 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ia/paper-4/12c/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Let

r = ∣ r_{1} - r_{2} ∣

and

r = (r_{1} - r_{2}) / r

. A differentiable central potential energy gives the internal forces

F_{12} = - \nabla_{r_{1}} U = - U^{'} (r) r, F_{21} = - \nabla_{r_{2}} U = + U^{'} (r) r .

There are no external forces. Differentiating the total kinetic energy and using Newton's second law therefore gives

\frac{d T}{d t} = m_{1} v_{1} \cdot \dot{v}_{1} + m_{2} v_{2} \cdot \dot{v}_{2} = F_{12} \cdot v_{1} + F_{21} \cdot v_{2} = - U^{'} (r) r \cdot (v_{1} - v_{2}) .

Since

\overset{r}{˙} = r \cdot (v_{1} - v_{2})

, the last expression is

- U^{'} (r) \overset{r}{˙} = - d U / d t

. Thus

\frac{d T}{d t} + \frac{d U}{d t} = 0, T + U = constant .

This conserves total mechanical energy, including the kinetic energy of centre-of-mass motion, not merely the relative kinetic energy.

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