MHB What is the height of the tennis ball after bouncing off a basketball?

[Ackbach]

Here is this week's POTW:

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(N.B. I will give credit for this problem to where it's due, but only after next week.)

A tennis ball with (small) mass $m_2$ sits atop a basketball with (large) mass $m_1$. The bottom of the basketball is a height $h$ above the ground, and the bottom of the tennis ball is a height $h+d$ above the ground. The balls are dropped. To what height does the tennis ball bounce? Note: Work in the approximation where $m_1$ is much larger than $m_2$, and assume that the balls bounce elastically.

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[Ackbach]

No one answered this week's POTW, which is due to David Morin from Harvard. His solution follows:

Spoiler

For simplicity, assume that the balls are separated by a very small distance, so that the relevant bounces happen a short time apart. This assumption isn't necessary, but it makes for a slightly cleaner solution.

Just before the basketball hits the ground, both balls are moving downward with speed (using $mv^2/2=mgh$)
\begin{equation}\label{Eq:First}
v=\sqrt{2gh}
\end{equation}
Just after the basketball hits the ground, it moves upward with speed $v$, while the tennis ball still moves downward with speed $v$. The relative speed is therefore $2v$. After the balls bounce off each other, the relative speed is still $2v$. (This is clear if you look at things in the frame of the basketball, which is essentially a brick wall.¹) Since the upward speed of the basketball essentially stays equal to $v$, the upward speed of the tennis ball is $2v+v=3v$. By conservation of energy, it will therefore rise to a height of $H=d+(3v)^2/(2g)$. But $v^2=2gh$, so we have
\begin{equation}\label{Eq:Second}
H=d+9h.
\end{equation}

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¹ It turns out that the relative speed is the same before and after any elastic collision, independent of what the masses are. This is easily seen by working in the center-of-mass frame, where the masses simply reverse their velocities.