# Proof 32: Momentum conservation law

The moment conservation law and Newton’s 3rd law as well as the idea of impulse seem equivalent (they all describe impacts/collisions). And in fact the laws can be derived from one another and expressed in terms of impulse.

**The momentum conservation law:**

First, rewrite Newton’s 2nd law in terms of momentum $\vec p=m\vec v$:

$$\sum \vec F=m\vec a=m\frac{\mathrm d\vec v}{\mathrm dt}=\frac{\mathrm d\overbrace{(m\vec v)}^{\vec p}}{\mathrm dt}=\frac{\mathrm d\vec p}{\mathrm dt}$$

Then use Newton’s laws to derive the momentum conservation law:

$$\begin{align}

\vec F_\text{A on B}&=-\vec F_\text{B on A}\qquad \leftarrow \text{Newton’s 3rd law}\\

\frac{\mathrm d\vec p_\text{A on B}}{\mathrm dt}&=-\frac{\mathrm d\vec p_\text{B on A}}{\mathrm dt}\qquad \leftarrow \text{Newton’s 2rd law}\\

\frac{\mathrm d\vec p_\text{A on B}}{\mathrm dt}+\frac{\mathrm d\vec p_\text{B on A}}{\mathrm dt}&=0 \quad\Leftrightarrow\quad\\

\end{align}\\

\frac{\mathrm d\vec p_\text{A on B}+\mathrm d\vec p_\text{B on A}}{\mathrm dt}=\frac{\mathrm d\left(\vec p_\text{A on B}+\vec p_\text{B on A}\right)}{\mathrm dt}=\frac{\mathrm d\sum \vec p}{\mathrm dt}=0

$$

$\frac{\mathrm d\sum \vec p}{\mathrm dt}$ is the change in the total momentum. That change is here shown to be zero, $\frac{\mathrm d\sum \vec p}{\mathrm dt}=0$, meaning **the total momentum will not change**. In other words, the total momentum we have before any moment in time will always be equal to the total momentum after that moment:

$$\sum \vec p_\text{before}=\sum \vec p_\text{after}\qquad\blacksquare$$

**The momentum conservation law in impulse terms:**

$$\begin{align}

\sum \vec p_\text{before}&=\sum \vec p_\text{after}\quad\Leftrightarrow\quad\\

\vec p_\text{A,before}+\vec p_\text{B,before}&=\vec p_\text{A,after}+\vec p_\text{B,after}\quad\Leftrightarrow\quad\\ \underbrace{\vec p_\text{A,after}-\vec p_\text{A,before}}_{\Delta \vec p_\text{A}}+\underbrace{\vec p_\text{B,after}-\vec p_\text{B,before}}_{\Delta \vec p_\text{B}}&=0\quad\Leftrightarrow\quad\\ \Delta \vec p_\text{A}+\Delta \vec p_\text{B}&= \vec J_\text{A}+\vec J_\text{B}= \sum \vec J=0\qquad \blacksquare

\end{align} $$

So, the sum of all impulse is always zero.

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