= Leaking Atwood machine
{title2=$v_{\rm empty}=\frac{gM_0}{\alpha}[-\epsilon-2\log(1-\epsilon/2)]$}
Two initially equal buckets of total <mass> $M_0$ share an ideal <Atwood machine>. If one loses sand at constant rate $\alpha$ and the sand leaves with zero initial velocity relative to that bucket, the leakage supplies no rocket thrust. The remaining masses satisfy $(M+M_0)\dot v=g(M_0-M)$ for upward velocity of the leaking side. Using $\dot M=-\alpha$ and integrating from rest gives $v=(g/\alpha)[M-M_0-2M_0\log((M+M_0)/(2M_0))]$.
If the initial sand fraction is $\epsilon$, emptying occurs at $M=(1-\epsilon)M_0$. This gives the displayed speed, which tends to zero for $\epsilon\to0$ and to $(gM_0/\alpha)(2\log2-1)$ for $\epsilon\to1^-$. The latter is a finite limiting result, not a physical claim about an exactly massless empty bucket. The derivation is an application of a <variable-mass system>, rather than of $d(Mv)/dt=F$ for a closed body.
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