Download Physics 103 Final Exam Solution

Physics 103 Final Exam
Solution
Point values are given for each problem. Within a problem, the points are not
necessarily evenly divided between the parts. The total is 85 points.
1. [5 points] Which would appear brighter to an observer on Earth: a single star,
1 kpc away, or a galaxy of 10,000,000,000 stars, 1 Mpc away? Justify your
answer. Assume that the single nearby star and all of the stars in the more
distant galaxy have exactly the same properties.
Intensity is proportional to r12 , and 1 Mpc is 1000 times farther than 1 kpc,
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so a single star 1 Mpc away would be 1,000,000
times as bright as a single
star 1 kpc away.
If you have 10,000,000,000 stars at 1 Mpc, their intensity would be
10,000,000,000
= 10, 000 times that of a single star at 1 kpc. Thus the
1,000,000
galaxy is brighter.
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2. [10 points] A galaxy is at redshift Z = 1.6.
(a) How fast is this galaxy moving away from us?
m
(Z + 1)2 − 1
(1.6 + 1)2 − 1
v=c
=c
= 0.742c = 2.2 × 108
2
2
s
(Z + 1) + 1
(1.6 + 1) + 1
(b) According to Hubble’s law, how far away is this galaxy?
v
= H0 d

d
=
108 ms
2.2 ×
v
=
km
H0
71 s


1 km
1000 m
= 3100 Mpc = 9.6 × 1025 m
Mpc
(c) A type Ia supernova goes off in the galaxy. With the distance calculated in
part (a) and knowledge of type Ia supernovae, we could make a prediction
about how bright the supernova will be when it is seen from the Earth.
Based on the conclusions drawn from distant supernova observations,
would you expect the actual brightness of the supernova at redshift 1.6
to be greater than or less than that predicted using the distance from the
Hubble law? What does this tell us about large-scale motion of galaxies
in the universe? (Very brief answers are acceptable.)
Recent observations of distant supernovae show that they are consistently
less bright than predicted by the Hubble law. This tells us that, on the
largest scales of the universe, galaxies are accelerating
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Sun
Diffraction
Grating
Wall of
Space Station
A
B
C
D
3. [5 points] An astronaut on the space station wants to get a suntan.
She knows that, in addition to visible light, the sun produces ultraviolet light,
and that exposing herself to ultraviolet light will give her a suntan. She replaced
one of the windows of the space station with a diffraction grating. Sunlight
shining into the window gets diffracted onto the wall opposite the window, with
different colors of light ending up at different points on the wall. The first order
spectra are spread out over the two locations indicated by long gray rectangles
in the diagram. Of course, the real spectra would have a rainbow of colors,
going in one direction or the other from red to violet, but the colors are not
shown in the figure.
The astronaut knows that the wavelengths of ultraviolet light are shorter than
the wavelengths of visible light. At which of the position(s), A, B, C, and/or
D, indicated in the figure should she locate herself to get a suntan?
Bright spots in a diffraction pattern are at positions
λ
y = n D,
d
where y is the distance from the center of the pattern. The important
point here is that y is proportional to wavelength λ. The wavelength of
ultraviolet light is shorter than that of visible light, so the ultraviolet light
will be at relatively small distances from the center of the pattern. Thus
the astronaut should go to point B or C to get her suntan.
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4. [15 points] Imagine that you have discovered a new element, mysterium, which
has atomic mass 200. In your lab, you have a 1 m3 box filled with mysterium
gas at 300 K and 1×105 mN2 . Mysterium gas is diatomic. (Each molecule consists
of two mysterium atoms.)
(a) How many moles of molecules of gas are in the box?
nm =
PV
=
RT
P V = nm RT
#
1 × 105 mN2 1 m3
J
= 40 moles.
J
nm
8.314 K
(300 K)
"
(b) How many molecules of gas are in the box (give a number, not in moles)?
(40 moles)
6.02 × 1023 molecules
mole
= 2.4 × 1025 molecules
(c) What is the average kinetic energy per molecule?
3
3
E = kT =
2
2
1.38 × 10
−23
J
K
(300 K) = 6.21 × 10−21 J
(d) What is the root-mean-square speed of the molecules?
The atomic weight is 200, so the molecular weight of diatomic molecules is
400. This means that 1 mole of molecules has mass 400 g=0.400 kg, so the
0.400 kg
−25
mass of an individual molecule is m = 6.02×10
kg.
23 = 6.64 × 10
E=
r
v=
2E
=
m
s
1
mv 2
2
(2)(6.21 × 10−21 J)
6.64 × 10−25 kg
kg m2
J s2
= 136
m
s
This problem is continued on the next page....
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Continued from the previous page....
Imagine that all the molecules suddenly split into two, so that the two atoms
could move independently. Assume that this splitting happened without adding
or removing energy to the system, so that the total kinetic energy is the same
as before the split.
(e) What is the average energy per atom now?
The energy of each molecule, calculated above, was 6.21 × 10−21 J. If the
molecule splits into two atoms without energy added or removed, each atom
will get half this, or 3.10 × 10−21 J.
(f) What is the root-mean-square speed of the atoms?
The energy per atom is half the energy per molecule, but the mass per atom
was half the mass per molecule. From E = 21 mv 2 , these effects cancel out
and the speed is the same as calculated for molecules above, 136 ms .
(g) What is the temperature now?
This answer is surprising. From E = 32 kT , the temperature is proportional
to the energy per gas particle, and it doesn’t matter what type of particle
it is. The energy per particle dropped by half when the molecules split into
atoms, so the temperature dropped by half, to 150 K . Pretty weird, eh?
(h) What is the pressure now?
Use P V = nm RT : the number of moles nm has doubled but the
temperature T has halved, volume V is unchanged, and R is a constant.
Thus P is unchanged, at 1.0 × 105 mN2 .
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droplet path
20,000 V
0.1 m
droplet starts here
5. [15 points]
(a) There is a small, spherical droplet with charge q = −1.6 × 10−17 C, mass
2.00 × 10−15 kg and radius 1.00 × 10−6 m located at the plate as shown in
the figure. It is accelerated through voltage 20,000 V by a voltage placed
across the parallel plates. Assume that the space between the plates is
empty except for the droplet as it is passing through. What is the energy
of the droplet as it emerges from the top plate? (Ignore gravity. All
numbers in this problem are non-relativistic.)
The drop has charge 100e, so accelerating it through 20,000 V
gives an energy of (100)(20, 000) = 2 × 106 eV . Alternatively, use
E = qV = (1.6 × 10−17 C)(20, 000 V) VJC = 3.2 × 10−13 J.
(b) What is the velocity as it emerges from the top plate?
E=
r
v=
2E
=
m
s
1
mv 2
2
(2)(3.2 × 10−13 J)
2.00 × 10−15 kg
kg m2
J s2
= 17.9
m
.
s
This problem is continued on the next page....
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Continued from the previous page....
(c) What is the electric field between the plates if they are separated by 0.1 m?
E=
20, 000 V
N
V
V
=
= 200, 000
= 200, 000 .
d
0.1 m
m
C
(d) What is the force on the droplet while it is being accelerated?
−12
N
F = qE = (1.6 × 10−17 C) 200, 000 N
C = 3.2 × 10
(e) Imagine that the space between the plates is not empty but, instead, is
filled with a gas of viscosity 1.8 × 10−5 mkgs . What would be the speed of
the particle as it emerged from the top plate in this situation? (Continue
to ignore gravity. Assume this experiment is being done in outer space.)
As the drop accelerates, the viscous drag force increases until it balances
out the electrical force and the drop moves with constant speed. Thus:
6πavη
v
v
= qE
qE
=
6πaη
(1.6 × 10−17 )(200, 000)
=
6π(1 × 10−6 )(1.8 × 10−5 )
m
= 0.0094
s
The fact that this speed is much less than the one calculated for empty
space between the plates tells us that our assumption that the forces are
balanced is correct.
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6. [10 points] Our Galaxy is about 30 kiloparsecs (9.3 × 1020 m) across. How long
would it take a proton with energy 1 × 1013 MeV to cross the Galaxy in the rest
frame of...
(a) . . . the Galaxy?
1 × 1013 MeV
E
=
= 1.1 × 1010
2
mc
938 MeV
The Lorentz factor is so high that we might as well take the speed of the
proton to be c. The actual speed is v = 0.9999999999999999995c. The time
it takes to cross the galaxy is
E = γmc2
v=
d
t
⇒
⇒
t=
γ=
d
9.3 × 1020 m
= 3.1 × 1012 s = 100, 000 yr.
=
v
3.0 × 108 ms
(b) . . . the proton?
Due to time dilation, the time in the proton’s reference frame is a factor of
γ = 1.1 × 1010 shorter, which is 280 s.
You can also do this using events and a Lorentz transformation equation.
Let event A be the proton “coming into” the Galaxy and let event B be
the proton “leaving” the Galaxy. The Galaxy doesn’t really have an edge
at which something can come in and leave, but never mind that. In the
proton’s frame, both of these events happen at the same place (i.e., at the
proton), so ∆x0 = 0. We know ∆t from part (a), and we want to find ∆t0 .
Use
∆t = γ ∆t0 + c12 V ∆x0
∆t
∆t0
= γ (∆t0 + 0)
∆t0
.
=
γ
This gives 280 s, as above.
b
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A
B
A
B
Before collision
After collision
7. [10 points] Note: all the numbers in this problem are non-relativistic. You
should not use relativistic formulas.
There are two rubber balls with mass 1 kg each.
laboratory.
The are observed in a
Initially, ball A has velocity +10 ms (it moves in the +x direction)
Initially, ball B has velocity −10 ms (it moves in the −x direction)
The two balls collide and bounce off of each other. After the collision:
The final velocity of ball A is −10 ms (in the −x direction)
The final velocity of ball B is +10 ms (in the +x direction)
During the collision, as observed in the laboratory, energy is conserved: the
final total kinetic energy of the balls is the same as the initial kinetic energy.
(a) Someone rides by on a bicycle moving in the +x direction at +10 ms . What
are the initial and final velocities of balls A and B in the bicyclist’s reference
frame?
Initial speed of ball A: v 0 = v − V = 10 ms − 10 ms = 0
Initial speed of ball B: v 0 = v − V = −10 ms − 10 ms = −20 ms
Final speed of ball A: v 0 = v − V = −10 ms − 10 ms = −20 ms
Final speed of ball B: v 0 = v − V =
10 ms − 10 ms = 0
(b) What are the initial and final kinetic energies of each of the balls in this
reference frame? Is kinetic energy conserved according to the observations
made by the bicyclist?
Initial kinetic energy of ball A:
Initial kinetic energy of ball B:
Final kinetic energy of ball A:
Final kinetic energy of ball B:
K
K
K
K
=
=
=
=
1
2
2 mv
1
2
2 mv
1
2
2 mv
1
2
2 mv
=
=
=
=
0
200 J
200 J
0
Total initial kinetic energy was 200 J. Total final kinetic energy was 200 J.
Energy was conserved.
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8. [5 points] Recall the example from class of a pole vaulter running through a
barn. In the reference frame of the Earth, the barn is 5 m long and at rest. It
has a front door and a back door. Initially the back door is closed and the front
door is open.
The pole is 10 m long viewed at rest. The pole vaulter runs fast enough that the
pole is length contracted to 5 m so that it barely fits into the barn (in the barn’s
reference frame). The fact that the pole is in the barn is proven by leaving the
back door of the barn closed until the pole is fully in the barn and by closing
the front door of the barn just as the pole has fully gone into the barn. Then
the back door is immediately opened, so that the pole vaulter can continue on
his way.
So, in the barn frame, the opening of the back door and the closing of the front
door happen at essentially the same time.
Can these two events, the opening of the back door and the closing of the front
door, be causally connected? Briefly justify your answer.
In the barn reference frame, the spacetime separation of these events is
s2 = (c∆t)2 − (∆x)2 − (∆y)2 − (∆z)2 = 0 − (5 m)2 − 0 − 0 = −25 m2 .
A negative spacetime tells us that these are space-like events and
they cannot be causally connected.
More simply put, if two events happen at the same time in different places,
there is no way for a message to be sent from one event to the other event,
so it is impossible for one event to cause the other event.
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9. [10 points]
(a) Suppose that in Sherwood forest the trees are all very tall and their trunks
all have radius r = 1 m. The average number of trees per area of ground
is 0.005 trees
. If Robin Hood shoots an arrow in a random direction, how
m2
far, on average, will it travel before it strikes a tree?
Notes: (i) Neglect gravity, and assume that the arrow will keep following a
straight path until it hits
√ a tree. (ii) It is OK if you final answer is off by a
small factor like π or 2. Don’t sweat the small stuff. (iii) This situation
is similar to the calculation of the mean-free-path, `, in an ideal gas. The
formula for ` is given on the formula sheet, but simply plugging numbers
into the formula will not give the correct answer for this problem. (iv)
Assume the arrow is very small. If you prefer, instead of small arrows, you
can imagine Robin Hood to be throwing spherical beach balls with radius
1 m, and you can imagine that the tree trunks aver very narrow; this will
lead to the same answer.
l
r
Let σ = 0.005 trees
m2 be the area density of trees in the forest. Then the
average area per tree is A = σ1 = 200 m2 .
See the diagram. The arrow starts at the asterisk and goes for a length
l before hitting a tree. Any tree whose center is within distance r of the
arrow’s path gets hit. Thus we can think of the tree as sweeping out an
area of 2rl before hitting a tree. This is analogous to the derivation of the
mean free path length of an ideal gas, except it is in two dimensions and it
assumes the arrow has negligible size. To find a typical distance the arrow
travels, equate the area swept out with the area per tree, 2rl = A and solve
for l:
A
200 m2
l=
=
= 100 m.
2r
(2)(1 m)
This problem is continued on the next page....
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Continued from the previous page....
(b) I found this problem about Sherwood forest in a textbook on cosmology.
The textbook author didn’t say so, but she must have been trying to make
an analogy between the forest and something in cosmology. What might
have that analogy be?
Hint: Let l be the distance you calculated in part (a). Think about the
difference in what would happen to the arrows in an infinitely large forest
compared to a finite forest, much smaller than l in size.
This is an example of Olber’s paradox.
Imagine the universe to be infinite, homogeneous on large scales, and filled
with galaxies of some finite size. If we were to look out in any direction,
our line of sight would eventually intersect a galaxy. There would be no
dark sky at night.
This is like Robin Hood’s arrow: shot in any direction, it will eventually
hit a tree.
In reality, the universe is finite. The sky is dark at night. We can look in
many directions and our line of sight does not hit a galaxy.
This is like Robin Hood’s arrow shot in a very small forest. In most
directions, the arrow will fly out of the forest without hitting a tree.
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