Linear momentum of a system of particles

A firework shell climbs silently on a single parabolic arc, and then, at the top, it becomes a hundred glowing fragments scattering in every direction. Chase any one fragment with a stopwatch and its motion looks chaotic, unpredictable, hard to pin down. Yet step back and add up the momentum of every single fragment, at every instant, and something astonishing happens: the vector sum keeps moving exactly along the original parabola, as if the explosion never occurred. The same quiet bookkeeping is at work when a rifle kicks into a shoulder, when two ice skaters push off each other and glide apart, and when a rocket hisses exhaust downward and climbs upward against gravity's pull. This is the idea of linear momentum for a system of particles — not one object, but many, each with its own mass and velocity, all interacting through pushes, pulls, collisions and explosions that are strictly internal to the group. The question this article answers is simple to state and powerful to use: what happens to the total momentum, the vector sum $\vec P = \sum_i m_i\vec v_i$ , when internal forces are flying around but no external force acts? The answer — that total momentum is conserved, no matter how violent the internal chaos — is one of the most reliable tools in a physicist's kit, and it is the direct extension of the two-body momentum conservation already familiar from collisions. We will connect total system momentum to the velocity of the centre of mass, derive Newton's second law for a whole system in one clean line, and use it to solve a recoiling-gun problem and a multi-fragment explosion, exactly the kind of question that shows up in board exams and competitive papers alike.

Key takeaways

The total linear momentum of a system of $n$ particles is the vector sum of the individual momenta: $\vec P = \sum_i m_i\vec v_i$ .
This total momentum equals the total mass times the centre-of-mass velocity: $\vec P = M\vec V_{cm}$ — the whole system behaves, for momentum purposes, like one particle of mass $M$ at the centre of mass.
Newton's second law for a system reads $\dfrac{d\vec P}{dt} = \vec F_{ext}$ — only the net external force changes total momentum; internal forces always cancel in pairs by Newton's third law.
Conservation of linear momentum for a system: if $\vec F_{ext} = 0$ , then $\vec P$ stays constant in time, however violent the internal collisions or explosions.
Recoil (guns, rockets), explosions (fireworks, bomb fragments) and collisions between many bodies are all governed by the same single vector equation.
Momentum is a vector: components must be added along each axis separately, never magnitudes.
Key idea: the total momentum of any system of particles equals the total mass times the centre-of-mass velocity, and it changes only under a net external force, so with zero external force it stays exactly constant no matter how the particles push and collide among themselves.

The core insight

For a system of particles, $\vec P = \sum_i m_i\vec v_i = M\vec V_{cm}$ changes only because of the net external force through $d\vec P/dt = \vec F_{ext}$ , so internal forces alone — collisions, explosions, recoil — can never alter the total momentum.

Where it fits in the syllabus

This is part of the CBSE Class 11 chapter "System of Particles and Rotational Motion," built directly on top of the motion of the centre of mass. The prescribed scope asks students to state the total momentum of a system, connect it to the centre-of-mass velocity, and apply conservation of momentum when no net external force acts. It is a direct generalisation of the two-body momentum conservation already met while studying Work, Energy and Power, extended here to systems with any number of particles — three particles, a spray of shell fragments, or an entire rigid body.

The physical picture

Every particle in a system carries its own momentum, $m_i\vec v_i$ , and the total linear momentum is just their vector sum:

\vec P = \sum_{i=1}^{n} m_i\vec v_i

where: $m_i$ is the mass of the $i$ -th particle, $\vec v_i$ its velocity, and the sum runs over all $n$ particles in the system. The physical reading is that $\vec P$ folds every particle's motion into one single vector — it says nothing about where each particle is individually headed, only about the combined push of the whole crowd. Recall that the centre-of-mass velocity is $\vec V_{cm} = \dfrac{\sum_i m_i\vec v_i}{M}$ , where $M = \sum_i m_i$ is the total mass. Multiplying both sides by $M$ shows the numerator is exactly the total momentum, so:

\vec P = M\vec V_{cm}

where: $M$ is the total mass of the system and $\vec V_{cm}$ is the velocity of its centre of mass. The physical reading is striking: no matter how scattered or chaotic the individual particles are, the total momentum is identical to what you would compute by pretending the whole mass is concentrated at a single point moving with $\vec V_{cm}$ . This is why the firework shell's centre of mass keeps sailing along the original parabola even as fragments scatter in a hundred directions — the individual momenta cancel out in every direction except the one the centre of mass was already travelling in. Differentiating $\vec P = M\vec V_{cm}$ with respect to time and recalling $M\vec A_{cm} = \vec F_{ext}$ from the motion of the centre of mass gives Newton's second law written for an entire system:

\frac{d\vec P}{dt} = \vec F_{ext}

where: $\vec F_{ext}$ is the vector sum of all external forces acting on the system — gravity, an applied push, friction from outside — but never the forces particles inside the system exert on each other. The physical reading is the heart of this article: only outside forces can change the total momentum; whatever the particles do to one another internally is invisible to $\vec P$ .

Figure

Three fragments flying apart after an explosion with arrows showing their individual momenta pointing in different directions, alongside a single combined arrow labelled total momentum P equal to M times V c m following the same straight path the object had before it broke apart

Newton's second law for a system of particles

Derivation ANewton's second law for a system of particles

The goal is to show, starting only from Newton's second and third laws applied particle by particle, that the total momentum of a many-particle system changes only in response to the net external force — no matter how strong or numerous the internal forces between particles are. This is the algebraic engine behind conservation of momentum: it explains why internal collisions, explosions and recoils never touch the total momentum, rather than simply asserting it.

Step 1. Write Newton's second law for one particle. For the $i$ -th particle of mass $m_i$ , the net force on it is the sum of an external part $\vec F_i^{ext}$ (forces from outside the system, such as gravity or an applied push) and an internal part $\vec F_i^{int}$ (forces from every other particle inside the system, such as a collision push or an explosive shove). So $m_i\dfrac{d\vec v_i}{dt} = \vec F_i^{ext} + \vec F_i^{int}$ .

Step 2. Sum this equation over every particle in the system. Adding the left-hand sides gives $\sum_i m_i\dfrac{d\vec v_i}{dt} = \dfrac{d}{dt}\Big(\sum_i m_i\vec v_i\Big) = \dfrac{d\vec P}{dt}$ , since the total momentum is exactly $\vec P = \sum_i m_i\vec v_i$ . Adding the right-hand sides gives $\sum_i \vec F_i^{ext} + \sum_i \vec F_i^{int}$ , the total external force plus the total internal force.

Step 3. Show the internal forces cancel completely. Every internal force comes in an action–reaction pair: if particle $j$ pushes particle $k$ with force $\vec F_{jk}$ , then by Newton's third law particle $k$ pushes back on particle $j$ with $\vec F_{kj} = -\vec F_{jk}$ . When every such pair is added into the sum $\sum_i \vec F_i^{int}$ , each pair contributes $\vec F_{jk} + \vec F_{kj} = 0$ . Since this happens for every internal pair without exception, the entire sum of internal forces vanishes: $\sum_i \vec F_i^{int} = 0$ .

Step 4. Combine the results. With the internal forces gone, summing the equations of motion leaves $\dfrac{d\vec P}{dt} = \vec F_{ext}$ , where $\vec F_{ext} = \sum_i \vec F_i^{ext}$ is the net external force on the whole system. This is Newton's second law written for the system as a single unit.

Step 5. Read off the conservation law. If the net external force is zero at every instant, $\dfrac{d\vec P}{dt} = 0$ , which means $\vec P$ is constant in time — the law of conservation of linear momentum for a system of particles. Internal forces, however large, never appear in this final result because Step 3 destroyed them in pairs; only what pushes on the system from outside can ever change its total momentum.

Worked examples

Board level — total momentum and centre-of-mass velocity of a three-particle system

A system consists of three particles: $m_1 = 1\ \text{kg}$ moving at $\vec v_1 = (3\ \text{m/s})\hat i$ , $m_2 = 2\ \text{kg}$ moving at $\vec v_2 = (-1\ \text{m/s})\hat i$ , and $m_3 = 2\ \text{kg}$ moving at $\vec v_3 = (2\ \text{m/s})\hat j$ . Find the total momentum of the system and the velocity of its centre of mass.

Step 1. Write the total momentum as the vector sum of individual momenta: $\vec P = m_1\vec v_1 + m_2\vec v_2 + m_3\vec v_3$ .

Step 2. Substitute the numbers, keeping the $\hat i$ and $\hat j$ components separate: $\vec P = 1(3\hat i) + 2(-1\hat i) + 2(2\hat j) = (3-2)\hat i + 4\hat j = 1\hat i + 4\hat j\ \text{kg}\,\cdot\,\text{m/s}$ .

Step 3. Find the total mass: $M = m_1+m_2+m_3 = 1+2+2 = 5\ \text{kg}$ .

Step 4. Use $\vec P = M\vec V_{cm}$ to get the centre-of-mass velocity: $\vec V_{cm} = \dfrac{\vec P}{M} = \dfrac{1\hat i+4\hat j}{5} = 0.2\hat i + 0.8\hat j\ \text{m/s}$ , with magnitude $|\vec V_{cm}| = \sqrt{0.2^2+0.8^2} = \sqrt{0.68} \approx 0.82\ \text{m/s}$ . Answer: $\vec P = (1\hat i + 4\hat j)\ \text{kg}\,\cdot\,\text{m/s}$ and $\vec V_{cm} \approx 0.82\ \text{m/s}$ , directed mostly along $\hat j$ .

Competitive level — recoil of a gun, then a follow-up fragment check

A gun of mass $5\ \text{kg}$ is at rest and fires a bullet of mass $0.02\ \text{kg}$ with a muzzle speed of $400\ \text{m/s}$ . (a) Find the recoil speed of the gun. (b) If, instead, the bullet had split in flight one metre after leaving the barrel into two fragments of equal mass moving symmetrically at $30^\circ$ to the original line of fire, would the gun's recoil speed found in part (a) change?

Step 1. Before firing, the gun-bullet system is at rest, so the total initial momentum is zero: $\vec P_i = 0$ . The firing is driven entirely by internal forces — the explosive gas pressure between gun and bullet — with no significant external horizontal force during the brief firing interval, so momentum is conserved: $\vec P_i = \vec P_f$ .

Step 2. Write the final momentum as the sum of bullet and gun momenta and solve for the gun's velocity: $0 = m_{bullet}v_{bullet} + m_{gun}v_{gun}$ , so $0 = (0.02)(400) + (5)v_{gun}$ , giving $v_{gun} = -\dfrac{8}{5} = -1.6\ \text{m/s}$ . The negative sign shows the gun recoils opposite to the bullet, with speed $1.6\ \text{m/s}$ .

Step 3. For part (b), notice that the bullet splitting in flight is an event that happens after the gun has already recoiled and involves only the bullet's own fragments — it introduces no new force on the gun at all. The gun-bullet momentum balance at the instant of firing, found in Step 2, is completely unaffected by anything the bullet does downstream. Answer: the recoil speed is $1.6\ \text{m/s}$ , and it is unchanged by the later fragmentation, because that split is an internal event within the bullet's own trajectory and exerts no force back on the gun. Timing note: under two minutes for part (a) — the chain is $\vec P_i = \vec P_f = 0$ , then solve the single linear equation for $v_{gun}$ . Part (b) needs no new arithmetic at all: recognising which objects belong to which momentum-conservation system, and when an event is simply irrelevant to an already-closed system, is worth more marks under time pressure than re-deriving anything.

Common traps

Adding momenta as scalars (adding magnitudes) instead of as vectors — components along each axis must be added separately, and directions matter.
Believing conservation of momentum means every individual particle's momentum stays fixed — only the total (vector sum) is conserved when the net external force is zero; particles can gain or lose momentum freely through internal forces.
Forgetting the shortcut $\vec P = M\vec V_{cm}$ , which links total momentum directly to centre-of-mass velocity in either direction, once one of the two is known.
Applying conservation of momentum when a real external force — friction from the ground, an outside push, air resistance — is actually acting and cannot be neglected.
Treating an event that happens strictly inside one part of the system (like a bullet fragmenting after it has already left the gun) as if it could retroactively change a momentum balance that was already fixed by an earlier internal interaction.

Common trapTotal momentum conservation is a statement about the whole system, never about any one particle inside it

The single biggest mark-losing habit is applying "momentum is conserved" to a single particle picked out of a larger system, or assuming that because the total momentum is constant, each piece must be moving at a constant velocity too. Neither is true. Internal forces are perfectly free to speed up, slow down, or completely redirect any individual particle — a fragment can shoot off in a wild new direction, a bullet can gain enormous speed relative to a stationary gun — precisely because those changes are compensated by an equal and opposite change somewhere else in the system, by Newton's third law. Conservation of momentum is a bookkeeping statement about the vector sum, valid only when no external force acts on the system as a whole; it says nothing at all about the fate of any single member of that system.

You should be able to…

Write the total momentum of a system of particles as $\vec P = \sum_i m_i\vec v_i$ and evaluate it component-wise for a given set of masses and velocities.
Relate total system momentum to centre-of-mass velocity through $\vec P = M\vec V_{cm}$ , working in either direction.
Derive $d\vec P/dt = \vec F_{ext}$ from Newton's second and third laws applied to each particle, and explain why internal forces drop out.
State the law of conservation of linear momentum for a system of particles and identify correctly when it applies.
Solve recoil and explosion problems (guns, rockets, fragmenting shells) using vector addition of momenta.
Distinguish a statement about the total momentum of a system from a (false) claim about the momentum of any one particle within it.

Ask the tutor: want a walkthrough of a recoil, explosion or multi-particle momentum problem, or a check on why internal forces never touch the total momentum — ask the AI tutor.

If momentum is conserved during an explosion, why do the fragments fly off so fast compared to the original object?

Conservation of momentum only fixes the vector sum of the fragment momenta, not how that total gets shared out. The explosion supplies extra kinetic energy from a chemical or mechanical source, and that energy can send fragments flying at high individual speeds in many directions, as long as their momenta still add up to the momentum the object had just before it broke apart. A firework shell's fragments can each move far faster than the shell itself was travelling, yet their vector sum still equals the shell's original momentum, because the internal explosive forces that separated them appeared and cancelled in equal-and-opposite pairs.

Does conservation of momentum for a system of particles require the particles to stay close together, like in one object?

No. The law $d\vec P/dt = \vec F_{ext}$ makes no assumption about how spread out the particles are — it only cares whether a force is internal (between members of the system) or external (from outside it). Three separate billiard balls rolling on a table form a perfectly valid system of particles, and if no net external horizontal force acts on all three together (friction ignored), their combined momentum is conserved exactly as it would be for fragments of a single exploding shell. What matters is which objects you have chosen to include in your system, not their physical proximity.

How is this different from the momentum conservation I already used for two-body collisions?

It is the exact same law, just written for any number of particles instead of only two. Two-body collision problems are the special case $n=2$ of $\vec P = \sum_i m_i\vec v_i$ staying constant when there is no net external force — the same reasoning that gave you $m_1u_1+m_2u_2 = m_1v_1+m_2v_2$ for a collision now extends, term by term, to three particles, ten fragments, or an entire rigid body treated as a swarm of point masses. Nothing new is being assumed; the vector sum simply grows longer.

Why does the direction of the recoil in the gun example come out negative?

The sign is not a mathematical accident — it is the vector nature of momentum doing its job. Once you fix a positive direction, here the direction the bullet travels, conservation of momentum from a zero-momentum start forces the gun's momentum to be exactly opposite in sign so that the two add back up to zero. The negative sign in $v_{gun} = -1.6\ \text{m/s}$ is the calculation itself confirming what you would expect physically: the gun must recoil backward, away from the bullet's direction of travel, and the size of that recoil speed is set by the ratio of the masses, $m_{bullet}/m_{gun}$ , times the bullet's speed.