ARCHETYPE: The Bouncing Ball

Drop a tennis ball and listen. The bounces come faster and faster — tok… tok… tok-tok-toktoktok — a drumroll accelerating toward silence. Hidden in that sound is a lovely paradox worthy of Zeno: the ball bounces infinitely many times, yet finishes bouncing in a finite time. Each rebound launches it at a fixed fraction $e$ of its landing speed, each flight is $e$ times shorter than the last, and an infinite geometric series of shrinking flights adds up to a perfectly ordinary number of seconds. You can hear a convergent series. The number $e$ — the coefficient of restitution, owned by the inelastic-collisions module — is the whole personality of the ball: superball $e \approx 0.9$ , tennis ball $\approx 0.75$ , cricket ball $\approx 0.3$ , putty $0$ . This archetype is what happens when you let $e$ act repeatedly: speeds fall as $e^n$ , heights as $e^{2n}$ , and every "total" question — total distance travelled, total time to stop, average speed, impulse hammered into the floor — becomes a geometric series with a two-line closed form. JEE-Main and JEE-Advanced pose the bouncing ball in all these costumes, and it is a gift once you see the pattern: one ratio, memorized sums, clean answers. It is also the cleanest laboratory in mechanics for the difference between distance and displacement, and between a single impulsive event and the accumulated consequences of many.

Key takeaways

A ball dropped from $h_0$ lands at $v_0 = \sqrt{2gh_0}$ ; each bounce rebounds it at $e$ times the landing speed.
After $n$ bounces: speed $v_n = e^n v_0$ , height $h_n = e^{2n} h_0$ — speeds ladder down by $e$ , heights by $e^2$ .
Total distance ever travelled: $D = h_0\,\dfrac{1+e^2}{1-e^2}$ . Total time until bouncing stops: $T = \sqrt{\dfrac{2h_0}{g}}\;\dfrac{1+e}{1-e}$ .
Infinitely many bounces, finite time — a convergent geometric series you can hear.
Impulse delivered to the floor at the $n$ -th bounce: $J_n = m(1+e)\,v_{n-1}$ — the ball reverses, so speeds add.
Energy fraction lost per bounce: $1 - e^2$ ; the ball's mass never enters any of it.
Key idea: one fixed ratio per bounce turns every "total" question into a geometric series — $\sum e^n = \frac{e}{1-e}$ does all the work.

The core insight

Each bounce multiplies the speed by $e$ (so heights by $e^2$ and flight times by $e$ ); summing the resulting geometric series gives closed totals $D = h_0\frac{1+e^2}{1-e^2}$ and $T = \sqrt{2h_0/g}\,\frac{1+e}{1-e}$ — infinite bounces, finite distance and time.

Where this lives today

Sport is regulated in coefficients of restitution. Tennis balls must pass a drop test — released from 100 inches onto concrete, they must rebound between 53 and 58 inches, pinning $e$ to roughly $0.73$ – $0.76$ — and basketballs, footballs and golf balls all carry equivalent certifications; the R&A and USGA even cap the restitution of golf drivers, because a springier clubface is a longer drive. Engineering drop-tests run the archetype in reverse: phone designers measure rebound behaviour to infer how much energy the chassis absorbs per impact. And in physics research, "granular gases" — collections of beads whose collisions all have $e < 1$ — show clustering and cooling behaviour governed entirely by the $(1-e^2)$ energy tax per collision, the bouncing ball's series played by $10^6$ particles at once. The accelerating drumroll you hear from a dropped ping-pong ball is still the fastest classroom proof that an infinite series can have a finite sum.

The setup

A ball is dropped from height $h_0$ onto a rigid floor. Each impact has coefficient of restitution $e$ (rebound speed $= e \times$ landing speed). Find: heights and speeds after $n$ bounces; total distance; total time; impulse per bounce; when the bouncing effectively ends.

Figure

The bouncing ball — heights fall by e squared, flight times by e

The engine of the archetype is a single sentence: the bounce is a fixed-ratio machine. Whatever speed arrives, $e$ times it leaves. Everything else is free-fall kinematics between impacts, and free fall ties the three currencies together rigidly: height goes as speed squared ( $h = v^2/2g$ ), flight time goes as speed to the first power ( $t = 2v/g$ for up-and-down). So one ladder of speeds $v_0, ev_0, e^2v_0,\dots$ drags along a ladder of heights shrinking by $e^2$ and a ladder of times shrinking by $e$ . Ask for any total and you are summing a geometric series whose ratio you already know. Keep one boundary fact straight before the algebra: the first fall from $h_0$ happens once, but every height after it is visited twice — up, then down. Totals therefore look like "first term + 2 × (the series)", and forgetting the asymmetry is the most common structural slip.

The master solution pattern

The ladders. Landing speed from the initial drop: $v_0 = \sqrt{2gh_0}$ , taking time $t_0 = \sqrt{2h_0/g}$ . After the $n$ -th bounce:

v_n = e^n\,v_0, \qquad h_n = e^{2n}\,h_0, \qquad t_{\text{flight},n} = \frac{2v_n}{g} = 2e^n t_0 \qquad (1)

where: $v_n$ — rebound speed after bounce $n$ ; $h_n$ — the height it then reaches; $t_{\text{flight},n}$ — the full up-and-down time of that flight. The exponents are the exam: speeds $e^n$ , heights $e^{2n}$ (because $h \propto v^2$ ), times $e^n$ (because $t \propto v$ ).

Total distance. The first fall contributes $h_0$ ; every later height is climbed and descended:

D = h_0 + 2h_0e^2 + 2h_0e^4 + \cdots = h_0\left(1 + \frac{2e^2}{1-e^2}\right) = h_0\,\frac{1+e^2}{1-e^2} \qquad (2)

where: the geometric sum $e^2 + e^4 + \cdots = \dfrac{e^2}{1-e^2}$ has been used. For $e = \tfrac12$ : $D = h_0 \times \frac{1.25}{0.75} = \frac{5}{3}h_0$ — the entire infinite future of bouncing adds only two-thirds of the original drop.

Total time. First fall $t_0$ , then a full flight per bounce:

T = t_0 + \sum_{n=1}^{\infty} 2e^n t_0 = t_0\left(1 + \frac{2e}{1-e}\right) = \sqrt{\frac{2h_0}{g}}\;\frac{1+e}{1-e} \qquad (3)

where: for $e = \tfrac12$ the ball finishes its infinity of bounces in $T = 3t_0$ — three times the first-drop time, then silence. Infinitely many events, finite duration: the series converges because $e < 1$ .

Impulse on the floor. At the $n$ -th impact the ball arrives downward at $v_{n-1}$ and leaves upward at $e\,v_{n-1}$ — a reversal, so the magnitudes add:

J_n = m\left(v_{n-1} + e\,v_{n-1}\right) = m(1+e)\,v_{n-1} \qquad (4)

where: $J_n$ — momentum delivered to the floor. The energy lost in that impact is the fraction $(1-e^2)$ of the arriving kinetic energy — the two bookkeepings ( $1+e$ for impulse, $1-e^2$ for energy) live one factor apart, and mixing them up is a classic distractor.

Figure

Hearing a convergent series — flight times stack up to a finite total

asked	formula	series ratio
height after $n$ bounces	$h_n = e^{2n}h_0$	—
total distance	$D = h_0\frac{1+e^2}{1-e^2}$	$e^2$
total time	$T = t_0\frac{1+e}{1-e}$	$e$
average speed of the whole journey	$D/T$	both
impulse at bounce $n$	$m(1+e)e^{n-1}v_0$	$e$
energy lost at bounce $n$	$(1-e^2)\times\tfrac12 mv_{n-1}^2$	$e^2$

Derivation AThe two grand totals — distance and time as geometric series

We derive equations (2) and (3), the closed forms for total distance and total time, from the ladders (1) and the geometric-series sum. The strategy: write out the journey term by term, spot the fixed ratio, and use $x + x^2 + x^3 + \cdots = \frac{x}{1-x}$ for $|x|<1$ . Watch two boundary details — the first fall is a half-journey (down only), and the series for distance runs in $e^2$ while the series for time runs in $e$ .

Step 1. Lay out the itinerary. Fall from $h_0$ (distance $h_0$ , time $t_0$ ). Bounce 1: up to $h_1 = e^2h_0$ and back down (distance $2h_1$ , time $2e\,t_0$ ). Bounce 2: up-down at $h_2 = e^4h_0$ (distance $2h_2$ , time $2e^2t_0$ ). And so on, forever.

Step 2. Total distance. $D = h_0 + 2h_0(e^2 + e^4 + e^6 + \cdots)$ . The bracket is a geometric series with first term $e^2$ and ratio $e^2$ :

e^2 + e^4 + \cdots = \frac{e^2}{1-e^2} \quad\Longrightarrow\quad D = h_0\left(1 + \frac{2e^2}{1-e^2}\right) = h_0\,\frac{1-e^2+2e^2}{1-e^2} = h_0\,\frac{1+e^2}{1-e^2}

Why the ratio is $e^2$ : distance rides on heights, and heights ride on the square of speed. One physical ratio $e$ , two mathematical ratios — this is the fork in the road where errors happen.

Step 3. Total time. $T = t_0 + 2t_0(e + e^2 + e^3 + \cdots) = t_0\left(1 + \dfrac{2e}{1-e}\right) = t_0\,\dfrac{1+e}{1-e}$ . Why time is the $e$ -series: flight time is proportional to launch speed (first power), so it lades down by $e$ per bounce, not $e^2$ .

Step 4. Convergence and the paradox. Both series converge because $e < 1$ — each term a fixed fraction of the last. So the ball completes infinitely many bounces within the finite $T$ : the bounce count diverges, the bounce durations shrink fast enough to fit. After time $T$ the ball sits still (in this ideal model); the drumroll's accelerando is the terms of the series arriving. Verified: $e = \tfrac12$ , $h_0 = 5$ m, $g = 10$ : $t_0 = 1$ s, $v_0 = 10$ m/s; $D = 5 \times \frac{1.25}{0.75} = \frac{25}{3} = 8.33$ m — by hand: $5 + 2(1.25) + 2(0.3125) + \cdots = 5 + 2.5 + 0.625 + \cdots \to 8.33$ ✓; $T = 1 \times \frac{1.5}{0.5} = 3$ s — by hand: $1 + 2(0.5) + 2(0.25) + \cdots = 1 + 1 + 0.5 + 0.25\cdots \to 3$ ✓. Limits: $e \to 0$ : $D = h_0$ , $T = t_0$ (dead ball); $e \to 1$ : both diverge (perpetual bouncer).

Limiting cases

case	result	meaning
$e = 0$ (putty)	$D = h_0$ , $T = t_0$	lands and stays — no series at all
$e \to 1$ (superball limit)	$D, T \to \infty$	perpetual bouncing; series diverges at $e = 1$
$e = \tfrac12$	$D = \tfrac53 h_0$ , $T = 3t_0$	the standard exam numbers — memorize
large $n$	$h_n \to 0$ but never exactly 0	"stops bouncing" means below any threshold you name
net displacement	always $h_0$ downward	distance is the series; displacement is one number
ball of double mass	identical $D$ , $T$ , $h_n$	mass cancels everywhere except impulse and energy

Common trap“e is the fraction of height kept per bounce”

No — $e$ is the fraction of speed kept; the height fraction is $e^2$ . A ball dropped from 1.0 m that rebounds to 0.64 m has $e = \sqrt{0.64} = 0.8$ , not $0.64$ . The confusion propagates viciously through the series: use $e$ where $e^2$ belongs and your total distance answer is wrong in a way that looks structurally fine. The safe habit: build everything from the speed ladder $v_n = e^nv_0$ , then convert — heights by squaring ( $h = v^2/2g$ ), times linearly ( $t = 2v/g$ ). The measurement rule $e = \sqrt{h'/h}$ (from the restitution module) is the same statement read backward. A second cousin error hides in the impulse: the floor receives $m(1+e)v$ , because the ball reverses — not $m(1-e)v$ , which is merely the loss of speed magnitude. Energy lost, meanwhile, carries $(1-e^2)$ . Three different brackets — $(1+e)$ , $(1-e)$ , $(1-e^2)$ — for three different questions; label which one you are using and why.

Worked examples

1 · Board level — reading e from the rebound (full method)

A ball is dropped from $2.0\ \mathrm{m}$ onto a floor and rebounds to $1.28\ \mathrm{m}$ . Find the coefficient of restitution and the height after the second bounce.

Step 1. Heights ratio gives $e^2$ : $e^2 = \dfrac{h_1}{h_0} = \dfrac{1.28}{2.0} = 0.64$ , so $e = \boxed{0.80}$ .

Step 2. Each bounce multiplies height by $e^2 = 0.64$ : $h_2 = 0.64 \times 1.28 = \boxed{0.82\ \mathrm{m}}$ (two significant figures).

Step 3. Sanity: heights $2.0 \to 1.28 \to 0.82$ — a clean geometric ladder, ratio $0.64$ . No $g$ needed anywhere: ratios eat the constants.

2 · Board level — impulse and average force on the floor

A $0.20\ \mathrm{kg}$ ball is dropped from $5.0\ \mathrm{m}$ ( $g = 10\ \mathrm{m/s^2}$ ) and rebounds with $e = 0.50$ . The impact lasts $0.050\ \mathrm{s}$ . Find (a) the impulse on the floor, (b) the average normal force during contact.

Step 1. Landing speed: $v_0 = \sqrt{2\times10\times5.0} = 10\ \mathrm{m/s}$ . Rebound speed: $ev_0 = 5.0\ \mathrm{m/s}$ .

Step 2. Impulse on the ball (upward positive): $J = m(ev_0) - m(-v_0) = m(1+e)v_0 = 0.20 \times 1.5 \times 10 = \boxed{3.0\ \mathrm{kg\,m/s}}$ . The floor receives 3.0 kg·m/s downward.

Step 3. Average force. During contact, the ball feels the floor's $N$ up and gravity down: $(N - mg)\,\Delta t = J$ , so $N = \dfrac{3.0}{0.050} + 0.20\times10 = 60 + 2 = \boxed{62\ \mathrm{N}}$ — about 31 times the ball's weight.

Step 4. Note the bracket: $(1+e)$ , because the velocity reverses. The distractor $m(1-e)v_0 = 1.0\ \mathrm{kg\,m/s}$ (mere speed loss) is on every option list.

3 · Competitive level — total distance, total time, average speed (JEE-Main pattern, ~3 minutes)

A ball is dropped from $5.0\ \mathrm{m}$ with $e = 0.50$ , $g = 10\ \mathrm{m/s^2}$ . Find the total distance travelled, the total time until bouncing ceases, and the average speed over the whole journey.

Step 1. First-drop quantities: $t_0 = \sqrt{2h_0/g} = 1.0$ s, $v_0 = 10$ m/s.

Step 2. Distance, equation (2): $D = 5.0 \times \dfrac{1+0.25}{1-0.25} = 5.0 \times \dfrac{1.25}{0.75} = \boxed{8.3\ \mathrm{m}}$ ( $25/3$ m).

Step 3. Time, equation (3): $T = 1.0 \times \dfrac{1+0.5}{1-0.5} = \boxed{3.0\ \mathrm{s}}$ .

Step 4. Average speed $= D/T = \dfrac{25/3}{3} = \boxed{2.8\ \mathrm{m/s}}$ . (Average velocity is different: net displacement $5.0$ m down in $3.0$ s $= 1.7$ m/s downward — if asked, read which word the paper used.) Timing note: ~2.5 min with the closed forms; a full series-by-hand rebuild costs 8. Memorize the two brackets — distance $(1+e^2)/(1-e^2)$ , time $(1+e)/(1-e)$ — and which power of $e$ each carries.

4 · Competitive level — JEE-Advanced stretch: when is the bounce below 1%? (~4 minutes)

A ball bounces with $e = 0.80$ . After how many bounces does the rebound height first fall below 1% of the original drop height? (Use $\log_{10}2 = 0.301$ .)

Step 1. Condition: $h_n = e^{2n}h_0 < 0.01\,h_0$ , i.e. $(0.8)^{2n} < 10^{-2}$ .

Step 2. Take logs: $2n\log_{10}0.8 < -2$ . Now $\log_{10}0.8 = \log_{10}8 - 1 = 3(0.301) - 1 = -0.097$ .

Step 3. So $2n \times 0.097 > 2$ , giving $n > \dfrac{2}{0.194} = 10.3$ .

Step 4. First integer beyond: $n = \boxed{11}$ bounces. Cross-check the neighbourhood: $h_{10} = 0.8^{20} = 10^{-20\times0.097} = 10^{-1.94} \approx 1.15\%$ (still above), $h_{11} = 0.8^{22} \approx 10^{-2.13} \approx 0.74\%$ (below) ✓. Timing note: ~3.5 min. The only skill is remembering heights go as $e^{2n}$ — with $e^n$ you would get $n \approx 21$ and its matching wrong option. The supplied $\log 2$ is the examiner's fingerprint: they expect $\log 0.8 = 3\log 2 - 1$ .

Exam traps

$e$ versus $e^2$ . Speeds ladder by $e$ ; heights and energies by $e^2$ ; flight times by $e$ . Every formula in this archetype is one of these three ladders — identify which before substituting. Measured rebound heights give $e = \sqrt{h'/h}$ .

First fall counts once. Total-distance and total-time formulas have the asymmetric first term ( $h_0$ once, $t_0$ once) and doubled series afterward. Symmetric-looking answers like $h_0\frac{2}{1-e^2}$ are the fingerprints of a mis-set series.

Distance ≠ displacement. Total distance is the series $\frac{1+e^2}{1-e^2}h_0$ ; total displacement is just $h_0$ downward (start at the drop point, end on the floor). Average speed uses distance; average velocity uses displacement. The question's noun decides.

Impulse bracket. Floor impulse per bounce is $m(1+e)v$ — reversal adds magnitudes. $(1-e)$ is speed loss, $(1-e^2)$ is energy loss fraction. Three brackets, three meanings.

Gravity during contact. For "average force" with a stated contact time, the honest equation is $(N - mg)\Delta t = m(1+e)v$ . When $N \gg mg$ (short contacts) the $mg$ correction is small — but JEE marks the difference when the numbers make it visible (Example 2: 62 N vs 60 N).

Momentum of the ball alone is not conserved. The floor (Earth) absorbs $m(1+e)v$ per bounce. Momentum conservation applies to ball + Earth; energy bookkeeping applies to the ball with the $(1-e^2)$ tax. Neither "conservation" is a licence to skip the bounce physics.

Series validity. All closed totals assume $e$ is constant every bounce and the floor is rigid and level. Problems that change $e$ midway, or bounce on a moving platform, must be rebuilt term by term — the formula is not a spell.

Origins · 1 minZeno of Elea and how this was discovered

Zeno of Elea

c. 490–430 BC · Elea, Magna Graecia

Zeno never saw a tennis ball, but he owns its paradox. His most famous arguments — Achilles and the tortoise, the dichotomy — insist that infinitely many steps can never be completed, so motion itself must be an illusion. The bouncing ball is nature's cheerful rebuttal: it genuinely performs infinitely many bounces (in the ideal model), each a real event, and finishes the whole infinite agenda in $T = t_0\frac{1+e}{1-e}$ seconds, because the durations shrink geometrically and their sum converges. The mathematics that settles Zeno — the convergent geometric series — took two millennia to make rigorous, from Archimedes' quadratures to the 19th-century limit concept of Cauchy and Weierstrass. Every time a dropped ball rattles to rest, it completes what Zeno declared impossible, on a schedule you can compute in two lines.

History

450

BC — Zeno of Elea argues that completing infinitely many steps is impossible; motion stands accused.

250

BC — Archimedes sums a geometric series to measure a parabola — the tool that will acquit the bouncing ball.

1687

— Newton's Principia reports restitution experiments: rebound speeds a fixed fraction of impact speeds, ball by ball.

1821

— Cauchy's Cours d'analyse makes convergence rigorous; infinite bounces in finite time becomes respectable mathematics.

1965

— The Super Ball (

e \approx 0.9

) becomes a toy-store sensation — sixty bounces where a cricket ball manages five.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1, Ch. 9 (restitution; bouncing series in the exercises) · Irodov, Problems in General Physics (impulse-train variants) · For the underlying $e$ machinery: Inelastic Collisions & Coefficient of Restitution (this chapter).

Provenance. All results are standard. A verifier should recompute: the ladders (1) from $h = v^2/2g$ and $t = 2v/g$ ; the sums (2) and (3) against the term-by-term checks ( $e = \tfrac12$ , $h_0 = 5$ m, $g=10$ : $D = 25/3$ m, $T = 3$ s); Example 1's $e = 0.8$ , $h_2 = 0.82$ m; Example 2's $J = 3.0$ kg·m/s and $N = 62$ N; Example 4's $n = 11$ via $\log_{10}0.8 = -0.097$ .

Prerequisites. Inelastic Collisions & Coefficient of Restitution · Linear Momentum & Impulse · Free Fall & Motion Under Gravity. Leads to. Which Conservation Law? (technique) · Losing Contact in SHM (archetype).

Does the ball really bounce infinitely many times? That sounds impossible.

In the ideal model, yes — and it is not a contradiction, because the bounces get faster: bounce $n$ 's flight lasts $2e^nt_0$ , and the sum of all flights converges to a finite $T$ . Infinitely many events fit inside a finite time when their durations shrink geometrically, exactly as $1 + \frac12 + \frac14 + \cdots$ fits inside 2. In reality, the model expires once rebound heights shrink to the scale of the ball's surface roughness and deformation — after which the ball rattles, rolls, and sits. But for exam purposes and for any audible bounce, the geometric series is the physics.

Why do heights shrink faster than speeds?

Free fall prices height in speed squared: $h = v^2/2g$ . Cut the speed by the factor $e$ each bounce and the height necessarily falls by $e^2$ . Flight time, in contrast, is linear in launch speed ( $t = 2v/g$ ), so it shrinks by plain $e$ . One physical ratio, three ladders — $e$ for speeds and times, $e^2$ for heights and energies. Most wrong answers in this archetype are just a ladder mix-up.

Where does the lost energy go at each bounce?

Into the ball, mostly: the impact squashes it, and the un-squashing doesn't pay back in full — internal friction in the rubber/leather converts the shortfall ( $1-e^2$ of the arriving kinetic energy) to heat, plus a little sound (the tok you hear) and micro-deformation of the floor. Squash players know this as warming the ball: a cold squash ball has tiny $e$ and barely bounces; after ten rallies it is warmer, springier and measurably faster — a rare case where you can feel the energy ledger of this module.

Is momentum conserved when the ball bounces?

Not for the ball alone — its momentum reverses, a change of $m(1+e)v$ , delivered by the floor. Include the Earth and conservation is restored: the planet absorbs exactly the $m(1+e)v$ the ball hands over, and its resulting velocity change is of order $10^{-24}$ m/s — invisible forever. This is the standard pattern: conservation laws hold for closed systems, and "ball + Earth" is closed while "ball" is not. The exam-side moral: never invoke momentum conservation across a bounce off a fixed surface; use restitution ( $v' = ev$ ) instead.

Why doesn't the ball's mass appear in the heights, distances or times?

Because free fall is mass-blind (all bodies fall with $g$ ) and the restitution ratio $e$ is defined through speeds, not forces. Double the mass and the impact forces double, but the rebound-speed fraction — a property of the materials' energy return — stays put. Mass enters only when the question asks about momentum or energy in absolute terms: impulse $m(1+e)v$ and energy lost $(1-e^2)\tfrac12mv^2$ both scale with $m$ . Heights, times, and counts of bounces do not.

What happens if the floor is not rigid — say the ball bounces on sand or on a drum skin?

Then part of the energy story moves into the floor. Sand barely returns anything (effective $e \approx 0$ : craters, no rebound); a drum skin stores and returns energy but also steals some into sound and its own vibration, and can even fling the ball back harder in special timing cases (a moving surface can do work on the ball — the trampoline effect). The constant- $e$ series assumes the floor is passive and rigid; any moving or deformable surface must be handled bounce by bounce with the restitution law relative to the surface's motion.

How do I quickly estimate e for a real ball?

Drop it from a measured height $h$ next to a wall, film or watch the first rebound height $h'$ , and take $e = \sqrt{h'/h}$ . One drop suffices in principle; averaging three is lab-honest. Typical values worth carrying: superball 0.85–0.9, basketball ≈ 0.76 (inflated correctly), tennis ≈ 0.75, golf ball on concrete ≈ 0.8, cricket ball ≈ 0.3, squash ball cold ≈ 0.2. If you have only your ears: count the audible bounce interval halving — the time ladder is $e$ .