Past exam of the physics 1 paper of the University of Oxford / 2019 / 7

During $dt$, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the expelled positive mass $-dM$ has velocity $v-u$ to first order. Subtract the initial momentum $Mv$ from their final momentum to obtain $Mdv+udM$. Equating this to the external impulse gives the rocket equationWith $F=-Mg-kMv$ and $\dot{M}/M=-b$, this reduces to $\dot{v}+kv=ub-g$. Therefore, for initial velocity $v_0$ and $k>0$,For a launch from rest,The inequality makes the initial upward acceleration positive. Equality gives no upward launch from rest. In the zero-drag limit, $v=(ub-g)t$. These expressions apply while the prescribed burn law and constant-$g$ approximation remain valid.

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 areTheir sum gives $\dot{v}=g(M_0-M)/(M+M_0)$. Since $\dot{M}=-\alpha$, it follows thatIntegrate from $M_0$, where the velocity is zero:At emptying, $M=(1-ϵ)M_0$, givingThis is upward and positive for $0<ϵ<1$. For no initial sand, $ϵ=0$, the emptying time is zero and the speed is zero. As the dry bucket mass tends to zero,This finite limit refers to the idealized model; the endpoint itself would describe a massless empty bucket. For small $ϵ$, $v_{\mathrm{empty}}\sim g{M}_0{ϵ}^2/(4\alpha )$, consistent with initially zero acceleration that grows as mass is lost.