 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4/3a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

During a short interval

d t

, let the rocket's mass change by

d m < 0

and its speed by

d v

. The expelled mass

- d m

v - u

to first order. Conservation of linear momentum, including the impulse

F d t

, gives

(m + d m) (v + d v) + (- d m) (v - u) - m v = F d t .

Discarding the second-order product

d m d v

leaves

m d v + u d m = F d t,

m \frac{d v}{d t} + u \frac{d m}{d t} = F .

In deep space

F = 0

, so

d v = - u d m / m

. Burning from initial mass

m_{0}

m_{0} /2

increases the speed by

Δ v = - u \int_{m_{0}}^{m_{0} /2} \frac{d m}{m} = u lo g 2 .

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