 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/16a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

At the moving boundary,

Q = 4 π a^{2} \overset{a}{˙} .

Substitution into the result of part (b) gives the Rayleigh collapse equation

a \overset{a}{¨} + \frac{3}{2} \overset{a}{˙}^{2} = - \frac{p _{0}}{ρ} .

Treat

v = \overset{a}{˙}

a

, so that

\overset{a}{¨} = v d v / d a

. With

w = v^{2}

, the equation becomes

a \frac{d w}{d a} + 3 w = - \frac{2 p _{0}}{ρ} .

a^{3}

after division by

a

gives

\frac{d}{d a} (a^{3} w) = - \frac{2 p _{0}}{ρ} a^{2} .

Using

w (a_{0}) = 0

w (a) = \frac{2 p _{0}}{3 ρ} (\frac{a _{0}^{3}}{a ^{3}} - 1) .

The collapsing branch has negative radial velocity, so

\overset{a}{˙} = - \frac{2 p _{0}}{3 ρ} (\frac{a _{0}^{3}}{a ^{3}} - 1) .

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