OpenStax College Physics
Instructor Solutions Manual
Chapter 28
CHAPTER 28: SPECIAL RELATIVITY
28.2 SIMULTANEITY AND TIME DILATION
1.
(a) What is  if v  0.250c ? (b) If v  0.500c ?
Solution (a) Using the definition of  , where v  0.250c :
1 / 2
1 / 2
 v2 
 (0.250c) 2 
  1  2 
 1 
 1.0328

c2
 c 


(b) Again using the definition of  , now where v  0.500c :
 (0.500c) 2 
  1 

c2


1 / 2
 1.1547  1.15
2.
(a) What is  if v  0.100c ? (b) If v  0.900c ?
Solution
 (0.100c) 2 
 1.00504
(a)   1 

c2


Showing three digits difference from 1.
1 / 2
 (0.900c) 2 
(b)   1 

c2


1 / 2
 2.294  2.29
3.
Particles called  -mesons are produced by accelerator beams. If these particles travel
at 2.70 108 m/s and live 2.60 108 s when at rest relative to an observer, how long
do they live as viewed in the laboratory?
Solution
 v2 
 (2.70 108 m/s) 2 


γ  1  2   1 
 2.294, so that
8
2
 c 
 (3.00 10 m/s) 
Δt  γΔt 0  (2.294)(2. 60 10 8 s)  5.96 10 8 s
4.
Suppose a particle called a kaon is created by cosmic radiation striking the
atmosphere. It moves by you at 0.980c , and it lives 1.24 108 s when at rest relative
to an observer. How long does it live as you observe it?
Solution
 v2 
 (0.980c) 2 
  1  2 
 1 
 5.025, so that

c2
 c 


Δt  γΔt 0  (5.025)(1. 24  10 8 s)  6.23  10 8 s
5.
A neutral  -meson is a particle that can be created by accelerator beams. If one such
1/2
1 / 2
1/2
1 / 2
OpenStax College Physics
Instructor Solutions Manual
Chapter 28
16
16
particle lives 1.40  10 s as measured in the laboratory, and 0.840  10 s when at
rest relative to an observer, what is its velocity relative to the laboratory?
Solution
 v2 
Δt 1.40 10 16 s
Δt  Δt 0  γ 

 1.667  1  2 
Δt 0 8.40 10 17 s
 c 
1/ 2

1
v  c1  2 
 γ 

1 
 c 1 
2
 (1.67) 
1/2
v2 1
 1 2  2
c
γ
1/ 2
 0.800c
6.
A neutron lives 900 s when at rest relative to an observer. How fast is the neutron
moving relative to an observer who measures its life span to be 2065 s?
Solution
 v2 
Using t  t 0 , where   1  2 
 c 
 v2
t
2065 s
 

 2.2944  1  2
t 0
900 s
 c
1 / 2
, we see that:



1 / 2
.
1
 c2  v2 
c2


Squaring the equation gives   
2

c2  v2
 c

c2
Cross-multiplying gives c 2  v 2  2 , and solving for speed finally gives:

2
v c 
c2
2
2

1
 c1  2
 



1/ 2


1
 c 1 
2 
 (2.2944) 
1/ 2
 0.90003c  0.900c
7.
If relativistic effects are to be less than 1%, then  must be less than 1.01. At what
relative velocity is   1.01 ?
Solution
 v2
γ  1  2
 c
8.
If relativistic effects are to be less than 3%, then  must be less than 1.03. At what
relative velocity is   1.03 ?
Solution
 v2
γ  1  2
 c
9.
(a) At what relative velocity is   1.50 ? (b) At what relative velocity is   100 ?






1 / 2
1 / 2

1
 v  c1  2
 

1
 v  c1  2
 






1/ 2
1/ 2

1 
 c 1 
2 
 (1.01) 

1 
 c 1 
2 
 (1.03) 
1/ 2
 0.140c
1/ 2
 0.240c
OpenStax College Physics
Solution
Instructor Solutions Manual
 v2 
(a) γ  1  2 
 c 
1 / 2

1 
(b) v  1 
2 
 (100) 

1
 v  c1  2
 



1/ 2
Chapter 28

1 
 1 
2 
 (1.50) 
1/ 2
 0.745c
1/ 2
 0.99995c (to five digits to show difference from c.)
10.
(a) At what relative velocity is   2.00 ? (b) At what relative velocity is   10.0 ?
Solution
 v2 
(a) γ  1  2 
 c 
1 / 2

1 
(b) v  1 
2 
 (10.0) 
11.
1/ 2

1 
 v  c1  2 
  

1 
 1 
2
 (2.00) 
1/ 2
 0.866c
1/ 2
 0.995c
Unreasonable Results (a) Find the value of  for the following situation. An Earthbound observer measures 23.9 h to have passed while signals from a high-velocity
space probe indicate that 24.0 h have passed on board. (b) What is unreasonable
about this result? (c) Which assumptions are unreasonable or inconsistent?
Solution (a) Using the equation t  t 0 , we can solve for  :
t
23.9 h
 

 0.9958  0.996
t 0 24.0 h
(b)  cannot be less than 1.
(c) The earthbound observer must measure a longer time in the observer’s rest frame
than the ship bound observer. The assumption that time is longer in the moving
ship is unreasonable.
28.3 LENGTH CONTRACTION
12.
A spaceship, 200 m long as seen on board, moves by the Earth at 0.970c . What is its
length as measured by an Earth-bound observer?
Solution
 (0.970c) 2 
200 m
1 



L


200
m
2

v 2 1/ 2
c


(1  2 )
c
13.
How fast would a 6.0 m-long sports car have to be going past you in order for it to
appear only 5.5 m long?
L0
1/ 2
 48.6 m
OpenStax College Physics
Solution
L
L0
 

 v2
γ  1  2
 c
14.
Instructor Solutions Manual



Chapter 28
L0 6.0 m

 1.0909, so that
L 5. 5 m
1 / 2

1
 v  c1  2
 



1/ 2


1
 c 1 
2 
 (1.0909) 
1/ 2
 0.400c
(a) How far does the muon in Example 28.1 travel according to the Earth-bound
observer? (b) How far does it travel as viewed by an observer moving with it? Base
your calculation on its velocity relative to the Earth and the time it lives (proper time).
(c) Verify that these two distances are related through length contraction γ = 3.20 .
Solution Using the values given in Example 28.1:
(a) L0  vt  (0.950c)( 4.87  10 6 s)
 (0.950c)( 2.998  108 m/s )( 4.87  10 6 s)  1.386  10 3 m  1.39 km
(b) L0  vt  (0.950c)( 2.998  10 8 m/s )(1.52  10 6 s)  4.329  10 2 m  0.433 km
(c) L 
15.
L0 1.386  10 3 m

 4.33  10 2 m  0.433 km
γ
3.20
(a) How long would the muon in Example 28.1 have lived as observed on the Earth if
its velocity was 0.0500c ? (b) How far would it have traveled as observed on the
Earth? (c) What distance is this in the muon’s frame?
Solution All answers are reported to 4 significant figures to show the differences:
1 / 2
1 / 2
 v2 
 (0.0500c) 2 
(a) γ  1  2 
 1 
 1.00125

c2
 c 


Δt  γΔt 0  (1.00125)(1.52  10 6 s)  1.5219  10 6 s  1.522  10 6 s
(b) L0  vt  (0.0500c)t  (0.0500)( 2.998  10 8 m/s )(1.5219  10 6 s)  22.81 m
(c) L  vt 0  (0.0500)( 2.998  10 8 m/s )(1.52  10 6 s)  22.78 m
16.
(a) How long does it take the astronaut in Example 28.2 to travel 4.30 ly at 0.99944c
(as measured by the Earth-bound observer)? (b) How long does it take according to the
astronaut? (c) Verify that these two times are related through time dilation with
γ = 30.00 as given.
Solution
(a) Δt 
L0
4.30 ly
4.30 y


 4.303 y , to four digits to show any effect.
v
0.9994c 0.9994
L
4.30 ly
L 0.1433 ly
(b) L  0 
 0.1433 ly, so that Δt 0  
 0.1434 y
γ
30.0
v 0.99944c
Δt
4.303 y
(c) Δt  γΔt 0  γ 

 30.0
Δt 0 0.1434 y
OpenStax College Physics
17.
Instructor Solutions Manual
Chapter 28
(a) How fast would an athlete need to be running for a 100- m race to look 100 yd
long? (b) Is the answer consistent with the fact that relativistic effects are difficult to
observe in ordinary circumstances? Explain.
Solution (a) (100 m)(1.0937 yd/m )  109.37 yd , and since
L
L
109.37 yd
L 0 γ 0 
 1.0937
γ
L
100 yd
1 / 2
1/ 2
1/ 2


 v2 

1 
1
γ  1  2 
 v  c1  2   c 1 
 0.405c.
2 
  
 c 
 (1.0937) 
This is definitely a world record!
(b) Yes, the foot racer would need to travel at extraordinary speeds in order to
observe a length contraction of this size; therefore it is difficult to observe
relativistic effects in ordinary circumstances.
18.
Unreasonable Results (a) Find the value of  for the following situation. An
astronaut measures the length of her spaceship to be 25.0 m, while an Earth-bound
observer measures it to be 100 m. (b) What is unreasonable about this result? (c)
Which assumptions are unreasonable or inconsistent?
Solution
(a) γ 
L0 25.0 m

 0.250
L
100 m
(b)  must be  1
(c) The earthbound observer must measure a shorter length, so it is unreasonable to
assume a longer length.
19.
Unreasonable Results A spaceship is heading directly toward the Earth at a velocity of
0.800c . The astronaut on board claims that he can send a canister toward the Earth
at 1.20c relative to the Earth. (a) Calculate the velocity the canister must have relative
to the spaceship. (b) What is unreasonable about this result? (c) Which assumptions
are unreasonable or inconsistent?
Solution
(a) u 
u 
v  u

; u  1.20c; v  0.800c . Using 1  v 2 u  v  u  , solving for u :
2
c 
1  (vu  / c )

vu
uv
 vu 
u  2 u  v  u so that u  v  u 1  2  and u  
 10.0c
c
1  (uv/c 2 )
 c 
(b) The resulting speed of the canister is greater than c, an impossibility.
(c) It is unreasonable to assume that the canister will move toward the earth at 1.20c.
OpenStax College Physics
Instructor Solutions Manual
Chapter 28
28.4 RELATIVISTIC ADDITION OF VELOCITIES
20.
Suppose a spaceship heading straight towards the Earth at 0.750c can shoot a
canister at 0.500c relative to the ship. (a) What is the velocity of the canister relative
to the Earth, if it is shot directly at the Earth? (b) If it is shot directly away from the
Earth?
Solution
(a) u 
21.
Repeat the previous problem with the ship heading directly away from the Earth.
Solution
(a) u 
22.
If a spaceship is approaching the Earth at 0.100c and a message capsule is sent
toward it at 0.100c relative to the Earth, what is the speed of the capsule relative to
the ship?
Solution
Using the equation u 
v  u
0.750c  0.500c

 0.909c
2
1  (vu  / c ) 1  (0.750c)(0.500c) / c 2
0.750c  0.500c
(b) u 
 0.400c
1  (0.750c)( 0.500c) / c 2 


v  u
 0.750c  0.500c

  0.400c
2
1  (vu  / c ) 1  (0.750c)(0.500c) / c 2
 0.750c  0.500c
(b) u 
  0.909c
1  (0.750c)( 0.500c) / c 2 


v  u
, we can add the relativistic velocities:
1  (vu / c 2 )
v  u
0.100c  0.100c
u

 0.198c
2
1  (vu  / c ) 1  (0.100c)(0.100c) / c 2


23.
(a) Suppose the speed of light were only 3000 m/s . A jet fighter moving toward a
target on the ground at 800 m/s shoots bullets, each having a muzzle velocity of
1000 m/s . What are the bullets’ velocity relative to the target? (b) If the speed of light
was this small, would you observe relativistic effects in everyday life? Discuss.
Solution
(a) u 
24.
If a galaxy moving away from the Earth has a speed of 1000 km/s and emits 656 nm
light characteristic of hydrogen (the most common element in the universe). (a) What
wavelength would we observe on the Earth? (b) What type of electromagnetic
radiation is this? (c) Why is the speed of the Earth in its orbit negligible here?
v  u
800 m/s  1000 m/s

 1653 m/s  1.65 km/s
2
1  (vu /c ) 1  (800 m/s)(1000 m/s)/(3000 m/s) 2
(b) Yes, if the speed of light were this small, speeds that we can achieve in everyday
life would be larger than 1% of the speed of light and we could observe relativistic
effects much more often.


OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 28
1  u/c
1  (106 m/s)/(3 108 m/s)
 (656 nm)
 658 nm
1  u/c
1  (106 m/s)/(3 108 m/s)
(b) The radiation is visible red light.
2  r 2  (1.496  1011 m)
(c) v 

 2.97  10 4 m/s , so
7
(a) λobs  λs
t
3.16  10 s
4
v
2.97  10 m/s

 9.92  10 5 , which is negligible.
8
c 2.998  10 m/s
25.
Solution
A space probe speeding towards the nearest star moves at 0.250c and sends radio
information at a broadcast frequency of 1.00 GHz. What frequency is received on the
Earth?
1  u/c
1  0.250
 (1.00 Ghz)
 0.7746 Ghz  775 Mhz
1  u/c
1  0.250
c  f , so f obs  f s
26.
If two spaceships are heading directly towards each other at 0.800c , at what speed
must a canister be shot from the first ship to approach the other at 0.999c as seen by
the second ship?
Solution
v  u
, we get:
1  (vu /c 2 )
uv
uv
0.999c  0.800c
u  u 2  v  u  u 

 0.991c
2
c
1  (uv/c ) 1  [(0.999c)(0.800c)/c 2 ]
27.
Two planets are on a collision course, heading directly towards each other at 0.250c .
A spaceship sent from one planet approaches the second at 0.750c as seen by the
second planet. What is the velocity of the ship relative to the first planet?
Solution
v  0.250c, u  0.750c . Find u .
v  u
uv
0.750c  0.250c
u
 u 

 0.615c
2
2
1  (vu /c )
1  (uv/c ) 1  [(0.750c)(0.250c)/c 2 ]
28.
When a missile is shot from one spaceship towards another, it leaves the first at
0.950c and approaches the other at 0.750c . What is the relative velocity of the two
ships?
Solution
u '  0.950c, u  0.750c . Starting with u 
v  0.800c, u  0.999c . Find u . Starting with u 
uu 
v  u
, u  v 2  v  u ' and
2
c
1  (vu / c )
u  u
0.950c  0.750c

  0.696c .
2
(uu  / c )  1 (0.750c)(0.950c) / c 2  1
The velocity v is the speed measured by the second spaceship, so the minus sign
indicates the ships are moving apart from each other ( v is in the opposite direction
v


OpenStax College Physics
Instructor Solutions Manual
Chapter 28
as u ).
29.
What is the relative velocity of two spaceships if one fires a missile at the other at
0.750c and the other observes it to approach at 0.950c ?
Solution
u '  0.750c, u  0.950c . Find v .
v  u
u  u
0.750c  0.950c
u
v

 0.696c
2
2
1  (vu  / c )
(uu  / c )  1 (0.750c)(0.950c) / c 2   1
The velocity v is the speed measured by the second spaceship, so the ships are
moving towards each other ( v is in the same direction as u ).
30.
Near the center of our galaxy, hydrogen gas is moving directly away from us in its
orbit about a black hole. We receive 1900 nm electromagnetic radiation and know
that it was 1875 nm when emitted by the hydrogen gas. What is the speed of the
gas?
Solution
λobs
1  u/c 1900 nm


 1.0133. So
λs
1  u/c 1875 nm
(1.0133) 2  1
1  u / c  (1.0133) (1  u / c)  u / c 
 1.324  10  2  u  0.01324c
2
(1.0133)  1
2
31.
Solution
A highway patrol officer uses a device that measures the speed of vehicles by
bouncing radar off them and measuring the Doppler shift. The outgoing radar has a
frequency of 100 GHz and the returning echo has a frequency 15.0 kHz higher. What
is the velocity of the vehicle? Note that there are two Doppler shifts in echoes. Be
certain not to round off until the end of the problem, because the effect is small.
f obs
1  u/c
1  u/c

. At the vehicle, f obs  f s
. For the echo returning to the
fs
1  u/c
1  u/c
1  u/c
where the velocities of sources
1  u/c
are positive in both cases because the source is always moving away from the
f
f f
1  u/c 1  u/c 1  u/c

.
observer. Since f s  f obs , then obs  obs obs 
fs
f s f s
1  u/c 1  u/c 1  u/c
  f s
officer, we have a similar equation: f obs
Therefore,
 /f s

1  f obs
f s  f obs
 1.50 10 4 Hz
8
uc
c
 (2.998 10 m/s)
  22.5 m/s
 /f s

1  f obs
f s  f obs
2.00 1011 Hz  1.50 10 4 Hz
The vehicle is therefore travelling at a speed of 22.5 m/s toward the highway patrol
officer.
OpenStax College Physics
Instructor Solutions Manual
Chapter 28
32.
Prove that for any relative velocity v between two observers, a beam of light sent
from one to the other will approach at speed c (provided that v is less than c , of
course).
Solution
u  c , so u 
33.
Show that for any relative velocity v between two observers, a beam of light
projected by one directly away from the other will move away at the speed of light
(provided that v is less than c , of course).
Solution
u   c , so u 
v  u
vc
vc
c (v  c )



c
2
2
cv
1  (vu  / c ) 1  (vc / c ) 1  (v / c)
v  u
vc
vc
c (v  c )



  c.
2
2
cv
1  (vu  / c ) 1  (vc / c ) 1  (v / c)
Similarly u  c  u  c
34.
(a) All but the closest galaxies are receding from our own Milky Way Galaxy. If a galaxy
12.0 109 ly away is receding from us at 0.900c , at what velocity relative to us must we
send an exploratory probe to approach the other galaxy at 0.990c , as measured from
that galaxy? (b) How long will it take the probe to reach the other galaxy as measured
from the Earth? You may assume that the velocity of the other galaxy remains constant.
(c) How long will it then take for a radio signal to be beamed back? (All of this is possible in
principle, but not practical.)
Solution
Note that all answers to this problem are reported to 5 significant figures, to
distinguish the results.
v  u
uv
(a) u 
, so u  u  2  v  u  and
2
1  (vu  / c )
c
uv
0.990c  (0.900c)
u 

 0.99947c
2
1  (uv / c ) 1  [(0.990c)( 0.900c) / c 2 ]
(b) When the probe reaches the other galaxy, it will have traveled a distance (as seen
on Earth) of d  x0  vt because the galaxy is moving away from us. As seen from
d x  vt
Earth, u  0.9995c , so t   0
.
u
u
x
12.0  109 ly
(1 y)c
Now, ut  x0  vt and t  0 

 1.2064  1011 y
u  v 0.99947c  0.900c 1 ly
(c) The radio signal travels at the speed of light, so the return time t is given by
d x  vt
t   0
, assuming the signal is transmitted as soon as the probe reaches
c
c
the other galaxy. Using the numbers, we can determine the time:
1.20 1010 ly  (0.900c)(1 .2064 1011 y)
t 
 1.2058 1011 y
c
OpenStax College Physics
Instructor Solutions Manual
Chapter 28
28.5 RELATIVISTIC MOMENTUM
35.
Find the momentum of a helium nucleus having a mass of 6.68  1027 kg that is
moving at 0.200c .
Solution
p  mu 

mu
1  u
2
/ c2

1
2
(6.68  10 27 kg)(0.200) (3.00  108 m/s)
1  (0.200c)
2
/c
2

1
 4.09  10 19 kg  m/s
2
36.
What is the momentum of an electron traveling at 0.980c ?
Solution
p  mu 

mu
1  u
2

/ c2 
1
2
(9.1110 31 kg)(0.980) (3.00 108 m/s)
1  (0.980c)
2
/c
2

1
 1.35 10 21 kg  m/s
2
37.
(a) Find the momentum of a 1.00  10 9 kg asteroid heading towards the Earth at
30.0 km/s . (b) Find the ratio of this momentum to the classical momentum. (Hint: Use
the approximation that   1  (1/ 2)v 2 / c 2 at low velocities.)
Solution
(a) p  mu 
mu
1  u
2
/ c2

1

2
(1.00 10 9 kg)( 30.0 10 3 m/s)
1  (3.00 10
4

m/s) 2 / 3.00 108 m/s
 
2
1
2
 3.000000015 1013 kg  m/s


9
4
13
(b) pc  mu  1.00 10 kg (3.00 10 m/s)  3.00 10 kg  m/s, so
p
 1.000000005
pc
38.
(a) What is the momentum of a 2000 kg satellite orbiting at 4.00 km/s? (b) Find the
ratio of this momentum to the classical momentum. (Hint: Use the approximation that
  1  (1 / 2)v 2 / c 2 at low velocities.)
Solution
(a) For low velocities we can use the approximation   1 
1 u2
2 c2
OpenStax College Physics
Instructor Solutions Manual
Chapter 28
 1 u2 
mu
p  mu  1 
2 
 2c 

1 u3
(0.5)2000 kg  4.00 10 3 m/s
 mu  m 2  (2000 kg) 4.00 10 3 m/s 
2
2 c
3.00 108 m/s





3
 8.00 10 6 kg  m/s  7.1110 4 kg  m/s  8.00 10 6 kg  m/s
(b) The classical momentum, mu , is just the first term above, pc  8.00 10 6 kg  m/s
p 8.00 10 6 kg.m/s  7.11110 4 kg  m/s

pc
8.00 10 6 kg  m/s
 1
7.1110 4 kg  m/s
 1  8.89 10 11
6
8.00 10 kg  m/s
39.
What is the velocity of an electron that has a momentum of 3.04 10 21 kg  m/s ? Note
that you must calculate the velocity to at least four digits to see the difference from c.
Solution
Beginning with the equation p  mu 
mu
1  u
2
c
2

1
First cross-multiply and square both sides, giving 1 
2
Then, solving for u gives u 2 
m
1
2
 
/ p 2  1/ c 2
Finally, taking the square root gives u 

p

we can solve for the speed u .
2
u 2 m2 2
 2u
c2
p
p2
m2  p 2 / c2


.
m  p2 / c2
Taking the values for the mass of the electron and the speed of light to five significant
figures gives:
3.04 10 21 kg  m/s
u
1/2
9.1094 10 31 kg 2  3.34 10 21 kg  m/s / 2.9979 108 kg 2 2 




8
 2.988 10 m/s

40.
2



Find the velocity of a proton that has a momentum of 4.48  1019 kg  m/s.
OpenStax College Physics
Solution
Instructor Solutions Manual
mu
p  mu 
2
 u2
u
2
2
1.67  10
 27
2

2

2
2
kg
 
 p 2 1  u 2 / c 2  m 2u 2
1  u / c 
m   p / c   p
1
Chapter 28

2
u
p
m   p
2
2
/ c2

1
2
(4.48  10 19 kg  m/s)

 (4.48  10 19 kg  m/s) 2 / 3.00  10 8 m/s
 
2
1
2
 1.9997  10 8 m/s  2.00  10 8 m/s
41.
(a) Calculate the speed of a 1.00 - g particle of dust that has the same momentum as
a proton moving at 0.999c . (b) What does the small speed tell us about the mass of a
proton compared to even a tiny amount of macroscopic matter?
Solution (a) For the proton:
mu
p
1 u2 / c2
(1.673  10 27 kg)(0.999) (2.998  10 8 m/s)

 1.121 10 17 kg  m/s
2
1  (0.999)
So, for the particle:
u
pm
1   p / mc
2
p 1.12110-17 kg  m/s

 1.12110 8 m/s. Thus,
9
m
1.00 10 kg
1.121  10 8 m/s


1  1.121  10 8 m/s 2.998  10 8 m/s

2
 1.12  10 8 m/s
(b) The small speed tells us that the mass of a proton is substantially smaller than that
of even a tiny amount of macroscopic matter!
42.
(a) Calculate  for a proton that has a momentum of 1.00 kg  m/s. (b) What is its
speed? Such protons form a rare component of cosmic radiation with uncertain
origins.
Solution
(a) p  mu  u 
 p
Now, u   
m
2
u

1 u 2 / c2
2
 u2 
1  2 
 c 


p
1.00 kg  m/s

 5.977 10 26 m/s.
 27
m 1.673 10 kg
OpenStax College Physics
u2 
u
Instructor Solutions Manual
Chapter 28
 pc / m 2 so
c 2   p / m2 
p/m
1   p / mc
2

(5.977  10 26 m/s)

1  5.977  10 26 m/s 2.998  10 8 m/s

2
 2.998  10 8 m/s
p
5.997  10 26 m/s

 1.99  1018
8
mu 2.998  10 m/s
(b) As found in part (a), u  2.998  108 m/s  3.00  108 m/s  c
Finally,  
28.6 RELATIVISTIC ENERGY
43.
What is the rest energy of an electron, given its mass is 9.1110 31 kg ? Give your
answer in joules and MeV.
Solution
E0  mc 2  9.11  10 31 kg 3.00  10 8 m/s




2

 1 MeV 
 8.20  10 14 J  8.20  10 14 J 
  0.512 MeV
13
 1.60  10 J 
(according to the number of sig. fig. stated. The exact value is closer to 0.511 MeV.)
44.
Find the rest energy in joules and MeV of a proton, given its mass is 1.67 10 27 kg .
Solution
E0  mc 2  1.67  10 27 kg 3.00  10 8 m/s




2
 1.503  10 10 J

 1 MeV 
 1.503  10 10 J 
  939 MeV
13
 1.60  10 J 
45.
Solution
If the rest energies of a proton and a neutron (the two constituents of nuclei) are
938.3 and 939.6 MeV respectively, what is the difference in their masses in kilograms?
E 0  mc 2  E 0  mc 2 
m 


E 939.6 MeV  938.3 MeV  1.60  10 13 J/MeV

 2.3  10 30 kg
2
2
8
c
3.00  10 m/s


To two digits since the difference in rest mass energies is found to two digits.
46.
68
The Big Bang that began the universe is estimated to have released 10 J of energy.
How many stars could half this energy create, assuming the average star’s mass is
4.00 1030 kg ?
OpenStax College Physics
Solution
47.
Solution
Instructor Solutions Manual
Chapter 28
E  0.5001.00 10 68 J  

 5.56 10 50 kg
c 2  3.00 108 m/s 2 
M 5.56 10 50 kg


 1.39 10 20
30
m 4.00 10 kg
E  Mc 2  M 
A supernova explosion of a 2.00 1031 kg star produces 1.00 10 44 kg of energy. (a)
How many kilograms of mass are converted to energy in the explosion? (b) What is
the ratio m/ m of mass destroyed to the original mass of the star?
(a) E  mc 2  m 

E
1.00  10 44 J

c2
3.00  108 m/s



2
 1.111  10 27 kg  1.11 10 27 kg
m 1.111  10 kg

 5.56  10 5  5.56  10 5
31
m
2.00  10 kg
27
(b)
48.
(a) Using data from Table 7.1, calculate the mass converted to energy by the fission of
1.00 kg of uranium. (b) What is the ratio of mass destroyed to the original mass,
m / m ?
Solution (a) From Table 7.1, the energy released from the nuclear fission of 1.00 kg of uranium
is E  8.0  1013 J . So, E 0  E released  mc 2 , we get


E
8.0  1013 J

 8.89  10 4 kg  0.89 g
2
2
8
c
3.00  10 m/s
(b) To calculate the ratio, simply divide by the original mass:
m 8.89  10 4 kg 

 8.89  10  4  8.9  10  4
2
m
1.00 kg 
 m 
49.
Solution


(a) Using data from Table 7.1, calculate the amount of mass converted to energy by
the fusion of 1.00 kg of hydrogen. (b) What is the ratio of mass destroyed to the
original mass, m / m ? (c) How does this compare with m / m for the fission of 1.00
kg of uranium?
(a) E  mc2  m 


E
6.4 1014 J

c2
3.00 108 m/s


2
 7.1103 kg
m 7.11  10 -3 kg

 7.1  10 3  7.1  10 3
m
1 kg
m
(c)
is greater for hydrogen.
m
(b)
50.
34
There is approximately 10 J of energy available from fusion of hydrogen in the
OpenStax College Physics
Instructor Solutions Manual
Chapter 28
33
world’s oceans. (a) If 10 J of this energy were utilized, what would be the decrease
in mass of the oceans? (b) How great a volume of water does this correspond to? (c)
Comment on whether this is a significant fraction of the total mass of the oceans.
Solution


E
11033 J
(a) E  mc  m  2 
c
3.00 108 m/s
2


2
 11016 kg
 1m 
  1  1013 m 3
(b) V  1.111  10 16 kg 
 1000 kg 
(c) Since approximately 2/3 of the surface of the earth is covered with water, the
surface area of water on Earth is:
2
2
8
Asurface   4r 2  π6.376  10 6 m   3.41  1014 m 2 .
3
3
So the volume of water found in part (b) would be equivalent to a thickness of
V
1  1013 m 3
water of: t  
 2.9  10  2 m  2.9 cm.
14
3
A 3.41  10 m
Therefore, no, this mass is not a significant fraction of the total mass of the
oceans.


3
51.
A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a
massless particle. (a) If all the lost mass is converted into the electron’s kinetic energy,
find  for the electron. (b) What is the electron’s velocity?
Solution
2
2
(a) m  c  (  1)me c  m   (  1)me  me  me
m   me 105.7 MeV  0.511 MeV
 

 207.85  208
me
0.511 MeV

(b)   1  (v / c )
2
2

1 / 2
1/ 2

1 
 v  c1  2 
  


1
 c 1 
2
 207.85 
1/ 2
 0.999988c
(six digits used to show difference from c.)
52.
A  -meson is a particle that decays into a muon and a massless particle. The  meson has a rest mass energy of 139.6 MeV, and the muon has a rest mass energy of
105.7 MeV. Suppose the  -meson is at rest and all of the missing mass goes into the
muon’s kinetic energy. How fast will the muon move?
Solution
KErel  mc 2  (m  m )c 2  (  1)m c 2
Solving for  will give us a way of calculating the speed of the muon. From the
equation above, we see that:
OpenStax College Physics
 
Instructor Solutions Manual
m  m
m
m  m  m
1 
m
Now, use   1  v 2 c 2 
1
v  c 1
53.
1

2
2
or 1 
Chapter 28
m 139.6 MeV

 1.32072.
m 105.7 MeV

v2
1
 2 , so
2
c

1
 0.6532c
(1.32072) 2
 c 1
(a) Calculate the relativistic kinetic energy of a 1000-kg car moving at 30.0 m/s if the
speed of light were only 45.0 m/s. (b) Find the ratio of the relativistic kinetic energy to
classical.
Solution
(a)
  1  v c
2
2

1
2
 30.0 m/s 2 
 1 
2
 45.0 m/s  
1/2
 1.342;
KE    1mc 2  0.3421000 kg 45.0 m/s   6.92  10 5 J
2
(b) KE c 
1
KE
mv 2  (0.5)(1000 kg)(30.0 m/s) 2  4.50  10 5 J;
 1.54
2
KE c
54.
Alpha decay is nuclear decay in which a helium nucleus is emitted. If the helium
nucleus has a mass of 6.80 1027 kg and is given 5.00 MeV of kinetic energy, what is
its velocity?
Solution
KE    1mc 2  mc 2  KE  mc 2   
KE  mc 2
mc 2
2

5.00 MeV 1.60 10 13 J/MeV   6.80  10 27 kg 3.00  108 m/s 

6.80 10
  1  v c
2
2

1
 27

kg 3.00 108 m/s
1/ 2
2

1 
 v  c1  2 
  

2


1
 c 1 
2
 1.0013 
 1.0013;
1/ 2
 0.0511c
55.
(a) Beta decay is nuclear decay in which an electron is emitted. If the electron is given
0.750 MeV of kinetic energy, what is its velocity? (b) Comment on how the high
velocity is consistent with the kinetic energy as it compares to the rest mass energy of
the electron.
Solution
KE  mc 2
(a) KE    1mc   
mc 2
2

0.750 MeV 1.60  10 13 J/MeV  

9.11 10
 2.4636;
31

9.11 10
31
kg 3.00  10 8 m/s

kg 3.00  10 8 m/s

2

2
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Chapter 28
1/ 2



1 
1
  1 v c
 v  c1  2   c 1 
 0.914c
2


2
.
4636
  


(b) The rest mass energy of an electron is 0.511 MeV, so the kinetic energy is
approximately 150% of the rest mass energy. The electron should be traveling
close to the speed of light.
 
56.
2
2

1
1/ 2
2
A positron is an antimatter version of the electron, having exactly the same mass.
When a positron and an electron meet, they annihilate, converting all of their mass
into energy. (a) Find the energy released, assuming negligible kinetic energy before
the annihilation. (b) If this energy is given to a proton in the form of kinetic energy,
what is its velocity? (c) If this energy is given to another electron in the form of kinetic
energy, what is its velocity?
Solution (a) E  mc 2  2me c 2  (2)(9.11  10 31 kg)(3.00  10 8 m/s) 2  1.64  10 13 J
KE  mc 2
mc 2
(b) KE    1mc 2   
1.64 10 J   1.67 10 kg 3.00 10

1.67 10 kg 3.00 10 m/s 
13
- 27
-27
  1  v c
2
2

1

1 
 v  c1  2 
  
27
 27

1 
 v  c 1 
2
 3.00  
57.
8

2
 1.00109;


1
 c 1 
2
 2.4636  
1.64 10 J  1.67 10 kg 3.00 10
(c)  
1.67 10 kg 3.00 10 m/s 
13
m/s
2
8
1/ 2
2
8
8
m/s

2
1/ 2
 0.0467c
2
 3.00
1/ 2
 0.943c
16
What is the kinetic energy in MeV of a -meson that lives 1.40 10 s as measured in
16
the laboratory, and 0.840 10 s when at rest relative to an observer, given that its
rest energy is 135 MeV?
Solution
  1.667 from the result of Exercise 28.5. Thus,
KE    1mc 2  (1.667  1)(135 MeV)  90.0 MeV
58.
Find the kinetic energy in MeV of a neutron with a measured life span of 2065 s, given
its rest energy is 939.6 MeV, and rest life span is 900s.
Solution From Exercise 28.6, we know that  = 2.2944, so we can determine the kinetic energy
of the neutron: KE rel  (  1)mc 2  2.2944  1939.6 MeV   1216 MeV.
59.
(a) Show that ( pc) 2 /( mc2 ) 2   2  1 . This means that at large velocities pc  mc2 . (b)
OpenStax College Physics
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Chapter 28
Is E  pc when   30.0 , as for the astronaut discussed in the twin paradox?
Solution
(a) E  p c  m c   m c , so that p c    1m c and
2
2 2
2 4
2
2 4
2 2
(b) Yes:   30.0   2  1  899 
2
2 4
2

pc 

mc 
2 2
  2 1
pc
 29.98  E  pc , to within 1 part in
mc 2


 1  mc 2 
mc 2
1 1 
4

E

pc
1

1  

pc
,
Since

5
.
56

10


2  899 
 pc 2
 2  pc 
2
2



60.
One cosmic ray neutron has a velocity of 0.250c relative to the Earth. (a) What is the
neutron’s total energy in MeV? (b) Find its momentum. (c) Is E  pc in this situation?
Discuss in terms of the equation given in part (a) of the previous problem.
Solution
v  0.250c
(a) E  mc2 , where  
E
1
1  v
2
c2

so E 
1
1  0.250 
2
(939 MeV)  970 MeV
(0.250) E
 243 MeV/ c
c
c
(c) No. This is true when pc  mc 2 which is not the case for this situation. Here
pc  mc 2 .
(b) p  mc 

61.
What is  for a proton having a mass energy of 938.3 MeV accelerated through an
effective potential of 1.0 TV (teravolt) at Fermilab outside Chicago?
Solution
KE  qV  (  1)mc 2 so that
qV  mc 2
mc 2
1.60  10 19 C 1.00  1012 V 1 1.60  10 13 J/MeV  938.3 MeV 

 1.07  10 3
938.3 MeV 
γ

62.

 

(a) What is the effective accelerating potential for electrons at the Stanford Linear
Accelerator, if   1.00 105 for them? (b) What is their total energy (nearly the same
as kinetic in this case) in GeV?
Solution (a) KE  (  1)mc 2  1.00  10 5  19.11  10 31 kg 3.00  10 8 m/s 2  8.199  10 9 J
KE 8.199  10 9 J

 5.12  1010 V  51.2GV
q
1.60  10 19 C
 1 GeV 
(b) E  mc 2  1.00  10 5 0.511 MeV 
  51.1 GeV
 1000 MeV 
KE  qV  V 
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Chapter 28
63.
(a) Using data from Table 7.1, find the mass destroyed when the energy in a barrel of
crude oil is released. (b) Given these barrels contain 200 liters and assuming the
density of crude oil is 750 kg/m 3 , what is the ratio of mass destroyed to original mass,
m / m ?
Solution
E
5.90 109 J
(a) E  mc  m  2 
c
3.00 108 m/s
2




2
 6.56 108 kg
(b) m  200 L1 m3 /1000 L750 kg/m 3   150 kg.
Therefore,
64.
Δm
 4.37 10 10
m
(a) Calculate the energy released by the destruction of 1.00 kg of mass. (b) How
many kilograms could be lifted to a 10.0 km height by this amount of energy?
Solution (a) E released  mc 2  1.00 kg 2.998  10 8 m/s   8.988  1016 J  8.99  1016 J
(b) m 
65.
PE
8.99  1016 J

 9.17  1011 kg
2
3
gh
9.80 m/s 10.0  10 m



A Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate
charged particles such as protons. (a) What is the velocity of a proton accelerated
by such a potential? (b) An electron?
Solution (a) KE  qV  (  1)mc 2
qV  mc 2
mc 2
1.60  10 19 C 5.00  10 7 V 1 MeV 1.60  10 13 J/MeV  938.3 MeV 

938.3 MeV 
 1.0532
γ


  1  v c
2
2

1

1/ 2
2



1 
 v  c1  2 
  


1
 c 1 
2
 1.0532  
1/ 2
 0.314c
OpenStax College Physics
(b)  
1.60 10
Instructor Solutions Manual
19


Chapter 28


C 5.00  10 7 V 1 MeV 1.60  10 13 J/MeV  0.511 MeV 
0.511 MeV 
 98.85


1
v  c 1 
2
 98.85 
1/ 2
 0.99995c
(Five digits used to show difference from c .)
66.
Suppose you use an average of 500 kW  h of electric energy per month in your home.
(a) How long would 1.00 g of mass converted to electric energy with an efficiency of
38.0% last you? (b) How many homes could be supplied at the 500 kW  h per month
rate for one year by the energy from the described mass conversion?
Solution (a) The monthly energy use is 500 kW  h 3600 s/hr   1.80 109 J
The energy available is 0.381.00 10-3 kg 3.00 108 m/s   3.42 1013 J.
2
Thus,
3.42  1013 J
 1.9  104 month  1.58  103 years
1.8  109 J/month
(b) 1.58  103 homes
67.
(a) A nuclear power plant converts energy from nuclear fission into electricity with an
efficiency of 35.0%. How much mass is destroyed in one year to produce a continuous
1000 MW of electric power? (b) Do you think it would be possible to observe this mass
loss if the total mass of the fuel is 104 kg ?
Solution (a)
 3600s  24h  365.24d 
  3.156 1016 J  0.350mc 2
E  1000 10 6 J/s 


 1h  1d  1y 


E
3.156 1016 J

 1.00 kg
0.350c 2 0.350 3.00 108 m/s 2
(b) This much mass would be measurable, but probably not observable just by looking
because it is 0.01% of the total mass.
m
68.


Nuclear-powered rockets were researched for some years before safety concerns
became paramount. (a) What fraction of a rocket’s mass would have to be destroyed
to get it into a low Earth orbit, neglecting the decrease in gravity? (Assume an orbital
altitude of 250 km, and calculate both the kinetic energy (classical) and the
gravitational potential energy needed.) (b) If the ship has a mass of 1.00  105 kg (100
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tons), what total yield nuclear explosion in tons of TNT is needed?
Solution (a)
r  rE  rO  6.37  10 6 m  2.50  10 4 m  6.62  10 6 m
 mv 2 
GMm
r 
KE  
2r
 2r 
6.67  10 11 N  m 2 /kg 2 5.98  10 24 kg 1.00  10 5 kg

 3.012  1012 J
2(6.62  10 6 m)








PE  mgrO  1.00  10 5 kg 9.80 m/s 2 2.50  10 5 m  2.45  1011 J;
E tot  KE  PE  3.257  10 J
12
E  mc 2  m 
Thus,
(b)
69.
Solution
E
3.257  1012 J

 3.618  10 5 kg
2
2
8
c
3.00  10 m/s


Δm 3.618  10 5 kg

 3.62  10 10
m
1.00  10 5 kg
3.257  1012 J
 775 tons of TNT
4.20  10 9 J/tonTNT
26
The Sun produces energy at a rate of 4.00 10 W by the fusion of hydrogen. (a) How
many kilograms of hydrogen undergo fusion each second? (b) If the Sun is 90.0%
hydrogen and half of this can undergo fusion before the Sun changes character, how
long could it produce energy at its current rate? (c) How many kilograms of mass is
the Sun losing per second? (d) What fraction of its mass will it have lost in the time
found in part (b)?
4.00 1026 J/s
 6.25 1011 kg/s  6.3 1011 kg/s
6.4 1014 J/kg
1.99  10 30 kg
1.43  1018 s
18
(b) 0.450
 1.43  10 s 
 4.5  1010 y
11
7
6.25  10 kg
3.1536  10 s/y
(a)
E
4.00  10 26 J
(c) E  mc  m  2 
c
3.00  108 m/s
4.44 109 kg/s 1.43 1018 s  0.32%
(d)
1.99 1030 kg
2


2
 4.44  10 9 kg
70.
Unreasonable Results A proton has a mass of 1.67 1027 kg . A physicist measures
the proton’s total energy to be 50.0 MeV. (a) What is the proton’s kinetic energy? (b)
What is unreasonable about this result? (c) Which assumptions are unreasonable or
inconsistent?
Solution
(a) E  mc2 , so that  
E
50.0 MeV

 0.0532 .
2
938.3 MeV
mc
OpenStax College Physics
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
Chapter 28

KE    1mc2   9.47  10 1 938.3 MeV    888.3 MeV
(b) The kinetic energy, KE, must be greater than or equal to 0. Here it is negative.
(c) The proton’s total energy must be  mc 2  938.3 MeV
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