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Special relativity problem (momentum and velocity)

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle travelling at 8 E7 m/s is accelerated so that its momentum doubles. What is the final velocity of the particle?


    2. Relevant equations
    p=ym0v
    y= 1 / sqr(1-(v/c)^2)
    p2= 2(p1)
    Where p2 is the final momentum and p1 the initial

    3. The attempt at a solution
    For the initial velocity gamma=1,0376
    p1 is therefore, calculated with m0=1 (I put it this way thinking it was not relevant to the problem, and it was meant to be eliminated when finding p2) 83008000 kg m/s
    I tried to compare the two equations for the relativistic momentum but the best i came up with was that (v2)x(y2)=166016000 m/s
    Given that the result written on the book is 1,5 E8 m/s, y2 must be 1,107
    because y2= 2y1v1 / v2
    But for a velocity of 1,5 E8 m/s (0,5c), the dilatation factor is 1,154
    I really don't know, I keep trying the same method, I can't see a different way to resolve it

    Sorry for the bad english, not my lenguage
    Thanks to everybody in advance
     
    Last edited: Sep 19, 2009
  2. jcsd
  3. Sep 19, 2009 #2

    Doc Al

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    Staff: Mentor

    That can't be right. What is the speed?
     
  4. Sep 19, 2009 #3
    uops, I'm sorry, i meant E7
    anyway I used 8 E7 in the equations
    Also, edited in the first post, thanks for pointing that out
     
  5. Sep 19, 2009 #4

    Doc Al

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    Staff: Mentor

    Assuming your arithmetic is correct (I didn't check) write the left hand side entirely in terms of v2. Then solve for v2.
     
  6. Sep 20, 2009 #5
    2y1v1 is = 166011471 m/s, I think it's correct now

    so

    2y1v1 = 166011471 m/s = k

    y2= 1 / sqr(1-(v2/c)2)

    v2 / sqr(1-(v2/c)2) = k

    v2^2 = k^2 - k^2(v2/c)^2

    (k^2/c^2 +1)v2 = k^2

    v2 = sqr( k^2 / (k^2/c^2 +1))

    which is 145254607 m/s, it's so close to 150000000 m/s

    Is this an error caused by bad arithmetic or something else, more important?
     
    Last edited: Sep 20, 2009
  7. Sep 20, 2009 #6

    Doc Al

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    Staff: Mentor

    It's not an error. The book just rounded off to 2 digits.
     
  8. Sep 20, 2009 #7
    Seems like I was too pessimistic then :)
    Thank you very, very much for your help
     
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