A proton of total energy E collides elastically with a second proton at rest in the laboratory. After the collision the two protons follow trajectories which are disposed symmetrically at angles ±Φ/2 to the direction of the incident particle. By considering the motion in the laboratory frame, or otherwise, show that
cos Φ = E – E0/E + 3E0
where E, is the rest mass energy of the proton.
What is the value of Φ when the first proton is accelerated from rest through a potential difference of 1.5 x 109V before colliding with the second proton?