Rack up the balls, line up the cue, and smash the break. In the instant the cue ball strikes the pack the whole tidy triangle explodes outward — fifteen balls flying off in fifteen directions. It looks like chaos, but hidden inside it is one of the most reliable bookkeeping laws in all of physics. Add up the "quantity of motion" of every ball before the break, add it up again after, and the total is exactly the same. Nothing about the collision — the clatter, the spin, the balls that stop dead — changes that running total. That conserved "quantity of motion" is momentum, and learning to track it turns a mess of colliding objects into a problem you can solve on the back of an envelope.

This page builds the whole toolkit around momentum in one sweep: what momentum is, why Newton's second law is really a statement about it, how a force acting over time (an impulse) changes it, why the total momentum of an isolated system can never change, where the special point called the centre of mass comes from, and finally how all of this lets you predict the outcome of a collision — whether the objects bounce apart or stick together.

Momentum: mass times velocity

The momentum of an object is its mass times its velocity. It is a vector — it points the way the object is moving — and we write it

A heavy lorry rolling slowly and a light bullet flying fast can carry the same momentum, because momentum trades mass against speed. The SI unit is the kilogram-metre per second (\text{kg·m/s}), which has no cosy name of its own. Momentum captures something velocity alone misses: how hard an object is to stop. It takes far more to halt a 40{,}000\ \text{kg} lorry at 2\ \text{m/s} than a jogger at the same speed, even though their velocities are identical — because the lorry's momentum is enormous.

In one dimension — objects moving along a single line, which is where we will spend most of this page — momentum is just a signed number: positive one way, negative the other. Keeping those signs honest is the single most important habit in every collision problem below.

Newton's second law, restated: \mathbf{F} = d\mathbf{p}/dt

You have probably met Newton's second law as \mathbf{F} = m\mathbf{a}. But that is a special case. Newton himself framed the law in terms of momentum: the net force on a body equals the rate of change of its momentum.

If the mass is constant it slides out of the derivative and you recover the familiar \mathbf{F} = m\,d\mathbf{v}/dt = m\mathbf{a}. But the momentum form is more general: it still works when the mass itself changes with time — a rocket burning and ejecting fuel, a raindrop growing as it falls, a railway wagon scooping up coal. That is why physicists treat \mathbf{F} = d\mathbf{p}/dt as the real statement of the second law and \mathbf{F} = m\mathbf{a} as the everyday shortcut.

Impulse: force acting over time

Rearrange the force law and integrate over a stretch of time. Whatever a force does to momentum between times t_1 and t_2 is captured by the impulse \mathbf{J}:

This is the impulse–momentum theorem, and it is beautifully practical: the change in an object's momentum equals the area under its force-versus-time graph. A small force applied for a long time can produce the same momentum change as a huge force applied for a split second — the same area under the curve. When a force is roughly constant, the integral collapses to a product you can do in your head:

This is exactly why cars have airbags and crumple zones. In a crash your momentum must change from "fast" to "zero" — that momentum change \Delta p is fixed by your mass and speed and nothing can alter it. But \Delta p = F\,\Delta t, so the force on you depends on how long the stop takes. Hit a rigid dashboard and you stop in milliseconds — a monstrous force. Sink into an airbag and you stop over a tenth of a second — the same momentum change spread over more time, so a far gentler force. A boxer rolling with a punch, a gymnast bending their knees on landing, a cricketer drawing the hands back to take a catch: all of them are stretching \Delta t to shrink F.

Worked example: impulse as the area under a pulse

Suppose a ball of mass m = 0.20\ \text{kg} is struck by a bat. The contact force rises and falls, but a data logger tells us it averages F_{\text{avg}} = 150\ \text{N} over a contact time of \Delta t = 8.0\ \text{ms} = 0.0080\ \text{s}. The impulse is the area of that force pulse:

By the impulse–momentum theorem this equals the ball's change in momentum, \Delta p = 1.2\ \text{kg·m/s}. If the ball started at rest, its speed leaving the bat is

Notice the units line up: a newton-second (\text{N·s}) is exactly a kilogram-metre per second (\text{kg·m/s}), because a newton is a \text{kg·m/s}^2 and multiplying by seconds cancels one of the seconds. Impulse and momentum genuinely have the same units — they are the same kind of thing.

The centre of mass

When many particles move at once, there is one special point that behaves with astonishing simplicity: the centre of mass (COM). For particles of mass m_i at positions \mathbf{r}_i, it is the mass-weighted average position,

Heavier particles pull the balance point towards themselves; the COM sits closer to the big masses. Now differentiate \mathbf{R} twice and multiply by the total mass. All the internal forces cancel in third-law pairs, leaving only the external force, and you get a clean, remarkable result:

In words: the centre of mass moves as if the entire mass were concentrated there and the total external force acted on that single point. Throw a spinning wrench across the room and every bit of it tumbles in a complicated way — but its centre of mass traces a perfect, smooth parabola, exactly like a single thrown pebble. A firework shell arcs up as one point; at the top it bursts, but the glowing fragments still fly so that their COM continues along the very same parabola the unexploded shell would have followed (until air resistance and the ground intervene). Internal explosions cannot budge the centre of mass — only outside forces can.

Worked example: finding a centre of mass

Put a 3\ \text{kg} mass at x = 0\ \text{m} and a 1\ \text{kg} mass at x = 4\ \text{m} on a line. Where is the centre of mass?

It sits at x = 1\ \text{m} — much closer to the heavy 3\ \text{kg} mass, as expected. Add a third mass, say 2\ \text{kg} at x = 10\ \text{m}, and just extend the same weighted average:

The recipe never changes: multiply each mass by its position, add them up, and divide by the total mass.

Collisions: watch the momentum, watch the centre of mass

A collision is any brief, forceful interaction between objects. During it the objects push hugely on each other, but those are internal forces — equal and opposite pairs that cancel. So as long as no big external force acts during the brief contact (gravity and friction are usually negligible over those few milliseconds), the total momentum is conserved. The figure below shows the simplest case: a heavy block catching up with a light one and sticking to it. Reveal it step by step and watch the marked centre of mass — it glides at a constant velocity the whole time, before, during, and after the crash, exactly as M\mathbf{a}_{\text{com}} = \mathbf{F}_{\text{ext}} = 0 demands.

Perfectly inelastic collisions: they stick together

In a perfectly inelastic collision the objects lock together and move off as one. Only one conservation law applies — momentum — and that alone pins down the answer. With masses m_1, m_2 and initial velocities u_1, u_2, the combined lump moves at a single final velocity v:

That final velocity is exactly the velocity of the centre of mass — which is why the stuck-together block in the figure simply continues at the COM speed. Concretely, a 2\ \text{kg} block at 3\ \text{m/s} hitting a stationary 1\ \text{kg} block gives

Check the kinetic energy, though. Before: \tfrac12(2)(3)^2 = 9\ \text{J}. After: \tfrac12(3)(2)^2 = 6\ \text{J}. Three joules have vanished — turned into heat, sound, and permanent deformation as the blocks crunched together. Momentum survived; kinetic energy did not. That is the signature of an inelastic collision.

Elastic collisions: both p and kinetic energy survive

At the opposite extreme is the elastic collision, in which the objects bounce apart with no loss of kinetic energy — think of two hardened steel balls, or (very nearly) billiard balls, or the collisions of ideal gas molecules. Now you have two conservation laws working at once, momentum and kinetic energy:

Two equations, two unknown final velocities — so the outcome is fully determined. Solving them (the algebra is a satisfying exercise) gives, for a target initially at rest (u_2 = 0):

These formulas hide three lovely special cases. If the masses are equal, then v_1 = 0 and v_2 = u_1: the incoming ball stops dead and hands its entire velocity to the target — exactly the crisp "stun shot" every pool player knows. If a heavy ball strikes a tiny one, it barely slows and knocks the light one off at nearly 2u_1. And if a light ball strikes a massive wall (m_2 \gg m_1), it bounces straight back at almost its original speed, v_1 \approx -u_1 — a ball rebounding elastically off a floor.