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Phase-Space of a bouncing ball

  1. Feb 12, 2012 #1
    Hey,

    The phase-space, a graph of momentum against position, shows a trajectory of a particular system and any point on this trajectory gives a microstate of a particular macrostate; given the Energy of the system is constant... I think this is roughly true, correct me where I'm wrong please!

    However the question concerns a ball bouncing between two walls placed at positions ±q, the ball interacts elastically with the walls and travels at a constant velocity. Therefore the energy is constant and magnitude of the momentum is conserved.

    So I reckon the Phase-Space of such a system would simply be a straight line from -q to +q at a particular momentum p and also the same line at -p (for the ball bouncing back in the opposite direction)

    Would this be correct? If not any help would be appreciated!

    Cheers,
    Tom
     
  2. jcsd
  3. Feb 12, 2012 #2

    Redbelly98

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    Yes, your description is correct. :smile:
     
  4. Feb 12, 2012 #3
    Woo! I like being correct, Thanks!
     
  5. Feb 12, 2012 #4

    Redbelly98

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    You're welcome!

    If you'd like to take it a step farther, think about how the phase-space diagram is modified for a deformable ball: when it hits the wall, it actually slows to a stop down over a short distance, then rebounds (speeds up over the same short distance) with the same velocity.
     
  6. Feb 12, 2012 #5
    Hmm that may be similar to the next problem on my work sheet which asks to consider inelastic collisions.

    Would the phase-space trajectory, over the short distance, rapidly decline to a zero momentum and then rapidly rise to the same momentum but negative (or opposite sign)?
     
  7. Feb 12, 2012 #6

    Redbelly98

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    Yes. So those two horizontal line segments would be connected.

    Inelastic conditions are a different question though.
     
  8. Feb 12, 2012 #7
    Cool, in the inelastic condition it's losing energy and assumed to be over an infinitesimally small time interval - so I think we just get lines from -q to +q which occur over a number of momenta values that are ever decreasing.
     
  9. Feb 12, 2012 #8

    Redbelly98

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    Yup, you got it.
     
  10. Feb 12, 2012 #9
    Cheers man, thank again!
     
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