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Special Relativity 2-D Collision Problem

  1. Apr 12, 2012 #1
    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
     
    Last edited: Apr 12, 2012
  2. jcsd
  3. Apr 12, 2012 #2
    Never mind, I think I got the answer. The components of relativistic momentum are just equal.
     
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