Relativistic scattering
a) a proton of total energy E collides elastically with a second proton at rest in the lab. After the collision the two protons follow trajectories which are set symmetrically at angles + phi/2 to the direction of the incident particle. Consider the motion of the particles in the lab frame and derive the relation cos phi = (E  E0)/(E + 3E0) where E0 is the rest mass energy of the proton.
cos phi + 1 = 2cos^2 phi/2
I'm having trouble with the math. I know the energy before is just E + E0 but I'm having troubles with the after portion.
b) A proton is accelerated from rest through a potential difference 1.5x10^9 V before colliding with the second proton. what is the proton's energy and what is the value of phi.
for part b I said delta PE =  delta KE
delta PE = qU where q is the charge and U is the potential
and so U = delta PE/q
since we can solve for energy now, I said:
phi = cos^1((E  E0)/(E + 3E0))?
Thanks for taking the time to help ^_^
