 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4/9a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

The rod imposes a mechanical constraint, and its force is perpendicular to the wrecking ball's instantaneous velocity, so it does no work. In the constant parallel Newtonian gravitational field, the ball falls through the vertical distance

l - l cos θ_{0}

B

and

W

therefore gives

\frac{1}{2} m v_{1}^{2} = m g l (1 - cos θ_{0}),

and hence

v_{1} = 2 g l (1 - cos θ_{0}) .

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