 Solution

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

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

Put

ϵ = l / R

and

a = 1 - cos θ_{0}

. Then

\frac{r _{B}}{R} = (1 + 2 a ϵ + 2 a ϵ^{2})^{1/2} .

\frac{R}{r _{B}} = 1 - a ϵ + (\frac{3}{2} a^{2} - a) ϵ^{2} + O (ϵ^{3}),

v_{2}^{2} = 2 g R [a ϵ + (a - \frac{3}{2} a^{2}) ϵ^{2} + O (ϵ^{3})] .

Since

v_{1}^{2} = 2 g R a ϵ

\frac{v _{2}}{v _{1}} = 1 + \frac{1}{2} (1 - \frac{3}{2} a) ϵ + O (ϵ^{2}) = 1 + (- \frac{1}{4} + \frac{3}{4} cos θ_{0}) \frac{l}{R} + O (\frac{l ^{2}}{R ^{2}}) .

Thus the constants in the stated Big O notation expansion are

A = - \frac{1}{4}, B = \frac{3}{4} .

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