The shock speed is
$$
V_s=\dot R=\frac{C}{3}t^{-2/3}=\frac{R}{3t}.
$$
Apply the <Strong-shock Rankine-Hugoniot conditions> in the shock frame and transform the downstream velocity back to the ambient-medium frame. With the <shock compression ratio>
$$
\chi=\frac{\gamma+1}{\gamma-1},
$$
the post-shock values are
$$
\boxed{
\rho_{\rm ps}=\chi\rho_0,
\qquad
u_{\rm ps}=\frac{2}{\gamma+1}V_s,
\qquad
p_{\rm ps}=\frac{2}{\gamma+1}\rho_0V_s^2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 4 / a / Solution

{title2=$\chi$}

The shock compression ratio is the downstream-to-upstream density ratio. For a strong perfect-gas shock,
$$
\chi=\frac{\rho_1}{\rho_0}=\frac{\gamma+1}{\gamma-1}.
$$
In the upstream rest frame, the downstream velocity is $2V_s/(\gamma+1)$ and its pressure is $2\rho_0V_s^2/(\gamma+1)$.


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