For an <oblique shock>, boost parallel to the front by the upstream tangential speed. The transformed upstream flow is normal, so part b applies. Inviscid momentum balance has no tangential stress and therefore makes the tangential velocity continuous. Transforming back gives
$$
\boxed{u_{y1}=u_{y2}},
$$
while all normal velocity, density, and pressure relations from part b remain unchanged with $M_x$ the normal Mach number.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 314 / 2 / c / Solution

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An oblique shock is inclined relative to the upstream velocity. In an inviscid perfect gas, the normal velocity satisfies the normal-shock jump conditions while the tangential velocity is continuous.


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