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Relative Momentum

  1. May 27, 2012 #1
    I have been getting rather confused with rest mass, m0. here's the problem.

    A particle intially has a speed of 0.58c. A) At what speed does it momentum increase by 1%. B) At 10%, and C) At 100%.

    I have set it up as p=γm0v....
    then γ=1/ sqrt[1=(0.58)^2

    but dont know what to do any further. Thanks
     
  2. jcsd
  3. May 28, 2012 #2

    tiny-tim

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    welcome to pf!

    hi dlucia! welcome to pf! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)

    what's the difficulty?

    you have momentum = p = po/√(1 - (v/c)2),

    so just put p/po = A) 1.01 B) 1.1 C) 2, and solve :wink:
     
  4. May 28, 2012 #3
    Shouldn't it be

    [itex]\frac{(v/c) γ(v/c)}{0.58 γ(0.58)}[/itex] = 1.01

    [itex]\frac{(v/c) γ(v/c)}{0.58 γ(0.58)}[/itex] = 1.1

    [itex]\frac{(v/c) γ(v/c)}{0.58 γ(0.58)}[/itex] = 2.0

    Chet
     
    Last edited: May 28, 2012
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