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Energy and Momentum-Relativity?

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    An antiproton p_o has the same rest energy as a proton. It is created in the reaction p+p->p+p+p+p_o. In an experiment, protons at rest in the laboratory are bombarded with protons of kintetic energy E_k, which must be great enough so that kintetic energy equal to 2mc^2 can be converted into the rest energy of the two particles. In the frame of the laboratory, the total kinetic energy cannot be converted into rest energy because of conversation of momentum. However, in the zero-momentum reference frame in which the two initial protons are moving toward each other with equal speed u, the total kintetic energy can be converted into rest energy.
    (a) Find the speed of each proton u such that the total kinetic energy in the zero-momentum frame is 2mc^2.

    2. Relevant equations



    3. The attempt at a solution

    [tex]E_k=\gamma mc^2-mc^2=2mc^2\rightarrow \gamma =3 \rightarrow 1-\frac{u^2}{c^2}=\frac{1}{9}\Rightarrow u=\frac{2\sqrt{2}}{3}c[/tex]
    1. The problem statement, all variables and given/known data



    Is it this simple, or am i missing something here?

    Thnx!
     
  2. jcsd
  3. Feb 6, 2010 #2
    Not quite. You are right that each proton has the energy [itex]E_k=m_pc^2(\gamma-1)[/itex] but you want [itex]E_k=mc^2[/itex], not [itex]E_k=2mc^2[/itex].

    EDIT: I should clarify here. You have two particles of equal mass and speed that gives you a total energy [itex]E_k=2mc^2[/itex], but to find the velocity of one (and thus the velocity of the other), you need half of this energy.
     
  4. Feb 6, 2010 #3
    Oh, yeah, that's right! That's an easily fixable detail, i was more concerned that the general approach was erroneous!

    Thnx!
     
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