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E=mc^2 question

  1. Nov 28, 2007 #1
    What is the momentum of a particle whose total energy is four times its rest energy? Give your answer as a multiple of mc.

    Well the rest energy of an object is E=mc^2
    so the total energy for this particle would be E= 4(mc^2)
    in order to achieve this Energy value the Lorentz factor would have to be equal to 1/4.
    Beyond that i do not understand how to relate the Energy to the relative momentum of the particle

    Relativist Momentum
    [tex]p = \frac{mv}{\sqrt{1 - v^2/c^2}}[/tex]
     
  2. jcsd
  3. Nov 28, 2007 #2

    Ben Niehoff

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    So, you know that the Lorentz factor is 1/4...use that to find v in terms of c. Then you can plug this into the expression for p.
     
  4. Nov 28, 2007 #3

    Ben Niehoff

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    There is a potentially easier method using the relation

    [tex]E^2 = (mc^2)^2 + (pc)^2[/tex]

    if you have learned this equation. However, you seem to be halfway there already using the other method, so you might as well do that (or you can use both methods and verify that they give the same answer).
     
  5. Nov 28, 2007 #4

    Doc Al

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    An easier way is to use the expression for total energy of a particle in terms of rest energy and momentum. Do you know it? (E = mc^2 is a special case of that more general expression.)

    Edit: Ben just gave it to you while I was typing. :wink:
     
  6. Nov 28, 2007 #5
    so using the expression that Ben gave me I would solve for p then?

    Edit: Or if i wanted to continue on my original path i would solve for v and then plug that into the equation for relative momentum, and solve for it in terms of mc.
    i calculate the v to be v= 0.97c

    p= mv/(square root(1-v^2/c^2))
    p= (m * .97c)/(square root(1 - (0.97)^2)
    p= .97mc/(1/4)
    p= 3.87mc
    ? is that correct?
     
    Last edited: Nov 28, 2007
  7. Nov 28, 2007 #6

    Doc Al

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    Perfectly correct. (Be sure to do it the other way, just for the practice. And to see how much easier it is.)
     
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