 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-4/38a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

In the shock rest frame the upstream and downstream speed magnitudes are

U

and

U - V

. Denote the downstream density by

ρ_{2}

, and write

v_{i} = 1/ ρ_{i}

j = ρ_{1} U = ρ_{2} (U - V), p_{1} + j^{2} v_{1} = p_{2} + j^{2} v_{2},

e_{1} + p_{1} v_{1} + \frac{1}{2} j^{2} v_{1}^{2} = e_{2} + p_{2} v_{2} + \frac{1}{2} j^{2} v_{2}^{2}, e_{i} = \frac{p _{i} v _{i}}{γ - 1} .

These follow by balancing the respective fluxes across a stationary infinitesimal control volume around the shock wave. The mass relation also gives

V = j (v_{1} - v_{2})

, and momentum gives

j^{2} (v_{1} - v_{2}) = p_{2} - p_{1}

e_{2} - e_{1} = p_{1} v_{1} - p_{2} v_{2} + \frac{1}{2} (p_{2} - p_{1}) (v_{1} + v_{2}) = \frac{1}{2} (p_{1} + p_{2}) (v_{1} - v_{2}) .

Substitute the given perfect-gas internal energy and collect the coefficients of

v_{1}, v_{2}

[(γ + 1) p_{2} + (γ - 1) p_{1}] v_{2} = [(γ - 1) p_{2} + (γ + 1) p_{1}] v_{1} .

Hence

v_{1} - v_{2} = \frac{2 ( p _{2} - p _{1} )}{ρ _{1} [( γ + 1 ) p _{2} + ( γ - 1 ) p _{1} ]} .

Finally

V^{2} = j^{2} (v_{1} - v_{2})^{2} = (p_{2} - p_{1}) (v_{1} - v_{2})

gives

V^{2} = \frac{( p _{2} - p _{1} ) ^{2}}{ρ _{1} [( γ + 1 ) p _{2} /2 + ( γ - 1 ) p _{1} /2 ]} .

For the advancing piston choose the compressive branch

p_{2} > p_{1}

U > V > 0

; the algebraic squared relation alone does not specify that branch.

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