Conservation of momentum

A rifle kicks your shoulder the instant the bullet leaves the barrel. A stationary raft drifts backward the moment you jump off it forward. A firecracker at rest bursts into fragments that fly off in every direction, yet somehow their combined push still adds up to nothing. None of these need a fresh law of physics — they are all the same bookkeeping rule wearing different costumes: the law of conservation of linear momentum. The rule is almost suspiciously simple. Add up the momentum ( $mass \times velocity$ ) of everything in a system, and as long as no outside force interferes, that total never changes — no matter how violent the collision or explosion happening inside the system. Bullets, rafts, firecrackers, rockets, two carts colliding on a track: internal pushes and pulls can shuffle momentum around between the pieces, but they can never create or destroy the total. This single idea, grown directly out of Newton's second and third laws, turns messy problems — a gun firing, two vehicles crumpling together, a rocket spitting out exhaust — into a one-line equation you can solve without knowing any of the ugly details of how the force acted, only that it acted between two bodies and nothing pushed from outside. It is one of the most quietly powerful tools in all of mechanics, and it is the CBSE Class 11 topic this article builds from the ground up: what momentum is, why its total is conserved, how to derive that fact, and how to use it on recoiling guns, colliding carts, and recoiling rockets.

Key takeaways

Linear momentum is $p = mv$ , a vector; impulse $J = F\Delta t$ equals the change in momentum (impulse-momentum theorem).
Newton's second law in its true form is $F_{net} = dp/dt$ ; for constant mass this reduces to the familiar $F = ma$ .
Newton's third law says the force A exerts on B is equal and opposite to the force B exerts on A — but the two forces act on different bodies, never the same one.
Conservation of momentum follows directly from the second and third laws: if the net external force on a system is zero, the total momentum $\sum p$ stays constant.
For two colliding bodies: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ , applied along a chosen positive direction.
Recoil of a gun and the thrust of a rocket are both explosions of a system starting at rest (or at some initial momentum) whose pieces must add back to the same total.
Key idea: because internal action-reaction pairs cancel in pairs, the total momentum of a system is unchanged unless a net external force acts on it, giving $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ for collisions, recoils, and explosions.

The core insight

When the net external force on a system is zero, internal action-reaction forces cancel in the sum and the total momentum $\sum mv$ before an interaction exactly equals the total after it.

Where it fits in the syllabus

This is core CBSE Class 11 Physics, Unit III — Laws of Motion, sitting directly after Newton's three laws and before equilibrium of a particle. The syllabus asks you to state the law of conservation of linear momentum, derive it from the second and third laws, and apply $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ to one-dimensional problems such as recoil, collisions, and explosions — qualitatively including the rocket. Detailed energy analysis of elastic versus inelastic collisions (kinetic energy before/after) belongs to the later Work, Energy and Power chapter; here the focus stays strictly on momentum bookkeeping.

The physical picture

Momentum measures "quantity of motion" — a heavy truck crawling at $5\ \text{m/s}$ can be harder to stop than a scooter racing at $30\ \text{m/s}$ , because momentum weighs both mass and velocity together.

\vec{p} = m\vec{v}

where: $m$ is mass in kilograms and $\vec{v}$ is velocity in metres per second, so $\vec{p}$ is a vector in kilogram-metres per second ( $\text{kg m/s}$ ), pointing the same way as the velocity. Doubling either the mass or the speed doubles the momentum. Newton's second law connects momentum directly to force: force is not just "mass times acceleration," it is the rate at which momentum changes.

\vec{F}_{net} = \frac{d\vec{p}}{dt}

where: for a body of constant mass this becomes the familiar $\vec{F}_{net} = m\vec{a}$ , since $d(mv)/dt = m(dv/dt) = ma$ . The unit of force, the newton, is defined from this: $1\ \text{N} = 1\ \text{kg m/s}^2$ , the force that changes a $1\ \text{kg}$ mass's velocity by $1\ \text{m/s}$ every second. A force does not need to act instantly — a small force acting for a long time can change momentum just as much as a large force acting briefly. That trade-off is impulse.

\vec{J} = \vec{F}\,\Delta t = \Delta \vec{p} = m\vec{v}_f - m\vec{v}_i

where: $\vec{F}$ is the (average) net force, $\Delta t$ the time it acts, and $\Delta \vec{p}$ the resulting change in momentum — this is the impulse-momentum theorem. It is why a cricketer draws her hands back while catching a ball: stretching $\Delta t$ reduces the peak force $F$ for the same impulse $\Delta p$ , so the catch does not sting. Newton's third law is the other half of the story: every force comes in an equal-and-opposite pair, but the two forces of a pair act on two different bodies, never on the same one — which is exactly why they never cancel within a single body's free-body diagram, yet always cancel when you add up the whole system's momentum.

Figure

Two carts on a track before and after a head on collision showing momentum arrows equal on each side of the equals sign

The three applications built on this law

Recoil. A gun and bullet at rest together have zero total momentum. After firing, the bullet's forward momentum must be exactly balanced by the gun's backward momentum, so the total stays zero.
Collisions. Two bodies meet; whatever their combined momentum was just before, it is exactly the same just after — whether they bounce apart or stick together.
Explosions and rockets. A body at rest (or moving) breaks apart or ejects mass; the vector sum of the momenta of all the pieces (including the ejected exhaust, for a rocket) still equals the original total.

Derivation AConservation of momentum from Newton's second and third laws

The goal here is to show, using nothing beyond the second and third laws already learned, that whenever two bodies interact through internal forces alone — with no outside push or pull on the pair — the total momentum of the two-body system measured before the interaction is exactly equal to the total momentum measured after it, however complicated or brief the interaction was. This is what lets you skip past the details of a collision entirely and jump straight from "before" to "after."

Step 1. Set up two bodies, mass $m_1$ and $m_2$ , interacting only with each other — for instance during a collision — for a short time $\Delta t$ . By Newton's third law, the force body 1 exerts on body 2, call it $\vec{F}_{12}$ , is equal and opposite to the force body 2 exerts on body 1, $\vec{F}_{21}$ , so $\vec{F}_{12} = -\vec{F}_{21}$ . Crucially these two forces act on two different bodies.

Step 2. Apply Newton's second law to each body separately over the same interval $\Delta t$ . For body 1, $\vec{F}_{21} = \dfrac{\Delta \vec{p}_1}{\Delta t} = \dfrac{m_1\vec{v}_1 - m_1\vec{u}_1}{\Delta t}$ . For body 2, $\vec{F}_{12} = \dfrac{\Delta \vec{p}_2}{\Delta t} = \dfrac{m_2\vec{v}_2 - m_2\vec{u}_2}{\Delta t}$ . Here $u$ denotes velocity before the interaction and $v$ after.

Step 3. Substitute the third-law relation $\vec{F}_{12} = -\vec{F}_{21}$ from Step 1 into the two expressions from Step 2. This gives $\dfrac{m_2\vec{v}_2 - m_2\vec{u}_2}{\Delta t} = -\dfrac{m_1\vec{v}_1 - m_1\vec{u}_1}{\Delta t}$ . The $\Delta t$ is the same brief interval for both bodies (they interact simultaneously), so it cancels from both sides, leaving $m_2\vec{v}_2 - m_2\vec{u}_2 = -(m_1\vec{v}_1 - m_1\vec{u}_1)$ .

Step 4. Rearrange by moving every "after" term to one side and every "before" term to the other: $m_1\vec{v}_1 + m_2\vec{v}_2 = m_1\vec{u}_1 + m_2\vec{u}_2$ . The left side is the total momentum of the system after the interaction; the right side is the total momentum before. They are equal, so the total momentum did not change — provided no external force acted on the pair during $\Delta t$ . If a net external force $\vec{F}_{ext}$ did act, it would add an extra term $\vec{F}_{ext}\Delta t$ to the balance, which is precisely why the law is stated for zero net external force.

Worked examples

Board level - recoil velocity of a rifle

A rifle of mass $4.0\ \text{kg}$ fires a bullet of mass $20\ \text{g}$ ( $0.020\ \text{kg}$ ) at a muzzle velocity of $400\ \text{m/s}$ . Find the recoil velocity of the rifle.

Step 1. Before firing, the rifle and bullet are at rest together, so the total momentum of the system is zero: $p_{before} = 0$ .

Step 2. After firing, take the bullet's direction as positive. Let $V$ be the rifle's recoil velocity. Conservation of momentum gives $0 = m_{bullet}v_{bullet} + m_{rifle}V$ .

Step 3. Solve for $V$ : $V = -\dfrac{m_{bullet}v_{bullet}}{m_{rifle}} = -\dfrac{0.020 \times 400}{4.0} = -2.0\ \text{m/s}$ . The rifle recoils backward at $2.0\ \text{m/s}$ — the heavy rifle moves slowly because it carries the same $8.0\ \text{kg m/s}$ of momentum magnitude as the light, fast bullet, just in the opposite direction.

Competitive level - perfectly inelastic collision of two trolleys

A trolley of mass $2.0\ \text{kg}$ moving at $3.0\ \text{m/s}$ collides head-on with a stationary trolley of mass $1.0\ \text{kg}$ , and the two lock together on impact. Find their common velocity just after the collision, and check whether kinetic energy was also conserved.

Step 1. No external horizontal force acts during the brief collision (the track is smooth and the collision is fast), so momentum is conserved. Momentum before: $p_{before} = 2.0 \times 3.0 + 1.0 \times 0 = 6.0\ \text{kg m/s}$ .

Step 2. After collision the combined mass $3.0\ \text{kg}$ moves at one common velocity $v$ , so $p_{after} = (2.0+1.0)v = 3.0v$ . Setting $p_{before}=p_{after}$ gives $3.0v = 6.0$ , so $v = 2.0\ \text{m/s}$ in the original direction.

Step 3. Kinetic energy check: $KE_{before} = \tfrac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}$ ; $KE_{after} = \tfrac{1}{2}(3.0)(2.0)^2 = 6.0\ \text{J}$ . Common velocity is $2.0\ \text{m/s}$ , and $3.0\ \text{J}$ of kinetic energy is lost to heat and sound even though momentum stayed exactly conserved — a reminder that momentum conservation never guarantees kinetic energy conservation in a sticking collision. Timing note: this two-body, one-dimensional pattern should take under two minutes — write the "before" momentum, write the "after" momentum with the unknown velocity, equate, solve. Only add the energy check if the question explicitly asks whether the collision was elastic; do not do it by default, as it costs extra time for no marks in a pure momentum question.

Common traps

Pairing action and reaction on the same body. The third-law pair for a book resting on a table is the book's weight and the table's normal force only if you (wrongly) think they cancel as a pair — they do not, because they act on the same body and happen to be equal by equilibrium, not by Newton's third law. The true third-law partner of the table pushing up on the book is the book pushing down on the table — a force on a different body.
Applying momentum conservation across a system with an unbalanced external force. Friction, gravity along an incline, or a wall pushing on the system all remove or add momentum; conservation applies strictly over the brief instant of a collision or explosion when internal forces dominate, not over the entire subsequent motion.
Dropping vector signs. Momentum is a vector; in one-dimensional problems, a chosen positive direction must be used consistently — a velocity in the opposite sense must carry a negative sign or the total will be wrong.
Assuming kinetic energy is also conserved. Momentum is conserved in every isolated collision, elastic or not; kinetic energy is conserved only in an elastic collision, as the worked example above shows explicitly.

Common trapNewton's third law pairs never act on the same body, and momentum conservation never implies energy conservation

The single biggest mark-loser in this chapter is misidentifying an action-reaction pair as two forces acting on one object, such as weight and normal reaction on a resting book — those merely balance each other by equilibrium, they are not a third-law pair. A genuine third-law pair always acts on two different bodies and can never cancel within one body's free-body diagram. The second biggest error is assuming that because momentum is conserved in a collision, kinetic energy must be too — it is only conserved in the special case of an elastic collision; a sticking (perfectly inelastic) collision always loses some kinetic energy even while momentum stays exactly the same.

You should be able to…

State the law of conservation of linear momentum and the exact condition, zero net external force, under which it holds.
Derive the law from Newton's second and third laws for a two-body interaction.
Correctly identify genuine Newton's third law force pairs, distinguishing them from forces that merely balance on the same body.
Apply $m_1u_1+m_2u_2 = m_1v_1+m_2v_2$ to recoil, collision, and explosion or rocket problems in one dimension, handling vector signs correctly.
Explain why momentum can be conserved in a collision even when kinetic energy is not.

Ask the tutor: stuck on setting up the before-and-after momentum equation for a recoil or collision problem — open the AI tutor on this page and walk through it step by step.

Why does momentum stay conserved even in a violent collision like a car crash?

Because the huge forces the two vehicles exert on each other during the crash are internal to the system of "both cars," and by Newton's third law those internal forces are equal and opposite and act for exactly the same brief time. Adding up the impulses they deliver to each other, they cancel out of the total, no matter how large they are individually. External forces — road friction, air resistance — do act too, but over the tiny fraction of a second the crash actually takes, they are negligible compared to the crushing forces between the cars, so momentum is conserved to a very good approximation during the collision itself.

If I jump forward off a stationary skateboard, does the skateboard really move backward with equal momentum?

Yes, exactly, as long as no outside force interferes during the push-off. Before you jump, you and the skateboard together have zero total momentum. The instant you push off, your legs exert a backward force on the board and the board exerts an equal forward force on you (Newton's third law), for the same short time. Your forward momentum and the board's backward momentum must add to zero, since the total was zero before and internal forces cannot change it. This is exactly the same physics as rifle recoil, just with you as the bullet.

How does a rocket push itself forward in empty space where there is nothing to push against?

A rocket does not push against the air or anything external — it works by momentum conservation alone. The rocket burns fuel and ejects hot gas backward at very high speed; that ejected gas carries backward momentum, so by conservation of the total momentum of the rocket-plus-exhaust system, the rocket itself must gain an equal forward momentum. It is really a continuous version of recoil: instead of one bullet, the rocket fires a steady stream of exhaust "bullets," and each bit of ejected gas gives the rocket a further forward kick, all without needing anything outside to push against.

In the trolley collision example, where did the missing 3 joules of kinetic energy go?

It was converted into other forms of energy during the collision — mostly heat generated by the deformation and friction at the point of impact, plus a little sound (the bang of the collision) and possibly permanent deformation of the coupling mechanism that locks the trolleys together. Momentum has no equivalent "loss channel": it can only be transferred between the colliding bodies, never converted into heat or sound, which is exactly why momentum stays exactly conserved in every collision while kinetic energy is conserved only in the special elastic case where nothing gets deformed or heated.