Chapter 36 Solutions - galileo.harvard.edu

Transcription

CHAPTER 36: The Special Theory of Relativity
Responses to Questions
1.
No. The train is an inertial reference frame, and the laws of physics are the same in all inertial
reference frames, so there is no experiment you can perform inside the train car to determine if you
are moving.
2.
The fact that you instinctively think you are moving is consistent with the relativity principle applied
to mechanics. Even though you are at rest relative to the ground, when the car next to you creeps
forward, you are moving backward relative to that car.
3.
As long as the railroad car is traveling with a constant velocity, the ball will land back in his hand.
4.
The relativity principle refers only to inertial reference frames. Neither the reference frame of the
Earth nor the reference frame of the Sun is inertial. Either reference frame is valid, but the laws of
physics will not be the same in each of the frames.
5.
The starlight would pass at c, regardless of your spaceship’s speed. This is consistent with the
second postulate of relativity which states that the speed of light through empty space is independent
of the speed of the source or the observer.
6.
It deals with space-time (sometimes called “the fabric of space-time”) and the actual passage of time
in the reference frame, not with the mechanical workings of clocks. Any measurement of time
(heartbeats or decay rates, for instance) would be measured as slower than normal when viewed by
an observer outside the moving reference frame.
7.
Time actually passes more slowly in the moving reference frames, according to observers outside
the moving frames.
8.
This situation is an example of the “twin paradox” applied to parent-child instead of to twins. This
might be possible if the woman was traveling at high enough speeds during her trip. Time would
have passed more slowly for her and she could have aged less than her son, who stayed on Earth.
(Note that the situations of the woman and son are not symmetric; she must undergo acceleration
during her journey.)
9.
No, you would not notice any change in your heartbeat, mass, height, or waistline, because you are
in the inertial frame of the spaceship. Observers on Earth, however, would report that your heartbeat
is slower and your mass greater than if you were at rest with respect to them. Your height and
waistline will depend on your orientation with respect to the motion. If you are “standing up” in the
spaceship such that your height is perpendicular to the direction of travel, then your height would not
change but your waistline would shrink. If you happened to be “lying down” so that your body is
parallel to the direction of motion when the Earth observers peer through the telescope, then you
would appear shorter but your waistline would not change.
10. Yes. However, at a speed of only 90 km/hr, v/c is very small, and therefore γ is very close to one, so
the effects would not be noticeable.
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427
Physics for Scientists & Engineers with Modern Physics, 4 th Edition
Instructor Solutions Manual
11. Length contraction and time dilation would not occur. If the speed of light were infinite, v/c would
be zero for all finite values of v, and therefore γ would always be one, resulting in t
t0 and
l
l 0.
12. The effects of special relativity, such as time dilation and length contraction, would be noticeable in
our everyday activities because everyday speeds would no longer be so small compared to the speed
of light. There would be no “absolute time” on which we would all agree, so it would be more
difficult, for instance, to plan to meet friends for lunch at a certain time! In addition, 25 m/s would
be the limiting speed and nothing in the universe would move faster than that.
13. Both the length contraction and time dilation formulas include the term 1 v 2 c 2 . If c were not
the limiting speed in the universe, then it would be possible to have a situation with v > c. However,
this would result in a negative number under the square root, which gives an imaginary number as a
result, indicating that c must be the limiting speed.
14. Mr. Tompkins appears shrunk in the horizontal direction, since that is the direction of his motion,
and normal size in the vertical direction, perpendicular to his direction of motion. This length
contraction is a result of the fact that, to the people on the sidewalk, Mr. Tompkins is in a moving
frame of reference. If the speed of light were only 20 mi/h, then the amount of contraction, which
depends on γ, would be enough to be noticeable. Therefore, Mr. Tompkins and his bicycle appear
very skinny. (Compare to the chapter-opening figure, which is shown from Mr. Tompkin’s
viewpoint. In this case, Mr. Tompkins sees himself as “normal” but all the objects moving with
respect to him are contracted.)
mv
15. No. The relativistic momentum of the electron is given by p
mv
. At low speeds
1 v 2 c2
(compared to c) this reduces to the classical momentum, p = mv. As v approaches c, γ approaches
infinity so there is no upper limit to the electron’s momentum.
16. No. To accelerate a particle with nonzero rest mass up to the speed of light would require an infinite
amount of energy, and so is not possible.
17. No. E = mc² does not conflict with the principle of conservation of energy as long as it is understood
that mass is a form of energy.
18. Yes, mass is a form of energy so technically it is correct to say that a spring has more mass when
compressed. However, the change in mass of the spring is very small and essentially negligible.
19. “Energy can be neither created nor destroyed.” Mass is a form of energy, and mass can be
“destroyed” when it is converted to other forms of energy. The total amount of energy remains
constant.
20. Technically yes, the notion that velocities simply add is wrong. However, at everyday speeds, the
relativistic equations reduce to classical ones, so our ideas about velocity addition are essentially
true for velocities that are low compared to the speed of light.
428
Chapter 36
The Special Theory of Relativity
Solutions to Problems
1.
You measure the contracted length. Find the rest length from Eq. 36-3a.
l
38.2 m
l0
72.5m
2
2
2
1 v c
1 0.850
2.
We find the lifetime at rest from Eq. 36-1a.
t0
2
t 1 v c
2
6
4.76 10 s
2.70 108 m s
3.00 108 m s
1
2
3.
The numerical values and
graph were generated in a
spreadsheet. The graph is
shown also. The spreadsheet
used for this problem can be
found on the Media Manager,
with filename
“PSE4_ISM_CH36.XLS,” on
tab “Problem 36.3.”
4.
The measured distance is the contracted length. Use Eq. 36-3a.
2
l
5.
l0 1 v c
v
1
2
48.5 ly
t 1 v 2 c2
c 1
t0
t
2
c 1
2.60 10 8 s
4.40 10 8 s
2
0.807c
2.42 108 m s
The speed is determined from the length contraction relationship, Eq. 36-3a.
l
7.
135 ly
2.80 108 m s
3.00 108 m s
The speed is determined from the time dilation relationship, Eq. 36-1a.
t0
6.
2
2.07 10 6 s
l 0 1 v2 c2
v
c 1
l
l0
2
c 1
35ly
56ly
2
0.78c
2.3 108 m s
The speed is determined from the length contraction relationship, Eq. 36-3a. Then the time is found
from the speed and the contracted distance.
l
l 0 1 v 2 c2
429
v
c 1
l
l0
2
l
v
; t
l
8.
2
2
25ly
65ly
c 1
25y c
c 0.923
27 y
The speed is determined from the length contraction relationship, Eq. 36-3a.
l
9.
25ly
l
l0
c 1
l 0 1 v2 c2
v
2
l
l0
c 1
c 1
0.900
2
0.436 c
The change in length is determined from the length contraction relationship, Eq. 36-3a. The speed is
very small compared to the speed of light.
l
l 0 1 v2 c2
l
l0
2
1 v c
v2
1 2
c
2
1/ 2
1
2
1
So the percent decrease is 6.97 10
8
v2
c2
11.2 103 m s
3.00 108 m s
1
2
1
2
1 6.97 10
10
%.
10. (a) The measured length is the contracted length. We find the rest length from Eq. 36-3a.
l
4.80m
l0
7.39 m
2
2
2
1 v c
1 0.760
Distances perpendicular to the motion do not change, so the rest height is 1.35m .
(b) The time in the spacecraft is the rest time, found from Eq. 36-1a.
t0
t 1 v 2 c2
20.0s
1
0.760
2
13.0s
(c) To your friend, you moved at the same relative speed: 0.760 c .
(d) She would measure the same time dilation: 13.0s .
11. (a) We use Eq. 36-3a for length contraction with the contracted length 99.0% of the rest length.
2
l
l0 1 v c
2
v
c 1
l
l0
2
c 1
0.990
2
0.141c
(b) We use Eq. 36-1a for time dilation with the time as measured from a relative moving frame
1.00% greater than the rest time.
2
t0
1
t0
t 1 v c
v c 1
c 1
t
1.0100
We see that a speed of 0.14 c results in about a 1% relativistic effect.
2
2
2
0.140 c
12. (a) To an observer on Earth, 18.6 ly is the rest length, so the time will be the distance divided by
the speed.
18.6 ly
l0
tEarth
19.58 yr 19.6 yr
v
0.950 c
(b) The time as observed on the spacecraft is shorter. Use Eq. 36-1a.
t0
t 1 v 2 c2
19.58yr
1
0.950
2
6.114 yr
6.11yr
430
Chapter 36
(c) To the spacecraft observer, the distance to the star is contracted. Use Eq. 36-3a.
l 0 1 v 2 c2
l
18.6 ly
1
0.950
2
5.808 ly
5.81 ly
(d) To the spacecraft observer, the speed of the spacecraft is their observed distance divided by
their observed time.
5.808 ly
l
v
0.950 c
t0
6.114 yr
13. (a) In the Earth frame, the clock on the Enterprise will run slower. Use Eq. 36-1a.
t0
t 1 v 2 c2
5.0yr
1
0.74
2
3.4 yr
(b) Now we assume the 5.0 years is the time as measured on the Enterprise. Again use Eq. 36-1a.
5.0 yr
t0
t0
t 1 v 2 c2
t
7.4 yr
2
1 v 2 c2
1 0.74
14. We find the speed of the particle in the lab frame, and use that to find the rest frame lifetime and
distance.
xlab
1.00 m
v
2.941 108 m s 0.9803 c
tlab 3.40 10 9 s
(a) Find the rest frame lifetime from Eq. 36-1a.
t0
tlab 1 v 2 c 2
3.40 10 9 s
1
0.9803
2
6.72 10 10 s
(b) In its rest frame, the particle will travel the distance given by its speed and the rest lifetime.
x0 v t0 2.941 108 m s 6.72 10 10 s
0.198m
This could also be found from the length contraction relationship:
x0
xlab
1 v2 c2
.
15. Since the number of particles passing per second is reduced from N to N / 2, a time T0 must have
elapsed in the particles’ rest frame. The time T elapsed in the lab frame will be greater, according to
Eq. 36-1a. The particles moved a distance of 2cT0 in the lab frame during that time.
T0
x
2cT0
4
T0 T 1 v 2 c 2
T
; v
v
0.894 c
5c
2
2
T0
T
1 v c
1 v 2 c2
16. The dimension along the direction of motion is contracted, and the other two dimensions are
unchanged. Use Eq. 36-3a to find the contracted length.
l
l 0 1 v 2 c2 ; V
l l0
2
l0
3
1 v 2 c2
2.0m
3
1
0.80
2
4.8m3
17. The vertical dimensions of the ship will not change, but the horizontal dimensions will be contracted
according to Eq. 36-3a. The base will be contracted as follows.
l base
l 1 v 2 c2
l 1
0.95
2
0.31l
0.50l
75.52 .
2.0l
1.936l is unchanged. The horizontal
When at rest, the angle of the sides with respect to the base is given by
The vertical component of l vert
2l sin
2l sin75.52
cos
1
431
component, which is 2 l cos
2l
0.50l 1 v 2 c2
l horizontal
1
4
0.50l at rest, will be contracted in the same way as the base.
0.50l 1
2
0.95
0.156l
Use the Pythagorean theorem to find the length of the leg.
l 2horizontal
l leg
l 2vert
2
0.156l
2
1.936l
1.942l
1.94l
18. In the Earth frame, the average lifetime of the pion will be dilated according to Eq. 36-1a. The speed
of the pion will be the distance moved in the Earth frame times the dilated time.
d
d
v
1 v 2 c2
t
t0
v
1
c
c t0
d
1
2
1
c
3.00 10 m s 2.6 10 8 s
1
0.95 c
2
8
25m
19. We take the positive direction in the direction of the Enterprise. Consider the alien vessel as
reference frame S, and the Earth as reference frame S . The velocity of the Earth relative to the alien
vessel is v
0.60 c. The velocity of the Enterprise relative to the Earth is ux 0.90 c. Solve for
the velocity of the Enterprise relative to the alien vessel, u x , using Eq. 36-7a.
ux v
0.90c 0.60c
ux
0.65c
vux
1
0.60 0.90
1
c2
We could also have made the Enterprise as reference frame S, with v
0.90 c, and the velocity of
the alien vessel relative to the Earth as ux 0.60 c. The same answer would result.
Choosing the two spacecraft as the two reference frames would also work. Let the alien vessel be
reference frame S, and the Enterprise be reference frame S . Then we have the velocity of the Earth
0.60 c, and the velocity of the Earth relative to the Enterprise as
relative to the alien vessel as ux
ux
0.90 c. We solve for v, the velocity of the Enterprise relative to the alien vessel.
ux v
.60c
0.90c
ux ux
ux
v
0.65c
vux
ux ux
0.90c .60c
1
1
1
c2
c2
c2
20. The Galilean transformation is given in Eq. 36-4.
(a)
x, y , z
x
vt , y , z
25m
30m s 3.5s ,20m,0
130m,20m,0
(b)
x, y , z
x
vt , y , z
25m
30m s 10.0s ,20m,0
325m,20m,0
21. (a) The person’s coordinates in S are found using Eq. 36-6, with x
t
25 m , y
20 m , z
0 , and
8
3.5 s. We set v 1.80 10 m/s.
x
x
y
1 v 2 c2
y
20 m ; z
1.8 108 m/s 3.5 s
25m
vt
8
1
1.8 10 m/s
z
2
8
3.0 10 m/s
2
820 m
0
432
Chapter 36
(b) We repeat part (a) using the time t
x
x
1 v 2 c2
y
y
1.8 108 m/s 10.0 s
25m
vt
1.8 108 m/s
1
20 m ; z
10.0 s.
z
2
3.0 108 m/s
2280 m
2
0
22. We determine the components of her velocity in the S frame using Eq. 36-7, where
ux u y 1.10 108 m/s and v 1.80 108 m/s . Then using trigonometry we combine the
components to determine the magnitude and direction.
ux v
1.10 108 m/s 1.80 108 m/s
ux
1 vux / c 2 1 1.80 108 m/s 1.10 108 m/s / 3.00 108 m/s
uy
u
1.10 108 m/s
u y 1 v 2 c2
1 vux / c
2
ux2 u 2y
tan
1
uy
1
1.8 108 m/s
8
2
3.0 108 m/s
8
8
1.80 10 m/s 1.10 10 m/s / 3.00 10 m/s
2.38 108 m/s
tan
ux
1
1
2
7.21 107 m/s
2.38 108 m/s
7.21 107 m/s
2
2.38 108 m/s
2
2
2
7.21 107 m/s
2.49 108 m/s
16.9
23. (a) We take the positive direction to be the direction of motion of spaceship 1. Consider spaceship
2 as reference frame S, and the Earth reference frame S . The velocity of the Earth relative to
spaceship 2 is v 0.60 c. The velocity of spaceship 1 relative to the Earth is ux 0.60 c. Solve
for the velocity of spaceship 1 relative to spaceship 2, u x , using Eq. 36-7a.
ux v
0.60 c 0.60 c
ux
0.88 c
vux
1
0.60 0.60
1
c2
(b) Now consider spaceship 1 as reference frame S. The velocity of the Earth relative to spaceship
1 is v
0.60 c. Solve for the
0.60 c. The velocity of spaceship 2 relative to the Earth is ux
velocity of spaceship 2 relative to spaceship 1, u x , using Eq. 36-7a.
ux v
0.60 c 0.60 c
ux
0.88 c
vux
1
0.60 0.60
1
c2
As expected, the two relative velocities are the opposite of each other.
24. (a) The Galilean transformation is given in Eq. 36-4.
x x vt x vt 100m 0.92 3.00 108 m s 1.00 10 6 s
376m
(b) The Lorentz transformation is given in Eq. 36-6. Note that we are given t, the clock reading in
frame S.
vx
t vx
t
t
t
2
c
c2
x
x
vt
x
v
t
vx
c2
x
v ct
c
vx
c
433
1
1 0.92
2
100m
0.92
1 0.92 2 3.00 108 m s 1.00 10 6 s
0.92 100m
316m
25. (a) We take the positive direction in the direction of the first spaceship. We choose reference frame
S as the Earth, and reference frame S as the first spaceship. So v 0.61c. The speed of the
second spaceship relative to the first spaceship is ux 0.87 c. We use Eq. 36-7a to solve for the
speed of the second spaceship relative to the Earth, u.
ux v
0.87c 0.61c
ux
0.97 c
vux
1
0.61 0.87
1
c2
(b) The only difference is now that ux
0.87 c.
ux v
0.87c 0.61c
ux
0.55 c
vux
1
0.61
0.87
1
c2
The problem asks for the speed, which would be 0.55c
26. We assume that the given speed of 0.90c is relative to the planet that you are approaching. We take
the positive direction in the direction that you are traveling. Consider your spaceship as reference
frame S, and the planet as reference frame S . The velocity of the planet relative to you is
v
0.90 c. The velocity of the probe relative to the planet is ux 0.95 c. Solve for the velocity of
the probe relative to your spaceship, u x , using Eq. 36-7a.
ux v
0.95c 0.90c
ux
0.34c
vux
1
0.90 0.95
1
c2
27. We set frame S as the frame at rest with the spaceship. In this frame the module has speed
u u y 0.82c. Frame S is the frame that is stationary with respect to the Earth. The spaceship, and
therefore frame S moves in the x-direction with speed 0.76c in this frame, or v 0.76c. We use
Eq. 36-7a and 36-7b to determine the components of the module velocity in frame S. Then using
trigonometry we combine the components to determine the speed and direction of travel.
ux
ux v
1 vux / c 2
0 0.76c
1 0
u
ux2
0.76c
u 2y
2
0.533c
2
u y 1 v 2 c2
0.82c 1 0.762
0.533c
1 vux / c 2
1 0
u
0.533c
0.93c ;
tan 1 y tan 1
35
ux
0.76c
0.76c ; u y
28. The velocity components of the particle in the S frame are ux
u cos
and u y
u sin . We find the
components of the particle in the S frame from the velocity transformations given in Eqs. 36-7a and
36-7b. Those transformations are for the S frame moving with speed v relative to the S frame. We
can find the transformations from the S frame to the S frame by simply changing v to –v and primed
to unprimed variables.
ux
ux
v
1 vux c 2
ux
ux
v
1 vux c 2
; uy
u y 1 v2 c2
1 vux c 2
uy
u y 1 v2 c2
1 vu x c 2
434
Chapter 36
uy 1 v2 c2
1 vu x c 2
uy
tan
ux
ux
uy 1 v2 c2
v
ux
1 vu x c
u sin 1 v 2 c 2
u cos
v
v
sin 1 v 2 c 2
cos
vu
2
29. (a) In frame S the horizontal component of the stick length will be contracted, while the vertical
component remains the same. We use the trigonometric relations to determine the x- and ycomponents of the length of the stick. Then using Eq. 36-3a we determine the contracted length
of the x-component. Finally, we use the Pythagorean theorem to determine stick length in
frame S .
lx
l 0 cos
l
l x2
; ly
l y2
l 0 sin
ly ; lx
l 0 2 cos2
1 v 2 c2
l x 1 v 2 c2
l 02 sin 2
l 0 cos
l0 1
1 v 2 c2
v cos
2
c
(b) We calculate the angle from the length components in the moving frame.
tan
1
ly
lx
tan
l 0 sin
1
l 0 cos
2
1 v c
2
tan
tan
1
2
1 v c
2
tan
1
tan
30. (a) We choose the train as frame S and the Earth as frame S. Since the guns fire simultaneously in
S , we set these times equal to zero, that is tA tB 0. To simplify the problem we also set the
location of gunman A equal to zero in frame S when the guns were fired, xA 0. This places
gunman B at xB 55.0m. Use Eq. 36-6 to determine the time that each gunman fired his
weapon in frame S.
vxA
v 0
tA
tA
0
0
c2
c2
tB
tB
vxB
c2
1
1
8
35.0 m/s 3.00 10 m/s
2
35m/s 55.0 m
0
8
3.00 10 m/s
2.14 10
2
14
s
Therefore, in Frame S, A fired first.
(b) As found in part (a), the difference in time is 2.14 10 14 s .
(c) In the Earth frame of reference, since A fired first, B was struck first. In the train frame, A is
moving away from the bullet fired toward him, and B is moving toward the bullet fired toward
him. Thus B will be struck first in this frame as well.
31. We set frame S as the frame moving with the observer. Frame S is the frame in which the two light
bulbs are at rest. Frame S is moving with velocity v with respect to frame S . We solve Eq. 36-6 for
the time t in terms of t, x, and v. Using the resulting equation we determine the time in frame S that
each bulb is turned on, given that in frame S the bulbs are turned on simultaneously at tA tB 0.
Taking the difference in these times gives the time interval as measured by the observing moving
with velocity v.
x
x
x vt
x
vt
t
t
vx
=
c2
t
v x
c2
vt
t 1
v2
c2
vx
c2
t
vx
c2
t
t
vx
c2
435
tA
tA
vx A
c2
0
v 0
c2
0 ; tB
tB
vxB
c2
0
vl
c2
vl
c2
vl
c2
According to the observer, bulb B turned on first.
t
tB
tA
32. We set up the two frames such that in frame S, the first object is located at the origin and the second
object is located 220 meters from the origin, so xA 0 and xB 220 m. We set the time when event
A occurred equal to zero, so tA 0 and tB 0.80 s. We then set the location of the two events in
frame S equal, and using Eq. 36-6 we solve for the velocity.
xA x B
0 220m
xA x B
xA vt A
xB vtB ; v
2.5 108 m/s
tA tB 0 0.88 s
33. From the boy’s frame of reference, the pole remains at rest with respect to him. As such, the pole
will always remain 12.0 m long. As the boy runs toward the barn, relativity requires that the
(relatively moving) barn contract in size, making the barn even shorter than its rest length of 10.0 m.
Thus it is impossible, in the boy’s frame of reference, for the barn to be longer than the pole. So
according to the boy, the pole will never completely fit within the barn.
In the frame of reference at rest with respect to the barn, it is possible for the pole to be shorter than
the barn. We use Eq. 36-3a to calculate the speed that the boy would have to run for the contracted
length of the pole, l, to equal the length of the barn.
l
l 0 1 v 2 c2
v
c 1 l 2 l 02
c 1
10.0 m
2
12.0 m
2
0.5528c
If persons standing at the front and back door of the barn were to close both doors exactly when the
pole was completely inside the barn, we would have two simultaneous events in the barn’s rest frame
S with the pole completely inside the barn. Let us set the time for these two events as tA tB 0. In
frame S these two events occur at the front and far side of the barn, or at xA 0 and xB 10.0m.
Using Eq. 36-6, we calculate the times at which the barn doors close in the boy’s frame of reference.
vx A
v 0
tA
tA
0
0
c2
c2
0.5528 10.0m
vxB
1
0
2.211 10 8 s
2
8
2
c
3.00
10
m/s
1 0.5528
Therefore, in the boy’s frame of reference the far door of the barn closed 22.1 ns before the front
door. If we multiply the speed of the boy by this time difference, we calculate the distance the boy
traveled between the closing of the two doors.
x vt 0.5528 3.00 108 m/s 2.211 10 8 s 3.67 m.
tB
tB
We use Eq. 36-3a to determine the length of the barn in the boy’s frame of reference.
l l 0 1 v2 c2 10.0 m 1 0.55282 8.33 m
Subtracting the distance traveled between closing the doors from the length of the pole, we find the
length of the barn in the boy’s frame of reference.
l 0,pole
x 12.0 m 3.67 m 8.33 m l barn
Therefore, in the boy’s frame of reference, when the front of the pole reached the far door it was
closed. Then 22.1 ns later, when the back of the pole reached the front door, that door was closed.
In the boy’s frame of reference these two events are not simultaneous.
436
Chapter 36
34. The momentum of the proton is given by Eq. 36-8.
1.67 10 27 kg 0.75 3.00 108 m s
mv
p
mv
2
1 v 2 c2
1 0.75
5.7 10
19
kg m s
35. (a) We compare the classical momentum to the relativistic momentum.
p classical
mv
2
1 v 2 c2
1 0.10
0.995
mv
prelativistic
1 v 2 c2
The classical momentum is about 0.5% in error.
(b) We again compare the two momenta.
p classical
mv
1 v2 c2
1 0.60
mv
prelativistic
2
0.8
1 v2 c2
The classical momentum is
20% in error.
36. The momentum at the higher speed is to be twice the initial momentum. We designate the initial
state with a subscript “0”, and the final state with a subscript “f”.
mvf
vf2
2
2
2
0.26 c
1 vf c
pf
vf2
1 vf2 c 2
2
4
4
0.29c 2
2
2
2
2
mv
v
p0
1 vf c
0
1 0.26
0
1 v02 c 2
1 v02 c 2
vf2
0.29 2
c
1.29
vf
0.47 c
37. The two momenta, as measured in the frame in which the particle was initially at rest, will be equal
to each other in magnitude. The lighter particle is designated with a subscript “1”, and the heavier
particle with a subscript “2”.
m1v1
m2 v2
p1 p2
2
2
1 v1 c
1 v22 c 2
v12
1 v12 c 2
v1
0.90 c
m2
m1
2
v22
1 v22 c 2
6.68 10
1.67 10
27
kg
27
kg
2
0.60 c
1
2
0.60
2
9.0c 2
0.95 c
38. We find the proton’s momenta using Eq. 36-8.
mp v1
mp 0.45 c
p0.45
0.5039mp c ; p0.80
2
v12
1
0.45
1
c2
mp v2
mp 0.98 c
p0.98
4.9247mp c
2
2
v2
1
0.98
1
c2
mp v2
1
v22
c2
mp 0.80 c
1
0.80
2
1.3333mp c
437
(a)
p2
(b)
p2
p1
p1
p1
p1
1.3333mpc 0.5039mpc
100
0.5039mpc
4.9247mp c 1.3333mpc
100
1.3333mpc
100 164.6
160%
100 269.4
270%
39. The rest energy of the electron is given by Eq. 36-12.
E
mc 2
31
9.11 10
8.20 10
1.60 10
13
14
kg 3.00 108 m s
J
J MeV
2
8.20 10
14
J
0.511MeV
40. We find the loss in mass from Eq. 36-12.
200 MeV 1.60 10 13 J MeV
E
m
2
c2
3.00 108 m s
3.56 10
28
kg
4 10
28
kg
41. We find the mass conversion from Eq. 36-12.
8 1019 J
E
m
900 kg
2
c2
3.00 108 m s
42. We calculate the mass from Eq. 36-12.
m
E
c2
1
mc 2
c2
27
8
1 1.6726 10 kg 2.9979 10 m s
c2
1.6022 10 13 J MeV
2
938.2 MeV c 2
43. Each photon has momentum 0.50 MeV/c. Thus each photon has mass 0.50 MeV. Assuming the
photons have opposite initial directions, then the total momentum is 0, and so the product mass will
not be moving. Thus all of the photon energy can be converted into the mass of the particle.
Accordingly, the heaviest particle would have a mass of 1.00MeV c2 , which is 1.78 10
30
kg. .
44. (a) The work is the change in kinetic energy. Use Eq. 36-10b. The initial kinetic energy is 0.
1
W
K K final
1 mc 2
1 938.3MeV 1.39 104 MeV
1 0.9982
13.9GeV
(b) The momentum of the proton is given by Eq. 36-8.
1
p
mv
938.3MeV c 2 0.998 c
2
1 0.998
1.48 104 MeV c
14.8GeV c
45. We find the energy equivalent of the mass from Eq. 36-12.
E
mc2
1.0 10 3 kg 3.00 108 m s
2
9.0 1013 J
We assume that this energy is used to increase the gravitational potential energy.
E
9.0 1013 J
E mgh
m
9.2 109 kg
hg
1.0 103 m 9.80m s2
438
Chapter 36
46. The work is the change in kinetic energy. Use Eq. 36-10b. The initial kinetic energy is 0.
W1
1 mc 2 ; W2 K 0.99 c K 0.90 c
1 mc 2
1 mc 2
0.90
0.99
0.90
1
W2
W1
0.99
1 mc
2
1 mc
0.90
2
2
0.99
1 mc
0.90
0.90
0.90
1
1 0.99
2
1 0.902
1
1
1 0.902
3.7
1
47. The kinetic energy is given by Eq. 36-10.
1 mc 2
K
mc 2
1
2
2
1 v c
3
c
4
v
2
0.866 c
48. The total energy of the proton is the kinetic energy plus the mass energy. Use Eq. 36-13 to find the
momentum.
E K mc 2 ;
pc
pc
2
E2
K2
mc 2
2
K
2 K mc 2
p 1638MeV c
2
mc 2
K 1 2
mc 2
mc 2
K
2
K2
950 MeV
2 K mc 2
1 2
938.3MeV
950 MeV
1638MeV
1.6GeV c
49. We find the speed in terms of c. The kinetic energy is given by Eq. 36-10 and the momentum by Eq.
36-8.
2.80 108 m s
v
0.9333 c
3.00 108 m s
p
1
1 mc 2
K
1 0.93332
1
mv
1 938.3MeV
938.3MeV c 2 0.9333 c
2
1674.6 MeV
2439 MeV c
1.67GeV
2.44GeV c
1 0.9333
50. We use Eq. 36-10 to find the speed from the kinetic energy.
1
1 mc 2
K
2
1 v c
v
c 1
1
K
mc 2
2
2
1 mc 2
1
c 1
1.25MeV
1
0.511MeV
1
2
0.957 c
51. Since the proton was accelerated by a potential difference of 125 MV, its potential energy decreased
by 125 MeV, and so its kinetic energy increased from 0 to 125 MeV. Use Eq. 36-10 to find the
speed from the kinetic energy.
K
1 mc 2
1
2
1 v c
2
1 mc 2
439
v
1
c 1
K
mc 2
2
1
c 1
1
0.470 c
2
125MeV
1
938.3MeV
52. We let M represent the rest mass of the new particle. The initial energy is due to both incoming
particles, and the final energy is the rest energy of the new particle. Use Eq. 36-11 for the initial
energies.
2m
E 2 mc 2
Mc 2
M 2 m
1 v 2 c2
We assumed that energy is conserved, and so there was no loss of energy in the collision.
The final kinetic energy is 0, so all of the kinetic energy was lost.
Klost
Kinitial
1
1 mc 2
2
2
1 v c
2
1 2mc 2
53. Since the electron was accelerated by a potential difference of 28 kV, its potential energy decreased
by 28 keV, and so its kinetic energy increased from 0 to 28 MeV. Use Eq. 36-10 to find the speed
from the kinetic energy.
1
1 mc 2
K
2
1 v c
v
1
c 1
K
mc 2
2
1 mc 2
2
1
c 1
1
0.32 c
2
0.028 MeV
1
0.511MeV
54. We use Eqs. 36-11 and 36-13 in order to find the mass.
E2
m
p2c2
m2c 4
p 2c 2 K 2
2 Kc 2
K mc 2
2
K2
2 Kmc 2
2
121MeV c c 2 45MeV
2 45MeV c 2
The particle is most likely a probably a
0
m 2c 4
2
140 MeV c 2
2.5 10
28
kg
meson.
55. (a) Since the kinetic energy is half the total energy, and the total energy is the kinetic energy plus
the rest energy, the kinetic energy must be equal to the rest energy. We also use Eq. 36-10.
K 12 E 12 K mc 2
K mc 2
K
1 mc 2
mc 2
1
2
1 v 2 c2
(b) In this case, the kinetic energy is half the rest energy.
1
3
K
1 mc 2 12 mc 2
2
1 v 2 c2
3
4
v
v
c
5
9
0.866 c
c
0.745 c
440
Chapter 36
56. We use Eq. 36-10 for the kinetic energy and Eq. 36-8 for the momentum.
1
1 mc 2
K
2
1 v c
1
1 mc 2
2
8.15 107 m s
3.00 108 m s
1
1 938.3MeV
2
36.7 MeV
p
2
1 mc v c
c 1 v 2 c2
mv
mv
2
1 v c
2
1
c
8.15 107 m s
3.00 108 m s
938.3MeV
8.15 107 m s
3.00 108 m s
1
2
265MeV c
Evaluate with the classical expressions.
Kc
1
2
mv 2
1
2
mc 2
v
c
2
1
2
938.3MeV
1 2 v
8.15 107 m s
mc
938.3MeV
c
c
3.00 108 m s
Calculate the percent error.
Kc K
34.6 36.7
errorK
100
100
5.7%
K
36.7
pc p
255 265
errorp
100
100
3.8%
p
265
pc
2
8.15 107 m s
3.00 108 m s
mv
34.6 MeV
255MeV c
57. (a) The kinetic energy is found from Eq. 36-10.
K
1
1 mc
1 v 2 c2
1
1 mc 2
1 0.182
1 1.7 104 kg 3.00 108 m s
2
2.541 1019 J 2.5 1019 J
(b) Use the classical expression and compare the two results.
K
1
2
1
2
mv
1.7 104 kg
2.479 1019 J
% error
0.18 3.00 108 m s
2.541 1019 J
100
2.541 1019 J
2
2.479 1019 J
2.4%
The classical value is 2.4% too low.
58. The kinetic energy of 998 GeV is used to find the speed of the protons. Since the energy is 1000
times the rest mass, we expect the speed to be very close to c. Use Eq. 36-10.
1
1 mc 2
K
2
1 v c
v
c 1
1
K
mc 2
2
1
c 1
2
1 mc 2
1
998GeV
1
0.938GeV
2
c to 7 sig. fig.
441
B
mv
rqv
2
998GeV
1 1.673 10
0.938GeV
K
1 mc
mc 2
rq
mv
rq
27
kg 3.00 108 m s
1.0 103 m 1.60 10
19
3.3T
C
59. By conservation of energy, the rest energy of the americium nucleus is equal to the rest energies of
the other particles plus the kinetic energy of the alpha particle.
mAm c 2
mNp m c 2 K
mNp
mAm
K
c2
m
241.05682 u 4.00260 u
5.5MeV
1u
2
c
931.49 MeV c 2
237.04832 u
60. (a) For a particle of non-zero mass, we derive the following relationship between kinetic energy
and momentum.
K mc 2 ;
E
K
2
2 K mc
2
pc
2
E2
K mc 2
2mc 2
2
pc
2
mc 2
0
2
mc 2
4 mc 2
2
2
K2
4 pc
2 K mc 2
2
K
2
For the kinetic energy to be positive, we take the positive root.
2mc 2
4 mc 2
2
4 pc
2
mc 2
K
mc 2
2
pc
2
2
If the momentum is large, we have the following relationship.
K
mc 2
mc 2
2
pc
2
pc mc 2
Thus there should be a linear relationship between kinetic energy and momentum for large
values of momentum.
If the momentum is small, we use the binomial expansion to derive the classical relationship.
mc
mc
mc
2
2
2 2
mc 1
pc
1
2
pc
mc 2
2
mc
2
2
mc
2
1
pc
mc 2
2
p2
2m
Thus we expect a quadratic relationship for
small values of momentum. The adjacent
graph verifies these approximations.
(b) For a particle of zero mass, the relationship is
simply K pc. See the included graph. The
spreadsheet used for this problem can be
found on the Media Manager, with filename
“PSE4_ISM_CH36.XLS,” on tab “Problem
36.60.”
m 0
m 0
K
K
2
p
442
Chapter 36
61. All of the energy, both rest energy and kinetic energy, becomes electromagnetic energy. We use Eq.
36-11. Both masses are the same.
1
1
2
2
2
Etotal E1 E2
105.7 MeV
1mc
2 mc
1
2 mc
1 0.432
1 0.552
243.6 MeV
240 MeV
62. We use Eqs. 36-11 and 36-13.
E
mc 2 ;
K
K2
p
pc
2
E2
mc 2
2
2
mc 2
K
2
mc 2
K2
2 K mc 2
2 K mc 2
c
63. (a) We assume the mass of the particle is m, and we are given that the velocity only has an xcomponent, u x . We write the momentum in each frame using Eq. 36-8, and we use the velocity
transformation given in Eq. 36-7. Note that there are three relevant velocities: u x , the velocity
in reference frame S; u x , the velocity in reference frame S ; and v, the velocity of one frame
relative to the other frame. There is no velocity in the y or z directions, in either frame. We
1
, and also use Eq. 36-11 for energy.
reserve the symbol for
1 v2 c2
mux
px
; py
1 ux2 c 2
ux v
1 vux c 2
ux
1 ux2 c 2
0
ux v
; uy
1 vux c 2
ux
mux
px
0 ; pz
; py
0 since u y
uy
0 ; pz
1 vux c 2
1 v 2 c2
0 since uz
Substitute the expression for u x into the expression for px .
ux v
m
1 vux c 2
ux v
mux
px
m
2
2
2
1 vu x c 2
1 ux c
ux v
1
1 2
c 1 vu c 2 2
0 ; uz
v
1 vux c
1 vux c
1
1 vu x c 2
m ux
1 2
vux
c2
vu x
c2
1 vu x c
2 2
u x2
c2
v
1 vu x c
m ux
2u x v
c2
v2
c2
1
vux
c2
2
2 2
v
2
c2
1 vu x c 2
2 2
2
c2
v
2
ux
ux
2 2
m ux
1
2
0
1 v 2 c2
1
1 vux c
ux
1 vux c 2
0
x
m
uz
2
v
ux
v
2
c2
v
ux2
c2
v2
c2
443
m ux
v
mux
mv
1 ux2 c 2
1 ux2 c 2
1 v 2 c 2 1 ux2 c 2
1 v 2 c2
mc 2
mux
2
x
1 u c
2
2
x
1 u c
v
c2
2
mc 2
px
2
x
1 u c
1 v 2 c2
2
v
c2
1 v 2 c2
1 v 2 c2
It is obvious from the first few equations of the problem that p y
E
mc 2
mc 2
1 u x2 c 2
0 and pz
py
2
1 vu x c
2
1 vu x c
2 2
2 2
ux
v
mc 1 vu x c
1 vu x c
E
2
mc
ux
2 2
v
0.
2
c2
1 vu x c 2
2
mc 2
2
pz
mc 2
ux v
1
2
c 1 vu c 2
x
1
vE c 2
px
2
2
2
x
mvu x
1 u c
mvu x
1 v 2 c 2 1 u x2 c 2
2
1 u x2 c 2
1 v 2 c2
c2
px v
1 v 2 c2
(b) We summarize these results, and write the Lorentz transformation from Eq. 36-6, but solved in
terms of the primed variables. That can be easily done by interchanged primed and unprimed
quantities, and changing v to v.
px vE c 2
E px v
px
; py py ; py py ; E
2
2
1 v c
1 v2 c2
x
x vt
; y
1 v2 c2
y ; z
z ; t
t vx c 2
1 v2 c2
These transformations are identical if we exchange p x with x, p y with y, p z with z, and E c 2
with t (or E c with ct).
64. The galaxy is moving away from the Earth, and so we use Eq. 36-15b.
f 0 f 0.0987 f 0
f 0.9013 f 0
f
f0
c v
c v
1
v
f f0
1+ f f 0
2
2
c
1 0.90132
c
1 0.90132
0.1035 c
65. For source and observer moving towards each other, use Eq. 36-14b.
c v
1 v c
1 0.70
f f0
f0
95.0 MHz
226 MHz
c v
1 v c
1 0.70
230 MHz
444
Chapter 36
66. We use Eq. 36-15a, and assume that v
c v
c v
0
0
1 vc
1 vc
0
1 v c 1 v 2 c2
c.
1 vc 1 vc
1 vc 1 vc
0
1/ 2
1 v c
0
0
0
0
1
1 v 2 c2
1 v c
v
c
0
v c
0
0
67. (a) We apply Eq. 36-14b to determine the received/reflected frequency f. Then we apply this same
equation a second time using the frequency f as the source frequency to determine the Dopplershifted frequency f . We subtract the initial frequency from this Doppler-shifted frequency to
obtain the beat frequency. The beat frequency will be much smaller than the emitted frequency
when the speed is much smaller than the speed of light. We then set c v c and solve for v.
c v
c v
c v c v
c v
f f0
f
f
f0
f0
c v
c v
c v c v
c v
f beat
f
f0
f0
c v
c v
f0
3.00 108 m/s 6670 Hz
v
c v
c v
f0
2v
c v
f0
2v
c
v
cf beat
2 f0
27.8m/s
2 36.0 109 Hz
(b) We find the change in velocity and solve for the resulting change in beat frequency. Setting
the change in the velocity equal to 1 km/h we solve for the change in beat frequency.
cf beat
c f beat
2 f0 v
v
v
f beat
2 f0
2 f0
c
f beat
2 36.0 109 Hz 1km/h
1m/s
3.600km/h
8
3.00 10 m/s
70 Hz
68. We consider the difference between Doppler-shifted frequencies for atoms moving directly towards
the observer and atoms moving directly away. Use Eqs. 36-14b and 36-15b.
f
f0
c v
c v
f0
c v
c v
f0
c v
c v
c v
c v
2v
f0
c2
f0
v2
2v c
1 v 2 c2
We take the speed to be the rms speed of thermal motion, given by Eq. 18-5. We also assume that
the thermal energy is much less than the rest energy, and so 3kT mc2 .
1/ 2
3kT
v
3kT
f
3kT
3kT
3kT
2
1
2
2
2
2
m
c
mc
f0
mc
mc
mc 2
We evaluate for a gas of H atoms (not H 2 molecules) at 550 K. Use Appendix F to find the mass.
v
vrms
f
f0
3kT
2
mc 2
2
3 1.38 10
1.008u 1.66 10
23
27
J K 550 K
8
kg u 3.00 10 m s
2
2.5 10
5
445
69. At the North Pole the clock is at rest, while the clock on the equator travels the circumference of the
Earth each day. We divide the circumference of the Earth by the length of the day to determine the
speed of the equatorial clock. We set the dilated time equal to 2.0 years and solve for the change in
rest times for the two clocks.
6
2 R 2 6.38 10 m
v
464 m/s
T
24 hr 3600s/hr
t0,eq
t
1 v 2 / c2
t0,pole
t
t0,pole
1 0
t0,eq
t 1 v2 / c2
t0,eq
t0,pole
t 1
t 1
v2
2c 2
t
v2
2c 2
t
2.0 yr 464 m/s
v2
t 2
2c
2
3.156 107 s/yr
2 3.00 108 m/s
75 s
2
70. We take the positive direction in the direction of the motion of the second pod. Consider the first
pod as reference frame S, and the spacecraft as reference frame S . The velocity of the spacecraft
relative to the first pod is v 0.60 c. The velocity of the first pod relative to the spacecraft is
ux 0.50 c. Solve for the velocity of the second pod relative to the first pod, u x , using Eq. 36-7a.
ux v
0.50c 0.60c
ux
0.846 c
vux
1
0.60 0.50
1
c2
71. We treat the Earth as the stationary frame, and the airplane as the moving frame. The elapsed time in
the airplane will be dilated to the observers on the Earth. Use Eq. 36-1a.
2 rEarth
2 rEarth
tEarth
; tplane tEarth 1 v 2 c 2
1 v 2 c2
v
v
t
tEarth
tplane
2 rEarth
1
v
1 v 2 c2
1m s
3.6 km h
6.38 106 m 1300 km h
8
3.00 10 m s
2 rEarth
1
v
1
1
2
v2
c2
rEarth v
c2
8.0 10 8 s
2
72. (a) To travelers on the spacecraft, the distance to the star is contracted, according to Eq. 36-3a.
This contracted distance is to be traveled in 4.6 years. Use that time with the contracted
distance to find the speed of the spacecraft.
v
xEarth 1 v 2 c 2
tspacecraft
xspacecraft
v
tspacecraft
1
c
1
c tspacecraft
xEarth
2
1
c
1
4.6ly
4.3ly
2
0.6829 c
0.68 c
446
Chapter 36
(b) Find the elapsed time according to observers on Earth, using Eq. 36-1a.
tspaceship
4.6y
tEarth
6.3y
2
2
1 v c
1 0.68292
Note that this agrees with the time found from distance and speed.
xEarth
4.3ly
tEarth
6.3yr
v
0.6829 c
73. (a) We use Eq. 36-15a. To get a longer wavelength than usual means that the object is moving
away from the Earth.
1.0702 1
c v
1.070
c v 0.067c
0
0
c v
1.0702 1
(b) We assume that the quasar is moving and the Earth is stationary. Then we use Eq. 16-9b.
f0
c
c
1
f
1.070 0
v 0.070 c
0 1 v c
1 vc
0 1 v c
74. We assume that some kind of a light signal is being transmitted from the astronaut to Earth, with a
frequency of the heartbeat. That frequency will then be Doppler shifted, according to Eq. 36-15b.
We express the frequencies in beats per minute.
f 02 f 2
602 302
c v
f f0
v c 2
c
0.60 c
c v
f
f 02
602 302
75. (a) The velocity components of the light in the S frame are ux
0 and u y
c. We transform those
velocities to the S frame according to Eq. 36-7.
ux
ux v
1 vux c 2
tan
(b) u
1
uy
tan
ux
ux2 u 2y
0 v
1 0
v2
1
v ; uy
c 1 v2 c2
v
c2 1 v 2 c2
u y 1 v2 c2
c 1 v2 c2
1 0
1 vux c 2
tan
v2
1
c 1 v2 c2
c2
1
v2
c2 v 2
c
(c) In a Galilean transformation, we would have the following.
ux
ux
v
v ; uy
uy
c ; u
v2
c2
c
;
tan
1
c
v
76. We take the positive direction as the direction of motion of rocket A. Consider rocket A as reference
frame S, and the Earth as reference frame S . The velocity of the Earth relative to rocket A is
v
0.65 c. The velocity of rocket B relative to the Earth is ux 0.85 c. Solve for the velocity of
rocket B relative to rocket A, u x , using Eq. 36-7a.
ux v
0.85c 0.65c
ux
0.45c
vux
1
0.65 0.85
1
c2
Note that a Galilean analysis would have resulted in ux 0.20c.
447
77. (a) We find the speed from Eq. 36-10.
1
1 mc 2
K
2
1 v c
v
1
14,001
c 1
2
1 mc 2
2
14,000mc 2
2
c
1
c
2 14,001
2
2
3.00 108 m s
c
1
1
0.77 m s
2 14,001
2
14,001
(b) The tube will be contracted in the rest frame of the electron, according to Eq. 36-3a.
c v
l 1 v 2 c2
l0
3.0 103 m
1
1
1
14,001
2
0.21m
78. The electrostatic force provides the radial acceleration. We solve that relationship for the speed of
the electron.
1 e2 melectron v 2
Felectrostatic Fcentripetal
4 0 r2
r
v
4
8.99 109 N m 2 C 2
e2
1
0
melectron r
31
9.11 10
kg
1.60 10
19
0.53 10
10
C
2
2.18 106 m s
m
0.0073 c
Because this is much less than 0.1c, the electron is not relativistic.
79. The minimum energy required would be the energy to produce the pair with no kinetic energy, so the
total energy is their rest energy. They both have the same mass. Use Eq. 36-12.
E
2mc 2
2 0.511MeV
1.022 MeV 1.64 10
13
J
80. The wattage times the time is the energy required. We use Eq. 36-12 to calculate the mass.
75W 3.16 107 s 1000g
Pt
2
E Pt mc
m
2.6 10 5 g
2
2
8
c
1kg
3.00 10 m s
81. Use Eqs. 36-13, 36-8, and 36-11.
E2
p 2c2
dE
dp
1
2
m2c4
p 2c2
E
m2c4
1/ 2
p 2c 2
2 pc 2
m 2c 4
pc 2
E
1/ 2
pc 2
E
mvc 2
mc 2
v
82. The kinetic energy available comes from the decrease in rest energy.
K mnc2 mpc2 mec2 mvc2 939.57MeV 938.27MeV 0.511MeV 0
0.79MeV
83. (a) We find the rate of mass loss from Eq. 36-12.
E mc 2
E
m c2
m
t
1
c2
E
t
4 1026 J s
8
3.00 10 m s
2
4.44 109 kg s
4 109 kg s
448
Chapter 36
(b) Find the time from the mass of the Sun and the rate determined in part (a).
5.98 1024 kg
mEarth
t
4.27 107 y 4 107 y
m t
4.44 109 kg s 3.156 107 s y
(c) We find the time for the Sun to lose all of its mass at this same rate.
1.99 1030 kg
mSun
t
1.42 1013 y 1 1013 y
m t
4.44 109 kg s 3.156 107 s y
84. Use Eq. 36-8 for the momentum to find the mass.
mv
p
mv
1 v 2 c2
2.24 108 m s
3.00 108 m s
22
2
3.07 10 kg m s 1
p 1 v 2 c2
m
9.12 10 31 kg
v
2.24 108 m s
This particle has the mass of an electron, and a negative charge, so it must be an electron.
85. The total binding energy is the energy required to provide the increase in rest energy.
E
2mp+e
mHe c 2
2mn
2 1.00783u
4.00260 u c 2
2 1.00867 u
931.5MeV c 2
u
28.32 MeV
86. The momentum is given by Eq. 36-8, and the energy is given by Eq. 36-11 and Eq. 36-13.
P
mv
mc 2 v
c2
Ev
c2
v
pc 2
E
pc 2
m2c4
pc
p 2c 2
m2c2
87. (a) The magnitudes of the momenta are equal. We use Eq. 36-8.
2
mv
1 mc v c
1 938.3MeV 0.985
p
mv
1 v 2 c2 c 1 v 2 c2 c
1 0.9852
5.36GeV c
2.86 10
18
5.36GeV c
1c
3.00 108 m s
p2
5356 MeV c
1.602 10 10 J GeV
1GeV
kg m s
(b) Because the protons are moving in opposite directions, the vector sum of the momenta is 0.
(c) In the reference frame of one proton, the laboratory is moving at 0.985c. The other
proton is moving at 0.985c relative to the laboratory. We find the speed of one proton
relative to the other, and then find the momentum of the moving proton in the rest frame of the
other proton by using that relative velocity.
0.985 c 0.985 c
v ux
ux
0.9999 c
vux
1 0.985 0.985
1
c2
449
p
2
1 mc ux c
c 1 u x2 c 2
mux
mux
2
x
1 u c
62.1GeV c
3.31 10
17
2
62.1GeV c
938.3MeV
1
c
2 0.985
1 0.9852
1
1c
3.00 108 m s
2 0.985
1 0.9852
62081MeV c
2
1.602 10 10 J GeV
1GeV
kg m s
88. We find the loss in mass from Eq. 36-12.
E
484 103 J
m
5.38 10
2
c2
3.00 108 m s
12
kg
Two moles of water has a mass of 36 10 3 kg. Find the percentage of mass lost.
5.38 10 12 kg
36 10 3 kg
1.49 10
10
1.5 10 8 %
89. Use Eq. 36-10 for kinetic energy, and Eq. 36-12 for rest energy.
K
1 mEnterprise c 2 mconverted c 2
mconverted
1
2
1 v c
2
1 mEnterprise
1
1 0.10
2
1 6 109 kg
3 107 kg
90. We set the kinetic energy of the spacecraft equal to the rest energy of an unknown mass. Use Eqs.
36-10 and 36-12.
K
1 mshipc 2 mc 2
m
1 mship
1
2
2
1
1 mship
2
1 1.8 105 kg
7.2 104 kg
1 v c
1 0.70
From the Earth’s point of view, the distance is 35 ly and the speed is 0.70c. That data is used to
calculate the time from the Earth frame, and then Eq. 36-1a is used to calculate the time in the
spaceship frame.
35y c
d
t
50 y ; t0
t 1 v 2 c2
50 y 1 0.702 36 y
v
0.70c
91. We assume one particle is moving in the negative direction in the laboratory frame, and the other
particle is moving in the positive direction. We consider the particle moving in the negative
direction as reference frame S, and the laboratory as reference frame S . The velocity of the
laboratory relative to the negative-moving particle is v 0.85 c, and the velocity of the positivemoving particle relative to the laboratory frame is ux 0.85 c. Solve for the velocity of the positivemoving particle relative to the negative-moving particle, u x .
ux v
0.85c 0.85c
ux
0.987 c
vux
1
0.85 0.85
1
c2
450
Chapter 36
92. We consider the motion from the reference frame of the spaceship. The passengers will see the trip
distance contracted, as given by Eq. 36-3a. They will measure their speed to be that contracted
distance divided by the year of travel time (as measured on the ship). Use that speed to find the work
done (the kinetic energy of the ship).
v
l 0 1 v 2 c2
t0
l
t0
v
c
1
c t0
l0
1
W
1
1 mc 2
K
1
1
0.9887
1
2
1 3.2 104 kg 3.00 108 m s
2
0.9887 c
2
1.0ly
6.6ly
1 mc 2
v 2 c2
1
1
2
1.6 1022 J
93. The kinetic energy is given by Eq. 36-10.
1
1 mc 2
K
1 v 2 c2
1
1 mc 2
1
0.98
1 14,500 kg 3.00 108 m s
2
2
5.3 1021 J
5.3 1021 J
1020 J
We compare this with annual U.S. energy consumption:
53.
The spaceship’s kinetic energy is over 50 times as great.
94. The pi meson decays at rest, and so the momentum of the muon and the neutrino must each have the
same magnitude (and opposite directions). The neutrino has no rest mass, and the total energy must
be conserved. We combine these relationships using Eq. 36-13.
pv 2 c 2
Ev
E
E
m c2
1/ 2
mv 2 c 4
m c2
Ev
p 2c 2
pc
pv c ; p
pv
p 2c 2
m 2c 4
p
m 2c 4
1/ 2
1/ 2
m c2
pv c
2
pc
p 2c 2
p 2c 2
m 2c 4
1/ 2
pc
m 2c 4
Solve for the momentum.
m 2c4
2m c 2 pc
p 2c 2
p 2c 2
m 2c 4
m c2
m c2 m c2
pc
m 2c 2
m 2c2
2m
Write the kinetic energy of the muon using Eqs. 36-11 and 36-13.
K
E m c2 ; E
E Ev m c 2 pc
K
m c2
pc
m 2c2 m 2c2
2m
2m m c 2 m c 2
m 2c 2 m 2c 2
2m
2m
2m
2
2m m
m
2
m
2
c2
m
2m
2
2m m
2m
m
2
c2
m
m
2
c2
2m
451
95. (a) The relative speed can be calculated in either frame, and will be the same value in both frames.
The time as measured on the Earth will be longer than the time measured on the spaceship, as
given by Eq. 36-1a.
tspaceship
tspaceship
xEarth
v
; tEarth
2
2
2
tEarth
1 v c
xEarth
1
c tEarth
tEarth
2
xEarth
c
2
tspaceship
2
tEarth
xEarth
c
2
2
2
tspaceship
2
2
xEarth
2
2
tEarth
tspaceship
6.0 y
2.50 y
6.5y
c
(b) The distance as measured by the spaceship will be contracted.
xspaceship
tspaceship
xEarth
2.50 y
v
xspaceship
xEarth
6.0ly
tEarth
tspaceship
tEarth
6.5y
2.3ly
This is the same distance as found using the length contraction relationship.
96. (a) To observers on the ship, the period is non-relativistic. Use Eq. 14-7b.
T
m
k
2
1.88kg
84.2 N m
2
0.939s
(b) The oscillating mass is a clock. According to observers on Earth, clocks on the spacecraft run
slow.
0.939s
T
TEarth
2.15s
2
2
2
1 v c
1 0.900
97. We use the Lorentz transformations to derive the result.
x
c t
x
2
vt
x
x
2
c
2
x
c
t
1
1 v 2 c2
2
; t
x
v x
2c t
c
c2
v2
t
1
c2 1 v 2 c2
1 v 2 c2
1 v 2 c2
2
1 v c
2
vx
c2
t
t
2
v x
c2
t
2
v t
c t
2
x
v t
v x
c
v
c
2
t
2
2
2
2
c t
t
v x
c
v x
c2
2
x
v t
2
x
2
2 xv t
v t
2
2
1
x
1 v2 c2
c t
2
2
x
x
2
2
452
Chapter 36
98. We assume that the left edge of the glass is even with point A when the flash of light is emitted.
There is no loss of generality with that assumption. We do the calculations in the frame of reference
in which points A and B are at rest, and the glass is then moving to the right with speed v.
If the glass is not moving, we would have this “no motion” result.
distance in glass distance in vacuum
d
tv 0 tglass tvacuum
speed in glass
speed in vacuum
vglass
l
d
c
d
l d nd l d nd l d l n 1 d
cn
c
c
c
c
c
If the index of refraction is n 1, then the glass will have no effect on the light, and the time would
simply be the distance divided by the speed of light.
distance in glass distance in vacuum d l d d l d l
tn 1 tglass tvacuum
speed in glass
speed in vacuum
c
c
c
c
Now, let us consider the problem from a relativistic point of view. The speed of light in the glass
will be the relativistic sum of the speed of light in stationary glass, c n , and the speed of the glass, v,
by Eq. 36-7a. We define to simplify further expressions.
vn
vn
c
c
1
1
v
v
c
c
c
c
n
n
vlight
cv
v
v
v
n
n
in glass
1
1
1
1
nc 2
nc
nc
nc
The contracted width of the glass, from the Earth frame of reference, is given by Eq. 36-3a.
d
d moving d 1 v 2 c 2
glass
We assume the light enters the block when the left edge of the block is at point A, and write simple
equations for the displacement of the leading edge of the light, and the leading edge of the block. Set
them equal and solve for the time when the light exits the right edge of the block.
c
d
xlight vlight t
t ; xright
vt ;
n
in glass
edge
c
d
d
n
xlight xright
tglass
vtglass
tglass
n
c nv
edge
Where is the front edge of the block when the light emerges? Use tglass
d
n
c nv
with either
expression – for the leading edge of the light, or the leading edge of the block.
cd
n
cd
xlight vlight tglass
n
c
nv
c
nv
in glass
xright
d
vtglass
d
v
d
edge
d
n
c nv
The part of the path that is left, l
c nv vdn
c nv
cd
c nv
cd
, will be traveled at speed c by the light. We express
c nv
that time, and then find the total time.
cd
l
c nv
tvacuum
c
453
ttotal
tglass
tvacuum
tglass
d
cd
c nv
c
l
n
c nv
l
c
d
n
c nv
n 1 d c v
l
c
c
c v
We check this for the appropriate limiting cases.
n 1 d c v l
n 1 d c c l
l
Case 1: ttotal
c
c
c v c
c
c c c
v c
This result was expected, because the speed of the light would always be c.
n 1 d c v l
n 1 d
l n 1 d
l
1
Case 2: ttotal
c
c
c v c
c
c
v 0
This result was obtained earlier in the solution.
n 1 d c v l
l
Case 3: ttotal
c
c
c v c
n 1
This result was expected, because then there is no speed change in the glass.
99. The spreadsheet used for this
problem can be found on the
Media Manager, with filename
1.2
K (10 17 J)
1.0
Classical
Relativistic
0.8
0.6
0.4
0.2
0.0
0
0.2
0.4
0.6
v /c
0.8
1
100. (a) We use Eq. 36-98. Since there is motion in two dimensions, we have
v2
1 x2
c
dp y
dp
dpx
;
0
px
mvx p0 ;
F
dt
dt
dt
Use the component equations to obtain expressions for v x2 and v 2y .
Fˆj
F
mv x
p0
mv y
v
2
y
vx
Ft
2 2
F t
p0
m
vy
c2
m2c2
p02
2 2
m
v x2
Ft
m
v 2y
F 2t 2
2 2
m
p02
v2
1 x2
2
m
c
py
v 2y
v x2
c2
F 2t 2
v x2
1
m2
c2
v 2y
c2
Ft
p02
.
mv y
c2
m2c2
v 2y
p02
v 2y
c2
v x2
F 2t 2
Substitute the expression for v 2y into the expression for v x2 .
454
Chapter 36
v
2
x
2
0
p
c2
v 2y
m2c 2
v x2 m 2 c 2
c2
2
0
p
p02
p02 m 2 c 2
v x2 m 4 c 4
v x2 m 2c 2 p02
v x2 m 2 c 2
v x2 p02
F 2t 2
m 2c 2
F 2t 2 v x2
p02 m 2c 2
F 2t 2
F 2t 2 v x2
v x2 F 2t 2 p02
p02c 2
m2c 4
p02
p02
p02 m 2 c 4
v x2 F 2t 2m 2c 2
v x2
m2c2
m2c 2
F 2t 2
v x2 F 2t 2
c2
F 2t 2
vx
p02m 2c 4
p02 F 2t 2v x2
p0c
2 2
mc
p02
F 2t 2
1/ 2
Use the expression for v x to solve for v y .
v 2y
F 2t 2
c2
m2c 2
F t
vy
v x2
m2c 2
m2c2
F 2t 2
F 2t 2
c 2 m 2c 2
2 2
c2
p02
m 2c 2
F 2t 2
F 2t 2 m 2 c 2
p02c 2
p02
F 2t 2
p02 c 2
p02 F 2t 2
F 2t 2
c2 m2c 2
2 2
F t
m 2c 2
F 2t 2
F 2t 2 m 2 c 2
p02
F 2t 2
Ftc
m2c 2
p02
F 2t 2
1/ 2
The negative sign comes from taking the negative square root of the previous equation. We
know that the particle is moving down.
(b) See the graph. We are
plotting v x c and v y c .
0.8
vx
(- vy)
0.6
v /c
The spreadsheet used for this
problem can be found on the
Media Manager, with filename
1.0
0.4
0.2
0.0
0
1
2
t ( s)
3
4
5
(c) The path is not parabolic, because the v x is not constant. Even though there is no force in the xdirection, as the net speed of the particle increases, increases. Thus v x must decrease as time
elapses in order for p x to stay constant.
455

Chapter 36 Solutions - galileo.harvard.edu

Transcription

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