= Solution
For a kick parallel or antiparallel to star 1's original <velocity>, the immediate post-explosion relative speed is $|v_0\pm v^\pm|$. Thus
$$
E'=\frac12\mu'v_0^2
\left[\left(1\pm\frac{v^\pm}{v_0}\right)^2-2\frac{M'}{M_0}\right].
$$
A positive-energy, hyperbolically escaping pair therefore satisfies precisely
$$
\boxed{\left|1\pm\frac{v^\pm}{v_1+v_2}\right|>
\sqrt{2\frac{M_{\mathrm{NS}}+M_2}{M_1+M_2}}.}
$$
The absolute value is essential when a sufficiently large opposite kick reverses the relative tangential <velocity>.
\b[Marginal case.] Strictly speaking, a pair is gravitationally bound only for $E'<0$. If unbound includes marginal escape, replace the printed strict inequality by $\geq$. For a concrete counterexample to an unrestricted “if and only if” with $>$, take no kick and $\Delta M=M_0/2$. Both sides equal one and the <energy> is zero: this is not a bound ellipse, although the printed strict inequality is false.
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