1. (a) Using formula for relativistic addition of velocities, we have
(measuring velocities to the right as positive)
Velocity of spaceship B relative to A =
0.6c  (0.6c)
1  (0.6c)(0.6c) / c 2
= - 1.2 c/(1+ 0.36) = - 0.88 c (moving to left) =  say
(b) Let proper length of spaceship B in its own rest frame be LB
Then LB as seen by spaceship A = ( 1  v2 / c2 )LB =40 m
Where v is relative velocity of A and B = -0.88 c from part (a)
So LB = 40/ 1  v 2 / c 2 m = 40 / 1  (0.88)2 m = 84.2 m
(c) LA as seen by spaceship B = ( 1  v2 / c2 )L A = 150.0 1  (0.88)2
= 71.2 m
2. (a) K.E. of electron =  e mec2  mec2  ( e  1)mec2
K.E. of proton =  p mpc2  mpc2  ( p  1)mpc2
Equating these, we get
Now  e 
 p 1 (
1
1  (0.75c)2 / c 2
me
)( e  1)
mp
 1.512
so p – 1 = (0.511/938.3) 0.512 = 2.288 . 10 -4
This is a very small number so p ~ 1, so up << c
1
1
 1  .2.(u p / c)2  .......
2
1  (u p / c)2
Expanding,  p 
So, (u p / c)2  2.288.104
1.512. 10 -2 c
or up =
(b) Momentum of electron = emeue
Momentum of proton = pmpup
Equating these , we get
 pu p  (
me
0.511
) eue  (
)(1.512)(0.75c)  6.175.10 4 c
mp
938.3
(u p / c)
1  (u p / c)
2
 6.175.10 4
This is satisfied to a good approximation if up = 6.175. 10 -4 c