Conservation of Momentum

Rack up a triangle of snooker balls and smash the white into them. The pack scatters in every direction, the white slows or stops — chaos. Yet buried in that chaos is one quantity that comes out of the collision exactly as big as it went in. Add up momentum p = m v for every ball before the break, add it up again afterwards, and the two totals are identical. The balls swapped their motion around between themselves, but not one scrap of total momentum was created or destroyed.

This is the conservation of momentum, and it is one of the deepest rules in all of physics. It is what lets a rocket climb with nothing to push against, what makes a gun kick, what a crash investigator uses to reconstruct a pile-up from the skid marks, and how physicists weigh particles they can never even see. This page is about why momentum is conserved, exactly when it is, and how to use it to solve real collisions.

The law: total momentum stays the same

Consider a closed, isolated system — a collection of objects with no external resultant force acting on them (nothing from outside pushing or pulling the group as a whole). For such a system the total momentum after any interaction equals the total momentum before it:

\sum p_{\text{before}} = \sum p_{\text{after}}.

For the everyday case of two bodies colliding — masses m_1 and m_2, with initial velocities u_1, u_2 and final velocities v_1, v_2 — this reads

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.

Every collision problem in this topic is, at heart, just this one equation with different numbers put in. Notice what it does not mention: it says nothing about how hard the objects hit, how long they were in contact, or whether they bounced, stuck, or shattered. Whatever violence happens in between, this book-keeping balances.

In a closed, isolated system — one with no external resultant force — the total linear momentum is constant: \sum p_{\text{before}} = \sum p_{\text{after}}.
For two colliding bodies this is m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.
Momentum is a vector, so the sum obeys signs (in 1D) or components (in 2D or 3D): momentum is conserved independently in each direction.
It holds for any interaction — collisions, explosions, recoil — whether or not kinetic energy is conserved.

Why it is true: Newton's third law

Conservation of momentum is not a lucky coincidence spotted in experiments — it follows directly from Newton's third law. When two bodies interact, body 1 pushes on body 2 with a force \vec{F}_{1 \to 2}, and body 2 pushes back on body 1 with an equal and opposite force:

\vec{F}_{1 \to 2} = -\,\vec{F}_{2 \to 1}.

They are in contact for exactly the same time \Delta t, so each delivers an impulse \vec{F}\,\Delta t to the other — and those impulses are equal and opposite too:

\vec{F}_{1 \to 2}\,\Delta t = -\,\vec{F}_{2 \to 1}\,\Delta t.

But impulse is change in momentum, \vec{F}\,\Delta t = \Delta \vec{p}. So the momentum body 2 gains is exactly the momentum body 1 loses:

\Delta \vec{p}_2 = -\,\Delta \vec{p}_1 \quad\Longrightarrow\quad \Delta \vec{p}_1 + \Delta \vec{p}_2 = 0.

If the change in the total momentum is zero, the total momentum does not change — which is precisely the conservation law. Every internal push comes with its equal-and-opposite partner, so internal forces can only shuffle momentum from one body to another, never change the sum. Only a force from outside the system can do that — which is why the law needs "no external resultant force."

Momentum is a vector — mind the signs

Because velocity is a vector, so is momentum, and this is the single most important habit in every conservation problem. In one dimension, choose a positive direction and give anything moving the other way a minus sign — then carry those signs faithfully through the arithmetic. A 2\ \text{kg} ball going right at 3\ \text{m/s} has momentum +6\ \text{kg·m/s}; the same ball going left at 3\ \text{m/s} has -6\ \text{kg·m/s}. Add momenta as signed numbers and the total can happily come out positive, negative, or zero.

In two dimensions — a glancing snooker shot, two cars meeting at a junction — momentum is conserved separately in each direction. Resolve every velocity into x and y components and apply the law twice:

\sum p_x^{\text{before}} = \sum p_x^{\text{after}}, \qquad \sum p_y^{\text{before}} = \sum p_y^{\text{after}}.

The two equations never mix, because a force along x cannot change the momentum along y. This is exactly why an explosion starting from rest sends its fragments flying in balanced, opposite directions: the total momentum was zero before, so it must stay zero, and every piece heading one way is matched by pieces heading the other so that all the arrows cancel.

Worked examples

Example 1 — two trolleys that stick (a perfectly inelastic collision). A 2\ \text{kg} trolley rolls at 3\ \text{m/s} into a stationary 1\ \text{kg} trolley; they couple together and move off as one. Conservation of momentum gives

m_1 u_1 + m_2 u_2 = (m_1 + m_2)\,v \;\Rightarrow\; (2)(3) + (1)(0) = (3)\,v \;\Rightarrow\; v = \frac{6}{3} = 2\ \text{m/s}.

When two objects stick together, they share one final velocity, so the whole right-hand side collapses to (m_1 + m_2)v — the tidiest case to solve.

Example 2 — a recoiling cannon. A 1000\ \text{kg} cannon, at rest, fires a 5\ \text{kg} ball forwards at 200\ \text{m/s}. Before firing the total momentum is zero, so it must be zero afterwards. Taking the ball's direction as positive,

0 = m_{\text{ball}} v_{\text{ball}} + m_{\text{gun}} v_{\text{gun}} = (5)(200) + (1000)\,v_{\text{gun}},

v_{\text{gun}} = -\frac{1000}{1000} = -1\ \text{m/s}.

The minus sign says the cannon recoils backwards. The ball and the gun carry equal and opposite momentum (1000\ \text{kg·m/s} each), but because the gun is 200\times heavier it moves 200\times slower — a gentle shove, not a bullet's speed.

Example 3 — an explosion from rest. A compressed spring holds two trolleys together at rest; released, it flings them apart. The 2\ \text{kg} trolley leaves at 3\ \text{m/s} to the left. How fast does the 3\ \text{kg} trolley go? Total momentum stays zero:

0 = m_1 v_1 + m_2 v_2 = (2)(-3) + (3)\,v_2 \;\Rightarrow\; v_2 = \frac{6}{3} = 2\ \text{m/s (to the right)}.

The two fragments carry 6\ \text{kg·m/s} each in opposite directions, so the arrows cancel and the total is still zero — as it must be.

Example 4 — jumping off a boat. A 60\ \text{kg} person stands on a stationary 240\ \text{kg} boat and jumps off the back at 3\ \text{m/s}. With the system (person + boat) starting at rest,

0 = (60)(3) + (240)\,v_{\text{boat}} \;\Rightarrow\; v_{\text{boat}} = -\frac{180}{240} = -0.75\ \text{m/s}.

The boat glides backwards at 0.75\ \text{m/s} — the same reason a rowing boat lurches when you step off it onto the jetty.

Example 5 — a head-on collision, keeping the signs. A 2\ \text{kg} cart moving right at 4\ \text{m/s} meets a 3\ \text{kg} cart moving left at 2\ \text{m/s}; they stick. Taking right as positive,

(2)(+4) + (3)(-2) = (5)\,v \;\Rightarrow\; 8 - 6 = 5v \;\Rightarrow\; v = \frac{2}{5} = 0.4\ \text{m/s}.

The combined cart drifts slowly to the right: the faster-but-lighter cart just wins. Drop a sign here and you would get 2.8\ \text{m/s} — completely wrong. The interactive below lets you feel exactly this trade-off.

See it: two carts collide and couple

Two carts approach, collide, and stick together. Set each cart's mass and velocity (a negative velocity means it moves to the left). The coloured arrows are the momenta p = mv; watch the total-momentum arrow before the collision and the total-momentum arrow after — they are always the same length and direction, however you set the sliders. The stuck-together carts leave at exactly the velocity needed to carry that unchanged total, v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}.

These four slips wreck conservation-of-momentum questions:

The system must have no external resultant force. Momentum is only conserved for a closed, isolated system. If an outside force acts during the interaction — friction dragging on a trolley, gravity on a ball mid-flight, a hand still holding a cart — then that force delivers an outside impulse and the total momentum changes. Always ask: is anything from outside pushing on this group? (Over a very short collision, large internal forces usually swamp friction, which is why we often treat a quick smash as isolated even on a rough floor.)
Momentum is a VECTOR — use signs. Pick a positive direction and give leftward/downward velocities a minus sign. A rebound reverses the sign, which is a bigger change than merely stopping. Adding speeds without their signs is the most common mistake in the whole topic.
Momentum is conserved; kinetic energy usually is NOT. In a collision where objects stick or crumple (an inelastic collision), momentum is still perfectly conserved, but kinetic energy is not — some is lost to heat, sound and deformation. Only in a special elastic collision is kinetic energy conserved too. Never assume \tfrac12 m v^2 balances just because mv does.
Count the whole system. The gun and the ball, both trolleys, the person and the boat — conservation applies to the total. Analysing one object alone (which does gain or lose momentum) and expecting its momentum to be conserved is a category error: only the sum over the closed system is fixed.

A rocket in the vacuum of space has no air, no ground, nothing external to push on — and yet it accelerates. Conservation of momentum is the whole answer. Rocket-plus-fuel is a closed system, so its total momentum is fixed. When the engine hurls a slug of hot exhaust gas backwards at high speed, that gas carries a lump of backward momentum; to keep the total unchanged, the rest of the rocket must gain exactly the same momentum forwards. The rocket is, quite literally, throwing part of itself out of the back to drive the rest forwards — an explosion that never stops.

The very same book-keeping runs a gun's recoil, a squid jetting water, a firework bursting into a balanced sphere of sparks, and a pair of ice skaters who push apart and glide off in opposite directions from a standstill. Start from zero total momentum, and every scrap of motion you create one way must be paid for with equal momentum the other way. Nature keeps a perfectly balanced ledger.