Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic speed involving particle decay

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle of rest mass M0 is at rest in the laboratory when it decays into three identical particles, each of rest mass m. Two of the particles have velocity
    u1=-4/5c i(vector)
    u2=-3/5c j(vector)
    Calculate the direction and speed of particle 3

    2. Relevant equations

    3. The attempt at a solution

    The starting momentum = 0 since it is at rest, so the momentum of the three particles should add up to 0.
    p1=m/(sqrt(1-(-4/5)^2)) -i(vector)
    p2=m/(sqrt(1-(-3/5)^2)) -j(vector)
    p3=m/(sqrt(1-(v/c)^2)) ij(vector)

    the total momentum is p1+p2+p3
    I calculated m/.6 -i + m/.8 -j + m/(sqrt(1-v/c)^2)) ij
    I'm not sure how to break this up to find v.
    I had tried to use u=(sqrt(ux^2+uy^2+uz^2)) but that = 1 for this.
    I think to use that equation I need to break down the ij vector of particle 3
    To do this it should be atanx=(3/4), but the answer is actually atanx=(9/16) I'm not sure why it's squared. Likely I'm forgetting something very basic that messed up the whole problem.

    Main points of interest:
    find out i and j vectors of particle 3
    find out why the units are squared as in atanx=(9/16) and is that 9/25 (3/5)^2? if so that doesn't compute out using sin 29 = (9/25)/h

    M/m should be (gamma)1*m+(gamma)2*m+(gamma)3*m since they all have different gammas.
    Last edited: Mar 3, 2009
  2. jcsd
  3. Mar 3, 2009 #2
    My momentum equations were missing velocity. i only had mass in the numerator. p1=4/3 i(vector) p2=3/4 j(vector)
    p3=m (sqrt ((4/3)^2+(3/4)^2))

    much thanks to Bekki from yahoo answers :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook