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E-P invariant on relativistic mecahnics problem

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle of mass [itex]m[/itex] moving at speed [itex]\frac{3}{5}c[/itex] collides with an identical particle at rest, and forms a new particle of mass [itex]M[/itex] which moves off at speed [itex]v[/itex]. Find v.


    2. Relevant equations
    E-P invariant: [itex]E_1^2-p_1c^2=E_2^2-p_2^2c^2=\mathrm{const.}[/itex]
    Momentum: [itex]p^2=m^2c^2\left({\gamma_v^2-1}\right)[/itex]
    Energy: [itex]E=\sum_i\gamma_im_ic^2[/itex]


    3. The attempt at a solution
    There is a single particle after the collision, so [itex]E_2^2-p_2^2c^2=M^2c^4[/itex]
    where [itex]E_2=\gamma_vMc^2[/itex]
    and [itex]p_2^2=M^2c^2\left({\gamma_v^2-1}\right)[/itex]

    so

    [tex]
    \gamma_v^2M^2c^4-M^2c^4\left({\gamma_v^2-1}\right)^2=M^2c^4\\
    \gamma_v^2-\left({\gamma_v^4-2\gamma_v^2+1}\right)=1\\
    \left({\gamma_v^2-1}\right)\left({\gamma_v^2-2}\right)=0
    [/tex]

    Reject the [itex]\gamma_v^2=1[/itex] solution, since this would mean [itex]v=0[/itex].

    The [itex]\gamma_v^2=2[/itex] solution gives [itex]v=\frac{\sqrt{2}}{2}c[/itex], which is incorrect - the answer is [itex]v=\frac{1}{3}c[/itex].

    I expect there's something fundamentally wrong with my method, but not sure what? Thank you!
     
  2. jcsd
  3. Apr 10, 2012 #2

    vela

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    You shouldn't have squared the factor ##(\gamma_v^2-1)## in the first line. With this approach, all you'll end up with is M=M, so it's not helpful.
    You haven't used any information from before the collision. Use conservation of energy and momentum.
     
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