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Homework Help: Momentum & relativistic collision

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A 3.000 u (1 u = 931.5 MeV/c2) object moving to the right through a laboratory at 0.8c collides with a 4.000 u object moving to the left through the laboratory at 0.6c. Afterward there are two objects, one of which is a 6.000 u object at rest.

    A) Determine the mass and speed of the other object
    B) Determine the change in kinetic energy of this collision

    2. Relevant equations
    1 u = 931.5 MeV/c2


    KE = -mc^2 = KEf - KEi

    3. The attempt at a solution

    A) γ2794.5Mev+γ3726MeV=5589Mev+γm4c2

    In this case, wouldn't there be two equations? One for the momentum (γumu)conserved and the other for energy (γumc^2) conserved

    B) KE = KEf - KEi

    Thank you for any help
  2. jcsd
  3. Feb 5, 2013 #2


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    Right. Alternatively, you can consider 4-vectors, where momentum and energy conservation are just two components of the more general enery-momentum conservation.
  4. Feb 5, 2013 #3
    γm1u1 +γm2u2=γm3u3+γm4u4


    Then, this would be the equation for momentum conserved. But how do we solve for m4?
    I know if we have the equations for momentum and energy then we divide momentum over energy
  5. Feb 5, 2013 #4


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    How does your m3 carry momentum? It is at rest.
    You can get (different) mass<->momentum-relations for both conserved quantities.
  6. Feb 5, 2013 #5
    Okay, then it would be


    When I divide these two equations would it be this

  7. Feb 6, 2013 #6


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    How (and why) do you divide the equations?
    You can calculate those gamma factors by the way.

    You should get two equations with two unknowns.
  8. Feb 7, 2013 #7


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    You should have subscripts on the gamma factors since they're all different.
    ##\gamma_1 m_1 c^2 + \gamma_2 m_2 c^2 = \gamma_3 m_3 c^2 +\gamma_4 m_4 c^2##

    This one is wrong still.

    I don't see how you got this at all.
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