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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

user6760
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6 Answers6

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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.

Rasmus Faber
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Gert
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  • @Gert were you able to observe any permanent deformation in the cup? I suspect not, and it isn't necessary to explain inelasticity in the collision. Kinetic energy is also lost to vibration, which would just as much affect a metal cup as a paper one, and indeed the cup is only a small fraction of the total mass of the system compared to the water so I think you'd find that the fluid dynamics - in particular the dynamics of the air-water interface - are more important factors in explaining the transfers of energy in the experiment (though these don't necessarily need explaining) – Will Jun 29 '20 at 09:50
  • @Will were you able to observe any permanent deformation in the cup? As clearly stated, YES. In one instance the cup actually BROKE! I think you'd find that the fluid dynamics - in particular the dynamics of the air-water interface - are more important factors in explaining the transfers of energy in the experiment Not in line with my observations, I'm afraid. – Gert Jun 29 '20 at 13:38
  • @Gert sorry, I skimmed over your notes about the bottom of the cup. It remains irrelevant unless you've repeated the experiment with a more robust cup that doesn't deform. I guarantee you still won't see one of those bounce elastically. And unless you can produce the phenomenon with an empty cup (besides the ball) or one filled with ice it would be rash to dismiss the role of fluid dynamics and an extrapolation from your observations to claim you've eliminated it. – Will Jun 29 '20 at 14:28
  • .It is claimed that this would not work with a tin cup. Did anyone try? – Philip Roe Jun 29 '20 at 16:21
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    @PhilipRoe Not that I know of. I plan to do it. – Gert Jun 29 '20 at 16:42
  • @Will Of course the permanent deformation is RELEVANT: that work enters the E equation. The bottom acts as a 'soft cushion', decelerating the ensemble more gently. It's no coincidence the OP used plastic party cups: they too deform easily. I'm not dismissing the role of fluid dynamics (straw man!), it's a different way of looking at things. I just wish you'd put a bit of 'meat on these bones', otherwise it's just talk and a waste of breath. – Gert Jun 29 '20 at 16:48
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    @Gert Excellent. My prediction is that the ball will rise higher/ – Philip Roe Jun 29 '20 at 16:54
  • @PhilipRoe On the basis of the E equation I'm inclined to agree: $W\approx 0$ for a sturdy can. – Gert Jun 29 '20 at 19:00
  • I've created a chat room to preserve the comments here. These are good comments, suggesting changes and improvements and pointing out possible shortcomings of the answer, but there are a lot of them and I wanted to delete the ones that have already been addressed. – David Z Jun 29 '20 at 20:09
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    What's the lemon for? – F Chopin Jun 30 '20 at 00:49
  • @Gert you've now made three different appeals to the importance of permanent deformation which you're throwing around interchangeably. Your answer opens by arguing that it's crucial for an inelastic collision and therefore the main reason the cup and water lose kinetic energy; your comments have argued that it's an important part of the E equation, even though the final inequality in your answer eliminates it, saying "it is small"; and then you claim the importance of deformation on the deceleration profile of the cup, which you don't mention in the answer. Which is it? – Will Jun 30 '20 at 05:14
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    @Will I'm getting VERY tired of answering a multitude of questions while trying to organise tests with a steel can at the same time. Please formulate your own answer if you don't like mine. Best regards. – Gert Jun 30 '20 at 05:27
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    @Karl: Size comparison, then lemon tea! ;-) – Gert Jun 30 '20 at 05:27
  • @Gert, it's the same one question, about improving your answer by correcting its erroneous first sentence. I'm not going to stand in your way if you're about to learn the same point experimentally. Your answer already contains good content which you're under no obligation to improve, whether in response to comments or new experimental data, so I'm sorry you've found my comments tiring. – Will Jun 30 '20 at 06:24
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    A Lemon Tea my dear Watson. – Russell McMahon Jun 30 '20 at 07:09
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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.

atarasenko
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  • 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! – Gert Jun 28 '20 at 16:07
  • 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. – Gert Jun 28 '20 at 16:09
  • And your (dismissed) energy balance doesn't take into account the work done to permanently deform the cup, as witnessed by me. – Gert Jun 28 '20 at 16:55
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    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 – atarasenko Jun 28 '20 at 17:37
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    The law of energy conservation is certainly valid if you take all energies into account. Which you don't do! – Gert Jun 28 '20 at 17:40
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    You are correct and I'm wrong. I edited my answer. Thank you for pointing out my error. +1 from me. – Gert Jun 29 '20 at 06:11
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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.

Archisman Panigrahi
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6

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.

Cleonis
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  • Interesting as your idea is, it too is very different from the 'open cup and ball' set-up described by the OP. – Gert Jun 28 '20 at 14:46
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    @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.) – Cleonis Jun 28 '20 at 17:11
  • I'm convinced that, using a soft cylinder, or one with a soft, deformable bottom, the ball will be shot up. – Gert Jun 28 '20 at 17:18
  • 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. – Gert Jun 28 '20 at 17:38
  • I posted a reply to part of your reply. – Gert Jun 29 '20 at 09:04
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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.

Philip Roe
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    Water is compressible ?? – Gert Jun 29 '20 at 06:55
  • @Gert How do you think sound propagates through water? – Philip Roe Jun 29 '20 at 13:48
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    My point was that liquids in general have very limited compressibility, which is why we can use the continuity equation on them. – Gert Jun 29 '20 at 14:05
  • You should do some 'back-of-an-envelope' calcs with your model. – Gert Jun 29 '20 at 14:08
  • @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. – Philip Roe Jun 29 '20 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.

Gert
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