Consider a set of particles with position vectors $\mathbf{r}_{i}(t)$ and masses $m_{i}$, where $i=1,2, \ldots, N$. Particle $i$ experiences an external force $\mathbf{F}_{i}$ and an internal force $\mathbf{F}_{i j}$ from particle $j$, for each $j \neq i$. Stating clearly any assumptions you need, show that

$$\frac{d \mathbf{P}}{d t}=\mathbf{F} \quad \text { and } \quad \frac{d \mathbf{L}}{d t}=\mathbf{G}$$

where $\mathbf{P}$ is the total momentum, $\mathbf{F}$ is the total external force, $\mathbf{L}$ is the total angular momentum about a fixed point $\mathbf{a}$, and $\mathbf{G}$ is the total external torque about $\mathbf{a}$.

Does the result $\frac{d \mathbf{L}}{d t}=\mathbf{G}$ still hold if the fixed point $\mathbf{a}$ is replaced by the centre of mass of the system? Justify your answer.

Suppose now that the external force on particle $i$ is $-k \frac{d \mathbf{r}_{i}}{d t}$ and that all the particles have the same mass $m$. Show that

$$\mathbf{L}(t)=\mathbf{L}(0) e^{-k t / m}$$