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Why - momentum must be in the wall?

  1. Jan 25, 2009 #1
    I was watching MIT Classical Mechanics Lecture16. On completely elastic collision, (at around 24min), professor Walter Lewin shown that when a tennis ball with mass m and speed v hitting the wall, it bouced back with speed v.
    (Cut from the lecture)
    ...Kinetic energy is conserved.
    All the kinetic energy is in the tennis ball; nothing is in the wall.
    The wall has an infinitely large mass, but the momentum of this tennis ball has changed by an amount 2 mv.

    That momentum must be in the wall--it's nonnegotiable, because momentum must be conserved.

    So now here you see in front of your eyes a case that the wall has momentum, but it has no kinetic energy.

    That the wall has momentum 2 mv, it's nonnegotiable.
    It must have momentum, and yet it has no kinetic energy...

    How can that be? The wall is NOT moving, how can it has momentum?
    I velocity for wall is zero, there is NO kinetic energy nor momentum, correct? But then why did he say that? Was it a trick question?
  2. jcsd
  3. Jan 25, 2009 #2

    Vanadium 50

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    v = p/m. If m is infinite, and p is finite, what is v?

    (If you object to "infinite", recognize that the wall is attached to the Earth and plug the Earth's mass in - what is v? Would you notice a change of velocity of that magnitude?)
  4. Jan 26, 2009 #3
    I'm really confused by this. The lecturer said the ball bounces from the wall with twice the momentum it had when it hit the wall?
  5. Jan 26, 2009 #4

    Doc Al

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    No, its change in momentum equals twice its original momentum.

    Original momentum of ball: +mv
    Final momentum of ball: -mv
    Change in momentum of ball: -2mv
    Change in momentum of wall: +2mv
  6. Jan 26, 2009 #5
    Your feet are rigid with the ground, and the ground is rigid with the wall.
    When you throw the ball, it's got +mv, you incur -mv for your troubles.
    When the ball hits the wall and stops, the ball looses mv and is at rest, and the wall/floor/you system gains back it's lost mv.
    But the ball is compressed so it recoils back with -mv, giving the wall a push for +mv.
    When you finally catch the ball, it relinquishes it's mv momentum into you, setting us right back where we started.
  7. Jan 26, 2009 #6
    And if we were to be fussy about this which is probably necessary to clear up any confusion there are two main points which are relevant
    1.Even ignoring gravity air resistance and so on the ball cannot bounce back with velocity v.That would be an example of a perfectly elastic collision and such collisions do not occur with macroscopic objects(they can and do with atomic scale objects for example gas atoms provided that the energy of collision is not too high)
    2.The wall and everything it is attached to does move in order to conserve momentum and therefore it does pick up kinetic energy.Because of the difference in mass the energy may be vanishingly small but it is not zero and can only become zero if the wall etc had an infinite mass which it doesnt.
  8. Jan 26, 2009 #7
    Thanks, Doc Al. I think I get it: since the original momentum of the ball was +mv the original momentum of the wall was automatically -mv. The collision causes both signs to be reversed, and the difference between the initial and final is 2. The reason we reason that the ball has undergone a change in momentum of minus 2mv is that, to go from +mv to 0mv an equal and opposite momentum of -mv had to be given to it, and then to go from 0mv to -mv another equal and opposit momentum of -mv had to be applied. Therefore the change in momentum is -2mv. It's all about the vectors. Please correct me if I screwed that up somewhere.
  9. Jan 26, 2009 #8

    Doc Al

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    You have it exactly right.
  10. Jan 26, 2009 #9
    Thank you for the replies.
    I think mathematically, I can work out the answer.
    Simply from the fact that it is an elastic collision (assuming zero energy loss) And using conservation of energy and conservation of momentum.

    v1' = (m1-m2)/(m1+m2)v1
    v2' = 2 m1/(m1+m2)v1
    and if m2 approaches infinity, v1' = -v1 and v2'=0

    I even played with assuming m2 is 100x m1 to see how these terms vary.

    However, it is interesting and difficult concept to grasp that the change in momentum absorbed by the wall is 2m1v1 when the wall velocity (v2') is zero. That is, it has zero final kinetic energy but it gain 2m1v1 momentum.
  11. Jan 26, 2009 #10
    Those of you who are just starting Walter Lewin's online lectures are in for an amazing experience. Don't stop at 8.01.
  12. Jan 26, 2009 #11
    Aha! Great! Thanks much!
  13. Jan 26, 2009 #12
    I need to add a few extra points of clarification.
    1.An elastic collision is one where there is no loss of kinetic energy not one where there is no loss of energy.Energy is always conserved.
    2.The collision between the ball and the wall is not elastic.If you could make a ball that bounced elastically with the floor you could drop it from a fixed height and it would bounce forever.
    3.The wall etc does move and does get a bit of kinetic energy.
    The professor should have made it clear that he was making simplifying assumptions and I am guessing that he was going on to derive the basic ideal gas equation where it can be assumed that at low enough temperatures gas molecules do indeed collide elastically.
  14. Jan 26, 2009 #13
    Correction: there is no loss of kinetic energy though the statement that there is no loss of energy is also true just not relative.
  15. Jan 26, 2009 #14
    Read post number 9 and you will see why my reference to energy is relevant.
  16. Jan 26, 2009 #15
    Thank you Dadface. That is clear, my issue was with the wall and I now see where and why I was confused.

    This another clarification/correction you wrote made me realized that I put down "zero energy loss" on post 9. You are absolutely correct. I can visualize the detail much better now.

    Thanks again and thank you all.
  17. Jan 26, 2009 #16


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    The main issue I have with the original problem statement is that the wall has infinite mass. This is createing all sort of problems, for example what is the momemtum of an object with infinite mass?

    Change this to a wall with a very large but finite amount of mass and everything works. You could have the wall attached to the earth, and then treat the path of the ball as a sub orbit and realize that angular momentum of ball, wall, and earth are preserved despite any energy losses in the collision.
  18. Jan 26, 2009 #17
    The original momentum of wall isn't -mv . From an inertial frame that records the velocity of the ball to be v, the wall just rests like a puppy with zero velocity as per the given statements by walter. So zero momentum.
  19. Jan 27, 2009 #18
    I see your point. "Original" is certainly not the right way to put what I meant. Which was: that during the collision the wall expresses a momentum equal to the balls, opposite in direction to that of the ball. My reasoning was that the wall only manifests momentum in direct response to being impinged upon by the ball. However, I will let Doc Al sort this out.
  20. Jan 27, 2009 #19
    I agree with Jeff Reid.Infinities,often accompanied by zeros,pop up in many areas of physics and if we were to take these literally it leads to contradictions and conceptual difficulties.Usually if we regard something as being infinite or, zero,we are doing nothing more than making simplifying assumptions.As an example when the ball hits the wall we can to an excellent approximation ignore the velocity change and kinetic energy etc of the wall.
  21. Jan 27, 2009 #20

    Doc Al

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    You did garble it up a bit when you said that the original momentum of the wall was "-mv", which I should have corrected. (Sorry about that.) The key point, which I think you understand, is that the net change in momentum of the "ball + wall" due to the collision is zero. Thus the change in the ball's momentum must be equal and opposite to the change in the wall's momentum.

    Let's assume, as sganesh88 stated, that the initial speed and momentum of the wall is exactly zero. Recapping what I said earlier, this time adding the initial and final momenta of the wall:

    Original momentum of ball: +mv
    Final momentum of ball: -mv
    Original momentum of wall: 0
    Final momentum of wall: +2mv
    Change in momentum of ball: -2mv
    Change in momentum of wall: +2mv

    Make sense?
    Last edited: Jan 27, 2009
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