Digits of π from Elastic Collisions

Galperin's Billiard

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Speed

Side view

Light block m₁ = 1 kg

Heavy block m₂ = 1 kg

Collisions

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Guided Inquiry

1 · Blocks, a wall and elastic impacts

Here’s the whole setup. A light block of mass m₁ rests on a frictionless floor, sitting between an immovable wall and a heavy block of mass m₂ sliding towards it.

Every impact — block on block and block on wall — is perfectly elastic, so the total kinetic energy never changes. Your job sounds simple: count every impact until no further contact is possible.

2 · Equal masses

Let’s start with the easiest case: the two blocks have the same mass. How many collisions do you think happen in total?

3 · A hundred times heavier

Now let’s make it lopsided. The heavy block is now 100 times the light one, and nothing else changes. Compared with the 3 collisions you just counted, how many do you expect now?

4 · Ten thousand times heavier

Let’s push it further. The mass ratio is now 10 000. The counts so far are 3 and 31. What do you think comes next?

5 · What the count measures

Here’s a lovely way to see what’s going on. Rescale the state to (√m₁·v₁, √m₂·v₂). Energy conservation pins that point on a fixed circle, and each impact reflects it off one of two lines meeting at a wedge angle θ. So what do you think the count equals?

Listen

Controls

▾

Mass exponent N

Selecting an N resets the scene immediately.

Heavy block initial speed 1.00

0.20 m/s2.00 m/s

Applies before a run starts (or on next ↻ Reset). The collision count is independent of v₀.

Live State lab frame

▾

Kinetic energy0.00 J

Last event—

Statuspaused

Block velocities v₁, v₂ are shown as color-coded arrows on the canvas.

1D elastic collision (block–block)

Primes denote post-collision values. Conserves both momentum m₁v₁+m₂v₂ and kinetic energy ½m₁v₁²+½m₂v₂².

Wall bounce (light block ↔ immovable wall)

The wall has effectively infinite mass, so v₁ simply reverses sign.

Galperin's geometric reduction

In rescaled coordinates uᵢ = √mᵢ·vᵢ each collision is a reflection in a wedge of angle θ = arctan(√(m₁/m₂)). Π = ⌈π/θ⌉−1 equals ⌊π/θ⌋ whenever π/θ is not an integer; at N=0, θ = π/4 exactly, so Π = 4−1 = 3. For large N, θ ≈ 10⁻ᴺ and Π = ⌊π·10ᴺ⌋.

Halting condition

Once the light block is no longer heading toward the wall and the heavy block is receding at least as fast, no further collision is possible.

The Physics

Galperin's billiard · what this sim shows, the physics behind it, and every control

At a glance

A heavy block slides toward a light block resting between it and a wall. Every impact is perfectly elastic and the floor has no friction, so the sim counts impacts until no more can happen. Explaining that count takes a change of coordinates that turns collisions into reflections.

What you'll learn

How the number of collisions depends on the mass ratio m₂/m₁.
How the heavy block's starting speed affects the run.
How rescaling velocities by √m turns each impact into a reflection.
How to count collisions by counting mirrors in a wedge of angle θ.

The set-up at the start. The heavy block moves left; the light block is trapped between it and the wall.

What's on screen

Light block (blue): m₁ = 1 kg, at rest at the start.
Heavy block (pink): m₂ = 100ᴺ kg, released leftward at speed v₀.
Velocity arrows, labelled v₁ and v₂ in m/s: leftward is negative.
Collisions badge: every impact counts, block–block and block–wall alike.
Live State panel: kinetic energy and the type of the last impact.

See itRun the equal-mass case, N = 0. Watch for: the light block shuttling between the wall and the heavy block, with the Collisions badge counting each impact.

The controls

Press ⌖ on a row to see where that control sits in the sim.

Control	What it changes	Range
Mass exponent N	The heavy block's mass, m₂ = 100ᴺ kg. The light block stays 1 kg. Changing N resets the run.	N = 0–3 opens at N = 0 (1 kg)
Heavy block initial speed	How fast the heavy block moves left when released. It sets the starting kinetic energy. Takes effect when a run starts.	0.20–2.00 m/s opens at 1.00 m/s
Collisions badge	The running count of impacts, block–block and wall.	readout
Live State panel	Total kinetic energy and the type of the last event (block–block, wall bounce or escaping).	panel opens collapsed
▶ Play / ⏸ Pause	Sends the heavy block in, or freezes both blocks between impacts.	top bar
Speed	How fast the run plays on screen. Only the playback rate changes, not the physics.	0.5×–4× opens at 1×
↻ Reset	Puts both blocks back at the start and zeroes the count. It keeps your N and v₀.	top bar

Fixed: m₁ = 1 kg, frictionless floor, perfectly elastic impacts, a fixed wall on the left.

One impact barely changes the heavy block

A block–block impact conserves momentum and kinetic energy. A wall impact only reverses the light block's velocity.

Block–block: momentum and kinetic energy are both conserved.

Wall bounce: the wall does not move, so v₁ just flips sign.

Rearranged, a block–block impact changes the heavy block's velocity by v₂′ − v₂ = [2m₁/(m₁ + m₂)] (v₁ − v₂). At m₂ = 100 m₁ that factor is 2/101 ≈ 0.02: one impact moves v₂ by about 2% of the relative velocity. The light block must bounce back and forth many times to turn the heavy block around.

Key idea. Impacts only pass energy back and forth between the blocks. None is lost.

See itRelease the 100 kg block at 0.50 m/s (N = 1) and pause after the first impact. Watch for: v₂ barely changing (−0.50 to −0.49 m/s) while the light block's arrow is about twice as long as the heavy block's.

A point on a circle holds both velocities

Describe the state by one point with coordinates u₁ = √m₁·v₁ and u₂ = √m₂·v₂.

Rescaled velocities. The sum of squares is twice the kinetic energy.

Since the kinetic energy is ½(u₁² + u₂²), conserving it keeps the point on a circle around the origin. Each impact then reflects the point across a line through the origin:

a wall impact flips u₁, a reflection in the u₂-axis;
a block–block impact keeps the part of u along (√m₁, √m₂), the part fixed by the momentum, and flips the rest.

These two mirror lines meet at the angle θ = arctan √(m₁/m₂).

See itRun N = 2 at 2.00 m/s. Watch for: the kinetic energy staying at 20,000 J through every impact.

Unfolding the wedge turns the count into mirrors

Reflect the picture instead of the point. The point then moves in a straight line, and the mirrors fan out at θ, 2θ, 3θ, …

N = 0: θ = π/4, and a straight path crosses three mirrors.

Every mirror the path crosses is one impact. It crosses each mirror whose angle kθ is below π; the mirror at kθ = π is parallel to the path. So the number of impacts is the number of whole steps of θ below π:

The ceiling form matters when π/θ is a whole number. For equal masses π/θ = 4 exactly, and the count is 3, not 4.

See itRun N = 1 at 2.00 m/s, where θ = 0.0997 rad and π/θ ≈ 31.5. Watch for: the counter stopping at ⌈31.5⌉ − 1 = 31.

Digits of π appear with every factor of 100 in mass

For m₂/m₁ = 100ᴺ, tan θ = 10⁻ᴺ. For N ≥ 1, θ is then almost exactly 10⁻ᴺ, and π/θ is just above π·10ᴺ.

N v₀

These sliders move the sim too.

A hundred times the mass gives about ten times the collisions, because the count is about π/θ ≈ π√(m₂/m₁).

See itRun N = 3, a million-kilogram block, at 2.00 m/s. Watch for: the counter climbing and stopping at 3141.

Starting speed sets the clock, not the count

The run ends when no impact is possible. That happens once both of these hold:

The light block is not heading for the wall, and the heavy block is moving away at least as fast.

Multiply every velocity by k: energies grow by k², times shrink by 1/k, and the order of the impacts stays the same.
In the circle picture only the radius changes. The mirrors, and the angle θ between them, depend on the masses alone.
The run ends without friction: the energy is still there, but the heavy block can no longer be caught.

See itRun N = 1 at 2.00 m/s, then at 1.00 m/s. Watch for: the same count both times, with the kinetic energy four times smaller the second time.

Where it matters

Galperin's result. Galperin published this link between collision counts and π in 2003, in a paper titled "Playing pool with π".
Billiards and hard-particle gases. The rescale-and-unfold trick turns a few hard particles moving on a line into one point bouncing in a polygon. That is a standard way to study such systems.
Not a way to compute π. N = 10 would need a 10²⁰ kg block and perfectly elastic impacts.

Go deeper: why the digits are exact

Since arctan x ≈ x − x³/3, π/θ ≈ π·10ᴺ + (π/3)·10⁻ᴺ for N ≥ 1. That is only about 10⁻ᴺ above π·10ᴺ.
The count ⌈π/θ⌉ − 1 equals ⌊π·10ᴺ⌋ as long as this small push does not carry π/θ past the next whole number. Computing it directly shows it does not for N = 1 to 7. The sim stops at N = 3, where 3141 impacts already play out in about 30 s at 1× and v₀ = 2.00 m/s.

Summary

In one line. The number of collisions is ⌈π/θ⌉ − 1, with θ = arctan √(m₁/m₂): the number of mirrors a straight path crosses once the collisions are unfolded.

Key points

Rescaling by √m puts the state on a circle and turns each impact into a reflection.
Two mirrors meet at θ, set by the mass ratio alone.
For m₂/m₁ = 100ᴺ the count is ⌊π·10ᴺ⌋: 3, 31, 314, 3141.
The starting speed changes the energy and the duration, not the count.

Common mix-ups

"The blocks stop colliding because energy is lost." No energy is lost. The run ends when v₁ ≥ 0 and v₂ ≥ v₁, so nothing is left to hit.
"The count is the mass ratio." A hundred times the mass gives about ten times the collisions.
"A faster heavy block means more collisions." v₀ does not enter θ, so it does not enter the count.

Things to try on your own

Set N = 1, open Live State and watch Last event through the first few impacts. Watch how block–block and wall impacts take turns.
Work out π/θ for N = 2 from θ = arctan(0.01), then compare ⌈π/θ⌉ − 1 with the counter.
Run N = 2 at 0.20 m/s and at 2.00 m/s. Compare the kinetic energies and the final counts.