The bounces that add up to a stop

A table-tennis ball dropped on a hard table makes a sound everyone recognises: a few distinct clicks, then clicks coming faster and faster until they blur into a buzz, then silence. The buzz is not the ball settling into some new kind of motion. It is the same bouncing, with each bounce a fixed fraction lower than the last and each gap between clicks a fixed fraction shorter. Followed to its logical end, the rule says the ball bounces infinitely many times. It also says the ball stops, at a definite moment a few seconds after it was dropped, and both statements are true of the same idealised ball. That a process with no last step can nevertheless be over is the whole content of a convergent series, and a bouncing ball is the most tangible one there is.

One number per impact #

What happens during an impact is complicated. The ball and the floor deform, a contact patch grows and shrinks, elastic waves run through both bodies, and some of the energy goes into heat and sound. Collisions are easier than forces precisely because none of that has to be known: momentum conservation and one extra number summarise the whole event. For a ball on a floor that is massive enough not to move, momentum conservation says only that the floor absorbs whatever impulse the ball needs, and the extra number — the coefficient of restitution $e$ — says how fast the ball leaves compared with how fast it arrived:

$$v_{\text{out}} = e\,v_{\text{in}}.$$

A perfectly elastic ball has $e = 1$ and bounces to the height it fell from for ever. A lump of putty has $e$ near zero and does not bounce at all. Everything real lies between, and the whole of this essay follows from the single assumption that $e$ is the same at every bounce — which, as the argument about Newton’s cradle found, is a rule smuggled in rather than a law, and which will fail in a precise way before the end.

With $e$ fixed, the bounces follow at once. A ball leaving the floor at speed $u$ rises for a time $u/g$ and falls for the same time, so it is in the air for $2u/g$, and it arrives back at the speed $u$ it left with. It then leaves at $eu$. So the flights after the first fall last

$$\frac{2ev_1}{g},\ \frac{2e^2v_1}{g},\ \frac{2e^3v_1}{g},\ \dots$$

where $v_1$ is the speed of the first impact. The heights of the bounces fall by $e^2$ each time, and the durations by $e$.

A series with a finite sum #

The total time from release to rest is the first fall plus every flight:

$$t_{\text{total}} = t_1 + \frac{2v_1}{g}\left(e + e^2 + e^3 + \cdots\right) = t_1\,\frac{1+e}{1-e},$$

since $t_1 = v_1/g$ and the geometric series sums to $e/(1-e)$. For a ball dropped from a metre with $e = 0.8$ that is 0.452 seconds times nine: 4.06 seconds. The sum has infinitely many terms and a finite value.

The sequence of bounce times climbs towards its limit and never reaches it, as the partial sums of any convergent series do. What is unusual about this series is that its limit is a physical event. At 4.064 seconds the ball has completed every one of its bounces, and after 4.064 seconds it is lying on the floor. There is no last bounce before it stops — any bounce has infinitely many smaller ones after it — and there is nothing paradoxical about that either. Zeno’s arrow, which must cross half the distance and then half the remainder without end, arrives because infinitely many ever-shorter intervals can fit into a finite time, and the ball’s bounces are the same infinity, heard.

The sound of the ball is the series made audible. The interval between clicks shrinks by the factor $e$ each time, so the clicks accelerate steadily, each one closer to the last in the same proportion, and the pitch of the buzz they merge into rises without limit as the final moment approaches. Measuring the ratio of successive intervals from a recording of the sound is one of the simplest ways to measure a coefficient of restitution: the ratio is $e$ directly, with no need to know the drop height, the ball’s mass or anything about the floor.

Why a pendulum losing energy never stops and a ball does #

The finite total time looks at first as though it must follow simply from the ball losing energy, and it does not. Plenty of systems lose a fixed fraction of their energy each cycle and never stop.

A pendulum with a little air resistance, or a mass on a spring with a dashpot, loses a fixed fraction of its amplitude each swing too — exponential decay is exactly that. Its amplitude after $n$ swings is some constant to the power $n$, a geometric sequence just like the bounce heights. But its motion never quite comes to rest: the amplitude falls towards zero and never reaches it, and a lightly damped oscillator is still swinging, in principle, after any finite time.

The difference is in how long each cycle lasts. A pendulum’s period does not depend on its amplitude — that is the property of small oscillations that made the pendulum a clock — so every swing takes the same time, and infinitely many swings take infinitely long. A bouncing ball’s period is proportional to its speed. Halve the speed and the flight time halves. So the cycles shrink as the motion shrinks, their durations form a geometric series rather than a constant one, and the total converges.

That is the general rule, and it has nothing to do with bouncing: a motion whose period shrinks in proportion to its amplitude, and which loses a fixed fraction per cycle, stops in a finite time. A pendulum whose losses are to dry friction at its pivot rather than to air — a loss that removes a fixed amount per swing rather than a fixed fraction — also stops in finite time, for a different reason: the amplitude falls by a constant each swing and reaches zero after a finite number of them. Static friction is what then holds it. Only the combination of a fixed period and a proportional loss produces the endless exponential tail, and it is the combination that linear models are built on.

The floor’s infinitely many kicks #

Each impact is a brief, violent push from the floor, and the bouncing sequence can be read as a sequence of those pushes rather than as a sequence of flights.

At impact number $k$ the ball arrives at speed $v_k$ and leaves at $e v_k$, so its momentum changes by $m v_k (1 + e)$, and that is the impulse the floor delivers. Between impacts the floor delivers nothing. So a scale under the floor, if it could respond fast enough, would read a series of spikes separated by silences, with the spikes getting smaller and closer together.

Averaged over time, the spikes give something exact. The total impulse of every bounce is $m v_1(1+e)/(1-e)$, which is precisely $mg$ times the time the ball took to stop. That equality is not a coincidence of the series. The ball began at rest and ended at rest, so its total change of momentum is zero, and the floor’s total impulse must cancel gravity’s, which is $mg$ multiplied by the elapsed time. After the bouncing ends the floor simply supports the ball, delivering $mg$ continuously. From the moment of release, the average reading of the scale is exactly one weight by the time the ball is still, and remains one weight for ever after.

That is the same accounting that makes a falling chain weigh three times as much as the part that has landed: a scale does not read the weight of what is resting on it but the rate at which it is delivering momentum, and anything arriving brings momentum with it. A bouncing ball brings its momentum in lumps, and between lumps the scale reads nothing, but over the whole episode the lumps average to the weight. Over any one cycle the account does not balance, and the way it fails is instructive. The first impact delivers $(1+e)$ times the weight multiplied by the fall that preceded it — the floor has to stop the ball as well as relaunch it. Every later impact delivers only $(1+e)/2$ times the weight multiplied by the flight before it, because the ball arrives slower than it left. The early surplus and the later shortfalls cancel only over the whole infinite series, which is why the running average in the figure overshoots, oscillates and settles rather than sitting at one weight from the start.

How long and how far #

The total time depends on $e$ in a way that is worth seeing as a curve, because it runs away as the ball approaches perfect elasticity.

The total distance travelled is another geometric series. Each bounce rises to $e^2$ times the previous height and falls back, so the path length is the drop height times $(1 + e^2)/(1 - e^2)$. A ball with $e = 0.9$ travels nine and a half times its drop height in all before stopping, while spending nineteen first-falls doing it. Both diverge as $e \to 1$, as they must — a perfectly elastic ball bounces for ever — but the time diverges faster than the distance, because as $e$ approaches one the late bounces are low but not proportionally brief: height goes as the square of the speed and time only as the speed.

The same arithmetic describes anything that returns a fixed fraction of what it is given. A dropped ball on a stair, a stone skipping across water with a fixed loss per skip, and a ball bouncing between two walls that are moving — where the bounces gain rather than lose, and the series diverges instead of converging — are all geometric sequences in the speed, and each asks the same question about whether the sum of the times converges.

What an inelastic collision does to the energy #

Every bounce destroys a fraction $1 - e^2$ of the ball’s kinetic energy, and the destruction is worth tracking, because the floor does something surprising while it happens: nothing. The floor is stationary, so the force it exerts does no work on the ball during the impact; its point of application does not move. The energy the ball loses goes into the ball’s own interior — into vibrations of the rubber or the steel that ring on after the contact ends and eventually become heat — and a little into sound, which is the only part the listener collects.

So the floor delivers the whole of the impulse and none of the energy change, and the ball’s material decides how much energy survives. That is the division of labour that makes $e$ a property of the pair of materials rather than of the floor alone: a steel ball on a steel anvil returns nearly everything, the same ball on a wooden floor much less, because the wood’s own deformation absorbs energy that the anvil would have returned. In a gas of such grains, each collision removing a little of the relative motion, the same loss makes the gas cool itself into clumps, and its most extreme form is the three-grain version of this series, where the collisions between neighbours become infinitely many in a finite time.

Where the series stops being true #

The idealisation that produces the infinite series is that each impact is instantaneous. It is not, and the failure is quantitative.

An elastic ball pressed against a hard surface deforms over a contact patch whose size depends on the force, and Hertz worked out in 1881 how long such a contact lasts. For two elastic spheres the contact time goes as the impact speed to the power minus one-fifth: gentler impacts last longer, because a slower ball compresses the contact less and the restoring force, which grows faster than the compression, takes longer to turn it round.

The flight time falls in proportion to the speed and the contact time rises slowly as it falls, so they must cross. For a steel ball on steel with $e = 0.8$ they cross at an impact speed of about a millimetre and a half per second, which the series reaches at the thirty-sixth bounce, a hop about a tenth of a micrometre high. Below that the idea of separate bounces has no meaning: the ball is in contact for longer than it is in the air, the contacts overlap with the flights, and the motion becomes a small vibration of the ball on the floor, damped by the same losses that set $e$, which dies away on the timescale of the contact itself. The thirty-five bounces before it are described extremely well by the series. The infinitely many after it are a mathematical tail with no physical counterpart.

The crossing is not the only thing that goes wrong near the end. The coefficient of restitution itself is not constant; for most materials it rises towards one at very low impact speeds, where the deformation is small and entirely elastic, and falls at high speeds, where it becomes plastic. Surfaces are not smooth, so at hop heights comparable to their roughness the ball rattles on asperities rather than on a plane. And any real floor transmits vibrations, so a ball bouncing on it is being shaken as well as bouncing. Each of these acts well after the first few dozen bounces and changes nothing about the finite total time, which is fixed almost entirely by the first few flights.

The bounces too small to draw #

The first figure shows height against time and invites the eye to count bounces, and it cannot show the last ones: below a hop of a millimetre they are thinner than the line. The picture is honest about the first ten and silent about the rest, which is appropriate, since the rest are where the model fails.

Nor can any of these figures show where the energy went. The ball’s kinetic energy at each impact is a clean number, but the energy lost is spread over vibrations inside the ball that ring on for milliseconds after each contact — in a steel ball two centimetres across, at frequencies above a hundred kilohertz — and those vibrations are themselves a source of error in the next bounce, since a ball still ringing from the last impact does not meet the floor in the state the model assumes.

Still open: what sets a coefficient of restitution #

For all its usefulness, $e$ is a measured number rather than a derived one. For two ideal elastic bodies it would be exactly one; for real bodies the losses come from plastic flow in the contact region, from viscoelastic friction inside the material, from elastic waves launched into both bodies that never return in time to push them apart, and from adhesion as the surfaces separate. Theories exist for each mechanism separately. Predicting $e$ for a given pair of materials, a given geometry and a given impact speed from the materials’ properties alone — well enough to replace the measurement — has not been achieved in general, and for soft, rough or porous materials the measured values scatter widely from sample to sample. Ball manufacturers and sports regulators measure $e$ by dropping balls, because no calculation would be trusted to settle a dispute.

The natural next question about momentum is the reverse of this one: not how a known sequence of impacts adds up, but how to learn the force during an impact that lasts fifty microseconds and cannot be watched, from what comes out of it. The habit worth carrying from here is to ask whether the time per cycle depends on the size of the cycle. A proportional loss with a fixed period decays for ever; the same loss with a period that shrinks as the motion does is over in a finite time — and which of the two a system is decides whether “it keeps going, just smaller” is true.