Collision & Momentum Simulator.

Five pieces of real apparatus on one stage. Set the masses, choose how bouncy the contact is, and watch a ledger say exactly what was conserved and what was lost.

Mode

Rig

The apparatus — Linear air track, 2.00 m, frictionless

Press Run to release the gliders. Space runs and pauses, R resets, and the arrow keys nudge the selected body's velocity.

Velocity v₁

0.000 m/s

Velocity v₂

0.600 m/s

Total momentum

0.150 kg·m/s

Total kinetic energy

0.045 J

Kinetic energy lost

0.000 J

Restitution e

1.000

Impulse |J|

0.150 kg·m/s

Centre of mass

0.300 m/s

Conservation ledger — what was kept, and what was not

Quantity	Before	After	Change	Verdict
Total momentum	0.150 kg·m/s	—	—	pending
Total kinetic energy	0.045 J	—	—	pending

Run the collision to settle it.

Graph

Collision

Restitution e 0.50

Choose Partly to unlock this. Elastic pins it at 1; Inelastic pins it at 0.

Mass m₁ —

Velocity u₁ —

Mass m₂ —

Velocity u₂ —

Load a classic set-up

Momentum is mass times velocity, and it is a vector

Momentum is the product of a body’s mass and its velocity. Because velocity is a vector, so is momentum: direction is part of the value, and two equal momenta pointing opposite ways cancel exactly. This is not a technicality — it is the single most common source of wrong answers in the topic. Two 1 kg trolleys, one going right at 3 m/s and one going left at 3 m/s, have a total momentum of zero and a total kinetic energy of 9 J. Nothing about those two numbers is contradictory, and understanding why is most of the subject.

p = m v · F = dp/dt · J = F Δt = Δp

→Pick a positive direction before you write anything down, and commit to it for the whole problem.
→Kinetic energy is a scalar. It is never negative and it never cancels, which is why a zero-momentum system can still be full of energy.
→Newton’s second law in its original form is about momentum, not acceleration: force is the rate of change of momentum.

One pair of equations covers every 1D collision

Solve momentum conservation together with the definition of the coefficient of restitution and you get both final velocities directly. This single pair covers elastic, partly elastic and perfectly inelastic collisions — you never need a separate formula for the special cases, because setting e to 1 or 0 produces them. Committing this to memory instead of three separate results is the highest-value thing on this page.

v₁ = [m₁u₁ + m₂u₂ + m₂e(u₂−u₁)] / (m₁+m₂) · v₂ = [m₁u₁ + m₂u₂ + m₁e(u₁−u₂)] / (m₁+m₂)

→Set e = 1 and it becomes the elastic result; set e = 0 and both reduce to the centre-of-mass velocity.
→The velocities are signed. Feeding in a speed and hoping for the best is how the sign errors start.
→Equal masses and e = 1 make the two bodies swap velocities exactly — the classic air-track demonstration.

Typical coefficients of restitution

These are representative values at modest impact speeds, and every one of them drops as the impact gets harder. They are listed by contact rather than by material, because that is what restitution actually belongs to. A steel ball on a steel anvil and a steel ball on a wooden bench share a material and share almost nothing else.

→Hardened steel on hardened steel — 0.90 to 0.95. The Newton’s cradle case, and about as elastic as everyday objects get.
→Glass marble on glass — ≈ 0.90.
→Air-table puck rims — ≈ 0.80, set by the bulk elasticity of the plastic.
→Tennis ball on hard court — ≈ 0.75, and specified by the rules of the sport.
→Basketball on hardwood — 0.75 to 0.85, and pressure-dependent.
→Wooden ball on wood — ≈ 0.50.
→Vehicle structure in a barrier impact — 0.10 to 0.20, and deliberately low.
→Projectile embedding in a block — exactly 0, by definition.

Every safety system is buying time

Stopping a given mass from a given speed always needs the same impulse. No design changes that. What a crumple zone, an airbag, a seatbelt pretensioner and a collapsible steering column all do is stretch the time over which that impulse is delivered, and since the impulse is fixed, more time means less force. The numbers are stark: the same 1,400 kg car at 14 m/s pulls about 76 g stopping in 0.15 m and about 13 g stopping in 0.9 m.

F = M v (1 + e) / (2d / v) = M v² (1 + e) / 2d

→Notice that restitution appears: a car that bounces off the barrier suffers a larger momentum change than one that stops dead. Bouncy is worse.
→That is why vehicle structures are designed to deform permanently rather than spring back.
→In a two-vehicle crash both feel the same force. The lighter one feels the larger acceleration, which is what injures people.

Adding speeds instead of signed velocities

This is the most common error in the topic by a wide margin, and it is fatal because the answer still looks like a number. Momentum is a vector: choose a positive direction, write every velocity with its sign, and let the arithmetic handle the rest. Two identical trolleys approaching each other at 3 m/s have zero total momentum, and any method that gives 6 kg·m/s has quietly discarded the physics.

→A negative result for a final velocity is information, not an error — it means that body rebounded.
→Light into heavy always gives a negative v₁. Try it on the track above.
→Kinetic energy uses v², so it never carries a sign. That asymmetry is the point.

How to run a collision experiment

Pick a rig. The air track gives head-on collisions in a straight line; the air table adds a second dimension; the ballistic pendulum, the cradle and the crash rig each isolate one further idea.
Choose the collision type. Elastic pins the restitution at 1, Inelastic pins it at 0 and makes the bodies stick, and Partly unlocks the slider so you can set anything in between.
Set the masses and the starting velocities. A negative velocity means that body travels the other way — momentum is a vector, and that sign is the whole reason the totals come out right.
Press Run and watch the conservation ledger settle. Momentum reads conserved on every isolated run; kinetic energy only when the restitution is exactly 1.
Turn Friction on and run it again. The collision still conserves momentum, but the system stops doing so, because friction is an external force — which is what the word isolated is doing in the law.

Momentum is the quantity that survives

Two bodies meet, deform each other, push apart and leave at new speeds. Almost nothing about that process is simple — the contact lasts a few thousandths of a second, the forces peak at thousands of newtons, and the material behaviour in between is genuinely complicated. Yet one quantity comes out the far side untouched, every single time, and you can predict it without knowing anything at all about what happened during the contact.

That quantity is momentum, mass times velocity. It is a vector, so direction is part of the value, and two equal momenta pointing opposite ways sum to nothing. The reason it survives is Newton's third law: during contact, body 1 pushes body 2 exactly as hard as body 2 pushes back, for exactly the same length of time. The two impulses are equal and opposite, so whatever momentum one body gains the other loses, and the total cannot move.

Kinetic energy has no such guarantee. It is a scalar, it is never negative, and it can be converted into deformation, sound and heat without violating anything. That asymmetry — one quantity locked, the other free to drain away — is the entire subject, and it is what the ledger under the apparatus above is there to make visible.

Elastic, inelastic, and the coefficient of restitution

All collisions conserve momentum. They are classified by what happens to the kinetic energy, and there are three cases worth naming.

Elastic. Every joule comes back. The bodies separate exactly as fast as they approached. Two hardened steel balls come close, gas molecules genuinely achieve it, and nothing you can hold in your hand quite does.
Inelastic. Some energy goes into permanently deforming the bodies, into sound, and into heat. This is nearly everything that actually happens.
Perfectly inelastic. The bodies leave stuck together at the centre-of-mass velocity. This loses the most energy any collision can while still conserving momentum — and note that it does not lose all of it, unless the total momentum happened to be zero to begin with.

The single number that spans all three is the coefficient of restitution, defined along the line of impact as the separation speed divided by the approach speed. An e of 1 is elastic, 0 is perfectly inelastic. The energy lost has a clean closed form that depends only on e and the relative velocity: ΔKE = ½ μ v_rel² (1 − e²), where μ is the reduced mass m₁m₂/(m₁+m₂). That formula is derived entirely separately from the velocity equations, which is why the test suite behind this page checks one against the other rather than merely restating it.

The point most textbooks underplay is that e belongs to the contact, not to the material. It falls as impact speed rises, it shifts with temperature, and on the air track above it is decided by which bumper you bolt on — a steel spring, a rubber block or a velcro pad — rather than by what the glider is made of. The other rigs have no bolt-on bumper, and their contact is fixed by the hardware: puck rims at about 0.8, polished steel at 0.90 to 0.95, a projectile burying itself in wood at exactly 0.

Impulse, contact time, and why crumple zones work

Rewrite Newton's second law as F = Δp/Δt and you get the most practically useful statement in the whole topic: the force in a collision is the momentum change divided by the time taken to make it. The momentum change is fixed by the physics — a given mass arriving at a given speed and stopping has one and only one impulse, and no design can alter it. The time, however, is an engineering choice.

That is the whole argument for a crumple zone. It does not reduce the impulse and it is not there to absorb energy in any mysterious way; it is there to make the stop take longer by letting the structure fold through a greater distance. Because the impulse is fixed, a longer contact time means a proportionally smaller force. Try it on the crash rig above: a 1,400 kg car at 14 m/s stopping in 0.15 m pulls roughly 76 g, and the same impact through 0.9 m pulls roughly 13 g. Same car, same speed, same impulse — six times the crumple distance, one sixth of the force.

The F–t graph makes the trade explicit. The area under the curve is the impulse and it does not change as you drag the crumple slider; only the shape does, spreading flatter and wider or rising into a spike. Everything protective in a vehicle, from the airbag to the seatbelt pretensioner to the collapsible steering column, is doing the same thing to the same graph.

Two dimensions, and the 90-degree rule

An oblique collision looks harder and is not, because one idea carries all of it: a frictionless contact can only push along the line of centres. Resolve both velocities into a component along that line and a component perpendicular to it. The perpendicular components pass through untouched. The components along the line obey exactly the one-dimensional equations you already have. That is the entire method.

The impact parameter — the slider labelled b on the air table — is how far the target sits off the incoming centre line, and it is what sets the angle of that line. At b = 0 the line of centres is the direction of travel and the collision reduces exactly to the 1D case. When b equals the sum of the two radii the bodies barely touch, the line of centres is perpendicular to the motion, and almost nothing happens.

The famous consequence is that two equal masses in an elastic oblique collision always separate at exactly 90°. Write down conservation of momentum and of kinetic energy together for equal masses and the cross term is forced to zero, which for two non-zero velocities means the vectors are perpendicular. It is why a cue ball and an object ball leave a break at a right angle. Both conditions matter: change the masses or drop the restitution below 1 and the angle moves off 90° immediately, which is worth doing on the table above precisely to watch it fail.

What each rig is actually for

The five pieces of apparatus are not five decorations on one simulation. Each isolates something the others cannot show.

The air track is the control experiment: one dimension, effectively no friction, and the restitution is yours to choose. It is the only rig where you can hold everything constant and vary e alone.
The air table adds the second dimension and with it the impact parameter, the line of centres and the 90-degree rule.
The ballistic pendulum exists to be a trap. It is two phases with different rules — a perfectly inelastic collision, where momentum is conserved and energy is not, followed by a swing, where energy is conserved and momentum is not. Apply energy conservation across both and you overestimate the projectile speed by a factor of fifty or more. Almost every student does this once.
Newton's cradle is the cleanest demonstration that momentum and energy must both balance. One ball in gives one ball out because no other outcome satisfies both at once. Two out at half speed would conserve momentum and lose half the energy; one out at double speed would conserve energy and double the momentum. Neither is allowed.
The crash rig moves the interesting quantity from velocity to force, and is where the topic stops being academic.

One detail about the cradle is worth stating because the simulation depends on it: the balls do not collide simultaneously. They touch in sequence, and each contact is its own pairwise impulse. Treating the line as one simultaneous multi-body event gives the wrong answer, and it is also the case a general physics engine handles worst — which is why this page resolves the cradle analytically rather than leaving it to the solver that moves everything else.

The mistakes that account for nearly every wrong answer

Adding speeds instead of signed velocities. Momentum is a vector. A 1 kg trolley going right at 3 m/s and an identical one going left at 3 m/s have a total momentum of zero, not 6 kg·m/s. Pick a positive direction, commit to it, and let the minus signs do their work.
Thinking "inelastic" means momentum is lost. It means kinetic energy is lost. Momentum is conserved in every collision where no external force acts, however much energy disappears.
Using energy conservation across a collision. Only valid when e = 1. The ballistic pendulum is the standard trap, and running the two phases with the same law is the single most expensive error in this topic.
Forgetting that kinetic energy is a scalar. It never cancels the way momenta do. In an explosion the total momentum can stay at zero while the kinetic energy climbs from nothing — which is exactly what the explosion preset above shows.
Applying conservation to a system that is not isolated. This is what the Friction toggle is for. Switch it on and the collision itself still balances, because contact lasts a few hundredths of a second and friction is far too small to matter over that window. What fails is the system over the seconds either side. The word doing the work in the law is isolated, and the only way to really learn that is to break it on purpose.

Frequently Asked Questions

Is momentum conserved in an inelastic collision?

Yes — always, provided no external force acts. This is the commonest confusion in the whole topic, and it comes from the word inelastic sounding as though something goes missing. What is lost is kinetic energy, not momentum. In a perfectly inelastic collision the two bodies leave stuck together at the centre-of-mass velocity, and adding up mv before and after gives exactly the same number. Load the sticky preset on the air track above and watch the ledger: the momentum row reads conserved and the energy row does not.

What is the difference between an elastic and an inelastic collision?

Both conserve momentum. They differ only in what happens to kinetic energy. An elastic collision gives every joule back — the bodies separate exactly as fast as they approached, and the coefficient of restitution is 1. An inelastic one turns some of that energy into permanent deformation, sound and heat. A perfectly inelastic collision loses the most energy any collision can while still conserving momentum, and the bodies move off together. Nothing at everyday scale is perfectly elastic; two hardened steel balls come closest, at roughly e = 0.95.

What is the coefficient of restitution?

It is the ratio of separation speed to approach speed along the line of impact: e = (v₂ − v₁) / (u₁ − u₂). An e of 1 is perfectly elastic, 0 means the bodies stick together, and everything real sits in between. The point most often missed is that e is not a property of a material — it belongs to the contact. It falls as impact speed rises, it shifts with temperature, and on the air track above it is decided by which bumper you bolt on rather than by what the glider is made of.

Why do two equal masses separate at 90 degrees after an elastic collision?

Because a frictionless impact can only push along the line of centres. The struck body always leaves along that line, and the striker keeps whatever velocity it had at right angles to it. Write momentum and energy conservation together for equal masses and the cross term v₁·v₂ is forced to zero, which for two non-zero velocities means the vectors are perpendicular. It is why a cue ball and an object ball leave a break at a right angle — and it fails the moment the masses differ or the collision is not elastic, both of which you can try on the air table above.

How does a crumple zone reduce the force in a crash?

It does not change the impulse at all. Stopping a given mass from a given speed always needs the same change in momentum, so F × t is fixed before the design starts. A crumple zone increases t by letting the structure fold through a longer distance, and because the impulse is fixed, a longer contact time means a smaller force. The arithmetic is unforgiving: on the crash rig above, a 1,400 kg car at 14 m/s stopping in 0.15 m pulls about 76 g, and the same impact through 0.9 m pulls about 13 g. Same impulse, six times the distance, one sixth of the force.

Why does only one ball swing out of a Newton’s cradle?

Because one ball out is the only outcome that satisfies both conservation laws at once. If two balls left at half the speed, momentum would balance but kinetic energy would be halved; if one left at twice the speed, energy would balance but momentum would double. Lift two and exactly two leave, at the original speed. The other detail worth knowing is that the balls do not collide simultaneously — they touch in sequence, and the simulator above resolves each contact as its own pairwise impulse for exactly that reason, because treating the line as one simultaneous event gets the answer wrong.

Does friction break the conservation of momentum?

Not within the collision, but yes for the system over time. Momentum is conserved when no external force acts, and friction is external. During the impact itself — a few hundredths of a second while the bumpers are touching — friction is far too small to matter, so the collision still balances. Over the seconds either side of it, friction steadily drains momentum from the pair and hands it to the track and the Earth. That is what the Friction toggle above is for: it lets you break the law deliberately and see which half of the statement actually fails.

Collision & Momentum Simulator.

Momentum is mass times velocity, and it is a vector

Momentum is conserved because forces come in pairs

Elastic, inelastic and perfectly inelastic

The centre of mass sails straight through

One pair of equations covers every 1D collision

e compares how fast they leave with how fast they met

Impulse is the bridge between force and momentum

A 2D collision is a 1D collision plus a passenger

Typical coefficients of restitution

What is actually touching, on each piece of apparatus

A sense of scale, so a wrong answer looks wrong

Every safety system is buying time

Restitution is written into the rules of games

Anything that pushes something away is an explosion

How you measure a speed too fast to time

Adding speeds instead of signed velocities

Inelastic does not mean momentum disappears

Carrying energy conservation through an impact

Applying the law to a system that is not isolated

How to run a collision experiment

Momentum is the quantity that survives

Elastic, inelastic, and the coefficient of restitution

Impulse, contact time, and why crumple zones work

Two dimensions, and the 90-degree rule

What each rig is actually for

The mistakes that account for nearly every wrong answer

Frequently Asked Questions

Is momentum conserved in an inelastic collision?

What is the difference between an elastic and an inelastic collision?

What is the coefficient of restitution?

Why do two equal masses separate at 90 degrees after an elastic collision?

How does a crumple zone reduce the force in a crash?

Why does only one ball swing out of a Newton’s cradle?

Does friction break the conservation of momentum?

All tools

Momentum and collisions reference library

Momentum is mass times velocity, and it is a vector

Momentum is conserved because forces come in pairs

Elastic, inelastic and perfectly inelastic

The centre of mass sails straight through

One pair of equations covers every 1D collision

e compares how fast they leave with how fast they met

Impulse is the bridge between force and momentum

A 2D collision is a 1D collision plus a passenger

Typical coefficients of restitution

What is actually touching, on each piece of apparatus

A sense of scale, so a wrong answer looks wrong

Every safety system is buying time

Restitution is written into the rules of games

Anything that pushes something away is an explosion

How you measure a speed too fast to time

Adding speeds instead of signed velocities

Inelastic does not mean momentum disappears

Carrying energy conservation through an impact

Applying the law to a system that is not isolated

How to run a collision experiment

Momentum is the quantity that survives

Elastic, inelastic, and the coefficient of restitution

Impulse, contact time, and why crumple zones work

Two dimensions, and the 90-degree rule

What each rig is actually for

The mistakes that account for nearly every wrong answer

Frequently Asked Questions

Is momentum conserved in an inelastic collision?

What is the difference between an elastic and an inelastic collision?

What is the coefficient of restitution?

Why do two equal masses separate at 90 degrees after an elastic collision?

How does a crumple zone reduce the force in a crash?

Why does only one ball swing out of a Newton’s cradle?

Does friction break the conservation of momentum?

All tools