 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-4/10b/d/solution

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

L^{2} = J^{2} - (qβ)^{2} = J^{2} sin^{2} α

. Splitting kinetic energy into radial and angular parts therefore gives

E = \frac{1}{2} m \overset{r}{˙}^{2} + J^{2} sin^{2} α / (2 m r^{2}) + V (r)

, proving the stated effective potential.

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