HomeTextbookMechanics 7: Conservation of Momentum and Collisions

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Mechanics 7: Conservation of Momentum and Collisions — 2 Comments

  1. Hello Doc. This is a timely entry, thanks!

    We are teaching momentum right now, and ran into something interesting. In the case of a ball hitting a stationary wall (one-dimensional, perfectly elastic collision), the conservation of momentum equation ALONE does not work. One needs to add conservation of energy, in order for things to make sense. But the kid is not there yet: we have not talked about kinetic energy formula — we proceeded directly to momentum because we had covered m and v, touched upon vectors, so we were ready to mesh them together and introduce momentum, p.

    So we are here:

    m1v1 (initial) + m2v2 (initial) = m1v1 (final) + m2v2 (final)

    But the wall (2) is always stationary, so its v2 = 0, both in the initial and final states.
    It follows that
    m1v1 (initial) = m1v1 (final)
    Which only works in the scalar sense, as the velocity vector has changed direction.

    Yes, yes: Conservation of energy makes sense of it. But the child is not there yet, and we are looking for a way to make sense of momentum WITHOUT introducing energy into the picture. Is there a way to explain what happens in this perfectly elastic, one-dimensional, collision of a ball vs. stationary wall, so that it makes intuitive sense to the kid while demonstrating the conservation of momentum?

    Thanks,
    Syd

    • Syd — thanks for the comment. Glad you are finding my material useful.
      Sorry for the long delay in replying — I had to think about this question for a few days.

      The problem is that a ball hitting a wall is “strange math.” It works, but you have to “interpret” the results.

      Here’s the problem: It’s true that the wall has velocity v=0, but it’s also true that (compared to the ball) it has infinite mass. So, when you calculate its momentum, you have (infinity x 0). That’s troublesome for a direct calculation. The “full” answer requires calculus and limits, but we can approximate it.

      Let the wall’s mass be very large — say 1000 kg. Then let its velocity be something very small — say 0.001 m/s. Now, do the elastic collision calculation with these numbers, but interpret anything that comes out very small as being 0, and anything that comes out very large as being infinity. I did it, and I get just the answer you’re looking for. I let the ball have a mass of 3 kg and velocity 5 m/s. I got a final velocity for the ball of -4.97 and the wall 0.031.

      Let me know if you want further clarification.

      Hope this helps! Thanks for the questions —
      Doc

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