Quiz 8
Problem 1
(a) What is the speed v of an electron whose kinetic energy is 14,000 times its rest
energy? You can state the answer as the difference c  v .
(b) If the electrons travel in the lab through a tube 3 km long, how long is this tube in
the electrons’ reference frame?


Solution:
(a)
1
K  (  1)mc 2  (
1 v / c
2
2
 1)mc 2  14,000mc 2
2
2
c 1 
 1 
v  c 1 
 c 
 ,
2  14,001 
 14,001 
2
c 1 
cv  
  0.77m / s.
2  14,001 
(b)
l  l0
  1 2 
1  v / c  (3  10 m) 1  1  
   0.21m.
  14,001  
2
2
3
Problem 2 (Ch. 36, # 30)
In the old West, a marshal riding on a train traveling 35 m/s sees a duel between two men
standing on the Earth 55 m apart parallel to train. The marshal’s instruments indicate that
in his reference frame the two men fired simultaneously. (a) Which of the two men, the
first one the train passes (A) or the second one (B) fired first in Earth’s reference frame?
(b) How much earlier did he fire? (c) Who was struck first?
Solution:
(a) We choose the train as frame S’ and the Earth as frame S.
t A '  t B '  0, x A '  0, x B '  55.0m
vx ' 

t A    t A ' 2A   0
c 


vx ' 
1
35m / s  55m 

 0 
  2.14  10 14 s.
t B    t B ' 2B  
8
2 
8
2
c
(
3

10
m
/
s
)


1  (35m / s / 3  10 m / s) 

Therefore, in Frame S, A fired first.
(b) The difference in time is 2.14  10 14 s.
(c) In the Earth frame of reference, since A fired first, B was stuck 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 stuck first in this frame as well.
Problem 3
Two lumps of clay, each of (rest) mass m, collide head-on at (3 5)c. They stick
together.
What is the mass (M) of the composite lump?
Solution:

Momentum equals zero before and equals zero after the collision.
The energy of each lump before the collision:
mc2

1  (3 5) 2
5 2
mc
4
Conservation of energy



5 2 5 2
5
mc  mc  Mc2  M  m
4
4
2
Problem 4 (Ch 36, # 27)
A spacecraft traveling at 0.76c away from the Earth fires a module with a speed of
0.82c at right angle to its own direction of travel as seen by the spacecraft. What is the
speed of the module and its direction of travel (relative to the spacecraft’s direction), as
seen by an observer on Earth?
[Hint: Using Lorentz
transformations obtain the equations for relativistic addition of

velocities.]
Solution:
u x ' v
ux 
 0.76c;
1  uxv / c2
uy 
uy ' 1  v2 / c2
1  uxv / c2
 0.533c;
u  u x  u y  0.93c;   tan 1
2
2
uy
ux
 tan 1
0.533c
 35.
0.76c