Solution
= Solution
Momentum conservation and the density ratio from part (b) give
$$
p_1-p_0
=\rho_0V^2-\frac32\rho_0\left(\frac{2V}{3}\right)^2
=\frac13\rho_0V^2.
$$
For $\gamma=3$, energy conservation becomes
$$
\frac{3p_0}{2\rho_0}+\frac12V^2
=\frac{p_1}{\rho_0}+\frac{2}{9}V^2.
$$
Eliminating $\rho_0V^2$ between these equations yields $p_1=4p_0$. Hence
$$
\boxed{\frac{p_1}{p_0}=4}.
$$