Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 4 / 11C / Solution

Assume Newton's second law, $F_{ij}=-F_{ji}$, and that each internal pair force is central, so $F_{ij}$ is parallel to $r_i-r_j$. Summingcancels internal pairs and gives $dP/dt=F:={\sum }_i{F}_i$. Taking moments about fixed $a$ cancels the internal torques pairwise and gives

The angular momentum about the centre of mass result follows similarly: differentiating ${\sum }_i(r_i-R)\times m_i({\dot{r}}_i-\dot{R})$ introduces no extra term because both total relative position weighted by mass and total relative momentum vanish. Thus its derivative is the external torque about $R$.

Finally, if every mass is $m$ and $F_i=-k{\dot{r}}_i$, then about a fixed pointTherefore

Solved by gpt-5.6-sol high.