= Leaking bucket under constant tension
{title2=$K=Tx\ln(L/x)$}
Let a massless bucket approach a wall under constant <tension> $T>0$, with remaining distance $x$ and remaining sand <mass> $m=Mx/L$. Assume <zero-relative-velocity mass shedding> horizontally. Then $m\ddot x=-T$ and, from rest at $x=L$, the <speed> satisfies $u^2=(2TL/M)\ln(L/x)$. Its <kinetic energy> is $K=Tx\ln(L/x)$ and its <momentum> magnitude is $p=x\sqrt{(2MT/L)\ln(L/x)}$. Differentiating gives $x_K=L/e$ and $x_p=L/\sqrt e$: the <kinetic energy> peaks later, closer to the wall, because its dependence on the increasing <speed> is stronger. Both tend to zero as the remaining <mass> vanishes, despite divergent ideal <speed>.
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