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Collision, conservation of momentum and energy

  1. Aug 26, 2012 #1
    In a collision the colliding particles will in general have different velocities at different instants of time. Overall this means that the particles can easily cover different distances during a collision and thus have different amounts of work done on them. My question is:
    Will the difference in the distance covered by our two particles be in accordance with the difference of the change in kinetic energy between our two particles?
    This is probably a weird question, but I'm just trying to get a good intuition for what happens in terms of forces on the microscopic level during a collision. And with the above way of thinking it would more or less seem that the relative changes in kinetic energy of the particles are decided by conservation of momentum, and that just sounds a bit weird.
     
  2. jcsd
  3. Aug 26, 2012 #2
    During a collision, which means the bodies are in contact, the bodies move as essentially a single body, thus covering the same distances.
     
  4. Aug 27, 2012 #3
    But that doesn't make sense for consider a collision between a ball and a very heavy object (say the earth). Then the ball regains all its kinetic energy during the collision, and if the ball and earth do the same work on each other during the collision then the ball must at some point do negative work on the earth - how should that make sense when using a model with a massless spring between our two objects?
    Can you try to explain in detail what happens on a microscopic level in terms of forces for a collision between, let's say for instance, two objects of same mass, where one lies still initially. I think its hard to picture what happens in terms of forces, even if you put a spring between the objects.
     
  5. Aug 27, 2012 #4
    Why is that? Even in a fully elastic collision the entire kinetic energy is not regained, with the only exception of the second body completely immovable.

    I do not understand what you are saying here.

    We have to imagine both objects as springs. When they get in contact with each other, both will contract till their mutual velocity is zero. At this time the forces exerted upon each other will be maximal (and equal in magnitude, of course). Then they will start expansion and eventually separate, when the forces drop to zero.
     
  6. Aug 27, 2012 #5
    But often with heavy body you make the approximation that all kinetic is regained for the light body - why is that? Because that would essentially mean that no work has been on the light body in total - right? No total work => no change in kinetic energy. But then that would mean that negative work has been done on the heavy body - how should that be possible? You can see that negative work must have been done, since during the collision at first positive work is done on it as our particle collides with it. This positive work must be gained back afterwards.
    Now you will probably say - aha, he's oblivious to the fact that if the body is heavy and thus practically immovable, then the work done it will be very little. But then that gives a contradiction to what you earlier said about the bodies always doing the same amount of work on each other- since they move with same velocity.
     
  7. Aug 27, 2012 #6
    You are ignoring the fact that the bodies deform during the collision. How do you define velocity, or even position, for a continuously deforming body? Those are things that are meaningfully defined for rigid bodies. Which is not the case during a collision.
     
  8. Aug 27, 2012 #7
    So what you are saying is more or less that everyday things like forces and so on, really lose their meaning on a microscopic level, and thus I should not try to understand a collision in terms of forces, but rather realize that energy conservation is a deep concept that can always be applied with meaning?
     
  9. Aug 27, 2012 #8
    In classical mechanics you can use forces even at the microscopic level (assuming quantum effects are still negligible). But you cannot treat bodies as rigid at that level. You cannot use a single force vector, a single velocity vector, etc. You will have to use vector fields. Which is quite a bit more complicated even if you model your bodies as 1D springs. And the beauty of the laws of conservation is that they stay intact despite all those complications - so use them whenever possible.
     
  10. Aug 27, 2012 #9

    A.T.

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    What about two equal masses, one at rest in a fully elastic 1D collision?
     
  11. Aug 27, 2012 #10
    In this case the energy is fully transferred from one body to another. Not regained.
     
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