 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/8b/a/solution

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

L = \frac{m}{2} (\overset{y}{˙}_{1}^{2} + \overset{y}{˙}_{3}^{2}) + \frac{M}{2} \overset{y}{˙}_{2}^{2} - \frac{k}{2} [(y_{1} - y_{2})^{2} + (y_{2} - y_{3})^{2}] - g (m y_{1} + M y_{2} + m y_{3}) .

T = 2 m + M

Y = \frac{m y _{1} + M y _{2} + m y _{3}}{T} .

All internal forces cancel, so

\ddot{Y} = - g

and

Y (t) = A + \dot{Y} (0) t - \frac{1}{2} g t^{2} .

If the molecule is released from rest, the middle term vanishes.

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