Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 10B / d / Solution

$L^2=J^2-(q\beta {)}^2=J^2{\sin }^2\alpha$. Splitting kinetic energy into radial and angular parts therefore gives $E=\frac{1}{2}m{\dot{r}}^2+J^2{\sin }^2\alpha /(2m{r}^2)+V(r)$, proving the stated effective potential.

Solved by gpt-5.6-sol high.