 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-4/9c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

R = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}} .

F_{21} = - F_{12}

, so

(m_{1} + m_{2}) \ddot{R} = m_{1} \ddot{x}_{1} + m_{2} \ddot{x}_{2} = F_{12} + F_{21} = 0.

Thus

\dot{R}

is constant. For

r = x_{1} - x_{2}

\ddot{r} = (\frac{1}{m _{1}} + \frac{1}{m _{2}}) F_{12} .

Therefore

μ \ddot{r} = F_{12}, μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}},

where

μ

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