Leaking Atwood machine ($v_{\mathrm{empty}}=\frac{g{M}_0}{\alpha }[-ϵ-2\log (1-ϵ/2)]$)

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 $ϵ$, emptying occurs at $M=(1-ϵ)M_0$. This gives the displayed speed, which tends to zero for $ϵ\to 0$ and to $(g{M}_0/\alpha )(2\log 2-1)$ for $ϵ\to 1^{-}$. 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.