 Solution

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

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

A central potential is

V (r)

. Then

\dot{L} = x \times (- \nabla V) = 0

. Since

L \cdot x = 0

, the trajectory lies in the fixed plane perpendicular to

L

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