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Is momentum conserved when an object bounces back against a wall? The wall doesn’t move, but the object moves in the opposite direction. Assume this is an ideal, elastic collision.

If, initially, the momentum of the system came from the ball, and after the collision, the momentum of the system was also from the ball, but in opposite directions, then momentum would not have been conserved?

I have a feeling this is wrong, so anybody clarify this for me?

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lightweaver
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    This link explains it perfectly. – Arpan Banerjee Apr 09 '15 at 11:54
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    I'm sure this is a duplicate, though after a quick search I can't find an obvious duplicate. The answer is that the wall does move. The wall is connected to the Earth, and when the ball hits the wall it makes the Earth move to conserve momentum. However the mass of the Earth is so great the velocity change of the Earth is immeasurably small. – John Rennie Apr 09 '15 at 12:03

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Is momentum conserved when an object bounces back against a wall? The wall doesn’t move, but the object moves in the opposite direction. Assume this is an ideal, elastic collision.

I have a feeling this is wrong, so anybody clarify this for me?

Yes, it is wrong. You are wrong if you think that in an ideal, elastic collision the velocity of the bouncing object is exactly the same.

And also you are wrong if you think that the wall doesn't move. You can't see it move but it wouldn't only if it had infinite mass, which is impossible.

Suppose a mass of 1 kg hits a wall of 10, 000 kg at $v_0=10$ m/s. It has momentum 1*10 = 10 kg m/s. The formula for an elastic collision tells you that the wall will absorb momentum $ \frac {20}{10 001} = 2$ g m/s and the ball will keep $v'=9.9980002$ m/s.

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  • Will the ball still have the same kinetic energy? Say this is not a wall but the floor. Will it bounce to the same height? – lightweaver Apr 09 '15 at 12:36
  • @stygian John Rennie's comment answers your query. The earth will also get a very small velocity. For all practical (and ideal) purposes, you can say it will bounce to almost the same height. – Arpan Banerjee Apr 09 '15 at 12:38
  • @svetlana And also, shouldn’t the wall absorb a momentum of 20 kg ms^-1? I thought 0.002 was the speed? – lightweaver Apr 09 '15 at 12:39
  • @ArpanBanerjee So a ball losing height after bouncing for a while applies even in perfectly elastic collisions? – lightweaver Apr 09 '15 at 12:40
  • @stygian, No , conservation of KE and momentum refer to the system and not to a single body. If you sum up the KE of the object and the wall (I'm sure you can do it) you'll find it is still 50 J. The momentum of the wall must be divided by its mass 0.002/ 10 0000, that's the speed. Is everything clear? –  Apr 09 '15 at 12:47
  • Hmm. Thanks. One last thing. With an inelastic collision, say with a putty ball that just sticks onto the ground upon impact, are their (common) velocities 0, or are they also moving very minutely (as in this case for the wall/floor)? – lightweaver Apr 09 '15 at 12:57
  • @stygian, it is always the same, the only difference is that in an inelastic collision some Ke of the system is lost, and transformed in heat etc., doing work to crush the putty ball. Note that Ke is lost but the sum of all types of energy is conserved –  Apr 09 '15 at 13:00
  • In inelastic collision they'll move with a common velocity of $\frac{m}{m+M}v$ where $m$ is mass of ball and $M$ is mass of wall+earth. – Arpan Banerjee Apr 09 '15 at 13:01