 Solution

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

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

Let

C

be the centre of the Earth and let

M

be its mass. Since

C W = R

and

CS = R + l

CSB

gives the ball's initial distance from

C

r_{B} = CB = (R + l)^{2} + l^{2} - 2 l (R + l) cos θ_{0} .

g = GM / R^{2}

- GM m / r

\frac{1}{2} m v_{2}^{2} = GM m (\frac{1}{R} - \frac{1}{r _{B}}) .

Therefore

v_{2} = [2 g R^{2} (\frac{1}{R} - \frac{1}{( R + l ) ^{2} + l ^{2} - 2 l ( R + l ) cos θ _{0}})]^{1/2} .

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