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A ball of mass $m$ and speed $v$ strikes a wall perpendicularly and rebounds with undiminished speed. If the time of collision is $\Delta{t}$, what is the average force exerted by the ball on the wall?

My thoughts were that we would simply use $\Delta{p}=-mv$ and continue on, but the given solution is

The change in momentum is $\Delta{p}=(-mv)-mv$; the average force is $F=\Delta{p}/\Delta{t}=-2mv/\Delta{t}$.

Why is $\Delta{p}$ equal to $-2mv$ instead of $-mv$? Is it because the ball bounces off the wall instead of staying but going to rest in $\Delta{t}$? I ask this because this is the only difference between this problem and some others, so it is probably why. I don't understand why the ball bouncing off the wall makes the change in momentum momentum $-2mv$, then.

For example, if the problem statement was instead "A car of mass $m$ and speed $v$ collides with a wall, and is brought to rest in a time of $\Delta{t}$, would $\Delta{p}=-mv$ this time?

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    $\begingroup$ You seem to fully answer your question within your post. What exactly are you looking for in an answer? "Yes"? $\endgroup$ – Dvij D.C. May 30 at 1:23
  • $\begingroup$ "I don't understand why the ball bouncing off the wall makes the momentum −2mv, then." - have you misread the solution (or misstated it)? It's the change in momentum that equals $-2mv$, not the final momentum, correct? $\endgroup$ – Alfred Centauri May 30 at 1:29
  • $\begingroup$ Yeah, the change in momentum. $\endgroup$ – David Dong May 30 at 1:31
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    $\begingroup$ In this case, we can't conserve momentum because there is an external force acting on the wall which is preventing it from moving. $\endgroup$ – SarGe May 30 at 1:41
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Assuming direction of velocity of the ball initially to be positive:

Before Collision:
$$p_i= mv$$

After Collision:
$$p_f = -mv$$ Since speed, that is magnitude of velocity, is same but direction of velocity is opposite.

So, $$\Delta p=p_f-p_i=-2mv$$

For example, if the problem statement was instead "A car of mass $m$ and speed $v$ collides with a wall, and is brought to rest in a time of $\Delta{t}$, would $\Delta{p}=-mv$ this time?

Yes, since the final velocity is a zero vector and so is the momentum of the car.

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The change in momentum is $-2mv$ because it not only adds the momentum $-mv$ (to the left), but it also adds $-mv$ to stop the ball in the first place (since it has a momentum of $mv$ to the right).

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