- **Special and General Relativity**
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- - **Relativistic scattering**
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Relativistic scatteringa) 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 ^_^ |

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