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?

• You seem to fully answer your question within your post. What exactly are you looking for in an answer? "Yes"? – Dvij D.C. May 30 at 1:23
• "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? – Alfred Centauri May 30 at 1:29
• Yeah, the change in momentum. – David Dong May 30 at 1:31
• In this case, we can't conserve momentum because there is an external force acting on the wall which is preventing it from moving. – SarGe May 30 at 1:41

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.

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).