Download Physics 103 Hour Exam #3 Solution Point values are given for each

Physics 103 Hour Exam #3
Solution
Point values are given for each problem. Within a problem, the points are not
necessarily evenly divided between the parts. The total is 50 points.
1. [15 points]
(a) What does the metric prefix Giga mean?
109
(b) What is 100 µs in seconds?
(100 µs)
1s
106 µs
= 10−4 s
(c) The Tevatron is a particle accelerator at Fermilab, near Chicago. It
accelerates protons up to energies of 1 TeV. What is this in nanoJoules?
(1 TeV)
1012 eV
TeV
1.6 × 10−19 J
eV
1
109 nJ
J
= 160 nJ
2. [10 points] Mars has two moons, Phobos and Deimos, both of which have nearly
circular orbits.
Phobos has mass 1.07 × 1016 kg; it orbits Mars once every 27,550 seconds
following an orbit with radius 9.377 × 106 m.
Deimos has mass 2.244 × 1015 kg; it orbits Mars once every 109,100 seconds
following an orbit with radius 2.346 × 107 m.
Use this information to calculate the mass of Mars.
(Hint: there is much more information here than you need.)
The solution here is written using the orbital radius and period of Phobos.
You could do the same calculation using the orbital radius and period of
Deimos, and you would get the same answer.
The speed of Phobos in its orbit is
v=
2πr
T
In order to keep Phobos in a circular orbit, the gravitational force must
v2
equal mp :
r
mp mm
r2
mp mm
G
r2
G
mm
v2
r
(2πr)2
= mp 2
T r
4π 2 r3
=
GT 2
2π(9.37 × 106 m)3
= 2
−
2
= mp
6.67×10 11 N m
kg2
=
6.42 × 1023 kg
2
(27 550 s)
kg m
N s2
3. [10 points] Positrons are particles which are identical to electrons (for example,
they have have the same mass as electrons), except that positrons are positively
charged. Positrons were first discovered in a 1933 study of cosmic rays by Carl
Anderson. He detected the positrons by using a cloud chamber in the presence
of a magnetic field. Charged particles leave visible tracks as they travel through
a cloud chamber. In the presence of a magnetic field, charged particles follow
curved paths, so their tracks are curved. One of the first images ever taken of a
positron is shown below. The positron track is the curved line in the left-center
part of the figure. It is partly above and partly below the horizontal band across
the middle of the figure.
This problem is continued on the next page....
Figure from Anderson, C. D., Physical Review, vol. 43, p. 491 (1933).
3
(a) How could Anderson tell that this particle was a positron rather than an
electron? Give a brief answer. (Hint: another way to ask this question
is: how would the track be different if it were due to an electron? Make a
reasonable assumption about how the apparatus was set up.)
If the particle were an electron, the force on it would have had
~
and hence
the opposite sign (because of the q in F~ = q~v × B),
the track would have curved in the opposite direction. In order for the
positively charged positron to curve in the direction shown in the picture,
the magnetic field must have been pointed into the page.
Carl Anderson was awarded the Nobel Prize in 1936 for this discovery.
(b) The dark horizontal band across the center of the figure is a 6 mm thick lead
plate. As the positron passed through the lead plate, it lost energy. Thus,
the positron had more energy on one side of the plate (on whichever side
it entered the chamber) and less energy on the other side. By considering
the differences between the curved tracks above and below the lead plate,
determine whether the positron had more energy in the upper half of the
chamber or the lower half of the chamber. In other words, did it start
near the top and go down to the bottom, or vice versa? (Assume that the
magnetic field is uniform throughout the apparatus.)
A charged particle traveling perpendicular to a uniform magnetic field
experiences circular motion. We can relate the speed of the particle to
the radius of its circular path:
F
qvB
qBr
m
= ma
v2
= m
r
= v
Hence its kinetic energy is
1
1
KE = mv 2 = m
2
2
qBr
m
2
=
q2 B 2 r2
2m
When the particle has higher energy, it has a larger radius of curvature, r;
that is, its path is “less curved.” In the figure on the previous page, the
path in the lower half of the chamber is less curved, hence the positron had
more energy in the lower half.
4
4. [15 points]
d=0.01m
V=5000V
Consider a Millikan-type oil drop experiment.
The drop has density 900 kg/m3 , radius 1.5×10−6 m, and mass 1.272×10−14 kg.
The air in the oil-drop experiment has temperature 273 K, pressure
1.013 × 105 Pa, and viscosity η = 1.83 × 10−5 Pa s.
(a) Initially the voltage is turned off, so there is no electric field. What is the
terminal velocity of the falling oil drop?
6πavη
v
= mg
mg
=
6πaη
(1.272 × 10−14 kg) 9.8 sm2
Pa m2
N s2
=
6π(1.5 × 10−6 m)(1.83 × 10−5 Pa s)
N
kg m
m
= 2.41 × 10−4
s
(b) Now the voltage is turned on, with a voltage of 5000 V applied across a
distance of d=0.01 m. The lower plate is at a higher voltage than the upper
plate, so the electric field points in an upward direction. The oil drop is
seen to move upward. What is the smallest possible upward velocity that
the oil drop could have? (Hint: what are the possible values of charge that
the oil drop could have? Further hint: what is the fundamental thing that
was learned from the Millikan oil drop experiment?)
The exam originally had an error in the figure above, giving the voltage as
500 V rather than 5000 V as specified in the problem. The figure has been
corrected, and the solution written assuming 5000 V.
The solution is given on the next page
Use the next page if you need extra work space....
5
This blank page is extra work space for problem 4.
qE
6π avη
mg
The electric field is E =
V
d
= 5 × 105
V
m
J
VC
Nm
J
= 5 × 105
N
C.
The particle is moving upward, so the Stokes force must be acting downward (since
it acts in the direction which slows the motion). Gravity also acts downward. The
electric force must be acting upward, and must equal the sum of the other forces.
6πavη + mg
6πavη
v
= qE
= qE − mg
qE − mg
=
6πaη
The smallest possible charge that any particle could have is q = e = 1.602 × 10−19 .
Plug in this value along with the known values of E, m, g, a, and η:
v =
qE − mg
6πaη
1.602 × 10−19 C
=
5 × 105
N
C
− 1.272 × 10−14 kg
9.8 sm2
6π(1.5 × 10−6 m)(1.83 × 10−5 Pa s)
= −8.62 × 10−5
N s2 kg m
Pa m2
N
m
s
That answer is no good, because we set the problem up in such a way that we’re
supposed to get a positive velocity. The negative velocity here means that the
particle is still moving downward. The electric force is not strong enough to
overcome gravity. More charge is needed, so try the next-smallest possibility,
q = 2e = 3.204 × 10−19 :
v =
qE − mg
6πaη
3.204 × 10−19 C
=
=
5 × 105
N
C
− 1.272 × 10−14 kg
6π(1.5 × 10−6 m)(1.83 × 10−5 Pa s)
6.86 × 10−5
m
s
That answer is positive, so it is acceptable.
Any larger charge (3e, 4e, etc.) would give a higher speed.
6
9.8 sm2
N s2 kg m
Pa m2
N