Answers · The physics underneath

Why is momentum conserved in a collision when kinetic energy is not?

In shortMomentum is conserved because the internal pushes in a collision cancel in pairs, while kinetic energy can be turned into heat and deformation.

Momentum is conserved in a collision because every push between the bodies appears twice with opposite signs, so the total cannot change unless something outside the system pushes. Kinetic energy is not conserved in general because it can be converted into heat and permanent deformation, which are internal motions that a simple model does not track.

p = m_1 v_1 + m_2 v_2 + \cdots + m_n v_n

The signed sum

In one dimension the total is a single number: each body contributes its mass times its velocity, with the sign of the velocity carrying the direction. A two kilogram lump moving right at three metres per second contributes $2 \times 3 = 6$ , and a one kilogram lump moving left at three metres per second contributes $1 \times (-3) = -3$ . The total is $6 - 3 = 3$ kilogram metres per second. During contact, if the first exerts force $F$ on the second, the second exerts $-F$ on the first. The change in the total over the contact time is $(F - F)\Delta t = 0$ . That cancellation is exact, not a small error that shrinks with a smaller time step. It happens between two terms of the same sum before any approximation is made.

A stepped calculation can therefore hold momentum to the last digit while holding energy only approximately. The two are not both approximate claims of the same kind: momentum's error is zero by construction, energy's is small only if the step is small and the model includes every place the energy can go.

\begin{aligned} p_{\text{before}} &= 2(3)+1(-3)=6-3=3,\\ v_{\text{after}} &= \frac{3}{2+1}=1,\\ K_{\text{before}} &= \tfrac12(2)(3)^2+\tfrac12(1)(-3)^2=9+4.5=13.5,\\ K_{\text{after}} &= \tfrac12(3)(1)^2=1.5,\\ \Delta K &= 1.5-13.5=-12. \end{aligned}

Take the two lumps as the system. Before contact the total momentum is $2(3)+1(-3)=3$ , and after they stick the combined mass is $2+1=3$ kilograms, so the common velocity is $3/3=1$ metre per second. The kinetic energy before is $\tfrac12(2)(3)^2+\tfrac12(1)(-3)^2=9+4.5=13.5$ joules. After it is $\tfrac12(3)(1)^2=1.5$ joules. Twelve joules have gone into warming and deforming the putty, and the momentum has not moved at all.

FIG 1

quantity	before	after	difference
momentum	3 kg m/s	3 kg m/s	0
kinetic energy	13.5 J	1.5 J	-12 J

The natural guess is that if momentum survives the impact, kinetic energy should survive it too, since both are conserved quantities. That treats conservation as a property of the word rather than of a particular expression. Momentum is a sum of first powers, and the internal forces cancel in that sum exactly. Kinetic energy is a sum of squares, and the microscopic motions into which it can leak also carry positive energy. A warm lump of putty has more internal energy and the same total momentum as a cold one. The two quantities are not two versions of the same guarantee: one is exact for an isolated system, the other is conditional on no energy leaving.

Where the energy goes

Heat is motion of parts too small to appear in the model. Those parts carry energy, which is a sum of squares, so motions in random directions add positive amounts. They also carry momentum, which is a signed sum, and in random directions it cancels to nothing. That is the asymmetry. Energy can leave the visible model into a channel that a scalar sum notices; momentum cannot leave through a channel that a signed sum ignores. It is not that momentum is more fundamental. It is that the thing energy leaks into is invisible to a signed sum.

When both are conserved

If the collision is elastic, no kinetic energy leaves, so both quantities hold. Then the system is confined to the intersection of two conditions, often a curve, and questions about the sequence of collisions become questions about that curve. In an inelastic collision only momentum remains as the reliable constraint. The assumption behind momentum conservation is that no net external impulse acts. A wall outside the system can take momentum away and hand it back reversed; include the wall and the Earth in the system and the total returns. For the two bodies alone it is not conserved.

A calculation of a closed system that reports momentum changing is reporting a bug, not a tolerance. A calculation that reports kinetic energy changing may simply be reporting heat.

Worth keeping

Momentum is a signed weighted sum, so the equal and opposite internal forces of a collision cancel exactly.
Kinetic energy can leave the visible bodies as heat and deformation because microscopic motion adds up in energy but cancels in momentum.
Momentum conservation for a system requires no net external impulse; a wall outside the system can carry momentum away.

Where this goes next

This question is one step of Motion and What Is Conserved, a course that works through its subject from the beginning. The page that teaches this part of it is What a Collision Keeps, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that page The whole course