1. A boy played the balls on the board of the spacecraft moving toward the Earth
with velocity v = 0.6c.
(i)
In his first game he threw the balls toward the wall of the
spacecraft so that they moved by the same line with equal
velocities u = (3/4)c (in the spacecraft reference frame). The first
ball collided with the wall 50 ns earlier than the second ball (in the
spacecraft reference frame). Find the proper distance between the
balls before their collision with the wall.
(ii)
In the second game the boy put an explosive into the balls and
again threw them one after another with equal velocities u = 0.99c
so that the distance between the balls in the spacecraft reference
frame was l= 120 m. At some moment the balls exploded
simultaneously in their reference frame. What was the time interval
between the explosions of the balls in the spacecraft reference
frame?
(iii) During the explosion the balls emitted the light of the wavelength
 ' 500nm (in the spacecraft reference frame) what is the
wavelength of the light observed on the Earth.
(iv)
The boy took another two balls with equal rest mass m and threw
one of them so that its kinetic energy was T. Find the momentum
of the ball.
(v)
The ball from the item (iv) collided with the second ball which was
initially at rest (in the spacecraft reference frame). Find the rest
mass and velocity of the balls (moving together) after collision.
Solution: (i) The distance in the spacecraft r. f. is x  ut , hence
l 0  x / 1  (v / c) 2  ut / 1  (v / c) 2  17m
(ii)
t 
t '(v / c 2 )x'
1 
2
, hencefort '  0, t  (v / c 2 )x' / 1   2 , butx'  l / 1   2 ,
hencet  (v / c 2 )l /(1   2 )  20s
(iii)    '
1 v / c
  ' / 2  250nm
1 v / c
(iv)

E  T  mc 2  m 2 c 4   pc    pc   T  mc 2
2
2
  mc 
2
2 2
 T (T  2mc 2 ) 
p  T (T  2mc 2 ) / c
(v) E  T  2mc 2 , E 2  p 2 c 2  M 2 c 4 , where M is the rest mass of the two
balls.  pc   T (T  2mc 2 ) hence
M 2 c 4  (T  2mc 2 ) 2  T (T  2mc 2 )  (T  2mc 2 )2mc 2 
2
M 
(T  2mc 2 )2m
c
Momentum of the new particle is equal to the momentum of the first particle:
p  T (T  2mc 2 ) / c 
V  pc 2 / E 
E
V
c2
T
c
T  2mc 2
2. As a result of the Compton scattering with an electron at rest the wavelength of the
photon with energy 0.3 MeV changed by 20%.
i) Find the kinetic energy of the electron after scattering.
ii)What is the momentum of the electron.
iii) What are the scattering angles of the photon and electron?
iv) The same photon is scattered by the electron moving in the opposite direction.
After scattering, the photon propagates in the opposite direction (180 degrees
scattering). What is the velocity of the electron if the frequency of the photon does not
change?
Solution:
i) mc 2  h  h ' E => T  E  mc 2  h  h ' ,  '  c /  '  c /(1.2 )   / 1.2 ,
hence T  h (1  1 / 1.2)  0.05MeV
ii)
p 2 c 2  E 2  m 2 c 4  (T  mc 2 ) 2  m 2 c 4  T 2  2Tmc 2  T (T  2mc 2 ) =>
pc  T (T  2mc 2 )  0.23MeV
iii) For the photon:  '  (h / mc)(1  cos  ) , hence
cos   1  ( ' ) /( h / mc)  1  (h 0.2) / mc 2  0.88
For the electron from the momentum conservation in the x direction we get
h /    h / 'cos  p cos => cos   h  h ' cos  0.3  0.3 * 0.88 / 1.2  0.35
pc
0.23
iv) Since the photon energy does not change, it's momentum also does not change by
value, hence the electron momentum also does not change by value and
pe= h / c  me v / 1  v 2 / c 2 => v  c[1  (m e c 2 / h ) 2 ]1/ 2  1.54 x1010 cm / s