= Solution
Write $f=M'/M_0$ for the retained fraction and $q=v^-/v_0$ for an opposite kick. With more than half the mass lost, $0<f<1/2$. A same-direction kick cannot bind the pair, because $1+v^+/v_0\geq1>\sqrt{2f}$. An opposite kick can reduce the relative speed enough to leave negative <energy>:
$$
\boxed{1-\sqrt{2f}<q<1+\sqrt{2f}.}
$$
Both endpoints are marginal zero-energy cases. Kicks somewhat larger than $v_0$ can still satisfy this interval, but then the direction of the relative orbital motion is reversed.
At $v^-=v_0=v_1+v_2$, the two surviving stars have identical instantaneous velocities, so their relative <velocity> and orbital <angular momentum> vanish. Their <energy> is nevertheless negative,
$$
E'=-\frac{GM_{\mathrm{NS}}M_2}{A}.
$$
They fall radially toward each other from rest in their new <centre-of-mass frame>. In the point-mass idealization the collision orbit has <semimajor axis> $A/2$ and zero pericentre; real finite-size stars collide or interact before reaching zero separation. It is not a stable circular binary. Zero relative <angular momentum>, rather than the negative <energy> alone, determines this plunge outcome.
\Image[/past-exam-of-the-astrophysics-course-of-the-university-of-oxford/2018-binary-kick-binding.png]
{title=Binding after instantaneous mass loss for parallel and opposite natal kicks}
{description=The retained mass fraction is $f=M'/M_0$ and the kick magnitude is normalized to the original relative speed. Blue regions have negative <orbital energy>. The curves mark marginal escape. At an opposite normalized kick of one, the relative <angular momentum> vanishes and the survivors plunge together.}
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