= Solution
{c}
The diagram makes $x$ the remaining distance to the wall: $x$ decreases from $L$ to $0$. With the endpoint condition $m(0)=0$, integrating the supplied <mass> gradient gives
$$
m(x)=\frac{Mx}{L}.
$$
Assume escaping sand has the same horizontal <velocity> as the bucket at release. There is then no horizontal recoil: the <momentum> flux of the outgoing sand cancels the $\dot m\dot x$ contribution to the bucket's <momentum> <derivative>. The <variable-mass system> equation is therefore
$$
m\ddot x=-T.
$$
Set $u=-\dot x\ge0$, the <speed> towards the wall. Since $\ddot x=u\,du/dx$, the initial condition $u(L)=0$ gives
$$
u\frac{du}{dx}=-\frac{TL}{Mx},\qquad
\boxed{u^2(x)=\frac{2TL}{M}\ln\frac Lx.}
$$
Consequently the remaining sand has <kinetic energy>
$$
\boxed{K(x)=\frac12m(x)u^2(x)=Tx\ln\frac Lx.}
$$
In terms of $z=x/L$, $K/(TL)=-z\ln z$. Its <derivative> is $-\ln z-1$ and its second <derivative> is $-1/z<0$, giving
$$
\boxed{x_K=\frac Le,\qquad K_{\max}=\frac{TL}{e}.}
$$
The magnitude of the remaining sand's <momentum> is
$$
\boxed{p(x)=m(x)u(x)=x\sqrt{\frac{2MT}{L}\ln\frac Lx}.}
$$
It is easier to maximize its square, $p^2/(2MTL)=-z^2\ln z$. Its <derivative> is $-z(2\ln z+1)$, giving the unique interior maximum
$$
\boxed{x_p=\frac L{\sqrt e},\qquad p_{\max}=\sqrt{\frac{MTL}{e}}.}
$$
Both <kinetic energy> and <momentum> tend to zero at $x\to0^+$, even though the idealized <speed> diverges. There is no finite-<mass> bucket left at the endpoint, so that divergence signals the limit of the idealization.
The <leaking bucket under constant tension> has $x_K=L/e<L/\sqrt e=x_p$: the <kinetic energy> maximum is closer to the wall. As the bucket loses <mass>, its increasing <speed> continues to boost $K=pu/2$ after $p$ has begun to fall. Equivalently, $p\propto z\sqrt{-\ln z}$ and $K\propto z(-\ln z)$, so <kinetic energy> has the stronger dependence on the growing <speed>. The work $T(L-x)$ is shared between the remaining sand and the sand already lost; it is not all the bucket's instantaneous <kinetic energy>.
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