Download PHY–302 K. Solutions for Problem set # 8. Textbook problem 7.16

PHY–302 K.
Solutions for Problem set # 8.
Textbook problem 7.16:
(a) Just before the hammer hits the nail, it has momentum
p = mv = 12 kg × 8.5 m/s = 102 N · s.
(1)
After hitting the nail, the hammer stops, so its momentum becomes
p′ = mv ′ = 0.
(2)
This change of momentum is due to impulse Inh of the nail’s force on the hammer Fnh ∆t,
∆p ≡ p′ − p = Inh ,
(3)
so that impulse is −102 N · s. By the third law of Newton, the hammer’s force on the nail
has the opposite impulse,
Fhn = −Fnh
=⇒
Ihn = −Inh = +102 N · s ≈ 100 N · s.
(4)
(b) We know the hammer’s force on the nail lasts for ∆t = 8.0 ms, but the force is not
constant during this time. The net impulse of a time-dependent force F (t) is the integral
I =
Z
F (t) dt
(5)
in terms of the average force, the net impulse is simply
I = F avg × ∆t.
(6)
Consequently, the average force of the hammer on the nail is
avg
Fhn
=
Ihn
102 N · s
=
≈ 13, 000 N ≈ 2900 lb.
∆t
8.0 ms
1
(7)
Textbook problem 7.5:
Consider the mass ∆m = µ∆t of incandescent gas ejected by the rocket during some short
time interval ∆t. The gases have backward velocity −ve and hence negative momentum
∆pgas = −ve × δmgas = −ve × µ × ∆t.
(8)
By momentum conservation, the rocket acquires the opposite momentum
∆procket = −∆pgas = +ve µ × ∆t.
(9)
This momentum grows with time, which corresponds to the rocket thrust force
F =
∆procket
= ve × µ.
∆t
(10)
For the rocket in question, the thrust force is
F = 4.0 · · · 104 m/s × 1500 kg/s = 6.0 · 107 N ≈ 6, 700 tons.
(11)
PS: The exhaust speed of 40,000 m/s used in this problem is completely unrealistic for a
chemically powered rocket. Indeed, the kinetic energy of the exhaust gases comes from the
chemical energy of the fuel and oxidizer, hence
ve2
K
E chem
=
=
,
2
m
m
(12)
The highest energy-to-mass ratio comes from liquid hydrogen fuel and liquid oxygen oxidizer,
which produce exhaust speed of about 4,450 m/s. Theoretically, it is possible to raise the
exhaust speed up to 5,300 m/s by using liquid fluorine oxidizer, but this is impractical
because fluorine is extremely corrosive, ignites on contact with almost anything, and the
combustion products are very toxic.
2
Textbook problem 7.4:
This is a recoil problem: The child throws a package forward while the recoil makes him and
the boat move back. The basic principle here is the momentum conservation. The external
forces acting on the package+child+boat system — the gravity and the buoyant force on the
boat — are vertical and have no horizontal components, so the net horizontal momentum of
the system is conserved,
Pnet = mpackage vpackage + mchild vchild + mboat vboat = const.
(13)
Before the child threw the package, everything was at rest so the net momentum was zero,
vpackage = vchild = vboat = 0
=⇒
Pnet = 0.
(14)
After the child threw the package, the net momentum of the package+child+boat system
remains zero, thus
mpackage vpackage + mchild vchild + mboat vboat = 0.
(15)
Moreover, the child and the boat have the same velocity, hence
mpackage vpackage + (mchild + mboat )vchild+boat = 0
(16)
and therefore
vchild+boat = −
mpackage vpackage
6.40 kg × 10.0 m/s
= −
≈ −0.901 m/s.
mchild + mboat
26.0 kg + 45.0 kg
3
(17)
Textbook problem 7.8:
This problem describes an inelastic collision of two boxcars. As in any collision, the net
momentum of the two cars is conserved,
m1 v1 + m2 v2 = m1 v1′ + m2 v2′ ,
(18)
where v1 and v2 denote the boxcars’ velocities just before the collision and v1′ and v2′ their
velocities just after the collision. For the collision in question, the two boxcars move together
after the collision, v1′ = v2′ , so this collision is totally inelastic.
In this problem, we know the velocities of both boxcars both before and after the collision,
but we know only one boxcar’s mass m1 and need to find the other boxcar’s mass m2 . So
starting with eq. (18), we put all terms containing m2 on one side and then solve the resulting
equation
m1 (v1 − v1′ ) = m2 (v2′ − v2 )
(19)
for the second boxcar’s mass:
m2 = m1 ×
15.0 m/s − 6.0 m/s
v1 − v1′
≈ 14, 000 kg.
= 9, 300 kg ×
′
v2 − v2
6.0 m/s − 0 m/s
(20)
Textbook problem 7.26:
In an elastic collision, both the net momentum and the net kinetic energy of the two colliding
bodies are conserved,
m1 v1 + m2 v2 = m1 v1′ + m2 v2′ ,
2
1
2 m 1 v1
+
2
1
2 m 2 v2
=
′2
1
2 m 1 v1
+
′2
1
2 m 2 v2 ,
(21)
(22)
and as explained in the textbook (page 176), together these two equations imply
v1 − v2 = v2′ − v1′
(23)
— the relative velocity of the two bodies changes its direction but keeps its magnitude,
v − rel = −vrel .
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For the two colliding balls in question, we know the velocities of both balls before the
collision and the velocity of the first ball after the collision. Hence, regardless of the balls’
masses, eq. (23) gives us the second ball’s velocity after the collision as
v2′ = v1′ + v2 − v2 = −3.7 m/s + 8.5 m/s − 0 m/s = 4.8 m/s.
(24)
Given this velocity, we can find the second ball’s mass from the momentum conservation
equation (21). Proceeding similarly to the previous problem 7.8, we write
m1 (v1 − v1′ ) = m2 (v2′ − v2 )
(25)
and hence
m2 = m1 ×
(+8.5/s) − (−3.7 m/s)
v1 − v1′
= 0.220 kg ×
≈ 0.56 kg.
′
v2 − v2
(+4.8 m/s) − 0 m/s
(26)
Non-textbook problem #1:
Here is the diagram of the collision:
y
~vtruck
30◦
~vcar
x
~v′ (car and truck)
In any collision, the net momentum vector just after the collision is equal to the net momen5
tum vector just before the collision,
m1~v1′ + m2~v2′ = m1~v1 + m2~v2 .
(27)
The collision between the car and the truck in question is totally inelastic — after the
collision ~vc′ = ~vt′ = ~v′ — so
(mt + mc )~v′ = mt~vt + mc~vc .
(28)
In components, this vector equation becomes
(mt + mc )vx′ = mt vtx + mc vcx ,
(29)
(mt + mc )vy′ = mt vty + mc vcy .
(30)
We know that after the collision, the truck and the car move due South. In our coordinate
system, this is the negative y direction, hence vx′ = 0 and vy′ < 0. On the other hand, given
the directions of the car and the truck just before the collision, their velocity vectors had
components
vtx = + sin(30◦ ) × vt ,
vty = − cos(30◦ ) × vt ,
vcx = −vt ,
vcy = 0.
(31)
Plugging all these components into eqs. (29) and (30), we obtain
+mt vt sin(30◦ ) − mc vc = 0,
(32)
−mt vt cos(30◦ ) + 0
(33)
< 0.
The second equation here does not tell us anything we do not know already, but the first
equation gives us the speed of the car before the collision:
vc =
mt
12, 000 lb
× 20 MPH × sin(30◦ ) = 40 MPH.
× vt sin(30◦ ) =
mc
3, 000 lb
6
(34)
Non-textbook problem #2:
All starts, planets, and large enough moons and asteroids are spheres or slightly oblate
spheroids, so by symmetry their respective centers of mass are located in their geometric
centers. In particularly, Pluto’s own CM is at Pluto’s center, and Charon’s CM is at Charon’s
center, L = 19600 km away from Pluto’s. Consequently, to find the CM of the Pluto–Charon
system, we may treat both bodies as point-like particles with all mass concentrated at the
respective CM’s. Hence, the system’s CM lies on the straight line connecting Pluto’s and
Charon’s centers, and its distance from Pluto’s center is
LP =
MC
1.9 · 1021 kg
× 19, 600 km = 2550 km. (35)
×L =
MP + M C
1.27 · 1022 kg + 1.9 · 1021 kg
This distance is larger than Pluto’s radius RP = 1137 km, so the system’s CM is above
Pluto’s surface by H = LP − RP ≈ 1410 km.
Non-textbook problem #3:
(a) The forces between the man and the boat can make them move relative to each other,
but they do not affect their net momentum. To change the net momentum, we need an
external force,
~ net = F
~ ext × ∆t.
∆P
(36)
The only horizontal external force acting on the man+boat system is the water drag on
the boat’s hull, but if the boat moves slowly this drag force is small and can be neglected.
Consequently, the horizontal component of the net momentum is conserved,
Pxnet = Mb vbx + Mm vmx = const.
(37)
Initially, the net momentum was zero because both the man and the boat were at rest.
Therefore, when the man walks on the boat, their net momentum (in the horizontal direction)
stays zero,
Pxnet = Mb vbx + Mm vmx = 0.
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(38)
The net momentum of the system is related to the motion of its center of mass
Xcm =
Mb X b + Mm X m
.
M b + Mm
(39)
Indeed, as the man and the boat move in x direction, the center of mass is displaced by
∆Xcm =
Mb ∆Xb + Mm ∆Xm
.
Mb + Mm
(40)
dividing both sides of this equation by ∆t and multiplying by the net mass Mb + Mm , we
get
(Mb + Mm )
∆Xb
∆Xm
∆Xcm
= Mb
+ Mm
= Mb vbx + Mm vmx = Pxnet .
∆t
∆t
∆t
(41)
The net momentum on the right hand side of this formula is zero and stays zero. Consequently, the left hand side must also vanish, thus
∆Xcm = 0
⇐⇒
Xcm = const.
(42)
In light of eq. (40), this means
Mb ∆Xb + Mm ∆Xm = 0
(43)
so when the man moves towards the pier, the boat moves back.
Clarification: To be precise, the Xb in eq. (39) is the boat’s own center of mass and likewise
Xm is man’s own CM, thus
system
Xcm
=
Mm
Mb
boat
man
× Xcm
+
× Xcm
Mm + Mb
Mm + M b
(44)
and hence the precise meaning of eq. (43) is
boat
man
Mb × ∆Xcm
+ Mm × ∆Xcm
= 0.
(45)
We do not know where the boat’s CM is relative to the boat’s stern or prow. But since the
boat does not change shape or rotate, the displacement of every particular part of the boat
8
is exactly the same,
∆X prow = ∆X stern = · · · = ∆X boat ,
(46)
and therefore the CM’s displacement is simply the boat’s displacement,
boat
∆Xcm
= ∆X boat .
(47)
As to the man, his legs and torso do not exactly move together as he stand up and walks from
the stern to the prow, but those differences are small compared to his overall displacement
from the stern to the prow. Therefore, we may neglects the man’s horizontal size (and the
fact that he moves one foot at a time) and approximate
man
= X man
Xcm
man
and ∆Xcm
= ∆X man .
(48)
Consequently, eq. (45) becomes simply
Mb × ∆X boat + Mm × ∆X man = 0.
(49)
(b) Eq. (49) gives us one relation between displacements of the man and the boat relative to
the pier. We also know that the man starts at the boat’s stern and ends at its prow, which
gives us his displacement relative to the boat, ∆X rel = L = 12 ft, the length of the boat. In
terms of the man’s and the boat’s displacements relative to the pier,
∆X rel = ∆X man − ∆X boat ,
(50)
∆X man − ∆X boat = L = 12 ft.
(51)
which tells us that
Together, eqs. (49) and (51) give us two linear equations for two unknowns, and all we need
9
to do is to solve them for the boat’s displacement. Combining the two equations, we have
∆X boat + L = ∆X man = −
Mb
× ∆X boat ,
Mm
(52)
hence
Mb
1+
Mm
× ∆X boat = −L,
(53)
−1
Mb
−L × 1 +
Mm
−1
70 lb
−12 ft × 1 +
140 lb
−1
3
−12 ft ×
2
−8 ft.
(54)
and therefore
×∆X
boat
=
=
=
=
Thus, while the man walks 12 feet from the stern to the prow, the boat moves back 8 feet
(i.e., 8 feet away from the pier).
And by the way, the man’s displacement relatively to the pier is only 12 ft − 8 ft = +4 ft.
Textbook problem 8.12:
(a) For constant angular acceleration α, the angular velocity ω changes with time according
to
ω(t) = ω0 + αt/
(55)
The angular velocity is related to the rotation rate as ω = 2πf (for ω in rad/s and f in RPS;
for f in RPM, there is an extra factor of 1 min/60 s). Hence for the wheel in question, the
initial angular velocity was
ω0 = 2π rad/s × 130 rev/min ×
1 min
= 13.6 rad/s ≡ 13.6 s−1 ,
60 s
(56)
while its final angular velocity at time tf = 4.0 s was
ωf = 2π rad/s × 280 rev/min ×
1 min
= 29.3 rad/s ≡ 29.3 s−1 .
60 s
10
(57)
Consequently, the wheel’s angular acceleration was
α =
ωf − ω0
29.3 rad/s − 13.6 rad/s
=
≈ 3.9 rad/s2 ≡ 3.9 s−2 .
tf
4.0 s
(58)
(b) The tangential acceleration of a point on a rotating wheel is related to the wheel’s angular
acceleration as
at = α × r
(59)
where r is the distance between the point in question and the axis of rotation. For a point
on wheel’s rim,
r = Rwheel =
wheel′ s diameter
,
2
hence for the wheel in question
at = αR = 3.9 s−2 ×
0.70 m
≈ 1.4 m/s2 .
2
(60)
But besides the tangential acceleration, a point on a rotating wheel also have the centripetal
acceleration
ac =
(ω × r)2
v2
=
= ω 2 × r.
r
r
(61)
At time t = 2.0 s, the wheel has angular velocity
ω = ω0 + αt = 13.6 rad/s + 3.9 rad/s2 × 2.0 s ≈ 21.5 rad/s ≡ 21.5 s−1 ,
(62)
hence the centripetal acceleration of the point on the rim is
ac = ω 2 × R = (21.5 s−1 )2 ×
0.70 m
≈ 160 m/s2 .
2
(63)
Note that this centripetal acceleration is much larger than the tangential acceleration (60).
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Textbook problem 8.18:
The linear distance traveled by a point on rim of a wheel of radius R is related to the angular
displacement of the wheel according to
s = R × ∆φ
(64)
provided we measure the ∆Φ in radians rather than some other units. For a constantly
accelerating wheel,
∆Φ = ωavg ∆t =
ω1 + ω2
× (t2 − t1 )
2
(65)
hence
s = R×
ω1 + ω2
× (t2 − t1 ),
2
(66)
and all we need to compute this distance is to convert the problem’s data into the appropriate
units. The initial rotation rate is f1 = 240 RPM = 4.0 RPS, hence
ω1 = 2π × f1 = 25.1 rad/s ≡ 25.1 s−1 .
(67)
Likewise, the final rotation rate f2 = 360 RPM = 6.0 RPS translates to
ω2 = 2πf2 = 37.7 rad/s ≡ 37.7 s−1 .
(68)
Finally, the wheel’s radius is one half of its diameter,
R =
33 cm
= 16.5 cm.
2
(69)
Plugging all these data and t2 − t1 = 6.5 s into eq. (66) gives us
s = 0.165 m ×
25.1 s−1 + 37.7 s−2
× 6.5 s = 23.7 m ≈ 34 m ≈ 110 ft.
2
12
(70)
Textbook problem 8.68:
As the spool rolls on the floor, it rotates with angular speed ω while its center moves at linear
velocity vc relative to the floor. The point at the spool’s rim has linear speed vrot = ω × R
relative to the spool’s center. At the bottom point where the spool touches the floor, this
rotational velocity is directed back, hence the velocity of the bottom point relative to the
floor is
v = −vrot + vc = −ωR + vc .
(71)
If the spool rolls without slipping, the bottom of the spool should have zero velocity relative
to the floor, hence
−ωR + vc = 0,
(72)
and therefore the angular velocity of the spool determines the linear velocity of is center
(relative to the floor) as
vc = ωR.
(73)
Consequently, the linear displacement of the spool’s center is related to the angular displacement of the spool as
∆Xc = vc × ∆t = Rω × ∆t = R × ∆φ.
(74)
Now consider the rope wound around this spool. Let ℓ denote the length of the un-wound
part of the rope. According to the textbook figure 8–50, the position Xm of the man holding
the spool’s end is related to ℓ and to the spool center’s position Xc as
Xm = Xc + ℓ.
(75)
When the spool rotates clockwise through angle ∆φ, a piece of rope of length R × ∆φ gets
un-wound from the spool, so ℓ increases by
∆ℓ = R × ∆φ.
(76)
At the same time, the spool’s center moves to the right through distance (74). Consequently,
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eq. (75) tells us that the man moves to the right through distance
∆Xm = ∆Xc + ∆ℓ = R∆φ + R∆φ = 2R∆φ.
(77)
Comparing this man’s displacement to the spools’ displacement (75) and the rope unwinding (76), we immediately see that
∆Xc = ∆ℓ =
1
2 ∆Xm .
(78)
In other words, when the man holding the rope walks through distance L, the spool moves
through half of that distance while 12 L length of rope gets un-wound from the spool.
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