# Proof of Conservation of Momentum

By Newton's Third Law of Motion,

$\mathbf{F}_{12}=-\mathbf{F}_{21}$

By Newton's Second Law of Motion,

\begin{aligned} \mathbf{F}_{12}&=-\mathbf{F}_{21}\\ \dot{\mathbf{p}}_{1}&=-\dot{\mathbf{p}}_{2}\\ \int_{t_1}^{t_2}\dot{\mathbf{p}}_{1}\,dt&=-\int_{t_1}^{t_2}\dot{\mathbf{p}}_{2}\,dt\\ \Delta\mathbf{p}_{1}&=-\Delta\mathbf{p}_{2}\\ \Delta(m\mathbf{v}_{1})&=-\Delta(m_2\mathbf{v}_{2})\\ m_1\Delta(\mathbf{v}_{1})&=-m_2\Delta(\mathbf{v}_{2})\\ m_1(\mathbf{v}_{1}-\mathbf{u}_{1})&=-m_2(\mathbf{v}_{2}-\mathbf{u}_{2})\\ m_1\mathbf{v}_{1}-m_1\mathbf{u}_{1}&=m_2\mathbf{u}_{21}-m_2\mathbf{v}_{2}\\ m_1\mathbf{v}_{1}+m_2\mathbf{v}_{2}&=m_2\mathbf{u}_{2}+m_1\mathbf{u}_{1}\\ \sum\mathbf{p}_{f}&=\sum\mathbf{p}_{i}\quad\blacksquare \end{aligned}

Note by Gandoff Tan
1 month, 3 weeks ago

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What happens when there are $3$ objects?

Staff - 1 month, 3 weeks ago

Alright, let's give this a try:

Let's say we have a bunch of masses, in a system with only internal forces. where mass $m_i$ exerts a force on $m_j$ denoted by $F_{ij}$. Thus, the net force is a sum of all the internal forces:

$F_{net} \ = \ \displaystyle\sum_{i, \ j} \ F_{ij}$

By Newton's Third Law, we will have $F_{ij} \ = \ -F_{ji}$ (if $m_i$ exterts $F_{ij}$ on $m_i$, then it follows that $m_j$ exerts an equal an opposite force on $m_i$. We also know that:

$F_{net} \ = \ \frac{d p_{total}}{dt}$

Thus, we can write our sum as a bunch of $F_{ij} \ + \ F_{ji} \ = \ F_{ij} \ - \ F_{ij} \ = \ 0$ terms, which just equates to $0$, giving us:

$F_{net} \ = \ \frac{d p_{total}}{dt} \ = \ 0$

This implies that total momentum does not change with time, thus momentum is conserved when there are only internal forces acting on the system.

I probably missed a lot of nuances, how did I do @Josh Silverman? :)

- 1 month, 3 weeks ago

It looks like a nose.

- 1 month, 1 week ago

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- 4 days, 10 hours ago