= Solution
Work in the rest frame of the planar <shock>, with nonzero <mass flux> $j=\rho u$. <Conservation of mass> gives $[j]=0$; the tangential and normal components of <balance of linear momentum> give $[jv]=0$ and $[p+ju]=0$. Thus $[v]=0$. For an <ideal gas>, <specific internal energy> is $e=c_vT=p/[(\gamma-1)\rho]$, and <specific enthalpy> is $e+p/\rho=\gamma p/[(\gamma-1)\rho]$. Total energy flux is $j[(u^2+v^2)/2+e+p/\rho]$. Its continuity, with $v$ already continuous, gives
$$
\boxed{[\rho u]=[v]=[p+\rho u^2]=\left[\frac{u^2}2+\frac{\gamma p}{(\gamma-1)\rho}\right]=0.}
$$
Put $c^2=\gamma p/\rho$ and $M=u/c$. Since $p=ju/(\gamma M^2)$, the normal component of <balance of linear momentum> says $u[1+1/(\gamma M^2)]$ is constant. Hence
$$
\boxed{\frac{u_+}{u_-}=\frac{M_+^2(1+\gamma M_-^2)}{M_-^2(1+\gamma M_+^2)}.}
$$
The energy equation says $u^2[1/2+1/((\gamma-1)M^2)]$ is constant, giving
$$
\boxed{\left(\frac{u_+}{u_-}\right)^2=\frac{M_+^2[2+(\gamma-1)M_-^2]}{M_-^2[2+(\gamma-1)M_+^2]}.}
$$
Set $a=M_-^2$, $b=M_+^2$, square the first relation and equate to the second. After clearing positive denominators the equation factors as
$$
\boxed{(b-a)\{b[2\gamma a-(\gamma-1)]-[2+(\gamma-1)a]\}=0.}
$$
The first factor is the unchanged state. For a genuine <shock> the second factor vanishes, yielding
$$
\boxed{M_+^2=\frac{2+(\gamma-1)M_-^2}{2\gamma M_-^2-(\gamma-1)}.}
$$
The physical selection principle is <entropy> increase in the direction of flow. For $u_\pm>0$, the upstream side is $x<0$ and the downstream side is $x>0$. A compressive <entropy>-admissible <shock> has $\boxed{M_->1,\quad M_+<1}$; interpreting the printed “ahead” and “behind” spatial labels must respect these explicit side assignments. The expansion-<shock> branch would violate the second law.
Substitution gives the <velocity> ratio
$$
\boxed{\frac{u_+}{u_-}=\frac{2+(\gamma-1)M_-^2}{(\gamma+1)M_-^2}<1.}
$$
Since $v$ is unchanged while the normal component decreases, $|v|/u$ increases. Thus an oblique flow bends \b[towards the <shock> plane], equivalently away from its normal. If $v=0$ there is no angular bending; the flow stays normal and simply slows. The tangential continuity, not a change in the tangential speed, causes the deflection.
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