= Solution
The escaping sand initially has the bucket's <velocity>, so it produces no relative-velocity thrust. This distinction is essential in a <leaking Atwood machine>. With upward <velocity> $v$ for the leaky side, the two bucket equations are
$$
M\dot v=T-Mg,\qquad M_0\dot v=M_0g-T.
$$
Their sum gives $\dot v=g(M_0-M)/(M+M_0)$. Since $\dot M=-\alpha$, it follows that
$$
\frac{dv}{dM}=\frac g\alpha\frac{M-M_0}{M+M_0}.
$$
Integrate from $M_0$, where the <velocity> is zero:
$$
v(M)=\frac g\alpha\left[M-M_0-2M_0\log\left(\frac{M+M_0}{2M_0}\right)\right].
$$
At emptying, $M=(1-\epsilon)M_0$, giving
$$
\boxed{v_{\rm empty}=\frac{gM_0}{\alpha}\left[-\epsilon-2\log(1-\epsilon/2)\right].}
$$
This is upward and positive for $0<\epsilon<1$. For no initial sand, $\epsilon=0$, the emptying time is zero and the <speed> is zero. As the dry bucket <mass> tends to zero,
$$
\boxed{\lim_{\epsilon\to1^-}v_{\rm empty}=\frac{gM_0}{\alpha}(2\log2-1).}
$$
This finite limit refers to the idealized model; the endpoint itself would describe a massless empty bucket. For small $\epsilon$, $v_{\rm empty}\sim gM_0\epsilon^2/(4\alpha)$, consistent with initially zero <acceleration> that grows as <mass> is lost.
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