Relativistic collision and conservation

[yeahhyeahyeah]

Hey, I'm pretty confused by this relativistic collision problem.

A particle of mass m moving along the x-axis collides elastically with a 2nd particle of identical mass at rest in the lab frame and scatters. Its final momentum makes an angle theta with the x-axis in the lab. If its initial kinetic energy K0 what is its final kinetic energy?


These are the two equations I am using, in addition to the rest mass invariance:

Conservation of relativistic momentum (in both x and y directions)
Conservation of relativistic energy

Some basic questions I have are -

I can't assume that the collision is symmetric right?
The angle of the first could be different than the angle that the second scatters at?

I end up with 3 equations ( the conservation of momentum in x, y, and conservation of energy) however I have 4 unknowns (the two angles, the two final energies)

Sorry, I would write them out but I'm not good at the LaTeX feature...

 

[jambaugh]

I think it is best to work in vector format. Ignoring the z direction:
P=(E,cp_x,cp_y)
Then conservation of energy-momentum is a single 3-dim eqn:
P_1+P_2 = P'_1+P'_2

You also have the constancy of masses (no mass exchange in the collision):
P\cdot P = E^2 - c^2p_x^2 -c^2 p_y^2 = m^2 c^4
(for both particles, before and after collision.)
That's 4 more equations for a grand total of 7. Apply the pre-collision mass equations to get a full set of initial conditions and you bring it down to 5 equations.

Unknowns:
Final momentum-energy of each particle (6) which leaves 1 free parameter (theta).

Have you applied the mass equations after collision? An implied constraint is that there was no exchange of rest mass between particles during the collision. That's part of the "elastic" assumption.

I am not clear but think your given initial kinetic energy doesn't include "rest energy" so the initial energy of the incoming particle is:
E_1 = K_0 + m\cdot c^2
Is that correct?

Finally, as I recall typically one works this problem by boosting to CoM (cent. of mass) frame so that total momentum before and after are zero. (You then have 3 eqns and 4 unknowns yielding 1 angle parameter.) You then must relate the CoM angle to the lab frame angle taking into account the Lorentz length contraction. But that's easier when you have specific values for initial conditions and may not be the best approach here.