Past exam of the physics 1 paper of the University of Oxford / 2022 / 9 / a / Solution

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 givesAssume 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 thereforeSet $u=-\dot{x}\ge 0$, the speed towards the wall. Since $\ddot{x}=udu/dx$, the initial condition $u(L)=0$ givesConsequently the remaining sand has kinetic energyIn terms of $z=x/L$, $K/(TL)=-z\ln z$. Its derivative is $-\ln z-1$ and its second derivative is $-1/z<0$, givingThe magnitude of the remaining sand's momentum isIt 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 maximumBoth kinetic energy and momentum tend to zero at $x\to 0^{+}$, 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.