Conservation of Linear Momentum

Momentum conservation is physics' favourite detective. In 1930, beta-decay experiments showed momentum apparently vanishing; Wolfgang Pauli bet on the ledger — something invisible must be carrying the missing amount — and invented the neutrino decades before any instrument could see one. The same wager runs today at the LHC: analysts sum the transverse momenta of everything a collision produces, and when the vectors fail to close, the imbalance is the signature of an escaping invisible particle — it is literally called "missing momentum", and it is the standard search channel for dark-matter candidates. At more familiar scales, every rocket is conservation with a throttle (momentum out the nozzle, equal and opposite momentum into the vehicle — the variable-mass module makes it quantitative), spacecraft dock by exchanging millimetre-per-second momentum increments computed to the gram, and crash investigators reconstruct pre-impact speeds from wreckage paths by running this article's equations backwards — the law is admissible evidence.

Draw a boundary around the bodies you care about — that boundary is the system, and the single most consequential act in any momentum problem is where you draw it. Every force in the problem is now one of two kinds. Internal forces act between members of the system: the bullet on the gun and the gun on the bullet, the two carts squeezing each other's buffers. By Newton's third law these come in equal-and-opposite pairs at every instant, so their contributions to the system's total momentum cancel identically — each pair is a private transfer between members, moving momentum around inside the boundary the way passing cash around a family changes no one's household total. External forces reach across the boundary from outside — gravity, friction from the ground, a wall. Only these can change the total.

Watch the law work on the skateboard. Before the throw: everything at rest, total momentum zero. During the throw your hands push the bag and the bag pushes back — an internal pair, cancelling. After: bag momentum left, your momentum right, total still zero. The distribution changed dramatically; the total never moved. Now redraw the boundary around the bag alone: suddenly your push is an external force, and the bag's momentum duly changes by exactly the impulse you delivered — the previous article's law. Same physics, two bookkeepings; the system choice decides which law you get to use. The vector fine print matters just as much. Momentum conservation is three independent statements, one per direction. A shell arcing through the sky feels gravity — a vertical external force — so its vertical momentum is being openly spent and repaid on schedule. But nothing external acts horizontally: the horizontal component of the system's momentum is conserved through the entire flight, including through the instant the shell explodes into fragments. And during the explosion itself — a millisecond of internal violence — even gravity's impulse ($mg\,\Delta t$, with tiny $\Delta t$) is negligible, so all components are conserved across that instant. This two-tier reading — conserved-forever along force-free directions, conserved-across-the-bang for everything — unlocks most of the chapter's problems.

Define the system's total momentum as the vector sum over its members:

where: each $\vec v_i$ is measured in the same (ground) frame, and the sum is a genuine vector sum — momenta at angles add tip-to-tail, and opposite momenta cancel. Two 1-kg balls at $\pm 5$ m/s: individually formidable, jointly zero. This capacity to cancel is what makes the total worth tracking. The previous article's impulse bookkeeping, summed over all members, gives the system law (derivation A):

where: $\vec F_{\text{ext,net}}$ — the net external force; the internal forces have already cancelled out of the sum, exactly and always. Read it physically: the system's total momentum ledger records only cross-boundary transactions. It is the single-body law scaled up, with the outside world in the role of "the force". Set the external force to zero and the ledger freezes — the conservation law itself:

where: "before" and "after" bracket any internal event whatsoever — a spring release, a collision, an explosion, a bullet embedding. And the impulsive fine print: if a nonzero external force $F_{\text{ext}}$ acts but the event lasts only $\Delta t$, the damage to the ledger is $F_{\text{ext}}\Delta t$ — negligible when $\Delta t$ is milliseconds and the internal transfers are large. That is why conservation legally crosses a bullet-into-block impact even on a table with friction and gravity present. The most-used special case starts from rest. If $\vec P = 0$ before, it is zero after, so two flying-apart pieces satisfy:

where: the minus sign says opposite directions — recoil. Speeds come out in inverse mass ratio: the light bullet flies, the heavy rifle nudges. Derivation B adds the energy postscript exams love: the kinetic energies also split in inverse mass ratio, so the lighter fragment carries almost all the energy — which is why the bullet, not the rifle, does the damage.

Reading the fine print like an examiner

"Isolated" is a per-direction, per-timescale judgment. A cart on a frictionless track: horizontal momentum conserved for all time (normal force and gravity are vertical). A shell in flight: horizontal conserved always; vertical only across the brief explosion. Two cars crashing on a real road: friction exists, but over the 0.1 s crunch its impulse is a rounding error against the collision impulses — conserve across the crash, then let friction take over the aftermath.

Relative-velocity wording. "The bullet leaves the barrel at 400 m/s relative to the gun" is a JEE trap: if the gun recoils at $V$ backward, the bullet's ground speed is $400 - V$. Conservation must be written in one frame (the ground), so convert relative velocities first.

What conservation cannot do. It is one vector equation. It will never tell you the forces (impulse's job), the energy lost (the KE audit's job), or two unknown speeds from one component (elastic collisions need the energy equation too — next articles). Knowing which question the law cannot answer is half the technique module's dispatch.

1 · Board level — rifle recoil, and where the energy went (CBSE/ICSE pattern, 3 marks)

A $4.0\ \mathrm{kg}$ rifle fires a $20\ \mathrm{g}$ bullet at $400\ \mathrm{m/s}$. Find the rifle's recoil speed, and compare the kinetic energies of bullet and rifle.

Step 1. System = rifle + bullet; horizontal external forces negligible during the shot. Before: $P = 0$.

Step 2. After: $0 = (0.020)(400) - (4.0)V$, so $V = \dfrac{8.0}{4.0} = \boxed{2.0\ \mathrm{m/s}}$ backward.

Step 3. Energies: bullet $\tfrac12(0.020)(400)^2 = \boxed{1600\ \mathrm{J}}$; rifle $\tfrac12(4.0)(2.0)^2 = \boxed{8.0\ \mathrm{J}}$.

Step 4. The ratio is $200 = m_{\text{rifle}}/m_{\text{bullet}}$ — equal momenta, so KE splits inversely with mass ($p^2/2m$). This is why the shoulder survives what the target does not, and it earns the reasoning mark.

2 · Board level — jumping off the boat (IGCSE/ICSE pattern)

A $50\ \mathrm{kg}$ girl standing on a $200\ \mathrm{kg}$ stationary boat jumps horizontally at $4.0\ \mathrm{m/s}$ (relative to the water). Find the boat's velocity, and state the assumption that makes conservation valid.

Step 1. System = girl + boat; before the jump, $P = 0$. Assumption to state: water resistance is negligible during the brief jump (its impulse over a fraction of a second is tiny).

Step 2. After: $0 = (50)(4.0) + (200)V$, so $V = -\dfrac{200}{200} = \boxed{1.0\ \mathrm{m/s}}$, opposite to the jump.

Step 3. Audit with equation (4): speed ratio $4.0/1.0 = 4 = 200/50$ ✓ — inverse mass ratio, as a from-rest split must be.

3 · Competitive level — the shell splits at the top (JEE-Main pattern, ~3 minutes)

A shell of mass $3m$ moving horizontally at speed $v$ at the top of its trajectory explodes into two pieces: $m$ drops with zero initial velocity (falls vertically from rest), and $2m$ continues. Find the velocity of the $2m$ piece just after the explosion, and the system's kinetic-energy gain.

Step 1. At the top, the shell's velocity is horizontal ($v$). During the millisecond explosion, gravity's impulse is negligible → conserve the full momentum vector across the bang.

Step 2. Horizontal: $3mv = m(0) + 2m\,v_2 \Rightarrow v_2 = \boxed{\dfrac{3v}{2}}$, horizontal, same direction. Vertical: $0 = 0 + 2m\,v_{2y} \Rightarrow v_{2y} = 0$ ✓ (the surviving piece stays horizontal).

Step 3. Energy audit: before, $\tfrac12(3m)v^2 = 1.5mv^2$; after, $\tfrac12(2m)(3v/2)^2 = 2.25mv^2$. Gain $= \boxed{0.75\,mv^2}$ — supplied by the explosive, not conjured. (An answer with less KE after a from-inside explosion is a sign error.)

Step 4. MCQ audit: total momentum after $= 2m \times 3v/2 = 3mv$ ✓. Timing note: ~3 min. "Drops dead / falls vertically from rest" translates to "this fragment's velocity is zero just after the bang" — momentum $3mv$ must be carried entirely by the remainder. The classic variant "one piece retraces its path at speed $v$" gives $2m$'s speed as $2v$ by the same one-liner. Never re-derive: the pattern is (total before) = (survivors' sum).

4 · Competitive level — three fragments, vector closure (NEET/JEE pattern, ~3.5 minutes)

A bomb at rest explodes into three fragments. Two, each of mass $m$, fly off perpendicular to each other, each at speed $v$. The third has mass $2m$. Find its velocity.

Step 1. From rest: $\vec p_1 + \vec p_2 + \vec p_3 = 0$, so $\vec p_3 = -(\vec p_1 + \vec p_2)$ — magnitude and direction both fixed. Set fragment 1 along $+x$, fragment 2 along $+y$: $\vec p_1 = mv\,\hat x$, $\vec p_2 = mv\,\hat y$.

Step 2. Sum: $|\vec p_1 + \vec p_2| = \sqrt{(mv)^2 + (mv)^2} = \sqrt2\,mv$, along the $45°$ bisector.

Step 3. So $\vec p_3$ has magnitude $\sqrt2\,mv$ pointing opposite the bisector ($135°$ from each fragment... more precisely $135°$ from $+x$ measured the long way: direction $(-\hat x -\hat y)/\sqrt2$). Speed: $v_3 = \dfrac{\sqrt2\,mv}{2m} = \boxed{\dfrac{v}{\sqrt2}}$, at $\boxed{135°}$ to each of the two equal fragments.

Step 4. Audit: components — $p_{3x} = -mv$ cancels $p_1$ ✓, $p_{3y} = -mv$ cancels $p_2$ ✓. The three momentum vectors close a right-angled triangle, as the diagram demands. Timing note: ~3.5 min drawn, ~90 s if you go straight to components. For three-fragment closures always work with momentum components, never speeds — masses differ, so the momentum triangle closes while the velocity triangle does not. The distractor $v/2$ (dividing speed sums instead of momentum sums) is on every version of this MCQ.

Conserving a non-isolated component. Momentum along an incline during sliding, vertical momentum over a whole projectile flight — external forces act there. Name the direction and check who pushes along it.

Ball-off-wall "conservation". One body interacting with something outside your system does not conserve its own momentum. Either include the wall/Earth or use the impulse law instead.

Relative-velocity muzzle speeds. "$u$ relative to the gun/boat" must be converted to the ground frame ($u - V$) before conservation is written. The JEE follow-up "find the bullet's ground speed" is testing exactly this.

Scalar addition of momenta. Momenta at angles add as vectors — components or closure diagrams. Adding magnitudes is the manufactured wrong answer in every 2-D explosion MCQ.

"Momentum conserved ⇒ KE conserved." Explosions gain KE (internal stores), sticking collisions lose it — both conserve momentum perfectly. The two books are audited separately, and the KE-change question is usually worth its own mark.

Signs in recoil. Set one positive direction and keep it: $0 = m_1v_1 - m_2v_2$ needs the minus because you wrote both as speeds. Mixing signed velocities with unsigned speeds double-counts the direction.

Forgetting conservation is legal across brief events despite friction/gravity. The reverse trap: students refuse to conserve across a bullet-embedding "because friction acts". Check impulses: $F_{\text{ext}}\Delta t$ over milliseconds is nothing against the collision's internal transfers.

Which $g$/frame bookkeeping. As ever: the paper's $g$; and all velocities in one frame — the ground unless explicitly told otherwise.

Texts. NCERT Physics Class 11, Ch. 5 (Laws of Motion — conservation of momentum) · HC Verma, Concepts of Physics Vol. 1, Ch. 9 · Morin, Introduction to Classical Mechanics, Ch. 4 · Feynman Lectures Vol. I, Ch. 10 for the conservation-as-symmetry preview.

Provenance. All results are standard. A verifier should recompute: the two-body cancellation in derivation A; the recoil split and $KE \propto 1/m$ in derivation B; Example 1 ($V = 2.0$ m/s, $1600$ J vs $8.0$ J, ratio 200); Example 2 ($1.0$ m/s, ratio audit 4:1); Example 3 ($v_2 = 3v/2$, KE gain $0.75mv^2$); Example 4 ($v_3 = v/\sqrt2$ at $135°$, components cancelling); the FAQ's friction-impulse estimate.

Prerequisites. Linear Momentum & Impulse · Newton's Third Law · Vector Components. Leads to. Elastic Collisions in 1D · Inelastic Collisions & Coefficient of Restitution · Explosions, Recoil & Fragments (archetype) · Conservation of Linear Momentum (UG deep dive).