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Relativistics kinetic energy

  1. Mar 19, 2016 #1
    1. The problem statement, all variables and given/known data
    At what speed the kinetic energy of particle equal n times it's rest energy

    2. Relevant equations
    i use this equation
    ##T=(\gamma - 1)m_{0}c^{2} ##

    3. The attempt at a solution
    ##T=n(m_{0}c^{2})##

    ##n=\gamma -1##

    ##\gamma =n+1##

    ##\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=n+1##

    ##\frac{1}{1-\frac{v^{2}}{c^{2}}}=n^{2}+2n+1##

    ##\frac{1}{n^{2}+2n+1}=1-\frac{v^{2}}{c^{2}}##

    ##\frac{v^{2}}{c^{2}}=1-\frac{1}{n^{2}+2n+1}##

    ##\frac{v^{2}}{c^{2}}=\frac{n^{2}+2n+1-1}{n^{2}+2n+1}##

    ##v=\sqrt{\frac{n^{2}+2n}{n^{2}+2n+1}} ~~ c##

    this is what i get.
    Is that correct? isn't there
    other solution?
     
  2. jcsd
  3. Mar 19, 2016 #2
    pl. check again the relations;
    what is the the expression for relativistic kinetic energy?
     
  4. Mar 19, 2016 #3

    blue_leaf77

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    Is there something strange with ##T=(\gamma - 1)m_{0}c^{2} ##? At least it is consistent with the classical kinetic energy for small velocities.
     
  5. Mar 19, 2016 #4
    I checked it and it is correct
    Or i can write it in this form
    ##E_{k} = \frac{m_{0}c^{2}}{\sqrt{1 - (v/c)^{2})}} - m_{0}c^{2}##
     
  6. Mar 19, 2016 #5

    ehild

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    Gold Member

    It is correct.
     
  7. Mar 19, 2016 #6
    Ok. Thank you all for your help
     
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