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Suppose I throw a ball horizontally towards a wall with momentum $\vec p$. Let it collides with the wall and then rebound back towards me with momentum $-\vec p$. Since the wall remains stationary, the initial momentum is $\vec p$ and the final momentum is $-\vec p$. But since there is no external force on the ball+wall system, shouldn't the total momentum be conserved? I am ignoring gravity here. Say we're doing this experiment in a freely falling elevator and the wall is that of the elevator.

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    $\begingroup$ You should take a closer look at your assumption that the wall remains stationary. $\endgroup$
    – my2cts
    Apr 14, 2022 at 16:10
  • $\begingroup$ The wall in a freely falling elevator will certainly move; stationary walls which are anchored to the ground will not move. However, the effect seen may be negligible if the mass of the elevator is much higher than the ball $\endgroup$
    – Righter
    Apr 14, 2022 at 16:22
  • $\begingroup$ "Conservation of momentum" is only true when you study a closed system (i.e., a system that is not subject to any external force.) When you are standing on the Earth, and you throw a ball at a wall that is anchored to the Earth, what is the boundary that encloses the entire system (i.e., the boundary through which no forces cross?) $\endgroup$ Apr 14, 2022 at 16:32
  • $\begingroup$ There is an external force preventing the wall from collapsing or displacing, and making it rigid. $\endgroup$
    – Cross
    Apr 14, 2022 at 16:39
  • $\begingroup$ Does this answer your question? Where is the momentum going? $\endgroup$
    – user50229
    Apr 14, 2022 at 16:52

2 Answers 2

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The elastic collision formula is: $$ v_m=\frac{m-M}{m+M}u_m +\frac{2 M}{m+M}u_M $$ $$ v_M=\frac{2 m}{m+M}u_m+\frac{ M-m}{m+M}u_M $$ where $u_i$ is the initial velocity of object $i$ and $v_i$ is the final velocity of object $i$, and $m$ and $M$ are the masses of the two objects. This formula is derived from the conservation of momentum and the conservation of energy. So we know that with any solution of these equations we are guaranteed that both energy and momentum are conserved.

For $u_M=0$ if we take the limit as $M \rightarrow \infty$ we get $$\lim_{M\rightarrow \infty} v_m = -u_m$$ $$ \lim_{M\rightarrow \infty} v_M =0$$

So momentum is still conserved, but for a very large mass the change in velocity approaches 0.

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I think @my2cts 's comment has the answer. Let me elaborate. Suppose I throw a ball horizontally towards a wall of a freely falling elevator (so as to ignore gravity) with momentum $\vec p$. Let it collides with the wall and then rebound back towards me with momentum $-\vec p$. Since the total initial momentum is $\vec p$ and the final momentum of the ball is $-\vec p$, and there is no external force on the ball+elevator system, for the conservation of momentum to hold, the momentum gained by the wall must be $2\vec p$. However, since the mass of the elevator is huge, its velocity will be imperceptibly small.

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    $\begingroup$ Were you in contact with the elevator when you threw the ball? Did you recoil into it? $\endgroup$
    – DJohnM
    Apr 14, 2022 at 21:23

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