Chapter 26
RELATIVITY
Problems
1. Strategy Compute the times required for the optical signal and the train to reach the station and find the
difference.
Solution The time required for the optical signal to reach the station, as measured by an observer at rest relative
to the station (the stationmaster) is
d
1.0 km

 3.3 s.
c 3.00  105 km s
Measured by the same observer, the time needed for the train to arrive is
d
d
1.0 km


 5.6 s.
v 0.60c 0.60  3.00 105 km s
The difference in the arrival times is 5.56 s  3.33 s  2.2 s .
13. (a) Strategy The observers on Earth see the contracted heights of the occupants of the
spaceship, since the heights are along the direction of motion. Use Eq. (26-4).
Solution Compute the approximate heights of the occupants as viewed by others
on the ship.
L
L
0.50 m
L  0 , so L0   L 

 2m .
2 2

1 v c
1  0.972
(b) Strategy The widths of the occupants are perpendicular to their direction of motion, so the widths are not
contracted.
Solution The others on the ship see the same widths as the observers on Earth, 0.50 m.
21. Strategy The observer on the ground measures the proper length between the towers as L0  3.0 km. The
distance is contracted for a passenger on the train. Use Eq. (26-4).
Solution
(a) Relative to the passenger on the train, the distance between the towers is
L
L  0  L0 1  v2 c2  (3.0 km) 1  0.802  1.8 km.

The time interval measured by the passenger is t 
L
1.8 km

 7.5  106 s  7.5 s .
v 0.80(3.00 105 km s)
(b) The observer on the ground measures the time interval
L
3.0 km
t  0 
 1.3  105 s  13 s .
v
0.80(3.00 105 km s)
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College Physics
Chapter 26: Relativity
31. Strategy The speed of electron B in any frame of reference in which electron A is at rest is the same as the speed
of electron B relative to electron A. The coordinate system relative to the lab is chosen so that objects traveling
west have a positive velocity. Then the velocity of electron B relative to the lab is vBL  54 c and the velocity of
electron A relative to the lab is vAL  53 c, so vLA  vAL   53 c. Use Eq. (26-5).
Solution
vBA 
 
  
4c 3c
5
5
 5
 c
2
3
4
13
1 5  5
1  vBL vLA c
vBL  vLA
The speed of electron B in a frame of reference in which electron A is at rest is
5 c
13
.
39. Strategy Since no energy is lost to the environment, the change in the kinetic energy of the two lumps can be
equated to a change in the rest energy of the system.
Solution E0  (m)c2 , so
1mv21m v 2
1 1
2 2 2
 2
.
2
2
2
c
c
c
The nonrelativistic form of the kinetic energy is used since the speeds are very small compared to c.
Using m1  m2  1.00 kg and v1  v2  30.0 m s gives
m 
m 
E0

1 (1.00
2
K1  K 2
kg)(30.0 m s) 2  12 (1.00 kg)(30.0 m s) 2
(3.00 10 m s)
8
2
 1.00 1014 kg.
The mass of the system increased by 1.00 1014 kg .
47. Strategy The object is moving at 1.80 3.00  0.600 times c. Use Eq. (26-8).
Solution Find the kinetic energy of the object.
K  (  1)mc2  [(1  v2 c2 )1 2  1]mc2  [(1  0.6002 )1 2  1](0.12 kg)(3.00 108 m s)2  2.7 1015 J
57. Strategy Use Eq. (26-11) with the condition that K  E0 .
Solution Show that ( pc)2  K 2  2KE0 reduces to K  p 2 /(2m) for nonrelativistic speeds.
( pc)2  p 2c2  K 2  2KE0  K ( K  2E0 )
If K  E0 , then K  2 E0  2 E0 . Solve for K and simplify.
p 2 c2  2KE0 , so K 
p2c2 p2c2
p2


.
2 E0
2mc2 2m
247
Chapter 26: Relativity
College Physics
73. Strategy If the collision is elastic, the total kinetic energy before and after the collision should be the same. Use
Eq. (26-8).
Solution The total kinetic energy before the collision is
Kp  Kn  ( p  1)mpc2  ( n  1)mn c2  [(1  0.702 )1 2  1]mpc2  [(1  02 c2 )1 2  1]mn c2  0.4mpc2 .
After the collision, the total kinetic energy is
Kp  Kn  ( p  1)mp c2  ( n  1)mn c2  [(1  0.632 )1 2  1]mpc2  [(1  0.1282 )1 2  1](14mp )c2  0.4mpc 2 .
The kinetic energy is conserved, so the collision is elastic.
248