 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-2/4c/i/solution

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

L = u b

, while at closest approach the velocity is transverse and

L = wp

. Thus

u = wp / b

\frac{u ^{2}}{2} - \frac{k}{R} = \frac{w ^{2}}{2} - \frac{k}{p} .

Under

u^{2} ≫ 2 k / R

, neglect the initial potential term and substitute

u = wp / b

\frac{w ^{2} p ^{2}}{2 b ^{2}} = \frac{w ^{2}}{2} - \frac{k}{p} .

Consequently

\frac{w ^{2}}{k} = \frac{2 b ^{2}}{p ( b ^{2} - p ^{2} )} .

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