Collisions are easier than forces, and momentum is the reason

Two balls collide. During the fraction of a millisecond they are touching, they deform, the contact area grows and shrinks, pressure waves run through both bodies, and the force between them rises to an enormous peak and falls away again. Nobody knows the shape of that force curve. For almost any real collision, nobody ever will.

And it does not matter. The velocities afterwards can be worked out anyway, exactly, with no information about the interior of the collision at all. That is not a convenience — it is one of the most powerful moves in physics, and it works for a reason worth being precise about.

The ignorance is the feature #

Newton’s third law says the two bodies push on each other with equal and opposite forces at every instant. Whatever that unknown force curve looks like, ball A feels $F(t)$ and ball B feels $-F(t)$, at the same times, for the same duration.

Force is the rate of change of momentum. So over the whole collision, the momentum handed to A is the integral of $F(t)$, and the momentum handed to B is the integral of $-F(t)$ — the same number with the opposite sign, whatever the integral happens to be.

$$\Delta p_A = -\Delta p_B \quad\Longrightarrow\quad \Delta(p_A + p_B) = 0.$$

The unknown function cancels against itself. This is why the method works and why it feels like cheating: the conclusion depends only on the antisymmetry of the interaction, not on its size, shape or duration. A gentle push over a second and a violent bang over a microsecond give the same conclusion, because both are covered by the same cancellation.

The habit generalises well beyond collisions. Whenever a problem contains a complicated middle and a simple beginning and end, the productive question is which quantity has the same value at both ends. That is exactly the move used to find the height of a projectile at the top of its arc without following the flight, and the move used to get a pendulum’s speed at the bottom without solving its equation of motion.

Two equations and one number #

Momentum conservation is one equation, and a collision has two unknown final velocities. Something else is needed.

The extra ingredient is the coefficient of restitution $e$, defined as the ratio of the separation speed afterwards to the approach speed before:

$$e = \frac{v_2 - v_1}{u_1 - u_2}.$$

With $e = 1$ the collision is elastic and kinetic energy is conserved as well. With $e = 0$ the bodies leave together and as much energy is lost as momentum conservation permits. Real collisions sit in between: about 0.9 for a superball, 0.6 for a tennis ball on a hard court, near zero for a lump of wet clay.

Together the two relations pin down both final velocities, and the results are worth having in front of the eye rather than in a formula.

The equal-mass elastic case is the famous one. The velocities are exchanged outright. It looks like the first ball passed its identity to the second, and it is the reason a well-struck cue ball stops on contact when it hits the object ball square. Nothing exotic is happening: it is simply the only pair of final velocities that keeps both the momentum sum and the energy sum unchanged.

The mass-ratio limits are instructive in both directions. A light ball hitting a heavy one bounces back at nearly the same speed; the heavy one takes on a tiny velocity. A heavy ball hitting a light one carries on almost unaffected and flings the light one away at nearly twice its own speed — which is how a bat drives a ball faster than the bat itself is moving, and how a spacecraft steals speed from a planet in a gravity assist.

That factor of two survives into places the mechanics never anticipated. It sets the maximum energy a neutron can lose to a nucleus in one bounce, and hence why moderators in reactors are made of light elements: hydrogen, being the same mass as a neutron, takes the whole of it in a single hit, exactly as in the equal-mass figure.

The energy that goes missing #

Momentum is conserved in every collision. Kinetic energy is conserved in almost none.

This asymmetry between the two conservation laws is not an accident of bookkeeping. Momentum is conserved because the interaction is antisymmetric, which is a statement about the force. Kinetic energy is conserved only if none of it is converted into something else, which is a statement about the materials. Deformation, heating, sound and permanent damage are all places for it to go, and every one of them is invisible to the momentum argument.

Where it goes is into the disordered motion of enormous numbers of molecules — kinetic energy that still exists but is no longer usable, because it is now distributed among particles moving in every direction at once. That conversion is one-directional in practice: two blocks never spontaneously cool down and spring apart. The counting argument for why it does not happen is entropy, and a perfectly inelastic collision is one of the cleanest everyday examples of an irreversible process.

The maximum possible loss is a fixed fraction, and it is fixed by the momentum law rather than by the material. The bodies cannot lose all their kinetic energy unless the total momentum is zero, because they must still be carrying that momentum afterwards, and carrying momentum requires moving. The energy that survives is exactly the energy of the centre-of-mass motion.

Where the lost energy goes #

Saying that kinetic energy is “lost” is a bookkeeping statement that hides the mechanism, and the mechanism is worth naming.

During contact the bodies deform, and deformation stores energy exactly as a compressed spring does. In a perfectly elastic collision the stored energy is entirely returned as the bodies spring apart, and that is what $e = 1$ means physically. In a real one, part of the deformation is permanent, part drives internal vibration, and part becomes the disordered molecular motion that is heat.

So the coefficient of restitution is a summary of how well a material returns stored elastic energy — which is why it depends on temperature, on impact speed, and on how long the contact lasts. A squash ball is nearly dead when cold and lively when warm, and the difference is entirely a change in how much of the deformation energy comes back.

The frame in which it is simplest #

There is a viewpoint from which every collision looks the same, and it makes the whole family of results obvious rather than calculated.

Ride along with the centre of mass. In that frame the total momentum is zero by construction — that is what the centre-of-mass frame means — so before the collision the two bodies approach with equal and opposite momenta, and afterwards they must leave with equal and opposite momenta too.

An elastic collision in that frame is therefore nothing but a reversal: both bodies keep their speeds and turn around. The inelastic case is the other extreme: both stop. Everything in between is the same reversal scaled down by $e$. Every result above is that one sentence, transformed back into whatever frame the observer happens to occupy.

That the answer becomes trivial in one frame and messy in another is a hint about what the messiness was. It was never physics; it was an unfortunate choice of coordinates — the same lesson as choosing axes to suit a slope rather than to suit the room, and a rehearsal for the much stronger version of the idea in relativity, where the choice of frame changes the description and leaves the physics alone.

The rehearsal is closer than it looks. Momentum and energy remain conserved at relativistic speeds, with the definitions adjusted, and the centre-of-momentum frame remains the one in which collisions are easiest to think about — it is the frame every particle physicist quotes collision energies in. What changes is that mass stops being conserved separately, and the energy that goes missing from a perfectly inelastic collision reappears as extra mass in the combined object. The classical picture calls that heat and stops; the relativistic one weighs it — and what “at the same time” means for the two bodies turns out to depend on who is watching.

What the conservation law costs #

The method’s whole appeal is that it never asks what happened during the collision. The price is that it asks something else instead, quietly, and getting that question wrong invalidates everything downstream.

The question is: what is the system? Momentum is conserved for an isolated collection of bodies, and nothing in the arithmetic identifies which bodies belong to the collection. That is supplied by judgement, before any equation is written.

The failure mode is easy to produce. A ball thrown at a wall comes back with its momentum reversed, which is a change of $2mv$ and looks like a flat violation. The rescue is to include the wall, the building and the planet it is bolted to — and then the momentum balances, because the Earth recoils. For a 150-gram ball at ten metres per second the Earth’s recoil is about $5\times10^{-25}$ metres per second, which is not merely unmeasured but unmeasurable by an enormous margin.

That is worth sitting with. The conservation law is saved, in the most common everyday case, by a bookkeeping entry that no experiment will ever check. The entry is not arbitrary — it is forced by the same law applied consistently — but the law’s reputation for being verified constantly is partly built on cases where the balancing term is a rounding error in a quantity nobody can access.

The same cost appears in a more useful guise whenever a “loss” turns up. Energy that vanishes from a collision has not vanished; it has moved into a part of the system the accounting was not tracking — the disordered motion of $10^{23}$ molecules. Momentum that appears from nowhere in a rocket calculation has come from exhaust that was left out. In every case the conservation law is intact and the boundary was drawn wrongly, and the law itself gives no warning, because it is an identity rather than a check.

There is a deeper reason the boundary matters, and it explains why these particular quantities are the conserved ones. Noether’s theorem, proved in 1918, ties each conservation law to a symmetry: momentum is conserved because the laws of physics are the same here as a metre to the left, angular momentum because they are the same after a rotation, and energy because they are the same tomorrow as today. A wall breaks the first symmetry for anything that treats the wall as external, which is exactly why the momentum appeared not to balance.

And the converse has teeth. In an expanding universe there is no symmetry under translation in time, and energy is correspondingly not globally conserved — light stretched by the expansion loses energy that goes nowhere at all. The most secure-looking law on this page turns out to hold because of a symmetry, and to fail precisely where the symmetry does.

The instrument built out of the energy loss #

The lost energy is usually presented as the price of the method. It is also the basis of the first instrument that could measure the speed of a bullet, and the arrangement is worth following because it uses each conservation law exactly where the other one fails.

Benjamin Robins described it in 1742: fire the bullet into a heavy block hanging on strings, and measure how far the block swings. The device has two stages and they must be treated differently.

Stage one, the impact. The bullet embeds itself, which is as inelastic as a collision gets — most of its kinetic energy is gone into heat and deformation, and any calculation that assumed otherwise would be wrong by a factor of a hundred. But momentum is conserved, because the collision is over long before the strings have taken up any load, so $mv = (m+M)V$ gives the block’s speed immediately after.

Stage two, the swing. Now the collision is finished and there is nothing left to lose. Energy is conserved as the block rises, so $\tfrac12(m+M)V^2 = (m+M)gh$ converts the speed into a height that can be read with a ruler.

Putting them together gives the bullet’s speed from three measurable quantities and a swing:

$$v = \frac{m+M}{m}\sqrt{2gh}.$$

Nothing in it requires knowing anything about the impact. Robins measured musket balls at around 500 metres per second at a time when there was no other way to find out, and the mismatch between that number and what ballistic theory then predicted is what began the study of air resistance at high speed.

Why a reactor is full of water #

The two-equation solution has a consequence that decides the design of every thermal reactor, and it follows from a single case of the elastic formula.

A neutron slowing down loses the most energy when it hits something of its own mass. In a head-on elastic collision the incoming body leaves with a fraction $(m_1-m_2)/(m_1+m_2)$ of its velocity, which is zero when the masses are equal — the two simply exchange velocities, and the neutron can stop dead in one hit. Against something much heavier the fraction approaches $-1$: the neutron bounces back with almost the speed it arrived at, having transferred nothing.

Fission neutrons come out at a couple of million electronvolts and have to be brought down to the thermal energies at which they are most likely to cause the next fission, a factor of about $10^8$. Against hydrogen, whose nucleus is a single proton of nearly the neutron’s mass, that takes on the order of twenty collisions. Against carbon it takes something over a hundred. Against uranium itself it takes a couple of thousand, which is why the fuel cannot moderate itself.

So a reactor is full of water, or graphite, or heavy water, and the choice between them is a trade between how fast each slows a neutron and how many it absorbs on the way. The whole distinction between reactor types is downstream of one term in a formula derived from two conservation laws and no knowledge of what happens during the collision.

Where the model stops #

The picture draws two objects as blocks moving along a line. Four assumptions are hiding in that.

One dimension. Real collisions happen in a plane or in space, and there momentum conservation gives two or three equations while the unknowns multiply faster. Two spheres colliding in a plane have four unknown velocity components and only three equations even with restitution supplied; the missing information is the impact parameter, how far off-centre the hit was. This is why snooker requires skill rather than arithmetic.

No rotation. An off-centre hit puts angular momentum into both bodies, and any energy in spin is energy not in translation. The point-particle model has no way to represent spin, so it silently mis-assigns that energy to losses — the same blindness that makes a free-body diagram unable to tell sliding from toppling.

Instantaneous contact. The derivation assumed no other force acted during the collision. Gravity is acting throughout, and over a millisecond it contributes so little momentum that ignoring it is safe. Over a long, soft contact — a car crumpling, a parachutist landing — it is not automatically safe, and the impulse from external forces has to be carried.

A single number for the material. The coefficient of restitution is not a constant. It falls with impact speed, changes with temperature, and depends on both bodies rather than one. A golf ball’s restitution against a driver face is regulated precisely because it is a property of the pair.

What impulse buys back #

Momentum conservation says nothing about how long a collision takes, and that silence is where the engineering lives.

The change in momentum is fixed by the initial and final states. The force is that change divided by the contact time. So a collision that is made to last ten times longer produces one-tenth the peak force, for exactly the same momentum change.

Every safety device is an application of that sentence. A crumple zone, an airbag, a climbing rope with some stretch in it, a gymnastics mat, a boxer riding a punch — all of them leave the momentum unchanged and stretch the duration. None of them reduce what has to be absorbed; they reduce the rate at which it is absorbed, and bodies fail to rates rather than to totals.

The reverse trade is just as deliberate. A hammer, a punch press and a nail gun all exist to make the contact time as short as possible, because a large peak force is exactly what is wanted when the intention is to deform something.

The impulse view also settles a question the velocity view leaves open: what a “hard” or “soft” collision means. It is not about the materials being stiff. It is about the contact duration relative to everything else in the problem, and a collision counts as instantaneous whenever it is brief compared with the timescale of the other forces present — a pendulum’s period, say, or the flight time of a thrown object. Outside that separation of timescales, the tidy before-and-after picture in the figures stops being available at all.

The ladder from here #

Later rungs on this anchor: two-dimensional collisions and the impact parameter. Rotational momentum, and the collisions that put energy into spin. Rutherford scattering, where the same conservation laws applied to a collision nobody could watch revealed the atomic nucleus. The rocket equation, which is a continuous collision with the exhaust. Gravity assists, where the “collision” is with a planet, the encounter is a turn rather than an impact, and the planet’s loss is unmeasurable. Newton’s cradle taken seriously, which turns out to require wave propagation through the balls rather than a sequence of pairwise collisions. And the collisions that make a gas exert pressure, where this anchor stops being about two objects and becomes the foundation of thermodynamics.

Huygens worked out elastic collisions in the 1650s, decades before Newton’s laws were written down. The conservation law came first; the force law was the explanation that arrived afterwards.