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

We build from the ground: what an arrow means, what "momentum" is as an arrow, what "rate of change" looks like, and only then why the total holds still.

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Step 1 — What is an arrow (a vector), and what is momentum?

Before we can push arrows around, we need the two arrows that matter here.

• $\vec{v}$ — the velocity arrow: points where the object moves, length = how fast (metres per second).
• $m$ — the mass: a plain number (kilograms), no direction. It says how hard to shove the object is.

WHAT we did: named momentum as a stretched velocity arrow. WHY: momentum, not velocity, is the quantity that gets conserved — because Newton's laws are written about how momentum changes. PICTURE: below, the thin velocity arrow $\vec{v}$ gets stretched into the thick momentum arrow $\vec{p}=m\vec{v}$.

Here $m\vec{v}$ means: same direction as $\vec{v}$, length multiplied by $m$. The symbol $m$ is doing the stretching; the arrow $\vec{v}$ is doing the pointing.

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Step 2 — Two particles that only push each other

New symbols, each earned:

• ${\vec{F}}_{12}$ — the force on 1, from 2 (read the subscript "on–from").
• ${\vec{F}}_{21}$ — the force on 2, from 1.

WHAT: we set the stage — two objects, two forces between them. WHY: momentum conservation is a statement about a closed room. If anything reaches in from outside, the story changes (we handle that in Step 8). PICTURE: two balls, each with a force arrow pointing at the other one.

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Step 3 — Newton's 2nd law: force = how fast the momentum arrow changes

WHY this tool, not plain $\vec{F}=m\vec{a}$? Because we want a statement about momentum $\vec{p}$, and this form says the force directly is the rate of change of $\vec{p}$. That is exactly the quantity we are trying to track. (See Newton's Second Law (momentum form).)

WHAT: we linked each push to each object's momentum change. PICTURE: force arrow on 1 sits right on top of the "which way 1's momentum grows" arrow — same arrow.

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Step 4 — Newton's 3rd law: the two pushes are mirror images

WHAT: we said the two forces are equal length, opposite direction. WHY: this is the engine of the whole proof. Without it, the two forces would not cancel and momentum could drift. (See Newton's Third Law.) PICTURE: two arrows of identical length pointing dead opposite. If you slid one on top of the other tip-to-tail, they'd land you back where you started — they sum to nothing.

Read the minus sign literally: "${\vec{F}}_{21}$ is ${\vec{F}}_{12}$ turned around."

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Step 5 — Add the two equations so the twin forces meet

We stack the two Step-3 equations and add their left sides and their right sides:

$\underset{\text{the two pushes}}{\underbrace{{\vec{F}}_{12}+{\vec{F}}_{21}}}=\underset{\text{the two momentum-change rates}}{\underbrace{\frac{d{\vec{p}}_1}{dt}+\frac{d{\vec{p}}_2}{dt}}}$

WHAT: we glued the two equations into one, and bundled the two momenta into a single total arrow $\vec{P}={\vec{p}}_1+{\vec{p}}_2$. WHY: to make the action–reaction twins sit side by side on the left, ready to cancel. PICTURE: tip-to-tail addition — ${\vec{p}}_1$ then ${\vec{p}}_2$ — giving one total arrow $\vec{P}$; and separately the two force arrows brought together.

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Step 6 — The forces cancel; the total holds still

Put Step 4 into the left side of Step 5:

${\vec{F}}_{12}+{\vec{F}}_{21}={\vec{F}}_{12}+(-{\vec{F}}_{12})=\vec{0}$

• ${\vec{F}}_{12}$ and $-{\vec{F}}_{12}$ — same length, opposite way.
• $\vec{0}$ — the zero arrow: no length, no direction. Nothing.

So the right side must also be zero:

$\frac{d}{dt}({\vec{p}}_1+{\vec{p}}_2)=\vec{0}$

$\boxed{{\vec{p}}_1+{\vec{p}}_2=\vec{P}=\text{constant}}$

WHAT: the twin forces annihilated, forcing the total-momentum change-rate to zero. WHY: zero change-rate is the mathematical fingerprint of "stays the same." PICTURE: the two force arrows collapse into a dot ($\vec{0}$); the total momentum arrow $\vec{P}$ sits unchanged before and after.

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Step 7 — Watch it in a real trade: the collision

Numbers make it concrete. Cart 1 ($3$ kg at $4$ m/s) hits stationary cart 2 ($1$ kg) and they stick (a perfectly inelastic collision, see Elastic vs Inelastic Collisions).

$m_1{v}_1+m_2{v}_2=(m_1+m_2)v^{\prime }$ $(3)(4)+(1)(0)=(4)v^{\prime }\Rightarrow v^{\prime }=3\text{m/s}$

• Before: total momentum arrow = $12\text{kg⋅m/s}$ to the right (all in cart 1).
• After: $12\text{kg⋅m/s}$ still, now shared across the $4$ kg pair.

WHAT: momentum got redistributed — cart 1 gave some to cart 2 — but the total arrow's length is untouched. WHY: exactly Step 6 in action. The contact forces are an internal action–reaction pair. PICTURE: the total arrow before = the total arrow after, even though the pieces regrouped.

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Step 8 — Edge cases: what if the room isn't closed, or nothing moves?

We must cover every scenario so you never meet an unshown one.

Case A — an outside push exists. Then each object feels an extra external force ${\vec{F}}^{\text{ext}}$. The internal pair still cancels, but the leftovers don't: $\frac{d\vec{P}}{dt}={\vec{F}}_{\text{net}}^{\text{ext}}$ If ${\vec{F}}_{\text{net}}^{\text{ext}}\ne \vec{0}$, the total arrow grows or turns. Momentum is not conserved. (This is the Impulse–Momentum Theorem view.)

Case B — everything at rest to start. Then $\vec{P}=\vec{0}$ before. Conservation forces $\vec{P}=\vec{0}$ after too — so any pieces that fly apart must have momentum arrows that cancel (a gun's bullet vs. its recoil; an exploding shell). Equal-and-opposite, guaranteed.

Case C — 2D motion. An arrow splits into an $x$-part and a $y$-part. Since $\vec{P}$ is constant as an arrow, each part is separately constant. Conserve $x$ and $y$ independently.

WHAT: we swept the three ways reality departs from the clean proof. WHY: to hand you a rule that never breaks: check whether the outside push is zero first. PICTURE: three mini-panels — (A) an outside arrow tilts the total; (B) two opposite arrows from a rest start; (C) a slanted arrow broken into $x$ and $y$ each held fixed.

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The one-picture summary

Everything above compressed into one diagram: equal-and-opposite pushes ⇒ they add to the zero arrow ⇒ the total momentum arrow is frozen.

The total-momentum arrow $\vec{P}$ is the same length and direction before and after, no matter how the pieces rearrange — because the only forces inside the room came in cancelling pairs. Link back: $\vec{P}=M{\vec{v}}_{cm}$, so the Centre of Mass Motion glides on unchanged.

Recall Feynman retelling — the whole walkthrough in plain words

Picture two kids on ice, palms together. Each kid's "motion-money" is their momentum arrow: how heavy they are times how fast they slide. When they shove off, the third law says each shoves the other exactly as hard, exactly the opposite way — mirror arrows. Newton's second law says a shove is the speed at which your motion-money changes. Add both kids' books: their two shoves are mirror images, so together they add to nothing. And if the total's change-rate is nothing, the total is frozen. So the kids fly apart, one left one right, but the two motion-monies added up are the same as before — zero if they started still. The only way to change the grand total is a push from outside the ice — a wall, a rope, gravity. No outside push, no change. That's the whole story, drawn.

Recall Where exactly did Newton's 3rd law get used?

Step 4 (the mirror arrows) fed into Step 6, turning ${\vec{F}}_{12}+{\vec{F}}_{21}$ into the zero arrow.

Recall What single condition must you check before claiming momentum is conserved?

Is the net external force zero during that interval? (Step 8, Case A.)

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Connections

• Newton's Third Law — the mirror-arrow engine (Step 4)
• Newton's Second Law (momentum form) — push = momentum change-rate (Step 3)
• Elastic vs Inelastic Collisions — Step 7's energy caveat
• Impulse–Momentum Theorem — Step 8 Case A, integral view
• Centre of Mass Motion — the frozen $\vec{P}$ as steady ${\vec{v}}_{cm}$
• Rocket Equation — Step 8 Case B with changing mass