= Solution
Put $m_A=M=m$ and $m_B=2m$. At reversal the transferred <mass> is $q_r=0.45m$, leaving A with $1.55m$ and, after capture of that exhaust, B with $2.45m$. The first burn gives
$$
v_{Ar}=u_0\ln\frac2{1.55}=u_0\ln\frac{40}{31},\qquad
v_{Br}=-\frac{1.55}{2.45}v_{Ar}=-\frac{31}{49}u_0\ln\frac{40}{31}.
$$
After reversal A emits its remaining exhaust away from B, so B receives no further fuel and eventually coasts at this <velocity>. Applying the <rocket equation> with reversed thrust to A gives
$$
v_{Af}=v_{Ar}-u_0\ln1.55=u_0\ln\frac{800}{961}\simeq-0.18336u_0,\qquad v_{Bf}\simeq-0.16126u_0.
$$
Both now move toward negative x, but A moves faster in that direction. The asymptotic closing <speed> is
$$
\boxed{v_{Bf}-v_{Af}\simeq0.0221u_0>0:\quad\text{A can catch B.}}
$$
This conclusion does not require a burn rate or an initial separation: after a finite burn, the positive eventual closing <speed> closes any finite remaining gap. Those additional inputs would be required to calculate the meeting time. Any pre-reversal exhaust still in transit affects the timing, not B's eventual <momentum> after collecting that fixed amount.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014-exhaust-collection-speeds.png]
{title=Signed <velocities> during exhaust collection and thrust reversal}
{description=Original Newtonian calculations versus the fraction of original fuel consumed. Left: the equal-speed event in the ideal negligible-flight-time collection model; the later dashed extension is conditional on collection and can violate ballistic interception. Right: reversing thrust at 45 percent produces a small positive eventual closing speed.}
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