Momentum & Collision Lab

Simulate 1D cart collisions to explore conservation of momentum, kinetic energy changes, and the coefficient of restitution.

Worked example — Momentum & Collision Lab

Coefficient (e)	1.00
Initial Momentum (Pi)	3.00 kg·m/s
Final Momentum (Pf)	3.00 kg·m/s
Initial Energy (Ki)	10.50 J
Final Energy (Kf)	10.50 J

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Momentum always survives the crash; energy need not 🖖

In any closed system, momentum is always conserved because of Newton's third law: the force Cart 1 exerts on Cart 2 is equal and opposite to the force Cart 2 exerts on Cart 1. Kinetic energy, however, is only conserved in perfectly elastic collisions (e = 1). In real-world macroscopic collisions, energy is always dissipated into thermal energy, sound, and material deformation (e < 1). On the subatomic scale, collisions between gas molecules are perfectly elastic, which is why gases don't eventually settle to the floor as cold puddles of static matter.

The one number that sets the bounce 🖖

The coefficient of restitution e is this tool's master dial. It is simply the separation speed divided by the approach speed: e = 1 is a perfect rebound (a superball), while e = 0 means the carts stick together like wet clay. You can measure it at home — drop a ball and divide bounce height by drop height, which equals e². A basketball that returns to about 64% of its drop height has e ≈ 0.8.

Voyager 2 played billiards with Jupiter 🖖

A spacecraft gravity assist obeys these exact rules with nothing ever touching. In the planet's reference frame the flyby is perfectly elastic (e = 1): the probe leaves at the same speed it arrived, merely redirected. But in the Sun's frame it can pick up almost twice the planet's orbital speed. Voyager 2 gained roughly 10 km/s at Jupiter this way — borrowed from the planet's orbital motion, which slowed by an utterly negligible amount in return.

COLLISIONS — MOMENTUM ALWAYS SURVIVES, ENERGY OFTEN DOES NOT

Which Collision Are You Modelling?

Every collision here conserves momentum: m₁u₁ + m₂u₂ is the same before and after, always. What separates the cases is how much kinetic energy survives, and one number decides that — the coefficient of restitution e, the separation speed divided by the approach speed. Fix e first; then the mass ratio tells you who ends up going where.

Perfectly elastic (e = 1) — kinetic energy survives intact e = 1, K′ = K

Perfectly inelastic (e = 0) — they stick, and lose the most e = 0, v₁ = v₂

Partially elastic (0 < e < 1) — the real-world middle e = (v₂ − v₁)/(u₁ − u₂)

Heavy strikes light — the small one is launched m₁ ≫ m₂ ⇒ v₂ → 2u₁

Light strikes heavy — it bounces straight back m₁ ≪ m₂ ⇒ v₁ → −u₁

Perfectly elastic (e = 1) — kinetic energy survives intact

What you know: Nothing deforms permanently and nothing heats up: the carts separate exactly as fast as they approached. Momentum and kinetic energy are both conserved.

Restitution: e = 1, K′ = K

Worked example: 2 kg at +3 m/s meets 2 kg at −3 m/s with e = 1 → they leave at −3 and +3 m/s. The velocities swap, and all 18 J is still there.

Open this case: Billiard balls bounce

Perfectly elastic (e = 1) — kinetic energy survives intact. Equal masses at e = 1: the carts trade velocities and keep every joule. Nothing deforms permanently and nothing heats up: the carts separate exactly as fast as they approached. Momentum and kinetic energy are both conserved. — Equal masses at e = 1: the carts trade velocities and keep every joule.

Perfectly inelastic (e = 0) — they stick, and lose the most

What you know: The carts leave with one shared velocity, so the separation speed is zero. Momentum still balances, but the kinetic-energy loss is as large as it can possibly be.

Restitution: e = 0, v₁ = v₂

Worked example: 2 kg at 3 m/s hits a stationary 3 kg with e = 0 → both move off at 1.2 m/s, and 5.4 J of the original 9 J is gone: 60% lost.

Open this case: Inelastic stick

Perfectly inelastic (e = 0) — they stick, and lose the most. At e = 0 the carts move off together, and the missing energy has gone into damage. The carts leave with one shared velocity, so the separation speed is zero. Momentum still balances, but the kinetic-energy loss is as large as it can possibly be. — At e = 0 the carts move off together, and the missing energy has gone into damage.

Partially elastic (0 < e < 1) — the real-world middle

What you know: The carts do separate, but more slowly than they approached. e is the ratio between the two: e = (v₂ − v₁)/(u₁ − u₂).

Restitution: e = (v₂ − v₁)/(u₁ − u₂)

Worked example: 3 kg at +4 m/s meets 2 kg at −2 m/s with e = 0.5 → v₁ = +0.4 and v₂ = +3.4 m/s. The 6 m/s approach comes back as 3 m/s, and 16.2 J of the 28 J is lost.

Open this case: Real bumper bounce

Partially elastic (0 < e < 1) — the real-world middle. With 0 < e < 1 they part more slowly than they met, and the difference leaves as heat. The carts do separate, but more slowly than they approached. e is the ratio between the two: e = (v₂ − v₁)/(u₁ − u₂). — With 0 < e < 1 they part more slowly than they met, and the difference leaves as heat.

Heavy strikes light — the small one is launched

What you know: An elastic collision with m₁ much larger than m₂. The heavy cart barely notices the impact; the light one leaves faster than anything arrived.

Restitution: m₁ ≫ m₂ ⇒ v₂ → 2u₁

Worked example: 8 kg at 3 m/s hits a stationary 1 kg elastically → the heavy cart only slows to 2.33 m/s while the light one leaves at 5.33 m/s, near the 2u₁ ceiling.

Open this case: Heavy hits light

Heavy strikes light — the small one is launched. With m₁ ≫ m₂ the heavy cart rolls on almost unchanged and the light one is flung forward. An elastic collision with m₁ much larger than m₂. The heavy cart barely notices the impact; the light one leaves faster than anything arrived. — With m₁ ≫ m₂ the heavy cart rolls on almost unchanged and the light one is flung forward.

Light strikes heavy — it bounces straight back

What you know: An elastic collision with m₁ much smaller than m₂. The incoming cart reverses direction; the heavy one hardly moves.

Restitution: m₁ ≪ m₂ ⇒ v₁ → −u₁

Worked example: 1 kg at 3 m/s hits a stationary 8 kg elastically → the light cart rebounds at −2.33 m/s and the heavy one creeps forward at 0.67 m/s.

Open this case: Light hits heavy

Light strikes heavy — it bounces straight back. With m₁ ≪ m₂ the light cart rebounds at nearly its arrival speed and the heavy one hardly stirs. An elastic collision with m₁ much smaller than m₂. The incoming cart reverses direction; the heavy one hardly moves. — With m₁ ≪ m₂ the light cart rebounds at nearly its arrival speed and the heavy one hardly stirs.

References (1)

Insight block 3 — the gravity assist as an elastic collision with a planet: G. A. Flandro, "Solar electric low-thrust missions to Jupiter with swingby continuation to the outer planets." Journal of Spacecraft and Rockets 5(9), 1029–1033, 1968 — the swingby trajectories that became the Voyager Grand Tour.

Problem solved in full

A 2 kg cart meeting a 3 kg cart elastically 5 steps
A 2 kg cart at 3 m/s meets a 3 kg cart at −1 m/s, elastically. Find both final velocities from the two conservation laws — and then see what the same collision costs if the carts stick together instead.
1. Momentum first, because it holds whatever the collision does. Signs carry the directions, so the second cart's −1 subtracts.
  
  $p : m_{1} u_{1} + m_{2} u_{2} = 2 (3) + 3 (- 1) = 3 kg m/s$
2. Kinetic energy is the second constraint, and it is the one that only holds when the collision is elastic. 10.5 J goes in.
  
  $K : \frac{1}{2} (2) (3)^{2} + \frac{1}{2} (3) (- 1)^{2} = 9 + 1.5 = 10.5 J$
3. Solving two equations, one of them quadratic, is avoidable. For an elastic collision the relative velocity simply reverses — the carts separate as fast as they approached — and that linear statement replaces the energy equation exactly.
  
  $e = 1 : v_{2} - v_{1} = - (u_{2} - u_{1}) = 4 with m_{1} v_{1} + m_{2} v_{2} = 3$
4. Two linear equations then give both velocities directly. The light cart bounces backwards; the heavy one is pushed forwards but ends up slower than the light one arrived.
  
  $⟹ v_{1} = - 1.8 m/s, v_{2} = 2.2 m/s$
5. Now set the restitution to zero and the carts move off together at the momentum-weighted average, 0.6 m/s, carrying 0.9 J.
  
  $e = 0 : v = \frac{3}{5} = 0.6, K_{f} = \frac{1}{2} (5) (0.6)^{2} = 0.9 J ⟹ 91.4% lost$
Answer

The tool prints 3.00 kg·m/s before and after, and 10.50 J before and after — momentum and energy both conserved, which is what elastic means. Drag the restitution slider to 0 and watch the second pair diverge while the first does not: momentum still reads 3.00, but the energy drops to 0.9 J. 91.4% of the kinetic energy is gone, into heat and sound and permanent deformation, and momentum does not notice. That asymmetry is the useful one — momentum is conserved in every collision there is, energy only in the idealised ones, and a crumple zone is a device for making the second number as large as possible.

$e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} : e = 1 \Rightarrow K conserved, e = 0 \Rightarrow 91.4% lost$

Learning path

Energy instead of forces

By the 42Math team · Updated 2026-07-27

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Billiard balls bounce - Equal masses, e=1: velocities swap exactly (+3 → -3 and -3 → +3) — the classic elastic billiard-ball exchange.
Inelastic stick - m1=2 kg at 3 m/s strikes stationary m2=3 kg, e=0: they stick together and move off at 1.2 m/s.
Heavy hits light - m1=8 kg hits stationary m2=1 kg elastically: the heavy cart barely slows (3→2.33 m/s) while the light one shoots off at 5.33 m/s.
Light hits heavy - m1=1 kg hits stationary m2=8 kg elastically: the light cart bounces straight back (3→-2.33 m/s) while the heavy one barely creeps forward (0→0.67 m/s).
Real bumper bounce - e=0.5, m1=3 kg at 4 m/s meets m2=2 kg at -2 m/s: the partial bounce dissipates ~58% of the kinetic energy, a realistic bumper-car collision.