Conservation of linear momentum

A gun fires and kicks back into your shoulder. A stationary billiard ball leaps away the instant it is struck. In every such event the forces involved are messy, brief and hard to measure directly — yet physics predicts the outcome exactly, without ever writing down those forces. The trick is the principle of conservation of linear momentum: for a system left alone by outside forces, whatever momentum goes in comes out, redistributed among the pieces but never lost or gained. This is a core result of Unit 3, Laws of Motion, in ISC Class 11 Physics, built directly on Newton's second and third laws. Here we derive the law from first principles, apply it to recoil and to a stick-together collision, and flag the traps that cost the most marks: forgetting momentum is a vector, and reaching for the wrong system boundary.

The core insight

If no external force acts on a system, its total linear momentum stays constant, however complicated the internal forces between its parts may be.

The core idea

The principle rests on nothing more exotic than Newton's third law applied to a pair of interacting bodies.

Statement. For a system of two bodies of masses $m_1, m_2$ moving with velocities $\vec u_1, \vec u_2$ before an interaction (collision, explosion, push-off) and $\vec v_1, \vec v_2$ afterward, if no external force acts on the system:

m_1\vec u_1 + m_2\vec u_2 = m_1\vec v_1 + m_2\vec v_2

where: $m_1, m_2$ are the masses, $\vec u_1, \vec u_2$ are the velocities just before the interaction, and $\vec v_1, \vec v_2$ are the velocities just after. The left side is the total momentum before, the right side the total momentum after — they must be equal for an isolated system.

Why "no external force" matters. Momentum is conserved for the system as a whole, not for any one body inside it. Body A's own momentum can change hugely (a bullet leaving a gun barrel), because the force from B on A is a real, unbalanced force on A alone. What stays fixed is the sum $\vec p_A + \vec p_B$ , because the internal forces A and B exert on each other are equal and opposite and so cancel in the total.

Vector, not scalar. Momentum $\vec p = m\vec v$ points the same way as velocity. Two bodies moving toward each other must have their momenta given opposite signs before adding — treating them as plain magnitudes is the single most common error in this topic.

Where it is used. The commonest applications at this level are recoil of a gun, explosion of a shell into fragments, and one-dimensional collisions of two bodies, including the case where they stick together after impact. A full quantitative treatment of elastic and inelastic collisions, with kinetic-energy bookkeeping, belongs to the Work, Energy and Power unit — here the focus is squarely on the conservation law itself and how to apply it correctly.

Figure

Before and after view of a bullet leaving a gun with equal and opposite momenta shown as arrows summing to zero

Derivation from Newton's laws

Derivation ATotal momentum of an isolated two-body system is constant

The goal is to show, using only Newton's second and third laws and no extra assumption about the size, shape or duration of the interaction force, that the total momentum of two bodies A and B interacting solely with each other stays fixed throughout — before, during and after any collision, push-off or explosion between them.

Step 1. Isolate the system. Let A and B interact only with each other; no external force from outside the pair acts on either. Let $\vec F_{on\,A}$ be the force B exerts on A, and $\vec F_{on\,B}$ the force A exerts on B.

Step 2. Apply Newton's third law. Every action has an equal and opposite reaction, so at every instant during the interaction, $\vec F_{on\,A} = -\vec F_{on\,B}$ . This holds regardless of how the force varies in time — spring-like, impulsive, explosive, or anything else.

Step 3. Apply Newton's second law to each body. $\vec F_{on\,A} = d\vec p_A/dt$ and $\vec F_{on\,B} = d\vec p_B/dt$ , where $\vec p = m\vec v$ is momentum. Adding the two equations and using Step 2, $\dfrac{d\vec p_A}{dt} + \dfrac{d\vec p_B}{dt} = \vec F_{on\,A} + \vec F_{on\,B} = 0$ .

Step 4. Read off the conclusion. Since $d(\vec p_A + \vec p_B)/dt = 0$ , the total momentum $\vec p_A + \vec p_B$ does not change with time — it is exactly the same before the interaction begins as after it ends, whatever complicated forces acted in between. This is why the law is so powerful: it needs no knowledge of the interaction's details, only that it is internal to the system chosen.

Worked example

A gun of mass $4\ \text{kg}$ fires a bullet of mass $0.02\ \text{kg}$ with a muzzle velocity of $300\ \text{m/s}$ . Find the recoil velocity of the gun.

Step 1. Choose the system and check it is isolated. Take gun + bullet together. The explosive force during firing acts entirely between gun and bullet, with no external horizontal force, so momentum is conserved for this system.

Step 2. Write the conservation equation. Before firing, both are at rest, so total initial momentum is zero: $m_{gun}v_{gun} + m_{bullet}v_{bullet} = 0$ .

Step 3. Substitute and solve. Taking the bullet's forward direction as positive, $4v_{gun} + (0.02)(300) = 0$ , so $v_{gun} = -6/4 = -1.5\ \text{m/s}$ . The gun recoils at $1.5\ \text{m/s}$ , opposite to the bullet's motion.

Competitive stretch

A block of mass $2\ \text{kg}$ moving at $6\ \text{m/s}$ collides head-on with a stationary $4\ \text{kg}$ block on a frictionless surface and they stick together. Find their common velocity.

Step 1. The surface is frictionless, so no external horizontal force acts on the two-block system during contact — momentum is conserved.

Step 2. Initial momentum: $p_i = (2)(6) + (4)(0) = 12\ \text{kg}\,\cdot\,\text{m/s}$ . After sticking, combined mass is $6\ \text{kg}$ moving at common velocity $v$ , so $p_f = 6v$ .

Step 3. Equate: $6v = 12 \implies v = 2\ \text{m/s}$ . The combined blocks move off at $2\ \text{m/s}$ , found without ever needing the large, brief contact force between the blocks.

Common traps

Adding momenta as plain magnitudes instead of signed vector quantities — opposite directions need opposite signs.
Invoking conservation when an external force (friction, an outside push) clearly acts on the chosen system without accounting for it.
Mixing up momentum conservation with kinetic energy conservation: momentum is conserved in every isolated collision, elastic or inelastic, but kinetic energy only in perfectly elastic ones.
Picking an inconsistent system boundary — analysing the bullet alone and forgetting the gun makes momentum look "not conserved" simply because the gun's force is now external to that shrunk system.

Ask the tutor: Want to try a few more recoil, explosion, or stick-together collision problems using momentum conservation?

Why is momentum conserved even though kinetic energy is often lost in a collision?

Momentum conservation comes straight from Newton's third law: the internal forces two colliding bodies exert on each other are always equal and opposite, at every instant, so their effect on total momentum always cancels. Kinetic energy has no such guarantee — some of it can convert to heat, sound or permanent deformation during contact. So momentum survives every isolated collision, elastic or not, while kinetic energy only survives the special case of a perfectly elastic one.

If the gun recoils, doesn't that mean momentum was created out of nothing?

No new momentum is created. Before firing, the gun-bullet system has zero total momentum since both are at rest. After firing, the bullet carries forward momentum and the gun carries an equal and opposite backward momentum, so the two exactly cancel and the total is still zero. The chemical energy of the propellant supplies the kinetic energy of both pieces, but momentum was always zero and stays zero — it is simply redistributed between bullet and gun.

How do I decide which objects to include in my system before applying conservation?

Include every object between which the interaction forces act, and check that nothing outside that group exerts a significant force during the interaction. For a gun-bullet problem, that means gun plus bullet, not the bullet alone, because the recoil force from the bullet acts on the gun. For a block sliding on a frictionless surface, the surface need not be included since it exerts no horizontal force. Getting this boundary wrong is the most common reason momentum "appears" to fail.