Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 65 / 1 / c / Solution

Assume $\gamma >1$. The upstream ideal gas is at rest in the laboratory frame and has mass density ${\rho }_1=C{R}^{-\beta }$. In the shock frame its incident speed is $V$. The strong-shock limit of the Rankine-Hugoniot conditions for a perfect gas gives compression $\chi =(\gamma +1)/(\gamma -1)$ and downstream shock-frame speed $V/\chi$. Transforming back to the laboratory frame gives $u_2=V(1-1/\chi )=2V/(\gamma +1)$.

The normal momentum jump yieldswhere the cold upstream pressure has been neglected. Thus the boundary values in the normalization of part (b) areThe velocity jump uses the relative speed $V-u_2$, not the laboratory speed $u_2$. This distinction is essential to all three shock wave values.