 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-4/17c/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

In the station frame, intersecting the signal trajectories with the endpoint world lines gives

u t_{A}^{'} = v t_{A}^{'} + \frac{L}{2 γ}, - u t_{B}^{'} = v t_{B}^{'} - \frac{L}{2 γ} .

Since

u > v

, both receptions occur at positive times,

t_{A}^{'} = \frac{L}{2 γ ( u - v )}, t_{B}^{'} = \frac{L}{2 γ ( u + v )} .

The inverse Lorentz transformation then gives

t_{A} = γ t_{A}^{'} (1 - uv / c^{2}) = \frac{L ( 1 - uv / c ^{2} )}{2 ( u - v )}, t_{B} = γ t_{B}^{'} (1 + uv / c^{2}) = \frac{L ( 1 + uv / c ^{2} )}{2 ( u + v )} .

Subtracting and simplifying yields

t_{B} - t_{A} = - \frac{Lv ( 1 - u ^{2} / c ^{2} )}{u ^{2} - v ^{2}}, t_{B}^{'} - t_{A}^{'} = - \frac{Lv}{γ ( u ^{2} - v ^{2} )} .

For

u < c

, both differences are negative: the rear receives first in both frames, but by different time intervals.

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