Why does a ping pong ball bounce higher when it is dropped together with a cup of water?

I've confirmed the experiment, using a McD_n_lds paper drinks cup and a beer can hollow plastic ball of about $5\mathrm{g}$, of about the same diameter as a ping pong ball (PPB):

The observed effect depends largely on the cup being soft and permanently deformable (like an object made of blutack or playdough), so its collision with Earth is inelastic. A stiff, hard cup (made of steel e.g.) would not work the same way here. The inelastic collision of the ensemble causes kinetic energy of cup and water, post-collision, to be small.

The PPB bounces back quite high (from a quarter-filled cup) and the cup of water loses quite little water and doesn't really bounce at all. It's quite a sight to behold! A simple model can be set up a follows.

We can write with Conservation of Energy (the collision is clearly not elastic - as evidenced by the permanent deformation of the bottom of the cup):

$$(M+m)gH=mgh+W+\Delta Q+K_{M+m}$$

where:

• $M$ is the mass of water plus cup and $m$ is the mass of the PPB
• $H$ is the height from which the cup, water and PPB are dropped and $h$ is the rebound height of the PPB, after the ensemble hits the Earth
• $W$ the work done on the cup's bottom
• $\Delta Q$ heat energy dissipated by various non-conservative forces
• $K_{M+m}$ the kinetic energy of water and cup, post-collision with Earth.

Trouble is, we don't know the value of $W+\Delta Q+K_{M+m}$. Direct observation suggests it is small, so we can write:

$$(M+m)gH\geq mgh$$

Or:

$$\boxed{h \leq H\Big(\frac{M+m}{m}\Big)}$$

If $M\gg m$ we can further approximate:

$$h \leq \frac{M}{m}H$$

I wanted to confirm experimentally the effect of $M$ on $h$.

Using a nearly empty cup, one half-filled and one filled completely I can confirm increased $M$ increases $h$.

Some further experiments are planned.

As mentioned in the comments above, the ball in the cup is similar to Galilean Cannon. The maximum height to which the ball can bounce $h_{max}$ can be estimated using the law of energy conservation: $$(m+M)gH=mgh+E_{cup}+E_{water}+E_{heat},$$ where $m$ is the mass of the ball, $M$ is the mass of cup+water, $H$ is the initial height from which the ball was thrown, $E_{cup}$, $E_{water}$ and $E_{heat}$ are the energy of cup, water and heat (due to dissipation). The maximum height corresponds to $E_{cup}=E_{water}=E_{heat}=0$. $$h_{max}=\frac{m+M}{m}H$$

As compared to the result by @Gert, for $M\gg m$, $h_{max}$ is proportional to $M$ not $M^2$. The latter would contradict the conservation of energy.

Recall that if a ball normally hits a wall elastically, its velocity will be exactly reversed.

Suppose the whole system hits the ground with speed $v$. Now, as the cup and the water hits the soft mat, their speed quickly reduces, and may start moving upward (depending on how soft the mat is) before the ping-pong ball is affected by a reaction force. Suppose the speed of the cup (and the bottom part of water) becomes $u$, along the upwards direction.

Let's go to the cups frame. Now the ball (and the top level of water) is hitting it with speed $u + v$. If the cup were much (actually infinitely) heavier than the ball, the ball would rebound at speed $u + v$ in this frame (the cup acts like a wall). Since the cup itself was moving upward at speed $u$, the upward velocity of the ball in ground frame will be $2 u + v$.

Now in the actual experiment, the collisions are not elastic, the cup's velocity does not change instantaneously, and the cup is not so heavy compared to the ball. So the final upward velocity of the ball be less than $2u + v$, but the above argument shows why it is greater than $v$.

Why Energy Conservation still holds: Since the cup and most of the water does not bounce back to their initial position, their initial potential energy is available to be converted into the extra kinetic energy of the ball, and the energy absorbed by the mat and water.

As mentioned in the comments, this is similar to a Galilean cannon.