Discussion
Air approaching an obstacle at \( M = 1.5 \) crosses a normal shock. The Mach number with which it emerges is nearest to
Answer: B
Substituting into \( M_2^2 = (1 + 0.2 M_1^2)/(1.4 M_1^2 - 0.2) \), the numerator is \( 1 + 0.45 = 1.45 \) and the denominator \( 3.15 - 0.2 = 2.95 \), giving \( M_2^2 = 0.4915 \) and \( M_2 = 0.701 \). The weaker shock leaves the flow closer to sonic than the Mach 2 case did.
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