s) = 9.16 cm.
(c) The radius of the helical path is given by Eq. 32-10, except that we use the perpendicular
velocity component, so
R=
(9.11×10−31 kg)(2.81×106 m/s) sin(65.5◦ )
mv⊥
=
= 3.20 cm
|q|B
(1.60×10−19 C)(455×10−6 T)
R
~ = i b d~l × B.
~ d~l has two components, those parallel to the path, say d~x and those
P32-12 F
a
perpendicular, say d~y. Then the integral can be written as
Z b
Z b
~ =
~ +
~
F
d~x × B
d~y × B.
a
a
R
~ is constant, and can be removed from the integral. b d~x = ~l, a vector that points from a to
But B
a
Rb
b. a d~y = 0, because there is no net motion perpendicular to ~l.
P32-13 qvy B = Fx = m dvx /dt; −qvx B = Fy = m dvy /dt. Taking the time derivative of the
second expression and inserting into the first we get
m d2 vy
qvy B = m −
,
qB dt2
which has solution vy = −v sin(mt/qB), where v is a constant. Using the second equation we find
that there is a similar solution for vx , except that it is out of phase, and so vx = v cos(mt/qB).
Integrating,
Z
Z
qBv
x = vx dt = v cos(mt/qB) =
sin(mt/qB).
m
Similarly,
Z
Z
qBv
y = vy dt = −v sin(mt/qB) =
cos(mt/qB).
m
This is the equation of a circle.
P32-14
~ = îdx + ĵdy + k̂dz. B
~ is uniform, so that the integral can be written as
dL
I
I
I
I
~
~
~
~
~
F = i (îdx + ĵdy + k̂dz) × B = iî × B dx + iĵ × B dy + ik̂ × B dz,
but since
H
dx =
H
dy =
H
dz = 0, the entire expression vanishes.
104
P32-15
The current pulse provides an impulse which is equal to
Z
Z
Z
F dt = BiL dt = BL i dt = BLq.
This gives an initial velocity of v0 = BLq/m, which will cause the rod to hop to a height of
h = v02 /2g = B 2 L2 q 2 /2m2 g.
Solving for q,
q=
(0.013 kg) p
m p
2gh =
2(9.8 m/s2 )(3.1 m) = 4.2 C.
BL
(0.12 T)(0.20 m)
P32-16
P32-17 The torque on a current carrying loop depends on the orientation of the loop; the
maximum torque occurs when the plane of the loop is parallel to the magnetic field. In this case the
magnitude of the torque is from Eq. 32-34 with sin θ = 1—
τ = N iAB.
The area of a circular loop is A = πr2 where r is the radius, but since the circumference is C = 2πr,
we can write
C2
A=
.
4π
The circumference is not the length of the wire, because there may be more than one turn. Instead,
C = L/N , where N is the number of turns.
Finally, we can write the torque as
τ = Ni
L2
iL2 B
B
=
,
4πN 2
4πN
which is a maximum when N is a minimum, or N = 1.
~ = i dL
~ × B;
~ the direction of dF
~ will be upward and somewhat toward the center. L
~
P32-18 dF
~ are a right angles, but only the upward component of dF
~ will survive the integration as the
and B
central components will cancel out by symmetry. Hence
Z
F = iB sin θ dL = 2πriB sin θ.
P32-19
The torque on the cylinder from gravity is
τ g = mgr sin θ,
where r is the radius of the cylinder. The torque from magnetism needs to balance this, so
mgr sin θ = N iAB sin θ = N i2rLB sin θ,
or
i=
mg
(0.262 kg)(9.8 m/s2 )
=
= 1.63 A.
2N LB
2(13)(0.127 m)(0.477 T)
105
E33-1 (a) The magnetic field from a moving charge is given by Eq. 33-5. If the protons are
moving side by side then the angle is φ = π/2, so
B=
µ0 qv
4π r2
and we are interested is a distance r = d. The electric field at that distance is
E=
1 q
,
4π0 r2
where in both of the above expressions q is the charge of the source proton.
On the receiving end is the other proton, and the force on that proton is given by
~ = q(E
~ + ~v × B).
~
F
The velocity is the same as that of the first proton (otherwise they wouldn’t be moving side by side.)
This velocity is then perpendicular to the magnetic field, and the resulting direction for the cross
~ Then for balance,
product will be opposite to the direction of E.
E
1 q
4π0 r2
1
0 µ0
= vB,
µ0 qv
= v
,
4π r2
= v2 .
We can solve this easily enough, and we find v ≈ 3 × 108 m/s.
(b) This is clearly a relativistic speed!
E33-2 B = µ0 i/2πd = (4π ×10−7 T · m/A)(120 A)/2π(6.3 m) = 3.8×10−6 T. This will deflect the
compass needle by as much as one degree. However, there is unlikely to be a place on the Earth’s
surface where the magnetic field is 210 µT. This was likely a typo, and should probably have been
21.0 µT. The deflection would then be some ten degrees, and that is significant.
E33-3 B = µ0 i/2πd = (4π×10−7 T · m/A)(50 A)/2π(1.3×10−3 m) = 37.7×10−3 T.
E33-4 (a) i = 2πdB/µ0 = 2π(8.13×10−2 m)(39.0×10−6 T)/(4π×10−7 T · m/A) = 15.9 A.
(b) Due East.
E33-5
Use
B=
µ0 i
(4π×10−7 N/A2 )(1.6 × 10−19 C)(5.6 × 1014 s−1 )
=
= 1.2×10−8 T.
2πd
2π(0.0015 m)
E33-6 Zero, by symmetry. Any contributions from the top wire are exactly canceled by contributions from the bottom wire.
E33-7 B = µ0 i/2πd = (4π×10−7 T · m/A)(48.8 A)/2π(5.2×10−2 m) = 1.88×10−4 T.
~ = q~v × B.
~ All cases are either parallel or perpendicular, so either F = 0 or F = qvB.
F
(a) F = qvB = (1.60×10−19 C)(1.08×107 m/s)(1.88×10−4 T) = 3.24×10−16 N. The direction of
~
F is parallel to the current.
(b) F = qvB = (1.60×10−19 C)(1.08×107 m/s)(1.88×10−4 T) = 3.24×10−16 N. The direction of
~ is radially outward from the current.
F
(c) F = 0.
106
E33-8 We want B1 = B2 , but with opposite directions. Then i1 /d1 = i2 /d2 , since all constants
cancel out. Then i2 = (6.6 A)(1.5 cm)/(2.25 cm) = 4.4 A, directed out of the page.
E33-9 For a single long straight wire, B = µ0 i/2πd but we need a factor of “2” since there are
two wires, then i = πdB/µ0 . Finally
i=
πdB
π(0.0405 m)(296, µT)
=
= 30 A
µ0
(4π×10−7 N/A2 )
E33-10 (a) The semi-circle part contributes half of Eq. 33-21, or µ0 i/4R. Each long straight wire
contributes half of Eq. 33-13, or µ0 i/4πR. Add the three contributions and get
µ0 i 2
(4π×10−7 N/A2 )(11.5 A) 2
Ba =
+1 =
+ 1 = 1.14×10−3 T.
4R π
4(5.20×10−3 m)
π
The direction is out of the page.
(b) Each long straight wire contributes Eq. 33-13, or µ0 i/2πR. Add the two contributions and
get
µ0 i
(4π×10−7 N/A2 )(11.5 A)
=
= 8.85×10−4 T.
Ba =
πR
π(5.20×10−3 m)
The direction is out of the page.
E33-11 z 3 = µ0 iR2 /2B = (4π ×10−7 N/A2 )(320)(4.20 A)(2.40×10−2 m)2 /2(5.0×10−6 T) = 9.73×
10−2 m3 . Then z = 0.46 m.
E33-12 The circular part contributes a fraction of Eq. 33-21, or µ0 iθ/4πR. Each long straight
wire contributes half of Eq. 33-13, or µ0 i/4πR. Add the three contributions and get
B=
µ0 i
(θ − 2).
4πR
The goal is to get B = 0 that will happen if θ = 2 radians.
E33-13 There are four current segments that could contribute to the magnetic field. The straight
segments, however, contribute nothing because the straight segments carry currents either directly
toward or directly away from the point P .
~ can be found by starting with
That leaves the two rounded segments. Each contribution to B
Eq. 33-21, or µ0 iθ/4πb. The direction is out of the page.
There is also a contribution from the top arc; the calculations are almost identical except that
this is pointing into the page and r = a, so µ0 iθ/4πa. The net magnetic field at P is then
µ0 iθ 1 1
B = B1 + B2 =
−
.
4π
b a
E33-14 For each straight wire segment use Eq. 33-12. When the length of wire is L, the distance
to the center is W/2; when the length of wire is W the distance to the center is L/2. There are four
terms, but they are equal in pairs, so
!
µ0 i
4L
4W
p
B =
+ p
,
4π W L2 /4 + W 2 /4 L L2 /4 + W 2 /4
√
2
2µ0 i
L
W2
2µ0 i L2 + W 2
√
=
+
=
.
π
WL
π L2 + W 2 W L W L
107
E33-15 We imagine the ribbon conductor to be a collection of thin wires, each of thickness dx
and carrying a current di. di and dx are related by di/dx = i/w. The contribution of one of these
thin wires to the magnetic field at P is dB = µ0 di/2πx, where x is the distance from this thin wire
to the point P . We want to change variables to x and integrate, so
Z
Z
Z
µ0 i dx
µ0 i
dx
B = dB =
=
.
2πwx
2πw
x
The limits of integration are from d to d + w, so
µ0 i
B=
ln
2πw
d+w
d
.
E33-16 The fields from each wire are perpendicular at P . Each contributes
B0 =
√ an amount
√
02
µ0 i/2πd, but since
√ they are perpendicular there is a net field of magnitude B = 2B = 2µ0 i/2πd.
Note that a = 2d, so B = µ0 i/πa.
(a) B = (4π×10−7 T · m/A)(115 A)/π(0.122 m) = 3.77×10−4 T. The direction is to the left.
(b) Same numerical result, except the direction is up.
E33-17 Follow along with Sample Problem 33-4.
Reversing the direction of the second wire (so that now both currents are directed out of the
page) will also reverse the direction of B2 . Then
µ0 i
1
1
B = B1 − B2 =
−
,
2π b + x b − x
µ0 i (b − x) − (b + x)
=
,
2π
b2 − x2
µ0 i
x
=
.
2
π
x − b2
~ survives, and must point to the
E33-18 (b) By symmetry, only the horizontal component of B
right.
(a) The horizontal component of the field contributed by the top wire is given by
µ0 i b/2
µ0 ib
µ0 i
sin θ =
=
,
2πh
2πh h
π(4R2 + b2 )
p
since h is the hypotenuse, or h = R2 + b2 /4. But there are two such components, one from the
top wire, and an identical component from the bottom wire.
B=
E33-19
loop,
(a) We can use Eq. 33-21 to find the magnetic field strength at the center of the large
µ0 i
(4π×10−7 T · m/A)(13 A)
=
= 6.8×10−5 T.
2R
2(0.12 m)
(b) The torque on the smaller loop in the center is given by Eq. 32-34,
B=
~ × B,
~
~τ = N iA
but since the magnetic field from the large loop is perpendicular to the plane of the large loop, and
the plane of the small loop is also perpendicular to the plane of the large loop, the magnetic field is
~ × B|
~ = AB. Consequently, the magnitude of the
in the plane of the small loop. This means that |A
torque on the small loop is
τ = N iAB = (50)(1.3 A)(π)(8.2×10−3 m)2 (6.8×10−5 T) = 9.3×10−7 N · m.
108
E33-20 (a) There are two contributions to the field. One is from the circular loop, and is given
by µ0 i/2R. The other is from the long straight wire, and is given by µ0 i/2πR. The two fields are
out of the page and parallel, so
µ0 i
B=
(1 + 1/π).
2R
(b) The two components are now at right angles, so
B=
µ0 i p
1 + 1/π 2 .
2R
The direction is given by tan θ = 1/π or θ = 18◦ .
E33-21 The force per meter for any pair of parallel currents is given by Eq. 33-25, F/L = µ0 i2 /2πd,
where d is the separation. The direction of the force is along the line connecting the intersection of
the currents with the perpendicular plane. Each current
three forces; two are at right
p experiences √
~ 12 + F
~ 14 |/L = F 2 + F 2 /L = 2µ0 i2 /2πa. The third force
angles and equal in magnitude, so |F
12
14
√
points parallel to this sum, but d = a, so the resultant force is
√
√
2µ0 i2
µ0 i2
4π ××10−7 N/A2 (18.7 A)2 √
F
=
+ √ =
( 2 + 1/ 2) = 6.06×10−4 N/m.
L
2πa
2π(0.245 m)
2π 2a
It is directed toward the center of the square.
E33-22 By symmetry we expect the middle wire to have a net force of zero; the two on the outside
will each be attracted toward the center, but the answers will be symmetrically distributed.
For the wire which is the farthest left,
F
µ0 i2 1
1
1
1
4π ××10−7 N/A2 (3.22 A)2
1 1 1
=
+
+
+
=
1+ + +
= 5.21×10−5 N/m.
L
2π
a 2a 3a 4a
2π(0.083 m)
2 3 4
For the second wire over, the contributions from the two adjacent wires should cancel. This
leaves
F
µ0 i2 1
1
4π ××10−7 N/A2 (3.22 A)2 1 1
= 2.08×10−5 N/m.
=
+ + =
+
L
2π
2a 3a
2π(0.083 m)
2 3
E33-23
(a) The force on the projectile is given by the integral of
~ = i d~l × B
~
dF
over the length of the projectile (which is w). The magnetic field strength can be found from adding
together the contributions from each rail. If the rails are circular and the distance between them is
small compared to the length of the wire we can use Eq. 33-13,
B=
µ0 i
,
2πx
where x is the distance from the center of the rail. There is one problem, however, because these
are not wires of infinite length. Since the current stops traveling along the rail when it reaches the
projectile we have a rod that is only half of an infinite rod, so we need to multiply by a factor of
1/2. But there are two rails, and each will contribute to the field, so the net magnetic field strength
between the rails is
µ0 i
µ0 i
B=
+
.
4πx 4π(2r + w − x)
109
In that last term we have an expression that is a measure of the distance from the center of the
lower rail in terms of the distance x from the center of the upper rail.
The magnitude of the force on the projectile is then
Z r+w
F = i
B dx,
r
Z
µ0 i2 r+w 1
1
=
+
dx,
4π r
x 2r + w − x
µ0 i2
r+w
=
2 ln
4π
r
The current through the projectile is down the page; the magnetic field through the projectile is
~ = i~l × B,
~ is to the right.
into the page; so the force on the projectile, according to F
(b) Numerically the magnitude of the force on the rail is
(450×103 A)2 (4π×10−7 N/A2 )
(0.067 m) + (0.012 m)
F =
ln
= 6.65×103 N
2π
(0.067 m)
The speed of the rail can be found from either energy conservation so we first find the work done
on the projectile,
W = F d = (6.65×103 N)(4.0 m) = 2.66×104 J.
This work results in a change in the kinetic energy, so the final speed is
p
p
v = 2K/m = 2(2.66×104 J)/(0.010 kg) = 2.31×103 m/s.
E33-24 The contributions from the left end and the right end of the square cancel out. This leaves
the top and the bottom. The net force is the difference, or
(4π×10−7 N/A2 )(28.6 A)(21.8 A)(0.323 m)
1
1
F =
−
,
2π
(1.10×10−2 m) (10.30×10−2 m)
=
3.27×10−3 N.
E33-25 The magnetic force on the upper wire near the point d is
FB =
µ0 ia ib L
µ0 ia ib L µ0 ia ib L
≈
−
x,
2π(d + x)
2πd
2πd2
where x is the distance from the equilibrium point d. The equilibrium magnetic force is equal to the
force of gravity mg, so near the equilibrium point we can write
x
FB = mg − mg .
d
There is then a restoring force against small perturbations of magnitude mgx/d which corresponds
to a spring constant of k = mg/d. This would give a frequency of oscillation of
f=
1 p
1 p
k/m =
g/d,
2π
2π
which is identical to the pendulum.
E33-26 B = (4π×10−7 N/A2 )(3.58 A)(1230)/(0.956m) = 5.79×10−3 T.
110
E33-27 The magnetic field inside an ideal solenoid is given by Eq. 33-28 B = µ0 in, where n is
the turns per unit length. Solving for n,
n=
B
(0.0224 T)
=
= 1.00×103 /m−1 .
µ0 i
(4π×10−7 N/A2 )(17.8 A)
The solenoid has a length of 1.33 m, so the total number of turns is
N = nL = (1.00×103 /m−1 )(1.33 m) = 1330,
and since each turn has a length of one circumference, then the total length of the wire which makes
up the solenoid is (1330)π(0.026 m) = 109 m.
E33-28 From the solenoid we have
B s = µ0 nis = µ0 (11500/m)(1.94 mA) = µ0 (22.3A/m).
From the wire we have
µ0 iw
µ0 (6.3 A)
µ0
=
= (1.002 A)
2πr
2πr
r
These fields are at right angles, so we are interested in when tan(40◦ ) = B w /B s , or
Bw =
r=
(1.002 A)
= 5.35×10−2 m.
tan(40◦ )(22.3 A/m)
E33-29 Let u = z − d. Then
B
=
=
=
µ0 niR2
2
d+L/2
Z
du
,
2 + u2 ]3/2
[R
d−L/2
d+L/2
u
µ0 niR2
√
,
2
R2 R2 + u2 d−L/2
µ0 ni
2
d + L/2
p
R2 + (d + L/2)2
!
d − L/2
−p
R2 + (d − L/2)2
.
If L is much, much greater than R and d then |L/2 ± d| >> R, and R can be ignored in the
denominator of the above expressions, which then simplify to
!
µ0 ni
d + L/2
d − L/2
p
−p
.
B =
2
R2 + (d + L/2)2
R2 + (d − L/2)2
!
µ0 ni
d + L/2
d − L/2
p
=
−p
.
2
(d + L/2)2
(d − L/2)2
= µ0 in.
It is important that we consider the relative size of L/2 and d!
E33-30 The net current in the loop is 1i0 + 3i0 + 7i0 − 6i0 = 5i0 . Then
H
~ · d~s = 5µ0 i0 .
B
E33-31 (a) The path is clockwise, so a positive current is into page. The net current is 2.0 A out,
H
~ · d~s = −µ0 i0 = −2.5×10−6 T · m.
so B
H
~ · d~s = 0.
(b) The net current is zero, so B
111
E33-32 Let R0 be the radius of the wire. On the surface of the wire B0 = µ0 i/2πR0 .
Outside the wire we have B = µ0 i/2πR, this is half B0 when R = 2R0 .
Inside the wire we have B = µ0 iR/2πR02 , this is half B0 when R = R0 /2.
E33-33
looks like
(a) We don’t want to reinvent the wheel. The answer is found from Eq. 33-34, except it
µ0 ir
.
2πc2
(b) In the region between the wires the magnetic field looks like Eq. 33-13,
B=
B=
µ0 i
.
2πr
This is derived on the right hand side
H of page 761.
~ · d~s = µ0 i, where i is the current enclosed. Our Amperian
(c) Ampere’s law (Eq. 33-29) is B
loop will still be a circle centered on the axis of the problem, so the left hand side of the above
equation will reduce to 2πrB, just like in Eq. 33-32. The right hand side, however, depends on
the net current enclosed which is the current i in the center wire minus the fraction of the current
enclosed in the outer conductor. The cross sectional area of the outer conductor is π(a2 − b2 ), so
the fraction of the outer current enclosed in the Amperian loop is
i
π(r2 − b2 )
r 2 − b2
=i 2
.
2
2
π(a − b )
a − b2
The net current in the loop is then
i−i
r 2 − b2
a2 − r2
=
i
,
a2 − b2
a2 − b2
so the magnetic field in this region is
B=
µ0 i a2 − r2
.
2πr a2 − b2
(d) This part is easy since the net current is zero; consequently B = 0.
H
~ · d~s = µ0 i, where i is the current enclosed. Our
E33-34 (a) Ampere’s law (Eq. 33-29) is B
Amperian loop will still be a circle centered on the axis of the problem, so the left hand side of the
above equation will reduce to 2πrB, just like in Eq. 33-32. The right hand side, however, depends
on the net current enclosed which is the fraction of the current enclosed in the conductor. The cross
sectional area of the conductor is π(a2 − b2 ), so the fraction of the current enclosed in the Amperian
loop is
π(r2 − b2 )
r 2 − b2
i
=
i
.
π(a2 − b2 )
a2 − b2
The magnetic field in this region is
B=
µ0 i r2 − b2
.
2πr a2 − b2
(b) If r = a, then
B=
µ0 i a2 − b2
µ0 i
=
,
2πa a2 − b2
2πa
which is what we expect.
112
If r = b, then
µ0 i b2 − b2
= 0,
2πb a2 − b2
B=
which is what we expect.
If b = 0, then
B=
µ0 i r2 − 02
µ0 ir
=
2πr a2 − 02
2πa2
which is what I expected.
E33-35 The magnitude of the magnetic field due to the cylinder will be zero at the center of
the cylinder and µ0 i0 /2π(2R) at point P . The magnitude of the magnetic field field due to the
wire will be µ0 i/2π(3R) at the center of the cylinder but µ0 i/2πR at P . In order for the net
field to have different directions in the two locations the currents in the wire and pipe must be in
different direction. The net field at the center of the pipe is µ0 i/2π(3R), while that at P is then
µ0 i0 /2π(2R) − µ0 i/2πR. Set these equal and solve for i;
i/3 = i0 /2 − i,
or i = 3i0 /8.
E33-36 (a) B = (4π×10−7 N/A2 )(0.813 A)(535)/2π(0.162 m) = 5.37×10−4 T.
(b) B = (4π×10−7 N/A2 )(0.813 A)(535)/2π(0.162 m + 0.052 m) = 4.07×10−4 T.
E33-37 (a) A positive particle would experience a magnetic force directed to the right for a
magnetic field out of the page. This particle is going the other way, so it must be negative.
(b) The magnetic field of a toroid is given by Eq. 33-36,
µ0 iN
,
2πr
while the radius of curvature of a charged particle in a magnetic field is given by Eq. 32-10
mv
R=
.
|q|B
B=
We use the R to distinguish it from r. Combining,
R=
2πmv
r,
µ0 iN |q|
so the two radii are directly proportional. This means
R/(11 cm) = (110 cm)/(125 cm),
so R = 9.7 cm.
P33-1
The field from one coil is given by Eq. 33-19
B=
µ0 iR2
.
2(R2 + z 2 )3/2
There are N turns in the coil, so we need a factor of N . There are two coils and we are interested
in the magnetic field at P , a distance R/2 from each coil. The magnetic field strength will be twice
the above expression but with z = R/2, so
B=
2µ0 N iR2
8µ0 N i
=
.
2
2
3/2
2(R + (R/2) )
(5)3/2 R
113
P33-2
(a) Change the limits of integration that lead to Eq. 33-12:
B
=
=
=
Z
L
dz
,
+ d2 )3/2
0
L
z
µ0 id
,
4π (z 2 + d2 )1/2 0
µ0 id
L
.
4π (L2 + d2 )1/2
µ0 id
4π
(z 2
(b) The angle φ in Eq. 33-11 would always be 0, so sin φ = 0, and therefore B = 0.
P33-3 This problem is the all important derivation of the Helmholtz coil properties.
(a) The magnetic field from one coil is
B1 =
µ0 N iR2
.
2(R2 + z 2 )3/2
The magnetic field from the other coil, located a distance s away, but for points measured from the
first coil, is
µ0 N iR2
B2 =
.
2(R2 + (z − s)2 )3/2
The magnetic field on the axis between the coils is the sum,
B=
µ0 N iR2
µ0 N iR2
+
.
2(R2 + z 2 )3/2
2(R2 + (z − s)2 )3/2
Take the derivative with respect to z and get
3µ0 N iR2
3µ0 N iR2
dB
=−
z−
(z − s).
2
2
5/2
2
dz
2(R + z )
2(R + (z − s)2 )5/2
At z = s/2 this expression vanishes! We expect this by symmetry, because the magnetic field will
be strongest in the plane of either coil, so the mid-point should be a local minimum.
(b) Take the derivative again and
d2 B
dz 2
3µ0 N iR2
15µ0 N iR2 2
+
z
2
2
5/2
2(R + z )
2(R2 + z 2 )5/2
3µ0 N iR2
15µ0 N iR2
−
+
(z − s)2 .
2
2
5/2
2
2(R + (z − s) )
2(R + (z − s)2 )5/2
= −
We could try and simplify this, but we don’t really want to; we instead want to set it equal to zero,
then let z = s/2, and then solve for s. The second derivative will equal zero when
−3(R2 + z 2 ) + 15z 2 − 3(R2 + (z − s)2 ) + 15(z − s)2 = 0,
and is z = s/2 this expression will simplify to
30(s/2)2 = 6(R2 + (s/2)2 ),
4(s/2)2 = R2 ,
s = R.
114
P33-4 (a) Each of the side of the square is a straight wire segment of length a which contributes
a field strength of
a
µ0 i
p
,
B=
4πr a2 /4 + r2
where r is the distance to the point on the axis of the loop, so
p
r = a2 /4 + z 2 .
This field is not parallel to the z axis; the z component is Bz = B(a/2)/r. There are four of these
contributions. The off axis components cancel. Consequently, the field for the square is
B
a/2
µ0 i
a
p
,
4πr a2 /4 + r2 r
=
4
=
µ0 i
a2
p
,
2
2πr
a2 /4 + r2
=
=
µ0 i
a2
p
,
2π(a2 /4 + z 2 ) a2 /2 + z 2
4µ0 i
a2
√
.
2
+ 4z ) 2a2 + 4z 2
π(a2
(b) When z = 0 this reduces to
B=
4µ0 i a2
4µ0 i
√
=√
.
π(a2 ) 2a2
2 πa
P33-5 (a) The polygon has n sides. A perpendicular bisector of each side can be drawn to the
center and has length x where x/a = cos(π/n). Each side has a length L = 2a sin(π/n). Each of the
side of the polygon is a straight wire segment which contributes a field strength of
B=
µ0 i
L
p
,
2
4πx L /4 + x2
This field is parallel to the z axis. There are n of these contributions. The off axis components
cancel. Consequently, the field for the polygon
B
= n
= n
= n
µ0 i
L
p
,
4πx L2 /4 + x2
µ0 i
2
p
tan(π/n),
2
4π L /4 + x2
µ0 i 1
tan(π/n),
2π a
since (L/2)2 + x2 = a2 .
(b) Evaluate:
lim n tan(π/n) = lim n sin(π/n) ≈ nπ/n = π.
n→∞
n→∞
Then the answer to part (a) simplifies to
B=
µ0 i
.
2a
115
P33-6 For a square loop of wire we have four finite length segments each contributing a term
which looks like Eq. 33-12, except that L is replaced by L/4 and d is replaced by L/8. Then at the
center,
L/4
16µ0 i
µ0 i
p
.
=√
B=4
2
2
4πL/8 L /64 + L /64
2πL
For a circular loop R = L/2π so
µ0 i
πµ0
=
.
2R
L
√
Since 16/ 2 π > π, the square wins. But only by some 7%!
B=
P33-7 We want to use the differential expression in Eq. 33-11, except that the limits of integration are going to be different. We have four wire segments. From the top segment,
3L/4
µ0 i
d
√
B1 =
,
4π z 2 + d2 −L/4
!
−L/4
µ0 i
3L/4
p
=
−p
.
4πd
(3L/4)2 + d2
(−L/4)2 + d2
For the top segment d = L/4, so this simplifies even further to
µ0 i √ √
B1 =
2(3 5 + 5) .
10πL
The bottom segment has the same integral, but d = 3L/4, so
µ0 i √ √
B3 =
2( 5 + 5) .
30πL
By symmetry, the contribution from the right hand side is the same as the bottom, so B2 = B3 ,
and the contribution from the left hand side is the same as that from the top, so B4 = B1 . Adding
all four terms,
√ √
2µ0 i √ √
3 2(3 5 + 5) + 2( 5 + 5) ,
B =
30πL
√
2µ0 i √
=
(2 2 + 10).
3πL
P33-8 Assume a current ring has a radius r and a width dr, the charge on the ring is dq = 2πσr dr,
where σ = q/πR2 . The current in the ring is di = ω dq/2π = ωσr dr. The ring contributes a field
dB = µ0 di/2r. Integrate over all the rings:
Z R
B=
µ0 ωσr dr/2r = µ0 ωR/2 = µωq/2πR.
0
P33-9
B = µ0 in and mv = qBr. Combine, and
i=
P33-10
mv
(5.11×105 eV/c2 )(0.046c)
=
= 0.271 A.
µ0 qrn
(4π×10−7 N/A2 )e(0.023 m)(10000/m)
This shape is a triangle with area A = (4d)(3d)/2 = 6d2 . The enclosed current is then
i = jA = (15 A/m2 )6(0.23 m)2 = 4.76 A
The line integral is then
µ0 i = 6.0×10−6 T · m.
116
P33-11 Assume that B does vary as the picture implies. Then the line integral along the path
~ · ~l on the right is not zero, while it is along the three other sides.
shown Hmust be nonzero, since B
~
~
Hence B · dl is non zero, implying some current passes through the dotted path. But it doesn’t,
~ cannot have an abrupt change.
so B
P33-12 (a) Sketch an Amperian loop which is a rectangle which enclosed N wires, has a vertical
H
~ · d~l = µ0 N i. Evaluate the integral
sides with height h, and horizontal sides with length L. Then B
~ is perpendicular to ~h, and then
along the four sides. The vertical side contribute nothing, since B
~ · ~h = 0. If the integral is performed in a counterclockwise direction (it must, since the sense
B
of integration was determined by assuming the current is positive), we get BL for each horizontal
section. Then
1
µ0 iN
= µ0 in.
B=
2L
2
(b) As a → ∞ then tan−1 (a/2R) → π/2. Then B → µ0 i/2a. If we assume that i is made up of
several wires, each with current i0 , then i/a = i0 n.
P33-13 Apply Ampere’s law with an Amperian loop that is a circle centered on the center of
the wire. Then
I
I
I
~ · d~s = B ds = B ds = 2πrB,
B
~ is tangent to the path and B is uniform along the path by symmetry. The current
because B
enclosed is
Z
ienc = j dA.
This integral is best done in polar coordinates, so dA = (dr)(r dθ), and then
ienc
=
Z
r
Z
(j0 r/a) rdr dθ,
Z r
2πj0 /a
r2 dr,
0
=
2π
0
0
=
2πj0 3
r .
3a
When r = a the current enclosed is i, so
i=
2πj0 a2
3i
or j0 =
.
3
2πa2
The magnetic field strength inside the wire is found by gluing together the two parts of Ampere’s
law,
2πrB
B
2πj0 3
r ,
3a
µ0 j0 r2
,
3a
2
µ0 ir
.
2πa3
= µ0
=
=
117
P33-14 (a) According to Eq. 33-34, the magnetic field inside the wire without a hole has magnitude
B = µ0 ir/2πR2 = µ0 jr/2 and is directed radially. If we superimpose a second current to create the
hole, the additional field at the center of the hole is zero, so B = µ0 jb/2. But the current in the
remaining wire is
i = jA = jπ(R2 − a2 ),
so
B=
µ0 ib
.
2π(R2 − a2 )
118
~ ·A
~ = (42×10−6 T)(2.5 m2 ) cos(57◦ ) = 5.7×10−5 Wb.
E34-1 ΦB = B
E34-2 |E| = |dΦB /dt| = A dB/dt = (π/4)(0.112 m)2 (0.157 T/s) = 1.55 mV.
E34-3
(a) The magnitude of the emf induced in a loop is given by Eq. 34-4,
dΦB ,
|E| = N dt = N (12 mWb/s2 )t + (7 mWb/s)
There is only one loop, and we want to evaluate this expression for t = 2.0 s, so
|E| = (1) (12 mWb/s2 )(2.0 s) + (7 mWb/s) = 31 mV.
(b) This part isn’t harder. The magnetic flux through the loop is increasing when t = 2.0 s. The
induced current needs to flow in such a direction to create a second magnetic field to oppose this
increase. The original magnetic field is out of the page and we oppose the increase by pointing the
other way, so the second field will point into the page (inside the loop).
By the right hand rule this means the induced current is clockwise through the loop, or to the
left through the resistor.
E34-4 E = −dΦB /dt = −A dB/dt.
(a) E = −π(0.16 m)2 (0.5 T)/(2 s) = −2.0×10−2 V.
(b) E = −π(0.16 m)2 (0.0 T)/(2 s) = 0.0×10−2 V.
(c) E = −π(0.16 m)2 (−0.5 T)/(4 s) = 1.0×10−2 V.
E34-5 (a) R = ρL/A = (1.69×10−8 Ω · m)[(π)(0.104 m)]/[(π/4)(2.50×10−3 m)2 ] = 1.12×10−3 Ω.
(b) E = iR = (9.66 A)(1.12×10−3 Ω) = 1.08×10−2 V. The required dB/dt is then given by
dB
E
=
= (1.08×10−2 V)/(π/4)(0.104 m)2 = 1.27 T/s.
dt
A
E34-6 E = −A ∆B/∆t = AB/∆t. The power is P = iE = E 2 /R. The energy dissipated is
E = P ∆t =
E 2 ∆t
A2 B 2
=
.
R
R∆t
E34-7 (a) We could re-derive the steps in the sample problem, or we could start with the end
result. We’ll start with the result,
di E = N Aµ0 n ,
dt
except that we have gone ahead and used the derivative instead of the ∆.
The rate of change in the current is
di
= (3.0 A/s) + (1.0 A/s2 )t,
dt
so the induced emf is
E
=
=
(130)(3.46×10−4 m2 )(4π×10−7 Tm/A)(2.2×104 /m) (3.0A/s) + (2.0A/s2 )t ,
(3.73×10−3 V) + (2.48×10−3 V/s)t.
(b) When t = 2.0 s the induced emf is 8.69×10−3 V, so the induced current is
i = (8.69×10−3 V)/(0.15 Ω) = 5.8×10−2 A.
119
E34-8 (a) i = E/R = N A dB/dt. Note that A refers to the area enclosed by the outer solenoid
where B is non-zero. This A is then the cross sectional area of the inner solenoid! Then
i=
1
di
(120)(π/4)(0.032 m)2 (4π×10−7 N/A2 )(220×102 /m) (1.5 A)
N Aµ0 n =
= 4.7×10−3 A.
R
dt
(5.3 Ω)
(0.16 s)
E34-9 P = Ei = E 2 /R = (A dB/dt)2 /(ρL/a), where A is the area of the loop and a is the cross
sectional area of the wire. But a = πd2 /4 and A = L2 /4π, so
L3 d2
P =
64πρ
dB
dt
2
=
(0.525 m)3 (1.1×10−3 m)2
(9.82×10−3 T/s)2 = 4.97×10−6 W.
64π(1.69×10−8 Ω · m)
E34-10 ΦB = BA = B(2.3 m)2 /2. EB = −dΦB /dt = −AdB/dt, or
EB = −
(2.3 m)2
[−(0.87 T/s)] = 2.30 V,
2
so E = (2.0 V) + (2.3 V) = 4.3 V.
E34-11 (a) The induced emf, as a function of time, is given by Eq. 34-5, E(t) = −dΦB (t)/dt
This emf drives a current through the loop which obeys E(t) = i(t)R Combining,
i(t) = −
1 dΦB (t)
.
R dt
Since the current is defined by i = dq/dt we can write
dq(t)
1 dΦB (t)
=−
.
dt
R dt
Factor out the dt from both sides, and then integrate:
1
dΦB (t),
R
Z
Z
1
dq(t) = −
dΦB (t),
R
1
q(t) − q(0) =
(ΦB (0) − ΦB (t))
R
dq(t)
= −
(b) No. The induced current could have increased from zero to some positive value, then decreased
to zero and became negative, so that the net charge to flow through the resistor was zero. This
would be like sloshing the charge back and forth through the loop.
E34-12 ∆P hiB = 2ΦB = 2N BA. Then the charge to flow through is
q = 2(125)(1.57 T)(12.2×10−4 m2 )/(13.3 Ω) = 3.60×10−2 C.
E34-13 The part above the long straight wire (a distance b − a above it) cancels out contributions
below the wire (a distance b − a beneath it). The flux through the loop is then
Z a
µ0 i
µ0 ib
a
ΦB =
b dr =
ln
.
2π
2a − b
2a−b 2πr
120
The emf in the loop is then
dΦB
µ0 b
E =−
=
ln
dt
2π
a
2a − b
[2(4.5 A/s2 )t − (10 A/s)].
Evaluating,
E=
4π×10−7 N/A2 (0.16 m)
ln
2π
(0.12 m)
2(0.12 m) − (0.16 m)
[2(4.5 A/s2 )(3.0 s)−(10 A/s)] = 2.20×10−7 V.
E34-14 Use Eq. 34-6: E = BDv = (55×10−6 T)(1.10 m)(25 m/s) = 1.5×10−3 V.
E34-15 If the angle doesn’t vary then the flux, given by
~ ·A
~
Φ=B
is constant, so there is no emf.
E34-16 (a) Use Eq. 34-6: E = BDv = (1.18T)(0.108 m)(4.86 m/s) = 0.619 V.
(b) i = (0.619 V)/(0.415 Ω) = 1.49 A.
(c) P = (0.619 V)(1.49 A) = 0.922 W.
(d) F = iLB = (1.49 A)(0.108 m)(1.18T) = 0.190 N.
(e) P = F v = (0.190 V)(4.86 m/s) = 0.923 W.
E34-17 The magnetic field is out of the page, and the current through the rod is down. Then Eq.
~ = iL
~ ×B
~ shows that the direction of the magnetic force is to the right; furthermore, since
32-26 F
everything is perpendicular to everything else, we can get rid of the vector nature of the problem
and write F = iLB. Newton’s second law gives F = ma, and the acceleration of an object from rest
results in a velocity given by v = at. Combining,
v(t) =
iLB
t.
m
E34-18 (b) The rod will accelerate as long as there is a net force on it. This net force comes from
F = iLB. The current is given by iR = E − BLv, so as v increases i decreases. When i = 0 the rod
stops accelerating and assumes a terminal velocity.
(a) E = BLv will give the terminal velocity. In this case, v = E/BL.
E34-19
E34-20 The acceleration is a = Rω 2 ; since E = BωR2 /2, we can find
a = 4E 2 /B 2 R3 = 4(1.4 V)2 /(1.2 T)2 (5.3×10−2 m)3 = 3.7×104 m/s2 .
E34-21 We will use the results of Exercise 11 that were worked out above. All we need to do is
find the initial flux; flipping the coil up-side-down will simply change the sign of the flux.
So
~ ·A
~ = (59 µT)(π)(0.13 m)2 sin(20◦ ) = 1.1×10−6 Wb.
ΦB (0) = B
121
Then using the results of Exercise 11 we have
q
N
(ΦB (0) − ΦB (t)),
R
950
=
((1.1×10−6 Wb) − (−1.1×10−6 Wb)),
85 Ω
= 2.5×10−5 C.
=
E34-22 (a) The flux through the loop is
Z vt Z
ΦB =
dx
0
The emf is then
E =−
a+L
dr
a
µ0 i
µ0 ivt a + L
=
ln
.
2πr
2π
a
µ0 iv a + L
dΦB
=−
ln
.
dt
2π
a
Putting in the numbers,
E=
(4π×10−7 N/A2 )(110 A)(4.86 m/s) (0.0102 m) + (0.0983 m)
ln
= 2.53×10−4 V.
2π
(0.0102 m)
(b) i = E/R = (2.53×10−4 V)/(0.415 Ω) = 6.10×10−4 A.
(c) P = iR2 R = (6.10×10−4 A)2 (0.415 Ω) = 1.54×10−7 W.
(d) F = Bil dl, or
Z a+L
µ0 i
µ0 i a + L
= il
ln
.
F = il
dr
2πr
2π
a
a
Putting in the numbers,
F = (6.10×10−4 A)
(4π×10−7 N/A2 )(110 A) (0.0102 m) + (0.0983 m)
ln
= 3.17×10−8 N.
2π
(0.0102 m)
(e) P = F v = (3.17×10−8 N)(4.86 m/s) = 1.54×10−7 W.
E34-23
(a) Starting from the beginning, Eq. 33-13 gives
B=
µ0 i
.
2πy
The flux through the loop is given by
ΦB =
Z
~ · dA,
~
B
but since the magnetic field from the long straight wire goes through the loop perpendicular to the
plane of the loop this expression simplifies to a scalar integral. The loop is a rectangular, so use
dA = dx dy, and let x be parallel to the long straight wire.
Combining the above,
Z D+b Z a µ0 i
ΦB =
dx dy,
2πy
D
0
Z D+b
µ0 i
dy
=
a
,
2π
y
D
µ0 i
D+b
=
a ln
2π
D
122
(b) The flux through the loop is a function of the distance D from the wire. If the loop moves
away from the wire at a constant speed v, then the distance D varies as vt. The induced emf is then
E
dΦB
,
dt
µ0 i
b
a
.
2π t(vt + b)
= −
=
The current will be this emf divided by the resistance R. The “back-of-the-book” answer is somewhat
different; the answer is expressed in terms of D instead if t. The two answers are otherwise identical.
E34-24 (a) The area of the triangle is A = x2 tan θ/2. In this case x = vt, so
ΦB = B(vt)2 tan θ/2,
and then
E = 2Bv 2 t tan θ/2,
(b) t = E/2Bv 2 tan θ/2, so
t=
(56.8 V)
= 2.08 s.
2(0.352 T)(5.21 m/s)2 tan(55◦ )
E34-25 E = N BAω, so
ω=
(24 V)
= 39.4 rad/s.
(97)(0.33 T)(0.0190 m2 )
That’s 6.3 rev/second.
E34-26 (a) The frequency of the emf is the same as the frequency of rotation, f .
(b) The flux changes by BA = Bπa2 during a half a revolution. This is a sinusoidal change, so
the amplitude of the sinusoidal variation in the emf is E = ΦB ω/2. Then E = Bπ 2 a2 f .
E34-27 We can use Eq. 34-10; the emf is E = BAω sin ωt, This will be a maximum when
sin ωt = 1. The angular frequency, ω is equal to ω = (1000)(2π)/(60) rad/s = 105 rad/s The
maximum emf is then
E = (3.5 T) [(100)(0.5 m)(0.3 m)] (105 rad/s) = 5.5 kV.
E34-28 (a) The amplitude of the emf is E = BAω, so
A = E/2πf B = (150 V)/2π(60/s)(0.50 T) = 0.798m2 .
(b) Divide the previous result by 100. A = 79.8 cm2 .
E34-29 dΦB /dt = A dB/dt = A(−8.50 mT/s).
(a) For this path
I
~ · d~s = −dΦB /dt = − − π(0.212 m)2 (−8.50 mT/s) = −1.20 mV.
E
(b) For this path
I
~ · d~s = −dΦB /dt = − − π(0.323 m)2 (−8.50 mT/s) = −2.79 mV.
E
123
(c) For this path
I
~ · d~s = −dΦB /dt = − − π(0.323 m)2 (−8.50 mT/s) − π(0.323 m)2 (−8.50 mT/s) = 1.59 mV.
E
H
~ · d~s = 2πrE.
E34-30 dΦB /dt = A dB/dt = A(−6.51 mT/s), while E
(a) The path of integration is inside the solenoid, so
E=
(0.022 m)(−6.51 mT/s)
−πr2 (−6.51 mT/s)
=
= 7.16×10−5 V/m.
2πr
2
(b) The path of integration is outside the solenoid, so
E=
E34-31
−πr2 (−6.51 mT/s)
(0.063 m)2 (−6.51 mT/s)
=
= 1.58×10−4 V/m
2πR
2(0.082 m)
The induced electric field can be found from applying Eq. 34-13,
I
~ · d~s = − dΦB .
E
dt
We start with the left hand side of this expression. The problem has cylindrical symmetry, so the
induced electric field lines should be circles centered on the axis of the cylindrical volume. If we
~ will be
choose the path of integration to lie along an electric field line, then the electric field E
parallel to d~s, and E will be uniform along this path, so
I
I
I
~
E · d~s = E ds = E ds = 2πrE,
where r is the radius of the circular path.
Now for the right hand side. The flux is contained in the path of integration, so ΦB = Bπr2 .
All of the time dependence of the flux is contained in B, so we can immediately write
2πrE = −πr2
dB
r dB
or E = −
.
dt
2 dt
What does the negative sign mean? The path of integration is chosen so that if our right hand fingers
curl around the path our thumb gives the direction of the magnetic field which cuts through the
path. Since the field points into the page a positive electric field would have a clockwise orientation.
Since B is decreasing the derivative is negative, but we get another negative from the equation
above, so the electric field has a positive direction.
Now for the magnitude.
E = (4.82×10−2 m)(10.7×10−3 T /s)/2 = 2.58×10−4 N/C.
The acceleration of the electron at either a or c then has magnitude
a = Eq/m = (2.58×10−4 N/C)(1.60×10−19 C)/(9.11×10−31 kg) = 4.53×107 m/s2 .
P34-1 The induced current is given by i = E/R. The resistance of the loop is given by R = ρL/A,
where A is the cross sectional area. Combining, and writing in terms of the radius of the wire, we
have
πr2 E
i=
.
ρL
124
The length of the wire is related to the radius of the wire because we have a fixed mass. The total
volume of the wire is πr2 L, and this is related to the mass and density by m = δπr2 L. Eliminating
r we have
mE
i=
.
ρδL2
The length of the wire loop is the same as the circumference, which is related to the radius R of the
loop by L = 2πR. The emf is related to the changing flux by E = −dΦB /dt, but if the shape of the
loop is fixed this becomes E = −A dB/dt. Combining all of this,
i=
mA dB
.
ρδ(2πR)2 dt
We dropped the negative sign because we are only interested in absolute values here.
Now A = πR2 , so this expression can also be written as
i=
mπR2 dB
m dB
=
.
ρδ(2πR)2 dt
4πρδ dt
~ ·A
~ = (76×10−3 T)(π/2)(0.037 m)2 cos(62◦ ) = 7.67×10−5 Wb. For
P34-2 For the lower surface B
−3
~
~
the upper surface B · A = (76×10 T)(π/2)(0.037 m)2 cos(28◦ ) = 1.44×10−4 Wb.. The induced emf
is then
E = (7.67×10−5 Wb + 1.44×10−4 Wb)/(4.5×10−3 s) = 4.9×10−2 V.
P34-3 (a) We are only interested in the portion of the ring in the yz plane. Then E = (3.32×
10−3 T/s)(π/4)(0.104 m)2 = 2.82×10−5 V.
~ field. Your right
(b) From c to b. Point your right thumb along −x to oppose the increasing B
fingers will curl from c to b.
P34-4 E ∝ N A, but A = πr2 and N 2πr = L, so E ∝ 1/N . This means use only one loop to
maximize the emf.
P34-5 This is a integral best performed in rectangular coordinates, then dA = (dx)(dy). The
~ · dA
~ = B dA. The flux is then
magnetic field is perpendicular to the surface area, so B
Z
Z
~ · dA
~ = B dA,
ΦB =
B
Z aZ a
=
(4 T/m · s2 )t2 y dy dx,
0
0
1 2
2 2
= (4 T/m · s )t
a a,
2
=
(2 T/m · s2 )a3 t2 .
But a = 2.0 cm, so this becomes
ΦB = (2 T/m · s2 )(0.02 m)3 t2 = (1.6×10−5 Wb/s2 )t2 .
The emf around the square is given by
dΦB
= −(3.2×10−5 Wb/s2 )t,
dt
and at t = 2.5 s this is −8.0×10−5 V. Since the magnetic field is directed out of the page, a positive
emf would be counterclockwise (hold your right thumb in the direction of the magnetic field and
your fingers will give a counter clockwise sense around the loop). But the answer was negative, so
the emf must be clockwise.
E =−
125
P34-6
then
(a) Far from the plane of the large loop we can approximate the large loop as a dipole, and
B=
µ0 iπR2
.
2x3
The flux through the small loop is then
ΦB = πr2 B =
µ0 iπ 2 r2 R2
.
2x3
(b) E = −dΦB /dt, so
3µ0 iπ 2 r2 R2
v.
2x4
(c) Anti-clockwise when viewed from above.
E=
P34-7
then
~ · dA
~ = B dA. The flux is
The magnetic field is perpendicular to the surface area, so B
Z
Z
~
~
ΦB = B · dA = B dA = BA,
since the magnetic field is uniform. The area is A = πr2 , where r is the radius of the loop. The
induced emf is
dΦB
dr
E =−
= −2πrB .
dt
dt
It is given that B = 0.785 T, r = 1.23 m, and dr/dt = −7.50×10−2 m/s. The negative sign indicate
a decreasing radius. Then
E = −2π(1.23 m)(0.785 T)(−7.50×10−2 m/s) = 0.455 V.
P34-8 (a) dΦB /dt = B dA/dt, but dA/dt is ∆A/∆t, where ∆A is the area swept out during one
rotation and ∆t = 1/f . But the area swept out is πR2 , so
|E| =
dΦB
= πf BR2 .
dt
(b) If the output current is i then the power is P = Ei. But P = τ ω = τ 2πf , so
τ=
P34-9
P
= iBR2 /2.
2πf
~ · A,so
~
(a) E = −dΦB /dt, and ΦB = B
E = BLv cos θ.
The component of the force of gravity on the rod which pulls it down the incline is FG = mg sin θ.
The component of the magnetic force on the rod which pulls it up the incline is FB = BiL cos θ.
Equating,
BiL cos θ = mg sin θ,
and since E = iR,
v=
E
mgR sin θ
= 2 2
.
BL cos θ
B L cos2 θ
(b) P = iE = E 2 /R = B 2 L2 v 2 cos2 θ/R = mgv sin θ. This is identical to the rate of change of
gravitational potential energy.
126
P34-10 Let the cross section of the wire be a.
(a) R = ρL/a = ρ(rθ + 2r)/a; with numbers,
R = (3.4×10−3 Ω)(2 + θ).
(b) ΦB = Bθr2 /2; with numbers,
ΦB = (4.32×10−3 Wb)θ.
(c) i = E/R = Bωr2 /2R = Baωr/2ρ(θ + 2), or
i=
Baαtr
.
ρ(αt2 + 4)
Take the derivative and set it equal to zero,
0=
4 − at2
,
(αt2 + 4)2
so at2 = 4, or θ = 12 at2 = 2 rad.
√
(d) ω = 2αθ, so
i=
q
2
(0.15 T)(1.2×10−6 m2 ) 2(12 rad/s )(2 rad)(0.24 m)
(1.7×10−8 Ω · m)(6 rad)
= 2.2 A.
P34-11 It does say approximate, so we will be making some rather bold assumptions here. First
we will find an expression for the emf. Since B is constant, the emf must be caused by a change in
the area; in this case a shift in position. The small square where B 6= 0 has a width a and sweeps
around the disk with a speed rω. An approximation for the emf is then E = Barω. This emf causes
a current. We don’t know exactly where the current flows, but we can reasonably assume that it
occurs near the location of the magnetic field. Let us assume that it is constrained to that region of
the disk. The resistance of this portion of the disk is the approximately
R=
1L
1 a
1
=
= ,
σA
σ at
σt
where we have assumed that the current is flowing radially when defining the cross sectional area of
the “resistor”. The induced current is then (on the order of)
E
Barω
=
= Barωσt.
R
1/(σt)
This current experiences a breaking force according to F = BIl, so
F = B 2 a2 rωσt,
where l is the length through which the current flows, which is a.
Finally we can find the torque from τ = rF , and
τ = B 2 a2 r2 ωσt.
127
P34-12 The induced electric field in the ring is given by Eq. 34-11: 2πRE = |dΦB /dt|. This
electric field will result in a force on the free charge carries (electrons?), given by F = Ee. The
acceleration of the electrons is then a = Ee/me . Then
a=
dΦB
e
.
2πRme dt
Integrate both sides with respect to time to find the speed of the electrons.
Z
Z
e
dΦB
a dt =
dt,
2πRme dt
Z
dΦB
e
v =
2πRme
,
e
=
∆ΦB .
2πRme
The current density is given by j = nev, and the current by iA = iπa2 . Combining,
i=
ne2 a2
∆P hiB .
2Rme
Actually, it should be pointed out that ∆P hiB refers to the change in flux from external sources. The
current induced in the wire will produce a flux which will exactly offset ∆P hiB so that the net flux
through the superconducting ring is fixed at the value present when the ring became superconducting.
P34-13 Assume that E does vary as the picture implies. Then the line integral along the path
~ · ~l on the right is not zero, while it is along the three other sides.
shown Hmust be nonzero, since E
~ · d~l is non zero, implying a change in the magnetic flux through the dotted path. But it
Hence E
~ cannot have an abrupt change.
doesn’t, so E
P34-14
The electric field a distance r from the center is given by
E=
πr2 dB/dT
r dB
=
.
2πr
2 dt
This field is directed perpendicular to the radial lines.
Define
h to be the distance from the center of the circle to the center of the rod, and evaluate
R
~ · d~s,
E= E
Z
dB
rh
E =
dx,
dt
2r
dB L
=
h.
dt 2
But h2 = R2 − (L/2)2 , so
E=
P34-15
dB L p 2
R − (L/2)2 .
dt 2
(a) ΦB = πr2 B av , so
E=
E
(0.32 m)
=
2(0.28 T)(120 π) = 34 V/m.
2πr
2
(b) a = F/m = Eq/m = (33.8 V/m)(1.6×10−19 C)/(9.1×10−31 kg) = 6.0×1012 m/s2 .
128
E35-1 If the Earth’s magnetic dipole moment were produced by a single current around the core,
then that current would be
i=
µ
(8.0 × 1022 J/T)
=
= 2.1×109 A
A
π(3.5 × 106 m)2
E35-2 (a) i = µ/A = (2.33 A · m2 )/(160)π(0.0193 m)2 = 12.4 A.
(b) τ = µB = (2.33 A · m2 )(0.0346 T) = 8.06×10−2 N · m.
E35-3 (a) Using the right hand rule a clockwise current would generate a magnetic moment which
would be into the page. Both currents are clockwise, so add the moments:
µ = (7.00 A)π(0.20 m)2 + (7.00 A)π(0.30 m)2 = 2.86 A · m2 .
(b) Reversing the current reverses the moment, so
µ = (7.00 A)π(0.20 m)2 − (7.00 A)π(0.30 m)2 = −1.10 A · m2 .
E35-4 (a) µ = iA = (2.58 A)π(0.16 m)2 = 0.207 A · m2 .
(b) τ = µB sin θ = (0.207 A · m2 )(1.20 T) sin(41◦ ) = 0.163 N · m.
E35-5
(a) The result from Problem 33-4 for a square loop of wire was
B(z) =
4µ0 ia2
.
π(4z 2 + a2 )(4z 2 + 2a2 )1/2
For z much, much larger than a we can ignore any a terms which are added to or subtracted from
z terms. This means that
4z 2 + a2 → 4z 2 and (4z 2 + 2a2 )1/2 → 2z,
but we can’t ignore the a2 in the numerator.
The expression for B then simplifies to
B(z) =
µ0 ia2
,
2πz 3
which certainly looks like Eq. 35-4.
(b) We can rearrange this expression and get
B(z) =
µ0
ia2 ,
2πz 3
where it is rather evident that ia2 must correspond to µ
~ , the dipole moment, in Eq. 35-4. So that
must be the answer.
E35-6 µ = iA = (0.2 A)π(0.08 m)2 = 4.02×10−3 A · m2 ; µ
~ = µn̂.
(a) For the torque,
~ = (−9.65×10−4 N · m)î + (−7.24×10−4 N · m)ĵ + (8.08×10−4 N · m)k̂.
~τ = µ
~ ×B
(b) For the magnetic potential energy,
~ = µ[(0.60)(0.25 T)] = 0.603×10−3 J.
U =µ
~ ·B
129
E35-7 µ = iA = iπ(a2 + b2 /2) = iπ(a2 + b2 )/2.
E35-8 If the distance to P is very large compared to a or b we can write the Law of Biot and
Savart as
~ = µ0 i ~s × ~r .
B
4π r3
~s is perpendicular to ~r for the left and right sides, so the left side contributes
B1 =
µ0 i
b
,
4π (x + a/2)2
and the right side contributes
B3 = −
µ0 i
b
.
4π (x − a/2)2
The top and bottom sides each contribute an equal amount
B2 = B4 =
µ0 i a sin θ
µ0 i a(b/2)
≈
.
4π x2 + b2 /4
4π x3
Add the four terms, expand the denominators, and keep only terms in x3 ,
B=−
µ0 i ab
µ0 µ
=−
.
3
4π x
4π x3
The negative sign indicates that it is into the page.
E35-9
(a) The electric field at this distance from the proton is
E=
1
4π(8.85×10−12 C2 /N
·
m2 )
(1.60×10−19 C)
= 5.14×1011 N/C.
(5.29×10−11 m)2
(b) The magnetic field at this from the proton is given by the dipole approximation,
B(z)
=
=
=
µ0 µ
,
2πz 3
(4π×10−7 N/A2 )(1.41×10−26 A/m2 )
,
2π(5.29×10−11 m)3
1.90×10−2 T
E35-10 1.50 g of water has (2)(6.02 × 1023 )(1.5)/(18) = 1.00 × 1023 hydrogen nuclei. If all are
aligned the net magnetic moment would be µ = (1.00×1023 )(1.41×10−26 J/T) = 1.41×10−3 J/T.
The field strength is then
B=
−3
µ0 µ
J/T)
−7
2 (1.41×10
=
(1.00×10
N/A
)
= 9.3×10−13 T.
4π x3
(5.33 m)3
E35-11 (a) There is effectively a current of i = f q = qω/2π. The dipole moment is then µ = iA =
(qω/2π)(πr2 ) = 12 qωr2 .
(b) The rotational inertia of the ring is mr2 so L = Iω = mr2 ω. Then
µ
(1/2)qωr2
q
=
=
.
L
mr2 ω
2m
130
E35-12 The mass of the bar is
3
m = ρV = (7.87 g/cm )(4.86 cm)(1.31 cm2 ) = 50.1 g.
The number of atoms in the bar is
N = (6.02×1023 )(50.1 g)/(55.8 g) = 5.41×1023 .
The dipole moment of the bar is then
µ = (5.41×1023 )(2.22)(9.27×10−24 J/T) = 11.6 J/T.
(b) The torque on the magnet is τ = (11.6 J/T)(1.53 T) = 17.7 N · m.
E35-13
The magnetic dipole moment is given by µ = M V , Eq. 35-13. Then
µ = (5, 300 A/m)(0.048 m)π(0.0055 m)2 = 0.024 A · m2 .
E35-14 (a) The original field is B0 = µ0 in. The field will increase to B = κm B0 , so the increase is
∆B = (κ1 − 1)µ0 in = (3.3×10−4 )(4π×10−7 N/A2 )(1.3 A)(1600/m) = 8.6×10−7 T.
(b) M = (κ1 − 1)B0 /µ0 = (κ1 − 1)in = (3.3×10−4 )(1.3 A)(1600/m) = 0.69 A/m.
E35-15 The energy to flip the dipoles is given by U = 2µB. The temperature is then
T =
2µB
4(1.2×10−23 J/T)(0.5 T)
=
= 0.58 K.
3k/2
3(1.38×10−23 J/K)
E35-16 The Curie temperature of iron is 770◦ C, which is 750◦ C higher than the surface temperature. This occurs at a depth of (750◦ C)/(30 C◦ /km) = 25 km.
E35-17 (a) Look at the figure. At 50% (which is 0.5 on the vertical axis), the curve is at
B0 /T ≈ 0.55 T/K. Since T = 300 K, we have B0 ≈ 165 T.
(b) Same figure, but now look at the 90% mark. B0 /T ≈ 1.60 T/K, so B0 ≈ 480 T.
(c) Good question. I think both fields are far beyond our current abilities.
E35-18 (a) Look at the figure. At 50% (which is 0.5 on the vertical axis), the curve is at B0 /T ≈
0.55 T/K. Since B0 = 1.8 T, we have T ≈ (1.8 T)/(0.55 T/K) = 3.3 K.
(b) Same figure, but now look at the 90% mark. B0 /T ≈ 1.60 T/K, so T ≈ (1.8 T)/(1.60 T/K) =
1.1 K.
E35-19 Since (0.5 T)/(10 K) = 0.05 T/K, and all higher temperatures have lower values of the
ratio, and this puts all points in the region near where Curie’s Law (the straight line) is valid, then
the answer is yes.
E35-20 Using Eq. 35-19,
µn =
Mr M
(108g/mol)(511×103 A/m)
VM
=
=
= 8.74×10−21 A/m2
N
Aρ
(10490 kg/m3 )(6.02×1023 /mol)
131
E35-21 (a) B = µ0 µ/2z 3 , so
B=
(4π×10−7 N/A2 )(1.5×10−23 J/T)
= 9.4×10−6 T.
2(10×10−9 m)3
(b) U = 2µB = 2(1.5×10−23 J/T)(9.4×10−6 T) = 2.82×10−28 J.
E35-22 ΦB = (43×10−6 T)(295, 000×106 m2 ) = 1.3×107 Wb.
E35-23 (a) We’ll assume, however, that all of the iron atoms are perfectly aligned. Then the
dipole moment of the earth will be related to the dipole moment of one atom by
µEarth = N µFe ,
where N is the number of iron atoms in the magnetized sphere. If mA is the relative atomic mass
of iron, then the total mass is
N mA
mA µEarth
m=
=
,
A
A µFe
where A is Avogadro’s number. Next, the volume of a sphere of mass m is
m
mA µEarth
V =
=
,
ρ
ρA µFe
where ρ is the density.
And finally, the radius of a sphere with this volume would be
1/3 1/3
3V
3µEarth mA
r=
=
.
4π
4πρµFe A
Now we find the radius by substituting in the known values,
r=
3(8.0×1022 J/T)(56 g/mol)
3
4π(14×106 g/m )(2.1×10−23 J/T)(6.0×1023 /mol)
!1/3
= 1.8×105 m.
(b) The fractional volume is the cube of the fractional radius, so the answer is
(1.8×105 m/6.4×106 )3 = 2.2×10−5 .
E35-24 (a) At magnetic equator Lm = 0, so
µ0 µ
(1.00×10−7 N/A2 )(8.0×1022 J/T)
=
= 31µT.
4πr3
(6.37×106 m)3
B=
There is no vertical component, so the inclination is zero.
(b) Here Lm = 60◦ , so
q
q
µ0 µ
(1.00×10−7 N/A2 )(8.0×1022 J/T)
2
B=
1
+
3
sin
L
=
1 + 3 sin2 (60◦ ) = 56µT.
m
4πr3
(6.37×106 m)3
The inclination is given by
arctan(B v /B h ) = arctan(2 tan Lm ) = 74◦ .
(c) At magnetic north pole Lm = 90◦ , so
B=
µ0 µ
2(1.00×10−7 N/A2 )(8.0×1022 J/T)
=
= 62µT.
3
2πr
(6.37×106 m)3
There is no horizontal component, so the inclination is 90◦ .
132
E35-25 This problem is effectively solving 1/r3 = 1/2 for r measured in Earth radii. Then
r = 1.26rE , and the altitude above the Earth is (0.26)(6.37×106 m) = 1.66×106 m.
E35-26 The radial distance from the center is r = (6.37×106 m) − (2900×103 m) = 3.47×106 m.
The field strength is
B=
µ0 µ
2(1.00×10−7 N/A2 )(8.0×1022 J/T)
=
= 380µT.
3
2πr
(3.47×106 m)3
E35-27 Here Lm = 90◦ − 11.5◦ = 78.5◦ , so
q
q
µ0 µ
(1.00×10−7 N/A2 )(8.0×1022 J/T)
2
1
+
3
sin
L
=
1 + 3 sin2 (78.5◦ ) = 61µT.
B=
m
4πr3
(6.37×106 m)3
The inclination is given by
arctan(B v /B h ) = arctan(2 tan Lm ) = 84◦ .
E35-28 The flux out the “other” end is (1.6×10−3 T)π(0.13 m)2 = 85µWb. The net flux through
the surface is zero, so the flux through the curved surface is 0 − (85µWb) − (−25µWb) = −60µWb..
The negative indicates inward.
E35-29 The total magnetic flux through a closed surface is zero. There is inward flux on faces
one, three, and five for a total of -9 Wb. There is outward flux on faces two and four for a total of
+6 Wb. The difference is +3 Wb; consequently the outward flux on the sixth face must be +3 Wb.
E35-30 The stable arrangements are (a) and (c). The torque in each case is zero.
E35-31 The field on the x axis between the wires is
µ0 i
1
1
B=
+
.
2π 2r + x 2r − x
H
~ · dA
~ = 0, we can assume the flux through the curved surface is equal to the flux through
Since B
the xz plane within the cylinder. This flux is
Z r µ0 i
1
1
ΦB = L
+
dx,
2r + x 2r − x
−r 2π
µ0 i
3r
r
= L
ln
− ln
,
2π
r
3r
µ0 i
= L
ln 3.
π
P35-1 We can imagine the rotating disk as being composed of a number of rotating rings of
radius r, width dr, and circumference 2πr. The surface charge density on the disk is σ = q/πR2 ,
and consequently the (differential) charge on any ring is
dq = σ(2πr)(dr) =
2qr
dr
R2
The rings “rotate” with angular frequency ω, or period T = 2π/ω. The effective (differential) current
for each ring is then
dq
qrω
di =
=
dr.
T
πR2
133
Each ring contributes to the magnetic moment, and we can glue all of this together as
Z
µ =
dµ,
Z
=
πr2 di,
Z R 3
qr ω
=
dr,
R2
0
qR2 ω
=
.
4
P35-2 (a) The sphere can be sliced into disks. The disks can be sliced into rings. Each ring has
some charge qi , radius ri , and mass mi ; the period of rotation for a ring is T = 2π/ω, so the current
in the ring is qi /T = ωqi /2π. The magnetic moment is
µi = (ωqi /2π)πri2 = ωqi ri2 /2.
Note that this is closely related to the expression for angular momentum of a ring: li = ωmi ri2 .
Equating,
µi = qi li /2mi .
If both mass density and charge density are uniform then we can write qi /mi = q/m,
Z
Z
µ = dµ = (q/2m) dl = qL/2m
For a solid sphere L = ωI = 2ωmR2 /5, so
µ = qωR2 /5.
(b) See part (a)
P35-3 (a) The orbital speed is given by K = mv 2 /2. The orbital radius is given by mv = qBr, or
r = mv/qB. The frequency of revolution is f = v/2πr. The effective current is i = qf . Combining
all of the above to find the dipole moment,
µ = iA = q
v
vr
mv 2
K
πr2 = q
=q
= .
2πr
2
2qB
B
(b) Since q and m cancel out of the above expression the answer is the same!
(c) Work it out:
M=
µ
(5.28×1021 /m3 )(6.21×10−20 J) (5.28×1021 /m3 )(7.58×10−21 J)
=
+
= 312 A/m.
V
(1.18 T)
(1.18 T)
P35-4 (b) Point the thumb or your right hand to the right. Your fingers curl in the direction of
the current in the wire loop.
~ has a component which is directed radially outward.
(c) In the vicinity of the wire of the loop B
~
Then B × d~s has a component directed to the left. Hence, the net force is directed to the left.
P35-5 (b) Point the thumb or your right hand to the left. Your fingers curl in the direction of the
current in the wire loop.
~ has a component which is directed radially outward.
(c) In the vicinity of the wire of the loop B
~ × d~s has a component directed to the right. Hence, the net force is directed to the right.
Then B
134
P35-6 (a) Let x = µB/kT . Adopt the convention that N+ refers to the atoms which have parallel
alignment and N− those which are anti-parallel. Then N+ + N− = N , so
N+ = N ex /(ex + e−x ),
and
N− = N e−x /(ex + e−x ),
Note that the denominators are necessary so that N+ + N− = N . Finally,
M = µ(N+ − N− ) = µN
ex − e−x
.
ex + e−x
(b) If µB kT then x is very small and e±x ≈ 1 ± x. The above expression reduces to
M = µN
(1 + x) − (1 − x)
µ2 B
= µN x =
.
(1 + x) + (1 − x)
kT
(c) If µB kT then x is very large and e±x → ∞ while e−x → 0. The above expression reduces
to
N = µN.
P35-7
(a) Centripetal acceleration is given by a = rω 2 . Then
a − a0
= r(ω0 + ∆ω)2 − rω02 ,
= 2rω0 ∆ω + r(∆ω0 )2 ,
≈ 2rω0 ∆ω.
(b) The change in centripetal acceleration is caused by the additional magnetic force, which has
magnitude FB = qvB = erωB. Then
∆ω =
a − a0
eB
=
.
2rω0
2m
Note that we boldly canceled ω against ω0 in this last expression; we are assuming that ∆ω is small,
and for these problems it is.
P35-8 (a) i = µ/A = (8.0×1022 J/T)/π(6.37×106 m)2 = 6.3×108 A.
(b) Far enough away both fields act like perfect dipoles, and can then cancel.
(c) Close enough neither field acts like a perfect dipole and the fields will not cancel.
P35-9
(a) B =
√
B h 2 + B v 2 , so
q
q
µ0 µ
µ0 µ
2 L + 4 sin2 L =
B=
cos
1 + 3 sin2 Lm .
m
m
4πr3
4πr3
(b) tan φi = B v /B h = 2 sin Lm / cos Lm = 2 tan Lm .
135
E36-1
The important relationship is Eq. 36-4, written as
ΦB =
(5.0 mA)(8.0 mH)
iL
=
= 1.0×10−7 Wb
N
(400)
E36-2 (a) Φ = (34)(2.62×10−3 T)π(0.103 m)2 = 2.97×10−3 Wb.
(b) L = Φ/i = (2.97×10−3 Wb)/(3.77 A) = 7.88×10−4 H.
E36-3 n = 1/d, where d is the diameter of the wire. Then
µ0 A
(4π×10−7 H/m)(π/4)(4.10×10−2 m)2
L
= µ0 n2 A = 2 =
= 2.61×10−4 H/m.
l
d
(2.52×10−3 m)2
E36-4 (a) The emf supports the current, so the current must be decreasing.
(b) L = E/(di/dt) = (17 V)/(25×103 A/s) = 6.8×10−4 H.
E36-5
(a) Eq. 36-1 can be used to find the inductance of the coil.
L=
EL
(3.0 mV)
=
= 6.0×10−4 H.
di/dt
(5.0 A/s)
(b) Eq. 36-4 can then be used to find the number of turns in the coil.
N=
iL
(8.0 A)(6.0×10−4 H)
=
= 120
ΦB
(40µWb)
E36-6 Use the equation between Eqs. 36-9 and 36-10.
ΦB
=
(4π×10−7 H/m)(0.81 A)(536)(5.2×10−2 m) (5.2×10−2 m) + (15.3×10−2 m)
ln
,
2π
(15.3×10−2 m)
=
1.32×10−6 Wb.
E36-7 L = κm µ0 n2 Al = κm µ0 N 2 A/l, or
L = (968)(4π×10−7 H/m)(1870)2 (π/4)(5.45×10−2 m)2 /(1.26 m) = 7.88 H.
E36-8 In each case apply E = L∆i/∆t.
(a) E = (4.6 H)(7 A)/(2×10−3 s) = 1.6×104 V.
(b) E = (4.6 H)(2 A)/(3×10−3 s) = 3.1×103 V.
(c) E = (4.6 H)(5 A)/(1×10−3 s) = 2.3×104 V.
E36-9 (a) If two inductors are connected in parallel then the current through each inductor will
add to the total current through the circuit, i = i1 + i2 , Take the derivative of the current with
respect to time and then di/dt = di1 /dt + di2 /dt,
The potential difference across each inductor is the same, so if we divide by E and apply we get
di/dt
di1 /dt di2 /dt
=
+
,
E
E
E
But
di/dt
1
= ,
E
L
136
so the previous expression can also be written as
1
1
1
=
+
.
Leq
L1
L2
(b) If the inductors are close enough together then the magnetic field from one coil will induce
currents in the other coil. Then we will need to consider mutual induction effects, but that is a topic
not covered in this text.
E36-10 (a) If two inductors are connected in series then the emf across each inductor will add to
the total emf across both, E = E1 + E2 ,
Then the current through each inductor is the same, so if we divide by di/dt and apply we get
E1
E2
E
=
+
,
di/dt
di/dt di/dt
But
E
= L,
di/dt
so the previous expression can also be written as
Leq = L1 + L2 .
(b) If the inductors are close enough together then the magnetic field from one coil will induce
currents in the other coil. Then we will need to consider mutual induction effects, but that is a topic
not covered in this text.
E36-11 Use Eq. 36-17, but rearrange:
τL =
t
(1.50 s)
=
= 0.317 s.
ln[i0 /i]
ln[(1.16 A)/(10.2×10−3 A)]
Then R = L/τL = (9.44 H)/(0.317 s) = 29.8 Ω.
E36-12 (a) There is no current through the resistor, so ER = 0 and then EL = E.
(b) EL = Ee−2 = (0.135)E.
(c) n = − ln(EL /E) = −ln(1/2) = 0.693.
E36-13
(a) From Eq. 36-4 we find the inductance to be
L=
N ΦB
(26.2×10−3 Wb)
=
= 4.78×10−3 H.
i
(5.48 A)
Note that ΦB is the flux, while the quantity N ΦB is the number of flux linkages.
(b) We can find the time constant from Eq. 36-14,
τL = L/R = (4.78×10−3 H)/(0.745 Ω) = 6.42×10−3 s.
The we can invert Eq. 36-13 to get
Ri(t)
t = −τL ln 1 −
,
E
(0.745 A)(2.53 A)
−3
= −(6.42×10 s) ln 1 −
= 2.42×10−3 s.
(6.00 V)
137
E36-14 (a) Rearrange:
E
E
−i
R
R
dt
L
= iR + L
=
=
di
,
dt
L di
,
R dt
di
.
E/R − i
(b) Integrate:
−
Z
0
t
R
dt
L
R
− t
L
E −t/τL
e
R
E 1 − e−t/τL
R
=
=
Z
i
di
,
i
−
E/R
0
i + E/R
ln
,
E/R
= i + E/R,
= i.
E36-15 di/dt = (5.0 A/s). Then
E = iR + L
di
= (3.0 A)(4.0 Ω) + (5.0 A/s)t(4.0 Ω) + (6.0 H)(5.0 A/s) = 42 V + (20 V/s)t.
dt
E36-16 (1/3) = (1 − e−t/τL ), so
τL = −
E36-17
(5.22 s)
t
=−
= 12.9 s.
ln(2/3)
ln(2/3)
We want to take the derivative of the current in Eq. 36-13 with respect to time,
di
E 1 −t/τL
E
=
e
= e−t/τL .
dt
R τL
L
Then τL = (5.0×10−2 H)/(180 Ω) = 2.78×10−4 s. Using this we find the rate of change in the current
when t = 1.2 ms to be
−3
−4
di
(45 V)
=
e−(1.2×10 s)/(2.78×10 s) = 12 A/s.
−2
dt
((5.0×10 H)
E36-18 (b) Consider some time ti :
EL (ti ) = Ee−ti /τL .
Taking a ratio for two different times,
EL (t1 )
= e(t2 −t1 )/τL ,
EL (t2 )
or
τL =
t2 − t1
(2 ms) − (1 ms)
=
= 3.58 ms
ln[EL (t1 )/EL (t2 )]
ln[(18.24 V)/(13.8 V)]
(a) Choose any time, and
E = EL et/τL = (18.24 V)e(1 ms)/(3.58 ms) = 24 V.
138
E36-19 (a) When the switch is just closed there is no current through the inductor. So i1 = i2 is
given by
E
(100 V)
i1 =
=
= 3.33 A.
R1 + R2
(10 Ω) + (20 Ω)
(b) A long time later there is current through the inductor, but it is as if the inductor has no
effect on the circuit. Then the effective resistance of the circuit is found by first finding the equivalent
resistance of the parallel part
1/(30 Ω) + 1/(20 Ω) = 1/(12 Ω),
and then finding the equivalent resistance of the circuit
(10 Ω) + (12 Ω) = 22 Ω.
Finally, i1 = (100 V)/(22 Ω) = 4.55 A and
∆V2 = (100 V) − (4.55 A)(10 Ω) = 54.5 V;
consequently, i2 = (54.5 V)/(20 Ω) = 2.73 A. It didn’t ask, but i2 = (4.55 A) − (2.73 A) = 1.82 A.
(c) After the switch is just opened the current through the battery stops, while that through the
inductor continues on. Then i2 = i3 = 1.82 A.
(d) All go to zero.
E36-20 (a) For toroids L = µ0 N 2 h ln(b/a)/2π. The number of turns is limited by the inner radius:
N d = 2πa. In this case,
N = 2π(0.10 m)/(0.00096 m) = 654.
The inductance is then
L=
(4π×10−7 H/m)(654)2 (0.02 m) (0.12 m)
ln
= 3.1×10−4 H.
2π
(0.10 m)
(b) Each turn has a length of 4(0.02 m) = 0.08 m. The resistance is then
R = N (0.08 m)(0.021 Ω/m) = 1.10 Ω
The time constant is
τL = L/R = (3.1×10−4 H)/(1.10 Ω) = 2.8×10−4 s.
E36-21 (I) When the switch is just closed there is no current through the inductor or R2 , so the
potential difference across the inductor must be 10 V. The potential difference across R1 is always
10 V when the switch is closed, regardless of the amount of time elapsed since closing.
(a) i1 = (10 V)/(5.0 Ω) = 2.0 A.
(b) Zero; read the above paragraph.
(c) The current through the switch is the sum of the above two currents, or 2.0 A.
(d) Zero, since the current through R2 is zero.
(e) 10 V, since the potential across R2 is zero.
(f) Look at the results of Exercise 36-17. When t = 0 the rate of change of the current is
di/dt = E/L. Then
di/dt = (10 V)/(5.0 H) = 2.0 A/s.
(II) After the switch has been closed for a long period of time the currents are stable and the
inductor no longer has an effect on the circuit. Then the circuit is a simple two resistor parallel
network, each resistor has a potential difference of 10 V across it.
139
(a) Still 2.0 A; nothing has changed.
(b) i2 = (10 V)/(10 Ω) = 1.0 A.
(c) Add the two currents and the current through the switch will be 3.0 A.
(d) 10 V; see the above discussion.
(e) Zero, since the current is no longer changing.
(f) Zero, since the current is no longer changing.
E36-22 U = (71 J/m3 )(0.022 m3 ) = 1.56 J. Then using U = i2 L/2 we get
p
p
i = 2U/L = 2(1.56 J)/(0.092 H) = 5.8 A.
E36-23 (a) L = 2U/i2 = 2(0.0253 J)/(0.062 A)2 = 13.2 H.
(b) Since the current is squared in the energy expression, doubling the current would quadruple
the energy. Then i0 = 2i0 = 2(0.062 A) = 0.124 A.
E36-24 (a) B = µ0 in and u = B 2 /2µ0 , or
u = µ0 i2 n2 /2 = (4π×10−7 N/A2 )(6.57 A)2 (950/0.853 m)2 /2 = 33.6 J/m3 .
(b) U = uAL = (33.6 J/m3 )(17.2×10−4 m2 )(0.853 m) = 4.93×10−2 J.
E36-25 uB = B 2 /2µ0 , and from Sample Problem 33-2 we know B, hence
uB =
(12.6 T)2
= 6.32×107 J/m3 .
2(4π×10−7 N/A2 )
E36-26 (a) uB = B 2 /2µ0 , so
uB =
(100×10−12 T)2
1
3
= 2.5×10−2 eV/cm .
2(4π×10−7 N/A2 ) (1.6×10−19 J/eV)
(b) x = (10)(9.46×1015 m) = 9.46×1016 m. Using the results from part (a) expressed in J/m3 we
find the energy contained is
U = (3.98×10−15 J/m3 )(9.46×1016 m)3 = 3.4×1036 J
E36-27 The energy density of an electric field is given by Eq. 36-23; that of a magnetic field is
given by Eq. 36-22. Equating,
0 2
E
2
=
E
=
1 2
B ,
2µ0
B
.
√
0 µ0
The answer is then
p
E = (0.50 T)/ (8.85×10−12 C2 /N · m2 )(4π×10−7 N/A2 ) = 1.5×108 V/m.
E36-28 The rate of internal energy increase in the resistor is given by P = i∆VR . The rate of
energy storage in the inductor is dU/dt = Li di/dt = i∆VL . Since the current is the same through
both we want to find the time when ∆VR = ∆VL . Using Eq. 36-15 we find
1 − e−t/τL = e−t/τL ,
ln 2 = t/τL ,
so t = (37.5 ms) ln 2 = 26.0 ms.
140
E36-29 (a) Start with Eq. 36-13:
= E(1 − e−t/τL )/R,
i
iR
1−
E
= e−t/τL ,
τL
=
−t
,
ln(1 − iR/E)
=
−(5.20×10−3 s)
,
ln[1 − (1.96×10−3 A)(10.4×103 Ω)/(55.0 V)]
=
1.12×10−2 s.
Then L = τL R = (1.12×10−2 s)(10.4×103 Ω) = 116 H.
(b) U = (1/2)(116 H)(1.96×10−3 A)2 = 2.23×10−4 J.
E36-30 (a) U = E∆q; q =
R
idt.
U
E 1 − e−t/τL dt,
R
2
E2 t + τL e−t/τL ,
R
0
E2
E 2 L −t/τL
t + 2 (e
− 1).
R
R
= E
=
=
Z
Using the numbers provided,
τL = (5.48 H)/(7.34 Ω) = 0.7466 s.
Then
U=
i
(12.2 V)2 h
(2 s) + (0.7466 s)(e−(2 s)/0.7466 s) − 1) = 26.4 J
(7.34 Ω)
(b) The energy stored in the inductor is UL = Li2 /2, or
Z 2
LE 2
1 − e−t/τL dt,
UL =
2
2R
= 6.57 J.
(c) UR = U − UL = 19.8 J.
E36-31
This shell has a volume of
V =
4π
(RE + a)3 − RE 3 .
3
Since a << RE we can expand the polynomials but keep only the terms which are linear in a. Then
V ≈ 4πRE 2 a = 4π(6.37×106 m)2 (1.6×104 m) = 8.2×1018 m3 .
The magnetic energy density is found from Eq. 36-22,
uB =
1 2
(60×10−6 T)2
B =
= 1.43×10−3 J/m3 .
2µ0
2(4π×10−7 N/A2 )
The total energy is then (1.43×10−3 J/m3 )(8.2eex18m3 ) = 1.2×1016 J.
141
E36-32 (a) B = µ0 i/2πr and uB = B 2 /2µ0 = µ0 i2 /8π 2 r2 , or
uB = (4π×10−7 H/m)(10 A)2 /8π 2 (1.25×10−3 m)2 = 1.0 J/m3 .
(b) E = ∆V /l = iR/l and uE = 0 E 2 /2 = 0 i2 (R/l)2 /2. Then
uE = (8.85×10−12 F/m)(10 A)2 (3.3×10−3 Ω/m)2 /2 = 4.8×10−15 J/m3 .
E36-33 i =
p
2U/L =
p
2(11.2×10−6 J)/(1.48×10−3 H) = 0.123 A.
E36-34 C = q 2 /2U = (1.63×10−6 C)2 /2(142×10−6 J) = 9.36×10−9 F.
E36-35 1/2πf =
√
LC so L = 1/4π 2 f 2 C, or
L = 1/4π 2 (10×103 Hz)2 (6.7×10−6 F) = 3.8×10−5 H.
E36-36 qmax 2 /2C = Limax 2 /2, or
p
√
imax = qmax / LC = (2.94×10−6 C)/ (1.13×10−3 H)(3.88×10−6 F) = 4.44×10−2 A.
E36-37 Closing a switch has the effect of “shorting” out the relevant circuit element, which
effectively removes it from the circuit. If S1 is closed we have τC = RC or C = τC /R, if instead S2
is closed we have τL = L/R or L = RτL , but if instead S3 is closed we have a LC circuit which will
oscillate with period
√
2π
T =
= 2π LC.
ω
Substituting from the expressions above,
T =
√
2π
= 2π τL τC .
ω
E36-38 The capacitors can be used individually, or in series, or in parallel. The four possible
capacitances are then 2.00µF, 5.00µF, 2.00µF + 5.00µF = 7.00µF, and (2.00µF )(5.00µF)(2.00µF +
5.00µF) = 1.43µF. The possible resonant frequencies are then
r
1
1
= f,
2π LC
s
1
1
= 1330 Hz,
2π (10.0 mH)(1.43µF )
s
1
1
= 1130 Hz,
2π (10.0 mH)(2.00µF )
s
1
1
= 712 Hz,
2π (10.0 mH)(5.00µF )
s
1
1
= 602 Hz.
2π (10.0 mH)(7.00µF )
142
p
p
E36-39 (a) k = (8.13 N)/(0.0021 m) = 3.87×103 N/m. ω = k/m = (3870 N/m)/(0.485 kg) =
89.3 rad/s.
(b) T = 2π/ω = 2π/(89.3 rad/s) = 7.03×10−2 s.
(c) LC = 1/ω 2 , so
C = 1/(89.3 rad/s)2 (5.20 H) = 2.41×10−5 F.
p
√
E36-40 The period of oscillation is T = 2π LC = 2π (52.2mH)(4.21µF) = 2.95 ms. It requires
one-quarter period for the capacitor to charge, or 0.736 ms.
E36-41 (a) An LC circuit oscillates so that the energy is converted from all magnetic to all
electrical twice each cycle. It occurs twice because once the energy is magnetic with the current
flowing in one direction through the inductor, and later the energy is magnetic with the current
flowing the other direction through the inductor.
The period is then four times 1.52µs, or 6.08µs.
(b) The frequency is the reciprocal of the period, or 164000 Hz.
(c) Since it occurs twice during each oscillation it is equal to half a period, or 3.04µs.
E36-42 (a) q = C∆V = (1.13×10−9 F)(2.87 V) = 3.24×10−9 C.
−9
(c) U =p
q 2 /2C = (3.24×10
C)2 /2(1.13×10−9 F) = 4.64×10−9 J.
p
(b) i = 2U/L = 2(4.64×10−9 J)/(3.17×10−3 H) = 1.71×10−3 A.
E36-43 (a) im = q m ω and q m = CV m . Multiplying the second expression by L we get Lq m =
V m /ω 2 . Combining, Lim ω = V m . Then
f=
ω
(50 V)
=
= 6.1×103 /s.
2π
2π(0.042 H)(0.031 A)
(b) See (a) above.
(c) C = 1/ω 2 L = 1/(2π6.1×103 /s)2 (0.042 H) = 1.6×10−8 F.
p
√
E36-44 (a) f = 1/2π LC = 1/2π (6.2×10−6 F)(54×10−3 H) = 275 Hz.
(b) Note that from Eq. 36-32 we can deduce imax = ωqmax . The capacitor starts with a charge
q = C∆V = (6.2×10−6 F)(34 V) = 2.11×10−4 C. Then the current amplitude is
p
√
imax = qmax / LC = (2.11×10−4 C)/ (6.2×10−6 F)(54×10−3 H) = 0.365 A.
p
√
E36-45 (a) ω = 1/ LC = 1/ (10×10−6 F)(3.0×10−3 H) = 5800 rad/s.
(b) T = 2π/ω = 2π/(5800 rad/s) = 1.1×10−3 s.
E36-46 f = (2×105 Hz)(1 + θ/180◦ ). C = 4π 2 /f 2 L, so
C=
4π 2
(9.9×10−7 F)
=
.
(2×105 Hz)2 (1 + θ/180◦ )2 (1 mH)
(1 + θ/180◦ )2
√
2
E36-47 (a) UE = UB /2 and UE + UB = U , so 3UE = U , or 3(q 2 /2C) = qmax
/2C, so q = qmax / 3.
(b) Solve q = qmax cos ωt, or
t=
√
T
arccos 1/ 3 = 0.152T.
2π
143
E36-48 (a) Add the contribution from the inductor and the capacitor,
U=
(3.83×10−6 C)2
(24.8×10−3 H)(9.16×10−3 A)2
+
= 1.99×10−6 J.
2
2(7.73×10−6 F)
p
(b) q m =p 2(7.73×10−6 F)(1.99×10−6 J) = 5.55×10−6 C.
(c) im = 2(1.99×10−6 J)/(24.8×10−3 H) = 1.27×10−2 A.
√
E36-49 (a) The frequency of such a system is given by Eq. 36-26, f = 1/2π LC. Note that
maximum frequency occurs with minimum capacitance. Then
s
r
f1
C2
(365 pF)
=
=
= 6.04.
f2
C1
(10 pF)
(b) The desired ratio is 1.60/0.54 = 2.96 Adding a capacitor in parallel will result in an effective
capacitance given by
C 1,eff = C1 + C add ,
with a similar expression for C2 . We want to choose C add so that
s
C 2,eff
f1
=
= 2.96.
f2
C 1,eff
Solving,
C 2,eff
C2 + C add
C add
= C 1,eff (2.96)2 ,
= (C1 + C add )8.76,
C2 − 8.76C1
=
,
7.76
(365 pF) − 8.76(10 pF)
=
= 36 pF.
7.76
The necessary inductance is then
L=
1
4π 2 f 2 C
=
1
4π 2 (0.54×106 Hz)2 (401×10−12 F)
= 2.2×10−4 H.
E36-50 The key here is that UE = C(∆V )2 /2. We want to charge a capacitor with one-ninth the
capacitance to have three times the potential difference. Since 32 = 9, it is reasonable to assume
that we want to take all of the energy from the first capacitor and give it to the second. Closing
S1 and S2 will not work, because the energy will be shared. Instead, close S2 until the capacitor
has completely discharged into the inductor, then simultaneously open S2 while closing S1 . The
inductor will then discharge into the second capacitor. Open S1 when it is “full”.
√
E36-51 (a) ω = 1/ LC.
qm =
p
im
= (2.0 A) (3.0×10−3 H)(2.7×10−6 F) = 1.80×10−4 C
ω
(b) dUC /dt = qi/C. Since q = q m sin ωt and i = im cos ωt then dUC /dt is a maximum when
sin ωt cos ωt is a maximum, or when sin 2ωt is a maximum. This happens when 2ωt = π/2, or
t = T /8.
(c) dUC /dt = q m im /2C, or
dUC /dt = (1.80×10−4 C)(2.0 A)/2(2.7×10−6 F) = 67 W.
144
−Rt/2L
E36-52 After
. Not only that, but t = N τ , where τ = 2π/ω 0 .
p only complete cycles q = qmax e
0
2
Finally, ω = (1/LC) − (R/2L) . Since the first term under the square root is so much larger√than
the second, we can ignore the effect of damping on the frequency, and simply use ω 0 ≈ ω = 1/ LC.
Then
√
√
q = qmax e−N Rτ /2L = qmax e−N πR LC/L = qmax e−N πR C/L .
p
p
Finally, πR C/L = π(7.22 Ω) (3.18 µF)/(12.3 H) = 1.15×10−2 . Then
N =5
N =5
N =5
:
:
:
q = (6.31µC)e−5(0.0115) = 5.96µC,
q = (6.31µC)e−10(0.0115) = 5.62µC,
q = (6.31µC)e−100(0.0115) = 1.99µC.
E36-53 Use Eq. 36-40, and since U ∝ q 2 , we want to solve e−Rt/L = 1/2, then
L
ln 2.
R
t=
E36-54 Start by assuming
that the presence of the resistance does not significantly change the
√
frequency. Then ω = 1/ LC, q = qmax e−Rt/2L , t = N τ , and τ = 2π/ω. Combining,
√
√
q = qmax e−N Rτ /2L = qmax e−N πR LC/L = qmax e−N πR C/L .
Then
p
p
L/C
(220mH)/(12µF)
R=−
ln(q/qmax ) = −
ln(0.99) = 8700 Ω.
Nπ
(50)π
It remains to be verified that 1/LC (R/2L)2 .
E36-55
be
The damped (angular) frequency is given by Eq. 36-41; the fractional change would then
p
p
ω − ω0
= 1 − 1 − (R/2Lω)2 = 1 − 1 − (R2 C/4L).
ω
Setting this equal to 0.01% and then solving for R,
s
r
4L
4(12.6×10−3 H)
R=
(1 − (1 − 0.0001)2 ) =
(1.9999×10−4 ) = 2.96 Ω.
C
(1.15×10−6 F)
P36-1
The inductance of a toroid is
L=
µ0 N 2 h b
ln .
2π
a
If the toroid is very large and thin then we can write b = a + δ, where δ << a. The natural log then
can be approximated as
b
δ
δ
ln = ln 1 +
≈ .
a
a
a
The product of δ and h is the cross sectional area of the toroid, while 2πa is the circumference,
which we will call l. The inductance of the toroid then reduces to
L≈
µ0 N 2 A
µ0 N 2 δ
=
.
2π a
l
But N is the number of turns, which can also be written as N = nl, where n is the turns per unit
length. Substitute this in and we arrive at Eq. 36-7.
145
P36-2 (a) Since ni is the net current per unit length and is this case i/W , we can simply write
B = µ0 i/W .
(b) There is only one loop of wire, so
L = φB /i = BA/i = µ0 iπR2 /W i = µ0 πR2 /W.
P36-3 Choose the y axis so that it is parallel to the wires and directly between them. The field
in the plane between the wires is
1
1
µ0 i
B=
+
.
2π d/2 + x d/2 − x
The flux per length l of the wires is
ΦB
Z
µ0 i d/2−a
1
1
= l
B dx = l
+
dx,
2π −d/2+a d/2 + x d/2 − x
−d/2+a
Z
µ0 i d/2−a
1
= 2l
dx,
2π −d/2+a d/2 + x
µ0 i d − a
= 2l
ln
.
2π
a
Z
d/2−a
The inductance is then
L=
φB
µ0 l d − a
=
ln
.
i
π
a
P36-4 (a) Choose the y axis so that it is parallel to the wires and directly between them. The
field in the plane between the wires is
µ0 i
1
1
B=
+
.
2π d/2 + x d/2 − x
The flux per length l between the wires is
Φ1
d/2−a
Z
µ0 i d/2−a
1
1
=
B dx =
+
dx,
2π −d/2+a d/2 + x d/2 − x
−d/2+a
Z
µ0 i d/2−a
1
= 2
dx,
2π −d/2+a d/2 + x
µ0 i d − a
= 2
ln
.
2π
a
Z
The field in the plane inside one of the wires, but still between the centers is
µ0 i
1
d/2 − x
B=
+
.
2π d/2 + x
a2
The additional flux is then
Φ2
=
=
Z
µ0 i d/2
1
d/2 − x
+
dx,
2π d/2−a d/2 + x
a2
d/2−a
µ0 i
d
1
2
ln
+
.
2π
d−a 2
2
Z
d/2
B dx = 2
146
The flux per meter between the axes of the wire is the sum, or
µ0 i
d 1
ΦB =
ln +
,
π
a 2
(4π×10−7 H/m)(11.3 A)
(21.8, mm) 1
=
ln
+
,
π
(1.3 mm)
2
=
1.5×10−5 Wb/m.
(b) The fraction f inside the wires is
d
1
d 1
f =
ln
+
/ ln +
,
d−a 2
a 2
(21.8, mm)
1
(21.8, mm) 1
=
+
/
+
,
(21.8, mm) − (1.3 mm) 2
(1.3 mm)
2
= 0.09.
(c) The net flux is zero for parallel currents.
P36-5
The magnetic field in the region between the conductors of a coaxial cable is given by
B=
µ0 i
,
2πr
~ is
so the flux through an area of length l, width b − a, and perpendicular to B
Z
Z
~ · dA
~ = B dA,
ΦB =
B
Z bZ l
µ0 i
=
dz dr,
a
0 2πr
µ0 il b
=
ln .
2π
a
We evaluated this integral is cylindrical coordinates: dA = (dr)(dz). As you have been warned so
many times before, learn these differentials!
The inductance is then
µ0 l b
ΦB
=
ln .
L=
i
2π
a
P36-6 (a) So long as the fuse is not blown there is effectively no resistance in the circuit. Then
the equation for the current is E = L di/dt, but since E is constant, this has a solution i = Et/L.
The fuse blows when t = imax L/E = (3.0 A)(5.0 H)/(10 V) = 1.5 s.
(b) Note that once the fuse blows the maximum steady state current is 2/3 A, so there must be
an exponential approach to that current.
P36-7 The initial rate of increase is di/dt = E/L. Since the steady state current is E/R, the
current will reach the steady state value in a time given by E/R = i = Et/L, or t = L/R. But that’s
τL .
P36-8 (a) U = 21 Li2 = (152 H)(32 A)2 /2 = 7.8×104 J.
(b) If the coil develops at finite resistance then all of the energy in the field will be dissipated as
heat. The mass of Helium that will boil off is
m = Q/Lv = (7.8×104 J)/(85 J/mol)/(4.00g/mol) = 3.7 kg.
147
P36-9
(a) B = (µ0 N i)/(2πr), so
u=
(b) U =
R
u dV =
R
B2
µ0 N 2 i2
=
.
2µ0
8π 2 r2
urdr dθ dz. The field inside the toroid is uniform in z and θ, so
U
=
b
µ0 N 2 i2
r dr,
8π 2 r2
a
hµ0 N 2 i2
b
ln .
4π
a
= 2πh
Z
(c) The answers are the same!
P36-10 The energy in the inductor is originally U = Li20 /2. The internal energy in the resistor
increases at a rate P = i2 R. Then
Z ∞
Z ∞
Li2
Ri20 τL
P dt = R
i20 e−2t/τL dt =
= 0.
2
2
0
0
P36-11
density is
(a) In Chapter 33 we found the magnetic field inside a wire carrying a uniform current
B=
µ0 ir
.
2πR2
The magnetic energy density in this wire is
uB =
1 2
µ0 i2 r2
B =
.
2µ0
8π 2 R4
We want to integrate in cylindrical coordinates over the volume of the wire. Then the volume
element is dV = (dr)(r dθ)(dz), so
Z
UB =
uB dV,
Z R Z l Z 2π
µ0 i2 r2
=
dθ dz rdr,
8π 2 R4
0
0
0
Z
µ0 i2 l R 3
=
r dr,
4πR4 0
µ0 i2 l
=
.
16π
(b) Solve
UB =
L 2
i
2
for L, and
L=
P36-12
2UB
µ0 l
=
.
2
i
8π
1/C = 1/C1 + 1/C2 and L = L1 + L2 . Then
√
1
= LC =
ω
r
C1 C2
(L1 + L2 )
=
C1 + C2
148
s
C2 /ω02 + C1 /ω02
1
=
.
C1 + C2
ω0
P36-13
(a) There is no current in the middle inductor; the loop equation becomes
L
d2 q
q
d2 q
q
+
+
L
+
= 0.
dt2
C
dt2
C
Try q = q m cos ωt as a solution:
−Lω 2 +
1
1
− Lω 2 +
= 0;
C
C
√
which requires ω = 1/ LC.
(b) Look at only the left hand loop; the loop equation becomes
L
q
d2 q
d2 q
+ + 2L 2 = 0.
2
dt
C
dt
Try q = q m cos ωt as a solution:
−Lω 2 +
√
which requires ω = 1/ 3LC.
P36-14
(b) (ω 0 − ω)/ω is the fractional shift; this can also be written as
ω0
ω−1
=
p
=
p
=
P36-15
1
− 2Lω 2 = 0;
C
1 − (LC)(R/2L)2 − 1,
1 − R2 C/4L − 1,
s
1−
(100 Ω)2 (7.3×10−6 F)
− 1 = −2.1×10−3 .
4(4.4 H)
We start by focusing on the charge on the capacitor, given by Eq. 36-40 as
q = q m e−Rt/2L cos(ω 0 t + φ).
After one oscillation the cosine term has returned to the original value but the exponential term has
attenuated the charge on the capacitor according to
q = q m e−RT /2L ,
where T is the period. The fractional energy loss on the capacitor is then
U0 − U
q2
= 1 − 2 = 1 − e−RT /L .
U0
qm
For small enough damping we can expand the exponent. Not only that, but T = 2π/ω, so
∆U
≈ 2πR/ωL.
U
149
P36-16
We are given 1/2 = e−t/2τL when t = 2πn/ω 0 . Then
ω0 =
2πn
πnR
2πn
=
=
.
t
2(L/R) ln 2
L ln 2
From Eq. 36-41,
ω 2 − ω 02
(ω − ω )(ω + ω 0 )
(ω − ω 0 )2ω 0
ω − ω0
ω
0
= (R/2L)2 ,
= (R/2L)2 ,
≈ (R/2L)2 ,
(R/2L)2
= ≈
,
2ω 02
2
(ln 2)
=
,
8π 2 n2
0.0061
=
.
n2
150
E37-1
The frequency, f , is related to the angular frequency ω by
ω = 2πf = 2π(60 Hz) = 377 rad/s
The current is alternating because that is what the generator is designed to produce. It does this
through the configuration of the magnets and coils of wire. One complete turn of the generator will
(could?) produce one “cycle”; hence, the generator is turning 60 times per second. Not only does
this set the frequency, it also sets the emf, since the emf is proportional to the speed at which the
coils move through the magnetic field.
E37-2 (a) XL = ωL, so
f = XL /2πL = (1.28×103 Ω)/2π(0.0452 H) = 4.51×103 /s.
(b) XC = 1/ωC, so
C = 1/2πf XC = 1/2π(4.51×103 /s)(1.28×103 Ω) = 2.76×10−8 F.
(c) The inductive reactance doubles while the capacitive reactance is cut in half.
√
E37-3 (a) XL = XC implies ωL = 1/ωC or ω = 1/ LC, so
p
ω = 1/ (6.23×10−3 H)(11.4×10−6 F) = 3750 rad/s.
(b) XL = ωL = (3750 rad/s)(6.23×10−3 H) = 23.4 Ω
(c) See (a) above.
E37-4 (a) im = E/XL = E/ωL, so
im = (25.0 V)/(377 rad/s)(12.7 H) = 5.22×10−3 A.
(b) The current and emf are 90◦ out of phase. When the current is a maximum, E = 0.
(c) ωt = arcsin[E(t)/E m ], so
ωt = arcsin
(−13.8 V)
= 0.585 rad.
(25.0 V)
and
i = (5.22×10−3 A) cos(0.585) = 4.35×10−3 A.
(d) Taking energy.
E37-5 (a) The reactance of the capacitor is from Eq. 37-11, XC = 1/ωC. The AC generator
from Exercise 4 has E = (25.0 V) sin(377 rad/s)t. So the reactance is
XC =
1
1
=
= 647 Ω.
ωC
(377 rad/s)(4.1µF)
The maximum value of the current is found from Eq. 37-13,
im =
(∆VC )max )
(25.0 V)
=
= 3.86×10−2 A.
XC
(647 Ω)
(b) The generator emf is 90◦ out of phase with the current, so when the current is a maximum
the emf is zero.
151
(c) The emf is -13.8 V when
ωt = arcsin
(−13.8 V)
= 0.585 rad.
(25.0 V)
The current leads the voltage by 90◦ = π/2, so
i = im sin(ωt − φ) = (3.86×10−2 A) sin(0.585 − π/2) = −3.22×10−2 A.
(d) Since both the current and the emf are negative the product is positive and the generator is
supplying energy to the circuit.
E37-6 R = (ωL − 1/omegaC)/ tan φ and ω = 2πf = 2π(941/s) = 5910 rad/s , so
R=
(5910 rad/s)(88.3×10−3 H) − 1/(5910 rad/s)(937×10−9 F)
= 91.5 Ω.
tan(75◦ )
E37-7
E37-8 (a) XL doesn’t change, so XL = 87 Ω.
−6
(b) XC =
p1/ωC = 1/2π(60/s)(70×10 F) = 37.9Ω.
2
2
(c) Z = (160 Ω) + (87 Ω − 37.9 Ω) = 167 Ω.
(d) im = (36 V)/(167 Ω) = 0.216 A.
(e) tan φ = (87 Ω − 37.9 Ω)/(160 Ω) = 0.3069, so
φ = arctan(0.3069) = 17◦ .
E37-9 A circuit is considered inductive if XL > XC , this happens when im lags E m . If, on the
other hand, XL < XC , and im leads E m , we refer to the circuit as capacitive. This is discussed on
page 850, although it is slightly hidden in the text of column one.
(a) At resonance, XL = XC . Since XL = ωL and XC = 1/ωC we expect that XL grows with
increasing frequency, while XC decreases with increasing frequency.
Consequently, at frequencies above the resonant frequency XL > XC and the circuit is predominantly inductive. But what does this really mean? It means that the inductor plays a major role in
the current through the circuit while the capacitor plays a minor role. The more inductive a circuit
is, the less significant any capacitance is on the behavior of the circuit.
For frequencies below the resonant frequency the reverse is true.
(b) Right at the resonant frequency the inductive effects are exactly canceled by the capacitive
effects. The impedance is equal to the resistance, and it is (almost) as if neither the capacitor or
inductor are even in the circuit.
E37-10 The net y component is XC − XL . The net x component is R. The magnitude of the
resultant is
p
Z = R2 + (XC − XL )2 ,
while the phase angle is
tan φ =
−(XC − XL )
.
R
E37-11 Yes.
p
At resonance ω = 1/ (1.2 H)(1.3×10−6 F) = 800 rad/s and Z = R. Then im = E/Z =
(10 V)/(9.6 Ω) = 1.04 A, so
[∆VL ]m = im XL = (1.08 A)(800 rad/s)(1.2 H) = 1000 V.
152
E37-12 (a) Let O = XL − XC and A = R, then H 2 = A2 + O2 = Z 2 , so
sin φ = (XL − XC )/Z
and
cos φ = R/Z.
E37-13 (a) The voltage across the generator is the generator emf, so when it is a maximum from
Sample Problem 37-3, it is 36 V. This corresponds to ωt = π/2.
(b) The current through the circuit is given by i = im sin(ωt − φ). We found in Sample Problem
37-3 that im = 0.196 A and φ = −29.4◦ = 0.513 rad.
For a resistive load we apply Eq. 37-3,
∆VR = im R sin(ωt − φ) = (0.196 A)(160Ω) sin((π/2) − (−0.513)) = 27.3 V.
(c) For a capacitive load we apply Eq. 37-12,
∆VC = im XC sin(ωt − φ − π/2) = (0.196 A)(177Ω) sin(−(−0.513)) = 17.0 V.
(d) For an inductive load we apply Eq. 37-7,
∆VL = im XL sin(ωt − φ + π/2) = (0.196 A)(87Ω) sin(π − (−0.513)) = −8.4 V.
(e) (27.3 V) + (17.0 V) + (−8.4 V) = 35.9 V.
E37-14 If circuit 1 and 2 have the same resonant frequency then L1 C1 = L2 C2 . The series
combination for the inductors is
L = L1 + L2 ,
The series combination for the capacitors is
1/C = 1/C1 + 1/C2 ,
so
C1 C2
L1 C1 C2 + L2 C2 C1
=
= L1 C1 ,
C1 + C2
C1 + C2
which is the same as both circuit 1 and 2.
LC = (L1 + L2 )
E37-15 (a) Z = (125 V)/(3.20 A) = 39.1 Ω.
(b) Let O = XL − XC and A = R, then H 2 = A2 + O2 = Z 2 , so
cos φ = R/Z.
Using this relation,
R = (39.1 Ω) cos(56.3◦ ) = 21.7 Ω.
(c) If the current leads the emf then the circuit is capacitive.
E37-16 (a) Integrating over a single cycle,
Z
1 T
sin2 ωt dt =
T 0
=
(b) Integrating over a single cycle,
Z
1 T
sin ωt cos ωt dt
T 0
Z
1 T 1
(1 − cos 2ωt) dt,
T 0 2
1
1
T = .
2T
2
1
T
= 0.
=
153
Z
0
T
1
sin 2ωtdt,
2
E37-17
The resistance would be given by Eq. 37-32,
R=
P av
(0.10)(746 W)
=
= 177 Ω.
2
irms
(0.650 A)2
This would not be the same as the direct current resistance of the coils of a stopped motor, because
there would be no inductive effects.
E37-18 Since irms = E rms /Z, then
P av = i2 rms R =
E37-19 (a) Z =
p
(160 Ω)2 + (177 Ω)2 = 239 Ω; then
P av =
(b) Z =
E 2 rms R
.
Z2
1 (36 V)2 (160 Ω)
= 1.82 W.
2
(239 Ω)2
p
(160 Ω)2 + (87 Ω)2 = 182 Ω; then
P av =
1 (36 V)2 (160 Ω)
= 3.13 W.
2
(182 Ω)2
p
E37-20 (a) Z = (12.2 Ω)2 + (2.30 Ω)2 = 12.4 Ω
(b) P av = (120 V)2 (12.2 Ω)/(12.4 Ω)2 = 1140 W.
(c) irms = (120 V)/(12.4 Ω) = 9.67 A.
√
E37-21 The rms value of any sinusoidal quantity is related to the maximum value by 2 v rms =
√
vmax . Since this factor of 2 appears in all of the expressions, we can conclude that if the rms values
are equal then so are the maximum values. This means that
(∆VR )max = (∆VC )max = (∆VL )max
or im R = im XC = im XL or, with one last simplification, R = XL = XC . Focus on the right hand
side of the last equality. If XC = XL then we have a resonance condition, and the impedance (see
Eq. 37-20) is a minimum, and is equal to R. Then, according to Eq. 37-21,
im =
Em
,
R
which has the immediate consequence that the rms voltage across the resistor is the same as the
rms voltage across the generator. So everything is 100 V.
√
E37-22 (a) The antenna is “in-tune” when the impedance is a minimum, or ω = 1/ LC. So
p
f = ω/2π = 1/2π (8.22×10−6 H)(0.270×10−12 F) = 1.07×108 Hz.
(b) irms = (9.13 µV)/(74.7 Ω) = 1.22×10−7 A.
(c) XC = 1/2πf C, so
VC = iXC = (1.22×10−7 A)/2π(1.07×108 Hz)(0.270 ×10−12 F) = 6.72×10−4 V.
154
E37-23 Assuming no inductors or capacitors in the circuit, then the circuit effectively behaves as
a DC circuit. The current through the circuit is i = E/(r + R). The power delivered to R is then
P = i∆V = i2 R = E 2 R/(r + R)2 . Evaluate dP/dR and set it equal to zero to find the maximum.
Then
r−R
dP
= E 2R
,
0=
dR
(r + R)3
which has the solution r = R.
E37-24 (a) Since P av = im 2 R/2 = E m 2 R/2Z 2 , then P av is a maximum
√ when Z is a minimum, and
vise-versa. Z is a minimum at resonance, when Z = R and f = 1/2π LC. When Z is a minimum
C = 1/4π 2 f 2 L = 1/4π 2 (60 Hz)2 (60 mH) = 1.2×10−7 F.
(b) Z is a maximum when XC is a maximum, which occurs when C is very small, like zero.
(c) When XC is a maximum P = 0. When P is a maximum Z = R so
P = (30 V)2 /2(5.0 Ω) = 90 W.
(d) The phase angle is zero for resonance; it is 90◦ for infinite XC or XL .
(e) The power factor is zero for a system which has no power. The power factor is one for a
system in resonance.
E37-25
(a) The resistance is R = 15.0 Ω. The inductive reactance is
XC =
1
1
=
= 61.3 Ω.
−1
ωC
2π(550 s )(4.72µF)
The inductive reactance is given by
XL = ωL = 2π(550 s−1 )(25.3 mH) = 87.4 Ω.
The impedance is then
Z=
q
2
(15.0 Ω)2 + ((87.4 Ω) − (61.3 Ω)) = 30.1 Ω.
Finally, the rms current is
E rms
(75.0 V)
=
= 2.49 A.
Z
(30.1 Ω)
(b) The rms voltages between any two points is given by
irms =
(∆V )rms = irms Z,
where Z is not the impedance of the circuit but instead the impedance between the two points in
question. When only one device is between the two points the impedance is equal to the reactance
(or resistance) of that device.
We’re not going to show all of the work here, but we will put together a nice table for you
Points
ab
bc
cd
bd
ac
Impedance Expression
Z=R
Z = XC
Z = XL
Z =p
|XL − XC |
Z = R2 + XC2
Impedance Value
Z = 15.0 Ω
Z = 61.3 Ω
Z = 87.4 Ω
Z = 26.1 Ω
Z = 63.1 Ω
(∆V )rms ,
37.4 V,
153 V,
218 V,
65 V,
157 V,
Note that this last one was ∆Vac , and not ∆Vad , because it is more entertaining. You probably
should use ∆Vad for your homework.
(c) The average power dissipated from a capacitor or inductor is zero; that of the resistor is
PR = [(∆VR )rms ]2 /R = (37.4 V)2 /(15.0Ω) = 93.3 W.
155
E37-26 (a) The energy stored in the capacitor is a function of the charge on the capacitor; although
the charge does vary with time it varies periodically and at the end of the cycle has returned to the
original values. As such, the energy stored in the capacitor doesn’t change from one period to the
next.
(b) The energy stored in the inductor is a function of the current in the inductor; although the
current does vary with time it varies periodically and at the end of the cycle has returned to the
original values. As such, the energy stored in the inductor doesn’t change from one period to the
next.
(c) P = Ei = E m im sin(ωt) sin(ωt − φ), so the energy generated in one cycle is
U
=
Z
T
P dt = E m im
Z
0
T
sin(ωt) sin(ωt − φ)dt,
0
= E m im
T
Z
sin(ωt) sin(ωt − φ)dt,
0
=
T
E m im cos φ.
2
(d) P = im 2 R sin2 (ωt − φ), so the energy dissipated in one cycle is
U
=
Z
T
2
P dt = im R
0
Z
T
sin2 (ωt − φ)dt,
0
= im 2 R
Z
T
sin2 (ωt − φ)dt,
0
T 2
im R.
2
=
(e) Since cos φ = R/Z and E m /Z = im we can equate the answers for (c) and (d).
E37-27
Apply Eq. 37-41,
∆V s = ∆V p
Ns
(780)
= (150 V)
= 1.8×103 V.
Np
(65)
E37-28 (a) Apply Eq. 37-41,
∆V s = ∆V p
Ns
(10)
= (120 V)
= 2.4 V.
Np
(500)
(b) is = (2.4 V)/(15 Ω) = 0.16 A;
ip = is
Ns
(10)
= (0.16 A)
= 3.2×10−3 A.
Np
(500)
E37-29 The autotransformer could have a primary connected between taps T1 and T2 (200 turns),
T1 and T3 (1000 turns), and T2 and T3 (800 turns).
The same possibilities are true for the secondary connections. Ignoring the one-to-one connections
there are 6 choices— three are step up, and three are step down. The step up ratios are 1000/200 = 5,
800/200 = 4, and 1000/800 = 1.25. The step down ratios are the reciprocals of these three values.
156
E37-30 ρ = (1.69×10−8 Ω · m)[1 − (4.3×10−3 /C◦ )(14.6◦ C)] = 1.58×10−8 Ω · m. The resistance of
the two wires is
ρL
(1.58×10−8 Ω · m)2(1.2×103 m)
R=
=
= 14.9 Ω.
A
π(0.9×10−3 m)2
P = i2 R = (3.8 A)2 (14.9 Ω) = 220 W.
E37-31 The supply current is
√
ip = (0.270 A)(74×103 V/ 2)/(220 V) = 64.2 A.
The potential drop across the supply lines is
∆V = (64.2 A)(0.62 Ω) = 40 V.
This is the amount by which the supply voltage must be increased.
E37-32 Use Eq. 37-46:
N p /N s =
p
(1000 Ω)/(10 Ω) = 10.
P37-1 (a) The emf is a maximum when ωt − π/4 = π/2, so t = 3π/4ω = 3π/4(350 rad/s) =
6.73×10−3 s.
(b) The current is a maximum when ωt − 3π/4 = π/2, so t = 5π/4ω = 5π/4(350 rad/s) =
1.12×10−2 s.
(c) The current lags the emf, so the circuit contains an inductor.
(d) XL = E m /im and XL = ωL, so
L=
Em
(31.4 V)
=
= 0.144 H.
im ω
(0.622 A)(350 rad/s)
P37-2 (a) The emf is a maximum when ωt − π/4 = π/2, so t = 3π/4ω = 3π/4(350 rad/s) =
6.73×10−3 s.
(b) The current is a maximum when ωt+π/4 = π/2, so t = π/4ω = π/4(350 rad/s) = 2.24×10−3 s.
(c) The current leads the emf, so the circuit contains an capacitor.
(d) XC = E m /im and XC = 1/ωC, so
C=
im
(0.622 A)
=
= 5.66×10−5 F.
E mω
(31.4 V)(350 rad/s)
P37-3 (a) Since the maximum values for the voltages across the individual devices are proportional to the reactances (or resistances) for devices in series (the constant of proportionality is the
maximum current), we have XL = 2R and XC = R.
From Eq. 37-18,
XL − XC
2R − R
tan φ =
=
= 1,
R
R
or φ = 45◦ .
(b) The impedance of the circuit, in terms of the resistive element, is
p
p
√
Z = R2 + (XL − XC )2 = R2 + (2R − R)2 = 2 R.
But E m = im Z, so Z = (34.4 V)/(0.320 A) = 108Ω. Then we can use our previous work to finds that
R = 76Ω.
157
P37-4 When the switch is open the circuit is an LRC circuit. In position 1 the circuit is an RLC
circuit, but the capacitance is equal to the two capacitors of C in parallel, or 2C. In position 2 the
circuit is a simple LC circuit with no resistance.
The impedance when the switch is in position 2 is Z2 = |ωL − 1/ωC|. But
Z2 = (170 V)/(2.82 A) = 60.3 Ω.
The phase angle when the switch is open is φ0 = 20◦ . But
tan φ0 =
Z2
ωL − 1/ωC
=
,
R
R
so R = (60.3 Ω)/ tan(20◦ ) = 166 Ω.
The phase angle when the switch is in position 1 is
tan φ1 =
ωL − 1/ω2C
,
R
so ωL − 1/ω2C = (166 Ω) tan(10◦ ) = 29.2 Ω. Equating the ωL part,
(29.2 Ω) + 1/ω2C
C
=
=
(−60.3 Ω) + 1/ωC,
1/2(377 rad/s)[(60.3 Ω) + (29.2 Ω)] = 1.48×10−5 F.
Finally,
L=
(−60.3Ω) + 1/(377 rad/s)(1.48×10−5 F)
= 0.315 H.
(377 rad/s)
P37-5 All three wires have emfs which vary sinusoidally in time; if we choose any two wires the
phase difference will have an absolute value of 120◦ . We can then choose any two wires and expect
(by symmetry) to get the same result. We choose 1 and 2. The potential difference is then
V1 − V2 = V m (sin ωt − sin(ωt − 120◦ )) .
We need to add these two sine functions to get just one. We use
1
1
sin α − sin β = 2 sin (α − β) cos (α + β).
2
2
Then
V1 − V 2
1
1
2V m sin (120◦ ) cos (2ωt − 120◦ ),
2
2
√
3
= 2V m (
) cos(ωt − 60◦ ),
2
√
=
3V m sin(ωt + 30◦ ).
=
P37-6 (a) cos φ = cos(−42◦ ) = 0.74.
(b) The current leads.
(c) The circuit is capacitive.
(d) No. Resonance circuits have a power factor of one.
(e) There must be at least a capacitor and a resistor.
(f) P = (75 V)(1.2 A)(0.74)/2 = 33 W.
158
p
√
P37-7 (a) ω = 1/ LC = 1/ (0.988 H)(19.3×10−6 F) = 229 rad/s.
(b) im = (31.3 V)/(5.12 Ω) = 6.11 A.
(c) The current amplitude will be halved when the impedance is doubled, or when Z = 2R. This
occurs when 3R2 = (ωL − 1/ωC)2 , or
3R2 ω 2 = ω 4 L2 − 2ω 2 L/C + 1/C 2 .
The solution to this quadratic is
2
ω =
2L + 3CR2 ±
√
9C 2 R4 + 12CR2 L
,
2L2 C
so ω1 = 224.6 rad/s and ω2 = 233.5 rad/s.
(d) ∆ω/ω = (8.9 rad/s)/(229 rad/s) = 0.039.
P37-8 (a) The current amplitude will be halved when the impedance is doubled, or when Z = 2R.
This occurs when 3R2 = (ωL − 1/ωC)2 , or
3R2 ω 2 = ω 4 L2 − 2ω 2 L/C + 1/C 2 .
The solution to this quadratic is
2
ω =
2L + 3CR2 ±
√
9C 2 R4 + 12CR2 L
,
2L2 C
Note that ∆ω = ω+ − ω− ; with a wee bit of algebra,
2
2
∆ω(ω+ + ω− ) = ω+
− ω−
.
Also, ω+ + ω− ≈ 2ω. Hence,
√
ω∆ω
≈
ω∆ω
≈
ω∆ω
≈
∆ω
ω
≈
≈
9C 2 R4 + 12CR2 L
,
2L2 C
√
ω 2 R 9C 2 R2 + 12LC
,
2L
√
ωR 9ω 2 C 2 R2 + 12
,
2L
p
R 9CR2 /L + 12
,
2Lω
√
3R
,
ωL
assuming that CR2 4L/3.
P37-9
P37-10
P37-11
Use Eq. 37-46.
(a) The resistance of this bulb is
R=
(∆V )2
(120 V)2
=
= 14.4 Ω.
P
(1000 W)
159
The power is directly related to the brightness; if the bulb is to be varied in brightness by a factor of
5 then it would have a minimum power of 200 W. The rms current through the bulb at this power
would be
p
p
irms = P/R = (200 W)/(14.4 Ω) = 3.73 A.
The impedance of the circuit must have been
Z=
E rms
(120 V)
=
= 32.2 Ω.
irms
(3.73 A)
The inductive reactance would then be
p
p
XL = Z 2 − R2 = (32.2 Ω)2 − (14.4 Ω)2 = 28.8 Ω.
Finally, the inductance would be
L = XL /ω = (28.8 Ω)/(2π(60.0 s−1 )) = 7.64 H.
(b) One could use a variable resistor, and since it would be in series with the lamp a value of
32.2 Ω − 14.4Ω = 17.8 Ω
would work. But the resistor would get hot, while on average there is no power radiated from a pure
inductor.
160
E38-1 The maximum value occurs where r = R; there Bmax = 12 µ0 0 R dE/dt. For r < R B is
half of Bmax when r = R/2. For r > R B is half of Bmax when r = 2R. Then the two values of r
are 2.5 cm and 10.0 cm.
E38-2 For a parallel plate capacitor E = σ/0 and the flux is then ΦE = σA/0 = q/0 . Then
id = 0
dΦE
dq
d
dV
=
= CV = C
.
dt
dt
dt
dt
E38-3 Use the results of Exercise 2, and change the potential difference across the plates of the
capacitor at a rate
id
(1.0 mA)
dV
=
=
= 1.0 kV/s.
dt
C
(1.0µF)
Provide a constant current to the capacitor
i=
d
dV
dQ
= CV = C
= id .
dt
dt
dt
E38-4 Since E is uniform between the plates ΦE = EA, regardless of the size of the region of
interest. Since j d = id /A,
id
1 dΦE
dE
jd =
= 0
= 0
.
A
A
dt
dt
E38-5 (a) In this case id = i = 1.84 A.
(b) Since E = q/0 A, dE/dt = i/0 A, or
dE/dt = (1.84 A)/(8.85×10−12 F/m)(1.22 m)2 = 1.40×1011 V/m.
(c) id = 0 dΦE /dt = 0 adE/dt. a here refers to the area of the smaller square. Combine this
with the results of part (b) and
id = ia/A = (1.84 A)(0.61 m/1.22 m)2 = 0.46 A.
(d)
H
~ · d~s = µ0 id = (4π×10−7 H/m)(0.46 A) = 5.78×10−7 T · m.
B
E38-6 Substitute Eq. 38-8 into the results obtained in Sample Problem 38-1. Outside the capacitor
ΦE = πR2 E, so
µ0 0 πR2 dE
µ0
B=
=
id .
2πr
dt
2πr
Inside the capacitor the enclosed flux is ΦE = πr2 E; but we want instead to define id in terms of
the total ΦE inside the capacitor as was done above. Consequently, inside the conductor
B=
E38-7
have
µ0 r 0 πR2 dE
µ0 r
=
id .
2πR2
dt
2πR2
Since the electric field is uniform in the area and perpendicular to the surface area we
Z
Z
Z
~ · dA
~ = E dA = E dA = EA.
ΦE = E
The displacement current is then
id = 0 A
dE
dE
= (8.85×10−12 F/m)(1.9 m2 )
.
dt
dt
161
(a) In the first region the electric field decreases by 0.2 MV/m in 4µs, so
id = (8.85×10−12 F/m)(1.9 m2 )
(−0.2×106 V/m)
= −0.84 A.
(4×10−6 s)
(b) The electric field is constant so there is no change in the electric flux, and hence there is no
displacement current.
(c) In the last region the electric field decreases by 0.4 MV/m in 5µs, so
id = (8.85×10−12 F/m)(1.9 m2 )
(−0.4×106 V/m)
= −1.3 A.
(5×10−6 s)
H
~ · d~s = 2πrB, where r is the distance from the
E38-8 (a) Because of the circular symmetry B
center of the circular plates. Not only that, but id = j d A = πr2 j d . Equate these two expressions,
and
B = µ0 rj d /2 = (4π×10−7 H/m)(0.053 m)(1.87×101 A/m)/2 = 6.23×10−7 T.
(b) dE/dt = id /0 A = j d /0 = (1.87×101 A/m)/(8.85×10−12 F/m) = 2.11×10−12 V/m.
E38-9 The magnitude of E is given by
E=
(162 V)
sin 2π(60/s)t;
(4.8×10−3 m)
Using the results from Sample Problem 38-1,
µ0 0 R dE Bm =
,
2
dt t=0
=
(4π×10−7 H/m)(8.85×10−12 F/m)(0.0321 m)
(162 V)
2π(60/s)
,
2
(4.8×10−3 m)
=
2.27×10−12 T.
E38-10 (a) Eq. 33-13 from page 764 and Eq. 33-34 from page 762.
(b) Eq. 27-11 from page 618 and the equation from Ex. 27-25 on page 630.
(c) The equations from Ex. 38-6 on page 876.
(d) Eqs. 34-16 and 34-17 from page 785.
E38-11 (a) Consider the path abef a. The closed line integral consists of two parts: b → e and
e → f → a → b. Then
I
~ · d~s = − dΦ
E
dt
can be written as
Z
~ · d~s +
E
b→e
Z
~ · d~s = − d Φabef .
E
dt
e→f →a→b
Now consider the path bcdeb. The closed line integral consists of two parts: b → c → d → e and
e → b. Then
I
~ · d~s = − dΦ
E
dt
can be written as
Z
b→c→d→e
~ · d~s +
E
Z
~ · d~s = − d Φbcde .
E
dt
e→b
162
These two expressions can be added together, and since
Z
Z
~
~ · d~s
E · d~s = −
E
e→b
b→e
we get
Z
e→f →a→b
~ · d~s +
E
Z
~ · d~s = − d (Φabef + Φbcde ) .
E
dt
b→c→d→e
The left hand side of this is just the line integral over the closed path ef adcde; the right hand side
is the net change in flux through the two surfaces. Then we can simplify this expression as
I
~ · d~s = − dΦ .
E
dt
(b) Do everything above again, except substitute B for E.
(c) If the equations were not self consistent we would arrive at different values of E and B
depending on how we defined our surfaces. This multi-valued result would be quite unphysical.
E38-12 (a) Consider the part on the left. It has a shared surface s, and the other surfaces l.
Applying Eq. I,
I
Z
Z
~
~
~
~
~ · dA.
~
ql /0 = E · dA = E · dA + E
s
l
~ is directed to the right on the shared surface.
Note that dA
Consider the part on the right. It has a shared surface s, and the other surfaces r. Applying Eq.
I,
I
Z
Z
~ · dA
~ = E
~ · dA
~ + E
~ · dA.
~
qr /0 = E
s
r
~ is directed to the left on the shared surface.
Note that dA
Adding these two expressions will result in a canceling out of the part
Z
~ · dA
~
E
s
since one is oriented opposite the other. We are left with
Z
Z
I
qr + ql
~ · dA
~ + E
~ · dA
~ = E
~ · dA.
~
= E
0
r
l
E38-13
E38-14 (a) Electric dipole is because the charges are separating like an electric dipole. Magnetic
dipole because the current loop acts like a magnetic dipole.
E38-15
A series LC circuit will oscillate naturally at a frequency
f=
ω
1
√
=
2π
2π LC
We will need to combine this with v = f λ, where v = c is the speed of EM waves.
We want to know the inductance required to produce an EM wave of wavelength λ = 550×10−9 m,
so
λ2
(550 × 10−9 m)2
L=
=
= 5.01 × 10−21 H.
4π 2 c2 C
4π 2 (3.00 × 108 m/s)2 (17 × 10−12 F)
This is a small inductance!
163
E38-16 (a) B = E/c, and B must be pointing in the negative y direction in order that the wave
be propagating in the positive x direction. Then Bx = Bz = 0, and
By = −Ez /c = −(2.34×10−4 V/m)/(3.00×108 m/s) = (−7.80×10−13 T) sin k(x − ct).
(b) λ = 2π/k = 2π/(9.72×106 /m) = 6.46×10−7 m.
E38-17 The electric and magnetic field of an electromagnetic wave are related by Eqs. 38-15
and 38-16,
(321µV/m)
E
B=
=
= 1.07 pT.
c
(3.00 × 108 m/s)
E38-18 Take the partial of Eq. 38-14 with respect to x,
∂ ∂E
∂x ∂x
∂2E
∂x2
∂ ∂B
,
∂x ∂t
∂2B
= −
.
∂x∂t
= −
Take the partial of Eq. 38-17 with respect to t,
∂ ∂B
∂t ∂x
∂2B
−
∂t∂x
−
∂ ∂E
,
∂t ∂t
2
∂ E
= µ0 0 2 .
∂t
= µ0 0
Equate, and let µ0 0 = 1/c2 , then
∂2E
1 ∂2E
=
.
∂x2
c2 ∂t2
Repeat, except now take the partial of Eq. 38-14 with respect to t, and then take the partial of Eq.
38-17 with respect to x.
E38-19 (a) Since sin(kx − ωt) is of the form f (kx ± ωt), then we only need do part (b).
(b) The constant E m drops out of the wave equation, so we need only concern ourselves with
f (kx ± ωt). Letting g = kx ± ωt,
∂2f
∂t2
2
2
∂ f ∂g
∂g 2 ∂t
∂g
∂t
ω
∂2f
,
∂x2
2
2
∂g
2∂ f
= c
,
∂g 2 ∂x
∂g
= c ,
∂x
= ck.
= c2
E38-20 Use the right hand rule.
E38-21 U = P t = (100×1012 W)(1.0×10−9 s) = 1.0×105 J.
E38-22 E = Bc = (28×10−9 T)(3.0×108 m/s) = 8.4 V/m. It is in the positive x direction.
164
E38-23 Intensity is given by Eq. 38-28, which is simply an expression of power divided by surface
area. To find the intensity of the TV signal at α-Centauri we need to find the distance in meters;
r = (4.30 light-years)(3.00×108 m/s)(3.15×107 s/year) = 4.06 × 1016 m.
The intensity of the signal when it has arrived at out nearest neighbor is then
I=
(960 kW)
P
2
=
= 4.63 × 10−29 W/m
4πr2
4π(4.06 × 1016 m)2
E38-24 (a) From Eq. 38-22, S = cB 2 /µ0 . B = B m sin ωt. The time average is defined as
1
T
Z
T
S dt =
0
cB m 2
µ0 T
T
Z
cos2 ωt dt =
0
cB m 2
.
2µ0
(b) S av = (3.0×108 m/s)(1.0×10−4 T)2 /2(4π×10−7 H/m) = 1.2×106 W/m2 .
E38-25 I = P/4πr2 , so
r=
p
P/4πI =
p
(1.0×103 W)/4π(130 W/m2 ) = 0.78 m.
E38-26 uE = 0 E 2 /2 = 0 (cB)2 /2 = B 2 /2µ0 = uB .
E38-27 (a) Intensity is related to distance by Eq. 38-28. If r1 is the original distance from the
street lamp and I1 the intensity at that distance, then
I1 =
P
.
4πr12
There is a similar expression for the closer distance r2 = r1 −162 m and the intensity at that distance
I2 = 1.50I1 . We can combine the two expression for intensity,
I2
P
4πr22
=
r12
=
r1
1.50I1 ,
P
= 1.50
,
4πr12
1.50r2 ,
√ 2
=
1.50 (r1 − 162 m).
The last line is easy enough to solve and we find r1 = 883 m.
(b) No, we can’t find the power output from the lamp, because we were never provided with an
absolute intensity reference.
√
E38-28 (a) E m = 2µ0 cI, so
p
E m = 2(4π×10−7 H/m)(3.00×108 m/s)(1.38×103 W/m2 ) = 1.02×103 V/m.
(b) B m = E m /c = (1.02×103 V/m)/(3.00×108 m/s) = 3.40×10−6 T.
E38-29 (a) B m = E m /c = (1.96 V/m)/(3.00×108 m/s) = 6.53×10−9 T.
(b) I = E m 2 /2µ0 c = (1.96 V)2 /2(4π×10−7 H/m)(3.00×108 m/s) = 5.10×10−3 W/m2 .
(c) P = 4πr2 I = 4π(11.2 m)2 (5.10×10−3 W/m2 ) = 8.04 W.
165
E38-30 (a) The intensity is
I=
(1×10−12 W)
P
=
= 1.96×10−27 W/m2 .
A
4π(6.37×106 m)2
The power received by the Arecibo antenna is
P = IA = (1.96×10−27 W/m2 )π(305 m)2 /4 = 1.4×10−22 W.
(b) The power of the transmitter at the center of the galaxy would be
P = IA = (1.96×10−27 W)π(2.3×104 ly)2 (9.46×1015 m/ly)2 = 2.9×1014 W.
E38-31
(a) The electric field amplitude is related to the intensity by Eq. 38-26,
I=
E 2m
,
2µ0 c
or
Em =
p
p
2µ0 cI = 2(4π×10−7 H/m)(3.00×108 m/s)(7.83µW/m2 ) = 7.68×10−2 V/m.
(b) The magnetic field amplitude is given by
Bm =
Em
(7.68 × 10−2 V/m)
=
= 2.56 × 10−10 T
c
(3.00 × 108 m/s)
(c) The power radiated by the transmitter can be found from Eq. 38-28,
P = 4πr2 I = 4π(11.3 km)2 (7.83µW/m2 ) = 12.6 kW.
E38-32 (a) The power incident on (and then reflected by) the target craft is P1 = I1 A = P0 A/2πr2 .
The intensity of the reflected beam is I2 = P1 /2πr2 = P0 A/4π 2 r4 . Then
I2 = (183×103 W)(0.222 m2 )/4π 2 (88.2×103 m)4 = 1.70×10−17 W/m2 .
(b) Use Eq. 38-26:
p
p
E m = 2µ0 cI = 2(4π×10−7 H/m)(3.00×108 m/s)(1.70×10−17 W/m2 ) = 1.13×10−7 V/m.
√
√
(c) B rms = E m / 2c = (1.13×10−7 V/m)/ 2(3.00×108 m/s) = 2.66×10−16 T.
E38-33 Radiation pressure for absorption is given by Eq. 38-34, but we need to find the energy
absorbed before we can apply that. We are given an intensity, a surface area, and a time, so
∆U = (1.1×103 W/m2 )(1.3 m2 )(9.0×103 s) = 1.3×107 J.
The momentum delivered is
p = (∆U )/c = (1.3×107 J)/(3.00×108 m/s) = 4.3×10−2 kg · m/s.
E38-34 (a) F/A = I/c = (1.38×103 W/m2 )/(3.00×108 m/s) = 4.60×10−6 Pa.
(b) (4.60×10−6 Pa)/(101×105 Pa) = 4.55×10−11 .
E38-35 F/A = 2P/Ac = 2(1.5×109 W)/(1.3×10−6 m2 )(3.0×108 m/s) = 7.7×106 Pa.
166
E38-36 F/A = P/4πr2 c, so
F/A = (500 W)/4π(1.50 m)2 (3.00×108 m/s) = 5.89×10−8 Pa.
E38-37 (a) F = IA/c, so
F =
(1.38×103 W/m2 )π(6.37×106 m)2
= 5.86×108 N.
(3.00×108 m/s)
E38-38 (a) Assuming MKSA, the units are
m C V sN
Ns
mFV N
=
= 2 .
s m m Am
s Vm m Cm
m s
(b) Assuming MKSA, the units are
A2 V N
A2 J N
1 J
J
=
=
= 2 .
N m Am
N Cm Am
sm m
m s
E38-39 We can treat the object as having two surfaces, one completely reflecting and the other
completely absorbing. If the entire surface has an area A then the absorbing part has an area f A
while the reflecting part has area (1 − f )A. The average force is then the sum of the force on each
part,
I
2I
F av = f A + (1 − f )A,
c
c
which can be written in terms of pressure as
F av
I
= (2 − f ).
A
c
E38-40 We can treat the object as having two surfaces, one completely reflecting and the other
completely absorbing. If the entire surface has an area A then the absorbing part has an area f A
while the reflecting part has area (1 − f )A. The average force is then the sum of the force on each
part,
I
2I
F av = f A + (1 − f )A,
c
c
which can be written in terms of pressure as
F av
I
= (2 − f ).
A
c
The intensity I is that of the incident beam; the reflected beam will have an intensity (1 − f )I.
Each beam will contribute to the energy density— I/c and (1 − f )I/c, respectively. Add these two
energy densities to get the net energy density outside the surface. The result is (2 − f )I/c, which is
the left hand side of the pressure relation above.
E38-41 The bullet density is ρ = N m/V . Let V = Ah; the kinetic energy density is K/V =
1
2
2 N mv /Ah. h/v, however, is the time taken for N balls to strike the surface, so that
P =
F
N mv
N mv 2
2K
=
=
=
.
A
At
Ah
V
167
E38-42 F = IA/c; P = IA; a = F/m; and v = at. Combine:
v = P t/mc = (10×103 W)(86400 s)/(1500 kg)(3×108 m/s) = 1.9×10−3 m/s.
E38-43 The force of radiation on the bottom of the cylinder is F = 2IA/c. The force of gravity
on the cylinder is
W = mg = ρHAg.
Equating, 2I/c = ρHg. The intensity of the beam is given by I = 4P/πd2 . Solving for H,
H=
8(4.6 W)
8P
= 4.9×10−7 m.
=
2
8
πcρgd
π(3.0×10 m/s)(1200 kg/m3 )(9.8 m/s2 )(2.6×10−3 m)2
E38-44 F = 2IA/c. The value for I is in Ex. 38-37, among other places. Then
F = (1.38×103 W/m2 )(3.1×106 m2 )/(3.00×108 m/s) = 29 N.
P38-1 For the two outer circles use Eq. 33-13. For the inner circle use E = V /d, Q = CV ,
C = 0 A/d, and i = dQ/dt. Then
i=
0 A dV
dE
dQ
=
= 0 A
.
dt
d dt
dt
The change in flux is dΦE /dt = A dE/dt. Then
I
~ · d~l = µ0 0 dΦE = µ0 i,
B
dt
so B = µ0 i/2πr.
P38-2 (a) id = i. Assuming ∆V = (174×103 V) sin ωt, then q = C∆V and i = dq/dt = Cd(∆V )/dt.
Combine, and use ω = 2π(50.0/s),
id = (100×10−12 F)(174×103 V)2π(50.0/s) = 5.47×10−3 A.
P38-3 (a) i = id = 7.63µA.
(b) dΦE /dt = id /0 = (7.63µA)/(8.85×10−12 F/m) = 8.62×105 V/m.
(c) i = dq/dt = Cd(∆V )/dt; C = 0 A/d; [d(∆V )/dt]m = E m ω. Combine, and
d=
0 A
0 AE m ω
(8.85×10−12 F/m)π(0.182 m)2 (225 V)(128 rad/s)
=
=
= 3.48×10−3 m.
C
i
(7.63µA)
R
R
P38-4 (a) q = i dt = α t dt = αt2 /2.
(b) E = σ/0 = q/0 A = αt2 /2πR2 0 .
(d) 2πrB = µ0 0 πr2 dE/dt, so
B = µ0 r(dE/dt)/2 = µ0 αrt/2πR2 .
(e) Check Exercise 38-10!
P38-5
~ = E ĵ and B
~ = B k̂. Then ~S = E
~ × B/µ
~ 0 , or
(a) E
~S = −EB/mu0 î.
Energy only passes through the yz faces; it goes in one face and out the other. The rate is P =
SA = EBa2 /mu0 .
(b) The net change is zero.
168
P38-6
(a) For a sinusoidal time dependence |dE/dt|m = ωE m = 2πf E m . Then
|dE/dt|m = 2π(2.4×109 /s)(13×103 V/m) = 1.96×1014 V/m · s.
(b) Using the result of part (b) of Sample Problem 38-1,
B=
1
1
(4π×10−7 H/m)(8.9×10−12 F/m)(2.4×10−2 m) (1.96×1014 V/m · s) = 1.3×10−5 T.
2
2
P38-7 Look back to Chapter 14 for a discussion on the elliptic orbit. On page 312 it is pointed
out that the closest distance to the sun is Rp = a(1 − e) while the farthest distance is Ra = a(1 + e),
where a is the semi-major axis and e the eccentricity.
The fractional variation in intensity is
∆I
I
Ip − Ia
,
Ia
Ip
− 1,
Ia
Ra 2
− 1,
Rp 2
≈
=
=
(1 + e)2
− 1.
(1 − e)2
=
We need to expand this expression for small e using (1 + e)2 ≈ 1 + 2e, and (1 − e)−2 ≈ 1 + 2e, and
finally (1 + 2e)2 ≈ 1 + 4e. Combining,
∆I
≈ (1 + 2e)2 − 1 ≈ 4e.
I
P38-8 The beam radius grows as r = (0.440 µrad)R, where R is the distance from the origin. The
beam intensity is
I=
P38-9
P
(3850 W)
=
= 4.3×10−2 W.
2
πr
π(0.440 µrad)2 (3.82×108 m)2
Eq. 38-14 requires
∂E
∂x
E m k cos kx sin ωt
Emk
∂B
,
∂t
= B m ω cos kx sin ωt,
= B m ω.
= −
Eq. 38-17 requires
∂E
∂t
µ0 0 E m ω sin kx cos ωt
µ0 0 E m ω
µ0 0
∂B
,
∂x
= B m k sin kx cos ωt,
= B m k.
= −
Dividing one expression by the other,
µ0 0 k 2 = ω 2 ,
169
or
ω
1
=c= √
k
µ0 0
.
Not only that, but E m = cB m . You’ve seen an expression similar to this before, and you’ll see
expressions similar to it again.
(b) We’ll assume that Eq. 38-21 is applicable here. Then
S
=
=
1
EmBm
=
sin kx sin ωt cos kx cos ωt,
µ0
µ0
E 2m
sin 2kx sin 2ωt
4µ0 c
is the magnitude of the instantaneous Poynting vector.
(c) The time averaged power flow across any surface is the value of
1
T
Z
T
Z
~S · dA
~ dt,
0
where T is the period of the oscillation. We’ll just gloss over any concerns about direction, and
assume that the ~S will be constant in direction so that we will, at most, need to concern ourselves
about a constant factor cos θ. We can then deal with a scalar, instead of vector, integral, and we
can integrate it in any order we want. We want to do the t integration first, because an integral over
sin ωt for a period T = 2π/ω is zero. Then we are done!
(d) There is no energy flow; the energy remains inside the container.
P38-10
(a) The electric field is parallel to the wire and given by
E = V /d = iR/d = (25.0 A)(1.00 Ω/300 m) = 8.33×10−2 V/m
(b) The magnetic field is in rings around the wire. Using Eq. 33-13,
B=
µ0 i
(4π×10−7 H/m)(25 A)
=
= 4.03×10−3 T.
2πr
2π(1.24×10−3 m)
(c) S = EB/µ0 , so
S = (8.33×10−2 V/m)(4.03×10−3 T)/(4π×10−7 H/m) = 267 W/m2 .
P38-11
(a) We’ve already calculated B previously. It is
B=
µ0 i
E
where i = .
2πr
R
The electric field of a long straight wire has the form E = k/r, where k is some constant. But
∆V = −
Z
~ · d~s = −
E
Z
b
E dr = −k ln(b/a).
a
In this problem the inner conductor is at the higher potential, so
k=
−∆V
E
=
,
ln(b/a)
ln(b/a)
170
and then the electric field is
E
.
r ln(b/a)
This is also a vector field, and if E is positive the electric field points radially out from the central
conductor.
(b) The Poynting vector is
~S = 1 E
~ × B;
~
µ0
~ is radial while B
~ is circular, so they are perpendicular. Assuming that E is positive the direction
E
of ~S is away from the battery. Switching the sign of E (connecting the battery in reverse) will flip
~ and B,
~ so ~S will pick up two negative signs and therefore still point away
the direction of both E
from the battery.
The magnitude is
E2
EB
S=
=
µ0
2πR ln(b/a)r2
(c) We want to evaluate a surface integral in polar coordinates and so dA = (dr)(rdθ). We have
already established that ~S is pointing away from the battery parallel to the central axis. Then we
can integrate
Z
Z
~S · dA
~ = S dA,
P =
Z b Z 2π
E2
=
dθ r dr,
2πR ln(b/a)r2
a
0
Z b
E2
=
dr,
a R ln(b/a)r
E2
=
.
R
(d) Read part (b) above.
E=
~ is oriented as rings around the cylinder. If the thumb is in the direction of current
P38-12 (a) B
~ is directed
then the fingers of the right hand grip ion the direction of the magnetic field lines. E
~
parallel to the wire in the direction of the current. S is found from the cross product of these two,
and must be pointing radially inward.
(b) The magnetic field on the surface is given by Eq. 33-13:
B = µ0 i/2πa.
The electric field on the surface is given by
E = V /l = iR/l
Then S has magnitude
i iR
i2 R
=
.
2πa l
2πal
R
~S · dA
~ is only evaluated on the surface of the cylinder, not the end caps. ~S is everywhere parallel
~
to
R dA, so the dot product reduces to S dA; S is uniform, so it can be brought out of the integral;
dA = 2πal on the surface.
Hence,
Z
~S · dA
~ = i2 R,
S = EB/µ0 =
as it should.
171
P38-13 (a) f = vlambda = (3.00×108 m/s)/(3.18 m) = 9.43×107 Hz.
~
(b) Bmust
be directed along the z axis. The magnitude is
B = E/c = (288 V/m)/(3.00×108 m/s) = 9.6×10−7 T.
(c) k = 2π/λ = 2π/(3.18 m) = 1.98/m while ω = 2πf , so
ω = 2π(9.43×107 Hz) = 5.93×108 rad/s.
(d) I = E m B m /2µ0 , so
I=
(288 V)(9.6×10−7 T)
= 110 W.
2(4π×10−7 H/m)
(e) P = I/c = (110 W)/(3.00×108 m/s) = 3.67×10−7 Pa.
~ is oriented as rings around the cylinder. If the thumb is in the direction of current
P38-14 (a) B
~ is directed
then the fingers of the right hand grip ion the direction of the magnetic field lines. E
~
parallel to the wire in the direction of the current. S is found from the cross product of these two,
and must be pointing radially inward.
(b) The magnitude of the electric field is
E=
V
Q
Q
it
=
=
=
.
d
Cd
0 A
0 A
The magnitude of the magnetic field on the outside of the plates is given by Sample Problem 38-1,
B=
µ0 0 R dE
µ0 0 iR
µ0 0 R
=
=
E.
2
dt
20 A
2t
~S has magnitude
S=
EB
0 R 2
=
E .
µ0
2t
Integrating,
Z
2
~S · dA
~ = 0 R E 2 2πRd = Ad 0 E .
2t
t
But E is linear in t, so d(E 2 )/dt = 2E 2 /t; and then
Z
~S · dA
~ = Ad d 1 0 E 2 .
dt 2
P38-15 (a) I = P/A = (5.00×10−3 W)/π(1.05)2 (633×10−9 m)2 = 3.6×109 W/m2 .
(b) p = I/c = (3.6×109 W/m2 )/(3.00×108 m/s) = 12 Pa
(c) F = pA = P/c = (5.00×10−3 W)/(3.00×108 m/s) = 1.67×10−11 N.
(d) a = F/m = F/ρV , so
a=
(1.67×10−11 N)
= 2.9×103 m/s2 .
4(4880 kg/m3 )(1.05)3 (633×10−9 )3 /3
172
P38-16
The force from the sun is F = GM m/r2 . The force from radiation pressure is
F =
2P A
2IA
=
.
c
4πr2 c
Equating,
A=
so
A=
4πGM m
,
2P/c
4π(6.67×10−11 N · m2 /kg2 )(1.99×1030 kg)(1650 kg)
= 1.06×106 m2 .
2(3.9×1026 W)/(3.0×108 m/s)
That’s about one square kilometer.
173
E39-1 Both scales are logarithmic; choose any data point from the right hand side such as
c = f λ ≈ (1 Hz)(3×108 m) = 3×108 m/s,
and another from the left hand side such as
c = f λ ≈ (1×1021 Hz)(3×10−13 m) = 3×108 m/s.
E39-2 (a) f = v/λ = (3.0×108 m/s)/(1.0×104 )(6.37×106 m) = 4.7×10−3 Hz. If we assume that
this is the data transmission rate in bits per second (a generous assumption), then it would take 140
days to download a web-page which would take only 1 second on a 56K modem!
(b) T = 1/f = 212 s = 3.5 min.
E39-3
(a) Apply v = f λ. Then
f = (3.0×108 m/s)/(0.067×10−15 m) = 4.5×1024 Hz.
(b) λ = (3.0×108 m/s)/(30 Hz) = 1.0×107 m.
E39-4 Don’t simply take reciprocal of linewidth! f = c/λ, so δf = (−c/λ2 )δλ. Ignore the negative,
and
δf = (3.00×108 m/s)(0.010×10−9 m)/(632.8×10−9 m)2 = 7.5×109 Hz.
E39-5 (a) We refer to Fig. 39-6 to answer this question. The limits are approximately 520 nm
and 620 nm.
(b) The wavelength for which the eye is most sensitive is 550 nm. This corresponds to to a
frequency of
f = c/λ = (3.00 × 108 m/s)/(550 × 10−9 m) = 5.45 × 1014 Hz.
This frequency corresponds to a period of T = 1/f = 1.83 × 10−15 s.
E39-6 f = c/λ. The number of complete pulses is f t, or
f t = ct/λ = (3.00×108 m/s)(430×10−12 s)/(520×10−9 m) = 2.48×105 .
E39-7 (a) 2(4.34 y) = 8.68 y.
(b) 2(2.2×106 y) = 4.4×106 y.
E39-8 (a) t = (150×103 m)/(3×108 m/s) = 5×10−4 s.
(b) The distance traveled by the light is (1.5×1011 m) + 2(3.8×108 m), so
t = (1.51×1011 m)/(3×108 m/s) = 503 s.
(c) t = 2(1.3×1012 m)/(3×108 m/s) = 8670 s.
(d) 1054 − 6500 ≈ 5400 BC.
E39-9 This is a question of how much time it takes light to travel 4 cm, because the light traveled
from the Earth to the moon, bounced off of the reflector, and then traveled back. The time to travel
4 cm is ∆t = (0.04 m)/(3 × 108 m/s) = 0.13 ns. Note that I interpreted the question differently than
the answer in the back of the book.
174
E39-10 Consider any incoming ray. The path of the ray can be projected onto the xy plane, the
xz plane, or the yz plane. If the projected rays is exactly reflected in all three cases then the three
dimensional incoming ray will be reflected exactly reversed. But the problem is symmetric, so it is
sufficient to show that any plane works.
Now the problem has been reduced to Sample Problem 39-2, so we are done.
E39-11 We will choose the mirror to lie in the xy plane at z = 0. There is no loss of generality
in doing so; we had to define our coordinate system somehow. The choice is convenient in that
any normal is then parallel to the z axis. Furthermore, we can arbitrarily define the incident ray
to originate at (0, 0, z1 ). Lastly, we can rotate the coordinate system about the z axis so that the
reflected ray passes through the point (0, y3 , z3 ).
The point of reflection for this ray is somewhere on the surface of the mirror, say (x2 , y2 , 0). This
distance traveled from the point 1 to the reflection point 2 is
q
p
d12 = (0 − x2 )2 + (0 − y2 )2 + (z1 − 0)2 = x22 + y22 + z12
and the distance traveled from the reflection point 2 to the final point 3 is
q
p
d23 = (x2 − 0)2 + (y2 − y3 )2 + (0 − z3 )2 = x22 + (y2 − y3 )2 + z32 .
The only point which is free to move is the reflection point, (x2 , y2 , 0), and that point can only
move in the xy plane. Fermat’s principle states that the reflection point will be such to minimize
the total distance,
q
q
d12 + d23 =
x22 + y22 + z12 +
x22 + (y2 − y3 )2 + z32 .
We do this minimization by taking the partial derivative with respect to both x2 and y2 . But
we can do part by inspection alone. Any non-zero value of x2 can only add to the total distance,
regardless of the value of any of the other quantities. Consequently, x2 = 0 is one of the conditions
for minimization.
We are done! Although you are invited to finish the minimization process, once we know that
x2 = 0 we have that point 1, point 2, and point 3 all lie in the yz plane. The normal is parallel to
the z axis, so it also lies in the yz plane. Everything is then in the yz plane.
E39-12 Refer to Page 442 of Volume 1.
E39-13 (a) θ1 = 38◦ .
(b) (1.58) sin(38◦ ) = (1.22) sin θ2 . Then θ2 = arcsin(0.797) = 52.9◦ .
E39-14 ng = nv sin θ1 / sin θ2 = (1.00) sin(32.5◦ )/ sin(21.0◦ ) = 1.50.
E39-15 n = c/v = (3.00×108 m/s)/(1.92×108 m/s) = 1.56.
E39-16 v = c/n = (3.00×108 m/s)/(1.46) = 2.05×108 m/s.
E39-17 The speed of light in a substance with index of refraction n is given by v = c/n. An
electron will then emit Cerenkov radiation in this particular liquid if the speed exceeds
v = c/n = (3.00 × 108 m/s)/(1.54) = 1.95×108 m/s.
175
E39-18 Since t = d/v = nd/c, ∆t = ∆n d/c. Then
∆t = (1.00029 − 1.00000)(1.61×103 m)/(3.00×108 m/s) = 1.56×10−9 s.
E39-19 The angle of the refracted ray is θ2 = 90◦ , the angle of the incident ray can be found by
trigonometry,
(1.14 m)
tan θ1 =
= 1.34,
(0.85 m)
or θ1 = 53.3◦ .
We can use these two angles, along with the index of refraction of air, to find that the index of
refraction of the liquid from Eq. 39-4,
n1 = n2
(sin 90◦ )
sin θ2
= 1.25.
= (1.00)
sin θ1
(sin 53.3◦ )
There are no units attached to this quantity.
E39-20 For an equilateral prism φ = 60◦ . Then
n=
sin[ψ + φ]/2
sin[(37◦ ) + (60◦ )]/2
=
= 1.5.
sin[φ/2]
sin[(60◦ )/2]
E39-21
E39-22 t = d/v; but L/d = cos θ2 =
p
1 − sin2 θ2 and v = c/n. Combining,
nL
n2 L
(1.63)2 (0.547 m)
q
t= p
= p
=
= 3.07×10−9 s.
2
2
2
2
c 1 − sin θ2
c n − sin θ1
(3×108 m/s) (1.632 ) − sin (24◦ )
E39-23 The ray of light from the top of the smokestack to the life ring is θ1 , where tan θ1 = L/h
with h the height and L the distance of the smokestack.
Snell’s law gives n1 sin θ1 = n2 sin θ2 , so
θ1 = arcsin[(1.33) sin(27◦ )/(1.00)] = 37.1◦ .
Then L = h tan θ1 = (98 m) tan(37.1◦ ) = 74 m.
E39-24 The length of the shadow on the surface of the water is
x1 = (0.64 m)/ tan(55◦ ) = 0.448 m.
The ray of light which forms the “end” of the shadow has an angle of incidence of 35◦ , so the ray
travels into the water at an angle of
(1.00)
sin(35◦ ) = 25.5◦ .
θ2 = arcsin
(1.33)
The ray travels an additional distance
x2 = (2.00 m − 0.64 m)/ tan(90◦ − 25.5◦ ) = 0.649 m
The total length of the shadow is
(0.448 m) + (0.649 m) = 1.10 m.
176
E39-25 We’ll rely heavily on the figure for our arguments. Let x be the distance between the
points on the surface where the vertical ray crosses and the bent ray crosses.
θ2
x
d app
θ1
d
In this exercise we will take advantage of the fact that, for small angles θ, sin θ ≈ tan θ ≈ θ In
this approximation Snell’s law takes on the particularly simple form n1 θ1 = n2 θ2 The two angles
here are conveniently found from the figure,
θ1 ≈ tan θ1 =
and
θ2 ≈ tan θ2 =
x
,
d
x
dapp
.
Inserting these two angles into the simplified Snell’s law, as well as substituting n1 = n and n2 = 1.0,
n1 θ 1
x
n
d
dapp
= n2 θ 2 ,
x
=
,
dapp
d
=
.
n
E39-26 (a) You need to address the issue of total internal reflection to answer this question.
(b) Rearrange
sin[ψ + φ]/2
n=
sin[φ/2]
/
and θ = (ψ + φ)/2 to get
θ = arcsin (n sin[φ/2]) = arcsin ((1.60) sin[(60◦ )/2]) = 53.1◦ .
E39-27 Use the results of Ex. 39-35. The apparent thickness of the carbon tetrachloride layer, as
viewed by an observer in the water, is
dc,w = nw dc /nc = (1.33)(41 mm)/(1.46) = 37.5 mm.
The total “thickness” from the water perspective is then (37.5 mm) + (20 mm) = 57.5 mm. The
apparent thickness of the entire system as view from the air is then
dapp = (57.5 mm)/(1.33) = 43.2 mm.
177
E39-28 (a) Use the results of Ex. 39-35. dapp = (2.16 m)/(1.33) = 1.62 m.
(b) Need a diagram here!
E39-29 (a) λn = λ/n = (612 nm)/(1.51) = 405 nm.
(b) L = nLn = (1.51)(1.57 pm) = 2.37 pm. There is actually a typo: the “p” in “pm” was
supposed to be a µ. This makes a huge difference for part (c)!
E39-30 (a) f = c/λ = (3.00×108 m/s)/(589 nm) = 5.09×1014 Hz.
(b) λn = λ/n = (589 nm)/(1.53) = 385 nm.
(c) v = f λ = (5.09×1014 Hz)(385 nm) = 1.96×108 m/s.
E39-31 (a) The second derivative of
p
p
L = a2 + x2 + b2 + (d − x)2
is
a2 (b2 + (d − 2)2 )3/2 + b2 (a2 + x2 )3/2
.
(b2 + (d − 2)2 )3/2 (a2 + x2 )3/2
This is always a positive number, so dL/dx = 0 is a minimum.
(a) The second derivative of
p
p
L = n1 a2 + x2 + n2 b2 + (d − x)2
is
n1 a2 (b2 + (d − 2)2 )3/2 + n2 b2 (a2 + x2 )3/2
.
(b2 + (d − 2)2 )3/2 (a2 + x2 )3/2
This is always a positive number, so dL/dx = 0 is a minimum.
E39-32 (a) The angle of incidence on the face ac will be 90◦ − φ. Total internal reflection occurs
when sin(90◦ − φ) > 1/n, or
φ < 90◦ − arcsin[1/(1.52)] = 48.9◦ .
(b) Total internal reflection occurs when sin(90◦ − φ) > nw /n, or
φ < 90◦ − arcsin[(1.33)/(1.52)] = 29.0◦ .
E39-33
(a) The critical angle is given by Eq. 39-17,
θc = sin−1
n2
(1.586)
= sin−1
= 72.07◦ .
n1
(1.667)
(b) Critical angles only exist when “attempting” to travel from a medium of higher index of
refraction to a medium of lower index of refraction; in this case from A to B.
E39-34 If the fire is at the water’s edge then the light travels along the surface, entering the
water near the fish with an angle of incidence of effectively 90◦ . Then the angle of refraction in
the water is numerically equivalent to a critical angle, so the fish needs to look up at an angle of
θ = arcsin(1/1.33) = 49◦ with the vertical. That’s the same as 41◦ with the horizontal.
178
E39-35 Light can only emerge from the water if it has an angle of incidence less than the critical
angle, or
θ < θc = arcsin 1/n = arcsin 1/(1.33) = 48.8◦ .
The radius of the circle of light is given by r/d = tan θc , where d is the depth. The diameter is twice
this radius, or
2(0.82 m) tan(48.8◦ ) = 1.87 m.
E39-36 The refracted angle is given by n sin θ1 = sin(39◦ ). This ray strikes the left surface with
an angle of incidence of 90◦ − θ1 . Total internal reflection occurs when
sin(90◦ − θ1 ) = 1/n;
but sin(90◦ − θ1 ) = cos θ1 , so we can combine and get tan θ = sin(39◦ ) with solution θ1 = 32.2◦ . The
index of refraction of the glass is then
n = sin(39◦ )/ sin(32.2) = 1.18.
E39-37 The light strikes the quartz-air interface from the inside; it is originally “white”, so if
the reflected ray is to appear “bluish” (reddish) then the refracted ray should have been “reddish”
(bluish). Since part of the light undergoes total internal reflection while the other part does not,
then the angle of incidence must be approximately equal to the critical angle.
(a) Look at Fig. 39-11, the index of refraction of fused quartz is given as a function of the
wavelength. As the wavelength increases the index of refraction decreases. The critical angle is a
function of the index of refraction; for a substance in air the critical angle is given by sin θc = 1/n.
As n decreases 1/n increases so θc increases. For fused quartz, then, as wavelength increases θc also
increases.
In short, red light has a larger critical angle than blue light. If the angle of incidence is midway
between the critical angle of red and the critical angle of blue, then the blue component of the light
will experience total internal reflection while the red component will pass through as a refracted ray.
So yes, the light can be made to appear bluish.
(b) No, the light can’t be made to appear reddish. See above.
(c) Choose an angle of incidence between the two critical angles as described in part (a). Using
a value of n = 1.46 from Fig. 39-11,
θc = sin−1 (1/1.46) = 43.2◦ .
Getting the effect to work will require considerable sensitivity.
E39-38 (a) There needs to be an opaque spot in the center of each face so that no refracted ray
emerges. The radius of the spot will be large enough to cover rays which meet the surface at less
than the critical angle. This means tan θc = r/d, where d is the distance from the surface to the
spot, or 6.3 mm. Since
θc = arcsin 1/(1.52) = 41.1◦ ,
then r = (6.3 mm) tan(41.1◦ ) = 5.50 mm.
(b) The circles have an area of a = π(5.50 mm)2 = 95.0 mm2 . Each side has an area of (12.6 mm)2 ;
the fraction covered is then (95.0 mm2 )/(12.6 mm)2 = 0.598.
E39-39 For u c the relativistic Doppler shift simplifies to
∆f = −f0 u/c = −u/λ0 ,
so
u = λ0 ∆f = (0.211 m)∆f.
179
E39-40 c = f λ, so 0 = f ∆λ + λ∆f . Then ∆λ/λ = −∆f /f . Furthermore, f0 − f , from Eq. 39-21,
is f0 u/c for small enough u. Then
∆λ
f − f0
u
=−
= .
λ
f0
c
E39-41
The Doppler theory for light gives
1 − u/c
1 − (0.2)
f = f0 p
= f0 p
= 0.82 f0 .
2
2
1 − u /c
1 − (0.2)2
The frequency is shifted down to about 80%, which means the wavelength is shifted up by an
additional 25%. Blue light (480 nm) would appear yellow/orange (585 nm).
E39-42 Use Eq. 39-20:
1 − u/c
1 − (0.892)
f = f0 p
= (100 Mhz) p
= 23.9 MHz.
2
2
1 − u /c
1 − (0.892)2
E39-43 (a) If the wavelength is three times longer then the frequency is one-third, so for the
classical Doppler shift
f0 /3 = f0 (1 − u/c),
or u = 2c.
(b) For the relativistic shift,
f0 /3
p
1 − u/c
,
= f0 p
1 − u2 /c2
1 − u2 /c2 = 3(1 − u/c),
c2 − u2 = 9(c − u)2 ,
0 = 10u2 − 18uc + 8c2 .
The solution is u = 4c/5.
E39-44 (a) f0 /f = λ/λ0 . This shift is small, so we apply the approximation:
λ0
(462 nm)
8
u=c
− 1 = (3×10 m/s)
− 1 = 1.9×107 m/s.
λ
(434 nm)
(b) A red shift corresponds to objects moving away from us.
E39-45 The sun rotates once every 26 days at the equator, while the radius is 7.0×108 m. The
speed of a point on the equator is then
v=
2πR
2π(7.0×108 m)
=
= 2.0×103 m/s.
T
(2.2×106 s)
This corresponds to a velocity parameter of
β = u/c = (2.0×103 m/s)/(3.0×108 m/s) = 6.7×10−6 .
This is a case of small numbers, so we’ll use the formula that you derived in Exercise 39-40:
∆λ = λβ = (553 nm)(6.7×10−6 ) = 3.7×10−3 nm.
180
E39-46 Use Eq. 39-23 written as
(1 − u/c)λ2 = λ20 (1 + u/c),
which can be rearranged as
u/c =
λ2 − λ20
(540 nm)2 − (620 nm)2
=
= −0.137.
2
2
λ + λ0
(540 nm)2 + (620 nm)2
The negative sign means that you should be going toward the red light.
E39-47 (a) f1 = cf /(c + v) and f2 = cf /(c − v).
∆f = (f2 − f ) − (f − f1 ) = f2 + f1 − 2f,
so
∆f
f
=
=
=
=
c
c
+
− 2,
c+v c−v
2v 2
,
c2 − v 2
2(8.65×105 m/s)2
,
(3.00×108 m/s)2 − (8.65×105 m/s)2
1.66×10−5 .
√
(b) f1 = f (c − u)/sqrtc2 − u2 and f2 = f (c + u)/ c2 − u2 .
∆f = (f2 − f ) − (f − f1 ) = f2 + f1 − 2f,
so
∆f
f
2c
− 2,
− u2
2(3.00×108 m/s)
=
√
=
p
− 2,
(3.00×108 m/s)2 − (8.65×105 m/s)2
=
c2
8.3×10−6 .
E39-48 (a) No relative motion, so every 6 minutes.
(b) The Doppler effect at this speed is
1 − u/c
1 − (0.6)
p
=p
= 0.5;
2
2
1 − u /c
1 − (0.6)2
this means the frequency is one half, so the period is doubled to 12 minutes.
(c) If C send the signal at the instant the signal from A passes, then the two signals travel together
to C, so C would get B’s signals at the same rate that it gets A’s signals: every six minutes.
E39-49
p
E39-50 The transverse Doppler effect is λ = λ0 / 1 − u2 /c2 . Then
p
λ = (589.00 nm)/ 1 − (0.122)2 = 593.43 nm.
The shift is (593.43 nm) − (589.00 nm) = 4.43 nm.
181
E39-51
The frequency observed by the detector from the first source is (Eq. 39-31)
p
f = f1 1 − (0.717)2 = 0.697f1 .
The frequency observed by the detector from the second source is (Eq. 39-30)
p
1 − (0.717)2
0.697f2
f = f2
=
.
1 + (0.717) cos θ
1 + (0.717) cos θ
We need to equate these and solve for θ. Then
0.697f1
1 + 0.717 cos θ
cos θ
θ
0.697f2
,
1 + 0.717 cos θ
= f2 /f1 ,
= (f2 /f1 − 1) /0.717,
= 101.1◦ .
=
Subtract from 180◦ to find the angle with the line of sight.
E39-52
P39-1 Consider the triangle in Fig. 39-45. The true position corresponds to the speed of light,
the opposite side corresponds to the velocity of earth in the orbit. Then
θ = arctan(29.8×103 m/s)/(3.00×108 m/s) = 20.500 .
P39-2
is
The distance to Jupiter from point x is dx = rj − re . The distance to Jupiter from point y
q
d2 = re2 + rj2 .
The difference in distance is related to the time according to
(d2 − d1 )/t = c,
so
p
(778×109 m)2 + (150×109 m)2 − (778×109 m) + (150×109 m)
= 2.7×108 m/s.
(600 s)
P39-3 sin(30◦ )/(4.0 m/s) = sin θ/(3.0 m/s). Then θ = 22◦ . Water waves travel more slowly in
shallower water, which means they always bend toward the normal as they approach land.
P39-4 (a) If the ray is normal to the water’s surface then it passes into the water undeflected.
Once in the water the problem is identical to Sample Problem 39-2. The reflected ray in the water
is parallel to the incident ray in the water, so it also strikes the water normal, and is transmitted
normal.
(b) Assume the ray strikes the water at an angle θ1 . It then passes into the water at an angle
θ2 , where
nw sin θ2 = na sin θ1 .
Once the ray is in the water then the problem is identical to Sample Problem 39-2. The reflected
ray in the water is parallel to the incident ray in the water, so it also strikes the water at an angle
θ2 . When the ray travels back into the air it travels with an angle θ3 , where
nw sin θ2 = na sin θ3 .
Comparing the two equations yields θ1 = θ3 , so the outgoing ray in the air is parallel to the incoming
ray.
182
P39-5
(a) As was done in Ex. 39-25 above we use the small angle approximation of
sin θ ≈ θ ≈ tan θ
The incident angle is θ; if the light were to go in a straight line we would expect it to strike a
distance y1 beneath the normal on the right hand side. The various distances are related to the
angle by
θ ≈ tan θ ≈ y1 /t.
The light, however, does not go in a straight line, it is refracted according to (the small angle
approximation to) Snell’s law, n1 θ1 = n2 θ2 , which we will simplify further by letting θ1 = θ, n2 = n,
and n1 = 1, θ = nθ2 . The point where the refracted ray does strike is related to the angle by
θ2 ≈ tan θ2 = y2 /t. Combining the three expressions,
y1 = ny2 .
The difference, y1 − y2 is the vertical distance between the displaced ray and the original ray as
measured on the plate glass. A little algebra yields
y1 − y2
= y1 − y1 /n,
= y1 (1 − 1/n) ,
n−1
= tθ
.
n
The perpendicular distance x is related to this difference by
cos θ = x/(y1 − y2 ).
In the small angle approximation cos θ ≈ 1 − θ2 /2. If θ is sufficiently small we can ignore the square
term, and x ≈ y2 − y1 .
(b) Remember to use radians and not degrees whenever the small angle approximation is applied.
Then
(1.52) − 1
x = (1.0 cm)(0.175 rad)
= 0.060 cm.
(1.52)
P39-6
(a) At the top layer,
n1 sin θ1 = sin θ;
at the next layer,
n2 sin θ2 = n1 sin θ1 ;
at the next layer,
n3 sin θ3 = n2 sin θ2 .
Combining all three expressions,
n3 sin θ3 = sin θ.
◦
(b) θ3 = arcsin[sin(50 )/(1.00029)] = 49.98◦ . Then shift is (50◦ ) − (49.98◦ ) = 0.02◦ .
P39-7 The “big idea” of Problem 6 is that when light travels through layers the angle that it
makes in any layer depends only on the incident angle, the index of refraction where that incident
angle occurs, and the index of refraction at the current point.
That means that light which leaves the surface of the runway at 90◦ to the normal will make an
angle
n0 sin 90◦ = n0 (1 + ay) sin θ
183
at some height y above the runway. It is mildly entertaining to note that the value of n0 is unimportant, only the value of a!
The expression
1
sin θ =
≈ 1 − ay
1 + ay
can be used to find the angle made by the curved path against the normal as a function of y. The
slope of the curve at any point is given by
dy
cos θ
= tan(90◦ − θ) = cot θ =
.
dx
sin θ
Now we need to know cos θ. It is
cos θ =
Combining
p
p
1 − sin2 θ ≈ 2ay.
√
2ay
dy
≈
,
dx
1 − ay
and now we integrate. We will ignore the ay term in the denominator because it will always be
small compared to 1. Then
Z
d
0
d
h
Z
dy
√
,
2ay
0
s
r
2h
2(1.7 m)
=
=
= 1500 m.
a
(1.5×10−6 m−1 )
dx =
P39-8 The energy of a particle is given by E 2 =pp2 c2 + m2 c4 . This energy is related to the mass
by E = γmc2 . γ is related to the speed by γ = 1/ 1 − u2 /c2 . Rearranging,
s
r
u
1
m2 c2
=
1− 2 = 1− 2
,
c
γ
p + m2 c2
s
p2
=
.
p2 + m2 c2
Since n = c/u we can write this as
s
n=
s
n=
s
(135 MeV)
(145 MeV)
2
= 1.37.
n=
s
(106 MeV)
(145 MeV)
2
= 1.24.
m2 c2
1+ 2 =
p
1+
mc2
pc
2
.
For the pion,
1+
For the muon,
1+
184
P39-9 (a) Before adding the drop of liquid project the light ray along the angle θ so that θ = 0.
Increase θ slowly until total internal reflection occurs at angle θ1 . Then
ng sin θ1 = 1
is the equation which can be solved to find ng .
Now put the liquid on the glass and repeat the above process until total internal reflection occurs
at angle θ2 . Then
ng sin θ2 = nl .
Note that ng < ng for this method to work.
(b) This is not terribly practical.
P39-10 Let the internal angle at Q be θQ . Then n sin θQ = 1, because it is a critical angle. Let
the internal angle at P be θP . Then θP + θQ = 90◦ . Combine this with the other formula and
q
1 = n sin(90 − θP ) = n cos θQ = n 1 − sin2 θP .
Not only that, but sin θ1 = n sin θP , or
1=n
p
1 − (sin θ1 )2 /n2 ,
which can be solved for n to yield
n=
q
1 + sin2 θ1 .
(b) The largest value of the sine function is one, so nmax =
√
2.
P39-11 (a) The fraction of light energy which escapes from the water is dependent on the critical
angle. Light radiates in all directions from the source, but only that which strikes the surface at an
angle less than the critical angle will escape. This critical angle is
sin θc = 1/n.
We want to find the solid angle of the light which escapes; this is found by integrating
Ω=
Z
2π
0
Z
θc
sin θ dθ dφ.
0
This is not a hard integral to do. The result is
Ω = 2π(1 − cos θc ).
There are 4π steradians in a spherical surface, so the fraction which escapes is
q
1
1
f = (1 − cos θc ) = (1 − 1 − sin2 θc ).
2
2
The last substitution
p is easy enough. We never needed to know the depth h.
(b) f = 12 (1 − 1 − (1/(1.3))2 ) = 0.18.
185
P39-12
to
(a) The beam of light strikes the face of the fiber at an angle θ and is refracted according
n1 sin θ1 = sin θ.
The beam then travels inside the fiber until it hits the cladding interface; it does so at an angle of
90◦ − θ1 to the normal. It will be reflected if it exceeds the critical angle of
n1 sin θc = n2 ,
or if
sin(90◦ − θ1 ) ≥ n2 /n1 ,
which can be written as
cos θ1 ≥ n2 /n1 .
but if this is the cosine, then we can use sin2 + cos2 = 1 to find the sine, and
q
sin θ1 ≤ 1 − n22 /n21 .
Combine this with the first equation and
θ ≤ arcsin
(b) θ = arcsin
q
n21 − n22 .
p
(1.58)2 − (1.53)2 = 23.2◦ .
P39-13 Consider the two possible extremes: a ray of light can propagate in a straight line
directly down the axis of the fiber, or it can reflect off of the sides with the minimum possible angle
of incidence. Start with the harder option.
The minimum angle of incidence that will still involve reflection is the critical angle, so
sin θc =
n2
.
n1
This light ray has farther to travel than the ray down the fiber axis because it is traveling at an
angle. The distance traveled by this ray is
L0 = L/ sin θc = L
n1
,
n2
The time taken for this bouncing ray to travel a length L down the fiber is then
t0 =
L0
L0 n1
L n21
=
=
.
v
c
c n2
Now for the easier ray. It travels straight down the fiber in a time
t=
The difference is
t0 − t = ∆t =
L
c
L
n1 .
c
n21
− n1
n2
=
Ln1
(n1 − n2 ).
cn2
(b) For the numbers in Problem 12 we have
∆t =
(350×103 m)(1.58)
((1.58) − (1.53)) = 6.02×10−5 s.
(3.00×108 m/s)(1.53)
186
P39-14
P39-15 We can assume the airplane speed is small compared to the speed of light, and use Eq.
39-21. ∆f = 990 Hz; so
|∆f | = f0 u/c = u/λ0 ,
hence u = (990/s)(0.12 m) = 119 m/s. The actual answer for the speed of the airplane is half this
because there were two Doppler shifts: once when the microwaves struck the plane, and one when the
reflected beam was received by the station. Hence, the plane approaches with a speed of 59.4 m/s.
187
E40-1 (b) Since i = −o, vi = di/dt = −do/dt = −vo .
(a) In order to change from the frame of reference of the mirror to your own frame of reference
you need to subtract vo from all velocities. Then your velocity is vo − v0 = 0, the mirror is moving
with velocity 0 − vo = −vo and your image is moving with velocity −vo − vo = −2vo .
E40-2 You are 30 cm from the mirror, the image is 10 cm behind the mirror. You need to focus
40 cm away.
E40-3 If the mirror rotates through an angle α then the angle of incidence will increase by an
angle α, and so will the angle of reflection. But that means that the angle between the incident
angle and the reflected angle has increased by α twice.
E40-4 Sketch a line from Sarah through the right edge of the mirror and then beyond. Sarah can
see any image which is located between that line and the mirror. By similar triangles, the image
of Bernie will be d/2 = (3.0 m)/2 = 1/5 m from the mirror when it becomes visible. Since i = −o,
Bernie will also be 1.5 m from the mirror.
E40-5 The images are fainter than the object. Several sample rays are shown.
E40-6 The image is displaced. The eye would need to look up to see it.
E40-7 The apparent depth of the swimming pool is given by the work done for Exercise 3925, dapp = d/n The water then “appears” to be only 186 cm/1.33 = 140 cm deep. The apparent
distance between the light and the mirror is then 250 cm + 140 cm = 390 cm; consequently the
image of the light is 390 cm beneath the surface of the mirror.
E40-8 Three. There is a single direct image in each mirror and one more image of an image in
one of the mirrors.
188
E40-9 We want to know over what surface area of the mirror are rays of light reflected from the
object into the eye. By similar triangles the diameter of the pupil and the diameter of the part of
the mirror (d) which reflects light into the eye are related by
(5.0 mm)
d
=
,
(10 cm)
(24 cm) + (10 cm)
which has solution d = 1.47 mm The area of the circle on the mirror is
A = π(1.47 mm)2 /4 = 1.7 mm2 .
E40-10 (a) Seven; (b) Five; and (c) Two. This is a problem of symmetry.
E40-11 Seven. Three images are the ones from Exercise 8. But each image has an image in the
ceiling mirror. That would make a total of six, except that you also have an image in the ceiling
mirror (look up, eh?). So the total is seven!
E40-12 A point focus is not formed. The envelope of rays is called the caustic. You can see a
similar effect when you allow light to reflect off of a spoon onto a table.
E40-13 The image is magnified by a factor of 2.7, so the image distance is 2.7 times farther from
the mirror than the object. An important question to ask is whether or not the image is real or
virtual. If it is a virtual image it is behind the mirror and someone looking at the mirror could see
it. If it were a real image it would be in front of the mirror, and the man, who serves as the object
and is therefore closer to the mirror than the image, would not be able to see it.
So we shall assume that the image is virtual. The image distance is then a negative number.
The focal length is half of the radius of curvature, so we want to solve Eq. 40-6, with f = 17.5 cm
and i = −2.7o
1
1
1
0.63
= +
=
,
(17.5 cm)
o −2.7o
o
which has solution o = 11 cm.
E40-14 The image will be located at a point given by
1
1
1
1
1
1
= − =
−
=
.
i
f
o
(10 cm) (15 cm)
(30 cm)
The vertical scale is three times the horizontal scale in the figure below.
189
E40-15 This problem requires repeated application of 1/f = 1/o + 1/i, r = 2f , m = −i/o, or
the properties of plane, convex, or concave mirrors. All dimensioned variables below (f, r, i, o) are
measured in centimeters.
(a) Concave mirrors have positive focal lengths, so f = +20; r = 2f = +40;
1/i = 1/f − 1/o = 1/(20) − 1/(10) = 1/(−20);
m = −i/o = −(−20)/(10) = 2; the image is virtual and upright.
(b) m = +1 for plane mirrors only; r = ∞ for flat surface; f = ∞/2 = ∞; i = −o = −10; the
image is virtual and upright.
(c) If f is positive the mirror is concave; r = 2f = +40;
1/i = 1/f − 1/o = 1/(20) − 1/(30) = 1/(60);
m = −i/o = −(60)/(30) = −2; the image is real and inverted.
(d) If m is negative then the image is real and inverted; only Concave mirrors produce real images
(from real objects); i = −mo = −(−0.5)(60) = 30;
1/f = 1/o + 1/i = 1/(30) + 1/(60) = 1/(20);
r = 2f = +40.
(e) If r is negative the mirror is convex; f = r/2 = (−40)/2 = −20;
1/o = 1/f − 1/i = 1/(−20) − 1/(−10) = 1/(20);
m = −(−10)/(20) = 0.5; the image is virtual and upright.
(f) If m is positive the image is virtual and upright; if m is less than one the image is reduced,
but only convex mirrors produce reduced virtual images (from real objects); f = −20 for convex
mirrors; r = 2f = −40; let i = −mo = −o/10, then
1/f = 1/o + 1/i = 1/o − 10/o = −9/o,
so o = −9f = −9(−20) = 180; i = −o/10 = −(180)/10 = −18.
(g) r is negative for convex mirrors, so r = −40; f = r/2 = −20; convex mirrors produce only
virtual upright images (from real objects); so i is negative; and
1/o = 1/f − 1/i = 1/(−20) − 1/(−4) = 1/(5);
m = −i/o = −(−4)/(5) = 0.8.
(h) Inverted images are real; only concave mirrors produce real images (from real objects);
inverted images have negative m; i = −mo = −(−0.5)(24) = 12;
1/f = 1/o + 1/i = 1/(24) + 1/(12) = 1/(8);
r = 2f = 16.
190
E40-16 Use the angle definitions provided by Eq. 40-8. From triangle OaI we have
α + γ = 2θ,
while from triangle IaC we have
β + θ = γ.
Combining to eliminate θ we get
α − γ = −2β.
Substitute Eq. 40-8 and eliminate s,
2
1 1
− =− ,
o
i
r
or
1
1
2
+
=
,
o −i
−r
which is the same as Eq. 40-4 if i → −i and r → −r.
E40-17 (a) Consider the point A. Light from this point travels along the line ABC and will be
parallel to the horizontal center line from the center of the cylinder. Since the tangent to a circle
defines the outer limit of the intersection with a line, this line must describe the apparent size.
(b) The angle of incidence of ray AB is given by
sin θ1 = r/R.
The angle of refraction of ray BC is given by
sin θ2 = r∗ /R.
Snell’s law, and a little algebra, yields
n1 sin θ1
r
n1
R
nr
= n2 sin θ2 ,
r∗
= n2 ,
R
= r∗ .
In the last line we used the fact that n2 = 1, because it is in the air, and n1 = n, the index of
refraction of the glass.
E40-18 This problem requires repeated application of (n2 − n1 )/r = n1 /o + n2 /i. All dimensioned
variables below (r, i, o) are measured in centimeters.
(a)
(1.5) − (1.0) (1.0)
−
= −0.08333,
(30)
(10)
so i = (1.5)/(−0.08333) = −18, and the image is virtual.
(b)
(1.0)
(1.5)
+
= −0.015385,
(10)
(−13)
so r = (1.5 − 1.0)/(−0.015385) = −32.5, and the image is virtual.
(c)
(1.5) − (1.0)
(1.5)
−
= 0.014167,
(30)
(600)
191
so o = (1.0)/(0.014167) = 71. The image was real since i > 0.
(d) Rearrange the formula to solve for n2 , then
1
1 1
n2
− 1i = n1
+
.
r
r
o
Substituting the numbers,
n2
1
1
−
(−20) (−20)
= (1.0)
1
1
+
(−20) (20)
,
which has any solution for n2 ! Since i < 0 the image is virtual.
(e)
(1.5) (1.0)
+
= −0.016667,
(10)
(−6)
so r = (1.0 − 1.5)/(−0.016667) = 30, and the image is virtual.
(f)
(1.0)
(1.0) − (1.5)
−
= 0.15,
(−30)
(−7.5)
so o = (1.5)/(0.15) = 10. The image was virtual since i < 0.
(g)
(1.0) − (1.5) (1.5)
−
= −3.81×10−2 ,
(30)
(70)
so i = (1.0)/(−3.81×10−2 ) = −26, and the image is virtual.
(h) Solving Eq. 40-10 for n2 yields
n2 = n1
so
n2 = (1.5)
1/o + 1/r
,
1/r − 1/i
1/(100) + 1/(−30)
= 1.0
1/(−30) − 1/(600)
and the image is real.
E40-19 (b) If the beam is small we can use Eq. 40-10. Parallel incoming rays correspond to an
object at infinity. Solving for n2 yields
n2 = n1
1/o + 1/r
,
1/r − 1/i
so if o → ∞ and i = 2r, then
n2 = (1.0)
1/∞ + 1/r
= 2.0
1/r − 1/2r
(c) There is no solution if i = r!
E40-20 The image will be located at a point given by
1
1
1
1
1
1
= − =
−
=
.
i
f
o
(10 cm) (6 cm)
(−15 cm)
192
E40-21
The image location can be found from Eq. 40-15,
1
1
1
1
1
1
= − =
−
=
,
i
f
o
(−30 cm) (20 cm)
−12 cm
so the image is located 12 cm from the thin lens, on the same side as the object.
E40-22 For a double convex lens r1 > 0 and r2 < 0 (see Fig. 40-21 and the accompanying text).
Then the problem states that r2 = −r1 /2. The lens maker’s equation can be applied to get
1
1
1
3(n − 1)
= (n − 1)
−
=
,
f
r1
r2
r1
so r1 = 3(n − 1)f = 3(1.5 − 1)(60 mm) = 90 mm, and r2 = −45 mm.
E40-23 The object distance is essentially o = ∞, so 1/f = 1/o + 1/i implies f = i, and the image
forms at the focal point. In reality, however, the object distance is not infinite, so the magnification
is given by m = −i/o ≈ −f /o, where o is the Earth/Sun distance. The size of the image is then
hi = ho f /o = 2(6.96×108 m)(0.27 m)/(1.50×1011 m) = 2.5 mm.
The factor of two is because the sun’s radius is given, and we need the diameter!
E40-24 (a) The flat side has r2 = ∞, so 1/f = (n − 1)/r, where r is the curved side. Then
f = (0.20 m)/(1.5 − 1) = 0.40 m.
(b) 1/i = 1/f − 1/o = 1/(0.40 m) − 1/(0.40 m) = 0. Then i is ∞.
E40-25 (a) 1/f = (1.5 − 1)[1/(0.4 m) − 1/(−0.4 m)] = 1/(0.40 m).
(b) 1/f = (1.5 − 1)[1/(∞) − 1/(−0.4 m)] = 1/(0.80 m).
(c) 1/f = (1.5 − 1)[1/(0.4 m) − 1/(0.6 m)] = 1/(2.40 m).
(d) 1/f = (1.5 − 1)[1/(−0.4 m) − 1/(0.4 m)] = 1/(−0.40 m).
(e) 1/f = (1.5 − 1)[1/(∞) − 1/(0.8 m)] = 1/(−0.80 m).
(f) 1/f = (1.5 − 1)[1/(0.6 m) − 1/(0.4 m)] = 1/(−2.40 m).
E40-26 (a) 1/f = (n − 1)[1/(−r) − 1/r], so 1/f = 2(1 − n)/r. 1/i = 1/f − 1/o so if o = r, then
1/i = 2(1 − n)/r − 1/r = (1 − 2n)/r,
so i = r/(1 − 2n). For n > 0.5 the image is virtual.
(b) For n > 0.5 the image is virtual; the magnification is
m = −i/o = −r/(1 − 2n)/r = 1/(2n − 1).
E40-27
According to the definitions, o = f + x and i = f + x0 . Starting with Eq. 40-15,
1 1
+
o
i
i+o
oi
2f + x + x0
(f + x)(f + x0 )
2f 2 + f x + f x0
f2
=
=
=
1
,
f
1
,
f
1
,
f
= f 2 + f x + f x0 + xx0 ,
= xx0 .
193
E40-28 (a) You can’t determine r1 , r2 , or n. i is found from
1
1
1
1
=
−
=
,
i
+10 +20
+20
the image is real and inverted. m = −(20)/(20) = −1.
(b) You can’t determine r1 , r2 , or n. The lens is converging since f is positive. i is found from
1
1
1
1
=
−
=
,
i
+10 +5
−10
the image is virtual and upright. m = −(−10)/(+5) = 2.
(c) You can’t determine r1 , r2 , or n. Since m is positive and greater than one the lens is
converging. Then f is positive. i is found from
1
1
1
1
=
−
=
,
i
+10 +5
−10
the image is virtual and upright. m = −(−10)/(+5) = 2.
(d) You can’t determine r1 , r2 , or n. Since m is positive and less than one the lens is diverging.
Then f is negative. i is found from
1
1
1
1
=
−
=
,
i
−10 +5
−3.3
the image is virtual and upright. m = −(−3.3)/(+5) = 0.66.
(e) f is found from
1
1
1
1
= (1.5 − 1)
−
=
.
f
+30 −30
+30
The lens is converging. i is found from
1
1
1
1
=
−
=
,
i
+30 +10
−15
the image is virtual and upright. m = −(−15)/(+10) = 1.5.
(f) f is found from
1
1
1
1
= (1.5 − 1)
−
=
.
f
−30 +30
−30
The lens is diverging. i is found from
1
1
1
1
=
−
=
,
i
−30 +10
−7.5
the image is virtual and upright. m = −(−7.5)/(+10) = 0.75.
(g) f is found from
1
1
1
1
= (1.5 − 1)
−
=
.
f
−30 −60
−120
The lens is diverging. i is found from
1
1
1
1
=
−
=
,
i
−120 +10
−9.2
the image is virtual and upright. m = −(−9.2)/(+10) = 0.92.
(h) You can’t determine r1 , r2 , or n. Upright images have positive magnification. i is found from
i = −(0.5)(10) = −5;
194
f is found from
1
1
1
1
=
+
=
,
f
+10 −5
−10
so the lens is diverging.
(h) You can’t determine r1 , r2 , or n. Real images have negative magnification. i is found from
i = −(−0.5)(10) = 5;
f is found from
1
1
1
1
=
+ =
,
f
+10 5
+3.33
so the lens is converging.
E40-29 o + i = 0.44 m = L, so
1
1 1
1
1
L
= + = +
=
,
f
o
i
o L−o
o(L − o)
which can also be written as o2 − oL + f L = 0. This has solution
p
p
(0.44 m) ± (0.44 m) − 4(0.11 m)(0.44 m)
L ± L2 − 4f L
o=
=
= 0.22 m.
2
2
There is only one solution to this problem, but sometimes there are two, and other times there are
none!
E40-30 (a) Real images (from real objects) are only produced by converging lenses.
(b) Since hi = −h0 /2, then i = o/2. But d = i+o = o+o/2 = 3o/2, so o = 2(0.40 m)/3 = 0.267 m,
and i = 0.133 m.
(c) 1/f = 1/o + 1/i = 1/(0.267 m) + 1/(0.133 m) = 1/(0.0889 m).
E40-31 Step through the exercise one lens at a time. The object is 40 cm to the left of a
converging lens with a focal length of +20 cm. The image from this first lens will be located by
solving
1
1
1
1
1
1
= − =
−
=
,
i
f
o
(20 cm) (40 cm)
40 cm
so i = 40 cm. Since i is positive it is a real image, and it is located to the right of the converging
lens. This image becomes the object for the diverging lens.
The image from the converging lens is located 40 cm - 10 cm from the diverging lens, but it is
located on the wrong side: the diverging lens is “in the way” so the rays which would form the image
hit the diverging lens before they have a chance to form the image. That means that the real image
from the converging lens is a virtual object in the diverging lens, so that the object distance for the
diverging lens is o = −30 cm.
The image formed by the diverging lens is located by solving
1
1
1
1
1
1
= − =
−
=
,
i
f
o
(−15 cm) (−30 cm)
−30 cm
or i = −30 cm. This would mean the image formed by the diverging lens would be a virtual image,
and would be located to the left of the diverging lens.
The image is virtual, so it is upright. The magnification from the first lens is
m1 = −i/o = −(40 cm)/(40 cm)) = −1;
195
the magnification from the second lens is
m2 = −i/o = −(−30 cm)/(−30 cm)) = −1;
which implies an overall magnification of m1 m2 = 1.
E40-32 (a) The parallel rays of light which strike the lens of focal length f will converge on the
focal point. This point will act like an object for the second lens. If the second lens is located a
distance L from the first then the object distance for the second lens will be L − f . Note that this
will be a negative value for L < f , which means the object is virtual. The image will form at a point
1/i = 1/(−f ) − 1/(L − f ) = L/f (f − L).
Note that i will be positive if L < f , so the rays really do converge on a point.
(b) The same equation applies, except switch the sign of f . Then
1/i = 1/(f ) − 1/(L − f ) = L/f (L − f ).
This is negative for L < f , so there is no real image, and no converging of the light rays.
(c) If L = 0 then i = ∞, which means the rays coming from the second lens are parallel.
E40-33 The image from the converging lens is found from
1
1
1
1
=
−
=
i1
(0.58 m) (1.12 m)
1.20 m
so i1 = 1.20 m, and the image is real and inverted.
This real image is 1.97 m − 1.20 m = 0.77 m in front of the plane mirror. It acts as an object
for the mirror. The mirror produces a virtual image 0.77 m behind the plane mirror. This image is
upright relative to the object which formed it, which was inverted relative to the original object.
This second image is 1.97 m + 0.77 m = 2.74 m away from the lens. This second image acts as an
object for the lens, the image of which is found from
1
1
1
1
=
−
=
i3
(0.58 m) (2.74 m)
0.736 m
so i3 = 0.736 m, and the image is real and inverted relative to the object which formed it, which was
inverted relative to the original object. So this image is actually upright.
E40-34 (a) The first lens forms a real image at a location given by
1/i = 1/f − 1/o = 1/(0.1 m) − 1/(0.2 m) = 1/(0.2 m).
The image and object distance are the same, so the image has a magnification of 1. This image is
0.3 m − 0.2 m = 0.1 m from the second lens. The second lens forms an image at a location given by
1/i = 1/f − 1/o = 1/(0.125 m) − 1/(0.1 m) = 1/(−0.5 m).
Note that this puts the final image at the location of the original object! The image is magnified by
a factor of (0.5 m)/(0.1 m) = 5.
(c) The image is virtual, but inverted.
196
E40-35 If the two lenses “pass” the same amount of light then the solid angle subtended by each
lens as seen from the respective focal points must be the same. If we assume the lenses have the
same round shape then we can write this as do /f o = de /f e . Then
de
fo
=
= mθ ,
do
fe
or de = (72 mm)/36 = 2 mm.
E40-36 (a) f = (0.25 m)/(200) ≈ 1.3 mm. Then 1/f = (n − 1)(2/r) can be used to find r;
r = 2(n − 1)f = 2(1.5 − 1)(1.3 mm) = 1.3 mm.
(b) The diameter would be twice the radius. In effect, these were tiny glass balls.
E40-37 (a) In Fig. 40-46(a) the image is at the focal point. This means that in Fig. 40-46(b)
i = f = 2.5 cm, even though f 0 6= f . Solving,
1
1
1
1
=
+
=
.
f
(36 cm) (2.5 cm)
2.34 cm
(b) The effective radii of curvature must have decreased.
E40-38 (a) s = (25 cm) − (4.2 cm) − (7.7 cm) = 13.1 cm.
(b) i = (25 cm) − (7.7 cm) = 17.3 cm. Then
1
1
1
1
=
−
=
o
(4.2 cm) (17.3 cm)
5.54 cm.
The object should be placed 5.5 − 4.2 = 1.34 cm beyond F1 .
(c) m = −(17.3)/(5.5) = −3.1.
(d) mθ = (25 cm)/(7.7 cm) = 3.2.
(e) M = mmθ = −10.
E40-39 Microscope magnification is given by Eq. 40-33. We need to first find the focal length
of the objective lens before we can use this formula. We are told in the text, however, that the
microscope is constructed so the at the object is placed just beyond the focal point of the objective
lens, then f ob ≈ 12.0 mm. Similarly, the intermediate image is formed at the focal point of the
eyepiece, so f ey ≈ 48.0 mm. The magnification is then
m=
−s(250 mm)
(285 mm)(250 mm)
=−
= 124.
f ob f ey
(12.0 mm)(48.0 mm)
A more accurate answer can be found by calculating the real focal length of the objective lens, which
is 11.4 mm, but since there is a huge uncertainty in the near point of the eye, I see no point in trying
to be more accurate than this.
P40-1 The old intensity is Io = P/4πd2 , where P is the power of the point source. With the
mirror in place there is an additional amount of light which needs to travel a total distance of 3d
in order to get to the screen, so it contributes an additional P/4π(3d)2 to the intensity. The new
intensity is then
In = P/4πd2 + P/4π(3d)2 = (10/9)P/4πd2 = (10/9)Io .
197
P40-2
(a) vi = di/dt; but i = f o/(o − f ) and f = r/2 so
vi =
d
dt
ro
2o − r
=−
r
2o − r
2
do
=−
dt
r
2o − r
2
vo .
(b) Put in the numbers!
vi = −
(15 cm)
2(75 cm) − (15 cm)
2
(5.0 cm/s) = −6.2×10−2 cm/s.
(c) Put in the numbers!
vi = −
(15 cm)
2(7.7 cm) − (15 cm)
2
(5.0 cm/s) = −70 m/s
(d) Put in the numbers!
vi = −
(15 cm)
2(0.15 cm) − (15 cm)
2
(5.0 cm/s) = −5.2 cm/s.
P40-3 (b) There are two ends to the object of length L, one of these ends is a distance o1 from
the mirror, and the other is a distance o2 from the mirror. The images of the two ends will be
located at i1 and i2 .
Since we are told that the object has a short length L we will assume that a differential approach
to the problem is in order. Then
L = ∆o = o1 − o2 and L0 = ∆i = i1 − i2 ,
Finding the ratio of L0 /L is then reduced to
L0
∆i
di
=
≈
.
L
∆o
do
We can take the derivative of Eq. 40-15 with respect to changes in o and i,
di do
+ 2 = 0,
i2
o
or
L0
di
i2
≈
= − 2 = −m2 ,
L
do
o
where m is the lateral magnification.
(a) Since i is given by
1
1
1
o−f
= − =
,
i
f
o
of
the fraction i/o can also be written
i
of
f
=
=
.
o
o(o − f )
o−f
Then
i2
L≈− 2 =−
o
198
f
o−f
2
P40-4 The left surface produces an image which is found from n/i = (n − 1)/R − 1/o, but since
the incoming rays are parallel we take o = ∞ and the expression simplifies to i = nR/(n − 1). This
image is located a distance o = 2R − i = (n − 2)R/(n − 1) from the right surface, and the image
produced by this surface can be found from
1/i = (1 − n)/(−R) − n/o = (n − 1)/R − n(n − 1)/(n − 2)R = 2(1 − n)/(n − 2)R.
Then i = (n − 2)R/2(n − 1).
P40-5 The “1” in Eq. 40-18 is actually nair ; the assumption is that the thin lens is in the air. If
that isn’t so, then we need to replace “1” with n0 , so Eq. 40-18 becomes
n
n − n0
n0
− 0 =
.
o
|i |
r1
A similar correction happens to Eq. 40-21:
n
n0
n − n0
+
=
−
.
|i0 |
i
r2
Adding these two equations,
n0
n0
+
= (n − n0 )
o
i
1
1
−
r1
r2
This yields a focal length given by
1
n − n0
=
f
n
P40-6
1
1
−
r1
r2
Start with Eq. 40-4
1 1
+
o
i
|f | |f |
+
o
i
1
1
+ 0
y y
=
=
1
,
f
|f |
,
f
= ±1,
where + is when f is positive and − is when f is negative.
The plot on the right is for +, that on the left for −.
Real image and objects occur when y or y 0 is positive.
199
.
.
P40-7 (a) The image (which will appear on the screen) and object are a distance D = o + i
apart. We can use this information to eliminate one variable from Eq. 40-15,
1 1
+
o
i
1
1
+
o D−o
D
o(D − o)
o2 − oD + f D
1
,
f
1
,
f
1
,
f
=
=
=
=
0.
This last expression is a quadratic, and we would expect to get two solutions for o. These solutions
will be of the form “something” plus/minus “something else”; the distance between the two locations
for o will evidently be twice the “something else”, which is then
p
p
d = o+ − o− = (−D)2 − 4(f D) = D(D − 4f ).
(b) The ratio of the image sizes is m+ /m− , or i+ o− /i− o+ . Now it seems we must find the actual
values of o+ and o− . From the quadratic in part (a) we have
p
D ± D(D − 4f )
D±d
=
,
o± =
2
2
so the ratio is
o−
=
o+
D−d
D+d
.
But i− = o+ , and vice-versa, so the ratio of the image sizes is this quantity squared.
P40-8 1/i = 1/f − 1/o implies i = f o/(o − f ). i is only real if o ≥ f . The distance between the
image and object is
of
o2
y =i+o=
+o=
.
o−f
o−f
This quantity is a minimum when dy/do = 0, which occurs when o = 2f . Then i = 2f , and y = 4f .
P40-9 (a) The angular size of each lens is the same when viewed from the shared focal point. This
means W1 /f1 = W2 /f2 , or
W2 = (f2 /f1 )W1 .
(b) Pass the light through the diverging lens first; choose the separation of the lenses so that
the focal point of the converging lens is at the same location as the focal point of the diverging lens
which is on the opposite side of the diverging lens.
(c) Since I ∝ 1/A, where A is the area of the beam, we have I ∝ 1/W 2 . Consequently,
I2 /I1 = (W1 /W2 )2 = (f1 /f2 )2
P40-10
The location of the image in the mirror is given by
1
1
1
= −
.
i
f
a+b
200
The location of the image in the plate is given by i0 = −a, which is located at b − a relative to the
mirror. Equating,
1
1
+
b−a b+a
2b
2
b − a2
b2 − a2
=
=
=
a =
=
1
,
f
1
,
f
2bf,
p
b2 − 2bf ,
p
(7.5 cm)2 − 2(7.5 cm)(−28.2 cm) = 21.9 cm.
P40-11 We’ll solve the problem by finding out what happens if you put an object in front of the
combination of lenses.
Let the object distance be o1 . The first lens will create an image at i1 , where
1
1
1
=
−
i1
f1
o1
This image will act as an object for the second lens.
If the first image is real (i1 positive) then the image will be on the “wrong” side of the second
lens, and as such the real image will act like a virtual object. In short, o2 = −i1 will give the correct
sign to the object distance when the image from the first lens acts like an object for the second lens.
The image formed by the second lens will then be at
1
i2
=
=
=
1
1
− ,
f2
o2
1
1
+ ,
f2
i2
1
1
1
+
− .
f2
f1
o1
In this case it appears as if the combination
1
1
+
f2
f1
is equivalent to the reciprocal of a focal length. We will go ahead and make this connection, and
1
1
1
f1 + f2
=
+
=
.
f
f2
f1
f1 f2
The rest is straightforward enough.
P40-12
(a) The image formed by the first lens can be found from
1
1
1
1
=
−
=
.
i1
f1
2f1
2f1
This is a distance o2 = 2(f1 + f2 ) = 2f2 from the mirror. The image formed by the mirror is at an
image distance given by
1
1
1
1
=
−
=
.
i2
f2
2f2
2f2
Which is at the same point as i1 !. This means it will act as an object o3 in the lens, and, reversing
the first step, produce a final image at O, the location of the original object. There are then three
images formed; each is real, same size, and inverted. Three inversions nets an inverted image. The
final image at O is therefore inverted.
201
P40-13
(a) Place an object at o. The image will be at a point i0 given by
1
1
1
= − ,
i0
f
o
or i0 = f o/(o − f ).
(b) The lens must be shifted a distance i0 − i, or
i0 − i =
fo
− 1.
o−f
(c) The range of motion is
∆i =
(0.05 m)(1.2 m)
− 1 = −5.2 cm.
(1.2 m) − (0.05 m)
P40-14 (a) Because magnification is proportional to 1/f .
(b) Using the results of Problem 40-11,
1
1
1
=
+ ,
f
f2
f1
so P = P1 + P2 .
P40-15 We want the maximum linear motion of the train to move no more than 0.75 mm on the
film; this means we want to find the size of an object on the train that will form a 0.75 mm image.
The object distance is much larger than the focal length, so the image distance is approximately
equal to the focal length. The magnification is then m = −i/o = (3.6 cm)/(44.5 m) = −0.00081.
The size of an object on the train that would produce a 0.75 mm image on the film is then
0.75 mm/0.00081 = 0.93 m.
How much time does it take the train to move that far?
t=
(0.93 m)
= 25 ms.
(135 km/hr)(1/3600 hr/s)
P40-16 (a) The derivation leading to Eq. 40-34 depends only on the fact that two converging
optical devices are used. Replacing the objective lens with an objective mirror doesn’t change
anything except the ray diagram.
(b) The image will be located very close to the focal point, so |m| ≈ f /o, and
hi = (1.0 m)
(16.8 m)
= 8.4×10−3 m
(2000 m)
(c) f e = (5 m)/(200) = 0.025 m. Note that we were given the radius of curvature, not the focal
length, of the mirror!
202
E41-1
In this problem we look for the location of the third-order bright fringe, so
θ = sin−1
mλ
(3)(554 × 10−9 m)
= sin−1
= 12.5◦ = 0.22 rad.
d
(7.7 × 10−6 m)
E41-2 d1 sin θ = λ gives the first maximum; d2 sin θ = 2λ puts the second maximum at the location
of the first. Divide the second expression by the first and d2 = 2d1 . This is a 100% increase in d.
E41-3 ∆y = λD/d = (512×10−9 m)(5.4 m)/(1.2×10−3 m) = 2.3×10−3 m.
E41-4 d = λ/ sin θ = (592×10−9 m)/ sin(1.00◦ ) = 3.39×10−5 m.
E41-5 Since the angles are very small, we can assume sin θ ≈ θ for angles measured in radians.
If the interference fringes are 0.23◦ apart, then the angular position of the first bright fringe
is 0.23◦ away from the central maximum. Eq. 41-1, written with the small angle approximation
in mind, is dθ = λ for this first (m = 1) bright fringe. The goal is to find the wavelength which
increases θ by 10%. To do this we must increase the right hand side of the equation by 10%, which
means increasing λ by 10%. The new wavelength will be λ0 = 1.1λ = 1.1(589 nm) = 650 nm
E41-6 Immersing the apparatus in water will shorten the wavelengths to λ/n. Start with d sin θ0 =
λ; and then find θ from d sin θ = λ/n. Combining the two expressions,
θ = arcsin[sin θ0 /n] = arcsin[sin(0.20◦ )/(1.33)] = 0.15◦ .
E41-7 The third-order fringe for a wavelength λ will be located at y = 3λD/d, where y is measured
from the central maximum. Then ∆y is
y1 − y2 = 3(λ1 − λ2 )D/d = 3(612×10−9 m − 480×10−9 m)(1.36 m)/(5.22×10−3 m) = 1.03×10−4 m.
E41-8 θ = arctan(y/D);
λ = d sin θ = (0.120 m) sin[arctan(0.180 m/2.0 m)] = 1.08×10−2 m.
Then f = v/λ = (0.25 m/s)/(1.08×10−2 m) = 23 Hz.
E41-9
A variation of Eq. 41-3 is in order:
ym =
1
m+
2
λD
d
We are given the distance (on the screen) between the first minima (m = 0) and the tenth minima
(m = 9). Then
λ(50 cm)
18 mm = y9 − y0 = 9
,
(0.15 mm)
or λ = 6×10−4 mm = 600 nm.
E41-10 The “maximum” maxima is given by the integer part of
m = d sin(90◦ )/λ = (2.0 m)/(0.50 m) = 4.
Since there is no integer part, the “maximum” maxima occurs at 90◦ . These are point sources
radiating in both directions, so there are two central maxima, and four maxima each with m = 1,
m = 2, and m = 3. But the m = 4 values overlap at 90◦ , so there are only two. The total is 16.
203
E41-11
This figure should explain it well enough.
E41-12 ∆y = λD/d = (589×10−9 m)(1.13 m)/(0.18×10−3 m) = 3.70×10−3 m.
E41-13 Consider Fig. 41-5, and solve it exactly for the information given. For the tenth bright
fringe r1 = 10λ + r2 . There are two important triangles:
r22 = D2 + (y − d/2)2
and
r12 = D2 + (y + d/2)2
Solving to eliminate r2 ,
p
p
D2 + (y + d/2)2 = D2 + (y − d/2)2 + 10λ.
This has solution
y = 5λ
r
4D2 + d2 − 100λ2
.
d2 − 100λ2
The solution predicted by Eq. 41-1 is
y0 =
10λ p 2
D + y 02 ,
d
or
0
y = 5λ
r
d2
4D2
.
− 100λ2
The fractional error is y 0 /y − 1, or
r
or
s
4D2
4D2
− 1,
+ d2 − 100λ2
4(40 mm)2
− 1 = −3.1×10−4 .
4(40 mm)2 + (2 mm)2 − 100(589×10−6 mm)2
E41-14 (a) ∆x = c/∆t = (3.00×108 m/s)/(1×10−8 s) = 3 m.
(b) No.
204
E41-15 Leading by 90◦ is the same as leading by a quarter wavelength, since there are 360◦ in a
circle. The distance from A to the detector is 100 m longer than the distance from B to the detector.
Since the wavelength is 400 m, 100 m corresponds to a quarter wavelength.
So a wave peak starts out from source A and travels to the detector. When it has traveled a
quarter wavelength a wave peak leaves source B. But when the wave peak from A has traveled
a quarter wavelength it is now located at the same distance from the detector as source B, which
means the two wave peaks arrive at the detector at the same time.
They are in phase.
E41-16 The first dark fringe involves waves π radians out of phase. Each dark fringe after that
involves an additional 2π radians of phase difference. So the mth dark fringe has a phase difference
of (2m + 1)π radians.
E41-17 I = 4I0 cos2
2πd
λ
sin θ , so for this problem we want to plot
2π(0.60 mm)
2
I/I0 = cos
sin θ = cos2 (6280 sin θ) .
(600×10−9 m)
E41-18 The resultant quantity will be of the form A sin(ωt + β). Solve the problem by looking at
t = 0; then y1 = 0, but x1 = 10, and y2 = 8 sin 30◦ = 4 and x2 = 8 cos 30 = 6.93. Then the resultant
is of length
p
A = (4)2 + (10 + 6.93)2 = 17.4,
and has an angle β given by
β = arctan(4/16.93) = 13.3◦ .
E41-19 (a) We want to know the path length difference of the two sources to the detector.
Assume the detector
is at x and the second
√
√ source is at y = d. The distance S1 D is x; the
distance S2 D is x2 + d2 . The difference is x2 + d2 − x. If this difference is an integral number
of wavelengths then we have a maximum; if instead it is a half integral number of wavelengths we
have a minimum. For part (a) we are looking for the maxima, so we set the path length difference
equal to mλ and solve for xm .
p
x2m + d2 − xm = mλ,
x2m + d2 = (mλ + xm )2 ,
x2m + d2 = m2 λ2 + 2mλxm + x2m ,
d 2 − m2 λ 2
xm =
2mλ
The first question we need to ask is what happens when m = 0. The right hand side becomes
indeterminate, so we need to go back to the first line in the above derivation. If m = 0 then d2 = 0;
since this is not true in this problem, there is no m = 0 solution.
In fact, we may have even more troubles. xm needs to be a positive value, so the maximum
allowed value for m will be given by
m2 λ2 < d2 ,
m < d/λ = (4.17 m)/(1.06 m) = 3.93;
but since m is an integer, m = 3 is the maximum value.
205
The first three maxima occur at m = 3, m = 2, and m = 1. These maxima are located at
x3
=
x2
=
x1
=
(4.17 m)2 − (3)2 (1.06 m)2
= 1.14 m,
2(3)(1.06 m)
(4.17 m)2 − (2)2 (1.06 m)2
= 3.04 m,
2(2)(1.06 m)
(4.17 m)2 − (1)2 (1.06 m)2
= 7.67 m.
2(1)(1.06 m)
Interestingly enough, as m decreases the maxima get farther away!
(b) The closest maxima to the origin occurs at x = ±6.94 cm. What then is x = 0? It is a local
minimum, but the intensity isn’t zero. It corresponds to a point where the path length difference is
3.93 wavelengths. It should be half an integer to be a complete minimum.
E41-20 The resultant can be written in the form A sin(ωt + β). Consider t = 0. The three
components can be written as
10 sin 0◦ = 0,
14 sin 26◦ = 6.14,
4.7 sin(−41◦ ) = −3.08,
0 + 6.14 − 3.08 = 3.06.
y1
y2
y3
y
=
=
=
=
x1
x2
x3
x
= 10 cos 0◦ = 10,
= 14 cos 26◦ = 12.6,
= 4.7 cos(−41◦ ) = 3.55,
= 10 + 12.6 + 3.55 = 26.2.
and
Then A =
p
(3.06)2 + (26.2)2 = 26.4 and β = arctan(3.06/26.2) = 6.66◦ .
E41-21 The order of the indices of refraction is the same as in Sample Problem 41-4, so
d = λ/4n = (620 nm)/4(1.25) = 124 nm.
E41-22 Follow the example in Sample Problem 41-3.
λ=
2dn
2(410 nm)(1.50)
1230 nm
=
=
.
m − 1/2
m − 1/2
m − 1/2
The result is only in the visible range when m = 3, so λ = 492 nm.
E41-23 (a) Light from above the oil slick can be reflected back up from the top of the oil layer
or from the bottom of the oil layer. For both reflections the light is reflecting off a substance with
a higher index of refraction so both reflected rays pick up a phase change of π. Since both waves
have this phase the equation for a maxima is
1
1
2d + λn + λn = mλn .
2
2
Remember that λn = λ/n, where n is the index of refraction of the thin film. Then 2nd = (m − 1)λ
is the condition for a maxima. We know n = 1.20 and d = 460 nm. We don’t know m or λ. It might
206
seem as if there isn’t enough information to solve the problem, but we can. We need to find the
wavelength in the visible range (400 nm to 700 nm) which has an integer m. Trial and error might
work. If λ = 700 nm, then m is
m=
2nd
2(1.20)(460 nm)
+1=
+ 1 = 2.58
λ
(700 nm)
But m needs to be an integer. If we increase m to 3, then
λ=
2(1.20)(460 nm)
= 552 nm
(3 − 1)
which is in the visible range. So the oil slick will appear green.
(b) One of the most profound aspects of thin film interference is that wavelengths which are
maximally reflected are minimally transmitted, and vice versa. Finding the maximally transmitted
wavelengths is the same as finding the minimally reflected wavelengths, or looking for values of m
that are half integer.
The most obvious choice is m = 3.5, and then
λ=
2(1.20)(460 nm)
= 442 nm.
(3.5 − 1)
E41-24 The condition for constructive interference is 2nd = (m − 1/2)λ. Assuming a minimum
value of m = 1 one finds
d = λ/4n = (560 nm)/4(2.0) = 70 nm.
E41-25 The top surface contributes a phase difference of π, so the phase difference because of the
thickness is 2π, or one complete wavelength. Then 2d = λ/n, or d = (572 nm)/2(1.33) = 215 nm.
E41-26 The wave reflected from the first surface picks up a phase shift of π. The wave which is
reflected off of the second surface travels an additional path difference of 2d. The interference will
be bright if 2d + λn /2 = mλn results in m being an integer.
m = 2nd/λ + 1/2 = 2(1.33)(1.21×10−6 m)/(585×10−9 m) + 1/2 = 6.00,
so the interference is bright.
E41-27 As with the oil on the water in Ex. 41-23, both the light which reflects off of the acetone
and the light which reflects off of the glass undergoes a phase shift of π. Then the maxima for
reflection are given by 2nd = (m − 1)λ. We don’t know m, but at some integer value of m we have
λ = 700 nm. If m is increased by exactly 12 then we are at a minimum of λ = 600 nm. Consequently,
2(1.25)d = (m − 1)(700 nm) and 2(1.25)d = (m − 1/2)(600 nm),
we can set these two expressions equal to each other to find m,
(m − 1)(700 nm) = (m − 1/2)(600 nm),
so m = 4. Then we can find the thickness,
d = (4 − 1)(700 nm)/2(1.25) = 840 nm.
207
E41-28 The wave reflected from the first surface picks up a phase shift of π. The wave which is
reflected off of the second surface travels an additional path difference of 2d. The interference will
be bright if 2d + λn /2 = mλn results in m being an integer. Then 2nd = (m − 1/2)λ1 is bright, and
2nd = mλ2 is dark. Divide one by the other and (m − 1/2)λ1 = mλ2 , so
m = λ1 /2(λ1 − λ2 ) = (600 nm)/2(600 nm − 450 nm) = 2,
then d = mλ2 /2n = (2)(450 nm)/2(1.33) = 338 nm.
E41-29 Constructive interference happens when 2d = (m − 1/2)λ. The minimum value for m is
m = 1; the maximum value is the integer portion of 2d/λ+1/2 = 2(4.8×10−5 m)/(680×10−9 m)+1/2 =
141.67, so mmax = 141. There are then 141 bright bands.
E41-30 (a) A half wavelength phase shift occurs for both the air/water interface and the water/oil
interface, so if d = 0 the two reflected waves are in phase. It will be bright!
(b) 2nd = 3λ, or d = 3(475 nm)/2(1.20) = 594 nm.
E41-31 There is a phase shift on one surface only, so the bright bands are given by 2nd = (m −
1/2)λ. Let the first band be given by 2nd1 = (m1 − 1/2)λ. The last bright band is then given by
2nd2 = (m1 + 9 − 1/2)λ. Subtract the two equations to get the change in thickness:
∆d = 9λ/2n = 9(630 nm)/2(1.50) = 1.89 µm.
E41-32 Apply Eq. 41-21: 2nd = mλ. In one case we have
2nair = (4001)λ,
in the other,
2nvac = (4000)λ.
Equating, nair = (4001)/(4000) = 1.00025.
E41-33
(a) We can start with the last equation from Sample Problem 41-5,
r
1
r = (m − )λR,
2
and solve for m,
r2
1
+
λR 2
In this exercise R = 5.0 m, r = 0.01 m, and λ = 589 nm. Then
m=
m=
(0.01 m)2
= 34
(589 nm)(5.0 m)
is the number of rings observed.
(b) Putting the apparatus in water effectively changes the wavelength to
(589 nm)/(1.33) = 443 nm,
so the number of rings will now be
m=
(0.01 m)2
= 45.
(443 nm)(5.0 m)
208
q
q
(10 − 12 )Rλ, while (1.27 cm) = (10 − 12 )Rλ/n. Divide one expression by
√
the other, and (1.42 cm)/(1.27 cm) = n, or n = 1.25.
E41-34 (1.42 cm) =
q
q
E41-35 (0.162 cm) = (n − 12 )Rλ, while (0.368 cm) = (n + 20 − 12 )Rλ. Square both expressions, the divide one by the other, and find
(n + 19.5)/(n − 0.5) = (0.368 cm/0.162 cm)2 = 5.16
which can be rearranged to yield
n=
19.5 + 5.16 × 0.5
= 5.308.
5.16 − 1
Oops! That should be an integer, shouldn’t it? The above work is correct, which means that there
really aren’t bright bands at the specified locations. I’m just going to gloss over that fact and solve
for R using the value of m = 5.308. Then
R = r2 /(m − 1/2)λ = (0.162 cm)2 /(5.308 − 0.5)(546 nm) = 1.00 m.
Well, at least we got the answer which is in the back of the book...
E41-36 Pretend the ship is a two point source emitter, one h above the water, and one h below
the water. The one below the water is out of phase by half a wavelength. Then d sin θ = λ, where
d = 2h, gives the angle for theta for the first minimum.
λ/2h = (3.43 m)/2(23 m) = 7.46×10−2 = sin θ ≈ H/D,
so D = (160 m)/(7.46×10−2 ) = 2.14 km.
E41-37 The phase difference is 2π/λn times the path difference which is 2d, so
φ = 4πd/λn = 4πnd/λ.
We are given that d = 100×10−9 m and n = 1.38.
(a) φ = 4π(1.38)(100×10−9 m)/(450×10−9 m) = 3.85. Then
(3.85)
I
= cos2
= 0.12.
I0
2
The reflected ray is diminished by 1 − 0.12 = 88%.
(b) φ = 4π(1.38)(100×10−9 m)/(650×10−9 m) = 2.67. Then
I
(2.67)
= cos2
= 0.055.
I0
2
The reflected ray is diminished by 1 − 0.055 = 95%.
E41-38 The change in the optical path length is 2(d − d/n), so 7λ/n = 2d(1 − 1/n), or
d=
7(589×10−9 m)
= 4.9×10−6 m.
2(1.42) − 2
209
E41-39 When M2 moves through a distance of λ/2 a fringe has will be produced, destroyed, and
then produced again. This is because the light travels twice through any change in distance. The
wavelength of light is then
2(0.233 mm)
= 588 nm.
λ=
792
E41-40 The change in the optical path length is 2(d − d/n), so 60λ = 2d(1 − 1/n), or
n=
1
1
=
= 1.00030.
−9
1 − 60λ/2d
1 − 60(500×10 m)/2(5×10−2 m)
P41-1 (a) This is a small angle problem, so we use Eq. 41-4. The distance to the screen is
2 × 20 m, because the light travels to the mirror and back again. Then
d=
λD
(632.8 nm)(40.0 m)
=
= 0.253 mm.
∆y
(0.1 m)
(b) Placing the cellophane over one slit will cause the interference pattern to shift to the left or
right, but not disappear or change size. How does it shift? Since we are picking up 2.5 waves then
we are, in effect, swapping bright fringes for dark fringes.
P41-2
The change in the optical path length is d − d/n, so 7λ/n = d(1 − 1/n), or
d=
7(550×10−9 m)
= 6.64×10−6 m.
(1.58) − 1
p
P41-3
(x + d/2)2 + y 2 . The distance from S2 to P is
p The distance from S1 to P is r1 =
2
2
r2 = (x − d/2) + y . The difference in distances is fixed at some value, say c, so that
r 1 − r2
− 2r1 r2 + r22
(r12 + r22 − c2 )2
2
2 2
(r1 − r2 ) − 2c2 (r12 + r22 ) + c4
(2xd)2 − 2c2 (2x2 + d2 /2 + 2y 2 ) + c4
4x2 d2 − 4c2 x2 − c2 d2 − 4c2 y 2 + c4
4(d2 − c2 )x2 − 4c2 y 2
r12
=
=
=
=
=
=
=
c,
c2 ,
4r12 r22 ,
0,
0,
0,
c2 (d2 − c2 ).
Yes, that is the equation of a hyperbola.
P41-4 The change in the optical path length for each slit is nt − t, where n is the corresponding
index of refraction. The net change in the path difference is then n2 t − n1 t. Consequently, mλ =
t(n2 − n1 ), so
(5)(480×10−9 m)
t=
= 8.0×10−6 m.
(1.7) − (1.4)
P41-5 The intensity is given by Eq. 41-17, which, in the small angle approximation, can be
written as
πdθ
Iθ = 4I0 cos2
.
λ
210
The intensity will be half of the maximum when
1
πd∆θ/2
= cos2
2
λ
or
π
πd∆θ
=
,
4
2λ
which will happen if ∆θ = λ/2d.
P41-6 Follow the construction in Fig. 41-10, except that one of the electric field amplitudes is
twice the other. The resultant field will have a length given by
p
E0 =
(2E0 + E0 cos φ)2 + (E0 sin φ)2 ,
p
= E0 5 + 4 cos φ,
so squaring this yields
I
P41-7
mum is
2πd sin θ
= I0 5 + 4 cos
,
λ
πd sin θ
= I0 1 + 8 cos2
,
λ
Im
πd sin θ
=
1 + 8 cos2
.
9
λ
We actually did this problem in Exercise 41-27, although slightly differently. One maxi2(1.32)d = (m − 1/2)(679 nm),
the other is
2(1.32)d = (m + 1/2)(485 nm).
Set these equations equal to each other,
(m − 1/2)(679 nm) = (m + 1/2)(485 nm),
and find m = 3. Then the thickness is
d = (3 − 1/2)(679 nm)/2(1.32) = 643 nm.
P41-8
(a) Since we are concerned with transmission there is a phase shift for two rays, so
2d = mλn
The minimum thickness occurs when m = 1; solving for d yields
d=
λ
(525×10−9 m)
=
= 169×10−9 m.
2n
2(1.55)
(b) The wavelengths are different, so the other parts have differing phase differences.
(c) The nearest destructive interference wavelength occurs when m = 1.5, or
λ = 2nd = 2(1.55)1.5(169×10−9 m) = 393×10−9 m.
This is blue-violet.
211
P41-9 It doesn’t matter if we are looking at bright are dark bands. It doesn’t even matter if we
concern ourselves with phase shifts. All that cancels out. Consider 2δd = δmλ; then
δd = (10)(480 nm)/2 = 2.4 µm.
P41-10
(a) Apply 2d = mλ. Then
d = (7)(600×10−9 m)/2 = 2100×10−9 m.
(b) When water seeps in it introduces an extra phase shift. Point A becomes then a bright fringe,
and the equation for the number of bright fringes is 2nd = mλ. Solving for m,
m = 2(1.33)(2100×10−9 m)/(600×10−9 m) = 9.3;
this means that point B is almost, but not quite, a dark fringe, and there are nine of them.
P41-11
(a) Look back at the work for Sample Problem 41-5 where it was found
r
1
rm = (m − )λR,
2
We can write this as
rm =
s
1
1−
2m
mλR
and expand the part in parentheses in a binomial expansion,
1 1 √
rm ≈ 1 −
mλR.
2 2m
We will do the same with
rm+1 =
r
rm+1 =
s
1
(m + 1 − )λR,
2
expanding
1+
1
2m
to get
rm+1 ≈
1+
1 1
2 2m
mλR
√
mλR.
Then
1 √
mλR,
2m
or
1p
∆r ≈
λR/m.
2
(b) The area between adjacent rings is found from the difference,
2
2
A = π rm+1
− rm
,
∆r ≈
and into this expression we will substitute the exact values for
1
A = π (m + 1 − )λR − (m −
2
= πλR.
rm and rm+1 ,
1
)λR ,
2
Unlike part (a), we did not need to assume m 1 in order to arrive at this expression; it is exact
for all m.
212
P41-12 The path length shift that occurs when moving the mirror as distance x is 2x. This means
φ = 2π2x/λ = 4πx/λ. The intensity is then
I = 4I0 cos2
213
2πx
λ
E42-1 λ = a sin θ = (0.022 mm) sin(1.8◦ ) = 6.91×10−7 m.
E42-2 a = λ/ sin θ = (0.10×10−9 m)/ sin(0.12×10−3 rad/2) = 1.7 µm.
E42-3 (a) This is a valid small angle approximation problem: the distance between the points
on the screen is much less than the distance to the screen. Then
θ≈
(0.0162 m)
= 7.5 × 10−3 rad.
(2.16 m)
(b) The diffraction minima are described by Eq. 42-3,
a sin θ = mλ,
a sin(7.5 × 10 rad) = (2)(441 × 10−9 m),
a = 1.18 × 10−4 m.
−3
E42-4 a = λ/ sin θ = (633×10−9 m)/ sin(1.97◦ /2) = 36.8 µm.
E42-5 (a) We again use Eq. 42-3, but we will need to throw in a few extra subscripts to
distinguish between which wavelength we are dealing with. If the angles match, then so will the sine
of the angles. We then have sin θa,1 = sin θb,2 or, using Eq. 42-3,
(1)λa
(2)λb
=
,
a
a
from which we can deduce λa = 2λb .
(b) Will any other minima coincide? We want to solve for the values of ma and mb that will be
integers and have the same angle. Using Eq. 42-3 one more time,
mb λ b
ma λa
=
,
a
a
and then substituting into this the relationship between the wavelengths, ma = mb /2. whenever mb
is an even integer ma is an integer. Then all of the diffraction minima from λa are overlapped by a
minima from λb .
E42-6 The angle is given by sin θ = 2λ/a. This is a small angle, so we can use the small angle
approximation of sin θ = y/D. Then
y = 2Dλ/a = 2(0.714 m)(593×10−9 m)/(420×10−6 m) = 2.02 mm.
E42-7 Small angles, so y/D = sin θ = λ/a. Then
a = Dλ/y = (0.823 m)(546×10−9 m)/(5.20×10−3 m/2) = 1.73×10−4 m.
E42-8 (b) Small angles, so ∆y/D = ∆mλ/a. Then
a = ∆mDλ/∆y = (5 − 1)(0.413 m)(546×10−9 m)/(0.350×10−3 m) = 2.58 mm.
(a) θ = arcsin(λ/a) = arcsin[(546×10−9 m)/(2.58 mm)] = 1.21×10−2◦ .
E42-9 Small angles, so ∆y/D = ∆mλ/a. Then
∆y = ∆mDλ/a = (2 − 1)(2.94 m)(589×10−9 m)/(1.16×10−3 m) = 1.49×10−3 m.
214
E42-10 Doubling the width of the slit results in a narrowing of the diffraction pattern. Since the
width of the central maximum is effectively cut in half, then there is twice the energy in half the
space, producing four times the intensity.
E42-11
(a) This is a small angle approximation problem, so
θ = (1.13 cm)/(3.48 m) = 3.25 × 10−3 rad.
(b) A convenient measure of the phase difference, α is related to θ through Eq. 42-7,
α=
π(25.2 × 10−6 m)
πa
sin(3.25 × 10−3 rad) = 0.478 rad
sin θ =
λ
(538 × 10−9 m)
(c) The intensity at a point is related to the intensity at the central maximum by Eq. 42-8,
Iθ
=
Im
sin α
α
2
=
sin(0.478 rad)
(0.478 rad)
2
= 0.926
E42-12 Consider Fig. 42-11; the angle with the vertical is given by (π − φ)/2. For Fig. 42-10(d)
the circle has wrapped once around onto itself so the angle with the vertical is (3π −φ)/2. Substitute
α into this expression and the angel against the vertical is 3π/2 − α.
Use the result from Problem 42-3 that tan α = α for the maxima. The lowest non-zero solution
is α = 4.49341 rad. The angle against the vertical is then 0.21898 rad, or 12.5◦ .
E42-13 Drawing heavily from Sample Problem 42-4,
αx λ
1.39
θx = arcsin
= arcsin
= 2.54◦ .
πa
10π
Finally, ∆θ = 2θx = 5.1◦ .
E42-14 (a) Rayleigh’s criterion for resolving images (Eq. 42-11) requires that two objects have an
angular separation of at least
1.22λ
1.22(540 × 10−9 )
−1
−1
θR = sin
= sin
= 1.34 × 10−4 rad
d
(4.90 × 10−3 m)
(b) The linear separation is y = θD = (1.34 × 10−4 rad)(163×103 m) = 21.9 m.
E42-15 (a) Rayleigh’s criterion for resolving images (Eq. 42-11) requires that two objects have
an angular separation of at least
1.22λ
1.22(562 × 10−9 )
−1
−1
θR = sin
= sin
= 1.37 × 10−4 rad.
d
(5.00 × 10−3 m)
(b) Once again, this is a small angle, so we can use the small angle approximation to find the
distance to the car. In that case θR = y/D, where y is the headlight separation and D the distance
to the car. Solving,
D = y/θR = (1.42 m)/(1.37 × 10−4 rad) = 1.04 × 104 m,
or about six or seven miles.
215
E42-16 y/D = 1.22λ/a; or
D = (5.20×10−3 m)(4.60×10−3 /m)/1.22(542×10−9 m) = 36.2 m.
E42-17
The smallest resolvable angular separation will be given by Eq. 42-11,
θR = sin−1
1.22λ
d
= sin−1
1.22(565 × 10−9 m)
(5.08 m)
= 1.36 × 10−7 rad,
The smallest objects resolvable on the Moon’s surface by this telescope have a size y where
y = DθR = (3.84 × 108 m)(1.36 × 10−7 rad) = 52.2 m
E42-18 y/D = 1.22λ/a; or
y = 1.22(1.57×10−2 m)(6.25×103 m)/(2.33 m) = 51.4 m
E42-19 y/D = 1.22λ/a; or
D = (4.8×10−2 m)(4.3×10−3 /m)/1.22(0.12×10−9 m) = 1.4×106 m.
E42-20 y/D = 1.22λ/a; or
d = 1.22(550×10−9 m)(160×103 m)/(0.30 m) = 0.36 m.
E42-21
Using Eq. 42-11, we find the minimum resolvable angular separation is given by
θR = sin
−1
1.22λ
d
−1
= sin
1.22(475 × 10−9 m)
(4.4 × 10−3 m)
= 1.32 × 10−4 rad
The dots are 2 mm apart, so we want to stand a distance D away such that
D > y/θR = (2 × 10−3 m)/(1.32 × 10−4 rad) = 15 m.
E42-22 y/D = 1.22λ/a; or
y = 1.22(500×10−9 m)(354×103 m)/(9.14 m/2) = 4.73×10−2 m.
E42-23 (a) λ = v/f . Now use Eq. 42-11:
(1450 m/s)
θ = arcsin 1.22
= 6.77◦ .
(25×103 Hz)(0.60 m)
(b) Following the same approach,
θ = arcsin 1.22
(1450 m/s)
(1×103 Hz)(0.60 m)
has no real solution, so there is no minimum.
216
E42-24 (a) λ = v/f . Now use Eq. 42-11:
(3×108 m/s)
θ = arcsin 1.22
= 0.173◦ .
(220×109 Hz)(0.55 m)
This is the angle from the central maximum; the angular width is twice this, or 0.35◦ .
(b) use Eq. 42-11:
(0.0157 m)
θ = arcsin 1.22
= 0.471◦ .
(2.33 m)
This is the angle from the central maximum; the angular width is twice this, or 0.94◦ .
E42-25
The linear separation of the fringes is given by
∆y
λ
λD
= ∆θ = or ∆y =
D
d
d
for sufficiently small d compared to λ.
E42-26 (a) d sin θ = 4λ gives the location of the fourth interference maximum, while a sin θ =
λ gives the location of the first diffraction minimum. Hence, if d = 4a there will be no fourth
interference maximum!
(b) Since d sin θmi = mi λ gives the interference maxima and a sin θmd = md λ gives the diffraction
minima, and d = 4a, then whenever mi = 4md there will be a missing maximum.
E42-27 (a) The central diffraction envelope is contained in the range
θ = arcsin
λ
a
This angle corresponds to the mth maxima of the interference pattern, where
sin θ = mλ/d = mλ/2a.
Equating, m = 2, so there are three interference bands, since the m = 2 band is “washed out” by
the diffraction minimum.
(b) If d = a then β = α and the expression reduces to
Iθ
sin2 α
,
α2
sin2 (2α)
= Im
,
2α2
2
sin α0
= 2I m
,
α0
= I m cos2 α
where α = 2α0 , which is the same as replacing a by 2a.
E42-28 Remember that the central peak has an envelope width twice that of any other peak.
Ignoring the central maximum there are (11 − 1)/2 = 5 fringes in any other peak envelope.
217
E42-29 (a) The first diffraction minimum is given at an angle θ such that a sin θ = λ; the order
of the interference maximum at that point is given by d sin θ = mλ. Dividing one expression by the
other we get d/a = m, with solution m = (0.150)/(0.030) = 5. The fact that the answer is exactly 5
implies that the fifth interference maximum is squelched by the diffraction minimum. Then there are
only four complete fringes on either side of the central maximum. Add this to the central maximum
and we get nine as the answer.
(b) For the third fringe m = 3, so d sin θ = 3λ. Then β is Eq. 42-14 is 3π, while α in Eq. 42-16
is
a
πa 3λ
α=
= 3π ,
λ d
d
so the relative intensity of the third fringe is, from Eq. 42-17,
2
(cos 3π)
P42-1
sin(3πa/d)
(3πa/d)
2
= 0.255.
y = mλD/a. Then
y = (10)(632.8×10−9 m)(2.65 m)/(1.37×10−3 m) = 1.224×10−2 m.
The separation is twice this, or 2.45 cm.
P42-2 If a λ then the diffraction pattern is extremely tight, and there is effectively no light at
P . In the event that either shape produces an interference pattern at P then the other shape must
produce an equal but opposite electric field vector at that point so that when both patterns from
both shapes are superimposed the field cancel.
But the intensity is the field vector squared; hence the two patterns look identical.
P42-3
(a) We want to take the derivative of Eq. 42-8 with respect to α, so
dIθ
dα
2
d
sin α
=
Im
,
dα
α
sin α
cos α sin α
= I m2
− 2
,
α
α
α
sin α
= I m 2 3 (α cos α − sin α) .
α
This equals zero whenever sin α = 0 or α cos α = sin α; the former is the case for a minima while the
latter is the case for the maxima. The maxima case can also be written as
tan α = α.
(b) Note that as the order of the maxima increases the solutions get closer and closer to odd
integers times π/2. The solutions are
α = 0, 1.43π, 2.46π, etc.
(c) The m values are m = α/π − 1/2, and correspond to
m = 0.5, 0.93, 1.96, etc.
These values will get closer and closer to integers as the values are increased.
218
P42-4
The outgoing beam strikes the moon with a circular spot of radius
r = 1.22λD/a = 1.22(0.69×10−6 m)(3.82×108 m)/(2 × 1.3 m) = 123 m.
The light is not evenly distributed over this circle.
If P0 is the power in the light, then
P0 =
Z
Iθ r dr dφ = 2π
R
Z
Iθ r dr,
0
where R is the radius of the central peak and Iθ is the angular intensity. For a λ we can write
α ≈ πar/λD, then
P0 = 2πIm
λD
πa
2 Z
0
π/2
sin2 α
dα ≈ 2πIm
α
λD
πa
2
(0.82).
Then the intensity at the center falls off with distance D as
2
Im = 1.9 (a/λD) P0
The fraction of light collected by the mirror on the moon is then
P1 /P0 = 1.9
(2 × 1.3 m)
(0.69×10−6 m)(3.82×108 m)
2
π(0.10 m)2 = 5.6×10−6 .
The fraction of light collected by the mirror on the Earth is then
P2 /P1 = 1.9
(2 × 0.10 m)
(0.69×10−6 m)(3.82×108 m)
2
π(1.3 m)2 = 5.6×10−6 .
Finally, P2 /P0 = 3×10−11 .
P42-5 (a) The ring is reddish because it occurs at the blue minimum.
(b) Apply Eq. 42-11 for blue light:
d = 1.22λ/ sin θ = 1.22(400 nm)/ sin(0.375◦ ) = 70 µm.
(c) Apply Eq. 42-11 for red light:
θ = arcsin (1.22(700 nm)/(70 µm)) ≈ 0.7◦ ,
which occurs 3 lunar radii from the moon.
P42-6 The diffraction pattern is a property of the speaker, not the interference between the speakers. The diffraction pattern should be unaffected by the phase shift. The interference pattern,
however, should shift up or down as the phase of the second speaker is varied.
P42-7 (a) The missing fringe at θ = 5◦ is a good hint as to what is going on. There should be
some sort of interference fringe, unless the diffraction pattern has a minimum at that point. This
would be the first minimum, so
a sin(5◦ ) = (440 × 10−9 m)
would be a good measure of the width of each slit. Then a = 5.05 × 10−6 m.
219
(b) If the diffraction pattern envelope were not present we could expect that the fourth interference maxima beyond the central maximum would occur at this point, and then
d sin(5◦ ) = 4(440 × 10−9 m)
yielding
d = 2.02 × 10−5 m.
(c) Apply Eq. 42-17, where β = mπ and
α=
πa mλ
πa
πa
sin θ =
=m
= mπ/4.
λ
λ d
d
Then for m = 1 we have
I1 = (7)
sin(π/4)
(π/4)
2
= 5.7;
2
= 2.8.
while for m = 2 we have
I2 = (7)
sin(2π/4)
(2π/4)
These are in good agreement with the figure.
220
E43-1 (a) d = (21.5×10−3 m)/(6140) = 3.50×10−6 m.
(b) There are a number of angles allowed:
θ
θ
θ
θ
θ
=
=
=
=
=
arcsin[(1)(589×10−9 m)/(3.50×10−6 m)] = 9.7◦ ,
arcsin[(2)(589×10−9 m)/(3.50×10−6 m)] = 19.5◦ ,
arcsin[(3)(589×10−9 m)/(3.50×10−6 m)] = 30.3◦ ,
arcsin[(4)(589×10−9 m)/(3.50×10−6 m)] = 42.3◦ ,
arcsin[(5)(589×10−9 m)/(3.50×10−6 m)] = 57.3◦ .
E43-2 The distance between adjacent rulings is
d=
(2)(612×10−9 m)
= 2.235×10−6 m.
sin(33.2◦ )
The number of lines is then
N = D/d = (2.86×10−2 m)/(2.235×10−6 m) = 12, 800.
E43-3 We want to find a relationship between the angle and the order number which is linear.
We’ll plot the data in this representation, and then use a least squares fit to find the wavelength.
The data to be plotted is
m
1
2
3
θ
17.6◦
37.3◦
65.2◦
sin θ
0.302
0.606
0.908
m
-1
-2
-3
θ
-17.6◦
-37.1◦
-65.0◦
sin θ
-0.302
-0.603
-0.906
On my calculator I get the best straight line fit as
0.302m + 8.33 × 10−4 = sin θm ,
which means that
λ = (0.302)(1.73 µm) = 522 nm.
E43-4 Although an approach like the solution to Exercise 3 should be used, we’ll assume that each
measurement is perfect and error free. Then randomly choosing the third maximum,
λ=
E43-5
d sin θ
(5040×10−9 m) sin(20.33◦ )
=
= 586×10−9 m.
m
(3)
(a) The principle maxima occur at points given by Eq. 43-1,
λ
sin θm = m .
d
The difference of the sine of the angle between any two adjacent orders is
sin θm+1 − sin θm = (m + 1)
λ
λ
λ
−m = .
d
d
d
Using the information provided we can find d from
d=
λ
(600 × 10−9 )
=
= 6 µm.
sin θm+1 − sin θm
(0.30) − (0.20)
221
It doesn’t take much imagination to recognize that the second and third order maxima were given.
(b) If the fourth order maxima is missing it must be because the diffraction pattern envelope has
a minimum at that point. Any fourth order maxima should have occurred at sin θ4 = 0.4. If it is a
diffraction minima then
a sin θm = mλ where sin θm = 0.4
We can solve this expression and find
a=m
(600 × 10−9 m)
λ
=m
= m1.5 µm.
sin θm
(0.4)
The minimum width is when m = 1, or a = 1.5 µm.
(c) The visible orders would be integer values of m except for when m is a multiple of four.
E43-6 (a) Find the maximum integer value of m = d/λ = (930 nm)/(615 nm) = 1.5, hence
m = −1, 0, +1; there are three diffraction maxima.
(b) The first order maximum occurs at
θ = arcsin(615 nm)/(930 nm) = 41.4◦ .
The width of the maximum is
δθ =
(615 nm)
= 7.87×10−4 rad,
(1120)(930 nm) cos(41.4◦ )
or 0.0451◦ .
E43-7 The fifth order maxima will be visible if d/λ ≥ 5; this means
λ≤
d
(1×10−3 m)
=
= 635×10−9 m.
5
(315 rulings)(5)
E43-8 (a) The maximum could be the first, and then
λ=
(1×10−3 m) sin(28◦ )
d sin θ
=
= 2367×10−9 m.
m
(200)(1)
That’s not visible. The first visible wavelength is at m = 4, then
λ=
d sin θ
(1×10−3 m) sin(28◦ )
=
= 589×10−9 m.
m
(200)(4)
The next is at m = 5, then
λ=
d sin θ
(1×10−3 m) sin(28◦ )
=
= 469×10−9 m.
m
(200)(5)
Trying m = 6 results in an ultraviolet wavelength.
(b) Yellow-orange and blue.
222
E43-9
A grating with 400 rulings/mm has a slit separation of
d=
1
= 2.5 × 10−3 mm.
400 mm−1
To find the number of orders of the entire visible spectrum that will be present we need only consider
the wavelength which will be on the outside of the maxima. That will be the longer wavelengths, so
we only need to look at the 700 nm behavior. Using Eq. 43-1,
d sin θ = mλ,
and using the maximum angle 90◦ , we find
m<
(2.5 × 10−6 m)
d
=
= 3.57,
λ
(700 × 10−9 m)
so there can be at most three orders of the entire spectrum.
E43-10 In this case d = 2a. Since interference maxima are given by sin θ = mλ/d while diffraction
minima are given at sin θ = m0 λ/a = 2m0 λ/d then diffraction minima overlap with interference
maxima whenever m = 2m0 . Consequently, all even m are at diffraction minima and therefore
vanish.
E43-11 If the second-order spectra overlaps the third-order, it is because the 700 nm second-order
line is at a larger angle than the 400 nm third-order line.
Start with the wavelengths multiplied by the appropriate order parameter, then divide both side
by d, and finally apply Eq. 43-1.
2(700 nm) > 3(400 nm),
2(700 nm)
3(400 nm)
>
,
d
d
sin θ2,λ=700 > sin θ3,λ=400 ,
regardless of the value of d.
E43-12 Fig. 32-2 shows the path length difference for the right hand side of the grating as d sin θ.
If the beam strikes the grating at ang angle ψ then there will be an additional path length difference
of d sin ψ on the right hand side of the figure. The diffraction pattern then has two contributions to
the path length difference, these add to give
d(sin θ + sin psi) = mλ.
E43-13
E43-14 Let d sin θi = λi and θ1 + 20◦ = θ2 . Then
sin θ2 = sin θ1 cos(20◦ ) + cos θ1 sin(20◦ ).
Rearranging,
sin θ2 = sin θ1 cos(20◦ ) +
q
1 − sin2 θ1 sin(20◦ ).
Substituting the equations together yields a rather nasty expression,
p
λ2
λ1
=
cos(20◦ ) + 1 − (λ1 /d)2 sin(20◦ ).
d
d
223
Rearranging,
2
(λ2 − λ1 cos(20◦ )) = d2 − λ21 sin2 (20◦ ).
Use λ1 = 430 nm and λ2 = 680 nm, then solve for d to find d = 914 nm. This corresponds to 1090
rulings/mm.
E43-15 The shortest wavelength passes through at an angle of
θ1 = arctan(50 mm)/(300 mm) = 9.46◦ .
This corresponds to a wavelength of
λ1 =
(1×10−3 m) sin(9.46◦ )
= 470×10−9 m.
(350)
The longest wavelength passes through at an angle of
θ2 = arctan(60 mm)/(300 mm) = 11.3◦ .
This corresponds to a wavelength of
λ2 =
(1×10−3 m) sin(11.3◦ )
= 560×10−9 m.
(350)
E43-16 (a) ∆λ = λ/R = λ/N m, so
∆λ = (481 nm)/(620 rulings/mm)(5.05 mm)(3) = 0.0512 nm.
(b) mm is the largest integer smaller than d/λ, or
mm ≤ 1/(481×10−9 m)(620 rulings/mm) = 3.35,
so m = 3 is highest order seen.
E43-17
The required resolving power of the grating is given by Eq. 43-10
R=
λ
(589.0 nm)
=
= 982.
∆λ
(589.6 nm) − (589.0 nm)
Our resolving power is then R = 1000.
Using Eq. 43-11 we can find the number of grating lines required. We are looking at the secondorder maxima, so
R
(1000)
N=
=
= 500.
m
(2)
E43-18 (a) N = R/m = λ/m∆λ, so
N=
(415.5 nm)
= 23100.
(2)(415.496 nm − 415.487 nm)
(b) d = w/N , where w is the width of the grating. Then
θ = arcsin
mλ
(23100)(2)(415.5×10−9 m)
= arcsin
= 27.6◦ .
d
(4.15×10−2 m)
224
E43-19 N = R/m = λ/m∆λ, so
N=
(656.3 nm)
= 3650
(1)(0.180 nm)
E43-20 Start with Eq. 43-9:
D=
E43-21
d sin θ/λ
tan θ
m
=
=
.
d cos θ
d cos θ
λ
(a) We find the ruling spacing by Eq. 43-1,
d=
mλ
(3)(589 nm)
=
= 9.98 µm.
sin θm
sin(10.2◦ )
(b) The resolving power of the grating needs to be at least R = 1000 for the third-order line; see
the work for Ex. 43-17 above. The number of lines required is given by Eq. 43-11,
N=
R
(1000)
=
= 333,
m
(3)
so the width of the grating (or at least the part that is being used) is 333(9.98 µm) = 3.3 mm.
E43-22 (a) Condition (1) is satisfied if
d ≥ 2(600 nm)/ sin(30◦ ) = 2400 nm.
The dispersion is maximal for the smallest d, so d = 2400 nm.
(b) To remove the third order requires d = 3a, or a = 800 nm.
E43-23 (a) The angles of the first three orders are
θ1
=
θ2
=
θ3
=
(1)(589×10−9 m)(40000)
= 18.1◦ ,
(76×10−3 m)
(2)(589×10−9 m)(40000)
= 38.3◦ ,
arcsin
(76×10−3 m)
(3)(589×10−9 m)(40000)
arcsin
= 68.4◦ .
(76×10−3 m)
arcsin
The dispersion for each order is
D1
=
D2
=
D3
=
(1)(40000)
360◦
= 3.2×10−2◦ /nm,
(76×106 nm) cos(18.1◦ ) 2π
(2)(40000)
360◦
= 7.7×10−2◦ /nm,
(76×106 nm) cos(38.3◦ ) 2π
(3)(40000)
360◦
= 2.5×10−1◦ /nm.
(76×106 nm) cos(68.4◦ ) 2π
(b) R = N m, so
R1
R2
R3
=
=
=
(40000)(1) = 40000,
(40000)(2) = 80000,
(40000)(3) = 120000.
225
E43-24 d = mλ/2 sin θ, so
d=
E43-25
(2)(0.122 nm)
= 0.259 nm.
2 sin(28.1◦ )
Bragg reflection is given by Eq. 43-12
2d sin θ = mλ,
where the angles are measured not against the normal, but against the plane. The value of d depends
on the family of planes under consideration, but it is at never larger than a0 , the unit cell dimension.
We are looking for the smallest angle; this will correspond to the largest d and the smallest m.
That means m = 1 and d = 0.313 nm. Then the minimum angle is
θ = sin−1
(1)(29.3 × 10−12 m)
= 2.68◦ .
2(0.313 × 10−9 m)
E43-26 2d/λ = sin θ1 and 2d/2λ = sin θ2 . Then
θ2 = arcsin[2 sin(3.40◦ )] = 6.81◦ .
E43-27
We apply Eq. 43-12 to each of the peaks and find the product
mλ = 2d sin θ.
The four values are 26 pm, 39 pm, 52 pm, and 78 pm. The last two values are twice the first two,
so the wavelengths are 26 pm and 39 pm.
E43-28 (a) 2d sin θ = mλ, so
d=
(3)(96.7 pm)
= 171 pm.
2 sin(58.0◦ )
(b) λ = 2(171 pm) sin(23.2◦ )/(1) = 135 pm.
E43-29 The angle against the face of the crystal is 90◦ − 51.3◦ = 38.7◦ . The wavelength is
λ = 2(39.8 pm) sin(38.7◦ )/(1) = 49.8 pm.
E43-30 If λ > 2d then λ/2d > 1. But
λ/2d = sin θ/m.
This means that sin θ > m, but the sine function can never be greater than one.
E43-31 There are too many unknowns. It is only possible to determine the ratio d/λ.
E43-32 A wavelength will be diffracted if mλ = 2d sin θ. The possible solutions are
λ3
λ4
=
=
2(275 pm) sin(47.8)/(3) = 136 pm,
2(275 pm) sin(47.8)/(4) = 102 pm.
226
E43-33
We use Eq. 43-12 to first find d;
d=
mλ
(1)(0.261 × 10−9 m)
=
= 1.45 × 10−10 m.
2 sin θ
2 sin(63.8◦ )
d is the spacing between the planes in Fig. 43-28; it correspond to half of the diagonal distance
between two cell centers. Then
(2d)2 = a20 + a20 ,
or
a0 =
√
2d =
√
2(1.45 × 10−10 m) = 0.205 nm.
E43-34 Diffraction occurs when 2d sin θ = mλ. The angles in this case are then given by
sin θ = m
(0.125×10−9 m)
= (0.248)m.
2(0.252×10−9 m)
There are four solutions to this equation. They are 14.4◦ , 29.7◦ , 48.1◦ , and 82.7◦ . They involve
rotating the crystal from the original orientation (90◦ − 42.4◦ = 47.6◦ ) by amounts
47.6◦ − 14.4◦
47.6◦ − 29.7◦
47.6◦ − 48.1◦
47.6◦ − 82.7◦
=
=
=
=
33.2◦ ,
17.9◦ ,
−0.5◦ ,
−35.1◦ .
P43-1 Since the slits are so narrow we only need to consider interference effects, not diffraction
effects. There are three waves which contribute at any point. The phase angle between adjacent
waves is
φ = 2πd sin θ/λ.
We can add the electric field vectors as was done in the previous chapters, or we can do it in a
different order as is shown in the figure below.
Then the vectors sum to
E(1 + 2 cos φ).
We need to square this quantity, and then normalize it so that the central maximum is the maximum.
Then
(1 + 4 cos φ + 4 cos2 φ)
I = Im
.
9
227
P43-2
(a) Solve φ for I = I m /2, this occurs when
3
√ = 1 + 2 cos φ,
2
or φ = 0.976 rad. The corresponding angle θx is
θx ≈
λφ
λ(0.976)
λ
=
=
.
2πd
2πd
6.44d
But ∆θ = 2θx , so
λ
.
3.2d
(b) For the two slit pattern the half width was found to be ∆θ = λ/2d. The half width in the
three slit case is smaller.
∆θ ≈
P43-3 (a) and (b) A plot of the intensity quickly reveals that there is an alternation of large
maximum, then a smaller maximum, etc. The large maxima are at φ = 2nπ, the smaller maxima
are half way between those values.
(c) The intensity at these secondary maxima is then
I = Im
(1 + 4 cos π + 4 cos2 π)
Im
=
.
9
9
Note that the minima are not located half-way between the maxima!
P43-4 Covering up the middle slit will result in a two slit apparatus with a slit separation of 2d.
The half width, as found in Problem 41-5, is then
∆θ = λ/2(2d), = λ/4d,
which is narrower than before covering up the middle slit by a factor of 3.2/4 = 0.8.
P43-5 (a) If N is large we can treat the phasors as summing to form a flexible “line” of length
N δE. We then assume (incorrectly) that the secondary maxima occur when the loop wraps around
on itself as shown in the figures below. Note that the resultant phasor always points straight up.
This isn’t right, but it is close to reality.
228
The length of the resultant depends on how many loops there are. For k = 0 there are none.
For k = 1 there are one and a half loops. The circumference of the resulting circle is 2N δE/3, the
diameter is N δE/3π. For k = 2 there are two and a half loops. The circumference of the resulting
circle is 2N δE/5, the diameter is N δE/5π. The pattern for higher k is similar: the circumference
is 2N δE/(2k + 1), the diameter is N δE/(k + 1/2)π.
The intensity at this “approximate” maxima is proportional to the resultant squared, or
(N δE)2
.
(k + 1/2)2 π 2
Ik ∝
but I m is proportional to (N δE)2 , so
Ik = I m
1
.
(k + 1/2)2 π 2
(b) Near the middle the vectors simply fold back on one another, leaving a resultant of δE. Then
Ik ∝ (δE)2 =
so
(N δE)2
,
N2
Im
,
N2
(c) Let α have the values which result in sin α = 1, and then the two expressions are identical!
Ik =
P43-6 (a) v = f λ, so δv = f δλ + λδf . Assuming δv = 0, we have δf /f = −δλ/λ. Ignore the
negative sign (we don’t need it here). Then
R=
λ
f
c
=
=
,
∆λ
∆f
λ∆f
and then
c
c
=
.
Rλ
N mλ
(b) The ray on the top gets there first, the ray on the bottom must travel an additional distance
of N d sin θ. It takes a time
∆t = N d sin θ/c
∆f =
to do this.
(c) Since mλ = d sin θ, the two resulting expression can be multiplied together to yield
(∆f )(∆t) =
c N d sin θ
= 1.
N mλ
c
This is almost, but not quite, one of Heisenberg’s uncertainty relations!
P43-7 (b) We sketch parallel lines which connect centers to form almost any right triangle similar
to the one shown in the Fig. 43-18. The triangle will have two sides which have integer
√ multiple
lengths of the lattice spacing a0 . The hypotenuse of the triangle will then have length h2 + k 2 a0 ,
where h and k are the integers. In Fig. 43-18 h = 2 while k = 1. The number of planes which cut
the diagonal is equal to h2 + k 2 if, and only if, h and k are relatively prime. The inter-planar spacing
is then
√
h2 + k 2 a0
a0
.
d=
=√
2
2
2
h +k
h + k2
229
(a) The next five spacings are then
h = 1,
k = 1,
d
h = 1,
k = 2,
d
h = 1,
k = 3,
d
h = 2,
k = 3,
d
h = 1,
k = 4,
d
√
= a0 / 2,
√
= a0 / 5,
√
= a0 / 10,
√
= a0 / 13,
√
= a0 / 17.
P43-8 The middle layer cells will also diffract a beam, but this beam will be exactly π out of phase
with the top layer. The two beams will then cancel out exactly because of destructive interference.
230
E44-1 (a) The direction of propagation is determined by considering the argument of the sine
function. As t increases y must decrease to keep the sine function “looking” the same, so the wave
is propagating in the negative y direction.
(b) The electric field is orthogonal (perpendicular) to the magnetic field (so Ex = 0) and the
direction of motion (so Ey = 0); Consequently, the only non-zero term is Ez . The magnitude of E
~ × B/µ
~ 0 , when B
~ points in the positive
will be equal to the magnitude of B times c. Since ~S = E
~ must point in the negative z direction in order that ~S point in the negative y
x direction then E
direction. Then
Ez = −cB sin(ky + ωt).
(c) The polarization is given by the direction of the electric field, so the wave is linearly polarized
in the z direction.
E44-2 Let one wave be polarized in the x direction and the other in the y direction. Then the net
electric field is given by E 2 = Ex2 + Ey2 , or
E 2 = E02 sin2 (kz − ωt) + sin2 (kz − ωt + β) ,
where β is the phase difference. We can consider any point in space, including z = 0, and then
average the result over a full cycle. Since β merely shifts the integration limits, then the result is
independent of β. Consequently, there are no interference effects.
E44-3 (a) The transmitted intensity is I0 /2 = 6.1×10−3 W/m2 . The maximum value of the electric
field is
p
p
E m = 2µ0 cI = 2(1.26×10−6 H/m)(3.00×108 m/s)(6.1×10−3 W/m2 ) = 2.15 V/m.
(b) The radiation pressure is caused by the absorbed half of the incident light, so
p = I/c = (6.1×10−3 W/m2 )/(3.00×108 m/s) = 2.03×10−11 Pa.
E44-4 The first sheet transmits half the original intensity, the second transmits an amount proportional to cos2 θ. Then I = (I0 /2) cos2 θ, or
p
p
θ = arccos 2I/I0 = arccos 2(I0 /3)/I0 35.3◦ .
E44-5 The first sheet polarizes the un-polarized light, half of the intensity is transmitted, so
I1 = 12 I0 .
The second sheet transmits according to Eq. 44-1,
I2 = I1 cos2 θ =
1
1
I0 cos2 (45◦ ) = I0 ,
2
4
and the transmitted light is polarized in the direction of the second sheet.
The third sheet is 45◦ to the second sheet, so the intensity of the light which is transmitted
through the third sheet is
1
1
I3 = I2 cos2 θ = I0 cos2 (45◦ ) = I0 .
4
8
231
E44-6 The transmitted intensity through the first sheet is proportional to cos2 θ, the transmitted
intensity through the second sheet is proportional to cos2 (90◦ − θ) = sin2 θ. Then
I = I0 cos2 θ sin2 θ = (I0 /4) sin2 2θ,
or
p
p
1
1
arcsin 4I/I0 = arcsin 4(0.100I0 )/I0 = 19.6◦ .
2
2
Note that 70.4◦ is also a valid solution!
θ=
E44-7 The first sheet transmits half of the original intensity; each of the remaining sheets transmits
an amount proportional to cos2 θ, where θ = 30◦ . Then
3
1
1
I
6
=
cos2 θ = (cos(30◦ )) = 0.211
I0
2
2
E44-8 The first sheet transmits an amount proportional to cos2 θ, where θ = 58.8◦ . The second
sheet transmits an amount proportional to cos2 (90◦ − θ) = sin2 θ. Then
I = I0 cos2 θ sin2 θ = (43.3 W/m2 ) cos2 (58.8◦ ) sin2 (58.8◦ ) = 8.50 W/m2 .
E44-9 Since the incident beam is unpolarized the first sheet transmits 1/2 of the original intensity.
The transmitted beam then has a polarization set by the first sheet: 58.8◦ to the vertical. The second
sheet is horizontal, which puts it 31.2◦ to the first sheet. Then the second sheet transmits cos2 (31.2◦ )
of the intensity incident on the second sheet. The final intensity transmitted by the second sheet
can be found from the product of these terms,
1
I = (43.3 W/m2 )
cos2 (31.2◦ ) = 15.8 W/m2 .
2
E44-10 θp = arctan(1.53/1.33) = 49.0◦ .
E44-11 (a) The angle for complete polarization of the reflected ray is Brewster’s angle, and is
given by Eq. 44-3 (since the first medium is air)
θp = tan−1 n = tan−1 (1.33) = 53.1◦ .
(b) Since the index of refraction depends (slightly) on frequency, then so does Brewster’s angle.
E44-12 (b) Since θr + θp = 90◦ , θp = 90◦ − (31.8◦ ) = 58.2◦ .
(a) n = tan θp = tan(58.2◦ ) = 1.61.
E44-13 The angles are between
θp = tan−1 n = tan−1 (1.472) = 55.81◦ .
and
θp = tan−1 n = tan−1 (1.456) = 55.52◦ .
E44-14 The smallest possible thickness t will allow for one half a wavelength phase difference for
the o and e waves. Then ∆nt = λ/2, or
t = (525×10−9 m)/2(0.022) = 1.2×10−5 m.
232
E44-15 (a) The incident wave is at 45◦ to the optical axis. This means that there are two
components; assume they originally point in the +y and +z direction. When they travel through
the half wave plate they are now out of phase by 180◦ ; this means that when one component is in
the +y direction the other is in the −z direction. In effect the polarization has been rotated by 90◦ .
(b) Since the half wave plate will delay one component so that it emerges 180◦ “later” than it
should, it will in effect reverse the handedness of the circular polarization.
(c) Pretend that an unpolarized beam can be broken into two orthogonal linearly polarized
components. Both are then rotated through 90◦ ; but when recombined it looks like the original
beam. As such, there is no apparent change.
E44-16 The quarter wave plate has a thickness of x = λ/4∆n, so the number of plates that can
be cut is given by
N = (0.250×10−3 m)4(0.181)/(488×10−9 m) = 371.
P44-1 Intensity is proportional to the electric field squared, so the original intensity reaching the
eye is I0 , with components I h = (2.3)2 I v , and then
I0 = I h + I v = 6.3I v or I v = 0.16I0 .
Similarly, I h = (2.3)2 I v = 0.84I0 .
(a) When the sun-bather is standing only the vertical component passes, while
(b) when the sun-bather is lying down only the horizontal component passes.
P44-2 The intensity of the transmitted light which was originally unpolarized is reduced to I u /2,
regardless of the orientation of the polarizing sheet. The intensity of the transmitted light which
was originally polarized is between 0 and I p , depending on the orientation of the polarizing sheet.
Then the maximum transmitted intensity is I u /2 + I p , while the minimum transmitted intensity is
I u /2. The ratio is 5, so
I u /2 + I p
Ip
5=
=1+2 ,
I u /2
Iu
or I p /I u = 2. Then the beam is 1/3 unpolarized and 2/3 polarized.
P44-3
Each sheet transmits a fraction
2
cos α = cos
2
θ
N
.
There are N sheets, so the fraction transmitted through the stack is
cos2
θ
N
N
.
We want to evaluate this in the limit as N → ∞.
As N gets larger we can use a small angle approximation to the cosine function,
1
cos x ≈ 1 − x2 for x 1
2
The the transmitted intensity is
1 θ2
1−
2 N2
233
2N
.
This expression can also be expanded in a binomial expansion to get
1 − 2N
1 θ2
,
2 N2
which in the limit as N → ∞ approaches 1.
The stack then transmits all of the light which makes it past the first filter. Assuming the light
is originally unpolarized, then the stack transmits half the original intensity.
P44-4 (a) Stack several polarizing sheets so that the angle between any two sheets is sufficiently
small, but the total angle is 90◦ .
(b) The transmitted intensity fraction needs to be 0.95. Each sheet will transmit a fraction
cos2 θ, where θ = 90◦ /N , with N the number of sheets. Then we want to solve
N
0.95 = cos2 (90◦ /N )
for N . For large enough N , θ will be small, so we can expand the cosine function as
cos2 θ = 1 − sin2 θ ≈ 1 − θ2 ,
so
0.95 ≈ 1 − (π/2N )2
which has solution N = π 2 /4(0.05) = 49.
N
≈ 1 − N (π/2N )2 ,
P44-5 Since passing through a quarter wave plate twice can rotate the polarization of a linearly
polarized wave by 90◦ , then if the light passes through a polarizer, through the plate, reflects off the
coin, then through the plate, and through the polarizer, it would be possible that when it passes
through the polarizer the second time it is 90◦ to the polarizer and no light will pass. You won’t see
the coin.
On the other hand if the light passes first through the plate, then through the polarizer, then is
reflected, the passes again through the polarizer, all the reflected light will pass through he polarizer
and eventually work its way out through the plate. So the coin will be visible.
Hence, side A must be the polarizing sheet, and that sheet must be at 45◦ to the optical axis.
P44-6
(a) The displacement of a ray is given by
tan θk = yk /t,
so the shift is
∆y = t(tan θe − tan θo ).
Solving for each angle,
θe
=
θo
=
1
◦
arcsin
sin(38.8 ) = 24.94◦ ,
(1.486)
1
◦
arcsin
sin(38.8 ) = 22.21◦ .
(1.658)
The shift is then
∆y = (1.12×10−2 m) (tan(24.94) − tan(22.21)) = 6.35×10−4 m.
(b) The e-ray bends less than the o-ray.
(c) The rays have polarizations which are perpendicular to each other; the o-wave being polarized
along the direction of the optic axis.
(d) One ray, then the other, would disappear.
234
P44-7 The method is outline in Sample Problem 44-24; use a polarizing sheet to pick out the o-ray
or the e-ray.
235
E45-1
(a) The energy of a photon is given by Eq. 45-1, E = hf , so
E = hf =
hc
.
λ
Putting in “best” numbers
hc =
(6.62606876×10−34 J · s)
(2.99792458×108 m/s) = 1.23984×10−6 eV · m.
(1.602176462×10−19 C)
This means that hc = 1240 eV · nm is accurate to almost one part in 8000!
(b) E = (1240 eV · nm)/(589 nm) = 2.11 eV.
E45-2 Using the results of Exercise 45-1,
λ=
(1240 eV · nm)
= 2100 nm,
(0.60 eV)
which is in the infrared.
E45-3 Using the results of Exercise 45-1,
E1 =
(1240 eV · nm)
= 3.31 eV,
(375 nm)
E2 =
(1240 eV · nm)
= 2.14 eV,
(580 nm)
and
The difference is ∆E = (3.307 eV) − (2.138 eV) = 1.17 eV.
E45-4 P = E/t, so, using the result of Exercise 45-1,
P = (100/s)
(1240 eV · nm)
= 230 eV/s.
(540 nm)
That’s a small 3.68×10−17 W.
E45-5 When talking about the regions in the sun’s spectrum it is more common to refer to
wavelengths than frequencies. So we will use the results of Exercise 45-1(a), and solve
λ = hc/E = (1240 eV · nm)/E.
The energies are between E = (1.0×1018 J)/(1.6×10−19 C) = 6.25 eV and E = (1.0×1016 J)/(1.6×
10−19 C) = 625 eV. These energies correspond to wavelengths between 198 nm and 1.98 nm; this is
the ultraviolet range.
E45-6 The energy per photon is E = hf = hc/λ. The intensity is power per area, which is energy
per time per area, so
P
E
nhc
hc n
I=
=
=
=
.
A
At
λAt
λA t
But R = n/t is the rate of photons per unit time. Since h and c are constants and I and A are equal
for the two beams, we have R1 /λ1 = R2 /λ2 , or
R1 /R2 = λ1 /λ2 .
236
E45-7 (a) Since the power is the same, the bulb with the larger energy per photon will emits
fewer photons per second. Since longer wavelengths have lower energies, the bulb emitting 700 nm
must be giving off more photons per second.
(b) How many more photons per second? If E1 is the energy per photon for one of the bulbs,
then N1 = P/E1 is the number of photons per second emitted. The difference is then
N1 − N2 =
P
P
P
−
=
(λ1 − λ2 ),
E1
E2
hc
or
N1 −N2 =
(130 W)
((700×10−9 m) − (400×10−9 m)) = 1.96×1020 .
(6.63×10−34 J·s)(3.00×108 m/s)
E45-8 Using the results of Exercise 45-1, the energy of one photon is
E=
(1240 eV · nm)
= 1.968 eV,
(630 nm)
The total light energy given off by the bulb is
E t = P t = (0.932)(70 W)(730 hr)(3600 s/hr) = 1.71×108 J.
The number of photons is
n=
Et
(1.71×108 J)
=
= 5.43×1026 .
E0
(1.968 eV)(1.6×10−19 J/eV)
E45-9 Apply Wien’s law, Eq. 45-4, λmax T = 2898 µm · K; so
T =
(2898 µm · K)
= 91×106 K.
(32×10−12 m)
Actually, the wavelength was supposed to be 32 µm. Then the temperature would be 91 K.
E45-10 Apply Wien’s law, Eq. 45-4, λmax T = 2898 µm · K; so
λ=
(2898 µm · K)
= 1.45 m.
(0.0020 K)
This is in the radio region, near 207 on the FM dial.
E45-11 The wavelength of the maximum spectral radiancy is given by Wien’s law, Eq. 45-4,
λmax T = 2898 µm · K.
Applying to each temperature in turn,
(a) λ = 1.06×10−3 m, which is in the microwave;
(b) λ = 9.4×10−6 m, which is in the infrared;
(c) λ = 1.6×10−6 m, which is in the infrared;
(d) λ = 5.0×10−7 m, which is in the visible;
(e) λ = 2.9×10−10 m, which is in the x-ray;
(f) λ = 2.9×10−41 m, which is in a hard gamma ray.
237
E45-12 (a) Apply Wien’s law, Eq. 45-4, λmax T = 2898 µm · K.; so
λ=
(2898 µm · K)
= 5.00×10−7 m.
(5800 K)
That’s blue-green.
(b) Apply Wien’s law, Eq. 45-4, λmax T = 2898 µm · K.; so
T =
(2898 µm · K)
= 5270 K.
(550×10−9 m)
E45-13 I = σT 4 and P = IA. Then
P = (5.67×10−8 W/m2 · K4 )(1900 K)4 π(0.5×10−3 m)2 = 0.58 W.
E45-14 Since I ∝ T 4 , doubling T results in a 24 = 16 times increase in I. Then the new power
level is
(16)(12.0 mW) = 192 mW.
E45-15
(a) We want to apply Eq. 45-6,
R(λ, T ) =
1
2πc2 h
.
5
hc/λkT
λ e
−1
We know the ratio of the spectral radiancies at two different wavelengths. Dividing the above
equation at the first wavelength by the same equation at the second wavelength,
λ51 ehc/λ1 kT − 1
,
3.5 = 5 hc/λ kT
2
λ2 e
−1
where λ1 = 200 nm and λ2 = 400 nm. We can considerably simplify this expression if we let
x = ehc/λ2 kT ,
because since λ2 = 2λ1 we would have
ehc/λ1 kT = e2hc/λ2 kT = x2 .
Then we get
5 2
1
x −1
1
3.5 =
=
(x + 1).
2
x−1
32
We will use the results of Exercise 45-1 for the exponents and then rearrange to get
T =
hc
(3.10 eV)
=
= 7640 K.
λ1 k ln(111)
(8.62×10−5 eV/K) ln(111)
(b) The method is the same, except that instead of 3.5 we have 1/3.5; this means the equation
for x is
1
1
=
(x + 1),
3.5
32
with solution x = 8.14, so then
T =
hc
(3.10 eV)
=
= 17200 K.
λ1 k ln(8.14)
(8.62×10−5 eV/K) ln(8.14)
238
E45-16 hf = φ, so
f=
φ
(5.32 eV)
=
= 1.28×1015 Hz.
h
(4.14×10−15 eV · s)
E45-17 We’ll use the results of Exercise 45-1. Visible red light has an energy of
E=
(1240 eV · nm)
= 1.9 eV.
(650 nm)
The substance must have a work function less than this to work with red light. This means that
only cesium will work with red light. Visible blue light has an energy of
E=
(1240 eV · nm)
= 2.75 eV.
(450 nm)
This means that barium, lithium, and cesium will work with blue light.
E45-18 Since K m = hf − φ,
K m = (4.14×10−15 eV · s)(3.19×1015 Hz) − (2.33 eV) = 10.9 eV.
E45-19
(a) Use the results of Exercise 45-1 to find the energy of the corresponding photon,
E=
hc
(1240 eV · nm)
=
= 1.83 eV.
λ
(678 nm)
Since this energy is less than than the minimum energy required to remove an electron then the
photo-electric effect will not occur.
(b) The cut-off wavelength is the longest possible wavelength of a photon that will still result in
the photo-electric effect occurring. That wavelength is
λ=
(1240 eV · nm)
(1240 eV · nm)
=
= 544 nm.
E
(2.28 eV)
This would be visible as green.
E45-20 (a) Since K m = hc/λ − φ,
Km =
(1240 eV · nm)
(4.2 eV) = 2.0 eV.
(200 nm)
(b) The minimum kinetic energy is zero; the electron just barely makes it off the surface.
(c) V s = K m /q = 2.0 V.
(d) The cut-off wavelength is the longest possible wavelength of a photon that will still result in
the photo-electric effect occurring. That wavelength is
λ=
(1240 eV · nm)
(1240 eV · nm)
=
= 295 nm.
E
(4.2 eV)
E45-21 K m = qV s = 4.92 eV. But K m = hc/λ − φ, so
λ=
(1240 eV · nm)
= 172 nm.
(4.92 eV + 2.28 eV)
239
E45-22 (a) K m = qV s and K m = hc/λ − φ. We have two different values for qV s and λ, so
subtracting this equation from itself yields
q(V s,1 − V s,2 ) = hc/λ1 − hc/λ2 .
Solving for λ2 ,
λ2
=
=
=
hc
,
hc/λ1 − q(V s,1 − V s,2 )
(1240 eV · nm)
,
(1240 eV · nm)/(491 nm) − (0.710 eV) + (1.43 eV)
382 nm.
(b) K m = qV s and K m = hc/λ − φ, so
φ = (1240 eV · nm)/(491 nm) − (0.710 eV) = 1.82 eV.
E45-23
(a) The stopping potential is given by Eq. 45-11,
h
φ
f− ,
e
e
V0 =
so
V0 =
(1240 eV · nm) (1.85 eV
−
= 1.17 V.
e(410 nm
e
(b) These are not relativistic electrons, so
p
p
p
v = 2K/m = c 2K/mc2 = c 2(1.17 eV)/(0.511×106 eV) = 2.14×10−3 c,
or v = 64200 m/s.
E45-24 It will have become the stopping potential, or
h
φ
f− ,
e
e
V0 =
so
V0 =
(2.49 eV)
(4.14×10−15 eV · m)
(6.33×1014 /s) −
= 0.131 V.
(1.0e)
(1.0e)
E45-25
E45-26 (a) Using the results of Exercise 45-1,
λ=
(1240 eV · nm)
= 62 pm.
(20×103 eV)
(b) This is in the x-ray region.
E45-27 (a) Using the results of Exercise 45-1,
E=
(1240 eV · nm)
= 29, 800 eV.
(41.6×10−3 nm)
(b) f = c/λ = (3×108 m/s)/(41.6 pm) = 7.21×1018 /s.
(c) p = E/c = 29, 800 eV/c = 2.98×104 eV/c.
240
E45-28 (a) E = hf , so
f=
(0.511×106 eV)
= 1.23×1020 /s.
(4.14×10−15 eV · s)
(b) λ = c/f = (3×108 m/s)/(1.23×1020 /s) = 2.43 pm.
(c) p = E/c = (0.511×106 eV)/c.
E45-29 The initial momentum of the system is the momentum of the photon, p = h/λ. This
momentum is imparted to the sodium atom, so the final speed of the sodium is v = p/m, where m
is the mass of the sodium. Then
v=
h
(6.63×10−34 J · s)
=
= 2.9 cm/s.
λm
(589×10−9 m)(23)(1.7×10−27 kg)
E45-30 (a) λC = h/mc = hc/mc2 , so
λC =
(1240 eV · nm)
= 2.43 pm.
(0.511×106 eV)
(c) Since Eλ = hf = hc/λ, and λ = h/mc = hc/mc2 , then
Eλ = hc/λ = mc2 .
(b) See part (c).
E45-31 The change in the wavelength of a photon during Compton scattering is given by Eq.
45-17,
h
λ0 = λ +
(1 − cos φ).
mc
We’ll use the results of Exercise 45-30 to save some time, and let h/mc = λC , which is 2.43 pm.
(a) For φ = 35◦ ,
λ0 = (2.17 pm) + (2.43 pm)(1 − cos 35◦ ) = 2.61 pm.
(b) For φ = 115◦ ,
λ0 = (2.17 pm) + (2.43 pm)(1 − cos 115◦ ) = 5.63 pm.
E45-32 (a) We’ll use the results of Exercise 45-1:
(1240 eV · nm)
= 2.43 pm.
(0.511×106 eV)
λ=
(b) The change in the wavelength of a photon during Compton scattering is given by Eq. 45-17,
λ0 = λ +
h
(1 − cos φ).
mc
We’ll use the results of Exercise 45-30 to save some time, and let h/mc = λC , which is 2.43 pm.
λ0 = (2.43 pm) + (2.43 pm)(1 − cos 72◦ ) = 4.11 pm.
(c) We’ll use the results of Exercise 45-1:
E=
(1240 eV · nm)
= 302 keV.
(4.11 pm)
241
E45-33
45-17,
The change in the wavelength of a photon during Compton scattering is given by Eq.
h
(1 − cos φ).
mc
We are not using the expression with the form ∆λ because ∆λ and ∆E are not simply related.
The wavelength is related to frequency by c = f λ, while the frequency is related to the energy
by Eq. 45-1, E = hf . Then
λ0 − λ =
∆E
= E − E 0 = hf − hf 0 ,
1
1
= hc
− 0 ,
λ λ
0
λ −λ
= hc
.
λ λ0
Into this last expression we substitute the Compton formula. Then
∆E =
h2 (1 − cos φ)
.
m
λ λ0
Now E = hf = hc/λ, and we can divide this on both sides of the above equation. Also, λ0 = c/f 0 ,
and we can substitute this into the right hand side of the above equation. Both of these steps result
in
∆E
hf 0
=
(1 − cos φ).
E
mc2
Note that mc2 is the rest energy of the scattering particle (usually an electron), while hf 0 is the
energy of the scattered photon.
E45-34 The wavelength is related to frequency by c = f λ, while the frequency is related to the
energy by Eq. 45-1, E = hf . Then
∆E
∆E
E
= E − E 0 = hf − hf 0 ,
1
1
= hc
−
,
λ λ0
λ0 − λ
,
= hc
λ λ0
∆λ
=
,
λ + ∆λ
But ∆E/E = 3/4, so
3λ + 3∆λ = 4∆λ,
or ∆λ = 3λ.
E45-35 The maximum shift occurs when φ = 180◦ , so
∆λm = 2
h
(1240 eV · nm)
=2
= 2.64×10−15 m.
mc
938 MeV)
E45-36 Since E = hf frequency shifts are identical to energy shifts. Then we can use the results
of Exercise 45-33 to get
(0.9999)(6.2 keV)
(0.0001) =
(1 − cos φ),
(511 keV)
which has solution φ = 7.4◦ .
(b) (0.0001)(6.2 keV) = 0.62 eV.
242
E45-37
45-17,
(a) The change in wavelength is independent of the wavelength and is given by Eq.
∆λ =
(1240 eV · nm)
hc
(1 − cos φ) = 2
= 4.85×10−3 nm.
2
mc
(0.511×106 eV)
(b) The change in energy is given by
∆E
hc hc
− ,
λf
λ
i
1
1
−
,
= hc
λi + ∆λ λi
1
1
= (1240 eV · nm)
−
= −42.1 keV
(9.77 pm) + (4.85 pm) (9.77 pm)
=
(c) This energy went to the electron, so the final kinetic energy of the electron is 42.1 keV.
E45-38 For φ = 90◦ ∆λ = h/mc. Then
∆E
E
=
=
hf 0
λ
=1−
,
hf
λ + ∆λ
h/mc
.
λ + h/mc
1−
But h/mc = 2.43 pm for the electron (see Exercise 45-30).
(a) ∆E/E = (2.43 pm)/(3.00 cm + 2.43 pm) = 8.1×10−11 .
(b) ∆E/E = (2.43 pm)/(500 nm + 2.43 pm) = 4.86×10−6 .
(c) ∆E/E = (2.43 pm)/(0.100 nm + 2.43 pm) = 0.0237.
(d) ∆E/E = (2.43 pm)/(1.30 pm + 2.43 pm) = 0.651.
E45-39 We can use the results of Exercise 45-33 to get
(0.10) =
(0.90)(215 keV)
(1 − cos φ),
(511 keV)
which has solution φ = 42/6◦ .
E45-40 (a) A crude estimate is that the photons can’t arrive more frequently than once every
10 − 8s. That would provide an emission rate of 108 /s.
(b) The power output would be
P = (108 )
(1240 eV · nm)
= 2.3×108 eV/s,
(550 nm)
which is 3.6×10−11 W!
E45-41 We can follow the example of Sample Problem 45-6, and apply
λ = λ0 (1 − v/c).
(a) Solving for λ0 ,
λ0 =
(588.995 nm)
= 588.9944 nm.
(1 − (−300 m/s)(3×108 m/s)
243
(b) Applying Eq. 45-18,
∆v = −
h
(6.6×10−34 J · s)
=−
= 3×10−2 m/s.
mλ
(22)(1.7×10−27 kg)(590×10−9 m)
(c) Emitting another photon will slow the sodium by about the same amount.
E45-42 (a) (430 m/s)/(0.15 m/s) ≈ 2900 interactions.
(b) If the argon averages a speed of 220 m/s, then it requires interactions at the rate of
(2900)(220 m/s)/(1.0 m) = 6.4×105 /s
if it is going to slow down in time.
P45-1 The radiant intensity is given by Eq. 45-3, I = σT 4 . The power that is radiated through
the opening is P = IA, where A is the area of the opening. But energy goes both ways through the
opening; it is the difference that will give the net power transfer. Then
P net = (I0 − I1 )A = σA T04 − T14 .
Put in the numbers, and
P net = (5.67×10−8 W/m2 ·K4 )(5.20 × 10−4 m2 ) (488 K)4 − (299 K)4 = 1.44 W.
P45-2
(a) I = σT 4 and P = IA. Then T 4 = P/σA, or
s
(100 W)
T = 4
= 3248 K.
(5.67×10−8 W/m2 · K4 )π(0.28×10−3 m)(1.8×10−2 m)
That’s 2980◦ C.
(b) The rate that energy is radiated off is given by dQ/dt = mC dT /dt. The mass is found from
m = ρV , where V is the volume. This can be combined with the power expression to yield
σAT 4 = −ρV CdT /dt,
which can be integrated to yield
∆t =
ρV C
(1/T23 − 1/T13 ).
3σA
Putting in numbers,
∆t
=
=
(19300kg/m3 )(0.28×10−3 m)(132J/kgC)
[1/(2748 K)3 − 1/(3248 K)3 ],
3(5.67×10−8 W/m2 · K4 )(4)
20 ms.
P45-3 Light from the sun will “heat-up” the thin black screen. As the temperature of the screen
increases it will begin to radiate energy. When the rate of energy radiation from the screen is equal
to the rate at which the energy from the sun strikes the screen we will have equilibrium. We need
first to find an expression for the rate at which energy from the sun strikes the screen.
The temperature of the sun is T S . The radiant intensity is given by Eq. 45-3, I S = σT S 4 . The
total power radiated by the sun is the product of this radiant intensity and the surface area of the
sun, so
P S = 4πrS 2 σT S 4 ,
244
where rS is the radius of the sun.
Assuming that the lens is on the surface of the Earth (a reasonable assumption), then we can
find the power incident on the lens if we know the intensity of sunlight at the distance of the Earth
from the sun. That intensity is
PS
PS
IE =
=
,
A
4πRE 2
where RE is the radius of the Earth’s orbit. Combining,
I E = σT S
4
rS
RE
2
The total power incident on the lens is then
P lens = I E Alens = σT S
4
rS
RE
2
πrl 2 ,
where rl is the radius of the lens. All of the energy that strikes the lens is focused on the image, so
the power incident on the lens is also incident on the image.
The screen radiates as the temperature increases. The radiant intensity is I = σT 4 , where T is
the temperature of the screen. The power radiated is this intensity times the surface area, so
P = IA = 2πri 2 σT 4 .
The factor of “2” is because the screen has two sides, while ri is the radius of the image. Set this
equal to P lens ,
2
rS
2πri 2 σT 4 = σT S 4
πrl 2 ,
RE
or
2
1
rS rl
T 4 = T S4
.
2
RE r i
The radius of the image of the sun divided by the radius of the sun is the magnification of the
lens. But magnification is also related to image distance divided by object distance, so
ri
i
= |m| = ,
rS
o
Distances should be measured from the lens, but since the sun is so far from the Earth, we won’t be
far off in stating o ≈ RE . Since the object is so far from the lens, the image will be very, very close
to the focal point, so we can also state i ≈ f . Then
ri
f
=
,
rS
RE
so the expression for the temperature of the thin black screen is considerably simplified to
T4 =
1 4
TS
2
2
rl
.
f
Now we can put in some of the numbers.
T =
1
21/4
(5800 K)
s
(1.9 cm)
= 1300 K.
(26 cm)
245
P45-4
The derivative of R with respect to λ is
−10
hc
π c2 h
hc
λ6 (e( λ k T ) − 1)
+
2 π c3 h2 e( λ k T )
hc
λ7 (e( λ k T ) − 1)2 k T
.
Ohh, that’s ugly. Setting it equal to zero allows considerable simplification, and we are left with
(5 − x)ex = 5,
where x = hc/λkT . The solution is found numerically to be x = 4.965114232. Then
λ=
(1240 eV · nm)
2.898×10−3 m · K
=
.
−5
(4.965)(8.62×10 eV/K)T
T
P45-5 (a) If the planet has a temperature T , then the radiant intensity of the planet will be
IσT 4 , and the rate of energy radiation from the planet will be
P = 4πR2 σT 4 ,
where R is the radius of the planet.
A steady state planet temperature requires that the energy from the sun arrive at the same rate
as the energy is radiated from the planet. The intensity of the energy from the sun a distance r
from the sun is
P sun /4πr2 ,
and the total power incident on the planet is then
P = πR2
P sun
.
4πr2
Equating,
4πR2 σT 4
T4
= πR2
=
P sun
,
4πr2
P sun
.
16πσr2
(b) Using the last equation and the numbers from Problem 3,
s
(6.96×108 m)
1
= 279 K.
T = √ (5800 K)
(1.5×1011 m)
2
That’s about 43◦ F.
P45-6 (a) Change variables as suggested, then λ = hc/xkT and dλ = −(hc/x2 kT )dx. Integrate
(note the swapping of the variables of integration picks up a minus sign):
Z
2πc2 h (hc/x2 kT )dx
I =
,
(hc/xkT )5
ex − 1
Z
2πk 4 T 4
x3 dx
=
,
h3 c2
ex − 1
2π 5 k 4 4
=
T .
15h3 c2
246
P45-7
(a) P = E/t = nhf /t = (hc/λ)(n/t), where n/t is the rate of photon emission. Then
n/t =
(100 W)(589×10−9 m)
= 2.96×1020 /s.
(6.63×10−34 J · s)(3.00×108 m/s)
(b) The flux at a distance r is the rate divided by the area of the sphere of radius r, so
s
(2.96×1020 /s)
r=
= 4.8×107 m.
4π(1×104 /m2 · s)
(c) The photon density is the flux divided by the speed of light; the distance is then
s
(2.96×1020 /s)
r=
= 280 m.
4π(1×106 /m3 )(3×108 m/s)
(d) The flux is given by
(2.96×1020 /s)
= 5.89×1018 /m2 · s.
4π(2.0 m)2
The photon density is
(5.89×1018 m2 · s)/(3.00×108 m/s) = 1.96×1010 /m3 .
P45-8
Momentum conservation requires
pλ = pe ,
while energy conservation requires
Eλ + mc2 = Ee .
Square both sides of the energy expression and
Eλ2 + 2Eλ mc2 + m2 c4
Eλ2 + 2Eλ mc2
2 2
pλ c + 2Eλ mc2
= Ee2 = p2e c2 + m2 c4 ,
= p2e c2 ,
= p2e c2 .
But the momentum expression can be used here, and the result is
2Eλ mc2 = 0.
Not likely.
P45-9
be
(a) Since qvB = mv 2 /r, v = (q/m)rB The kinetic energy of (non-relativistic) electrons will
K=
or
K=
1
1 q 2 (rB)2
mv 2 =
,
2
2
m
1 (1.6×10−19 C)
(188×10−6 T · m)2 = 3.1×103 eV.
2 (9.1×10−31 kg)
(b) Use the results of Exercise 45-1,
φ=
(1240 eV · nm)
− 3.1×103 eV = 1.44×104 eV.
(71×10−3 nm)
247
P45-10
P45-11
then
(a) The maximum value of ∆λ is 2h/mc. The maximum energy lost by the photon is
∆E
hc hc
− ,
λf
λ
i
1
1
= hc
−
,
λi + ∆λ λi
−2h/mc
= hc
,
λ(λ + 2h/mc)
=
where in the last line we wrote λ for λi . The energy given to the electron is the negative of this, so
Kmax =
2h2
.
mλ(λ + 2h/mc)
Multiplying through by 12 = (Eλ/hc)2 we get
Kmax =
2E 2
.
mc2 (1 + 2hc/λmc2 )
or
Kmax =
E2
.
mc2 /2 + E
(b) The answer is
Kmax =
(17.5 keV)2
= 1.12 keV.
(511 eV)/2 + (17.5 keV)
248
E46-1
(a) Apply Eq. 46-1, λ = h/p. The momentum of the bullet is
p = mv = (0.041 kg)(960 m/s) = 39kg · m/s,
so the corresponding wavelength is
λ = h/p = (6.63×10−34 J · s)/(39kg · m/s) = 1.7×10−35 m.
(b) This length is much too small to be significant. How much too small? If the radius of the
galaxy were one meter, this distance would correspond to the diameter of a proton.
E46-2 (a) λ = h/p and p2 /2m = K, then
λ= √
hc
2mc2 K
1.226 nm
(1240 eV · nm)
√
√ =
.
=p
K
2(511 keV) K
(b) K = eV , so
1.226 nm
=
λ= √
eV
r
1.5 V
nm.
V
√
E46-3 For non-relativistic particles λ = h/p and p2 /2m = K, so λ = hc/ 2mc2 K.
(a) For the electron,
(1240 eV · nm)
λ= p
= 0.0388 nm.
2(511 keV)(1.0 keV)
(c) For the neutron,
(1240 MeV · fm)
λ= p
2(940 MeV)(0.001 MeV)
= 904 fm.
(b) For ultra-relativistic particles K ≈ E ≈ pc, so
λ=
hc
(1240 eV · nm)
=
= 1.24 nm.
E
(1000 eV)
E46-4 For non-relativistic particles p = h/λ and p2 /2m = K, so K = (hc)2 /2mc2 λ2 . Then
K=
E46-5
(1240 eV · nm)2
= 4.34×10−6 eV.
2(5.11×106 eV)(589 nm)2
(a) Apply Eq. 46-1, p = h/λ. The proton speed would then be
v=
h
hc
(1240 MeV · fm)
=c 2 =c
= 0.0117c.
mλ
mc λ
(938 MeV)(113 fm)
This is good, because it means we were justified in using the non-relativistic equations. Then
v = 3.51×106 m/s.
(b) The kinetic energy of this electron would be
K=
1
1
mv 2 = (938 MeV)(0.0117)2 = 64.2 keV.
2
2
The potential through which it would need to be accelerated is 64.2 kV.
249
√
E46-6 (a) K = qV and p = 2mK. Then
p
p = 2(22)(932 MeV/c2 )(325 eV) = 3.65×106 eV/c.
(b) λ = h/p, so
λ=
(1240 eV · nm)
hc
=
= 3.39×10−4 nm.
pc
(3.65×106 eV/c)c
√
E46-7 (a) For non-relativistic particles λ = h/p and p2 /2m = K, so λ = hc/ 2mc2 K. For the
alpha particle,
(1240 MeV · fm)
λ= p
= 5.2 fm.
2(4)(932 MeV)(7.5 MeV)
(b) Since the wavelength of the alpha is considerably smaller than the distance to the nucleus
we can ignore the wave nature of the alpha particle.
E46-8 (a) For non-relativistic particles p = h/λ and p2 /2m = K, so K = (hc)2 /2mc2 λ2 . Then
K=
(1240 keV · pm)2
= 15 keV.
2(511keV)(10 pm)2
(b) For ultra-relativistic particles K ≈ E ≈ pc, so
E=
E46-9
(1240 keV · pm)
hc
=
= 124 keV.
λ
(10 pm)
The relativistic relationship between energy and momentum is
E 2 = p2 c2 + m2 c4 ,
and if the energy is very large (compared to mc2 ), then the contribution of the mass to the above
expression is small, and
E 2 ≈ p2 c2 .
Then from Eq. 46-1,
λ=
h
hc
hc
(1240 MeV · f m)
=
=
=
= 2.5×10−2 fm.
p
pc
E
(50×103 MeV)
E46-10 (a) K = 3kT /2, p =
λ
=
=
√
2mK, and λ = h/p, so
hc
h
=√
,
3mkT
3mc2 kT
(1240 MeV · f m)
p
= 74 pm.
3(4)(932MeV)(8.62×10−11 MeV/K)(291 K)
√
(b) pV = N kT ; assuming that each particle occupies a cube of volume d3 = V0 then the interparticle spacing is d, so
s
p
(1.38×10−23 J/K)(291 K)
3
d = V /N = 3
= 3.4 nm.
(1.01×105 Pa)
250
E46-11 p = mv and p = h/λ, so m = h/λv. Taking the ratio,
λv
me
=
= (1.813×10−4 )(3) = 5.439×10−4 .
m
λe ve
The mass of the unknown particle is then
m=
(0.511 MeV/c2 )
= 939.5 MeV.
(5.439×10−4 )
That would make it a neutron.
√
E46-12 (a) For non-relativistic particles λ = h/p and p2 /2m = K, so λ = hc/ 2mc2 K.
For the electron,
(1240 eV · nm)
λ= p
= 1.0 nm.
2(5.11×105 eV)(1.5 eV)
For ultra-relativistic particles K ≈ E ≈ pc, so for the photon
λ=
hc
(1240 eV · nm)
=
= 830 nm.
E
(1.5 eV)
(b) Electrons with energies that high are ultra-relativistic. Both the photon and the electron will
then have the same wavelength;
λ=
E46-13
hc
(1240 MeV · fm)
=
= 0.83 fm.
E
(1.5 GeV)
(a) The classical expression for kinetic energy is
√
p = 2mK,
so
λ=
h
hc
(1240 keV · pm)
=√
=p
= 7.76 pm.
2
p
2(511 keV)(25.0 keV)
2mc K
(a) The relativistic expression for momentum is
p
p
pc = sqrtE 2 − m2 c4 = (mc2 + K)2 − m2 c4 = K 2 + 2mc2 K.
Then
λ=
(1240 keV · pm)
hc
= 7.66 pm.
=p
pc
(25.0 keV)2 + 2(511 keV)(25.0 keV)
E46-14 We want to match the wavelength of the gamma to that of the electron. For the gamma,
λ = hc/Eγ . For the electron, K = p2 /2m = h2 /2mλ2 . Combining,
K=
Eγ2
h2
2
E
=
.
γ
2mh2 c2
2mc2
With numbers,
K=
(136keV)2
= 18.1 keV.
2(511 keV)
That would require an accelerating potential of 18.1 kV.
251
E46-15 First find the√wavelength of the neutrons. For non-relativistic particles λ = h/p and
p2 /2m = K, so λ = hc/ 2mc2 K. Then
(1240 keV · pm)
λ= p
2(940×103 keV)(4.2×10−3 keV)
= 14 pm.
Bragg reflection occurs when 2d sin θ = λ, so
θ = arcsin(14 pm)/2(73.2 pm) = 5.5◦ .
E46-16 This is merely a Bragg reflection problem. Then 2d sin θ = mλ, or
θ
θ
θ
= arcsin(1)(11 pm)/2(54.64 pm) = 5.78◦ ,
= arcsin(2)(11 pm)/2(54.64 pm) = 11.6◦ ,
= arcsin(3)(11 pm)/2(54.64 pm) = 17.6◦ .
E46-17 (a) Since sin 52◦ = 0.78, then 2(λ/d) = 1.57 > 1, so there is no diffraction order other
than the first.
(b) For an accelerating potential of 54 volts we have λ/d = 0.78. Increasing the potential will
increase the kinetic energy, increase the momentum, and decrease the wavelength. d won’t change.
The
√ kinetic energy is increased by a factor of 60/54 = 1.11, the momentum increases by a factor of
1.11 = 1.05, so the wavelength changes by a factor of 1/1.05 = 0.952. The new angle is then
θ = arcsin(0.952 × 0.78) = 48◦ .
E46-18 First find the√wavelength of the electrons. For non-relativistic particles λ = h/p and
p2 /2m = K, so λ = hc/ 2mc2 K. Then
(1240 keV · pm)
λ= p
2(511 keV)(0.380 keV)
= 62.9 pm.
This is now a Bragg reflection problem. Then 2d sin θ = mλ, or
θ
θ
θ
θ
θ
θ
θ
θ
θ
=
=
=
=
=
=
=
=
=
arcsin(1)(62.9
arcsin(2)(62.9
arcsin(3)(62.9
arcsin(4)(62.9
arcsin(5)(62.9
arcsin(6)(62.9
arcsin(7)(62.9
arcsin(8)(62.9
arcsin(9)(62.9
pm)/2(314
pm)/2(314
pm)/2(314
pm)/2(314
pm)/2(314
pm)/2(314
pm)/2(314
pm)/2(314
pm)/2(314
pm) = 5.74◦ ,
pm) = 11.6◦ ,
pm) = 17.5◦ ,
pm) = 23.6◦ ,
pm) = 30.1◦ ,
pm) = 36.9◦ ,
pm) = 44.5◦ ,
pm) = 53.3◦ ,
pm) = 64.3◦ .
But the odd orders vanish (see Chapter 43 for a discussion on this).
E46-19 Since ∆f · ∆t ≈ 1/2π, we have
∆f = 1/2π(0.23 s) = 0.69/s.
252
E46-20 Since ∆f · ∆t ≈ 1/2π, we have
∆f = 1/2π(0.10×10−9 s) = 1.6×1010 /s.
The bandwidth wouldn’t fit in the frequency allocation!
E46-21
Apply Eq. 46-9,
∆E ≥
h
4.14×10−15 eV · s)
=
= 7.6×10−5 eV.
2π∆t
2π(8.7×10−12 s)
This is much smaller than the photon energy.
E46-22 Apply Heisenberg twice:
∆E1 =
4.14×10−15 eV · s)
= 5.49×10−8 eV.
2π(12×10−9 s)
∆E2 =
4.14×10−15 eV · s)
= 2.86×10−8 eV.
2π(23×10−9 s)
and
The sum is ∆E transition = 8.4×10−8 eV.
E46-23 Apply Heisenberg:
∆p =
6.63×10−34 J · s)
= 8.8×10−24 kg · m/s.
2π(12×10−12 m)
E46-24 ∆p = (0.5 kg)(1.2 s) = 0.6 kg · m/s. The position uncertainty would then be
∆x =
E46-25
(0.6 J/s)
= 0.16 m.
2π(0.6 kg · m/s)
We want v ≈ ∆v, which means p ≈ ∆p. Apply Eq. 46-8, and
∆x ≥
h
h
≈
.
2π∆p
2πp
According to Eq. 46-1, the de Broglie wavelength is related to the momentum by
λ = h/p,
so
∆x ≥
λ
.
2π
E46-26 (a) A particle confined in a (one dimensional) box of size L will have a position uncertainty
of no more than ∆x ≈ L. The momentum uncertainty will then be no less than
∆p ≥
so
∆p ≈
h
h
≈
.
2π∆x
2πL
(6.63×10−34 J · s)
= 1×10−24 kg · m/s.
2π(×10−10 m)
253
(b) Assuming that p ≈ ∆p, we have
h
,
2πL
and then the electron will have a (minimum) kinetic energy of
p≥
E≈
or
E≈
h2
p2
≈
.
2
2m
8π mL2
(hc)2
(1240 keV · pm)2
=
= 0.004 keV.
8π 2 mc2 L2
8π 2 (511 keV)(100 pm)2
E46-27 (a) A particle confined in a (one dimensional) box of size L will have a position uncertainty of no more than ∆x ≈ L. The momentum uncertainty will then be no less than
∆p ≥
so
∆p ≈
h
h
≈
.
2π∆x
2πL
(6.63×10−34 J · s)
= 1×10−20 kg · m/s.
2π(×10−14 m)
(b) Assuming that p ≈ ∆p, we have
h
,
2πL
and then the electron will have a (minimum) kinetic energy of
p≥
E≈
or
E≈
p2
h2
≈
.
2
2m
8π mL2
(1240 MeV · fm)2
(hc)2
=
= 381 MeV.
8π 2 mc2 L2
8π 2 (0.511 MeV)(10 fm)2
This is so large compared to the mass energy of the electron that we must consider relativistic effects.
It will be very relativistic (381 0.5!), so we can use E = pc as was derived in Exercise 9. Then
E=
hc
(1240 MeV · fm)
=
= 19.7 MeV.
2πL
2π(10 fm)
This is the total energy; so we subtract 0.511 MeV to get K = 19 MeV.
E46-28 We want to find L when T = 0.01. This means solving
E
E
T = 16
1−
e−2kL ,
U0
U0
(5.0 eV)
(5.0 eV) −2k0 L
(0.01) = 16
1−
e
,
(6.0 eV)
(6.0 eV)
0
= 2.22e−2k L ,
ln(4.5×10−3 ) = 2(5.12×109 /m)L,
5.3×10−10 m = L.
254
E46-29
The wave number, k, is given by
k=
2π p
2mc2 (U0 − E).
hc
(a) For the proton mc2 = 938 MeV, so
k=
p
2π
2(938 MeV)(10 MeV − 3.0 MeV) = 0.581 fm−1 .
(1240 MeV · fm)
The transmission coefficient is then
(3.0 MeV)
(3.0 MeV) −2(0.581 fm−1 )(10 fm)
T = 16
1−
e
= 3.0×10−5 .
(10 MeV)
(10 MeV)
(b) For the deuteron mc2 = 2 × 938 MeV, so
k=
p
2π
2(2)(938 MeV)(10 MeV − 3.0 MeV) = 0.821 fm−1 .
(1240 MeV · fm)
The transmission coefficient is then
(3.0 MeV)
(3.0 MeV) −2(0.821 fm−1 )(10 fm)
T = 16
1−
e
= 2.5×10−7 .
(10 MeV)
(10 MeV)
E46-30 The wave number, k, is given by
k=
2π p
2mc2 (U0 − E).
hc
(a) For the proton mc2 = 938 MeV, so
k=
p
2π
2(938 MeV)(6.0 eV − 5.0 eV) = 0.219 pm−1 .
(1240 keV · pm)
We want to find T . This means solving
E
E
T = 16
1−
e−2kL ,
U0
U0
(5.0 eV)
(5.0 eV) −2(0.219×1012 )(0.7×10−9 )
= 16
1−
e
,
(6.0 eV)
(6.0 eV)
=
1.6×10−133 .
A current of 1 kA corresponds to
N = (1×103 C/s)/(1.6×10−19 C) = 6.3×1021 /s
protons per seconds. The time required for one proton to pass is then
t = 1/(6.3×1021 /s)(1.6×10−133 ) = 9.9×10110 s.
That’s 10104 years!
255
P46-1
We will interpret low energy to mean non-relativistic. Then
λ=
h
h
.
=√
p
2mn K
The diffraction pattern is then given by
d sin θ = mλ = mh/
p
2mn K,
where m is diffraction order while mn is the neutron mass. We want to investigate the spread by
taking the derivative of θ with respect to K,
mh
d cos θ dθ = − √
dK.
2 2mn K 3
Divide this by the original equation, and then
cos θ
dK
dθ = −
.
sin θ
2K
Rearrange, change the differential to a difference, and then
∆θ = tan θ
∆K
.
2K
We dropped the negative sign out of laziness; but the angles are in radians, so we need to multiply
by 180/π to convert to degrees.
P46-2
P46-3
We want to solve
E
T = 16
U0
E
1−
U0
e−2kL ,
for E. Unfortunately, E is contained in k since
k=
2π p
2mc2 (U0 − E).
hc
We can do this by iteration. The maximum possible value for
E
E
1−
U0
U0
is 1/4; using this value we can get an estimate for k:
(0.001) = 16(0.25)e−2kL ,
ln(2.5×10−4 ) = −2k(0.7 nm),
5.92/ nm = k.
Now solve for E:
E
= U0 − (hc)2 k 2 8mc2 π 2 ,
(1240 eV · nm)2 (5.92/nm)2
= (6.0 eV) −
,
8π 2 (5.11×105 eV)
= 4.67 eV.
256
Put this value for E back into the transmission equation to find a new k:
E
E
1−
e−2kL ,
T = 16
U0
U0
(4.7 eV)
(4.7 eV) −2kL
(0.001) = 16
1−
e
,
(6.0 eV)
(6.0 eV)
ln(3.68×10−4 ) = −2k(0.7 nm),
5.65/ nm = k.
Now solve for E using this new, improved, value for k:
= U0 − (hc)2 k 2 8mc2 π 2 ,
(1240 eV · nm)2 (5.65/nm)2
= (6.0 eV) −
,
8π 2 (5.11×105 eV)
= 4.78 eV.
E
Keep at it. You’ll eventually stop around E = 5.07 eV.
P46-4 (a) A one percent increase in the barrier height means U0 = 6.06 eV.
For the electron mc2 = 5.11×105 eV, so
k=
p
2π
2(5.11×105 eV)(6.06 eV − 5.0 eV) = 5.27 nm−1 .
(1240 eV · nm)
We want to find T . This means solving
E
E
T = 16
1−
e−2kL ,
U0
U0
(5.0 eV)
(5.0 eV)
= 16
1−
e−2(5.27)(0.7) ,
(6.06 eV)
(6.06 eV)
=
1.44×10−3 .
That’s a 16% decrease.
(b) A one percent increase in the barrier thickness means L = 0.707 nm.
For the electron mc2 = 5.11×105 eV, so
k=
p
2π
2(5.11×105 eV)(6.0 eV − 5.0 eV) = 5.12 nm−1 .
(1240 eV · nm)
We want to find T . This means solving
E
E
T = 16
1−
e−2kL ,
U0
U0
(5.0 eV)
(5.0 eV) −2(5.12)(0.707)
= 16
1−
e
,
(6.0 eV)
(6.0 eV)
=
1.59×10−3 .
That’s a 8.1% decrease.
(c) A one percent increase in the incident energy means E = 5.05 eV.
For the electron mc2 = 5.11×105 eV, so
k=
p
2π
2(5.11×105 eV)(6.0 eV − 5.05 eV) = 4.99 nm−1 .
(1240 eV · nm)
257
We want to find T . This means solving
E
E
T = 16
1−
e−2kL ,
U0
U0
(5.05 eV)
(5.05 eV) −2(4.99)(0.7)
= 16
1−
e
,
(6.0 eV)
(6.0 eV)
=
1.97×10−3 .
That’s a 14% increase.
P46-5
First, the rule for exponents
ei(a+b) = eia eib .
Then apply Eq. 46-12, eiθ = cos θ + i sin θ,
cos(a + b) + i sin(a + b) = (cos a + i sin a)(sin b + i sin b).
Expand the right hand side, remembering that i2 = −1,
cos(a + b) + i sin(a + b) = cos a cos b + i cos a sin b + i sin a cos b − sin a sin b.
Since the real part of the left hand side must equal the real part of the right and the imaginary part
of the left hand side must equal the imaginary part of the right, we actually have two equations.
They are
cos(a + b) = cos a cos b − sin a sin b
and
sin(a + b) = cos a sin b + sin a cos b.
P46-6
258
E47-1
(a) The ground state energy level will be given by
E1 =
(6.63 × 10−34 J · s)2
h2
=
= 3.1 × 10−10 J.
2
8mL
8(9.11 × 10−31 kg)(1.4 × 10−14 m)2
The answer is correct, but the units make it almost useless. We can divide by the electron charge to
express this in electron volts, and then E = 1900 MeV. Note that this is an extremely relativistic
quantity, so the energy expression loses validity.
(b) We can repeat what we did above, or we can apply a “trick” that is often used in solving
these problems. Multiplying the top and the bottom of the energy expression by c2 we get
E1 =
(hc)2
8(mc2 )L2
Then
(1240 MeV · fm)2
= 1.0 MeV.
8(940 MeV)(14 fm)2
(c) Finding an neutron inside the nucleus seems reasonable; but finding the electron would not.
The energy of such an electron is considerably larger than binding energies of the particles in the
nucleus.
E1 =
E47-2 Solve
En =
n2 (hc)2
8(mc2 )L2
for L, then
nhc
,
8mc2 En
(3)(1240 eV · nm)
= p
,
8(5.11×105 eV)(4.7 eV)
= 0.85 nm.
L =
√
E4 − E1
=
E47-3 Solve for E4 − E1 :
=
=
42 (hc)2
12 (hc)2
−
,
8(mc2 )L2
8(mc2 )L2
(16 − 1)(1240 eV · nm)2
,
8(5.11×105 )(0.253 nm)2
88.1 eV.
E47-4 Since E ∝ 1/L2 , doubling the width of the well will lower the ground state energy to
(1/2)2 = 1/4, or 0.65 eV.
E47-5 (a) Solve for E2 − E1 :
E2 − E1
=
=
=
22 h2
12 h2
−
,
8mL2
8mL2
(3)(6.63×10−34 J · s)2
,
8(40)(1.67×10−27 kg)(0.2 m)2
6.2×10−41 J.
(b) K = 3kT /2 = 3(1.38×10−23 J/K)(300 K)/2 = 6.21×10−21 . The ratio is 1×10−20 .
(c) T = 2(6.2×10−41 J)/3(1.38×10−23 J/K) = 3.0×10−18 K.
259
E47-6 (a) The fractional difference is (En+1 − En )/En , or
2
2
∆En
h2
2 h
2 h
=
(n + 1)2
−
n
/
n
,
En
8mL2
8mL2
8mL2
(n + 1)2 − n2
=
,
n2
2n + 1
=
.
n2
(b) As n → ∞ the fractional difference goes to zero; the system behaves as if it is continuous.
E47-7
Then
(a) We will take advantage of the “trick” that was developed in part (b) of Exercise 47-1.
En = n 2
(1240 eV · nm)2
(hc)2
2
=
(15)
= 8.72 keV.
8mc2 L
8(0.511 × 106 eV)(0.0985 nm)2
(b) The magnitude of the momentum is exactly known because E = p2 /2m. This momentum is
given by
√
p
pc = 2mc2 E = 2(511 keV)(8.72 keV) = 94.4 keV.
What we don’t know is in which direction the particle is moving. It is bouncing back and forth
between the walls of the box, so the momentum could be directed toward the right or toward the
left. The uncertainty in the momentum is then
∆p = p
which can be expressed in terms of the box size L by
r
√
n 2 h2
nh
∆p = p = 2mE =
=
.
2
4L
2L
(c) The uncertainty in the position is 98.5 pm; the electron could be anywhere inside the well.
E47-8 The probability distribution function is
P2 =
2
2πx
sin2
.
L
L
We want to integrate over the central third, or
Z L/6
2
2πx
P =
sin2
dx,
L
L
−L/6
Z
1 π/3
=
sin2 θ dθ,
π −π/3
= 0.196.
E47-9 (a) Maximum probability occurs when the argument of the cosine (sine) function is kπ
([k + 1/2]π). This occurs when
x = N L/2n
for odd N .
(b) Minimum probability occurs when the argument of the cosine (sine) function is [k + 1/2]π
(kπ). This occurs when
x = N L/2n
for even N .
260
E47-10 In Exercise 47-21 we show that the hydrogen levels can be written as
En = −(13.6 eV)/n2 .
(a) The Lyman series is the series which ends on E1 . The least energetic state starts on E2 . The
transition energy is
E2 − E1 = (13.6 eV)(1/12 − 1/22 ) = 10.2 eV.
The wavelength is
λ=
(1240 eV · nm)
hc
=
= 121.6 nm.
∆E
(10.2 eV)
(b) The series limit is
0 − E1 = (13.6 eV)(1/12 ) = 13.6 eV.
The wavelength is
λ=
E47-11
hc
(1240 eV · nm)
=
= 91.2 nm.
∆E
(13.6 eV)
The ground state of hydrogen, as given by Eq. 47-21, is
E1 = −
me4
(9.109 × 10−31 kg)(1.602 × 10−19 C)4
=
−
= 2.179 × 10−18 J.
820 h2
8(8.854 × 10−12 F/m)2 (6.626 × 10−34 J · s)2
In terms of electron volts the ground state energy is
E1 = −(2.179 × 10−18 J)/(1.602 × 10−19 C) = −13.60 eV.
E47-12 In Exercise 47-21 we show that the hydrogen levels can be written as
En = −(13.6 eV)/n2 .
(c) The transition energy is
∆E = E3 − E1 = (13.6 eV)(1/12 − 1/32 ) = 12.1 eV.
(a) The wavelength is
λ=
hc
(1240 eV · nm)
=
= 102.5 nm.
∆E
(12.1 eV)
(b) The momentum is
p = E/c = 12.1 eV/c.
E47-13 In Exercise 47-21 we show that the hydrogen levels can be written as
En = −(13.6 eV)/n2 .
(a) The Balmer series is the series which ends on E2 . The least energetic state starts on E3 . The
transition energy is
E3 − E2 = (13.6 eV)(1/22 − 1/32 ) = 1.89 eV.
The wavelength is
λ=
hc
(1240 eV · nm)
=
= 656 nm.
∆E
(1.89 eV)
261
(b) The next energetic state starts on E4 . The transition energy is
E4 − E2 = (13.6 eV)(1/22 − 1/42 ) = 2.55 eV.
The wavelength is
λ=
hc
(1240 eV · nm)
=
= 486 nm.
∆E
(2.55 eV)
(c) The next energetic state starts on E5 . The transition energy is
E5 − E2 = (13.6 eV)(1/22 − 1/52 ) = 2.86 eV.
The wavelength is
λ=
hc
(1240 eV · nm)
=
= 434 nm.
∆E
(2.86 eV)
(d) The next energetic state starts on E6 . The transition energy is
E6 − E2 = (13.6 eV)(1/22 − 1/62 ) = 3.02 eV.
The wavelength is
λ=
hc
(1240 eV · nm)
=
= 411 nm.
∆E
(3.02 eV)
(e) The next energetic state starts on E7 . The transition energy is
E7 − E2 = (13.6 eV)(1/22 − 1/72 ) = 3.12 eV.
The wavelength is
λ=
hc
(1240 eV · nm)
=
= 397 nm.
∆E
(3.12 eV)
E47-14 In Exercise 47-21 we show that the hydrogen levels can be written as
En = −(13.6 eV)/n2 .
The transition energy is
∆E =
hc
(1240 eV · nm)
=
= 10.20 eV.
λ
(121.6 nm)
This must be part of the Lyman series, so the higher state must be
En = (10.20 eV) − (13.6 eV) = −3.4 eV.
That would correspond to n = 2.
E47-15 The binding energy is the energy required to remove the electron. If the energy of the
electron is negative, then that negative energy is a measure of the energy required to set the electron
free.
The first excited state is when n = 2 in Eq. 47-21. It is not necessary to re-evaluate the constants
in this equation every time, instead, we start from
En =
E1
where E1 = −13.60 eV.
n2
Then the first excited state has energy
E2 =
(−13.6 eV)
= −3.4 eV.
(2)2
The binding energy is then 3.4 eV.
262
E47-16 rn = a0 n2 , so
n=
E47-17
p
(847 pm)/(52.9 pm) = 4.
(a) The energy of this photon is
E=
hc
(1240 eV · nm)
=
= 0.96739 eV.
λ
(1281.8 nm)
The final state of the hydrogen must have an energy of no more than −0.96739, so the largest
possible n of the final state is
p
n < 13.60 eV/0.96739 eV = 3.75,
so the final n is 1, 2, or 3. The initial state is only slightly higher than the final state. The jump
from n = 2 to n = 1 is too large (see Exercise 15), any other initial state would have a larger energy
difference, so n = 1 is not the final state.
So what level might be above n = 2? We’ll try
p
n = 13.6 eV/(3.4 eV − 0.97 eV) = 2.36,
which is so far from being an integer that we don’t need to look farther. The n = 3 state has energy
13.6 eV/9 = 1.51 eV. Then the initial state could be
p
n = 13.6 eV/(1.51 eV − 0.97 eV) = 5.01,
which is close enough to 5 that we can assume the transition was n = 5 to n = 3.
(b) This belongs to the Paschen series.
E47-18 In Exercise 47-21 we show that the hydrogen levels can be written as
En = −(13.6 eV)/n2 .
(a) The transition energy is
∆E = E4 − E1 = (13.6 eV)(1/12 − 1/42 ) = 12.8 eV.
(b) All transitions n → m are allowed for n ≤ 4 and m < n. The transition energy will be of the
form
En − Em = (13.6 eV)(1/m2 − 1/n2 ).
The six possible results are 12.8 eV, 12.1 eV, 10.2 eV, 2.55 eV, 1.89 eV, and 0.66 eV.
E47-19 ∆E = h/2π∆t, so
∆E = (4.14×10−15 eV · s)/2π(1×10−8 s) = 6.6×10−8 eV.
E47-20 (a) According to electrostatics and uniform circular motion,
mv 2 /r = e2 /4π0 r2 ,
or
v=
s
e2
==
4π0 mr
s
263
e4
e2
=
.
420 h2 n2
20 hn
Putting in the numbers,
v=
(1.6×10−19 C)2
2.18×106 m/s
=
.
2(8.85×10−12 F/m)(6.63×10−34 J · s)n
n
In this case n = 1.
(b) λ = h/mv,
λ = (6.63×10−34 J · s)/(9.11×10−31 kg)(2.18×106 m/s) = 3.34×10−10 m.
(c) λ/a0 = (3.34×10−10 m)/(5.29×10−11 ) = 6.31 ≈ 2π. Actually, it is exactly 2π.
E47-21 In order to have an inelastic collision with the 6.0 eV neutron there must exist a transition
with an energy difference of less than 6.0 eV. For a hydrogen atom in the ground state E1 = −13.6 eV
the nearest state is
E2 = (−13.6 eV)/(2)2 = −3.4 eV.
Since the difference is 10.2 eV, it will not be possible for the 6.0 eV neutron to have an inelastic
collision with a ground state hydrogen atom.
E47-22 (a) The atom is originally in the state n given by
p
n = (13.6 eV)/(0.85 eV) = 4.
The state with an excitation energy of 10.2 eV, is
p
n = (13.6 eV)/(13.6 eV − 10.2 eV) = 2.
The transition energy is then
∆E = (13.6 eV)(1/22 − 1/42 ) = 2.55 eV.
E47-23 According to electrostatics and uniform circular motion,
mv 2 /r = e2 /4π0 r2 ,
or
v=
s
e2
==
4π0 mr
s
e4
420 h2 n2
The de Broglie wavelength is then
λ=
h
20 hn
=
.
mv
me2
The ratio of λ/r is
λ
20 hn
=
= kn,
r
me2 a0 n2
where k is a constant. As n → ∞ the ratio goes to zero.
264
=
e2
.
20 hn
E47-24 In Exercise 47-21 we show that the hydrogen levels can be written as
En = −(13.6 eV)/n2 .
The transition energy is
∆E = E4 − E1 = (13.6 eV)(1/12 − 1/42 ) = 12.8 eV.
The momentum of the emitted photon is
p = E/c = (12.8 eV)/c.
This is the momentum of the recoiling hydrogen atom, which then has velocity
v=
p
pc
(12.8 eV)
=
c=
(3.00×108 m/s) = 4.1 m/s.
m
mc2
(932 MeV)
E47-25 The first Lyman line is the n = 1 to n = 2 transition. The second Lyman line is the
n = 1 to n = 3 transition. The first Balmer line is the n = 2 to n = 3 transition. Since the photon
frequency is proportional to the photon energy (E = hf ) and the photon energy is the energy
difference between the two levels, we have
fn→m =
Em − E n
h
where the En is the hydrogen atom energy level. Then
f1→3
E 3 − E1
,
h
E 3 − E2 + E2 − E1
E3 − E2
E2 − E 1
=
=
+
,
h
h
h
= f2→3 + f1→2 .
=
E47-26 Use
En = −Z 2 (13.6 eV)/n2 .
(a) The ionization energy of the ground state of He+ is
En = −(2)2 (13.6 eV)/(1)2 = 54.4 eV.
(b) The ionization energy of the n = 3 state of Li2+ is
En = −(3)2 (13.6 eV)/(3)2 = 13.6 eV.
E47-27
(a) The energy levels in the He+ spectrum are given by
En = −Z 2 (13.6 eV)/n2 ,
where Z = 2, as is discussed in Sample Problem 47-6. The photon wavelengths for the n = 4 series
are then
hc
hc/E4
λ=
=
,
En − E 4
1 − En /E4
265
which can also be written as
λ
=
=
=
16hc/(54.4 eV)
,
1 − 16/n2
16hcn2 /(54.4 eV)
,
n2 − 16
Cn2
,
n2 − 16
where C = hc/(3.4 eV) = 365 nm.
(b) The wavelength of the first line is the transition from n = 5,
λ=
(365 nm)(5)2
= 1014 nm.
(5)2 − (4)2
The series limit is the transition from n = ∞, so
λ = 365 nm.
(c) The series starts in the infrared (1014 nm), and ends in the ultraviolet (365 nm). So it must
also include some visible lines.
E47-28 We answer these questions out of order!
(a) n = 1.
(b) r = a0 = 5.29×10−11 m.
(f) According to electrostatics and uniform circular motion,
mv 2 /r = e2 /4π0 r2 ,
or
v=
s
e2
==
4π0 mr
s
e4
420 h2 n2
=
e2
.
20 hn
Putting in the numbers,
v=
(1.6×10−19 C)2
= 2.18×106 m/s.
2(8.85×10−12 F/m)(6.63×10−34 J · s)(1)
(d) p = (9.11×10−31 kg)(2.18×106 m/s) = 1.99×10−24 kg · m/s.
(e) ω = v/r = (2.18×106 m/s)/(5.29×10−11 m) = 4.12×1016 rad/s.
(c) l = pr = (1.99×10−24 kg · m/s)(5.29×10−11 m) = 1.05×10−34 J · s.
(g) F = mv 2 /r, so
F = (9.11×10−31 kg)(2.18×106 m/s)2 /(5.29×10−11 m) = 8.18×10−8 N.
(h) a = (8.18×10−8 N)/(9.11×10−31 kg) = 8.98×1022 m/s2 .
(i) K = mv 2 /r, or
K=
(9.11×10−31 kg)(2.18×106 m/s)2
= 2.16×10−18 J = 13.6 eV.
2
(k) E = −13.6 eV.
(j) P = E − K = (−13.6 eV) − (13.6 eV) = −27.2 eV.
266
E47-29 For each r in the quantity we have a factor of n2 .
(a) n is proportional to n.
(b) r is proportional to p
n2 .
(f) v is proportional to 1/r, or 1/n.
(d) p is proportional to v, or 1/n.
(e) ω is proportional to v/r, or 1/n3 .
(c) l is proportional to pr, or n.
(g) f is proportional to v 2 /r, or 1/n4 .
(h) a is proportional to F , or 1/n4 .
(i) K is proportional to v 2 , or 1/n2 .
(j) E is proportional to 1/n2 .
(k) P is proportional to 1/n2 .
E47-30 (a) Using the results of Exercise 45-1,
E1 =
(1240 eV · nm)
= 1.24×105 eV.
(0.010 nm)
(b) Using the results of Problem 45-11,
E2
(1.24×105 eV)2
=
= 40.5×104 eV.
mc2 /2 + E
(5.11×105 eV)/2 + (1.24×105 eV)
Kmax =
(c) This would likely knock the electron way out of the atom.
E47-31 The energy of the photon in the series limit is given by
E limit = (13.6 eV)/n2 ,
where n = 1 for Lyman, n = 2 for Balmer, and n = 3 for Paschen. The wavelength of the photon is
λlimit =
(1240 eV · nm) 2
n = (91.17 nm)n2 .
(13.6 eV)
The energy of the longest wavelength comes from the transition from the nearest level, or
E long =
(−13.6 eV) (−13.6 eV)
2n + 1
−
= (13.6 eV)
.
(n + 1)2
n2
[n(n + 1)]2
The wavelength of the photon is
λlong =
(1240 eV · nm)[n(n + 1)]2
[n(n + 1)]2
= (91.17 nm)
.
2
(13.6 eV)n
2n + 1
(a) The wavelength interval λlong − λlimit , or
∆λ = (91.17 nm)
n2 (n + 1)2 − n2 (2n + 1)
n4
= (91.17 nm)
.
2n + 1
2n + 1
For n = 1, ∆λ = 30.4 nm. For n = 2, ∆λ = 292 nm. For n = 3, ∆λ = 1055 nm.
(b) The frequency interval is found from
∆f =
E limit − E long
(13.6 eV)
1
(3.29×1015 /s)
=
=
.
−15
2
h
(4.14×10 eV · s) (n + 1)
(n + 1)2
For n = 1, ∆f = 8.23×1014 Hz. For n = 2, ∆f = 3.66×1014 Hz. For n = 3, ∆f = 2.05×1014 Hz.
267
E47-32
E47-33
(a) We’ll use Eqs. 47-25 and 47-26. At r = 0
ψ 2 (0) =
1 −2(0)/a0
1
e
=
= 2150 nm−3 ,
3
πa0
πa30
while
P (0) = 4π(0)2 ψ 2 (0) = 0.
(b) At r = a0 we have
ψ 2 (a0 ) = f rac1πa30 e−2(a0 )/a0 =
e−2
= 291 nm−3 ,
πa30
and
P (a0 ) = 4π(a0 )2 ψ 2 (a0 ) = 10.2 nm−1 .
E47-34 Assume that ψ(a0 ) is a reasonable estimate for ψ(r) everywhere inside the small sphere.
Then
e−2
0.1353
=
.
ψ2 =
πa30
πa30
The probability of finding it in a sphere of radius 0.05a0 is
Z 0.05a0
4
(0.1353)4πr2 dr
= (0.1353)(0.05)3 = 2.26×10−5 .
3
πa
3
0
0
E47-35 Using Eq. 47-26 the ratio of the probabilities is
e−2
P (a0 )
(a0 )2 e−2(a0 )/a0
=
= −4 = 1.85.
2
−2(2a
)/a
0
0
P (2a0 )
4e
(2a0 ) e
E47-36 The probability is
P
=
Z
1.016a0
a0
4r2 e−2r/a0
dr,
a30
Z
1 2.032 2 −u
u e du,
2 2
= 0.00866.
=
E47-37 If l = 3 then ml can be 0, ±1, ±2, or ±3.
(a) From Eq. 47-30, Lz = ml h/2π.. p
So Lz can equal 0, ±h/2π, ±h/π, or ±3h/2π.
(b) From Eq. 47-31, θ = arccos(ml / l(l + 1)), so θ can equal 90◦ , 73.2◦ , 54.7◦ , or 30.0◦ .
~ is given by Eq. 47-28,
(c) The magnitude of L
L=
p
√
h
l(l + 1)
= 3h/π.
2π
E47-38 The maximum possible value of ml is 5. Apply Eq. 47-31:
(5)
θ = arccos p
(5)(5 + 1)
268
= 24.1◦ .
E47-39 Use the hint.
∆p · ∆x =
r
∆p ∆x =
r
∆x
∆p · r
=
r
∆L · ∆θ
=
h
,
2π
h
,
2π
h
,
2π
h
.
2π
E47-40 Note that there is a typo in the formula; P (r) must have dimensions of one over length.
The probability is
Z ∞ 4 −r/a0
r e
dr,
P =
24a50
0
Z ∞
1
=
u4 e−u du,
24 0
= 1.00
What does it mean? It means that if we look for the electron, we will find it somewhere.
E47-41 (a) Find the maxima by taking the derivative and setting it equal to zero.
r(2a − r)(4a2 − 6ra + r2 ) −r
dP
=
e = 0.
dr
8a60
The solutions are r = 0, r = 2a, and 4a2 − 6ra + r2 = 0. The first two correspond to minima (see
Fig. 47-14). The other two are the solutions to the quadratic, or r = 0.764a0 and r = 5.236a0 .
(b) Substitute these two values into Eq. 47-36. The results are
P (0.764a0 ) = 0.981 nm−1 .
and
P (5.236a0 ) = 3.61 nm−1 .
E47-42 The probability is
P
5.01a0
=
Z
=
0.01896.
5.00a0
r2 (2 − r/a0 )2 e−r/a0
dr,
8a30
E47-43 n = 4 and l = 3, while ml can be any of
−3, −2, −1, 0, 1, 2, 3,
while ms can be either −1/2 or 1/2. There are 14 possible states.
E47-44 n must be greater than l, so n ≥ 4. |ml | must be less than or equal to l, so |ml | ≤ 3. ms
is −1/2 or 1/2.
E47-45
If ml = 4 then l ≥ 4. But n ≥ l + 1, so n > 4. We only know that ms = ±1/2.
269
E47-46 There are 2n2 states in a shell n, so if n = 5 there are 50 states.
E47-47 Each is in the n = 1 shell, the l = 0 angular momentum state, and the ml = 0 state. But
one is in the state ms = +1/2 while the other is in the state ms = −1/2.
E47-48 Apply Eq. 47-31:
(+1/2)
θ = arccos p
(1/2)(1/2 + 1)
and
(−1/2)
θ = arccos p
(1/2)(1/2 + 1)
= 54.7◦
= 125.3◦ .
E47-49 All of the statements are true.
E47-50 There are n possible values for l (start at 0!). For each value of l there are 2l + 1 possible
values for ml . If n = 1, the sum is 1. If n = 2, the sum is 1 + 3 = 4. If n = 3, the sum is 1 + 3 + 5 = 9.
The pattern is clear, the sum is n2 . But there are two spin states, so the number of states is 2n2 .
P47-1
We can simplify the energy expression as
E = E0 n2x + n2y + n2z
where E0 =
h2
.
8mL2
To find the lowest energy levels we need to focus on the values of nx , ny , and nz .
It doesn’t take much imagination to realize that the set (1, 1, 1) will result in the smallest value
for n2x + n2y + n2z . The next choice is to set one of the values equal to 2, and try the set (2, 1, 1).
Then it starts to get harder, as the next lowest might be either (2, 2, 1) or (3, 1, 1). The only way
to find out is to try. I’ll tabulate the results for you:
nx
1
2
2
3
2
ny
1
1
2
1
2
nz
1
1
1
1
2
n2x + n2y + n2z
3
6
9
11
12
Mult.
1
3
3
3
1
nx
3
3
4
3
4
ny
2
2
1
3
2
nz
1
2
1
1
1
n2x + n2y + n2z
14
17
18
19
21
Mult.
6
3
3
3
6
We are now in a position to state the five lowest energy levels. The fundamental quantity is
E0 =
(hc)2
(1240 eV · nm)2
=
= 6.02×10−6 eV.
2
2
8mc L
8(0.511×106 eV)(250 nm)2
The five lowest levels are found by multiplying this fundamental quantity by the numbers in the
table above.
P47-2 (a) Write the states between 0 and L. Then all states, odd or even, can be written with
probability distribution function
2
nπx
P (x) = sin2
,
L
L
270
we find the probability of finding the particle in the region 0 ≤ x ≤ L/3 is
P
Z
L/3
nπx
2
cos2
dx,
L
L
0
1
sin(2nπ/3)
1−
.
3
2nπ/3
=
=
(b) If n = 1 use the formula and P = 0.196.
(c) If n = 2 use the formula and P = 0.402.
(d) If n = 3 use the formula and P = 0.333.
(e) Classically the probability distribution function is uniform, so there is a 1/3 chance of finding
it in the region 0 to L/3.
P47-3 The region of interest is small compared to the variation in P (x); as such we can approximate the probability with the expression P = P (x)∆x.
(b) Evaluating,
P
=
=
=
2
4πx
sin2
∆x,
L
L
2
4π(L/8)
sin2
(0.0003L),
L
L
0.0006.
(b) Evaluating,
P
=
=
=
P47-4
2
4πx
sin2
∆x,
L
L
2
4π(3L/16)
sin2
(0.0003L),
L
L
0.0003.
(a) P = Ψ∗ Ψ, or
2
P = A20 e−2πmωx
/h
.
(b) Integrating,
1
Z
∞
= A20
r
h
2πmω
= A20
r
h p
pi,
2πmω
=
A20
e−2πmωx
2
/h
dx,
−∞
r
4
2mω
h
= A0 .
(c) x = 0.
P47-5
We will want an expression for
d2
ψ0 .
dx2
271
Z
∞
−∞
2
e−u du,
Doing the math one derivative at a time,
d2
d
d
ψ
=
ψ
0
0 ,
dx2
dx dx
2
d =
A0 (−2πmωx/h)e−πmωx /h ,
dx
2
2
= A0 (−2πmωx/h)2 e−πmωx /h + A0 (−2πmω/h)e−πmωx /h ,
2
= (2πmωx/h)2 − (2πmω/h) A0 e−πmωx /h ,
= (2πmωx/h)2 − (2πmω/h) ψ0 .
In the last line we factored out ψ0 . This will make our lives easier later on.
Now we want to go to Schrödinger’s equation, and make some substitutions.
h2 d 2
ψ 0 + U ψ0
8π 2 m dx2
(2πmωx/h)2 − (2πmω/h) ψ0 + U ψ0
−
−
h2
8π 2 m
−
h2
(2πmωx/h)2 − (2πmω/h) + U
8π 2 m
= Eψ0 ,
= Eψ0 ,
= E,
where in the last line we divided through by ψ0 . Now for some algebra,
U
h2
(2πmωx/h)2 − (2πmω/h) ,
8π 2 m
mω 2 x2
hω
= E+
−
.
2
4π
= E+
But we are given that E = hω/4π, so this simplifies to
mω 2 x2
2
U=
which looks like a harmonic oscillator type potential.
P47-6 Assume the electron is originally in the state n. The classical frequency of the electron is
f0 , where
f0 = v/2πr.
According to electrostatics and uniform circular motion,
mv 2 /r = e2 /4π0 r2 ,
or
v=
Then
f0 =
s
e2
==
4π0 mr
s
e4
e2
=
.
420 h2 n2
20 hn
e2 1 πme2
me4
−2E1
=
=
20 hn 2π 0 h2 n2
420 h3 n3
hn3
Here E1 = −13.6 eV.
Photon frequency is related to energy according to f = ∆Enm /h, where ∆Enm is the energy of
transition from state n down to state m. Then
E1 1
1
f=
,
−
h n2
m2
272
where E1 = −13.6 eV. Combining the fractions and letting m = n − δ, where δ is an integer,
f
=
=
=
≈
=
E1 m2 − n2
,
h m2 n2
−E1 (n − m)(m + n)
,
h
m2 n2
−E1 δ(2n + δ)
,
h (n + δ)2 n2
−E1 δ(2n)
,
h (n)2 n2
−2E1
δ = f0 δ.
hn3
P47-7 We need to use the reduced mass of the muon since the muon and proton masses are so
close together. Then
(207)(1836)
me = 186me .
m=
(207) + (1836)
(a) Apply Eq. 47-20 1/2:
aµ = a0 /(186) = (52.9 pm)/(186) = 0.284 pm.
(b) Apply Eq. 47-21:
Eµ = E1 (186) = (13.6 eV)(186) = 2.53 keV.
(c) λ = (1240 keV · pm)/(2.53 pm) = 490 pm.
P47-8
(a) The reduced mass of the electron is
m=
(1)(1)
me = 0.5me .
(1) + (1)
The spectrum is similar, except for this additional factor of 1/2; hence
λpos = 2λH .
(b) apos = a0 /(186) = (52.9 pm)/(1/2) = 105.8 pm. This is the distance between the particles,
but they are both revolving about the center of mass. The radius is then half this quantity, or
52.9 pm.
P47-9 This problem isn’t really that much of a problem. Start with the magnitude of a vector
in terms of the components,
L2x + L2y + L2z = L2 ,
and then rearrange,
L2x + L2y = L2 − L2z .
According to Eq. 47-28 L2 = l(l + 1)h2 /4π 2 , while according to Eq. 47-30 Lz = ml h/2π. Substitute
that into the equation, and
L2x + L2y = l(l + 1)h2 /4π 2 − m2l h2 /4π 2 = l(l + 1) − m2l
Take the square root of both sides of this expression, and we are done.
273
h2
.
4π 2
The maximum value for ml is l, while the minimum value is 0. Consequently,
q
q
p
L2x + L2y = l(l + 1) − m2l h/2π ≤ l(l + 1) h/2π,
and
q
P47-10
Then
L2x + L2y =
q
√
l(l + 1) − m2l h/2π ≥ l h/2π.
Assume that ψ(0) is a reasonable estimate for ψ(r) everywhere inside the small sphere.
ψ2 =
e−0
1
=
.
πa30
πa30
The probability of finding it in a sphere of radius 1.1×10−15 m is
Z
1.1×10−15 m
0
P47-11
Then
4πr2 dr
4 (1.1×10−15 m)3
=
= 1.2×10−14 .
3
πa0
3 (5.29×10−11 m)3
Assume that ψ(0) is a reasonable estimate for ψ(r) everywhere inside the small sphere.
ψ2 =
(2)2 e−0
1
=
.
32πa30
8πa30
The probability of finding it in a sphere of radius 1.1×10−15 m is
Z
1.1×10−15 m
0
P47-12
4πr2 dr
1 (1.1×10−15 m)3
=
= 1.5×10−15 .
8πa30
6 (5.29×10−11 m)3
(a) The wave function squared is
ψ2 =
e−2r/a0
πa30
The probability of finding it in a sphere of radius r = xa0 is
Z xa0
4πr2 e−2r/a0 dr
P =
,
πa30
0
Z x
=
4x2 e−2x dx,
0
= 1 − e−2x (1 + 2x + 2x2 ).
(b) Let x = 1, then
P = 1 − e−2 (5) = 0.323.
P47-13
We want to evaluate the difference between the values of P at x = 2 and x = 2. Then
P (2) − P (1) = 1 − e−4 (1 + 2(2) + 2(2)2 ) − 1 − e−2 (1 + 2(1) + 2(1)2 ) ,
=
5e−2 − 13e−4 = 0.439.
274
P47-14
Using the results of Problem 47-12,
0.5 = 1 − e−2x (1 + 2x + 2x2 ),
or
e−2x = 1 + 2x + 2x2 .
The result is x = 1.34, or r = 1.34a0 .
P47-15
The probability of finding it in a sphere of radius r = xa0 is
Z xa0 2
r (2 − r/a0 )2 e−r/a0 dr
P =
8a30
0
Z x
1
=
x2 (2 − x)2 e−x dx
8 0
= 1 − e−x (y 4 /8 + y 2 /2 + y + 1).
The minimum occurs at x = 2, so
P = 1 − e−2 (2 + 2 + 2 + 1) = 0.0527.
275
E48-1 The highest energy x-ray photon will have an energy equal to the bombarding electrons,
as is shown in Eq. 48-1,
hc
λmin =
eV
Insert the appropriate values into the above expression,
λmin =
(4.14 × 10−15 eV · s)(3.00 × 108 m/s)
1240 × 10−9 eV · m
=
.
eV
eV
The expression is then
1240 × 10−9 V · m
1240 kV · pm
=
.
V
V
So long as we are certain that the “V ” will be measured in units of kilovolts, we can write this as
λmin =
λmin = 1240 pm/V.
E48-2 f = c/λ = (3.00×108 m/s)/(31.1×10−12 m) = 9.646×1018 /s. Planck’s constant is then
h=
E
(40.0 keV)
=
= 4.14×10−15 eV · s.
f
(9.646×1018 /s)
E48-3 Applying the results of Exercise 48-1,
∆V =
(1240kV · pm)
= 9.84 kV.
(126 pm)
E48-4 (a) Applying the results of Exercise 48-1,
λmin =
(1240kV · pm)
= 35.4 pm.
(35.0 kV)
(b) Applying the results of Exercise 45-1,
λKβ =
(1240keV · pm)
= 49.6 pm.
(25.51 keV) − (0.53 keV)
(c) Applying the results of Exercise 45-1,
λKα =
(1240keV · pm)
= 56.5 pm.
(25.51 keV) − (3.56 keV)
E48-5 (a) Changing the accelerating potential of the x-ray tube will decrease λmin . The new
value will be (using the results of Exercise 48-1)
λmin = 1240 pm/(50.0) = 24.8 pm.
(b) λKβ doesn’t change. It is a property of the atom, not a property of the accelerating potential
of the x-ray tube. The only way in which the accelerating potential might make a difference is if
λKβ < λmin for which case there would not be a λKβ line.
(c) λKα doesn’t change. See part (b).
276
E48-6 (a) Applying the results of Exercise 45-1,
(1240keV · pm)
= 64.2 keV.
(19.3 pm)
∆E =
(b) This is the transition n = 2 to n = 1, so
∆E = (13.6 eV)(1/12 − 1/22 ) = 10.2 eV.
E48-7 Applying the results of Exercise 45-1,
∆Eβ =
(1240keV · pm)
= 19.8 keV.
(62.5 pm)
∆Eα =
(1240keV · pm)
= 17.6 keV.
(70.5 pm)
and
The difference is
∆E = (19.8 keV) − (17.6 keV) = 2.2 eV.
E48-8 Since Eλ = hf = hc/λ, and λ = h/mc = hc/mc2 , then
Eλ = hc/λ = mc2 .
or ∆V = Eλ /e = mc2 /e = 511 kV.
E48-9 The 50.0 keV electron makes a collision and loses half of its energy to a photon, then the
photon has an energy of 25.0 keV. The electron is now a 25.0 keV electron, and on the next collision
again loses loses half of its energy to a photon, then this photon has an energy of 12.5 keV. On the
third collision the electron loses the remaining energy, so this photon has an energy of 12.5 keV. The
wavelengths of these photons will be given by
λ=
(1240 keV · pm)
,
E
which is a variation of Exercise 45-1.
E48-10 (a) The x-ray will need to knock free a K shell electron, so it must have an energy of at
least 69.5 keV.
(b) Applying the results of Exercise 48-1,
λmin =
(1240kV · pm)
= 17.8 pm.
(69.5 kV)
(c) Applying the results of Exercise 45-1,
λKβ =
(1240keV · pm)
= 18.5 pm.
(69.5 keV) − (2.3 keV)
Applying the results of Exercise 45-1,
λKα =
(1240keV · pm)
= 21.3 pm.
(69.5 keV) − (11.3 keV)
277
E48-11 (a) Applying the results of Exercise 45-1,
EKβ =
(1240keV · pm)
= 19.7 keV.
(63 pm)
Again applying the results of Exercise 45-1,
EKβ =
(1240keV · pm)
= 17.5 keV.
(71 pm)
(b) Zr or Nb; the others will not significantly absorb either line.
E48-12 Applying the results of Exercise 45-1,
λKα =
(1240keV · pm)
= 154.5 pm.
(8.979 keV) − (0.951 keV)
Applying the Bragg reflection relationship,
d=
(154.5 pm)
λ
=
= 282 pm.
2 sin θ
2 sin(15.9◦ )
E48-13 Plot the data. The plot should look just
√ likepFig 48-4. Note that the vertical axis is
which is related to the wavelength according to f = c/λ.
√
f,
E48-14 Remember that the m in Eq. 48-4 refers to the electron,
the nucleus. This means
√ not p
that the constant C in Eq. 48-5 is the same for all elements. Since f = c/λ, we have
λ1
=
λ2
Z2 − 1
Z1 − 1
2
.
For Ga and Nb the wavelength ratio is then
2
λNb
(31) − 1
=
= 0.5625.
λGa
(41) − 1
E48-15 (a) The ground state question is fairly easy. The n = 1 shell is completely occupied by
the first two electrons. So the third electron will be in the n = 2 state. The lowest energy angular
momentum state in any shell is the s sub-shell, corresponding to l = 0. There is only one choice for
ml in this case: ml = 0. There is no way at this level of coverage to distinguish between the energy
of either the spin up or spin down configuration, so we’ll arbitrarily pick spin up.
(b) Determining the configuration for the first excited state will require some thought. We could
assume that one of the K shell electrons (n = 1) is promoted to the L shell (n = 2). Or we could
assume that the L shell electron is promoted to the M shell. Or we could assume that the L shell
electron remains in the L shell, but that the angular momentum value is changed to l = 1. The
question that we would need to answer is which of these possibilities has the lowest energy.
The answer is the last choice: increasing the l value results in a small increase in the energy of
multi-electron atoms.
E48-16 Refer to Sample Problem 47-6:
r1 =
a0 (1)2
(5.29×10−11 m)
=
= 5.75×10−13 m.
Z
(92)
278
E48-17 We will assume that the ordering of the energy of the shells and sub-shells is the same.
That ordering is
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p
< 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s.
If there is no spin the s sub-shell would hold 1 electron, the p sub-shell would hold 3, the d sub-shell
5, and the f sub-shell 7. inert gases occur when a p sub-shell has filled, so the first three inert gases
would be element 1 (Hydrogen), element 1 + 1 + 3 = 5 (Boron), and element 1 + 1 + 3 + 1 + 3 = 9
(Fluorine).
Is there a pattern? Yes. The new inert gases have half of the atomic number of the original inert
gases. The factor of one-half comes about because there are no longer two spin states for each set
of n, l, ml quantum numbers.
We can save time and simply divide the atomic numbers of the remaining inert gases in half:
element 18 (Argon), element 27 (Cobalt), element 43 (Technetium), element 59 (Praseodymium).
E48-18 The pattern is
2 + 8 + 8 + 18 + 18 + 32 + 32+?
or
2(12 + 22 + 22 + 33 + 33 + 42 + 42 + x2 )
The unknown is probably x = 5, the next noble element is probably
118 + 2 · 52 = 168.
E48-19 (a) Apply Eq. 47-23, which can be written as
En =
(−13.6 eV)Z 2
.
n2
For the valence electron of sodium n = 3,
s
(5.14 eV)(3)2
= 1.84,
Z=
(13.6 eV)
while for the valence electron of potassium n = 4,
s
(4.34 eV)(4)2
Z=
= 2.26,
(13.6 eV)
(b) The ratios with the actual values of Z are 0.167 and 0.119, respectively.
E48-20 (a) There are three ml states allowed, and two ms states. The first electron can be in
any one of these six combinations of M1 and m2 . The second electron, given no exclusion principle,
could also be in any one of these six states. The total is 36. Unfortunately, this is wrong, because
we can’t distinguish electrons. Of this total of 36, six involve the electrons being in the same state,
while 30 involve the electron being in different states. But if the electrons are in different states,
then they could be swapped, and we won’t know, so we must divide this number by two. The total
number of distinguishable states is then
(30/2) + 6 = 21.
(b) Six. See the above discussion.
279
E48-21 (a) The Bohr orbits are circular orbits of radius rn = a0 n2 (Eq. 47-20). The electron is
orbiting where the force is
e2
Fn =
,
4π0 rn2
and this force is equal to the centripetal force, so
mv 2
e2
=
.
rn
4π0 rn2
where v is the velocity of the electron. Rearranging,
s
e2
v=
.
4π0 mrn
The time it takes for the electron to make one orbit can be used to calculate the current,
s
q
e
e
e2
i= =
=
.
t
2πrn /v
2πrn 4π0 mrn
The magnetic moment of a current loop is the current times the area of the loop, so
s
e
e2
µ = iA =
πr2 ,
2πrn 4π0 mrn n
which can be simplified to
e
µ=
2
s
e2
rn .
4π0 mrn
But rn = a0 n2 , so
e
µ=n
2
s
a0 e2
.
4π0 m
This might not look right, but a0 = 0 h2 /πme2 , so the expression can simplify to
r
e
h2
eh
µ=n
=n
= nµB .
2 4π 2 m2
4πm
(b) In reality the magnetic moments depend on the angular momentum quantum number, not
the principle quantum number. Although the Bohr theory correctly predicts the magnitudes, it does
not correctly predict when these values would occur.
E48-22 (a) Apply Eq. 48-14:
Fz = µz
dBz
= (9.27×10−24 J/T)(16×10−3 T/m) = 1.5×10−25 N.
dz
(b) a = F/m, ∆z = at2 /2, and t = y/vy . Then
∆z =
F y2
(1.5×10−25 N)(0.82 m)2
=
= 3.2×10−5 m.
2mvy2
2(1.67×10−27 kg)(970 m/s)2
E48-23 a = (9.27×10−24 J/T)(1.4×103 T/m)/(1.7×10−25 kg) = 7.6×104 m/s2 .
280
E48-24 (a) ∆U = 2µB, or
∆U = 2(5.79×10−5 eV/T)(0.520 T) = 6.02×10−5 eV.
(b) f = E/h = (6.02×10−5 eV)(4.14×10−15 eV · s) = 1.45×1010 Hz.
(c) λ = c/f = (3×108 m/s)/(1.45×1010 Hz) = 2.07×10−2 m.
E48-25 The energy change can be derived from Eq. 48-13; we multiply by a factor of 2 because
the spin is completely flipped. Then
∆E = 2µz Bz = 2(9.27×10−24 J/T)(0.190 T) = 3.52×10−24 J.
The corresponding wavelength is
λ=
hc
(6.63×10−34 J · s)(3.00×108 m/s)
=
= 5.65×10−2 m.
E
(3.52×10−24 J)
This is somewhere near the microwave range.
E48-26 The photon has an energy E = hc/λ. This energy is related to the magnetic field in the
vicinity of the electron according to
E = 2µB,
so
B=
hc
(1240 eV · nm)
=
= 0.051 T.
2µλ
2(5.79×10−5 J/T)(21×107 nm)
E48-27 Applying the results of Exercise 45-1,
E=
(1240 eV · nm)
= 1.55 eV.
(800nm)
The production rate is then
R=
(5.0×10−3 W)
= 2.0×1016 /s.
(1.55 eV)(1.6×10−19 J/eV)
E48-28 (a) x = (3×108 m/s)(12×10−12 s) = 3.6×10−3 m.
(b) Applying the results of Exercise 45-1,
E=
(1240 eV · nm)
= 1.786 eV.
(694.4nm)
The number of photons in the pulse is then
N = (0.150J)/(1.786 eV)(1.6×10−19 J/eV) = 5.25×1017 .
E48-29 We need to find out how many 10 MHz wide signals can fit between the two wavelengths.
The lower frequency is
f1 =
c
(3.00 × 108 m/s)
=
= 4.29 × 1014 Hz.
λ1
700 × 10−9 m)
281
The higher frequency is
f1 =
c
(3.00 × 108 m/s)
=
= 7.50 × 1014 Hz.
λ1
400 × 10−9 m)
The number of signals that can be sent in this range is
f2 − f1
(7.50 × 1014 Hz) − (4.29 × 1014 Hz)
=
= 3.21 × 107 .
(10 MHz)
(10 × 106 Hz)
That’s quite a number of television channels.
E48-30 Applying the results of Exercise 45-1,
E=
(1240 eV · nm)
= 1.960 eV.
(632.8nm)
The number of photons emitted in one minute is then
N=
(2.3×10−3 W)(60 s)
= 4.4×1017 .
(1.960 eV)(1.6×10−19 J/eV)
E48-31 Apply Eq. 48-19. E1 3 − E1 1 = 2(1.2 eV).. The ratio is then
−5
n1 3
= e−(2.4 eV)/(8.62×10 eV/K)(2000 K) = 9×10−7 .
n1 1
E48-32 (a) Population inversion means that the higher energy state is more populated; this can
only happen if the ratio in Eq. 48-19 is greater than one, which can only happen if the argument of
the exponent is positive. That would require a negative temperature.
(b) If n2 = 1.1n1 then the ratio is 1.1, so
T =
E48-33
(−2.26 eV
= −2.75×105 K.
(8.62×10−5 eV/K) ln(1.1)
(a) At thermal equilibrium the population ratio is given by
N2
e−E2 /kT
= −E /kT = e−∆E/kT .
N1
e 1
But ∆E can be written in terms of the transition photon wavelength, so this expression becomes
N2 = N1 e−hc/λkT .
Putting in the numbers,
−5
N2 = (4.0×1020 )e−(1240 eV·nm)/(582 nm)(8.62×10 eV/K)(300 K)) = 6.62×10−16 .
That’s effectively none.
(b) If the population of the upper state were 7.0×1020 , then in a single laser pulse
E=N
hc
(6.63×10−34 J · s)(3.00×108 m/s)
= (7.0×1020 )
= 240 J.
λ
(582×10−9 m)
282
E48-34 The allowed wavelength in a standing wave chamber are λn = 2L/n. For large n we can
write
2L
2L 2L
λn+1 =
≈
− 2.
n+1
n
n
The wavelength difference is then
λ2
2L
∆λ = 2 = n ,
n
2L
which in this case is
(533×10−9 m)2
∆λ =
= 1.7×10−12 m.
2(8.3×10−2 m)
E48-35 (a) The central disk will have an angle as measured from the center given by
d sin θ = (1.22)λ,
and since the parallel rays of the laser are focused on the screen in a distance f , we also have
R/f = sin θ. Combining, and rearranging,
1.22f λ
.
d
(b) R = 1.22(3.5 cm)(515 nm)/(3 mm) = 7.2×10−6 m.
(c) I = P/A = (5.21 W)/π(1.5 mm)2 = 7.37×105 W/m2 .
(d) I = P/A = (0.84)(5.21 W)/π(7.2 µm)2 = 2.7×1010 W/m2 .
R=
E48-36
P48-1 Let λ1 be the wavelength of the first photon. Then λ2 = λ1 + 130 pm. The total energy
transfered to the two photons is then
hc hc
E1 + E2 =
+
= 20.0 keV.
λ1
λ2
We can solve this for λ1 ,
20.0 keV
1
1
=
+
,
hc
λ1
λ1 + 130 pm
2λ1 + 130 pm
=
,
λ1 (λ1 + 130 pm)
which can also be written as
λ21
λ1 (λ1 + 130 pm) = (62 pm)(2λ1 + 130 pm),
+ (6 pm)λ1 − (8060 pm2 ) = 0.
This equation has solutions
λ1 = 86.8 pm and − 92.8 pm.
Only the positive answer has physical meaning. The energy of this first photon is then
E1 =
(1240 keV · pm)
= 14.3 keV.
(86.8 pm)
(a) After this first photon is emitted the electron still has a kinetic energy of
20.0 keV − 14.3 keV = 5.7 keV.
(b) We found the energy and wavelength of the first photon above. The energy of the second
photon must be 5.7 keV, with wavelength
λ2 = (86.8 pm) + 130 pm = 217 pm.
283
P48-2
Originally,
1
γ=p
1−
(2.73×108 m/s)2 /(3×108 m/s)2
= 2.412.
The energy of the electron is
E0 = γmc2 = (2.412)(511 keV) = 1232 keV.
Upon emitting the photon the new energy is
E = (1232 keV) − (43.8 keV) = 1189 keV,
so the new gamma factor is
γ = (1189 keV)/(511 keV) = 2.326,
and the new speed is
p
v = c 1 − 1/(2.326)2 = (0.903)c.
P48-3 Switch to a reference frame where the electron is originally at rest.
Momentum conservation requires
0 = pλ + pe = 0,
while energy conservation requires
mc2 = Eλ + Ee .
Rearrange to
Ee = mc2 − Eλ .
Square both sides of this energy expression and
Eλ2 − 2Eλ mc2 + m2 c4
Eλ2 − 2Eλ mc2
p2λ c2 − 2Eλ mc2
= Ee2 = p2e c2 + m2 c4 ,
= p2e c2 ,
= p2e c2 .
But the momentum expression can be used here, and the result is
−2Eλ mc2 = 0.
Not likely.
P48-4 (a) In the Bohr theory we can assume that the K shell electrons “see” a nucleus with charge
Z. The L shell electrons, however, are shielded by the one electron in the K shell and so they “see”
a nucleus with charge Z − 1. Finally, the M shell electrons are shielded by the one electron in the
K shell and the eight electrons in the K shell, so they “see” a nucleus with charge Z − 9.
The transition wavelengths are then
1
∆E
E0 (Z − 1)2 1
1
=
=
−
,
λα
hc
hc
22
12
E0 (Z − 1)2 3
=
.
hc
4
and
1
λβ
=
∆E
E0
=
hc
hc
=
E0 (Z − 9)2 −8
.
hc
9
284
1
1
− 2
32
1
,
The ratio of these two wavelengths is
27 (Z − 1)2
λβ
=
.
λα
32 (Z − 9)2
Note that the formula in the text has the square in the wrong place!
P48-5
(a) E = hc/λ; the energy difference is then
1
1
∆E = hc
−
,
λ1
λ2
λ 2 − λ1
= hc
.
λ2 λ 1
hc
=
∆λ.
λ2 λ1
Since λ1 and λ2 are so close together we can treat the product λ1 λ2 as being either λ21 or λ22 . Then
∆E =
(1240 eV · nm)
(0.597 nm) = 2.1×10−3 eV.
(589 nm)2
(b) The same energy difference exists in the 4s → 3p doublet, so
∆λ =
(1139 nm)2
(2.1×10−3 eV) = 2.2 nm.
(1240 eV · nm)
P48-6 (a) We can assume that the K shell electron “sees” a nucleus of charge Z − 1, since the
other electron in the shell screens it. Then, according to the derivation leading to Eq. 47-22,
rK = a0 /(Z − 1).
(b) The outermost electron “sees” a nucleus screened by all of the other electrons; as such Z = 1,
and the radius is
r = a0
P48-7 We assume in this crude model that one electron moves in a circular orbit attracted to
the helium nucleus but repelled from the other electron. Look back to Sample Problem 47-6; we
need to use some of the results from that Sample Problem to solve this problem.
The factor of e2 in Eq. 47-20 (the expression for the Bohr radius) and the factor of (e2 )2 in Eq.
47-21 (the expression for the Bohr energy levels) was from the Coulomb force between the single
electron and the single proton in the nucleus. This force is
F =
e2
.
4π0 r2
In our approximation the force of attraction between the one electron and the helium nucleus is
F1 =
2e2
.
4π0 r2
The factor of two is because there are two protons in the helium nucleus.
There is also a repulsive force between the one electron and the other electron,
F2 =
e2
,
4π0 (2r)2
285
where the factor of 2r is because the two electrons are on opposite side of the nucleus.
The net force on the first electron in our approximation is then
F1 − F2 =
2e2
e2
−
,
4π0 r2
4π0 (2r)2
which can be rearranged to yield
F net =
e2
4π0 r2
2−
1
4
=
e2
4π0 r2
7
.
4
It is apparent that we need to substitute 7e2 /4 for every occurrence of e2 .
(a) The ground state radius of the helium atom will then be given by Eq. 47-20 with the
appropriate substitution,
0 h2
4
r=
= a0 .
2
πm(7e /4)
7
(b) The energy of one electron in this ground state is given by Eq. 47-21 with the substitution
of 7e2 /4 for every occurrence of e2 , then
E=−
m(7e2 /4)2
49 me4
=
−
.
840 h2
16 840 h2
We already evaluated all of the constants to be 13.6 eV.
One last thing. There are two electrons, so we need to double the above expression. The ground
state energy of a helium atom in this approximation is
E0 = −2
49
(13.6 eV) = −83.3eV.
16
(c) Removing one electron will allow the remaining electron to move closer to the nucleus. The
energy of the remaining electron is given by the Bohr theory for He+ , and is
E He+ = (4)(−13.60 eV) = 54.4 eV,
so the ionization energy is 83.3 eV - 54.4 eV = 28.9 eV. This compares well with the accepted value.
P48-8
Applying Eq. 48-19:
T =
(−3.2 eV)
= 1.0×104 K.
(8.62×10−5 eV/K) ln(6.1×1013 /2.5×1015 )
P48-9 sin θ ≈ r/R, where r is the radius of the beam on the moon and R is the distance to the
moon. Then
1.22(600×10−9 m)(3.82×108 m)
r=
= 2360 m.
(0.118 m)
The beam diameter is twice this, or 4740 m.
P48-10
(a) N = 2L/λn , or
N=
2(6×10−2 m)(1.75)
= 3.03×105 .
(694×10−9 )
286
(b) N = 2nLf /c, so
∆f =
c
(3×108 m/s)
=
= 1.43×109 /s.
2nL
2(1.75)(6×10−2 m)
Note that the travel time to and fro is ∆t = 2nL/c!
(c) ∆f /f is then
λ
(694×10−9 )
∆f
=
=
= 3.3×10−6 .
f
2nL
2(1.75)(6×10−2 m)
287
E49-1
(a) Equation 49-2 is
n(E) =
√
√
8 2πm3/2 1/2
8 2π(mc2 )3/2 1/2
E
=
E .
h3
(hc)3
We can evaluate this by substituting in all known quantities,
√
8 2π(0.511 × 106 eV)3/2 1/2
n(E) =
E
= (6.81 × 1027 m−3 · eV−3/2 )E 1/2 .
(1240 × 10−9 eV · m)3
Once again, we simplified the expression by writing hc wherever we could, and then using hc =
1240 × 10−9 eV · m.
(b) Then, if E = 5.00 eV,
n(E) = (6.81 × 1027 m−3 · eV−3/2 )(5.00 eV)1/2 = 1.52 × 1028 m−3 · eV−1 .
E49-2 Apply the results of Ex. 49-1:
n(E) = (6.81 × 1027 m−3 · eV−3/2 )(8.00 eV)1/2 = 1.93 × 1028 m−3 · eV−1 .
E49-3 Monovalent means only one electron is available as a conducting electron. Hence we need
only calculate the density of atoms:
ρNA
(19.3×103 kg/m3 )(6.02×1023 mol−1 )
N
=
=
= 5.90×1028 /m3 .
V
Ar
(0.197 kg/mol)
E49-4 Use the ideal gas law: pV = N kT . Then
p=
E49-5
N
kT = (8.49×1028 m3 )(1.38×10−23 J/ · K)(297 K) = 3.48×108 Pa.
V
(a) The approximate volume of a single sodium atom is
V1 =
(0.023 kg/mol)
= 3.93×10−29 m3 .
(6.02×1023 part/mol)(971 kg/m3 )
The volume of the sodium ion sphere is
4π
(98×10−12 m)3 = 3.94×10−30 m3 .
3
The fractional volume available for conduction electrons is
V2 =
V 1 − V2
(3.93×10−29 m3 ) − (3.94×10−30 m3 )
=
= 90%.
V1
(3.93×10−29 m3 )
(b) The approximate volume of a single copper atom is
V1 =
(0.0635 kg/mol)
= 1.18×10−29 m3 .
(6.02×1023 part/mol)(8960 kg/m3 )
The volume of the copper ion sphere is
4π
(96×10−12 m)3 = 3.71×10−30 m3 .
3
The fractional volume available for conduction electrons is
V2 =
V 1 − V2
(1.18×10−29 m3 ) − (3.71×10−30 m3 )
=
= 69%.
V1
(1.18×10−29 m3 )
(c) Sodium, since more of the volume is available for the conduction electron.
288
E49-6 (a) Apply Eq. 49-6:
h
i
−5
p = 1/ e(0.0730 eV)/(8.62×10 eV/K)(0 K) + 1 = 0.
(b) Apply Eq. 49-6:
h
i
−5
p = 1/ e(0.0730 eV)/(8.62×10 eV/K)(320 K) + 1 = 6.62×10−2 .
E49-7 Apply Eq. 49-6, remembering to use the energy difference:
h
i
−5
p = 1/ e(−1.1) eV)/(8.62×10 eV/K)(273 K) + 1 = 1.00,
h
i
−5
p = 1/ e(−0.1) eV)/(8.62×10 eV/K)(273 K) + 1 = 0.986,
h
i
−5
p = 1/ e(0.0) eV)/(8.62×10 eV/K)(273 K) + 1 = 0.5,
h
i
−5
p = 1/ e(0.1) eV)/(8.62×10 eV/K)(273 K) + 1 = 0.014,
h
i
−5
p = 1/ e(1.1) eV)/(8.62×10 eV/K)(273 K) + 1 = 0.0.
(b) Inverting the equation,
∆E
,
k ln(1/p − 1)
T =
so
T =
(0.1 eV)
= 700 K
(8.62×10−5 eV/K) ln(1/(0.16) − 1)
E49-8 The energy differences are equal, except for the sign. Then
1
e+∆E/kt
+1
+
1
e−∆E/kt
+1
e−∆E/2kt
e+∆E/2kt
+
e+∆E/2kt + e−∆E/2kt
e−∆E/2kt + e+∆E/2kt
e−∆E/2kt + e+∆E/2kt
e−∆E/2kt + e+∆E/2kt
E49-9
= ,
= ,
= 1.
The Fermi energy is given by Eq. 49-5,
EF =
h2
8m
3n
π
2/3
,
where n is the density of conduction electrons. For gold we have
3
n=
(19.3 g/cm )(6.02×1023 part/mol)
3
3
= 5.90×1022 elect./cm = 59 elect./nm
(197 g/mol)
The Fermi energy is then
(1240 eV · nm)2
EF =
8(0.511×106 eV)
3
3(59 electrons/nm )
π
289
!2/3
= 5.53 eV.
E49-10 Combine the results of Ex. 49-1 and Eq. 49-6:
√
C E
no = ∆E/kt
.
e
+1
Then for each of the energies we have
p
(6.81×1027 m−3 · eV−3/2 ) (4 eV)
no =
= 1.36×1028 /m3 · eV,
e(−3.06 eV)/(8.62×10−5 eV/K)(1000 K) + 1
p
(6.81×1027 m−3 · eV−3/2 ) (6.75 eV)
no =
= 1.72×1028 /m3 · eV,
e(−0.31 eV)/(8.62×10−5 eV/K)(1000 K) + 1
p
(6.81×1027 m−3 · eV−3/2 ) (7 eV)
no =
= 9.02×1027 /m3 · eV,
e(−0.06 eV)/(8.62×10−5 eV/K)(1000 K) + 1
p
(6.81×1027 m−3 · eV−3/2 ) (7.25 eV)
no =
= 1.82×1027 /m3 · eV,
e(0.19 eV)/(8.62×10−5 eV/K)(1000 K) + 1
p
(6.81×1027 m−3 · eV−3/2 ) (9 eV)
no =
= 3.43×1018 /m3 · eV.
e(1.94 eV)/(8.62×10−5 eV/K)(1000 K) + 1
E49-11 Solve
n2 (hc)2
8(mc2 )L2
for n = 50, since there are two electrons in each level. Then
En =
Ef =
(50)2 (1240 eV · nm)2
= 6.53×104 eV.
8(5.11×105 eV)(0.12 nm)2
E49-12 We need to be much higher than T = (7.06 eV)/(8.62×10−5 eV/K) = 8.2×104 K.
E49-13
Equation 49-5 is
h2
EF =
8m
3n
π
2/3
,
and if we collect the constants,
h2
EF =
8m
2/3
3
n3/2 = An3/2 ,
π
where, if we multiply the top and bottom by c2
2/3
2/3
(hc)2 3
(1240 × 10−9 eV · m)2 3
A=
=
= 3.65 × 10−19 m2 · eV.
8mc2 π
8(0.511 × 106 eV)
π
E49-14 (a) Inverting Eq. 49-6,
∆E = kT ln(1/p − 1),
so
∆E = (8.62×10−5 eV/K)(1050 K) ln(1/(0.91) − 1) = −0.209 eV.
Then E = (−0.209 eV) + (7.06 eV) = 6.85 eV.
(b) Apply the results of Ex. 49-1:
n(E) = (6.81 × 1027 m−3 · eV−3/2 )(6.85 eV)1/2 = 1.78 × 1028 m−3 · eV−1 .
(c) no = np = (1.78 × 1028 m−3 · eV−1 )(0.910) = 1.62 × 1028 m−3 · eV−1 .
290
E49-15 Equation 49-5 is
h2
EF =
8m
3n
π
2/3
,
and if we rearrange,
E F 3/2 =
3h3
√
n,
16 2πm3/2
Equation 49-2 is then
√
8 2πm3/2 1/2
3
n(E) =
E
= nE F −3/2 E 1/2 .
3
h
2
E49-16 ph = 1 − p, so
ph
=
=
=
1−
1
,
e∆E/kT + 1
e∆E/kT
,
e∆E/kT + 1
1
.
−∆E/kT
1+e
E49-17 The steps to solve this exercise are equivalent to the steps for Exercise 49-9, except now
the iron atoms each contribute 26 electrons and we have to find the density.
First, the density is
ρ=
m
(1.99×1030 kg)
=
= 1.84×109 kg/m3
4πr3 /3
4π(6.37×106 m)3 /3
Then
3
n
=
(26)(1.84×106 g/cm )(6.02×1023 part/mol)
3
= 5.1×1029 elect./cm ,
(56 g/mol)
=
5.1×108 elect./nm
3
The Fermi energy is then
(1240 eV · nm)2
EF =
8(0.511×106 eV)
3
3(5.1×108 elect./nm )
π
!2/3
= 230 keV.
E49-18 First, the density is
ρ=
m
2(1.99×1030 kg)
=
= 9.5×1017 kg/m3
3
4πr /3
4π(10×103 m)3 /3
Then
n = (9.5×1017 kg/m3 )/(1.67×10−27 kg) = 5.69×1044 /m3 .
The Fermi energy is then
(1240 MeV · fm)2
EF =
8(940 MeV)
3
3(5.69×10−1 /fm )
π
291
!2/3
= 137 MeV.
E49-19
E49-20 (a) E F = 7.06 eV, so
f=
3(8.62×10−5 eV · K)(0 K)
= 0,
2(7.06 eV)
(b) f = 3(8.62×10−5 eV · K)(300 K)/2(7.06 eV) = 0.0055.
(c) f = 3(8.62×10−5 eV · K)(1000 K)/2(7.06 eV) = 0.0183.
E49-21
Using the results of Exercise 19,
T =
2f E F
2(0.0130)(4.71 eV)
=
= 474 K.
3k
3(8.62×10−5 eV · K)
E49-22 f = 3(8.62×10−5 eV · K)(1235 K)/2(5.5 eV) = 0.029.
E49-23 (a) Monovalent means only one electron is available as a conducting electron. Hence we
need only calculate the density of atoms:
N
ρNA
(10.5×103 kg/m3 )(6.02×1023 mol−1 )
=
=
= 5.90×1028 /m3 .
V
Ar
(0.107 kg/mol)
(b) Using the results of Ex. 49-13,
E F = (3.65 × 10−19 m2 · eV)(5.90×1028 /m3 )2/3 = 5.5 eV.
(c) v =
p
2K/m, or
v=
p
2(5.5 eV)(5.11×105 eV/c2 ) = 1.4×108 m/s.
(d) λ = h/p, or
λ=
(6.63×10−34 J · s)
= 5.2×10−12 m.
(9.11×10−31 kg)(1.4×108 m/s)
E49-24 (a) Bivalent means two electrons are available as a conducting electron. Hence we need
to double the calculation of the density of atoms:
N
ρNA
2(7.13×103 kg/m3 )(6.02×1023 mol−1 )
=
=
= 1.32×1029 /m3 .
V
Ar
(0.065 kg/mol)
(b) Using the results of Ex. 49-13,
E F = (3.65 × 10−19 m2 · eV)(1.32×1029 /m3 )2/3 = 9.4 eV.
(c) v =
p
2K/m, or
v=
p
2(9.4 eV)(5.11×105 eV/c2 ) = 1.8×108 m/s.
(d) λ = h/p, or
λ=
(6.63×10−34 J · s)
= 4.0×10−12 m.
(9.11×10−31 kg)(1.8×108 m/s)
292
E49-25 (a) Refer to Sample Problem 49-5 where we learn that the mean free path λ can be
written in terms of Fermi speed v F and mean time between collisions τ as
λ = v F τ.
The Fermi speed is
p
p
v F = c 2EF /mc2 = c 2(5.51 eV)/(5.11×105 eV) = 4.64×10−3 c.
The time between collisions is
τ=
m
(9.11×10−31 kg)
=
= 3.74×10−14 s.
2
28
−3
ne ρ
(5.86×10 m )(1.60×10−19 C)2 (1.62×10−8 Ω · m)
We found n by looking up the answers from Exercise 49-23 in the back of the book. The mean free
path is then
λ = (4.64×10−3 )(3.00×108 m/s)(3.74×10−14 s) = 52 nm.
(b) The spacing between the ion cores is approximated by the cube root of volume per atom.
This atomic volume for silver is
V =
(108 g/mol)
(6.02×1023 part/mol)(10.5
3
g/cm )
= 1.71×10−23 cm3 .
The distance between the ions is then
l=
√
3
V = 0.257 nm.
The ratio is
λ/l = 190.
E49-26 (a) For T = 1000 K we can use the approximation, so for diamond
−5
p = e−(5.5 eV)/2(8.62×10 eV/K)(1000 K) = 1.4×10−14 ,
while for silicon,
−5
p = e−(1.1 eV)/2(8.62×10 eV/K)(1000 K) = 1.7×10−3 ,
(b) For T = 4 K we can use the same approximation, but now ∆E kT and the exponential
function goes to zero.
E49-27 (a) E − E F ≈ 0.67 eV/2 = 0.34 eV.. The probability the state is occupied is then
h
i
−5
p = 1/ e(0.34) eV)/(8.62×10 eV/K)(290 K) + 1 = 1.2×10−6 .
(b) E − E F ≈ −0.67 eV/2 = −0.34 eV.. The probability the state is unoccupied is then 1 − p, or
h
i
−5
p = 1 − 1/ e(−0.34) eV)/(8.62×10 eV/K)(290 K) + 1 = 1.2×10−6 .
E49-28 (a) E − E F ≈ 0.67 eV/2 = 0.34 eV.. The probability the state is occupied is then
h
i
−5
p = 1/ e(0.34) eV)/(8.62×10 eV/K)(289 K) + 1 = 1.2×10−6 .
293
E49-29
(a) The number of silicon atoms per unit volume is
3
n=
(6.02×1023 part/mol)(2.33 g/cm )
3
= 4.99×1022 part./cm .
(28.1 g/mol)
If one out of 1.0eex7 are replaced then there will be an additional charge carrier density of
3
3
4.99×1022 part./cm /1.0×107 = 4.99×1015 part./cm = 4.99×1021 m−3 .
(b) The ratio is
(4.99×1021 m−3 )/(2 × 1.5×1016 m−3 ) = 1.7×105 .
The extra factor of two is because all of the charge carriers in silicon (holes and electrons) are charge
carriers.
E49-30 Since one out of every 5×106 silicon atoms needs to be replaced, then the mass of phosphorus would be
1 30
m=
= 2.1×10−7 g.
5×106 28
E49-31 l =
p
3
1/1022 /m3 = 4.6×10−8 m.
E49-32 The atom density of germanium is
N
ρNA
(5.32×103 kg/m3 )(6.02×1023 mol−1 )
=
=
= 1.63×1028 /m3 .
V
Ar
(0.197 kg/mol)
The atom density of the impurity is
(1.63×1028 /m3 )/(1.3×109 ) = 1.25×1019 .
The average spacing is
l=
p
3
1/1.25×1019 /m3 = 4.3×10−7 m.
E49-33 The first one is an insulator because the lower band is filled and band gap is so large;
there is no impurity.
The second one is an extrinsic n-type semiconductor: it is a semiconductor because the lower
band is filled and the band gap is small; it is extrinsic because there is an impurity; since the impurity
level is close to the top of the band gap the impurity is a donor.
The third sample is an intrinsic semiconductor: it is a semiconductor because the lower band is
filled and the band gap is small.
The fourth sample is a conductor; although the band gap is large, the lower band is not completely
filled.
The fifth sample is a conductor: the Fermi level is above the bottom of the upper band.
The sixth one is an extrinsic p-type semiconductor: it is a semiconductor because the lower band
is filled and the band gap is small; it is extrinsic because there is an impurity; since the impurity
level is close to the bottom of the band gap the impurity is an acceptor.
E49-34 6.62×105 eV/1.1 eV = 6.0×105 electron-hole pairs.
E49-35 (a) R = (1 V)/(50×10−12 A) = 2×1010 Ω.
(b) R = (0.75 V)/(8 mA) = 90Ω.
294
E49-36 (a) A region with some potential difference exists that has a gap between the charged
areas.
(b) C = Q/∆V . Using the results in Sample Problem 49-9 for q and ∆V ,
C=
E49-37
n0 eAd/2
= 2κ0 A/d.
n0 ed2 /4κ0
(a) Apply that ever so useful formula
λ=
hc
(1240 eV · nm)
=
= 225 nm.
E
(5.5 eV)
Why is this a maximum? Because longer wavelengths would have lower energy, and so not enough
to cause an electron to jump across the band gap.
(b) Ultraviolet.
E49-38 Apply that ever so useful formula
E=
hc
(1240 eV · nm)
=
= 4.20 eV.
λ
(295 nm)
E49-39 The photon energy is
E=
hc
(1240 eV · nm)
=
= 8.86 eV.
λ
(140 nm)
which is enough to excite the electrons through the band gap. As such, the photon will be absorbed,
which means the crystal is opaque to this wavelength.
E49-40
P49-1
We can calculate the electron density from Eq. 49-5,
n
3/2
=
π
3
8mc2 E F
(hc)2
=
π
3
8(0.511×106 eV)(11.66 eV)
(1240 eV · nm)2
=
181 electrons/nm .
,
3/2
,
3
From this we calculate the number of electrons per particle,
3
(181 electrons/nm )(27.0 g/mol)
3
(2.70 g/cm )(6.02×1023 particles/mol)
which we can reasonably approximate as 3.
295
= 3.01,
P49-2
49-15,
At absolute zero all states below E F are filled, an none above. Using the results of Ex.
E av
1
n
=
EF
En(E) dE,
0
Z
3 −3/2 E F 3/2
EF
E dE,
2
0
3 −3/2 2 5/2
EF
EF ,
2
5
3
EF.
5
=
=
=
P49-3
Z
(a) The total number of conduction electron is
n=
(0.0031 kg)(6.02×1023 mol−1 )
= 2.94×1022 .
(0.0635 kg/mol)
The total energy is
E=
3
(7.06 eV)(2.94×1022 ) = 1.24×1023 eV = 2×104 J.
5
(b) This will light a 100 W bulb for
t = (2×104 J)/(100 W) = 200 s.
P49-4
(a) First do the easy part: nc = N c p(E c ), so
Nc
.
e(E c −E F )/kT + 1
Then use the results of Ex. 49-16, and write
nv = N v [1 − p(E v )] =
Nv
.
e−(E v −E F )/kT + 1
Since each electron in the conduction band must have left a hole in the valence band, then these two
expressions must be equal.
(b) If the exponentials dominate then we can drop the +1 in each denominator, and
Nc
(E
−E
c
F )/kT
e
=
Nv
−(E
−E F )/kT
v
e
,
Nc
Nv
= e(E c −2E F +E v )/kT ,
EF
=
1
(E c + E v + kT ln(N c /N v )) .
2
P49-5 (a) We want to use Eq. 49-6; although we don’t know the Fermi energy, we do know
the differences between the energies in question. In the un-doped silicon E − E F = 0.55 eV for the
bottom of the conduction band. The quantity
kT = (8.62×10−5 eV/K)(290 K) = 0.025 eV,
which is a good number to remember— at room temperature kT is 1/40 of an electron-volt.
296
Then
1
= 2.8×10−10 .
e(0.55 eV)/(0.025 eV) + 1
In the doped silicon E − E F = 0.084 eV for the bottom of the conduction band. Then
p=
p=
e(0.084
1
= 3.4×10−2 .
eV)/(0.025 eV) + 1
(b) For the donor state E − E F = −0.066 eV, so
p=
P49-6
1
= 0.93.
e(−0.066 eV)/(0.025 eV) + 1
(a) Inverting Eq. 49-6,
E − E F = kT ln(1/p − 1),
so
E F = (1.1 eV − 0.11 eV) − (8.62×10−5 eV/K)(290 K) ln(1/(4.8×10−5 ) − 1) = 0.74 eV
above the valence band.
(b) E − E F = (1.1 eV) − (0.74 eV) = 0.36 eV, so
p=
e(0.36
1
= 5.6×10−7 .
eV)/(0.025 eV) + 1
P49-7 (a) Plot the graph with a spreadsheet. It should look like Fig. 49-12.
(b) kT = 0.025 eV when T = 290 K. The ratio is then
if
e(0.5 eV)/(0.025 eV) + 1
= (−0.5 eV)/(0.025 eV)
= 4.9×108 .
ir
e
+1
P49-8
297
E50-1 We want to follow the example set in Sample Problem 50-1. The distance of closest
approach is given by
d
=
=
=
qQ
,
4π0 Kα
(2)(29)(1.60×10−19 C)2
4π(8.85×10−12 C2 /Nm2 )(5.30MeV)(1.60
× 10−13 J/MeV)
,
1.57×10−14 m.
That’s pretty close.
E50-2 (a) The gold atom can be treated as a point particle:
F
=
q1 q2
,
4π0 r2
=
(2)(79)(1.60×10−19 C)2
,
4π(8.85×10−12 C2 /Nm2 )(0.16×10−9 m)2
=
1.4×10−6 N.
(b) W = F d, so
d=
(5.3×106 eV)(1.6×10−19 J/eV)
= 6.06×10−7 m.
(1.4×10−6 N)
That’s 1900 gold atom diameters.
E50-3 Take an approach similar to Sample Problem 50-1:
K
=
=
=
qQ
,
4π0 d
(2)(79)(1.60×10−19 C)2
4π(8.85×10−12 C2 /Nm2 )(8.78×10−15 m)(1.60
2.6×107 eV.
E50-4 All are stable except
E50-5
× 10−19 J/eV)
88
Rb and
239
Pb.
We can make an estimate of the mass number A from Eq. 50-1,
R = R0 A1/3 ,
where R0 = 1.2 fm. If the measurements indicate a radius of 3.6 fm we would have
3
A = (R/R0 )3 = ((3.6 fm)/(1.2 fm)) = 27.
E50-6
E50-7 The mass number of the sun is
A = (1.99×1030 kg)/(1.67×10−27 kg) = 1.2×1057 .
The radius would be
R = (1.2×10−15 m)
p
3
1.2×1057 = 1.3×104 m.
298
,
E50-8 239 Pu is composed of 94 protons and 239 − 94 = 145 neutrons. The combined mass of the
free particles is
M = Zmp + N mn = (94)(1.007825 u) + (145)(1.008665 u) = 240.991975 u.
The binding energy is the difference
E B = (240.991975 u − 239.052156 u)(931.5 MeV/u) = 1806.9 MeV,
and the binding energy per nucleon is then
(1806.9 MeV)/(239) = 7.56 MeV.
E50-9 62 Ni is composed of 28 protons and 62 − 28 = 34 neutrons. The combined mass of the
free particles is
M = Zmp + N mn = (28)(1.007825 u) + (34)(1.008665 u) = 62.513710 u.
The binding energy is the difference
E B = (62.513710 u − 61.928349 u)(931.5 MeV/u) = 545.3 MeV,
and the binding energy per nucleon is then
(545.3 MeV)/(62) = 8.795 MeV.
E50-10 (a) Multiply each by 1/1.007825, so
m1H = 1.00000,
m12C = 11.906829,
and
m238U = 236.202500.
E50-11 (a) Since the binding energy per nucleon is fairly constant, the energy must be proportional
to A.
(b) Coulomb repulsion acts between pairs of protons; there are Z protons that can be chosen as
first in the pair, and Z − 1 protons remaining that can make up the partner in the pair. That makes
for Z(Z − 1) pairs. The electrostatic energy must be proportional to this.
(c) Z 2 grows faster than A, which is roughly proportional to Z.
E50-12 Solve
(0.7899)(23.985042) + x(24.985837) + (0.2101 − x)(25.982593) = 24.305
for x. The result is x = 0.1001, and then the amount
299
26
Mg is 0.1100.
E50-13 The neutron confined in a nucleus of radius R will have a position uncertainty on the
order of ∆x ≈ R. The momentum uncertainty will then be no less than
∆p ≥
h
h
≈
.
2π∆x
2πR
Assuming that p ≈ ∆p, we have
h
,
2πR
and then the neutron will have a (minimum) kinetic energy of
p≥
E≈
p2
h2
≈
.
2
2m
8π mR2
E≈
(hc)2
.
8π 2 mc2 R02 A2/3
But R = R0 A1/3 , so
For an atom with A = 100 we get
E≈
(1240 MeV · fm)2
= 0.668 MeV.
MeV)(1.2 fm)2 (100)2/3
8π 2 (940
This is about a factor of 5 or 10 less than the binding energy per nucleon.
E50-14 (a) To remove a proton,
E = [(1.007825) + (3.016049) − (4.002603)] (931.5 MeV) = 19.81 MeV.
To remove a neutron,
E = [(1.008665) + (2.014102) − (3.016049)] (931.5 MeV) = 6.258 MeV.
To remove a proton,
E = [(1.007825) + (1.008665) − (2.014102)] (931.5 MeV) = 2.224 MeV.
(b) E = (19.81 + 6.258 + 2.224)MeV = 28.30 MeV.
(c) (28.30 MeV)/4 = 7.07 MeV.
E50-15 (a) ∆ = [(1.007825) − (1)](931.5 MeV) = 7.289 MeV.
(b) ∆ = [(1.008665) − (1)](931.5 MeV) = 8.071 MeV.
(c) ∆ = [(119.902197) − (120)](931.5 MeV) = −91.10 MeV.
E50-16 (a) E B = (ZmH + N mN − m)c2 . Substitute the definition for mass excess, mc2 = Ac2 + ∆,
and
EB
(b) For
197
= Z(c2 + ∆H ) + N (c2 + ∆N ) − Ac2 − ∆,
= Z∆H + N ∆N − ∆.
Au,
E B = (79)(7.289 MeV) + (197 − 79)(8.071 MeV) − (−31.157 MeV) = 1559 MeV,
and the binding energy per nucleon is then
(1559 MeV)/(197) = 7.92 MeV.
300
E50-17 The binding energy of
63
Cu is given by
M = Zmp + N mn = (29)(1.007825 u) + (34)(1.008665 u) = 63.521535 u.
The binding energy is the difference
E B = (63.521535 u − 62.929601 u)(931.5 MeV/u) = 551.4 MeV.
The number of atoms in the sample is
n=
(0.003 kg)(6.02×1023 mol−1 )
= 2.87×1022 .
(0.0629 kg/mol)
The total energy is then
(2.87×1022 )(551.4 MeV)(1.6×10−19 J/eV) = 2.53×1012 J.
E50-18 (a) For ultra-relativistic particles E = pc, so
λ=
(1240 MeV · fm)
= 2.59 fm.
(480 MeV)
(b) Yes, since the wavelength is smaller than nuclear radii.
E50-19 We will do this one the easy way because we can. This method won’t work except when
there is an integer number of half-lives. The activity of the sample will fall to one-half of the initial
decay rate after one half-life; it will fall to one-half of one-half (one-fourth) after two half-lives. So
two half-lives have elapsed, for a total of (2)(140 d) = 280 d.
E50-20 N = N0 (1/2)t/t1/2 , so
N = (48×1019 )(0.5)(26)/(6.5) = 3.0×1019 .
E50-21 (a) t1/2 = ln 2/(0.0108/h) = 64.2 h.
(b) N = N0 (1/2)t/t1/2 , so
N/N0 = (0.5)(3) = 0.125.
(c) N = N0 (1/2)t/t1/2 , so
N/N0 = (0.5)(240)/(64.2) = 0.0749.
E50-22 (a) λ = (−dN/dt)/N , or
λ = (12/s)/(2.5×1018 ) = 4.8×10−18 /s.
(b) t1/2 = ln 2/λ, so
t1/2 = ln 2/(4.8×10−18 /s) = 1.44×1017 s,
which is 4.5 billion years.
301
E50-23
(a) The decay constant for
67
Ga can be derived from Eq. 50-8,
ln 2
ln 2
=
= 2.461×10−6 s−1 .
t1/2
(2.817×105 s)
λ=
The activity is given by R = λN , so we want to know how many atoms are present. That can be
found from
1 atom
1u
= 3.077×1022 atoms.
3.42 g
1.6605×10−24 g
66.93 u
So the activity is
R = (2.461×10−6 /s−1 )(3.077×1022 atoms) = 7.572×1016 decays/s.
(b) After 1.728×105 s the activity would have decreased to
R = R0 e−λt = (7.572×1016 decays/s)e−(2.461×10
−6
/s−1 )(1.728×105 s)
= 4.949×1016 decays/s.
E50-24 N = N0 e−λt , but λ = ln 2/t1/2 , so
N = N0 e− ln 2t/t1/2 = N0 (2)−t/t1/2 = N0
E50-25 The remaining
223
t/t1/2
1
.
2
is
N = (4.7×1021 )(0.5)(28)/(11.43) = 8.6×1020 .
The number of decays, each of which produced an alpha particle, is
(4.7×1021 ) − (8.6×1020 ) = 3.84×1021 .
E50-26 The amount remaining after 14 hours is
m = (5.50 g)(0.5)(14)/(12.7) = 2.562 g.
The amount remaining after 16 hours is
m = (5.50 g)(0.5)(16)/(12.7) = 2.297 g.
The difference is the amount which decayed during the two hour interval:
(2.562 g) − (2.297 g) = 0.265 g.
E50-27
(a) Apply Eq. 50-7,
R = R0 e−λt .
We first need to know the decay constant from Eq. 50-8,
λ=
ln 2
ln 2
=
= 5.618×10−7 s−1 .
t1/2
(1.234×106 s)
And the the time is found from
t
1
R
= − ln
,
λ R0
1
(170 counts/s)
= −
ln
,
(5.618×10−7 s−1 ) (3050 counts/s)
=
5.139×106 s ≈ 59.5 days.
302
Note that counts/s is not the same as decays/s. Not all decay events will be picked up by a detector
and recorded as a count; we are assuming that whatever scaling factor which connects the initial
count rate to the initial decay rate is valid at later times as well. Such an assumption is a reasonable
assumption.
(b) The purpose of such an experiment would be to measure the amount of phosphorus that is
taken up in a leaf. But the activity of the tracer decays with time, and so without a correction factor
we would record the wrong amount of phosphorus in the leaf. That correction factor is R0 /R; we
need to multiply the measured counts by this factor to correct for the decay.
In this case
−7 −1
5
R
= eλt = e(5.618×10 s )(3.007×10 s) = 1.184.
R0
E50-28 The number of particles of
n = (0.15)
147
Sm is
(0.001 kg)(6.02×1023 mol−1 )
= 6.143×1020 .
(0.147 kg/mol)
The decay constant is
λ = (120/s)/(6.143×1020 ) = 1.95×10−19 /s.
The half-life is
t1/2 = ln 2/(1.95×10−19 /s) = 3.55×1018 s,
or 110 Gy.
E50-29 The number of particles of
n0 =
239
Pu is
(0.012 kg)(6.02×1023 mol−1 )
= 3.023×1022 .
(0.239 kg/mol)
The number which decay is
h
i
n0 − n = (3.025×1022 ) 1 − (0.5)(20000)/(24100) = 1.32×1022 .
The mass of helium produced is then
m=
(0.004 kg/mol)(1.32×1022 )
= 8.78×10−5 kg.
(6.02×1023 mol−1 )
E50-30 Let R33 /(R33 + R32) = x, where x0 = 0.1 originally, and we want to find out at what
time x = 0.9. Rearranging,
(R33 + R32)/R33 = 1/x,
so
R32/R33 = 1/x − 1.
t/t1/2
Since R = R0 (0.5)
we can write a ratio
1
1
−1=
− 1 (0.5)t/t32 −t/t33 .
x
x0
Put in some of the numbers, and
ln[(1/9)/(9)] = ln[0.5]t
which has solution t = 209 d.
303
1
1
−
14.3 25.3
,
E50-31
E50-32 (a) N/N0 = (0.5)(4500)/(82) = 3.0×10−17 .
(b) N/N0 = (0.5)(4500)/(0.034) = 0.
E50-33
The Q values are
Q3
Q4
Q5
= (235.043923 − 232.038050 − 3.016029)(931.5 MeV) = −9.46 MeV,
= (235.043923 − 231.036297 − 4.002603)(931.5 MeV) = 4.68 MeV,
= (235.043923 − 230.033127 − 5.012228)(931.5 MeV) = −1.33 MeV.
Only reactions with positive Q values are energetically possible.
E50-34 (a) For the
14
C decay,
Q = (223.018497 − 208.981075 − 14.003242)(931.5 MeV) = 31.84 MeV.
For the 4 He decay,
Q = (223.018497 − 219.009475 − 4.002603)(931.5 MeV) = 5.979 MeV.
E50-35 Q = (136.907084 − 136.905821)(931.5 MeV) = 1.17 MeV.
E50-36 Q = (1.008665 − 1.007825)(931.5 MeV) = 0.782 MeV.
E50-37 (a) The kinetic energy of this electron is significant compared to the rest mass energy,
so we must use relativity to find the momentum. The total energy of the electron is E = K + mc2 ,
the momentum will be given by
p
p
pc =
E 2 − m2 c4 = K 2 + 2Kmc2 ,
p
=
(1.00 MeV)2 + 2(1.00 MeV)(0.511 MeV) = 1.42 MeV.
The de Broglie wavelength is then
λ=
hc
(1240 MeV · fm)
=
= 873 fm.
pc
(1.42 MeV)
(b) The radius of the emitting nucleus is
R = R0 A1/3 = (1.2 fm)(150)1/3 = 6.4 fm.
(c) The longest wavelength standing wave on a string fixed at each end is twice the length of
the string. Although the rules for standing waves in a box are slightly more complicated, it is a fair
assumption that the electron could not exist as a standing a wave in the nucleus.
(d) See part (c).
304
E50-38 The electron is relativistic, so
p
pc =
E 2 − m2 c4 ,
p
=
(1.71 MeV + 0.51 MeV)2 − (0.51 MeV)2 ,
= 2.16 MeV.
This is also the magnitude of the momentum of the recoiling 32 S. Non-relativistic relations are
K = p2 /2m, so
(2.16 MeV)2
K=
= 78.4 eV.
2(31.97)(931.5 MeV)
E50-39 N = mNA /Mr will give the number of atoms of
λ = ln 2/t1/2 will give the decay constant. Combining,
m=
198
Au; R = λN will give the activity;
Rt1/2 Mr
N Mr
=
.
NA
ln 2NA
Then for the sample in question
m=
(250)(3.7×1010 /s)(2.693)(86400 s)(198 g/mol)
= 1.02×10−3 g.
ln 2(6.02×1023 /mol)
E50-40 R = (8722/60 s)/(3.7×1010 /s) = 3.93×10−9 Ci.
E50-41 The radiation absorbed dose (rad) is related to the roentgen equivalent man (rem) by
the quality factor, so for the chest x-ray
(25 mrem)
= 29 mrad.
(0.85)
This is well beneath the annual exposure average.
Each rad corresponds to the delivery of 10−5 J/g, so the energy absorbed by the patient is
1
−5
(88 kg) = 1.28×10−2 J.
(0.029)(10 J/g)
2
E50-42 (a) (75 kg)(10−2 J/kg)(0.024 rad) = 1.8×10−2 J.
(b) (0.024 rad)(12) = 0.29 rem.
E50-43 R = R0 (0.5)t/t1/2 , so
R0 = (3.94 µCi)(2)(6.048×10
5
s)/(1.82×105 s)
= 39.4 µCi.
E50-44 (a) N = mNA /MR , so
N=
(2×10−3 g)(6.02×1023 /mol)
= 5.08×1018 .
(239 g/mol)
(b) R = λN = ln 2N/t1/2 , so
R = ln 2(5.08×1018 )/(2.411×104 y)(3.15×107 s/y) = 4.64×106 /s.
(c) R = (4.64×106 /s)/(3.7×1010 decays/s · Ci) = 1.25×10−4 Ci.
305
E50-45 The hospital uses a 6000 Ci source, and that is all the information we need to find the
number of disintegrations per second:
(6000 Ci)(3.7×1010 decays/s · Ci) = 2.22×1014 decays/s.
We are told the half life, but to find the number of radioactive nuclei present we want to know the
decay constant. Then
ln 2
ln 2
λ=
=
= 4.17×10−9 s−1 .
t1/2
(1.66×108 s)
The number of
60
Co nuclei is then
N=
R
(2.22×1014 decays/s)
=
= 5.32×1022 .
λ
(4.17×10−9 s−1 )
E50-46 The annual equivalent does is
(12×10−4 rem/h)(20 h/week)(52 week/y) = 1.25 rem.
E50-47 (a) N = mNA /MR and MR = (226) + 2(35) = 296, so
N=
(1×10−1 g)(6.02×1023 /mol)
= 2.03×1020 .
(296 g/mol)
(b) R = λN = ln 2N/t1/2 , so
R = ln 2(2.03×1020 )/(1600 y)(3.15×107 s/y) = 2.8×109 Bq.
(c) (2.8×109 )/(3.7×1010 ) = 76 mCi.
E50-48 R = λN = ln 2N/t1/2 , so
N=
(4.6×10−6 )(3.7×1010 /s)(1.28×109 y)(3.15×107 s/y)
= 9.9×1021 ,
ln 2
N = mNA /MR , so
m=
E50-49
(40 g/mol)(9.9×1021 )
= 0.658 g.
(6.02×1023 /mol)
We can apply Eq. 50-18 to find the age of the rock,
t1/2
NF
t =
ln 1 +
,
ln 2
NI
(4.47×109 y)
(2.00×10−3 g)/(206 g/mol)
=
ln 1 +
,
ln 2
(4.20×10−3 g)/(238 g/mol)
=
2.83×109 y.
306
E50-50 The number of atoms of
N=
238
U originally present is
(3.71×10−3 g)(6.02×1023 /mol)
= 9.38×1018 .
(238 g/mol)
The number remaining after 260 million years is
N = (9.38×1018 )(0.5)(260 My)/(4470 My) = 9.01×1018 .
The difference decays into lead (eventually), so the mass of lead present should be
m=
(206 g/mol)(0.37×1018 )
= 1.27×10−4 g.
(6.02×1023 /mol)
E50-51 We can apply Eq. 50-18 to find the age of the rock,
t1/2
NF
t =
ln 1 +
,
ln 2
NI
(150×10−6 g)/(206 g/mol)
(4.47×109 y)
=
ln 1 +
,
ln 2
(860×10−6 g)/(238 g/mol)
=
1.18×109 y.
Inverting Eq. 50-18 to find the mass of
40
K originally present,
NF
= 2t/t1/2 − 1,
NI
so (since they have the same atomic mass) the mass of
m=
40
K is
(1.6×10−3 g)
= 1.78×10−3 g.
2(1.18)/(1.28) − 1
E50-52 (a) There is an excess proton on the left and an excess neutron, so the unknown must be
a deuteron, or d.
(b) We’ve added two protons but only one (net) neutron, so the element is Ti and the mass
number is 43, or 43 Ti.
(c) The mass number doesn’t change but we swapped one proton for a neutron, so 7 Li.
E50-53 Do the math:
Q = (58.933200 + 1.007825 − 58.934352 − 1.008665)(931.5 MeV) = −1.86 MeV.
E50-54 The reactions are
201
Hg(γ, α)197 Pt,
Au(n, p)197 Pt,
196
Pt(n, γ)197 Pt,
198
Pt(γ, n)197 Pt,
196
Pt(d, p)197 Pt,
198
Pt(p, d)197 Pt.
197
307
E50-55 We will write these reactions in the same way as Eq. 50-20 represents the reaction of
Eq. 50-19. It is helpful to work backwards before proceeding by asking the following question: what
nuclei will we have if we subtract one of the allowed projectiles?
The goal is 60 Co, which has 27 protons and 60 − 27 = 33 neutrons.
1. Removing a proton will leave 26 protons and 33 neutrons, which is
unstable.
59
2. Removing a neutron will leave 27 protons and 32 neutrons, which is
stable.
59
3. Removing a deuteron will leave 26 protons and 32 neutrons, which is
stable.
Fe; but that nuclide is
Co; and that nuclide is
58
Fe; and that nuclide is
It looks as if only 59 Co(n)60 Co and 58 Fe(d)60 Co are possible. If, however, we allow for the possibility
of other daughter particles we should also consider some of the following reactions.
1. Swapping a neutron for a proton:
60
Ni(n,p)60 Co.
2. Using a neutron to knock out a deuteron:
61
Ni(n,d)60 Co.
3. Using a neutron to knock out an alpha particle:
63
4. Using a deuteron to knock out an alpha particle:
Cu(n,α)60 Co.
62
Ni(d,α)60 Co.
E50-56 (a) The possible results are 64 Zn, 66 Zn, 64 Cu, 66 Cu,
(b) The stable results are 64 Zn, 66 Zn, 61 Ni, and 67 Zn.
61
Ni,
63
Ni,
65
Zn, and
67
Zn.
E50-57
E50-58 The resulting reactions are
194
Pt(d,α)192 Ir,
196
Pt(d,α)194 Ir, and
198
Pt(d,α)196 Ir.
E50-59
E50-60 Shells occur at numbers 2, 8, 20, 28, 50, 82. The shells occur separately for protons and
neutrons. To answer the question you need to know both Z and N = A − Z of the isotope.
(a) Filled shells are 18 O, 60 Ni, 92 Mo, 144 Sm, and 207 Pb.
(b) One nucleon outside a shell are 40 K, 91 Zr, 121 Sb, and 143 Nd.
(c) One vacancy in a shell are 13 C, 40 K, 49 Ti, 205 Tl, and 207 Pb.
E50-61 (a) The binding energy of this neutron can be found by considering the Q value of the
reaction 90 Zr(n)91 Zr which is
(89.904704 + 1.008665 − 90.905645)(931.5 MeV) = 7.19 MeV.
89
(b) The binding energy of this neutron can be found by considering the Q value of the reaction
Zr(n)90 Zr which is
(88.908889 + 1.008665 − 89.904704)(931.5 MeV) = 12.0 MeV.
This neutron is bound more tightly that the one in part (a).
(c) The binding energy per nucleon is found by dividing the binding energy by the number of
nucleons:
(40×1.007825 + 51×1.008665 − 90.905645)(931.5 MeV)
= 8.69 MeV.
91
The neutron in the outside shell of 91 Zr is less tightly bound than the average nucleon in 91 Zr.
308
P50-1 Before doing anything we need to know whether or not the motion is relativistic. The
rest mass energy of an α particle is
mc2 = (4.00)(931.5 MeV) = 3.73 GeV,
and since this is much greater than the kinetic energy we can assume the motion is non-relativistic,
and we can apply non-relativistic momentum and energy conservation principles. The initial velocity
of the α particle is then
p
p
p
v = 2K/m = c 2K/mc2 = c 2(5.00 MeV)/(3.73 GeV) = 5.18×10−2 c.
For an elastic collision where the second particle is at originally at rest we have the final velocity
of the first particle as
v 1,f = v 1,i
m2 − m1
(4.00u) − (197u)
= (5.18×10−2 c)
= −4.97×10−2 c,
m2 + m1
(4.00u) + (197u)
while the final velocity of the second particle is
v 2,f = v 1,i
2m1
2(4.00u)
= (5.18×10−2 c)
= 2.06×10−3 c.
m2 + m1
(4.00u) + (197u)
(a) The kinetic energy of the recoiling nucleus is
K=
1
1
mv 2 = m(2.06×10−3 c)2 = (2.12×10−6 )mc2
2
2
= (2.12×10−6 )(197)(931.5 MeV) = 0.389 MeV.
(b) Energy conservation is the fastest way to answer this question, since it is an elastic collision.
Then
(5.00 MeV) − (0.389 MeV) = 4.61 MeV.
P50-2
The gamma ray carries away a mass equivalent energy of
mγ = (2.2233 MeV)/(931.5 MeV/u) = 0.002387 u.
The neutron mass would then be
mN = (2.014102 − 1.007825 + 0.002387)u = 1.008664 u.
P50-3 (a) There are four substates: mj can be +3/2, +1/2, -1/2, and -3/2.
(b) ∆E = (2/3)(3.26)(3.15×10−8 eV/T)(2.16 T) = 1.48×10−7 eV.
(c) λ = (1240 eV · nm)/(1.48×10−7 eV) = 8.38 m.
(d) This is in the radio region.
P50-4 (a) The charge density is ρ = 3Q/4πR3 . The charge on the shell of radius r is dq = 4πr2 ρ dr.
The potential at the surface of a solid sphere of radius r is
V =
q
ρr2
=
.
4π0 r
30
The energy required to add a layer of charge dq is
dU = V dq =
309
4πρ2 r4
dr,
30
which can be integrated to yield
U=
(b) For
239
4πρ2 R5
3Q2
=
.
30
20π0 R
Pu,
U=
3(94)2 (1.6×10−19 C)
= 1024×106 eV.
20π(8.85×10−12 F/m)(7.45×10−15 m)
(c) The electrostatic energy is 10.9 MeV per proton.
P50-5 The decay rate is given by R = λN , where N is the number of radioactive nuclei present.
If R exceeds P then nuclei will decay faster than they are produced; but this will cause N to decrease,
which means R will decrease until it is equal to P . If R is less than P then nuclei will be produced
faster than they are decaying; but this will cause N to increase, which means R will increase until it
is equal to P . In either case equilibrium occurs when R = P , and it is a stable equilibrium because
it is approached no matter which side is larger. Then
P = R = λN
at equilibrium, so N = P/λ.
P50-6 (a) A = λN ; at equilibrium A = P , so P = 8.88×1010 /s.
(b) (8.88×1010 /s)(1−e−0.269t ), where t is in hours. The factor 0.269 comes from ln(2)/(2.58) = λ.
(c) N = P/λ = (8.88×1010 /s)(3600 s/h)/(0.269/h) = 1.19×1015 .
(d) m = N Mr /NA , or
m=
P50-7
(1.19×1015 )(55.94 g/mol)
= 1.10×10−7 g.
(6.02×1023 /mol)
(a) A = λN , so
A=
ln 2mNA
ln 2(1×10−3 g)(6.02×1023 /mol)
=
= 3.66×107 /s.
t1/2 Mr
(1600)(3.15×107 s)(226 g/mol)
(b) The rate must be the same if the system is in secular equilibrium.
(c) N = P/λ = t1/2 P/ ln 2, so
m=
P50-8
(3.82)(86400 s)(3.66×107 /s)(222 g/mol)
= 6.43×10−9 g.
(6.02×1023 /mol) ln 2
The number of water molecules in the body is
N = (6.02×1023 /mol)(70×103 g)/(18 g/mol) = 2.34×1027 .
There are ten protons in each water molecule. The activity is then
A = (2.34×1027 ) ln 2/(1×1032 y) = 1.62×10−5 /y.
The time between decays is then
1/A = 6200 y.
310
P50-9 Assuming the 238 U nucleus is originally at rest the total initial momentum is zero, which
means the magnitudes of the final momenta of the α particle and the 234 Th nucleus are equal.
The α particle has a final velocity of
p
p
p
v = 2K/m = c 2K/mc2 = c 2(4.196 MeV)/(4.0026×931.5 MeV) = 4.744×10−2 c.
Since the magnitudes of the final momenta are the same, the 234 Th nucleus has a final velocity
of
(4.0026 u)
−2
(4.744×10 c)
= 8.113×10−4 c.
(234.04 u)
The kinetic energy of the
K=
234
Th nucleus is
1
1
mv 2 = m(8.113×10−4 c)2 = (3.291×10−7 )mc2
2
2
= (3.291×10−7 )(234.04)(931.5 MeV) = 71.75 keV.
The Q value for the reaction is then
(4.196 MeV) + (71.75 keV) = 4.268 MeV,
which agrees well with the Sample Problem.
P50-10
(a) The Q value is
Q = (238.050783 − 4.002603 − 234.043596)(931.5 MeV) = 4.27 MeV.
(b) The Q values for each step are
Q = (238.050783 − 237.048724 − 1.008665)(931.5 MeV) = −6.153 MeV,
Q = (237.048724 − 236.048674 − 1.007825)(931.5 MeV) = −7.242 MeV,
Q = (236.048674 − 235.045432 − 1.008665)(931.5 MeV) = −5.052 MeV,
Q = (235.045432 − 234.043596 − 1.007825)(931.5 MeV) = −5.579 MeV.
(c) The total Q for part (b) is −24.026 MeV. The difference between (a) and (b) is 28.296 MeV.
The binding energy for the alpha particle is
E = [2(1.007825) + 2(1.008665) − 4.002603](931.5 MeV) = 28.296 MeV.
P50-11 (a) The emitted positron leaves the atom, so the mass must be subtracted. But the
daughter particle now has an extra electron, so that must also be subtracted. Hence the factor
−2me .
(b) The Q value is
Q = [11.011434 − 11.009305 − 2(0.0005486)](931.5 MeV) = 0.961 MeV.
P50-12 (a) Capturing an electron is equivalent to negative beta decay in that the total number
of electrons is accounted for on both the left and right sides of the equation. The loss of the K shell
electron, however, must be taken into account as this energy may be significant.
(b) The Q value is
Q = (48.948517 − 48.947871)(931.5 MeV) − (0.00547 MeV) = 0.596 MeV.
311
P50-13
The decay constant for
90
Sr is
ln 2
ln 2
=
= 7.58×10−10 s−1 .
t1/2
(9.15×108 s)
λ=
The number of nuclei present in 400 g of
N = (400 g)
so the overall activity of the 400 g of
90
90
Sr is
(6.02×1023 /mol)
= 2.68×1024 ,
(89.9 g/mol)
Sr is
R = λN = (7.58×10−10 s−1 )(2.68×1024 )/(3.7×1010 /Ci · s) = 5.49×104 Ci.
This is spread out over a 2000 km2 area, so the “activity surface density” is
(5.49×104 Ci)
= 2.74×10−5 Ci/m2 .
(20006 m2 )
If the allowable limit is 0.002 mCi, then the area of land that would contain this activity is
(0.002×10−3 Ci)
= 7.30×10−2 m2 .
(2.74×10−5 Ci/m2 )
P50-14
(a) N = mNA /Mr , so
N = (2.5×10−3 g)(6.02×1023 /mol)/(239 g/mol) = 6.3×1018 .
(b) A = ln 2N/t1/2 , so the number that decay in 12 hours is
ln 2(6.3×1018 )(12)(3600 s)
= 2.5×1011 .
(24100)(3.15×107 s)
(c) The energy absorbed by the body is
E = (2.5×1011 )(5.2 MeV)(1.6×10−19 J/eV) = 0.20 J.
(d) The dose in rad is (0.20 J)/(87 kg) = 0.23 rad.
(e) The biological dose in rem is (0.23)(13) = 3 rem.
P50-15
(a) The amount of
238
N=
U per kilogram of granite is
(4×10−6 kg)(6.02×1023 /mol)
= 1.01×1019 .
(0.238 kg/mol)
The activity is then
A=
ln 2(1.01×1019 )
= 49.7/s.
(4.47×109 y)(3.15×107 s/y)
The energy released in one second is
E = (49.7/s)(51.7 MeV) = 4.1×10−10 J.
The amount of
232
Th per kilogram of granite is
N=
(13×10−6 kg)(6.02×1023 /mol)
= 3.37×1019 .
(0.232 kg/mol)
312
The activity is then
A=
ln 2(3.37×1019 )
= 52.6/s.
(1.41×1010 y)(3.15×107 s/y)
The energy released in one second is
E = (52.6/s)(42.7 MeV) = 3.6×10−10 J.
The amount of
40
K per kilogram of granite is
N=
(4×10−6 kg)(6.02×1023 /mol)
= 6.02×1019 .
(0.040 kg/mol)
The activity is then
A=
ln 2(6.02×1019 )
= 1030/s.
(1.28×109 y)(3.15×107 s/y)
The energy released in one second is
E = (1030/s)(1.32 MeV) = 2.2×10−10 J.
The total of the three is 9.9×10−10 W per kilogram of granite.
(b) The total for the Earth is 2.7×1013 W.
P50-16
(a) Since only a is moving originally then the velocity of the center of mass is
V =
ma va + mX (0)
ma
= va
.
mX + ma
ma + mX
No, since momentum is conserved.
(b) Moving to the center of mass frame gives the velocity of X as V , and the velocity of a as
va − V . The kinetic energy is now
K cm
1
mX V 2 + ma (va − V )2 ,
2 va2
m2a
m2X
=
mX
+
m
,
a
2
(ma + mX )2
(ma + mX )2
ma va2 ma mX + m2X
=
,
2 (ma + mX )2
mX
= K lab
.
ma + mX
=
Yes; kinetic p
energy is not conserved.
(c) va = 2K/m, so
p
va = 2(15.9 MeV)/(1876 MeV)c = 0.130c.
The center of mass velocity is
V = (0.130c)
(2)
= 2.83×10−3 c.
(2) + (90)
Finally,
K cm = (15.9 MeV)
(90)
= 15.6 MeV.
(2) + (90)
313
P50-17
Let Q = K cm in the result of Problem 50-16, and invert, solving for K lab .
P50-18
(a) Removing a proton from
209
Bi:
E = (207.976636 + 1.007825 − 208.980383)(931.5 MeV) = 3.80 MeV.
Removing a proton from
208
Pb:
E = (206.977408 + 1.007825 − 207.976636)(931.5 MeV) = 8.01 MeV.
(b) Removing a neutron from
209
Pb:
E = (207.976636 + 1.008665 − 208.981075)(931.5 MeV) = 3.94 MeV.
Removing a neutron from
208
Pb:
E = (206.975881 + 1.008665 − 207.976636)(931.5 MeV) = 7.37 MeV.
314
E51-1 (a) For the coal,
m = (1×109 J)/(2.9×107 J/kg) = 34 kg.
(b) For the uranium,
m = (1×109 J)/(8.2×1013 J/kg) = 1.2×10−5 kg.
E51-2 (a) The energy from the coal is
E = (100 kg)(2.9×107 J/kg) = 2.9×109 J.
(b) The energy from the uranium in the ash is
E = (3×10−6 )(100 kg)(8.2×1013 J) = 2.5×1010 J.
E51-3
(a) There are
(1.00 kg)(6.02×1023 mol−1 )
= 2.56×1024
(235g/mol)
atoms in 1.00 kg of 235 U.
(b) If each atom releases 200 MeV, then
(200×106 eV)(1.6×10−19 J/ eV)(2.56×1024 ) = 8.19×1013 J
of energy could be released from 1.00 kg of 235 U.
(c) This amount of energy would keep a 100-W lamp lit for
t=
(8.19×1013 J)
= 8.19×1011 s ≈ 26, 000 y!
(100 W)
E51-4 2 W = 1.25×1019 eV/s. This requires
(1.25×1019 eV/s)/(200×106 eV) = 6.25×1010 /s
as the fission rate.
E51-5 There are
atoms in 1.00 kg of
(1.00 kg)(6.02×1023 mol−1 )
= 2.56×1024
(235g/mol)
235
U. If each atom releases 200 MeV, then
(200×106 eV)(1.6×10−19 J/ eV)(2.56×1024 ) = 8.19×1013 J
of energy could be released from 1.00 kg of 235 U. This amount of energy would keep a 100-W lamp
lit for
(8.19×1013 J)
t=
= 8.19×1011 s ≈ 30, 000 y!
(100 W)
E51-6 There are
atoms in 1.00 kg of
(1.00 kg)(6.02×1023 mol−1 )
= 2.52×1024
(239g/mol)
239
Pu. If each atom releases 180 MeV, then
(180×106 eV)(1.6×10−19 J/ eV)(2.52×1024 ) = 7.25×1013 J
of energy could be released from 1.00 kg of
239
Pu.
315
E51-7 When the 233 U nucleus absorbs a neutron we are given a total of 92 protons and 142
neutrons. Gallium has 31 protons and around 39 neutrons; chromium has 24 protons and around
28 neutrons. There are then 37 protons and around 75 neutrons left over. This would be rubidium,
but the number of neutrons is very wrong. Although the elemental identification is correct, because
we must conserve proton number, the isotopes are wrong in our above choices for neutron numbers.
E51-8 Beta decay is the emission of an electron from the nucleus; one of the neutrons changes into
a proton. The atom now needs one more electron in the electron shells; by using atomic masses (as
opposed to nuclear masses) then the beta electron is accounted for. This is only true for negative
beta decay, not for positive beta decay.
E51-9 (a) There are
(1.0 g)(6.02×1023 mol−1 )
= 2.56×1021
(235g/mol)
atoms in 1.00 g of
235
U. The fission rate is
A = ln 2N/t1/2 = ln 2(2.56×1021 )/(3.5×1017 y)(365d/y) = 13.9/d.
(b) The ratio is the inverse ratio of the half-lives:
(3.5×1017 y)/(7.04×108 y) = 4.97×108 .
E51-10 (a) The atomic number of Y must be 92 − 54 = 38, so the element is Sr. The mass number
is 235 + 1 − 140 − 1 = 95, so Y is 95 Sr.
(b) The atomic number of Y must be 92 − 53 = 39, so the element is Y. The mass number is
235 + 1 − 139 − 2 = 95, so Y is 95 Y.
(c) The atomic number of X must be 92 − 40 = 52, so the element is Te. The mass number is
235 + 1 − 100 − 2 = 134, so X is 134 Te.
(d) The mass number difference is 235 + 1 − 141 − 92 = 3, so b = 3.
E51-11 The Q value is
Q = [51.94012 − 2(25.982593)](931.5 MeV) = −23 MeV.
The negative value implies that this fission reaction is not possible.
E51-12 The Q value is
Q = [97.905408 − 2(48.950024)](931.5 MeV) = 4.99 MeV.
The two fragments would have a very large Coulomb barrier to overcome.
E51-13
The energy released is
(235.043923 − 140.920044 − 91.919726 − 2×1.008665)(931.5 MeV) = 174 MeV.
E51-14 Since En > Eb fission is possible by thermal neutrons.
E51-15 (a) The uranium starts with 92 protons. The two end products have a total of 58 + 44 =
102. This means that there must have been ten beta decays.
(b) The Q value for this process is
Q = (238.050783 + 1.008665 − 139.905434 − 98.905939)(931.5 MeV) = 231 MeV.
316
E51-16 (a) The other fragment has 92 − 32 = 60 protons and 235 + 1 − 83 = 153 neutrons. That
element is 153 Nd.
(b) Since K = p2 /2m and momentum is conserved, then 2m1 K1 = 2m2 K2 . This means that
K2 = (m1 /m2 )K1 . But K1 + K2 = Q, so
K1
m2 + m1
= Q,
m2
or
K1 =
with a similar expression for K2 . Then for
while for
153
(c) For
while for
153
Ge
K=
(153)
(170 MeV) = 110 MeV,
(83 + 153)
K=
(83)
(170 MeV) = 60 MeV,
(83 + 153)
Nd
83
83
m2
Q,
m1 + m2
Ge,
2K
=
m
s
2K
=
m
s
v=
r
v=
r
Nd
2(110 MeV)
c = 0.053c,
(83)(931 MeV)
2(60 MeV)
c = 0.029c.
(153)(931 MeV)
E51-17 Since 239 Pu is one nucleon heavier than 238 U only one neutron capture is required. The
atomic number of Pu is two more than U, so two beta decays will be required. The reaction series
is then
238
U+n →
239
U →
239
Np →
239
U,
Np + β − + ν̄,
239
Pu + β − + ν̄.
239
E51-18 Each fission releases 200 MeV. The total energy released over the three years is
(190×106 W)(3)(3.15×107 s) = 1.8×1016 J.
That’s
(1.8×1016 J)/(1.6×10−19 J/eV)(200×106 eV) = 5.6×1026
fission events. That requires
m = (5.6×1026 )(0.235 kg/mol)/(6.02×1023 /mol) = 218 kg.
But this is only half the original amount, or 437 kg.
E51-19 According to Sample Problem 51-3 the rate at which non-fission thermal neutron capture
occurs is one quarter that of fission. Hence the mass which undergoes non-fission thermal neutron
capture is one quarter the answer of Ex. 51-18. The total is then
(437 kg)(1 + 0.25) = 546 kg.
317
E51-20 (a) Qeff = E/∆N , where E is the total energy released and ∆N is the number of decays.
This can also be written as
P t1/2
P t1/2 Mr
P
=
=
,
Qeff =
A
ln 2N
ln 2NA m
where A is the activity and P the power output from the sample. Solving,
Qeff =
(2.3 W)(29 y)(3.15×107 s)(90 g/mol)
= 4.53×10−13 J = 2.8 MeV.
ln 2(6.02×1023 /mol)(1 g)
(b) P = (0.05)m(2300 W/kg), so
m=
(150 W)
= 1.3 kg.
(0.05)(2300 W/kg)
E51-21 Let the energy released by one fission be E1 . If the average time to the next fission event
is tgen , then the “average” power output from the one fission is P1 = E1 /tgen . If every fission event
results in the release of k neutrons, each of which cause a later fission event, then after every time
period tgen the number of fission events, and hence the average power output from all of the fission
events, will increase by a factor of k.
For long enough times we can write
P (t) = P0 k t/tgen .
E51-22 Invert the expression derived in Exercise 51-21:
tgen /t (1.3×10−3 s)/(2.6 s)
P
(350)
=
= 0.99938.
k=
P0
(1200)
E51-23 Each fission releases 200 MeV. Then the fission rate is
(500×106 W)/(200×106 eV)(1.6×10−19 J/eV) = 1.6×1019 /s
The number of neutrons in “transit” is then
(1.6×1019 /s)(1.0×10−3 s) = 1.6×1016 .
E51-24 Using the results of Exercise 51-21:
P = (400 MW)(1.0003)(300 s)/(0.03 s) = 8030 MW.
E51-25
The time constant for this decay is
λ=
ln 2
= 2.50×10−10 s−1 .
(2.77×109 s)
The number of nuclei present in 1.00 kg is
N=
(1.00 kg)(6.02×1023 mol−1 )
= 2.53×1024 .
(238 g/mol)
The decay rate is then
R = λN = (2.50×10−10 s−1 )(2.53×1024 ) = 6.33×1014 s−1 .
The power generated is the decay rate times the energy released per decay,
P = (6.33×1014 s−1 )(5.59×106 eV)(1.6×10−19 J/eV) = 566 W.
318
E51-26 The detector detects only a fraction of the emitted neutrons. This fraction is
A
(2.5 m2 )
=
= 1.62×10−4 .
2
4πR
4π(35 m)2
The total flux out of the warhead is then
(4.0/s)/(1.62×10−4 ) = 2.47×104 /s.
The number of
239
Pu atoms is
N=
(2.47×104 /s)(1.34×1011 y)(3.15×107 s/y)
A
=
= 6.02×1022 .
λ
ln 2(2.5)
That’s one tenth of a mole, so the mass is (239)/10 = 24 g.
E51-27 Using the results of Sample Problem 51-4,
t=
so
t=
ln[R(0)/R(t)]
,
λ5 − λ8
ln[(0.03)/(0.0072)]
= 1.72×109 y.
(0.984 − 0.155)(1×10−9 /y)
E51-28 (a) (15×109 W · y)(2×105 y) = 7.5×104 W.
(b) The number of fissions required is
N=
The mass of
235
(15×109 W · y)(3.15×107 s/y)
= 1.5×1028 .
(200 MeV)(1.6×10−19 J/eV)
U consumed is
m = (1.5×1028 )(0.235kg/mol)/(6.02×10−23 /mol) = 5.8×103 kg.
E51-29
and then
If 238 U absorbs a neutron it becomes 239 U, which will decay by beta decay to first 239 Np
Pu; we looked at this in Exercise 51-17. This can decay by alpha emission according to
239
239
Pu →235 U + α.
E51-30 The number of atoms present in the sample is
N = (6.02×1023 /mol)(1000 kg)/(2.014g/mol) = 2.99×1026 .
It takes two to make a fusion, so the energy released is
(3.27 MeV)(2.99×1026 )/2 = 4.89×1026 MeV.
That’s 7.8×1013 J, which is enough to burn the lamp for
t = (7.8×1013 J)/(100 W) = 7.8×1011 s = 24800 y.
E51-31 The potential energy at closest approach is
U=
(1.6×10−19 C)2
= 9×105 eV.
4π(8.85×10−12 F/m)(1.6×10−15 m)
319
E51-32 The ratio can be written as
n(K1 )
=
n(K2 )
so the ratio is
s
r
K1 (K2 −K1 )/kT
e
,
K2