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Homework Help: Double the speed

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    A proton is accelerated from rest through a potential difference of 500MV. Classical mechanics indicates that quadrupling the potential difference would double the speed. Were a classical analysis valid, what speed would result from a 2000MV potential difference?

    2. Relevant equations


    3. The attempt at a solution
    We were first asked to find the original speed of the proton going through 500MV which is 0.76c Since 2000MV = 4*500MV and 4V means v is doubled, why can't I just double 0.76c to get the new speed for the proton accelerated through the 2000MV?
  2. jcsd
  3. Oct 22, 2012 #2


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    I believe that you are asked to do precisely that.
  4. Oct 22, 2012 #3
    Yet the answer is 2.07c :/
  5. Oct 22, 2012 #4
    the book solves it by 4*(the original KE, 8*10^-11 J) = .5 * m * v^2 then solving for v.......but I really just don't see why I can't logically just straight-up double the velocity...
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