Two snooker balls click and spring apart. A car folds itself around a lamp-post and stops dead. A lump of clay slaps into a wall and just… sticks. All three are collisions — bodies meeting and pushing on each other for a fleeting instant — yet they end in wildly different ways. The balls keep almost all their bounce; the car keeps none; the clay keeps none and travels on stuck to whatever it hit.

There is one thing every isolated collision has in common, and one thing that separates them. The shared law is the conservation of momentum: with no outside forces, the total momentum after equals the total momentum before, always. What differs is what happens to the kinetic energy — and that single question, "was the kinetic energy kept or lost?", is what sorts every collision into elastic or inelastic.

Momentum always survives — kinetic energy is the question

Take two bodies of mass m_1, m_2 approaching with velocities u_1, u_2 and leaving with velocities v_1, v_2. Because the system is isolated, momentum is conserved in every case:

This single equation is true whether the bodies bounce, crumple, or stick. But it is only one equation, and after a collision there are usually two unknown velocities — so on its own it is not enough to solve everything. The extra information comes from the kinetic energy, and that is exactly where collisions split into two families.

• In an elastic collision the total kinetic energy is also conserved. Nothing is lost to heat, sound or squashing — the store of motion is handed over intact. \tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2 = \tfrac12 m_1 v_1^2 + \tfrac12 m_2 v_2^2. Perfectly elastic collisions are an idealisation, but some come astonishingly close: gas molecules, many subatomic and nuclear collisions, and — near enough for a school problem — hard steel ball bearings or snooker balls.
• In an inelastic collision momentum is still conserved, but kinetic energy is not — some of it is turned into heat, sound and the permanent work of bending metal or squashing clay. Most everyday collisions are inelastic. The extreme case, where the bodies stick together and move off as one, is called perfectly inelastic, and it loses the most kinetic energy that momentum conservation will allow.

What is conserved in a collision

For two bodies colliding in an isolated system:

• Momentum is always conserved: m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 — in every collision, elastic or not.
• Kinetic energy is conserved only if the collision is elastic: \sum \tfrac12 m u^2 = \sum \tfrac12 m v^2. If it is inelastic, the total kinetic energy afterwards is less than before.
• A perfectly inelastic collision is one in which the bodies join and move with a single common velocity v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2} — the maximum possible loss of kinetic energy.

Perfectly inelastic: worked example (they stick together)

A 1200\ \text{kg} car travelling at 15\ \text{m/s} runs into the back of a stationary 800\ \text{kg} car; their bumpers lock and they move off together. Find (a) their common velocity and (b) the kinetic energy lost.

Step 1 — conserve momentum. Only the first car is moving, so u_2 = 0. They stick, so afterwards both share one velocity v:

Step 2 — kinetic energy before.

Step 3 — kinetic energy after (one lump of 2000\ \text{kg} moving at 9\ \text{m/s}):

Step 4 — the loss.

Forty per cent of the kinetic energy has gone — into crumpled metal, a bang, and warmth. Momentum, in contrast, is untouched: 2000 \times 9 = 18\,000\ \text{kg·m/s}, exactly what the moving car brought with it.

Elastic: worked example (the bounce that keeps its energy)

A 2\ \text{kg} ball moving at 3\ \text{m/s} strikes a stationary 1\ \text{kg} ball head-on in a perfectly elastic collision. Find both velocities afterwards.

Now we have two conservation laws — momentum and kinetic energy — so two equations for the two unknowns v_1, v_2:

Solving the pair (the trivial "no collision" answer v_1 = 3,\ v_2 = 0 is rejected) gives

Check both laws. Momentum: 2(1) + 1(4) = 6 ✓. Kinetic energy: \tfrac12(2)(1^2) + \tfrac12(1)(4^2) = 1 + 8 = 9\ \text{J} ✓ — every joule accounted for, so the collision really is elastic.

Is this collision elastic? Check the kinetic energy

Sometimes you are handed all four velocities and asked which kind of collision it was. The recipe is simple: momentum will balance either way, so the test is the kinetic energy — work out the total before and after and compare.

A 4\ \text{kg} trolley moving at 5\ \text{m/s} hits a stationary 4\ \text{kg} trolley. Afterwards the first moves at 1\ \text{m/s} and the second at 4\ \text{m/s} (same direction). Elastic or inelastic?

Momentum check first (it must balance for the data to be valid): 4(5) = 20 before, 4(1) + 4(4) = 20 after ✓.

Now the kinetic energy:

Kinetic energy has fallen from 50\ \text{J} to 34\ \text{J} — a loss of 16\ \text{J}. So although momentum is perfectly conserved, this collision is inelastic. Had the two totals been equal, it would have been elastic.

Explosions: a collision run backwards

An explosion is the mirror image of a perfectly inelastic collision. There, two bodies merge and lose kinetic energy; here, one body flies apart and gains kinetic energy — released from a chemical, elastic or nuclear store inside it. Yet momentum is still conserved: if the object started at rest, the total momentum was zero, so the fragments must fly off with momenta that cancel exactly.

A stationary 3\ \text{kg} firework bursts into a 1\ \text{kg} piece and a 2\ \text{kg} piece. The light piece shoots off at 4\ \text{m/s}. How fast, and which way, does the heavy piece go?

The minus sign says the 2\ \text{kg} piece flies off at 2\ \text{m/s} in the opposite direction — a small heavy chunk kicking back against a light fast one, just like a gun recoiling against its bullet. The kinetic energy, meanwhile, has climbed from nothing to \tfrac12(1)(4^2) + \tfrac12(2)(2^2) = 8 + 4 = 12\ \text{J}, drawn straight from the gunpowder. Momentum conserved; kinetic energy created from a store — the reverse of a crash.

Play with a collision

Set the mass and starting velocity of each cart, then pick the collision type. The figure works out the velocities after the collision (momentum is always conserved) and draws a bar for the total kinetic energy before and after. Watch the key pattern: switching from elastic to inelastic to stick together leaves the momentum untouched but makes the "after" energy bar shrink further and further — with the perfectly inelastic case losing the most.