Try to stop a rolling tennis ball with your hand and it is no trouble at all. Try to stop a rolling car with your hand and you will be flattened — even if, at that instant, both are creeping along at the very same slow speed. And a bullet is lighter than a tennis ball, yet nobody would stand in front of one. Clearly "how fast it's going" is not the whole story of how hard something is to stop. What matters is a blend of how much stuff is moving and how fast it moves — and that blend has a name: momentum.

You already know velocity and Newton's second law. This page takes both a step deeper. Momentum is the quantity of motion an object carries, and it turns out to be the key that unlocks the real form of Newton's second law — the one that explains crumple zones, airbags, and why a cricketer draws their hands back to catch a fast ball.

Momentum: the quantity of motion

Momentum, given the symbol p, is defined as mass times velocity:

With mass m in kilograms and velocity v in metres per second, momentum comes out in kilogram-metres per second, \text{kg·m/s}. A 1200\ \text{kg} car doing 20\ \text{m/s} has p = 1200 \times 20 = 24\,000\ \text{kg·m/s}; a 0.06\ \text{kg} tennis ball at the same speed has only p = 0.06 \times 20 = 1.2\ \text{kg·m/s}. The car carries twenty thousand times the motion, which is exactly why it is twenty thousand times harder to stop.

A tiny mass moving very fast can match a huge mass ambling slowly. The 0.004\ \text{kg} bullet leaving a rifle at 900\ \text{m/s} carries p = 0.004 \times 900 = 3.6\ \text{kg·m/s} — more than the tennis ball above, from a fraction of the mass. Momentum weighs the two ingredients together.

Momentum has a direction

Because velocity is a vector, so is momentum: a ball flying east at 5\ \text{m/s} and the same ball flying west at 5\ \text{m/s} have momenta that are equal in size but opposite in sign. This is not a technicality — it is the single most important habit in every momentum calculation. Pick one direction as positive; anything moving the other way gets a minus sign, and you must carry that minus sign through the arithmetic.

The pay-off appears the moment something rebounds. Stopping a ball removes its momentum; bouncing it back reverses its momentum, which is a far bigger change — as we will see, a change of nearly double. Keep the sign, and the physics does the work for you.

Newton's second law, in its true form

You met Newton's second law as F = m\,a. That is only a special case. Newton himself wrote the law in terms of momentum: the resultant force on an object equals the rate at which its momentum changes.

Read it as "force is how quickly the momentum is changing." If the mass stays constant, the only thing that can change is the velocity, and the equation collapses straight back into the familiar form:

because \Delta v / \Delta t is just the acceleration. So F = m\,a is F = \Delta p / \Delta t with the mass held fixed. The momentum version is more general, though: it still works when the mass itself is changing — a rocket burning fuel, a hopper filling with sand, a raindrop growing as it falls — situations where F = m\,a alone cannot cope.

Impulse: force acting over time

Multiply both sides of F = \Delta p / \Delta t by the time \Delta t and the rearrangement is worth a name of its own. The product force times the time it acts is called the impulse, and it equals the change in momentum:

Impulse is measured in newton-seconds, \text{N·s}. And here is a small delight: a newton-second is exactly the same unit as a kilogram-metre per second, because 1\ \text{N·s} = 1\ \tfrac{\text{kg·m}}{\text{s}^2} \times \text{s} = 1\ \text{kg·m/s}. That has to be true — impulse is a change in momentum, so they cannot have different units.

This is the equation that runs a whole slice of the world. To change an object's momentum by some fixed amount \Delta p, you can apply a big force for a short time, or a small force for a long time — the product, the impulse, is the same either way. That single trade-off is the idea behind every safety device you are about to meet.

Impulse is the area under a force–time graph

In a real collision the force is not constant — it surges to a peak and dies away again. Since F\,\Delta t is impulse, adding up force times little slices of time is exactly the area under the force–time graph, and that area is the total change in momentum.

The figure below is a collision that always delivers the same impulse (\Delta p = 6\ \text{N·s}, the shaded area). Drag the slider to stretch the collision out over a longer time. Because the area must stay fixed, a longer collision forces the peak down: the pulse gets wider but flatter. That falling peak is the whole point of a crumple zone or an airbag — spread the same momentum change over more time and the force drops.

Why a longer collision is a gentler one

Rearrange the impulse equation to make the force the subject:

The change in momentum \Delta p is usually fixed — a falling person must lose all their downward momentum, a caught ball must be brought to rest. So the only lever you can pull is \Delta t, the time the stop takes. Make the collision last longer and F shrinks in exact proportion. This one line explains a surprising amount of everyday engineering:

• Car crumple zones fold and collapse so the car takes longer to stop, stretching \Delta t and cutting the force on the passengers.
• Airbags let your head decelerate over tens of milliseconds instead of hitting a rigid dashboard in a couple of milliseconds — same \Delta p, far smaller force.
• Catching a ball, you draw your hands back as you catch, lengthening the time to stop the ball so it stings less.
• Landing a jump, you bend your knees to extend the stopping time and spare your joints; land stiff-legged and the same \Delta p arrives in a flash — a huge force, and a likely injury.
• Crash mats, packaging and running shoes all do the same trick: soften the stop by stretching it out in time.

Worked examples

Example 1 — momentum. A 0.5\ \text{kg} football is kicked at 18\ \text{m/s}. Its momentum is

Example 2 — impulse from a force. A rocket motor pushes with a steady 250\ \text{N} for 4\ \text{s}. The impulse it delivers is

so the rocket's momentum increases by 1000\ \text{kg·m/s}.

Example 3 — force from a change of momentum. A 0.15\ \text{kg} ball hits the floor moving down at 8\ \text{m/s} and is brought to rest in 0.02\ \text{s}. Taking downwards as positive, the change in momentum is

so the average force from the floor is

— the minus sign says it acts upwards, against the motion, with size 60\ \text{N}.

Example 4 — a rebound (mind the signs!). The same 0.15\ \text{kg} ball now bounces: it arrives downwards at 8\ \text{m/s} and leaves upwards at 6\ \text{m/s}, in the same 0.02\ \text{s} of contact. With downwards positive, the final velocity is -6\ \text{m/s}:

The change is bigger than in Example 3 — 2.1 against 1.2 — because reversing the momentum takes more than merely removing it. The force is F = -2.1 / 0.02 = -105\ \text{N}: nearly double the stopping force.

Example 5 — impulse as an area. A struck ball feels a force that rises straight from 0 to a peak of 400\ \text{N} and falls straight back to 0 over a total contact time of 0.03\ \text{s}. The force–time graph is a triangle, so the impulse is its area:

The ball's momentum therefore changes by 6\ \text{kg·m/s}.

Example 6 — the safety trade-off, in numbers. A 70\ \text{kg} person hits an airbag while moving at 14\ \text{m/s}. They must lose \Delta p = 70 \times 14 = 980\ \text{kg·m/s} of momentum whatever happens. Stop dead against the dashboard in 0.01\ \text{s} and the force is 980 / 0.01 = 98\,000\ \text{N} — bone-breaking. Let the airbag stretch the stop to 0.2\ \text{s} and the force falls to 980 / 0.2 = 4\,900\ \text{N}: twenty times gentler, for exactly the same change in momentum.