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I never did Physics in university and I consider that a mistake so I am correcting that now by teaching myself. To that extent I have been watching the MIT lecture videos by Walter Lewin and I am currently up to Lecture 16. The lecture is about Elastic Collisions.

In this lecture he poses this question which I invite you to watch (The link should take you to the correct time in the video but if it does not then the question is posed at 22m52s).

I suspect that the reason that the wall can have that momentum but no Kinetic Energy is that:

$$p = mv$$ $$KE = \frac{mv^2}{2}$$

But in this case m is extremely large and therefore v is going to be very very small even when it is squared. Thus, if it happened to be a completely elastic collision and the ball bounced off then the value of 'v' for the wall would be nothing and the laws of physics hold. So I think that the wall will have a velocity of zero in the question and thus doubling it is zero.

I suspect that in reality 'v' for the wall is not quite zero so the wall does get some small amount Kinetic energy. Therefore it will need to loose that energy somehow (otherwise you could keep throwing tennis balls against the wall until it exploded with energy). My suspicion is that walls loose this in vibrations and heat. I also suspect that is why floppy pieces of metal wobble around for a bit when you throw a ball against them.

I obviously cannot ask Walter Lewin if I got that correct so I am asking here. Am I on the right track or did I get that completely wrong? Thankyou so much for your help.

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If you are interested in motion of the wall, it is negligible, the wall velocity is zero. Now what the use of the wall momentum and energy, if the wall does not move? – Vladimir Kalitvianski Apr 14 '12 at 9:33
up vote 2 down vote accepted

You are right. Supposing that the wall (and in fact the whole Earth) is a perfectly rigid body, it will have both momentum and kinetic energy after the collision. And kinetic energy, expressed as huge mass - tiny velocity squared, will be negligible, while momentum expressed as huge mass - tiny velocity will not be negligible.

$$M \gg m, V \ll v$$

$$\frac{M V^2}{2} \ll \frac{m v^2}{2}$$

$$m v \approx M V$$

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Let me add a comment showing that the statements are not independent. It's useful to write the kinetic energy in terms of the momentum $P$, namely $E=(mv)^2/2m = p^2/2m$. Now, $p$ of the wall and the object are the same, up to the sign, due to the conservation of the momentum i.e. action and reaction. So you see that $p^2/2m$ is a decreasing function of $m$ for a fixed $p$, so the heavier object you have in the collision, the smaller kinetic energy it will carry. – Luboš Motl Apr 14 '12 at 10:04
Right! Good point, Luboš. – Pygmalion Apr 14 '12 at 11:42
Thankyou for the good answer Pygmalion and that is a clever rearrangement of the Kinetic Energy equation Lubos. I will keep that in mind for the future and try to remember to apply it in problems. – Robert Massaioli Apr 15 '12 at 22:05

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