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Conservation of energy and momentum transfer

  1. Dec 14, 2015 #1
    Let's say a tennis ball with velocity with only an horizontal component hits a vertical wall at rest.

    After collision, conservation of momentum tells that :

    [tex]m_{wall}v_{wall} = 2m_{ball}v_{ball}[/tex]

    Thus, the wall has now a (tiny) velocity and kinetic energy :

    [tex]v_{wall} = \frac{2m_{ball}v_{ball}}{m_{wall}} , K_{wall} = \frac{2m_{ball}^2v_{ball}^2}{m_{wall}}[/tex]

    while the kinetic energy of the ball didn't change. Where does the "new" energy in the wall come from ?
     
  2. jcsd
  3. Dec 14, 2015 #2

    mfb

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    This is not true. If the wall has a finite mass (if not, that equation does not make sense), the ball will be slower after the collision.
     
  4. Dec 14, 2015 #3

    Doc Al

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    You merely assumed that the final speed of the ball equals its initial speed. A good approximation, but an approximation nonetheless. In reality, it slows down.

    Also realize that the wall is most likely attached to something (the ground, for instance) making its effective mass much larger (>>) than that of the ball.

    (Edit: I see mfb making the same statement.)
     
  5. Dec 14, 2015 #4
    Thank you guys for answering. So you tell me a purely elastic collision in such a case is not physical, all right. Are there some conditions where purely ellastic collision is possible ? Or would it be a violation of conservation of energy (assuming no object can have an infinite mass...not sure this applies to black holes though...) ?

    Or maybe 100% reflection of light ? Is a 100% reflecting mirror physical ? Actually I was initially trying to understand why a 100% reflected beam of light on a mirror doesn't transfer energy to it.
     
  6. Dec 14, 2015 #5

    Doc Al

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    The issue here is not one of elastic vs inelastic collision. You can assume a perfectly elastic collision. (Unrealistic, of course, but why not?) That just means that the total KE will remain the same. If the wall moves, it takes with it some of that KE, leaving less for the ball. (Just depends on how exactly you want to calculate things.)
     
  7. Dec 14, 2015 #6
    But if the wall takes some of the KE of the ball with it, doesn't that imply that the speed of the ball after the collision is less than before the collision ? Thus implying a partially elastic collision only?

    Actually either the speed of the ball is less, or its mass. But that doesn't sound like something physical either.
     
  8. Dec 14, 2015 #7

    Doc Al

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    A perfectly elastic collision is where the total KE remains the same. Not the KE of each participant separately.
     
  9. Dec 14, 2015 #8
    Oh, right ! Big mistake for me there, thank you.
     
  10. Dec 14, 2015 #9

    mfb

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    Just a small side-remark:
    Black holes have a finite mass.
     
  11. Dec 14, 2015 #10
    Thanks for the precision. As I don't know anything about GR, I won't ask further questions regarding black holes. :D

    Concerning photons "bouncing" off a mirror, is it safe to assume their speed after collision is the same as before ? Not sure how this agrees with momentum conservation though, since I have heard photons have no "rest" mass. Probably too much out of classical physics knowledge as well?

    EDIT : ahhh yes I recall black holes have infinite density, not mass!
     
  12. Dec 14, 2015 #11

    jtbell

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    Yes.

    For a photon, p = E/c.

    An example of energy and momentum conservation involving photons that is commonly studied at the introductory undergraduate level, is Compton scattering. An incoming photon scatters off a stationary electron, and the two particles recoil in different directions.

    https://en.wikipedia.org/wiki/Compton_scattering#Derivation_of_the_scattering_formula

    Note that we have to use relativistic equations for energy and momentum, not classical.
     
    Last edited: Dec 15, 2015
  13. Dec 15, 2015 #12
    Thank you !
     
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