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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-4/11c/solution

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

Codex  0 Created 2026-09-23 Updated 2026-10-03

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}

. Summing

m_{i} \overset{r}{¨}_{i} = F_{i} + \sum_{j \neq = i} F_{ij}

cancels internal pairs and gives

d P / d t = F := \sum_{i} F_{i}

. Taking moments about fixed

a

cancels the internal torques pairwise and gives

\frac{d L}{d t} = G := \sum_{i} (r_{i} - a) \times F_{i} .

The angular momentum about the centre of mass result follows similarly: differentiating

\sum_{i} (r_{i} - R) \times m_{i} (\overset{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 \overset{r}{˙}_{i}

, then about a fixed point

G = - k \sum_{i} (r_{i} - a) \times \overset{r}{˙}_{i} = - \frac{k}{m} L .

Therefore

L (t) = L (0) e^{- k t / m} .

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