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Perfectly elastic/inelastic collision

  1. Mar 16, 2010 #1
    Pendulum mass m released from stop h above bottom of arc.
    within plane
    m hits M at bottom of arc.--how high does m+M go?
    Perfectly inelastic collision.
    Kinetic energy is not conserved--work is done (deformation, noise, temp etc)
    but momentum is conserved.
    how is this possible?
    If there is loss of energy, how can momentum be conserved?
    What am I missing?
  2. jcsd
  3. Mar 16, 2010 #2


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    Why not?

    It's possible because energy and momentum are two totally different things. Take another look at the definitions for momentum and kinetic energy, and you'll see that the condition for conservation of momentum imposes a particular constraint on the system (represented by an equation relating the quantities before and after the collision) The condition for conservation of kinetic energy (if it were met) would impose a totally different constraint on the system (represented by another, different equation that relates the quantities before and after the collision).
  4. Mar 16, 2010 #3
    In a perfectly inelastic collision, does
    mv^2/2 + mgh = =(m+M)V + (m+M) gH with H the height after low point ?

    If seems not if there is energy dissipated in inelastic colision.

    If the post collision Velocity is effected by the loss, than why isn't momentum effected"
  5. Mar 16, 2010 #4
    Hmmm, am I having difficulty as momentum is a vector quantity with energy a scaler?
  6. Mar 16, 2010 #5


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    I'm going to assume that you meant V2. No, this equation is not true. Mechanical energy is not conserved here.

    Of course! That's the definition of an inelastic collision -- a collision in which kinetic energy is lost.

    The simple answer is, "because, in this situation, it is possible for the velocity to change while the momentum does not." If that still doesn't make sense, maybe this will help. Conservation of momentum before and after the collision says that:

    [tex] mv = (m + M)V [/tex]​

    with the result that the velocity after is lower than the velocity before:

    [tex] \left(\frac{m}{m + M}\right)v = V [/tex]​

    Now, the kinetic energy before the collision is just (1/2)mv2. The kinetic energy after the collision is:

    [tex] \frac{1}{2}(m+M)V^2[/tex]

    If you substitute in the above expression for V, you end up with the result that the kinetic energy afterwards is:

    [tex] \frac{1}{2} \left(\frac{m^2}{m+M}\right)v^2 [/tex]​

    So, as you can see, as long as M is not zero, the final energy is always lower than the initial energy.

    Summary: momentum is always conserved. But, as I have shown above, if momentum is conserved, and the two masses stick together, then kinetic energy cannot also be conserved -- it is impossible. The final kinetic energy will always be lower. Therefore, if the two masses stick together, you know that an inelastic collision has occured. In an elastic collision, the first mass would have bounced back.
    Last edited: Mar 16, 2010
  7. Mar 16, 2010 #6


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    In general you have to be careful with that, although in this particular case, at the instant of collision, it's sort of a 1D problem, so it doesn't really matter too much.
  8. Mar 16, 2010 #7
    cepheid--thank you for your reply--I have a better understanding.
    It seems that one may have a whole range of inelastic collisions from near totally inelastic collisions with extensive energy loss and collisions with very little energy loss(hard billiard balls with a latch to connects them after collision)--assume no energy loss.
    .5mv^2 + mgh =.5(m+M)V^2 + (m+M)gH
    Would the above equation then be usable for calculating H(V=0)
    without using momentum conservation?
  9. Mar 16, 2010 #8
    From lack of response, I wonder if my model of elastic masses with frictionless silent rigid latches is acceptable in the context of this experiment. Any thoughts?
  10. Mar 17, 2010 #9


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    I am not sure if it's really realistic to think that you can "latch together" the two masses without energy being lost. The math shows that if the final mass is m+M, AND momentum is conserved, then energy is lost. So, one of these two conditions (energy cons. or momentum cons.) must be relaxed if the situation is to be logically consistent - they cannot both be true. I'm not sure which one would be violated (maybe both). It is possible that momentum would not be conserved in that situation, since each ball would then have an outside force acting on it (from the latch). That's the best I can think of right now.
  11. Mar 17, 2010 #10
    Thanx again.
    It sounds like I may have been proposing something like Maxwell's demon.
    I know it is not possible to latch the masses together without energy loss, but it may be possible to minimize this loss and therefore approach conservation of energy.
    If I can design it, maybe I should try to do the experiment!!
    In any event, be well.
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