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

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

Codex  0 2026-10-25

Choose positive x along the rocket's motion. If its instantaneous total mass is

μ

and it ejects a positive amount

d η = - d μ

, the exhaust velocity in the original inertial frame is

v - u_{0}

to first order. Conservation of momentum over that increment gives

μv = (μ - d η) (v + d v) + d η (v - u_{0}), μ d v = u_{0} d η = - u_{0} d μ .

Integrating the rocket equation from

μ = m + M

to

μ = m

gives

v_{f} = u_{0} ln \frac{m + M}{m} .

The constant exhaust speed is relative to the rocket, not the initial frame. Kinetic energy of rocket plus exhaust is not conserved because stored fuel energy is released. The calculation is Newtonian and requires resulting speeds small compared with the speed of light.

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