Question

A cosmic ray proton streaks through the lab with velocity 0.80c at an angle of 55

Answer 1

First break the proton's velocity into components.
vy = v sin(theta) =0.6553c
vx = v cos(theta) =0.4588c

Then use the velocity addition rule to subtract the speed of the reference frame from vx:
vx' = (vx - beta) / (1 - vx beta)
= (v cos(theta) - beta) / (1 - v cos(theta) beta)
= -0.4076c
The y component of the velocity is unchanged:
vy' = vy =0.6553c

So the speed in S' is:

v' = sqrt (vx'^2 + vy'^2)
= sqrt ( ((v cos(theta) - beta) / (1 - v cos(theta) beta))^2 + (v sin(theta))^2 )

= 0.79996c

and the direction is:
theta' = arctangent (v sin(theta) (1 - v cos(theta) beta) / (v cos(theta) - beta))

=-58.118 degrees