Elastic collision between protons

[missinghorn]

Homework Statement

a proton with kinetic energy 10GeV collides with a proton at rest. 1 proton goes off with an angle of 45 degrees relative to the incoming proton. Find the velocity of both protons after the collison. This is an ellastic collision.
E(initial) = 938 mev
i honestly don't know where to start off but here's my attempts at a answer

Homework Equations

The Attempt at a Solution

E=(MC^2)/(1-(v/c)^2)^1/2

V/c= (1-((mc^2)/(E^2)))^1/2 V= .99632c


V= v1 cos (30) + v2 cos(theta)
v1 sin (30) + v2 sin (theta)
E = E1 + E2
10.938 gev = .938 gev/ (1-(v1/c)^2))^1/2 + .938gev/(1-(v2/c)^2)^1/2
substitute and get the answer v1 is .00458
and v2 is .49780
however these answers don't work when put back into the equations..

 

[Count Iblis]

You forgot the gamma factors in the conservation of momentum equations.

 

[Count Iblis]

Also, you can solve the problem more easily using four-momentum algebra. You then only have to deal with one equation for hte energy of one proton after the collision, you don't have to consider a simultaneous set of equations for the energies and momenta of the two protons after the collision.

 

[missinghorn]

four momentum algebra..? can u start me off with the equations?
and also where have i missed the gamma..? because i can't seem to find it ..? unless my equations are wrong.