Exhaust collection by a second rocket ($v_{Bf}=-m_A{v}_{Af}/(m_B+M)$)

Two initially stationary rockets exchange a fuel load $M$: A ejects it with constant rearward relative speed $u_0$ and B captures all of it. A has final velocity $v_{Af}=u_0\ln [(m_A+M)/m_A]$. Since final momentum is zero, $(m_B+M)v_{Bf}=-m_A{v}_{Af}$. No elastic-capture assumption is needed. With negligible flight time, transferred mass $q$ gives the same balance with instantaneous masses $m_A+M-q$ and $m_B+q$; their opposite speeds are equal when those masses agree. Final complete collection is a condition on the model, not a guarantee that every ballistic exhaust trajectory intercepts B.

For exhaust moving freely from a rocket A toward a collector B behind it along a common axis, the exhaust must overtake B in the rearward direction. At negligible separation and flight time this gives the displayed condition. In the ideal simultaneous-collection solution, write $a=m_A+M-q$, $b=m_B+q$ and $v_A=u_0\ln [(m_A+M)/a]$, $v_B=-a{v}_A/b$. The criterion becomes $[(a+b)/b]\ln [(m_A+M)/a]<1$. A formal endpoint obtained by assuming all fuel is collected need not satisfy this interception test throughout a large burn.