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Relative velocities of decay products

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle of mass M decays into two identical particles each of mass m, where m = 0.3M. Prior to the decay, the particle of mass M has a total energy of 5Mc^2 in the laboratory reference frame. The velocities of the decay product are along the direction of motion M. Find the velocities of the decay products in the laboratory reference frame.


    2. Relevant equations

    E=[tex]\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}[/tex]


    3. The attempt at a solution

    I think I have the concept down in my head. I know I will get two different velocities for the two identical particles. I think it will similar to a doppler effect where the particle formed in the direction of motion gets a speed boost, whereas the particle formed in the direction opposite of motion has a slower speed. So far I have used the equation above to find that the initial particle is traveling at .98c. However, this is where I am confused because how do I figure out how fast the particles resulting from the decay of the original particle move?
     
  2. jcsd
  3. Sep 8, 2009 #2

    Astronuc

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    Staff: Mentor

    Hint: The velocities of the decay product are along the direction of motion M.

    Two particles of equal mass. Think conservation of momentum.
     
  4. Sep 8, 2009 #3
    Ya, that's what I was going with. So I set up MV=mv+mv. So MV=.6Mv. Since V=.98c and the M's all cancel, I am getting v=1.63c. Have I made an incorrect assumption or maybe some mathematical error?
     
  5. Sep 8, 2009 #4

    Astronuc

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    Staff: Mentor

    Realize that it's relativistic. And the velocities are vectors.

    In the COM, the inertial frame of the initial particle of M, the net momentum = 0. One particle moves forward in along the initial trajectory and the other particle moves backward.

    In COM frame, the net momentum must still be zero for the two particles.

    and don't forget [itex]\gamma{mv}[/itex]

    It's been a while since I've done relativistic mechanics, so I have to pull the cobwebs out of my belfrey.
     
  6. Sep 9, 2009 #5
    If I set it up so the momentum of the two particles is zero, then:

    [tex]p_{1}[/tex]+[tex]p_{2}[/tex]=0

    [tex]p_{1}[/tex]=-[tex]p_{2}[/tex]

    Using [tex]\vec{p}[/tex]=[tex]\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}[/tex]

    [tex]\frac{mu}{\sqrt{1-\frac{u^2}{c^2}}}[/tex]= -[tex]\frac{mu}{\sqrt{1-\frac{u^2}{c^2}}}[/tex]

    This is where I am confused. Is the u in the numerator the same as the u in the denominator. I do know that the u in the numerator on the left side is equal in magnitude as the u on the right side but opposite in sign. It seems like I need another equation. Can anybody help me out by pointing me in the right direction. Any input is much appreciated.
     
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