Law of Conservation of Linear Momentum

For a vector $\vec{s}$ , if $\vec{F} \cdot \vec{s} = 0$ , then:

\vec{F} \cdot \vec{s} = \frac{d \vec{p}}{d t} \cdot \vec{s} = \frac{d}{d t} (\vec{p} \cdot \vec{s})

$\vec{F} \cdot \vec{s} = 0$ implies $\vec{p} \cdot \vec{s}$ is a constant. Thus, momentum in the direction of $\vec{s}$ is conserved.

Torque, Angular Momentum, and Its Conservation

Angular momentum about the origin:

\vec{L} \equiv \vec{r} \times \vec{p}

Torque:

\vec{N} \equiv \frac{d \vec{L}}{d t} = \frac{d \vec{r}}{d t} \times \vec{p} + \vec{r} \times \frac{d \vec{p}}{d t}

\vec{N} = \vec{v} \times m \vec{v} + \vec{r} \times \vec{F} = 0 + \vec{r} \times \vec{F} = \vec{r} \times \vec{F}

If $\vec{N} \cdot \vec{s} = 0$ , then:

\frac{d}{d t} (\vec{L} \cdot \vec{s}) = 0

This implies conservation of angular momentum along the direction of zero torque.