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

WHAT are we proving?

WHY do we care? Because it lets us predict the after of a collision/explosion without knowing the messy details of the forces during contact. We never need to know how the force varied microsecond by microsecond — only the totals before and after.

HOW: derive it from scratch (two-body case)

Take two particles, 1 and 2, that interact only with each other (no outside forces). Particle 2 pushes on 1 with force $\vec{F}_{12}$, and 1 pushes on 2 with $\vec{F}_{21}$.

Step 1 — Newton's 2nd law for each, in momentum form. $\vec{F}_{12} = \frac{d\vec{p}_1}{dt}, \qquad \vec{F}_{21} = \frac{d\vec{p}_2}{dt}$ Why this step? Newton's 2nd law is really $\vec F = d\vec p/dt$ (force = rate of change of momentum). Each particle's momentum changes only because of the force on it.

Step 2 — Newton's 3rd law. $\vec{F}_{12} = -\vec{F}_{21}$ Why this step? The forces of an action–reaction pair are equal in magnitude, opposite in direction. This is the engine of the whole proof.

Step 3 — Add the two equations of Step 1. $\vec{F}_{12} + \vec{F}_{21} = \frac{d\vec{p}_1}{dt} + \frac{d\vec{p}_2}{dt} = \frac{d}{dt}\big(\vec{p}_1+\vec{p}_2\big)$ Why this step? Adding lets the action–reaction pair appear together so we can cancel it.

Step 4 — Use Step 2 on the left side. $\vec{F}_{12} + \vec{F}_{21} = \vec{F}_{12} + (-\vec{F}_{12}) = 0$

Step 5 — Conclusion. $\frac{d}{dt}\big(\vec{p}_1+\vec{p}_2\big) = 0 \;\;\Longrightarrow\;\; \vec{p}_1+\vec{p}_2 = \text{constant}$

Generalising to N particles (Forecast-then-Verify)

Total momentum: $\vec{P}=\sum_i \vec p_i$. Differentiate: $\frac{d\vec P}{dt}=\sum_i \frac{d\vec p_i}{dt}=\sum_i \vec F_i^{\text{net}}$ Each $\vec F_i^{\text{net}}=\underbrace{\sum_{j\ne i}\vec F_{ij}}_{\text{internal}}+\underbrace{\vec F_i^{\text{ext}}}_{\text{external}}$.

Every internal force $\vec F_{ij}$ has a partner $\vec F_{ji}=-\vec F_{ij}$, so the double sum of internal forces is zero. Left with: $\frac{d\vec P}{dt}=\sum_i \vec F_i^{\text{ext}} = \vec F^{\text{ext}}_{\text{net}}$

Worked Examples

Common Mistakes

Active Recall

Flashcards

Connections

• Newton's Third Law — the action–reaction core of this proof
• Newton's Second Law (momentum form) — $\vec F=d\vec p/dt$
• Elastic vs Inelastic Collisions — when KE is/ isn't conserved
• Centre of Mass Motion — $\vec P = M\vec v_{cm}$, so $\vec v_{cm}$ constant if isolated
• Impulse–Momentum Theorem — integral version, $\vec J=\Delta\vec p$
• Rocket Equation — variable-mass application of this principle

Concept Map

Hinglish (regional understanding)