 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/39c/b/solution

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

Work in the shock frame, and let the positive upstream and downstream speeds be

w_{0}

and

w_{1}

ρ_{0} w_{0} = ρ_{1} w_{1}, p_{0} + ρ_{0} w_{0}^{2} = p_{1} + ρ_{1} w_{1}^{2} .

Put

r = \frac{ρ _{1}}{ρ _{0}}, \frac{p _{1}}{p _{0}} = 1 + β .

w_{1} = w_{0} / r

w_{0}^{2} = \frac{p _{0}}{ρ _{0}} \frac{β r}{r - 1} . (1)

Applying the energy integral from part (a) on the two sides gives

\frac{γ}{γ - 1} \frac{p _{0}}{ρ _{0}} + \frac{1}{2} w_{0}^{2} = \frac{γ}{γ - 1} \frac{p _{0} ( 1 + β )}{ρ _{0} r} + \frac{1}{2} \frac{w _{0}^{2}}{r ^{2}} . (2)

Substitution of (1) into (2), followed by cancellation of

p_{0} / ρ_{0}

, yields

\frac{β ( r + 1 )}{2} = \frac{γ}{γ - 1} (1 + β - r) .

\frac{ρ _{1}}{ρ _{0}} = r = \frac{2 γ + ( γ + 1 ) β}{2 γ + ( γ - 1 ) β} .

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