= Solution
The source expression $E=\tfrac12\mu v^2-GMm/r$ is \b[total mechanical energy] in the centre-of-mass frame, not kinetic energy alone. Take $v_0$ as the relative speed at a large initial separation where the mutual potential is negligible. At closest approach, the radial relative velocity is zero, so <conservation of angular momentum> and <conservation of energy> give
$$
\mu b v_0=\mu r_cv_c,\qquad
\frac12\mu v_0^2=\frac12\mu v_c^2-\frac{GMm}{r_c}.
$$
Using $Mm/\mu=M+m$ eliminates $v_c$:
$$
b^2=r_c^2+\frac{2G(M+m)r_c}{v_0^2}.
$$
An impact occurs when $r_c\leq R$, where $R=r_M+r_m$. Since the right side increases with positive $r_c$, the collision threshold is
$$
\boxed{b_{\rm col}^2=R^2\left[1+\frac{2G(M+m)}{Rv_0^2}\right],\qquad
f=\frac{2G(M+m)}{Rv_0^2}.}
$$
The mutual <escape velocity> is the relative speed at contact required to reach infinite separation with zero residual speed. Setting total mechanical energy to zero gives
$$
\boxed{v_{\rm esc}=\sqrt{\frac{2G(M+m)}{r_M+r_m}},\qquad f=v_{\rm esc}^2/v_0^2.}
$$
This includes target recoil. If the given initial speed were instead at a finite non-negligible separation, its potential energy would have to be retained; $v_0$ would then not equal the asymptotic speed used in the focusing formula.
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