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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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