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## Homework Statement

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!

## Homework Equations

ρ = mv[itex]\gamma[/itex]

E = mc

^{2}[itex]\gamma[/itex]

## The Attempt at a Solution

By conservation of momentum, ρi = ρf, so:

0 = (3/4)mc + (4/3)mc + v

_{3}[itex]\gamma[/itex]

_{3}

v

_{3}[itex]\gamma[/itex]

_{3}= -(3/4)c - (4/3)c

After plugging in gamma to solve for v

_{3}:

v/sqrt(1-v

^{2}/c

^{2}) = -(25/12)c

v = +/- sqrt(c

^{2}/((144/625) + 1))

v = 0.9c

By conservation of energy, Ei = Ef, so:

Mc

^{2}= (5/4)mc

^{2}+ (5/3)mc

^{2}+ 2.294mc

^{2}where 2.294 is [itex]\gamma[/itex]

_{3}(found using v solved for above)

so M = 5.21m

M/m = 5.21

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