Special Relativity 2-D Collision Problem

wuphys

1. The problem statement, all variables and given/known data
This is my first time posting here, so please let me know if I do not have the proper format.

A particle of mass M is at rest in the laboratory when it decays into three identical particles, each of mass m. Two of the particles have velocities and directions as shown (one travels in the -x direction with a speed of 4c/5 and another travels in the -y direction with a speed of 3c/5. Calculate the direction and speed of the third particle (shown on the diagram moving with +x and +y components at some angle above the x axis) and find the ratio M/m.

I'm not sure (1) if the way I solved for v and the ratio is correct and (2) how to find the direction of the third particle. I originally solved it by saying the components of momentum are equal, but I don't think that would be correct because it is relativistic. Any help would be greatly appreciated!

2. Relevant equations

ρ = mv[itex]\gamma[/itex]
E = mc²[itex]\gamma[/itex]

3. The attempt at a solution

By conservation of momentum, ρi = ρf, so:
0 = (3/4)mc + (4/3)mc + v₃[itex]\gamma[/itex]₃
v₃[itex]\gamma[/itex]₃ = -(3/4)c - (4/3)c

After plugging in gamma to solve for v₃:
v/sqrt(1-v²/c²) = -(25/12)c
v = +/- sqrt(c²/((144/625) + 1))
v = 0.9c

By conservation of energy, Ei = Ef, so:
Mc² = (5/4)mc² + (5/3)mc² + 2.294mc² where 2.294 is [itex]\gamma[/itex]₃ (found using v solved for above)

so M = 5.21m
M/m = 5.21

wuphys

Never mind, I think I got the answer. The components of relativistic momentum are just equal.