Counting Clacks, Finding π

Two blocks sit on a smooth rail with a wall at the left end. The large block slides in, and the small one bounces between it and the wall. With a mass ratio of 100 there are 31 collisions; with 10,000, 314; with 1,000,000, 3141.

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Chapters · 8

01Two Blocks
02The Digits of π
03Two Conservation Laws
04A Circle
05Equal Arcs
06Why π
07The Same Geometry
08The Bench

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01

Two Blocks

The rail has no friction and every collision is perfectly elastic: no energy is lost. The small brass block on the left is at rest; the steel block on the right slides towards it at a steady speed. A collision between the blocks counts, and so does the small block hitting the wall.

With equal masses there are 3. The large block hands all its speed to the small one and stops; the small block bounces off the wall; then it hands the speed back. The large block leaves to the right, the small one stays put, and they never meet again.

02

The Digits of π

Make the large block 100 times heavier and there are 31 collisions. 10,000 times, 314. 1,000,000 times, 3141. Each factor of 100 adds a digit, and the digits are those of π = 3.14159….

At a million to one, nine in ten of the 3141 collisions crowd into less than 20 milliseconds, when the large block is closest to the wall. The model slows down there by itself and shows by how much at the top right; the counter misses none. Every moment of the model is the exact solution, not a step-by-step approximation.

03

Two Conservation Laws

When the blocks collide, the momentum m·v₁ + M·v₂ stays the same, and so does the kinetic energy ½m·v₁² + ½M·v₂². Two equations, two unknowns: the two speeds after the collision are fixed.

When the small block hits the wall, its speed stays the same and its direction reverses. Energy is kept; momentum is not, and the difference goes into the wall and the ground. The arrows show the two velocities.

04

A Circle

Draw the two velocities as one point on a plane: X = √M·v₂ across, Y = √m·v₁ up. Scaled this way, conservation of energy says X² + Y² stays the same, so the point can only sit on a circle.

Conservation of momentum says √M·X + √m·Y stays the same: a line of slope −√(M/m). When the blocks collide, the point jumps along that line to another spot on the circle; when the small block hits the wall, Y changes sign and the point jumps straight across. The figure below draws each jump as the model makes it, until the point lands in the narrow wedge on the right: the small block moving right, no faster than the large one, never to catch it again.

05

Equal Arcs

A wall jump is vertical; a jump from the blocks colliding is tilted from vertical by a fixed angle θ, with tan θ = √(m/M). Two jumps in a row meet at a point on the circle at angle θ. By the inscribed angle theorem, the arc between their other two ends is 2θ.

So the collision points line up round the circle at equal arcs of 2θ. From the start to the narrow wedge at the end there is room for ⌈π/θ⌉ − 1 of them, and that is the number of collisions.

06

Why π

With a mass ratio of 100ⁿ, √(m/M) = 10⁻ⁿ. For small angles arctan x is almost exactly x, so θ is a hair under 10⁻ⁿ. Then π/θ is a hair over π·10ⁿ, and the count is the whole-number part of π·10ⁿ: 3, 31, 314, 3141. The π comes from the circle: in steps of 2θ, half a turn takes π/θ of them.

The hair is about 10⁻ⁿ and usually changes nothing. Only if π had about n nines in a row just after its n-th digit would it add one to the count. No one has proved that this never happens.

07

The Same Geometry

Unfold the zigzag in the circle and it becomes a beam of light bouncing between two mirrors. With the mirrors at angle θ and the beam sent in along one of them, it bounces ⌈π/θ⌉ − 1 times.

Two reflections make one rotation. Grover’s search algorithm in quantum computing works the same way, turning the state by the same small angle at each step. A 2019 paper showed that the two blocks and Grover’s algorithm match exactly: finding one item among N takes about (π/4)·√N steps, growing with the square root just as the count of collisions does.

08

The Bench

Five mass ratios and one rail. Time can be slowed, and the collisions can be heard.

Try thisWith the large block a hundred million times heavier, where does the count stop?

What the model leaves out

The calculation is the ideal case: no friction, perfectly elastic collisions, rigid blocks, instant impacts and a wall that never moves. Real blocks lose a little energy and would not reach these counts. From start to finish the model is the exact solution: the time of each collision is solved for directly and the blocks move at constant speed between collisions, with no time steps. The large block is not to scale: at the same density, a block a million times heavier would be 100 times wider, so here its size only means heavier. The sound is synthesised, and collisions that come too close together merge.

A common misreading

π does not turn up because anything here is round, and it is not a coincidence. It comes from the energy circle in phase space: the count is how many steps of 2θ fit in half a turn. The count spells out decimal digits of π because the mass ratio is a power of 100; with any other ratio it is still ⌈π/θ⌉ − 1, it just no longer looks like π.

Sources

Data

The puzzle and its solution are due to Gregory Galperin: “Playing pool with π (the number π from a billiard point of view)”, Regular and Chaotic Dynamics 8(4), 2003. Grant Sanderson (3Blue1Brown) brought it to a wide audience in 2019 and returned to it in 2025; his videos are under Watch next.