Leaking bucket under constant tension ($K=Tx\ln (L/x)$)

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.