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Relativistic Momentum and energy conservation

  1. Aug 27, 2009 #1
    Hello everyonea. I was studying special relativity and i got stuck on an example. Here it is:'to an observer,two bodies of equal rest mass collide head on with equal but opposite velocities 0.8c and cohere. To a second observer,one body is initially at rest. Find the apparent velocity of the other moving mass before the collision and compare its initial energy in the 2 frames of reference.'
    i am trying to find their relative velocity but am finding a value greater than c. Can somebody help me on this or give some tips on how i should start. Thanks for understanding.
     
  2. jcsd
  3. Aug 27, 2009 #2

    tiny-tim

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    Hello nard! :smile:
    That shouldn't be possible. :confused:

    What formula are you using (from memory, I think it should be (u+v)/(1+uv))?
     
  4. Aug 27, 2009 #3

    Dale

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  5. Aug 27, 2009 #4
    This is the derivationhttp://en.wikipedia.org/wiki/Velocity-addition_formula" [Broken]
    Another mistake is that you didn't name yourself properly:smile:
     
    Last edited by a moderator: May 4, 2017
  6. Aug 27, 2009 #5
    I used lorentz transformations, i found a velocity which is less than c,however i stil dont know how to procede
     
  7. Aug 27, 2009 #6
  8. Aug 27, 2009 #7

    tiny-tim

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    Show us what you did. :smile:
     
  9. Sep 2, 2009 #8
    u'=(u-v)/(1-uv/c^2), where u' and u are speeds of the body in different frames and v their relative velocity.
    for viewer 1.
    P1+P2=P3 from momentum conservation but P1=-P2, then
    P3=0
    E1+E2=E3 from energy conservation but E1=E2;
    2E1=E3;
    for viewer 2.
    P1'=P3' since P2' is zero;
    E1'+moc^2=E3';
    whats next i dont know
     
  10. Sep 2, 2009 #9
    u'=(u-v)/(1-uv/c^2), where u' and u are speeds of the body in different frames and v their relative velocity.
    for viewer 1.
    P1+P2=P3 from momentum conservation but P1=-P2, then
    P3=0
    E1+E2=E3 from energy conservation but E1=E2;
    2E1=E3;
    for viewer 2.
    P1'=P3' since P2' is zero;
    E1'+moc2=E3';
    whats next i dont know
     
  11. Sep 2, 2009 #10

    tiny-tim

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    Hello nard! :smile:

    erm :redface: … you're not actually answering any of the question …
    To rewrite it, the question is, if the velocities in one frame are ±0.8c, what is one velocity in the frame in which the other velocity is 0? In other words (as you say), what is the relative velocity?

    And what is the energy in the two frames?

    (you're not asked to solve any collision problem)

    So what figures did you get? :smile:
     
  12. Sep 4, 2009 #11
    hello! the relative velocity in the frame where the two masses have equal velocities(.8c) and different directions is aproximatly(.97c).
    in the frame where one mass is at rest the velocity of the moving one is equal to the relative velocity(.97c).
    now if we compare the two energies in frame 1 E1=1.66moc^2
    and E1'=4.41moc^2
    now the remaining question is why the big difference in energy in the frames.
     
  13. Sep 4, 2009 #12

    tiny-tim

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    Hello nard! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Yes, that looks right :smile:

    (but I'd be inclined to do it a bit more accurately than .97 … it's the difference between that and 1.0 which is important, and you've only got that to one sig fig! :wink:)
    (one reason why exam questions often use 0.8c is that it uses the 3,4,5 Pythagorean triangle, so things tend to turn out as exact fractions … this one does, if you try it :wink:)

    "why the big difference?"

    'cos that's the way it is!

    what's worrying you about that? :smile:
     
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