CHAPTER 36: The Special Theory of Relativity … · CHAPTER 36: The Special Theory of Relativity Responses to Questions 1. ... Earth nor the reference frame of the Sun is inertial.

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427

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.

Physics for Scientists & Engineers with Modern Physics, 4th

Edition Instructor Solutions Manual


428

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 0

t t and

0.l l

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 2 2

1 .v c 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.)

15. No. The relativistic momentum of the electron is given by 2 2

.1

mvp mv

v c At low speeds

(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.

Chapter 36 The Special Theory of Relativity


429

Solutions to Problems

1. You measure the contracted length. Find the rest length from Eq. 36-3a.

0

2 2 2

38.2m72.5m

1 1 0.850v c

ll

2. We find the lifetime at rest from Eq. 36-1a.

28

2 2 6 6

0 8

2.70 10 m s1 4.76 10 s 1 2.07 10 s

3.00 10 m st t v c

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.

28

2 2

0 8

2.80 10 m s1 135 ly 1 48.5 ly

3.00 10 m sv cl l

5. The speed is determined from the time dilation relationship, Eq. 36-1a.

2 2

0

22 880

8

1

2.60 10 s1 1 0.807 2.42 10 m s

4.40 10 s

t t v c

tv c c c

t

6. The speed is determined from the length contraction relationship, Eq. 36-3a.

2 2

2 2 8

0

0

35ly1 1 1 0.78 2.3 10 m s

56lyv c v c c c

ll l

l

7. The speed is determined from the length contraction relationship, Eq. 36-3a. Then the time is found

from the speed and the contracted distance.

2 2

0 1 v cl l




430

2

2 20

0

25y25ly1 ; 27y

0.92325ly

1165ly

cv c t

v ccc

l l l

ll

l

8. The speed is determined from the length contraction relationship, Eq. 36-3a.

2

22 2

0

0

1 1 1 0.900 0.436v c v c c cl

l ll

9. 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.

2 2

0

1/ 2 22 2 3

2 2 101 12 22 2 8

0

1

11.2 10 m s1 1 1 1 1 6.97 10

3.00 10 m s

v c

v vv c

c c

l l

l

l

So the percent decrease is 86.97 10 % .

10. (a) The measured length is the contracted length. We find the rest length from Eq. 36-3a.

0

2 2 2

4.80m7.39m

1 1 0.760v c

ll

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. 22 2

0 1 20.0s 1 0.760 13.0st t v c

(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

22 2

0

0

1 1 1 0.990 0.141v c v c c cl

l ll

(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 2

2 2 00

11 1 1 0.140

1.0100

tt t v c v c c c

t

We see that a speed of 0.14 c results in about a 1% relativistic effect.

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.

0Earth

18.6 ly19.58yr 19.6yr

0.950t

v c

l

(b) The time as observed on the spacecraft is shorter. Use Eq. 36-1a.

22 2

0 1 19.58yr 1 0.950 6.114yr 6.11yrt t v c



431

(c) To the spacecraft observer, the distance to the star is contracted. Use Eq. 36-3a.

22 2

0 1 18.6 ly 1 0.950 5.808 ly 5.81 lyv cl l

(d) To the spacecraft observer, the speed of the spacecraft is their observed distance divided by

their observed time.

0

5.808 ly0.950

6.114 yrv c

t

l

13. (a) In the Earth frame, the clock on the Enterprise will run slower. Use Eq. 36-1a.

22 2

0 1 5.0yr 1 0.74 3.4yrt t v c

(b) Now we assume the 5.0 years is the time as measured on the Enterprise. Again use Eq. 36-1a.

2 2 00

2 2 2

5.0yr1 7.4yr

1 1 0.74

tt t v c t

v c

14. We find the speed of the particle in the lab frame, and use that to find the rest frame lifetime and

distance.

8lab

9

lab

1.00m2.941 10 m s 0.9803

3.40 10 s

xv c

t

(a) Find the rest frame lifetime from Eq. 36-1a.

22 2 9 10

0 lab 1 3.40 10 s 1 0.9803 6.72 10 st t v c

(b) In its rest frame, the particle will travel the distance given by its speed and the rest lifetime.

8 10

0 0 2.941 10 m s 6.72 10 s 0.198mx v t

This could also be found from the length contraction relationship: lab0

2 2.

1

xx

v c

15. Since the number of particles passing per second is reduced from N to N / 2, a time 0T 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 02cT in the lab frame during that time.

2 2 0 0 40 52 2

0

2 2

21 ; 0.894

1

1

T x cTT T v c T v v c c

TTv c

v c

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.

2 3 3 22 2 2 2 3

0 0 01 ; 1 2.0m 1 0.80 4.8mv c V v cl l l l l

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. 22 2

base 1 1 0.95 0.31v cl l l l

When at rest, the angle of the sides with respect to the base is given by 1 0.50cos 75.52 .

2.0

l

l

The vertical component of vert 2 sin 2 sin75.52 1.936l l l l is unchanged. The horizontal




432

component, which is 14

2 cos 2 0.50l l l at rest, will be contracted in the same way as the base.

22 2

horizontal 0.50 1 0.50 1 0.95 0.156v cl l l l

Use the Pythagorean theorem to find the length of the leg. 2 22 2

leg horizontal vert 0.156 1.936 1.942 1.94l l l l l l l

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.

2 2

0

2 28 8

0

1

1 10.95

3.00 10 m s 2.6 10 s1 125m

d dv v c

t t

v c c ctc

d

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 0.60 .v c The velocity of the Enterprise relative to the Earth is 0.90 .xu c Solve for

the velocity of the Enterprise relative to the alien vessel, ,xu using Eq. 36-7a.

2

0.90 0.600.65

1 0.60 0.901

x

x

x

u v c cu c

vu

c

We could also have made the Enterprise as reference frame S, with 0.90 ,v c and the velocity of

the alien vessel relative to the Earth as 0.60 .xu 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

relative to the alien vessel as 0.60 ,xu c and the velocity of the Earth relative to the Enterprise as

0.90 .xu c We solve for v, the velocity of the Enterprise relative to the alien vessel.

2 2 2

.60 0.90 0.65

0.90 .601 1 1

x x xx

x x x

u v c cu uu v c

vu u u c c

c c c

20. The Galilean transformation is given in Eq. 36-4.

(a) , , , , 25m 30m s 3.5s ,20m,0 130m,20m,0x y z x vt y z

(b) , , , , 25m 30m s 10.0s ,20m,0 325m,20m,0x y z x vt y z

21. (a) The person’s coordinates in S are found using Eq. 36-6, with 25 mx , 20 my , 0z , and

3.5 s.t We set 81.80 10 m/s.v

8

2 2 2 28 8

25m 1.8 10 m/s 3.5 s820m

1 1 1.8 10 m/s 3.0 10 m/s

20m ; 0

x vtx

v c

y y z z



433

(b) We repeat part (a) using the time 10.0 s.t

8

2 2 2 28 8

25m 1.8 10 m/s 10.0 s2280m

1 1 1.8 10 m/s 3.0 10 m/s

20m ; 0

x vtx

v c

y y z z

22. We determine the components of her velocity in the S frame using Eq. 36-7, where 81.10 10 m/sx yu u and 81.80 10 m/sv . Then using trigonometry we combine the

components to determine the magnitude and direction. 8 8

8

22 8 8 8

1.10 10 m/s 1.80 10 m/s2.38 10 m/s

1 / 1 1.80 10 m/s 1.10 10 m/s / 3.00 10 m/s

xx

x

u vu

vu c

2 28 8 82 2

7

22 8 8 8

1.10 10 m/s 1 1.8 10 m/s 3.0 10 m/s17.21 10 m/s

1 / 1 1.80 10 m/s 1.10 10 m/s / 3.00 10 m/s

y

y

x

u v cu

vu c

2 22 2 8 7 8

71 1

8

2.38 10 m/s 7.21 10 m/s 2.49 10 m/s

7.21 10 m/stan tan 16.9

2.38 10 m/s

x y

y

x

u u u

u

u

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 0.60 .v c The velocity of spaceship 1 relative to the Earth is 0.60 .xu c Solve

for the velocity of spaceship 1 relative to spaceship 2, ,xu using Eq. 36-7a.

2

0.60 0.600.88

1 0.60 0.601

x

x

x

u v c cu c

vu

c

(b) Now consider spaceship 1 as reference frame S. The velocity of the Earth relative to spaceship

1 is 0.60 .v c The velocity of spaceship 2 relative to the Earth is 0.60 .xu c Solve for the

velocity of spaceship 2 relative to spaceship 1, ,xu using Eq. 36-7a.

2

0.60 0.600.88

1 0.60 0.601

x

x

x

u v c cu c

vu

c

As expected, the two relative velocities are the opposite of each other.

24. (a) The Galilean transformation is given in Eq. 36-4. 8 6100m 0.92 3.00 10 m s 1.00 10 s 376mx x vt x vt

(b) The Lorentz transformation is given in Eq. 36-6. Note that we are given t, the clock reading in

frame S.

2 2

2

vx t vxt t t

c c

t vx v ct vxx x vt x v x

c c c




434

2 8 6

2

1 100m 0.92 1 0.92 3.00 10 m s 1.00 10 s 0.92 100m

1 0.92

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 0.61 .v c The speed of the

second spaceship relative to the first spaceship is 0.87 .xu c We use Eq. 36-7a to solve for the

speed of the second spaceship relative to the Earth, u.

2

0.87 0.610.97

1 0.61 0.871

x

x

x

u v c cu c

vu

c

(b) The only difference is now that 0.87 .xu c

2

0.87 0.610.55

1 0.61 0.871

x

x

x

u v c cu c

vu

c

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

0.90 .v c The velocity of the probe relative to the planet is 0.95 .xu c Solve for the velocity of

the probe relative to your spaceship, ,xu using Eq. 36-7a.

2

0.95 0.900.34

1 0.90 0.951

x

x

x

u v c cu c

vu

c

27. We set frame S as the frame at rest with the spaceship. In this frame the module has speed

0.82 .yu u c 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 0.76 .v c 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.

2 2 2

2 2

10 0.76 0.82 1 0.760.76 ; 0.533

1 / 1 0 1 / 1 0

yxx y

x x

u v cu v c cu c u c

vu c vu c

2 22 2 1 1 0.5330.76 0.533 0.93 ; tan tan 35

0.76

y

x y

x

u cu u u c c c

u c

28. The velocity components of the particle in the S frame are cosxu u and sin .yu u 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.

2 2 2 2

2 2 2 2

1 1 ;

1 1 1 1

y yx x

x x y y

x x x x

u v c u v cu v u vu u u u

vu c vu c vu c vu c



435

2 2

2 2 2 2 2 2 2

2

1

1 1 sin 1 sin 1tan

cos cos

1

y

xy y

xx x

x

u v c

vu cu u v c u v c v c

u vu u v u v v u

vu c

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 y-

components 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 .

2 2 2 2

0 0 0

22 2 2 2 2 2 2 2

0 0 0

cos ; sin ; 1 cos 1

cos 1 sin 1 cos

x y y x x

x y

v c v c

v c v c

l l l l l l l l

l l l l l l

(b) We calculate the angle from the length components in the moving frame.

1 1 1 10

2 2 2 2

0

sin tantan tan tan tan tan

cos 1 1

y

x v c v c

l l

l l

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 A B 0.t t To simplify the problem we also set the

location of gunman A equal to zero in frame S when the guns were fired, A 0.x This places

gunman B at B 55.0m.x Use Eq. 36-6 to determine the time that each gunman fired his

weapon in frame S.

AA A 2 2

14BB B 22 2 88

00 0

35m/s 55.0m10 2.14 10 s

3.00 10 m/s1 35.0m/s 3.00 10 m/s

vx vt t

c c

vxt t

c

Therefore, in Frame S, A fired first.

(b) As found in part (a), the difference in time is 142.14 10 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 A B 0.t t

Taking the difference in these times gives the time interval as measured by the observing moving

with velocity v.

2

2 2 2 2 2 2

= 1

xx x vt x vt

vx v x v vx t vx vxt t t vt t t t

c c c c c c




436

2 2 2 2 2

2

00 0 ; 0A B

A A B B

B A

vx v vx v vt t t t

c c c c c

vt t t

c

l l

l

According to the observer, bulb B turned on first.

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 A 0x and B 220 m.x We set the time when event

A occurred equal to zero, so A B0 and 0.80 s.t t We then set the location of the two events in

frame S equal, and using Eq. 36-6 we solve for the velocity.

8A BA B A B B

A B

0 220m ; 2.5 10 m/s

0 0.88 sA

x xx x x vt x vt v

t t

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. 2 22 2 2 2

0 01 1 1 10.0 m 12.0 m 0.5528v c v c c cl l l l

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 A B 0.t t In

frame S these two events occur at the front and far side of the barn, or at A 0x and B 10.0m.x

Using Eq. 36-6, we calculate the times at which the barn doors close in the boy’s frame of reference.

2 2

8

2 82

00 0

0.5528 10.0m10 2.211 10 s

3.00 10 m/s1 0.5528

AA A

BB B

vx vt t

c c

vxt t

c

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. 8 80.5528 3.00 10 m/s 2.211 10 s 3.67 m.x vt

We use Eq. 36-3a to determine the length of the barn in the boy’s frame of reference.

2 2 2

0 1 10.0 m 1 0.5528 8.33 mv cl l

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.

0,pole barn12.0 m 3.67 m 8.33 mxl l

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.



437

34. The momentum of the proton is given by Eq. 36-8.

27 8

19

2 2 2

1.67 10 kg 0.75 3.00 10 m s5.7 10 kg m s

1 1 0.75

mvp mv

v c

35. (a) We compare the classical momentum to the relativistic momentum.

2classical 2 2

relativistic

2 2

1 1 0.10 0.995

1

p mvv c

mvp

v c

The classical momentum is about 0.5% in error.

(b) We again compare the two momenta.

2classical 2 2

relativistic

2 2

1 1 0.60 0.8

1

p mvv c

mvp

v c

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”.

2f

f22 2 22 2

2ff ff

2 22 20 00 f

2 22 200

2 2

f f

0.261 12 4 4 0.29

1 1 0.26

11

0.29 0.47

1.29

mv v

cv cp vv cc

mv vp v c

v cv c

v c v 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”.

1 1 2 21 2 2 2 2 2

1 2

2 2 22 2 2721 2 2

2272 2 2 211 2

1

1 1

0.606.68 10 kg9.0

1.67 10 kg1 1 1 0.60

0.90 0.95

m v m vp p

v c v c

cv m vc

mv c v c

v c c

38. We find the proton’s momenta using Eq. 36-8.

p 1 p p 2 p

0.45 p 0.80 p2 2 2 2

1 2

2 2

p 2 p

0.98 p2 2

2

2

0.45 0.800.5039 ; 1.3333

1 0.45 1 0.801 1

0.984.9247

1 0.981

m v m c m v m cp m c p m c

v v

c c

m v m cp m c

v

c




438

(a) p p2 1

1 p

1.3333 0.5039100 100 164.6 160%

0.5039

m c m cp p

p m c

(b) p p2 1

1 p

4.9247 1.3333100 100 269.4 270%

1.3333

m c m cp p

p m c

39. The rest energy of the electron is given by Eq. 36-12.

22 31 8 14

14

13

9.11 10 kg 3.00 10 m s 8.20 10 J

8.20 10 J0.511MeV

1.60 10 J MeV

E mc

40. We find the loss in mass from Eq. 36-12.

13

28 28

22 8

200MeV 1.60 10 J MeV3.56 10 kg 4 10 kg

3.00 10 m s

Em

c

41. We find the mass conversion from Eq. 36-12.

19

22 8

8 10 J900kg

3.00 10 m s

Em

c

42. We calculate the mass from Eq. 36-12.

227 8

2 2

2 2 2 13

1.6726 10 kg 2.9979 10 m s1 1938.2 MeV

1.6022 10 J MeV

Em mc c

c c c

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 21.00MeV ,c which is 301.78 10 kg. .

44. (a) The work is the change in kinetic energy. Use Eq. 36-10b. The initial kinetic energy is 0.

2 4

final2

11 1 938.3MeV 1.39 10 MeV

1 0.998

13.9GeV

W K K mc

(b) The momentum of the proton is given by Eq. 36-8.

2 4

2

1938.3MeV 0.998 1.48 10 MeV 14.8GeV

1 0.998p mv c c c c

45. We find the energy equivalent of the mass from Eq. 36-12.

2

2 3 8 131.0 10 kg 3.00 10 m s 9.0 10 JE mc

We assume that this energy is used to increase the gravitational potential energy.

13

9

3 2

9.0 10 J 9.2 10 kg

1.0 10 m 9.80m s

EE mgh m

hg



439

46. The work is the change in kinetic energy. Use Eq. 36-10b. The initial kinetic energy is 0.

2 2 2

1 0.90 2 0.99 0.90 0.99 0.90

2 2 2 20.99 0.902 0.99 0.90

2

1 0.90 0.90

2

1 ; 1 1

1 1

1 1 1 0.99 1 0.90 3.711 1

11 0.90

c cW mc W K K mc mc

mc mcW

W mc

47. The kinetic energy is given by Eq. 36-10.

2 2

2 2

1 31 2 0.866

41K mc mc v c c

v c

48. The total energy of the proton is the kinetic energy plus the mass energy. Use Eq. 36-13 to find the

momentum.

2

2 2 22 2 2 2 2 2 2

22 2

;

2

938.3MeV2 1 2 950MeV 1 2 1638MeV

950MeV

1638MeV 1.6GeV

E K mc

pc E mc K mc mc K K mc

mcpc K K mc K

K

p c 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.

8

8

2

2

2

2

2.80 10 m s0.9333

3.00 10 m s

11 1 938.3MeV 1674.6MeV 1.67GeV

1 0.9333

1938.3MeV 0.9333 2439 MeV 2.44GeV

1 0.9333

v c

K mc

p mv c c c c

50. We use Eq. 36-10 to find the speed from the kinetic energy.

2 2

2 2

2 2

2

11 1

1

1 11 1 0.957

1.25MeV1 1

0.511MeV

K mc mcv c

v c c cK

mc

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.

2 2

2 2

11 1

1K mc mc

v c




440

2 2

2

1 11 1 0.470

125MeV1 1

938.3MeV

v c c cK

mc

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.

2 2

2 2

22 2

1

mE mc Mc M m

v c

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.

2 2

lost initial2 2

12 1 1 2

1K K mc mc

v c

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.

2 2

2 2

2 2

2

11 1

1

1 11 1 0.32

0.028MeV1 1

0.511MeV

K mc mcv c

v c c cK

mc

54. We use Eqs. 36-11 and 36-13 in order to find the mass.

22 2 2 2 4 2 2 2 2 4

2 222 2 22 28

2 2

2

121MeV 45MeV140MeV 2.5 10 kg

2 2 45MeV

E p c m c K mc K Kmc m c

c cp c Km c

Kc c

The particle is most likely a probably a 0 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.

2 21 12 2

2 2 342 2

11 2 0.866

1

K E K mc K mc

K mc mc v c cv c

(b) In this case, the kinetic energy is half the rest energy.

2 2 3 51

2 2 92 2

11 0.745

1K mc mc v c c

v c



441

56. We use Eq. 36-10 for the kinetic energy and Eq. 36-8 for the momentum.

2 2

2 2 27

8

7

2 8

2 2 2 2 27

8

1 11 1 1 938.3MeV

1 8.15 10 m s1

3.00 10 m s

36.7 MeV

8.15 10 m s938.3MeV

3.00 10 m s1 1265MeV

1 1 8.15 10 m s1

3.00 10 m s

K mc mcv c

mc v cmvp mv c

c cv c v c

Evaluate with the classical expressions.

22 72 21 1 1

c 2 2 2 8

72

c 8

8.15 10 m s938.3MeV 34.6MeV

3.00 10 m s

1 8.15 10 m s938.3MeV 255MeV

3.00 10 m s

vK mv mc

c

vp mv mc c

c c

Calculate the percent error.

c

c

34.6 36.7error 100 100 5.7%

36.7

255 265error 100 100 3.8%

265

K

p

K K

K

p p

p

57. (a) The kinetic energy is found from Eq. 36-10.

22 4 8

2 2 2

19 19

1 11 1 1 1.7 10 kg 3.00 10 m s

1 1 0.18

2.541 10 J 2.5 10 J

K mc mcv c

(b) Use the classical expression and compare the two results.

24 8 191 1

2 2

19 19

19

1.7 10 kg 0.18 3.00 10 m s 2.479 10 J

2.479 10 J 2.541 10 J%error 100 2.4%

2.541 10 J

K mv

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.

2 2

2 2

2 2

2

11 1

1

1 11 1 to 7 sig. fig.

998GeV1 1

0.938GeV

K mc mcv c

v c c cK

mc




442

27 8

2 2

3 19

998GeV1 1.673 10 kg 3.00 10 m s1

0.938GeV3.3T

1.0 10 m 1.60 10 C

Kmc

mv mv mcB

rqv rq rq

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.

2 2

Am Np

Np Am 2 2 2

5.5MeV 1u241.05682u 4.00260u 237.04832u

931.49 MeV

m c m m c K

Km m m

c c c

60. (a) For a particle of non-zero mass, we derive the following relationship between kinetic energy

and momentum.

2 2 222 2 2 2 2 2 2

2 22 2

22 2

; 2

2 4 42 0

2

E K mc pc E mc K mc mc K K mc

mc mc pcK K mc pc K

For the kinetic energy to be positive, we take the positive root.

2 22 22 22 2

2 4 4

2

mc mc pcK mc mc pc

If the momentum is large, we have the following relationship.

2 22 2 2K mc mc pc pc mc

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.

22 22 2 2 2

21

pcK mc mc pc mc mc

mc

2 22 2 1

2 2 1

2

pc pmc mc

mc m

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.”

p

K

0m

0m



443

61. All of the energy, both rest energy and kinetic energy, becomes electromagnetic energy. We use Eq.

36-11. Both masses are the same.

2 2 2

total 1 2 1 2 1 22 2

1 1105.7 MeV

1 0.43 1 0.55

243.6MeV 240MeV

E E E mc mc mc

62. We use Eqs. 36-11 and 36-13. 2 2 222 2 2 2 2 2 2

2 2

; 2

2

E K mc pc E mc K mc mc K K mc

K K mcp

c

63. (a) We assume the mass of the particle is m, and we are given that the velocity only has an x-

component, .xu 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: ,xu the velocity

in reference frame S; ,xu 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

reserve the symbol for 2 2

1,

1 v c and also use Eq. 36-11 for energy.

2 2

2 2

2 2 2 2 2 2

2 2

; 0 ; 01

1 1 ; 0 ; 0

1 1 1 1

; 0 since 0 ; 0 since 01

xx y z

x

x x x xx x y y z z

x x

xx y y z z

x

mup p p

u c

u v u v vu c vu cu u u u u u

vu c vu c v c v c

mup p u p u

u c

Substitute the expression for xu into the expression for .xp

2

22 2 2 2

22

22 22

2 22 2

2 2 22 2

2 2

2 22

2 2 2

2 2 2 2 2

1 1

11 11 1

11 1

1

1 11 1

1

21 2

x

x xxx

xx x x

x

x

x x

x x

xx x

x x

x

x

x x x x

u vm

vu c u vmup m

vu cu c u v u vvu c

c cvu cvu c vu c

u v m u vm

vu c u v u vvu c vu c

c cvu c

m u v m

vu vu u u v v

c c c c c

2 2 2

2 2 21

x

x x

u v

vu u v

c c c




444

2 2 2 2

2 2 2 2 2 2

2 2

2 22 2 2 2 2 2

2

2 2 2 2 2 2

1 1

1 1 1

1 1 1

1 1 1

x

x xx

x

xx

x x xx

mu mv

u c u cm u v

v c u c v c

mu mc v mc vp

c cu c u c u c p vE c

v c v c v c

It is obvious from the first few equations of the problem that 0y yp p and 0 .z zp p

2 2 2

2 2 2 2

22

22 22

2 22 2

2

2 2 2 22 2 2

2 2 2 2 2 2 22

2

2

2 2

1 11 1

11 1

1 11

1 1 11

1

x x x

x

x

x x

x

x xx x

xx

x

x

mc mc mcE

u c u v u vvu cc cvu cvu c vu c

mc mvu

u c u cmc vu c mc mvu

v c u c v cu vvu c

c

E p v

v c

(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

2

2 2 2 2

2

2 2 2 2

; ; ; 1 1

; ; ; 1 1

x xx y y y y

p vE c E p vp p p p p E

v c v c

x vt t vx cx y y z z t

v c v c

These transformations are identical if we exchange xp with x, yp with y, zp with z, and 2E c

with t (or E c with ct).

64. The galaxy is moving away from the Earth, and so we use Eq. 36-15b.

0 0 0

22

0

0 22

0

0.0987 0.9013

1 1 0.9013 0.1035

1 0.90131+

f f f f f

f fc vf f v c c c

c v f f

65. For source and observer moving towards each other, use Eq. 36-14b.

0 0

1 1 0.7095.0MHz 226MHz 230MHz

1 1 0.70

c v v cf f f

c v v c



445

66. We use Eq. 36-15a, and assume that .v c

0 0 0 0 2 2

1/ 22 2 0

0 0 0 0

0 0

1 11 11

1 1 1 1

1 1 1

v c v cc v v cv c

c v v c v c v c v c

vv c v c v c v c

c

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 Doppler-

shifted 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.

0 0 0

beatbeat 0 0 0 0 0

0

8

9

2 2

2

3.00 10 m/s 6670Hz27.8m/s

2 36.0 10 Hz

c v c v c v c v c vf f f f f f

c v c v c v c v c v

c v c v v v cff f f f f f f v

c v c v c v c f

v

(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.

beat beat 0beat

0 0

9

beat 8

2

2 2

2 36.0 10 Hz 1km/h 1m/s70Hz

3.600km/h3.00 10 m/s

cf c f f vv v f

f f c

f

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.

0 0 0 0 02 2 2 2

2 2

1

c v c v c v c v v v cf f f f f f

c v c v c v c v c v v c

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 23 .kT mc

1/ 2

rms 2 2 2 2

0

3 3 3 3 3 2 1 2

kT v kT f kT kT kTv v

m c mc f mc mc mc

We evaluate for a gas of H atoms (not 2H molecules) at 550 K. Use Appendix F to find the mass.

23

5

22 27 80

3 1.38 10 J K 550K32 2 2.5 10

1.008u 1.66 10 kg u 3.00 10 m s

f kT

f mc




446

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. 62 6.38 10 m2

464m/s24hr 3600s/hr

Rv

T

20,eq 2 2

0,eq 22 2

0,pole

0,pole

2

0,eq 0,pole 2

2 72

22 8

1 / 121 /

1 0

12

2.0yr 464m/s 3.156 10 s/yr 75 s

2 2 3.00 10 m/s

t vt t t v c t

cv c

tt t t

vt t t t

c

vt

c

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 0.60 .v c The velocity of the first pod relative to the spacecraft is

0.50 .xu c Solve for the velocity of the second pod relative to the first pod, ,xu using Eq. 36-7a.

2

0.50 0.600.846

1 0.60 0.501

x

x

x

u v c cu c

vu

c

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 2 2 2Earth EarthEarth plane Earth

22 2Earth Earth Earth1

Earth plane 2 2 2

6

8

28

2 2 ; 1 1

2 21 1 1 1

1m s6.38 10 m 1300km h

3.6km h8.0 10 s

3.00 10 m s

r rt t t v c v c

v v

r r v r vt t t v c

v v c c

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.

2 2spacecraft Earth

spacecraft spacecraft

2 2

spacecraft

Earth

1

1 10.6829 0.68

4.6ly11

4.3ly

x x v cv

t t

v c c c c

c t

x



447

(b) Find the elapsed time according to observers on Earth, using Eq. 36-1a.

spaceship

Earth2 2 2

4.6y6.3y

1 1 0.6829

tt

v c

Note that this agrees with the time found from distance and speed.

EarthEarth

4.3ly6.3yr

0.6829

xt

v 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. 2

0 0 2

1.070 11.070 0.067

1.070 1

c vc v c

c v

(b) We assume that the quasar is moving and the Earth is stationary. Then we use Eq. 16-9b.

00 0

0

1 1 1.070 0.070

1 1

f c cf v c v c

v c 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.

2 2 2 2

0

0 2 2 2 2

0

60 30 0.60

60 30

f fc vf f v c c c

c v f f

75. (a) The velocity components of the light in the S frame are 0xu and .yu c We transform those

velocities to the S frame according to Eq. 36-7.

2 2 2 2

2 2

2 2

2 2 21 1 1

2

1 10 ; 1

1 1 0 1 1 0

1tan tan tan 1

yxx y

x x

y

x

u v c c v cu v vu v u c v c

vu c vu c

u c v c c

u v v

(b) 2 2 2 2 2 2 2 2 21x yu u u v c v c v c v c

(c) In a Galilean transformation, we would have the following.

2 2 1 ; ; ; tanx x y y

cu u v v u u c u v c 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

0.65 .v c The velocity of rocket B relative to the Earth is 0.85 .xu c Solve for the velocity of

rocket B relative to rocket A, ,xu using Eq. 36-7a.

2

0.85 0.650.45

1 0.65 0.851

x

x

x

u v c cu c

vu

c

Note that a Galilean analysis would have resulted in 0.20 .xu c




448

77. (a) We find the speed from Eq. 36-10.

2 2 2

2 2

2 2

82 2

11 1 14,000

1

1 11

14,001 2 14,001

3.00 10 m s1 10.77m s

2 14,001 2 14,001

K mc mc mcv c

cv c c

cc v

(b) The tube will be contracted in the rest frame of the electron, according to Eq. 36-3a.

2

2 2 3

0

11 3.0 10 m 1 1 0.21m

14,001v cl l

78. The electrostatic force provides the radial acceleration. We solve that relationship for the speed of

the electron.

9 2 2

2 2

electronelectrostatic centripetal 2

0

2192

6

31 100 electron

8.99 10 N m C

1

4

1.60 10 C12.18 10 m s 0.0073

4 9.11 10 kg 0.53 10 m

e m vF F

r r

ev c

m r

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.

2 132 2 0.511MeV 1.022MeV 1.64 10 JE mc

80. The wattage times the time is the energy required. We use Eq. 36-12 to calculate the mass.

7

2 5

22 8

75W 3.16 10 s 1000g 2.6 10 g

1kg3.00 10 m s

PtE Pt mc m

c

81. Use Eqs. 36-13, 36-8, and 36-11. 1/ 2

2 2 2 2 4 2 2 2 4

2 2 21/ 2

2 2 2 4 212 2

2

E p c m c E p c m c

dE pc pc mvcp c m c pc v

dp E E mc

82. The kinetic energy available comes from the decrease in rest energy.

2 2 2 2

n p 939.57MeV 938.27MeV 0.511MeV 0 0.79MeVe vK m c m c m c m c

83. (a) We find the rate of mass loss from Eq. 36-12.

2 2

269 9

22 8

1 4 10 J s4.44 10 kg s 4 10 kg s

3.00 10 m s

E mc E m c

m E

t c t



449

(b) Find the time from the mass of the Sun and the rate determined in part (a). 24

7 7Earth

9 7

5.98 10 kg4.27 10 y 4 10 y

4.44 10 kg s 3.156 10 s y

mt

m t

(c) We find the time for the Sun to lose all of its mass at this same rate.

30

13 13Sun

9 7

1.99 10 kg1.42 10 y 1 10 y

4.44 10 kg s 3.156 10 s y

mt

m t

84. Use Eq. 36-8 for the momentum to find the mass.

2 2

28

22

82 2

31

8

1

2.24 10 m s3.07 10 kg m s 1

3.00 10 m s19.12 10 kg

2.24 10 m s

mvp mv

v c

p v cm

v

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.

2

p+e n He

22

2 2

931.5MeV2 1.00783u 2 1.00867u 4.00260u 28.32 MeV

u

E m m m c

cc

86. The momentum is given by Eq. 36-8, and the energy is given by Eq. 36-11 and Eq. 36-13.

2 2 2

2 2 2 4 2 2 2 2 2

mc v Ev pc pc pcP mv v

c c E m c p c m c p

87. (a) The magnitudes of the momenta are equal. We use Eq. 36-8.

2

2 2 2 2 2

10

8

18

938.3MeV 0.9851 15356MeV

1 1 1 0.985

1 1.602 10 J GeV5.36GeV 5.36GeV

3.00 10 m s 1GeV

2.86 10 kg m s

mc v cmvp mv c

c cv c v c

cc c

(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.

2

0.985 0.9850.9999

1 0.985 0.9851

x

x

x

c cv uu c

vu

c




450

22

2 2 2 2 2

2

10

8

17

2 0.985938.3MeV

1 0.9851 162081MeV

1 1 2 0.9851

1 0.985

1 1.602 10 J GeV62.1GeV 62.1GeV

3.00 10 m s 1GeV

3.31 10 kg m s

xxx

x x

mc u cmup mu c

c cu c u c

cc c

88. We find the loss in mass from Eq. 36-12.

3

12

22 8

484 10 J5.38 10 kg

3.00 10 m s

Em

c

Two moles of water has a mass of 336 10 kg. Find the percentage of mass lost.

12

10 8

3

5.38 10 kg1.49 10 1.5 10 %

36 10 kg

89. Use Eq. 36-10 for kinetic energy, and Eq. 36-12 for rest energy.

2 2

Enterprise converted

9 7

converted Enterprise2 2 2

1

1 11 1 6 10 kg 3 10 kg

1 1 0.10

K m c m c

m mv c

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. 2 2

ship

5 4

ship ship2 2 2

1

1 11 1 1 1.8 10 kg 7.2 10 kg

1 1 0.70

K m c mc

m m mv c

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.

2 2 2

0

35y50y ; 1 50y 1 0.70 36y

0.70

cdt t t v c

v c

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 0.85 ,v c and the velocity of the positive-

moving particle relative to the laboratory frame is 0.85 .xu c Solve for the velocity of the positive-

moving particle relative to the negative-moving particle, .xu

2

0.85 0.850.987

1 0.85 0.851

x

x

x

u v c cu c

vu

c



451

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). 2 2

0

2 20 0

0

0

1 1 1 0.9887

1.0ly11

6.6ly

v c vv c

t t cc t

ll

l

2 2

2 2

24 8 22

2

11 1

1

11 3.2 10 kg 3.00 10 m s 1.6 10 J

1 0.9887

W K mc mcv c

93. The kinetic energy is given by Eq. 36-10.

22 2 8

2 2 2

21

1 11 1 1 14,500kg 3.00 10 m s

1 1 0.98

5.3 10 J

K mc mcv c

We compare this with annual U.S. energy consumption: 21

20

5.3 10 J53.

10 J

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.

1/ 22 2 2 4

1/ 2 1/ 22 2 2 2 4 2 2 2 4

1/ 2 22 2 2 2 4 2 2 2 2 4

;

v v v v v

v v

E p c m c p c p p p

E E E m c p c m c p c p c m c pc

m c pc p c m c m c pc p c m c

Solve for the momentum.

2 2 2 2

2 4 2 2 2 2 2 2 42 2

m c m cm c m c pc p c p c m c pc

m

Write the kinetic energy of the muon using Eqs. 36-11 and 36-13.

2 2

2 2 2 2

2 2 2 2

2 2 2 2 2 2

22 2 2 2 2 2 2 2

;

2

2

2 2

2 2 2

2 2 2

vK E m c E E E m c pc

m c m cK m c pc m c m c m c

m

m m c m c m c m c

m m

m m m m m c m m m m c m m c

m m m




452

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.

spaceship spaceshipEarthEarth

2 2 2Earth

Earth

Earth

2 22 22 2Earth Earth

Earth spaceship Earth spaceship

2

EarthEarth spacesh

; 1

1

t txv t

t v c x

c t

x xt t t t

c c

xt t

c

2 2 2

ip 6.0y 2.50y 6.5y

(b) The distance as measured by the spaceship will be contracted.

spaceship spaceshipEarth

spaceship Earth

Earth spaceship Earth

2.50y 6.0ly 2.3ly

6.5y

x txv x x

t t t

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.

1.88kg

2 2 0.939s84.2 N m

mT

k

(b) The oscillating mass is a clock. According to observers on Earth, clocks on the spacecraft run

slow.

Earth2 2 2

0.939s2.15s

1 1 0.900

TT

v c

97. We use the Lorentz transformations to derive the result.

2 2

; vx v x

x x vt x x v t t t t tc c

2 222 2 2

2

22 2 22 2

22 22 2

2 2

2 2

11

1

v x v xc t x c t x v t c t x v t

c c

v x v xc t c t x x v t v t

c c

vc v t x

cv c

2 22 2 2 2 2

2 2

2 2

2 2 2 2

2 2

11 1

1

1

1

c v c t v c xv c

v cc t x c t x

v c



453

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.

0 glass vacuum

glass

distance in glass distance in vacuum

speed in glass speed in vacuum

1

v

d dt t t

v c

n dd d nd d nd d

c n c c c c c

l

ll l l

If the index of refraction is 1,n then the glass will have no effect on the light, and the time would

simply be the distance divided by the speed of light.

1 glass vacuum

distance in glass distance in vacuum

speed in glass speed in vacuumn

d d d dt t t

c c c c

l l l

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.

lightin glass

2

1 1

1 1 1 1

vn vnc cv v

c cc cn nvcv v v vn n

nc nc nc nc

The contracted width of the glass, from the Earth frame of reference, is given by Eq. 36-3a.

2 2

movingglass

1d

d d v c

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.

light light rightin glass edge

light right glass glass glassedge

; ;

c dx v t t x vt

n

c d d nx x t vt t

n c nv

Where is the front edge of the block when the light emerges? Use glass

d nt

c nv with either

expression – for the leading edge of the light, or the leading edge of the block.

light light glassin glass

right glassedge

c d n cdx v t

n c nv c nv

d c nv vdnd d d n cdx vt v

c nv c nv c nv

The part of the path that is left, ,cd

c nvl will be traveled at speed c by the light. We express

that time, and then find the total time.

vacuum

cd

c nvt

c

l




454

total glass vacuum glass

1

cd

c nvd n d nt t t t

c nv c c c nv

n d c v

c c c v

ll

l

We check this for the appropriate limiting cases.

Case 1: total

1 1

v c

n d n dc v c ct

c c c v c c c c c

l l l

This result was expected, because the speed of the light would always be c.

Case 2: total0

1 1 11

v

n d n d n dc vt

c c c v c c c

ll l

This result was obtained earlier in the solution.

Case 3: total1

1

n

n d c vt

c c c v c

l l

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



100. (a) We use Eq. 36-98. Since there is motion in two dimensions, we have 22

2 2

1.

1yx

vv

c c

0

ˆ ; 0 ; yx

x x y y

dpd dpF p mv p F p Ft mv

dt dt dt

pF j

Use the component equations to obtain expressions for 2

xv and 2 .yv

2 222 2 22 2 20 0 0

0 02 2 2 2 2 2 2 2

0

22 2 2 2 22

2 2 2 2 2

2 2

2 2 2

2 2 2 2

1

1

yyxx x x x

yxy y y

x

y

c vvp p p vmv p v v v p

m m m c c m c p

vFt F t F t vmv Ft v v

m m m c c

c vv F t

m c F t

Substitute the expression for 2

yv into the expression for 2.xv

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 0.2 0.4 0.6 0.8v /c

K (

10

17 J

) Classical

Relativistic



455

2 2

2 2 2

2 2 2 22 2 2 4 2 2 2

2 2 2 2

0 0 02 2 2 2 2 2 2 2 2 2 2 2 2

0 0 0

2 2 2 2 2 2 2 2 2 2 4 2 2 2

0 0

2 4 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2

0 0 0 0

x

y x

x

x x

x x x x x

c vc F t

m c F tc v m c F t vv p p p

m c p m c p m c p m c F t

v m c p m c F t p m c F t v

v m c v m c p v F t m c v F t p p m c p F t v

2 2 2 2 2 2 2 2 2 2 00 0 1/ 2

2 2 2 2 2

0

x x x x

p cv m c v p v F t p c v

m c p F t

Use the expression for xv to solve for .yv

2 22 0

2 2 2 2 22 202 2 2 2 2

2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 20 02 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

0 0

1/ 22 2 2 2 2

0

x

y

y

p cc

m c p F tc vv F t F t

m c F t m c F t

c m c p F t p c c m c F tF t F t

m c F t m c p F t m c F t m c p F t

Ftcv

m c p F t

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 xv c and .yv c

The spreadsheet used for this

problem can be found on the

Media Manager, with filename



(c) The path is not parabolic, because the xv is not constant. Even though there is no force in the x-

direction, as the net speed of the particle increases, increases. Thus xv must decrease as time

elapses in order for xp to stay constant.

0.0

0.2

0.4

0.6

0.8

1.0

0 1 2 3 4 5t ( s)

v/c

vx

(- vy)

CHAPTER 36: The Special Theory of Relativity … · CHAPTER 36: The Special Theory of Relativity Responses to Questions 1. ... Earth nor the reference frame of the Sun is inertial.

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