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Special Relativity kinetic energy

  1. Oct 31, 2013 #1
    1. The problem statement, all variables and given/known data
    In reference frame S', two protons, each moving at .500c, approach each other head-on. a) calculate the total kinetic energy of the two protons in frame S'. b) Calculate the total kinetic energy of the protons as seen in the reference frame S, which is moving with one of the protons.



    2. Relevant equations
    E = mc^2 + K
    K = γ1/2mv^2
    Proton rest mass = 938 MeV/c^2

    3. The attempt at a solution
    So for part A I just found the kinetic energy of one of the protons given by γ(1/2)mv^2
    (1.1547)(1/2)(938)(.25) * 2 = 270 MeV.. and the back of the book says 290 MeV.. but this textbook is known for having a lot of wrong answers, can anyone verify whether I am correct or incorrect?
     
  2. jcsd
  3. Oct 31, 2013 #2

    Simon Bridge

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    I get 290.22MeV Note:
    ##K=(\gamma-1)mc^2## because ##E=\gamma mc^2##
     
  4. Oct 31, 2013 #3
    what is incorrect about my method of using the equation K = γ1/2mv^2 to obtain the answer?
     
  5. Oct 31, 2013 #4

    Doc Al

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    It looks like you just took the non-relativistic expression for KE (1/2mv^2), which is only good for low speeds, and thought you could make it relativistically correct by slapping on a gamma. But that doesn't work. Simon Bridge gave the correct formula.
     
  6. Oct 31, 2013 #5

    Simon Bridge

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    DocAl is correct - ##K\neq \gamma\frac{1}{2}mv^2##
    What lead you to believe the equation would be correct?
     
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