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Relativistic kinetic energy comapred to newtonian

  1. Sep 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the speed of a particle whose relativistic kinetic energy is 40% greater than the Newtonian value for the same speed.

    Krel = relativistic kinetic energy
    Knew = Newtonian kinetic energy

    2. Relevant equations
    Krel = (gamma - 1)mc^2
    Knew = 0.5mv^2
    gamma = 1/sqrt(1-x)
    x = v^2 / c^2

    3. The attempt at a solution
    So I set it up as K(relativistic) = 1.4K(Newton).. because my problem was 40%.
    So (gamma - 1)mc^2 = 1.4 (0.5mv^2) and..
    (gamma - 1) = (0.7mv^2)/mc^2
    ... m's on top and bottom cancel out, then I replaced v^2/c^2 by x...
    (gamma - 1) = 0.7x
    1/sqrt(1-x) = 0.7x + 1
    1/(1-x) = (0.7x+1)^2
    1 = 0.49x^2 + 14x + 1 - 0.49x^3 - 1.4x^2 - x
    0 = x(-0.49x^2 - 1.09x + 0.4)

    Using the quadratic formula, I get either x = -3.12 or x = 0.9

    Since x = v^2 / c^2,
    v = sqrt(x)*c
    So I get v = 0.9c

    But the answer is 0.61c.

    Where did I go wrong?
  2. jcsd
  3. Sep 12, 2010 #2


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    Homework Helper

    One of the coefficients in this equation is wrong. (Just a little math mistake)
  4. Sep 12, 2010 #3
    Ohh man, how did I not catch that? I reworked this three times, with a calculator even, and didn't notice that little mistake.

    Thank you so much! I'm relieved it was just a small math mistake. You're a lifesaver :)
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