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Relativistic and non-relativisitic kinetic energy

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data
    at what speed does the expression for relativistic kinetic energy vary from the non-relativistic expression by 5%?


    2. Relevant equations
    Relativistic kinetic energy K=(gamma-1)mc^2
    Non-relativistic kinetic energy K=0.5mv^2


    3. The attempt at a solution
    I'm not sure how to attempt this!
    Should I be finding the difference? ie K(relativistic)-K(non-relativistic)=0.05??
     
  2. jcsd
  3. Dec 3, 2009 #2

    gabbagabbahey

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    Hi lola2000, welcome to PF!:smile:

    Close, you should be finding the relative difference |Krel-Knon-rel|/Knon-rel=0.05 .... make sense?
     
  4. Dec 3, 2009 #3
    I see, that makes sense.

    But when I do this I get
    0.05 = [(gamma-1)mc^2 - 0.5mc^2] / 0.5mv^2
    which simplifies to
    0.6mv^2 = (gamma-1)mc^2
    0.6v^2/c^2 +1 = gamma

    which is really nasty to solve! Is there a trick I have missed?
     
  5. Dec 3, 2009 #4

    gabbagabbahey

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    I assume this is a typo?

    Really?
     
  6. Dec 4, 2009 #5
    You are right!
    That was a typo - it should have been -0.5mv^2

    But I am still stuck with the algebra

    I have 0.525v^2/c^2 = gamma - 1

    How do I rearrange this??

    It is not simplifying!
     
  7. Dec 4, 2009 #6

    nrqed

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    I did not check the previous steps so I cannot guarantee this is the correct equation. But assuming it is, you just need to add 1 to both sides (to have gamma isolated). Then square both sides and rename [tex] v^2/c^2 = X [/tex] . Then you will have a quadratic equation for X. Solve, keep only the positive root. That will be [tex] v^2/c^2[/tex]. Then the answer is the square root of X.
     
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