Conservation of linear momentum — derivation from Newton's third law: Question bank

Before we start, four reminders that every answer leans on:

Definition The tools we use everywhere on this page

Momentum of one particle $p = m v$ is a vector — it has a direction, so along a chosen axis it can be positive or negative. (Here $m$ = mass, $v$ = velocity.)
Total momentum of the whole system is the vector sum over all particles, $P = \sum_{i} m_{i} v_{i}$ . Note the capital $P$ (whole system) versus the small $p$ (one particle).
Total mass $M = \sum_{i} m_{i}$ is the sum of all the individual masses. With it, the total momentum can be written $P = M v_{c m}$ , where $v_{c m}$ is the velocity of the system's centre of mass.
Frame velocity $u$ is the constant velocity of a new observer relative to the old one. Watching from a frame that moves at $u$ , every velocity looks smaller by $u$ , so the observed total momentum becomes $P - M u$ .
Total momentum stays constant only when the net external force on the system is zero. A system with zero net external force is called an isolated system.
Subscripts label which object: $m_{1}, v_{1}$ mean "mass and velocity of body 1"; $p_{2}$ means "momentum of body 2". For a gun-and-bullet problem we write $m_{b}, v_{b}$ for the bullet and $m_{g}, v_{g}$ for the gun. The letters after $F$ read as from→on: $F_{12}$ = force on body 1 from body 2. When we write $F_{ij}$ , the index $i$ is the body being pushed and $j$ is the body doing the pushing.

Figure 1 shows what "add momenta with signs" actually looks like — keep it in mind for the whole page.

Figure — Conservation of linear momentum — derivation from Newton's third law

Figure 1 — Gun-and-bullet from rest: signed momenta $+ 4$ and $- 4$ add to the initial zero.

Spot the error

Error hunt: "The gun and bullet start at rest, so after firing their speeds must be equal."

Wrong — their momenta are equal and opposite, not their speeds. Since

m_{b} v_{b} = m_{g} v_{g}

(momentum from rest sums to zero), the light bullet gets a huge speed and the heavy gun a small recoil speed.

Error hunt: "Two carts stick together, so I set both final velocities to

v_{1}^{'}

and

v_{2}^{'}

separately."

Wrong — sticking means one common final velocity

v^{'}

. Writing two unknowns invents freedom that the physics has removed.

Error hunt: "KE before = KE after, therefore momentum before = momentum after — same statement."

Wrong — they are different conservation laws. KE (

\frac{1}{2} m v^{2}

, a scalar) can be destroyed while momentum (

m v

, a vector) survives; equating them confuses a scalar with a vector.

Error hunt: "A bullet of mass

m_{b} = 0.01

kg leaves at

v_{b} = 400

m/s and the gun of mass

m_{g} = 2

kg recoils at

v_{g} = 2

m/s, so the total speed after is

402

m/s — but before it was

0

. Momentum broke!"

Wrong — you add momenta with signs along an axis, not speeds. Taking rightward as

+

, the bullet's momentum is

+ m_{b} v_{b} = + 4

kg·m/s and the gun's is

- m_{g} v_{g} = - 4

kg·m/s; their vector sum is

4 - 4 = 0

, exactly the starting value. Speeds are unsigned magnitudes and cannot be summed like this — momentum's direction is precisely the information that makes the total cancel (see Figure 1).

Error hunt: "A ball bounces off a wall and comes back, so momentum is conserved for the ball."

Wrong — the wall exerts an external force on the ball, reversing its momentum. Momentum is conserved only for the ball + wall + Earth system, not the ball alone.

Error hunt: "During the collision the contact force is enormous, so it must change the total momentum a lot."

Wrong — the contact force is internal to the two-body system; its action–reaction partner cancels it exactly. Big or small, internal forces move momentum around, never change the total (see Figure 2).

Error hunt: "The system loses KE in an inelastic hit, so it must lose momentum too."

Wrong — energy and momentum are independent. The lost KE becomes heat/sound; momentum has no such "leak" because internal forces still cancel in pairs.

Figure 2 — The internal action–reaction force pair is equal-and-opposite at every instant, so their impulses (areas) cancel and the total momentum stays steady.

Why questions

Why do internal forces never change the total momentum?

Because they occur in action–reaction pairs

F_{ij} = - F_{j i}

(force on

i

from

j

equals minus the force on

j

from

i

); summed over the whole system these opposite forces cancel to zero, leaving only external forces to alter

P

Why does the impulse of an internal force pair also cancel, not just the instantaneous force?

The impulse–momentum theorem says

\int F d t = Δ p

. For the pair,

\int F_{ij} d t + \int F_{j i} d t = \int (F_{ij} + F_{j i}) d t = \int 0 d t = 0

, because they are equal-and-opposite at every instant. So

Δ p_{i} + Δ p_{j} = 0

— whatever momentum

i

gains,

j

loses (see Impulse–Momentum Theorem and Figure 2).

Why must we conserve each coordinate axis separately in 2D and 3D?

Because

p = m v

is a vector, and

d P / d t = F^{ext}

holds component-by-component; a force along

x

cannot change the

y

-momentum.

Why is the recoil velocity of the gun negative in the worked example?

The minus sign means "opposite direction to the bullet." It is the physics of recoil, not a bookkeeping accident — dropping it would violate the zero starting momentum.

Why can we predict the after-state of a collision without knowing the force at every instant?

Because conservation compares only totals before and after; the messy time-history of the internal force integrates to equal-and-opposite impulses that cancel (see Impulse–Momentum Theorem).

Why does the two-body proof still work for a system of many bodies?

For

N

bodies,

d P / d t = \sum_{i} F_{i}^{ext} + \sum_{i \neq = j} F_{ij}

. Every internal term

F_{ij}

has a partner

F_{j i} = - F_{ij}

, so the whole double sum of internal forces cancels in pairs — leaving only the external forces, exactly as in the two-body case.

Why does the centre of mass of an isolated system move at constant velocity?

Because

P = M v_{c m}

(with

M

the total mass), and if

P

is constant with

M

fixed,

v_{c m}

must be constant too (elaborated in Centre of Mass Motion).

Why is the centre-of-mass frame special when analysing a collision?

It is the frame in which the total momentum is exactly zero, so the two bodies always approach and (for elastic hits) leave with equal-and-opposite momenta — the algebra becomes symmetric and simple. Conservation still holds; the frame just makes the bookkeeping cleanest.

Why does a rocket accelerate even in empty space with nothing to push against?

It pushes against its own ejected exhaust: the rocket + fuel system is isolated, so momentum gained forward equals momentum thrown backward (see Rocket Equation).

Why is Newton's second law written as

F = d p / d t

rather than

F = m a

in this derivation?

The momentum form is more fundamental and stays valid when mass changes; it also lets us add the two particle equations and recognise the sum as

d (p_{1} + p_{2}) / d t

, where

p_{1}, p_{2}

are the momenta of bodies 1 and 2.

Edge cases

Edge case: What is the total momentum of a system at rest?

Exactly zero, and it stays zero unless an external force acts. This is the key that unlocks explosion and recoil problems.

Edge case: A particle has zero velocity — does it contribute to total momentum?

No, its momentum

m v = 0

; but it still contributes mass, which matters for the common velocity after a sticky collision.

Edge case: Gravity acts on everything — is any collision on Earth truly isolated?

Over a very short collision time the gravitational impulse (

\int m g d t

) is negligible, so momentum is approximately conserved. Over long times gravity is external and does change

P

Edge case: Two identical objects moving toward each other at equal speeds collide and stick — what happens?

Their momenta are equal and opposite, summing to zero, so the stuck pair is at rest afterward. All the kinetic energy is lost, but momentum (

= 0

) is perfectly conserved.

Edge case: Can a collision increase the total kinetic energy?

Yes — a "super-elastic" event (like an explosion during contact, or a compressed spring releasing) adds KE from stored internal energy, yet momentum is still conserved.

Edge case: Does switching to a fast-moving observer ever break momentum conservation?

No — in any inertial frame the total is still constant (only its numerical value shifts by

M u

for a frame moving at frame-velocity

u

). Conservation is a statement about "constant in time," which no constant-velocity change of viewpoint can undo.

Connections

Newton's Third Law — why internal forces cancel in pairs
Newton's Second Law (momentum form) — the $F = d p / d t$ these traps rest on
Elastic vs Inelastic Collisions — the KE-conserved / KE-lost distinction
Centre of Mass Motion — constant $v_{c m}$ for isolated systems
Impulse–Momentum Theorem — why only before/after totals matter
Rocket Equation — pushing against your own exhaust

Question bank — Conservation of linear momentum — derivation from Newton's third law

True or false — justify

Spot the error

Why questions

Edge cases

Connections