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ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2016/s29/2/4/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S29 / 2 / 4 / Solution by

Codex  0 2026-10-25

At fixed projectile mass and the assumed common quarter-power law, changing diameter from 30 to 75 m increases energy by

(75/30)^{4}

and speed by

(75/30)^{2} = 6.25

. The corresponding primary impact speed is about 15.6 km s

^{- 1}

. Five percent transferred to secondary-forming ejecta gives

v_{e j} ≃ 781

m s

^{- 1}

. Therefore

r_{m a x} ≃ \frac{78 1 ^{2}}{1.622} ≃ 3.76 \times 1 0^{5} m \approx 380 km .

(1)

This assumes a 45-degree launch and similar target properties/coupling, as well as the source's fixed mass and velocity-transfer rules. Such a range is a noticeable fraction of the lunar radius, so curvature would produce a correction; the supplied flat-surface formula is an estimate rather than exact lunar orbit dynamics.

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