= Solution
In centre-of-mass and relative coordinates, the kinetic energy separates into centre-of-mass motion plus $\mu v^2/2$, while the mutual potential is $-GM_1M_2/r=-GM\mu/r$. In the centre-of-mass frame the orbital energy is therefore
$$
\boxed{E=\mu\left(\frac12v^2-\frac{GM}{r}\right)}.
$$
The force is central, so the specific relative angular momentum $h=r^2\dot\theta$ is conserved. With $u=1/r$, the radial equation becomes the <Binet equation>
$$
u''+u=\frac{GM}{h^2},
$$
whose solution is
$$
\boxed{r=\frac{l}{1+e\cos\theta},
\qquad l=\frac{h^2}{GM}}.
$$
Here $\theta$ is the <true anomaly> measured from periapsis and $e$ is the <orbital eccentricity>. Substitution into the energy, or evaluation at an apsis, gives
$$
\frac E\mu=-\frac{GM}{2a},
\qquad
l=a(1-e^2),
$$
and hence
$$
\boxed{E=-\frac{GM\mu}{2a}}.
$$
Immediately before the supernova the circular relative speed obeys $v^2=GM/a$. The impulsive <supernova kick in a binary star> changes it to
$$
v'^2=v^2(1+2\alpha\cos\psi+\alpha^2).
$$
Applying the energy formula just after the explosion at the unchanged separation $r=a$ gives
$$
-\frac{GM'}{2a'}
=\frac12v'^2-\frac{GM'}a.
$$
Using the original circular-speed relation,
$$
\boxed{\frac{M'}{a'}
=\frac{2M'}a-\frac Ma(1+2\alpha\cos\psi+\alpha^2)}.
$$
The post-explosion system is bound exactly when $1+2\alpha\cos\psi+\alpha^2<2M'/M$. The left side ranges from $(1-\alpha)^2$ to $(1+\alpha)^2$, so every kick direction remains bound if
$$
\boxed{M'>\frac12(1+\alpha)^2M},
$$
whereas every direction unbinds the stars if
$$
\boxed{M'<\frac12(1-\alpha)^2M}.
$$
For an unbound orbit, conservation of relative specific energy at infinity gives $V^2/2=v'^2/2-GM'/a$. Therefore
$$
\boxed{V=\left[(1+2\alpha\cos\psi+\alpha^2)
-\frac{2M'}M\right]^{1/2}v}.
$$
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