In short
When the net external force on a system of bodies is zero, the system's total momentum remains constant — no matter how violently the bodies interact internally. This follows directly from Newton's third law: internal forces come in equal-and-opposite pairs and cancel when you sum the momenta. The law is a vector equation: momentum is conserved independently in each direction.
A jawan at a firing range braces his 4 kg rifle against his shoulder and pulls the trigger. The 8 g bullet leaves the barrel at 700 m/s. The rifle slams backward into his shoulder.
Before the shot, both rifle and bullet were at rest. Total momentum of the system: zero. After the shot, the bullet carries momentum forward — 0.008 \times 700 = 5.6 kg·m/s to the right. For the total to remain zero, the rifle must carry exactly 5.6 kg·m/s to the left. That gives a recoil speed of 5.6 / 4 = 1.4 m/s.
Nobody injected momentum into this system. Nobody removed any. The gunpowder explosion was an internal force — it pushed the bullet forward and pushed the rifle backward with equal force for equal time. Forward momentum gained by the bullet exactly equals backward momentum gained by the rifle. The total stayed at zero — before, during, and after the shot.
Here is the remarkable thing: to calculate the rifle's recoil speed, you did not need to know the force of the explosion, how long it lasted, or how the pressure built up inside the barrel. All you needed was the total momentum before (zero) and the bullet's momentum after. Conservation of momentum lets you skip the messy details of the interaction and jump straight to the answer.
This is conservation of momentum. It is one of the most powerful principles in all of physics, and it follows from something you have already met — Newton's third law.
Why internal forces cannot change the total
Picture two blocks sitting on a frictionless surface with a compressed spring between them. When the spring is released, it pushes block A to the left and block B to the right. By Newton's third law, the force the spring exerts on A is exactly equal in magnitude and opposite in direction to the force it exerts on B.
These equal-and-opposite forces act for exactly the same duration — however long the spring takes to expand. Equal force acting for equal time produces equal impulse. Block A gains momentum to the left; block B gains the same amount of momentum to the right. One gains what the other loses. The total momentum of the two-block system does not change.
Think of momentum as money in a closed room. When A pushes B, A is transferring momentum to B. No new momentum is created, none is destroyed — it moves from one body to another. Internal forces are transactions between members of the system: individuals get richer or poorer, but the total cash in the room stays the same. Only a force from outside the system — an external force — can change the room's total.
If two children of different weights stand face to face on roller skates and push off each other, the lighter child rolls backward faster. Both started at rest, both pushed for the same duration, but the lighter one picks up more speed. The total momentum was zero before they pushed and is still zero after — the lighter child's backward momentum exactly balances the heavier child's forward momentum.
Deriving the law from Newton's third law
The intuition above can be made precise in four lines of algebra. Take two bodies, masses m_1 and m_2. Each experiences both an external force (gravity, friction, anything from outside the system) and the internal force from the other body.
Step 1. Write Newton's second law for each body separately.
Why: the total force on body 1 is the external force on it plus the internal force from body 2. Newton's second law says this total equals the rate of change of body 1's momentum. Same for body 2.
Step 2. Add the two equations.
Why: you want the rate of change of the total momentum \vec{p}_1 + \vec{p}_2. Adding the equations gives exactly that on the left side. The right side collects all forces — external and internal.
Step 3. Apply Newton's third law: \vec{F}_{21} = -\vec{F}_{12}.
Why: Newton's third law guarantees \vec{F}_{21} + \vec{F}_{12} = \vec{0}. Internal forces always cancel in the sum. Only external forces survive — they are the sole drivers of total momentum change.
Step 4. Set external forces to zero.
If \vec{F}_{\text{ext, total}} = \vec{0}, then:
Why: if the derivative of a quantity is zero, the quantity does not change with time. The total momentum is the same at every instant — before, during, and after any internal interaction.
This is the complete proof. Newton's third law eliminates internal forces; Newton's second law converts forces to momentum changes; and when external forces vanish, the total momentum is locked.
Equation (4) is actually the more general result — it says the total momentum changes only because of external forces, whether those forces are zero or not. The conservation law is the special case where external forces happen to be zero.
Conservation of Linear Momentum
For a system of bodies with no net external force, the total momentum is conserved:
Equivalently, the total momentum before any event equals the total momentum after:
where \vec{u} denotes velocities before and \vec{v} denotes velocities after the event.
Generalising to many bodies
The proof above used two bodies, but the argument extends to any number. In a system of N bodies, every internal force has a Newton's-third-law partner. When you sum the forces on all N bodies, the internal forces cancel in pairs — \vec{F}_{12} cancels \vec{F}_{21}, \vec{F}_{13} cancels \vec{F}_{31}, and so on. Only external forces survive:
A Diwali bomb at rest that shatters into twenty fragments is an N = 20 system. Every force between fragments is internal. If you neglect air resistance and gravity during the brief explosion, the total momentum of all twenty fragments is zero — the same as the bomb had before it exploded.
Choosing your system — internal versus external
The words "internal" and "external" depend entirely on where you draw the boundary around your system.
When a rifle fires, the gunpowder explosion is an internal force if your system includes both the rifle and the bullet. Momentum is conserved for that system. But if your system is just the bullet alone, the explosive force from the rifle is external — it changes the bullet's momentum from zero to mv. You have not broken any law; you have simply chosen a system for which the net external force is not zero.
The strategic rule: choose your system so that the troublesome forces become internal. If you want to use momentum conservation for two objects interacting, draw the system boundary around both. Every interaction between them is then internal and cancels automatically.
When does conservation of momentum apply?
Conservation of momentum requires the net external force to be zero — or negligible during the time interval you care about. Concretely:
- A gun fires a bullet: during the millisecond of firing, internal forces dominate. Momentum is conserved.
- Two marbles collide on a smooth floor: friction is negligible during the brief impact. Momentum is conserved.
- A batsman hits a cricket ball: during the ~1 ms of contact, gravity's impulse (0.16 \times 9.8 \times 0.001 \approx 0.002 kg·m/s) is negligible compared to the ball's momentum change (several kg·m/s). Momentum is conserved during the hit.
- A box slides across a rough floor and slows down: friction is a sustained external force acting throughout the motion. Momentum is not conserved for the box alone.
- A ball is thrown upward: gravity acts continuously. Momentum is not conserved for the ball alone. But if you include the Earth in your system, total momentum is conserved — though the Earth's velocity change is immeasurably tiny.
The pattern: for brief, violent interactions (collisions, explosions, firing), external forces rarely matter. For slow, sustained processes (sliding, orbiting, falling), external forces usually do matter — and you need to account for them.
Recoil and explosion — seeing conservation in action
Spring-loaded blocks
Two blocks — m_1 = 1 kg (red) and m_2 = 2 kg (dark) — sit on a frictionless surface with a compressed spring between them. Before release, both are at rest: \vec{p}_{\text{total}} = 0.
By conservation of momentum after the spring releases:
If the spring gives the 2 kg block a speed of 2 m/s to the right, the 1 kg block moves at 4 m/s to the left. The lighter block moves twice as fast — but both carry the same magnitude of momentum: 1 \times 4 = 2 \times 2 = 2 kg·m/s.
This is the same physics as the rifle: the lighter object (bullet or block A) moves much faster, but carries the same magnitude of momentum as the heavier one. The total is always zero.
The Diwali bomb
A firecracker at rest explodes into fragments. Before the explosion, the total momentum is zero. After, the fragments fly in every direction — but the vector sum of all their momenta is still exactly zero. Every kilogram-metre-per-second of forward momentum is balanced by an equal amount of backward momentum. Every upward component is balanced by a downward component. The explosion redistributes momentum violently, but the total accounting is perfect.
This is why a cannon mounted on wheels rolls backward when it fires: the cannonball carries momentum forward, and the cannon must carry an equal amount backward. Indian cannons at forts like Golconda and Daulatabad were mounted on heavy carriages specifically to absorb this recoil — the heavier the carriage, the slower the recoil speed.
The direction matters — a vector equation
Momentum conservation is not a statement about magnitudes alone. It is a vector equation. This means it holds independently in each direction:
If a firecracker at rest explodes into three pieces — one flying north-east, one flying east — the third piece must fly in the specific direction that makes both the total x-momentum and total y-momentum zero. You cannot determine that direction from magnitudes alone. You need the components.
This vector nature is what makes momentum conservation so powerful in two-dimensional collision problems, which you will meet in Collisions in Two Dimensions.
Explore the recoil yourself
A 1 kg block moves left at 1 m/s, so its momentum is -1 kg·m/s. A second block of mass m_2 recoils in the opposite direction. Conservation of momentum demands m_2 v_2 = 1 kg·m/s regardless of what m_2 is. Drag the red point below to change m_2 and watch: the recoil speed v_2 changes, but the momentum m_2 v_2 stays stubbornly flat at 1 kg·m/s. That flat line is conservation of momentum.
Worked examples
Example 1: Diwali bomb splits into two pieces
A 0.5 kg firecracker is lying at rest on the ground. It explodes into two pieces. One piece (mass 0.3 kg) flies horizontally to the east at 6 m/s. Find the velocity of the other piece. Neglect vertical forces during the brief explosion.
Step 1. Identify the system and the forces.
System: the entire firecracker (all fragments). During the explosion — a few milliseconds — the internal explosive forces are enormous compared to gravity. The net external impulse during that interval is negligible.
Why: the explosion lasts milliseconds, so F_{\text{ext}} \times \Delta t \approx 0. Momentum is conserved during the explosion.
Step 2. Write conservation of momentum.
Before the explosion: p_{\text{before}} = 0 (the firecracker is at rest).
After: m_1 v_1 + m_2 v_2 = 0, where m_1 = 0.3 kg, v_1 = +6 m/s (east), and m_2 = 0.5 - 0.3 = 0.2 kg.
Why: total momentum before (zero) must equal total momentum after. Take east as the positive direction. The unknown mass is the total minus the known piece.
Step 3. Solve for v_2.
Why: the negative sign means the second piece moves west — opposite to the first. The explosion pushes the two pieces apart, exactly as you would expect.
Step 4. Verify.
Result: The 0.2 kg piece flies west at 9 m/s.
The lighter piece moves faster — 9 m/s versus 6 m/s — but both carry the same magnitude of momentum: 0.3 \times 6 = 0.2 \times 9 = 1.8 kg·m/s. Conservation demands this: what one piece gains, the other must balance.
Example 2: Jumping off a boat
A 60 kg boy stands on a stationary 40 kg boat floating on still water. He jumps horizontally toward the shore at 3 m/s relative to the ground. Find the velocity of the boat just after he jumps. Neglect water resistance during the jump.
Step 1. Define the system and check external forces.
System: boy + boat. Before the jump, both are at rest: p_{\text{total}} = 0. Water resistance during the brief jump is negligible, so the only forces are internal — the boy's feet push the boat backward, the boat pushes the boy forward.
Why: internal forces between boy and boat obey Newton's third law and cancel in the total. With external forces negligible, momentum is conserved.
Step 2. Apply conservation of momentum. Take the direction toward shore as positive.
Why: total momentum before (zero) equals total momentum after.
Step 3. Solve for v_{\text{boat}}.
Why: the negative sign means the boat moves away from shore — opposite to the boy's jump. Exactly what happens when you jump off a real boat.
Result: The boat moves at 4.5 m/s away from the shore.
The boy is 1.5 times heavier than the boat, so the boat recoils 1.5 times faster than the boy jumps. If you have ever jumped off a small boat at a river ghat, you know this — the boat shoots backward and you barely make it to the bank.
Common confusions
-
"Momentum is always conserved in every situation." Careful — momentum of a system is conserved only when the net external force on that system is zero (or negligible during the event). When you catch a cricket ball, the ball's momentum changes because your hand exerts an external force on it. If you include the ball, your body, and the entire Earth in the system, then total momentum is conserved — but the Earth's velocity change is immeasurably small.
-
"If momentum is conserved, nothing slows down." Conservation of total momentum does not mean individual objects maintain their speeds. In an equal-mass head-on collision, one object stops completely while the other takes off. Individual momenta change dramatically; the total does not.
-
"Momentum conservation only works in elastic collisions." This is the most widespread misconception. Momentum is conserved in every collision — elastic, inelastic, or perfectly inelastic (where the objects stick together). What distinguishes elastic collisions is that kinetic energy is also conserved. Momentum conservation comes from Newton's third law, which holds regardless of how much energy goes to heat, sound, or deformation.
-
"A heavier object always carries more momentum." Not necessarily. An 8 g bullet at 700 m/s carries momentum 0.008 \times 700 = 5.6 kg·m/s. A 5 kg bowling ball rolling at 1 m/s carries only 5 kg·m/s. The bullet — 625 times lighter — has more momentum because it is moving so much faster. Momentum depends on both mass and velocity.
-
"If the total momentum is zero, nothing is moving." Zero total momentum can mean everything is at rest, but it can also mean objects are moving in opposite directions with equal and opposite momenta. The rifle-and-bullet system after the shot has zero total momentum, even though both are moving fast.
If you can apply conservation of momentum to recoil and explosion problems, you have the essential tool. The rest of this section is for readers who want to see why the law is so much deeper than it first appears.
When external forces act but momentum is still "approximately" conserved
In many real situations — a bat striking a cricket ball, two autorickshaws colliding, a bullet embedding in a wooden block — external forces (gravity, friction, normal forces) do act on the system. Momentum is not exactly conserved.
But it is approximately conserved during the collision itself, because the collision happens so fast that external forces do not have time to deliver significant impulse. A cricket ball is in contact with the bat for about 1 millisecond. During that \Delta t \approx 0.001 s, gravity exerts an impulse of mg \times \Delta t = 0.16 \times 9.8 \times 0.001 \approx 0.002 kg·m/s on the ball. Compare this to the ball's momentum change of several kg·m/s — the gravitational impulse is negligible. Momentum conservation during the hit is an excellent approximation.
The rule: if F_{\text{ext}} \times \Delta t \ll p_{\text{system}}, treat momentum as conserved during the event. After the event, external forces take over — the ball follows a parabolic trajectory under gravity, the wreckage slides to a halt under friction. But during the brief interaction itself, internal forces dominate and the total momentum is locked.
The deepest reason — symmetry of space
Conservation of momentum is not merely a consequence of Newton's third law. At the deepest level, it follows from a symmetry: the laws of physics are the same at every location in space. Set up an experiment in Delhi or in Chennai, or on the Moon — the same equations govern the motion.
This translational symmetry of space, combined with a powerful mathematical result called Noether's theorem (proved by Emmy Noether in 1915), directly implies that the total momentum of an isolated system is conserved.
Every conservation law in physics corresponds to a symmetry:
| Conservation law | Symmetry |
|---|---|
| Momentum | Space is the same everywhere (translational symmetry) |
| Energy | Physics is the same at all times (time symmetry) |
| Angular momentum | Physics is the same in all directions (rotational symmetry) |
Newton's third law is itself a consequence of translational symmetry. So the chain runs: symmetry of space \to Newton's third law \to conservation of momentum. The symmetry is the deeper reason.
Beyond Newton — a law that survives every revolution
Conservation of momentum is one of those rare laws that survives every revolution in physics. It holds in Einstein's special relativity (with a modified definition of momentum: \vec{p} = \gamma m \vec{v}, where \gamma grows with speed). It holds in quantum mechanics (the total momentum operator commutes with the Hamiltonian when there is translational symmetry). It holds in quantum field theory. Newton's laws are approximations that break down at high speeds and small scales — but conservation of momentum does not. It is, in a precise sense, more fundamental than the laws from which you derived it in this article.
Where this leads next
- Elastic Collisions — when both momentum and kinetic energy are conserved, you can solve for both final velocities completely.
- Inelastic Collisions — what happens when kinetic energy is not conserved, including the perfectly inelastic case where objects stick together.
- Collisions in Two Dimensions — applying momentum conservation as a vector equation to oblique collisions and scattering.
- Variable Mass Systems — Rockets — when mass enters or leaves the system, leading to the Tsiolkovsky rocket equation.
- Centre of Mass — Definition and Calculation — the special point that moves as if all external force acts on the total mass concentrated there.