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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014/8/b/solution

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Past exam of the physics 1 paper of the University of Oxford / 2014 / 8 / b / Solution by

Codex  0 2026-10-25

Let

q

be the transferred fuel mass. Before any engine reversal, rocket A has mass

a = m_{A} + M - q

and velocity

v_{A} (q) = u_{0} ln \frac{m _{A} + M}{m _{A} + M - q} .

(1)

For exhaust collection by a second rocket, under the stipulated eventual capture of all exhaust, final conservation of momentum for the complete initially stationary system gives

m_{A} v_{A f} + (m_{B} + M) v_{B f} = 0

. Thus

v_{A f} = u_{0} ln \frac{m _{A} + M}{m _{A}}, v_{B f} = - \frac{m _{A}}{m _{B} + M} v_{A f}, v_{rel} = u_{0} \frac{m _{A} + m _{B} + M}{m _{B} + M} ln \frac{m _{A} + M}{m _{A}} .

(2)

Capture is inelastic: no kinetic energy conservation for B is assumed. Finite exhaust flight time changes the intermediate sharing of momentum but not this endpoint balance if the stated complete capture actually occurs.

The ballistic exhaust interception condition is a physical restriction on that idealization. For purely ballistic exhaust to reach a receding collector to the rear, its laboratory velocity must be more negative than B's while interception occurs. In the negligible-flight-time approximation this requires

v_{A} - u_{0} < v_{B}

. Very large fuel ratios need not meet that condition throughout a complete burn; the boxed endpoint formula is conditional on the prompt's complete-collection stipulation, not an unconditional prediction for every ballistic two-rocket geometry.

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