 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-65/1/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Assume

γ > 1

. The upstream ideal gas is at rest in the laboratory frame and has mass density

ρ_{1} = C R^{- β}

. In the shock frame its incident speed is

V

. The strong-shock limit of the Rankine-Hugoniot conditions for a perfect gas gives compression

χ = (γ + 1) / (γ - 1)

V / χ

. Transforming back to the laboratory frame gives

u_{2} = V (1 - 1/ χ) = 2 V / (γ + 1)

The normal momentum jump yields

p_{2} + ρ_{2} (V - u_{2})^{2} = ρ_{1} V^{2}, p_{2} = \frac{2}{γ + 1} ρ_{1} V^{2},

where the cold upstream pressure has been neglected. Thus the boundary values in the normalization of part (b) are

U (1) = \frac{2}{γ + 1}, D (1) = \frac{γ + 1}{γ - 1}, P (1) = \frac{2}{γ + 1} .

V - u_{2}

u_{2}

. This distinction is essential to all three shock wave values.

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