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Relativistic Kinematics conservation of energy

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle with rest mass m_{0} and kinetic energy 3m_{0}c^{2} makes a completely inelastic collision with a stationary particle of rest mass 2m_{0}. What are the velocity and rest mass of the composite particle?

    2. Relevant equations

    [itex] T = (\gamma - 1)m_{0}c^{2} = E - m_{0}c^{2}\\
    E = \gamma m_{0}c^{2} [/itex]

    3. The attempt at a solution
    Inelastic means the particles don't bounce or separate. One hits the other and they move away together(as a new particle perhaps). Energy needs to be conserved, so the problem as I see it is;

    Incoming particle has total energy of the kinetic energy plus the rest mass energy,
    [itex] 3m_{0}c^2 + m_{0}c^{2}[/itex]

    This needs to equal the sum of the rest mass and kinetic energy of the new system,

    [itex] T + 2m_{0}c^{2}[/itex]

    If my approach is correct I would assume that the kinetic energy should drop to 2m_{0}c^{2}. Something seems wrong though and all my workings so far aren't coming out with much sense. A point in the right direction would be great.
     
  2. jcsd
  3. Apr 30, 2013 #2

    Doc Al

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    Staff: Mentor

    Don't forget the energy of the target particle, which counts towards the total energy of the system.

    What else is conserved besides the total energy?
     
  4. May 1, 2013 #3
    Besides energy, momentum is conserved. A formula that contains both would be;

    [itex] E^{2} = p^{2}c^{2} + m^{2}c^{4}[/itex]

    So total energy of the system equals the sum total of both particles rest mass plus the kinetic energy of the incident particle. The system after the collision must equal the same with the composite particle now moving.

    [itex] E_{total} = E_{incident rest} + E_{incident kinetic} + E_{target rest} = m_{0}c^{2} + 3m_{0}c^{2} + 2m_{0}c^{2} = 6m_{0}c^{2}\\

    6^{2}m_{0}^{2}c^{4} = E^{2}[/itex]

    And as [itex]p = \gamma m_{0}v[/itex] I can solve for v?
     
  5. May 1, 2013 #4

    Doc Al

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    Staff: Mentor

    Set up separate equations for momentum and energy conservation.
     
  6. May 1, 2013 #5
    Let M be the final rest mass of the combined particle and V be the final velocity of the combined particle. In terms of M and V, what is the final energy of the combined particle and what is the final momentum of the combined particle?
     
  7. May 2, 2013 #6
    I think I am starting to get it, the Momentum;

    -
     
  8. May 3, 2013 #7
    Sorry I still am not sure. I was starting to think I understood. Momentum would be [itex]\gamma v_{1}m_{0} = \gamma_{2} v_{2}M_{0}[/itex]

    V1 could be worked out with equation for kinetic energy which shows gamma = 4. Kind of confused as to the next step :/.
     
  9. May 3, 2013 #8
    You're missing the energy balance (aka, momentum balance in time-direction):
    [tex]\gamma m_{0}c+2m_{0}c= \gamma_{2}M_{0}c[/tex]
    Now you have two equations in two unknowns.
     
  10. May 3, 2013 #9
    So my equation would be the conservation of momentum, the one you have stated is the rest mass of incident and target particles equalling the rest mass of the composite? I think where I was getting confused was thinking that conservation of momentum would need to sum the kinetic and rest mass energies. If I have understood you correctly I shall attempt the question and get back to you.

    Thank you Chestermiller
     
  11. May 3, 2013 #10
    SUCCESS v2 has equalled 0.645c and M0 has equalled 4.58m0. Both the answers provided.

    I cant thank you enough.

    Cheers!
     
  12. May 3, 2013 #11
    That's not what my equation says. The equation I wrote was for the t-component of conservation of momentum, which is equivalent to the conservation of energy. You may be referring to the antiquated (and out-of-favor) concept of "relativistic mass"; my equation is also equivalent to saying that relativistic mass is conserved; but I want to strongly discourage from thinking of it in this way. Please try to get used to working with all four components of 4 momentum, and thinking in terms of 4D.
     
  13. May 4, 2013 #12
    Like I said even if it is not what you meant it led me to get the answer provided by my lecturer. I am familiar with the fact there are improved equations that require using the 4 vectors of each quantity. I have not been taught these yet however and we may build up to it for next years particle physics module.
     
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