Find the proper length of the rod if in the system S (at rest) it's velocity is c/2,
length l = 1 m and the angle between the rod and direction of it's motion is
450?
Solution:
1.
l x'  l x / 1   2 , where  v / c, l y'  l y , l x  l cos  , l y  l sin  ,
l '  l x'  l y'  l 2 cos 2  /(1   2 )  l 2 sin 2   l
2
2
1   2 sin 2 
 1.08m
1  2
2. Two unstable particles move by a straight line with velocity v=0.99 c. The
distance between them in the laboratory reference frame is l = 120 m. At some
moment the particles broke up simultaneously in their reference frame. What
is the time interval between the destruction of the particles in the lab reference
frame? Which particle broke up earlier?
Solution:
t '(v / c 2 )x'
t 
, hencefort '  0, t  (v / c 2 )x' / 1   2 , butx'  l / 1   2 ,
2
1 
hencet  (v / c 2 )l /(1   2 )  20 s  0
Hence the first particle will break up later.
3.
3 events A, B and C are shown on the x-t diagram. Find:
i) the time interval between events A and B in the reference frame where they
occurred in one place.
ii) the distance between the points where the events A and C occurred in the reference
frame where they are simultaneous.
Solution:
(i) ct ' AB   ct AB   x AB , x ' AB   ( x AB   ct AB ) , since x' AB  0 ,
  x AB / ct AB  3 / 5  0.6 ,   1 / 1   2  1 / 0.8 hence
ct ' AB   ct AB  x AB   (5  3 * 0.6) / 0.8  13ns
(ii) ct ' AC   ct AC  x AC , hence
  ct AC / x AC  0.6   1 / 1   2  1 / 0.8 ,hence
x' AC   (x AC  ct AC )  (5  0.6 * 3) / 0.8  4m
4. Two rulers with the proper lengths l0 move along the x axis in the opposite
directions. An observer connected with one of them found that the time
between the moments when the left and right ends of the rules coincide is
equal to  . Find the relative velocity of the rulers.
Solution:
In the reference frame connected with one of the rulers at the moment when the
left ends coincide x=0, the right end of the moving ruler has a coordinate
x r  l 0 1  u 2 / c 2 hence the distance between this end and the right end of the
ruler at rest is l 0  l 0 1  u 2 / c 2 = u , hence
(l0  u ) 2  l0 (1  u 2 / c 2 )  2l0 u  u 2 2  l0 u 2 / c 2 ,hence canceling u we
2l 
get u  2 20 2
(l0 / c   )
2
2
5. (i)Find momentum of the particle if its rest mass is m and kinetic energy - T.
(ii) A relativistic particle with the rest mass m and kinetic energy T collides
inelastically with the same particle at rest. Find the rest mass and velocity of
the new particle formed after collision.
Solution:
(i)

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
(ii) E  T  2mc 2 , E 2  p 2 c 2  M 2 c 4 , where M is the rest mass of the new
 pc 2
 T (T  2mc 2 ) hence
M 2 c 4  (T  2mc 2 ) 2  T (T  2mc 2 )  (T  2mc 2 )2mc 2 
particle.
(T  2mc2 )2m
c
Momentum of the new particle is equal to the momentum of the first particle:
E
p  T (T  2mc 2 ) / c  2 V 
c
T
V  pc 2 / E 
c
T  2mc 2
M
6. Find velocity of the particle with charge e and rest mass m and with zero
initial velocity accelerated by the potential difference V. Find the limiting
values of the velocity for (i) classical case when eV<<mc2 and
(ii)ultrarelativistic case when eV>>mc2.
Solution:
Kinetic energy T is equal to the work of the field eV hence:
T  E  mc 2  mc 2  mc 2  eV , where   1 / 1  v 2 / c 2 ,
Hence   1  eV / mc 2 , v  c 1  1 /  2 
(i) classical case eV<<mc2, v 
2eV
m
1  eV /( 2mc 2 )
[1  eV /( mc 2 )] 2
2eV
; (ii) ultrarelativistic case v ≈c
m