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Setup: an official ping pong ball is floating inside a party plastic cup filled with clean water, which is then dropped from a certain height onto a soft mat.

Observation: the ping pong ball shoots up to a height which is much higher than its initial position.

Question: why does the ping pong ball do that? Why didn't the water and soft mat absorb the kinetic energy? Is this an inelastic collision? enter image description here

PS: the first time it was an accident, the second time the soft mat and I were thrown out XD

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Jun 29 at 17:27
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    $\begingroup$ There is a similar effect when stacked balls are dropped . $\endgroup$ – rcgldr Jun 30 at 8:00
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    $\begingroup$ If Newton had been a college undergrad, this is the experiment he would have performed... have you tried substituting the liquid for beer? honey? $\endgroup$ – smci Jul 1 at 0:04
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    $\begingroup$ @smci: good idea, I think honey might be sticky though ;D $\endgroup$ – user6760 Jul 1 at 0:08
  • $\begingroup$ user6760: yes we know honey is sticky, we're trying to validate for multiple values of (viscosity/ η "eta") aka the ratio of shearing stress (F/A) to velocity gradient (∆vx/∆z or dvx/dz) in a fluid. Science... $\endgroup$ – smci Jul 1 at 0:10
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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):

Cup, ball and lemon

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.

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    $\begingroup$ @PhilipRoe Not that I know of. I plan to do it. $\endgroup$ – Gert Jun 29 at 16:42
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    $\begingroup$ @Gert Excellent. My prediction is that the ball will rise higher/ $\endgroup$ – Philip Roe Jun 29 at 16:54
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    $\begingroup$ What's the lemon for? $\endgroup$ – Karl Jun 30 at 0:49
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    $\begingroup$ @Karl: Size comparison, then lemon tea! ;-) $\endgroup$ – Gert Jun 30 at 5:27
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    $\begingroup$ A Lemon Tea my dear Watson. $\endgroup$ – Russell McMahon Jun 30 at 7:09
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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.

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  • $\begingroup$ My derivation starts from the idea the collision is INELASTIC, so kinetic and potential energy are NOT conserved. Hence the use of conservation of momentum. Your $E_{cup}=E_{water}=E_{heat}=0$ argument just makes me giggle! $\endgroup$ – Gert Jun 28 at 16:07
  • $\begingroup$ In the 'final analysis' I don't think it matters much: we're not here to create a precise model, although it would be possible to refine what we have and empirically verify it. $\endgroup$ – Gert Jun 28 at 16:09
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    $\begingroup$ The law of energy conservation is still valid: the work done to permanently deform the cup changes its energy and is partially turned into heat. See the first law of thermodynamics $\endgroup$ – atarasenko Jun 28 at 17:37
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    $\begingroup$ The law of energy conservation is certainly valid if you take all energies into account. Which you don't do! $\endgroup$ – Gert Jun 28 at 17:40
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    $\begingroup$ You are correct and I'm wrong. I edited my answer. Thank you for pointing out my error. +1 from me. $\endgroup$ – Gert Jun 29 at 6:11
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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.

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My hypothesis why the ping pong ball receives a large upward impulse:

The floating ping pong ball is displacing some water. The amount of displacement does not change much during the fall.

As the cup hits the floor the deceleration of the quantity of water gives a short pressure peak. Because of that pressure peak the water that is in contact with the ping pong ball is (briefly) exerting a much stronger force upon the ping pong ball. The water reflows, moving down along the walls of the cup, and moving up along the central axis. Thus the ping pong ball receives a large impulse.

It may even be that there is a secondary effect. It may be that the peak in the force exerted on the wall of the cup causes elastic deformation of the cup wall, and as the cup wall bounces back all of that motion focusses on the central axis of the cup, which is right where the ping pong ball is located.

It may well be that after kicking up the ping pong ball the water is left with little energy, so it remains in the cup. My guess is that without the ping pong ball the water will predominantly jump up along the central axis.


This suggests a comparison experiment.

This suggested setup will require some manufacturing. Instead of a cup (which is tapered) a cylinder must be used, and instead of a ball a second cylinder must be used (short, closed at both ends), this second cylinder must slide freely inside the first cilinder. I will refer to these two as 'the cylinder' and 'the piston'. (Of course the cylinder, like the cup, must be closed at one end)

Before release water should not be allowed to enter the gap between piston and the cylinder. (During the fall both will be weightless; not much water will penetrate the gap.)

Under those circumstances I don't expect the piston to bounce up, certainly not higher than the height of release.

The piston is flat underneath, so there is no opportunity for the water to reflow. I think it is the forced reflow that transmits the impulse to the ping pong ball, so I expect that when reflow is eliminated then the ooportunity for impulse transfer is removed.


In a comment and in an anwser it has been suggested that there is a similarity with the a Galilean cannon setup.
However, in the setup of this question the impulse is transferred to the ball by a fluid, which is incompressible. For comparison, imagine trying a Galilean cannon setup where both of the two balls are filled with water. That would not work, because the elasticity of the air in the balls is a crucial element. So, while there is some similarity, the differences are such that comparison with a Galilean cannon setup is not particularly helpful.

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  • $\begingroup$ Interesting as your idea is, it too is very different from the 'open cup and ball' set-up described by the OP. $\endgroup$ – Gert Jun 28 at 14:46
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    $\begingroup$ @Gert I'm not sure what the intention of your comment is. The very purpose of the comparison suggestion is to try a setup that is different. Specifically, the suggested comparison setup is designed to eliminate the very factor that I believe is crucial in the case of the cup-and-ball setup. If my explanation is correct the comparison setup will not display the shooting-up effect. (Independent of that: the comparison setup may be inconclusive anyway; too many uncontrollable factors.) $\endgroup$ – Cleonis Jun 28 at 17:11
  • $\begingroup$ I'm convinced that, using a soft cylinder, or one with a soft, deformable bottom, the ball will be shot up. $\endgroup$ – Gert Jun 28 at 17:18
  • $\begingroup$ Independent of that: the comparison setup may be inconclusive anyway; too many uncontrollable factors Nope. It's the job of the experimenter to eliminate most of these. Entirely feasible, is that. $\endgroup$ – Gert Jun 28 at 17:38
  • $\begingroup$ I posted a reply to part of your reply. $\endgroup$ – Gert Jun 29 at 9:04
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Assume that the water in the cup is compressible and inviscid, experiencing one-dimensional flow and thereby satisfying the one-dimensional Euler equations. Initial conditions, velocity =$\sqrt{gh}$ downward and pressure =1 atm, are both uniform. The bottom of the cup is struck from below in such a way that the velocity of the water is reduced and the pressure is increased, similarly to the well-known piston problem. This creates an upward-moving pressure wave inside the water, and produces a pressure gradient in the vertical direction. The pressure gradient creates an upward force on the PPB, instantaneously equal to the submerged volume times the magnitude of the gradient (Archimedes principle). This gives the PPB an initial acceleration, but only for a short period until the PPB leaves the water.

I believe that this has all the makings of a good explanation. but it is terribly hard to put numbers to. Even the decision to include compressibilty needs more justification than I have been able to muster. However, there are times when water at fairly low speeds must be considered compressible. An example is "water hammer" the noise made sometimes by domestic water pipes in response to the sudden closing of a tap. The speeds and decelerations involved might be quite similar.

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    $\begingroup$ Water is compressible ?? $\endgroup$ – Gert Jun 29 at 6:55
  • $\begingroup$ @Gert How do you think sound propagates through water? $\endgroup$ – Philip Roe Jun 29 at 13:48
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    $\begingroup$ My point was that liquids in general have very limited compressibility, which is why we can use the continuity equation on them. $\endgroup$ – Gert Jun 29 at 14:05
  • $\begingroup$ You should do some 'back-of-an-envelope' calcs with your model. $\endgroup$ – Gert Jun 29 at 14:08
  • $\begingroup$ @Gert. We can do the compressible continuity equation $\partial_t\rho+\nabla\cdot(\rho\vec{V})=0$ on anything. Water is not very compressible which it why the pressure changes will be large. If it was incompressible they would be infinite. The difficulty with "back of the envelope" calcs is too many events for which no simple model is available. What is the time scale on which the cup crumples?. How to describe the transient of the compression wave hitting the PPB?. These are order of magnitude uncertainties. DoD spends millions on computations of this complexity. $\endgroup$ – Philip Roe Jun 29 at 14:59
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This is a reaction to the otherwise fine answer of 'Cleonis'.

Here's is his set-up, as I understand it:

Cylinder and piston

The ensemble of cylinder, water and piston hits the Earth at $v_0$ because:

$$\frac12 v_0^2=gH$$

where $H$ is the drop height.

Due to the soft, inelastic cushion at the bottom of the cylinder, the restitution coefficient is $\text{zero}$ and the energy balance is:

$$(M+m)gH=mgh+\Sigma E$$

where $\Sigma E$ are various small energies described in my first post.

In the limit for $\Sigma E \to 0$, we get:

$$(M+m)H=mh$$

Note that a hole in the cylinder is needed as otherwise a partial vacuum between the 'escaping' cylinder and piston would arise.

Under those circumstances I don't expect the piston to bounce up, certainly not higher than the height of release.

So I believe this to be wrong.

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  • $\begingroup$ Have you lost a square in the first equation? $\endgroup$ – Ruslan Jun 30 at 14:27
  • $\begingroup$ @Ruslan Yes, edited. Thank you. $\endgroup$ – Gert Jun 30 at 14:29

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