Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 4 / 11B / b / Solution

Now $\dot{m}=\beta$ and $\dot{M}=\alpha -\beta$, soFor $\alpha >0$, set $w=u\beta -\alpha v$. Then $\dot{w}=-\alpha w/M$ and $w(0)=u\beta$. For $\alpha \ne \beta$, integration givesConsequentlyThis solution of the rocket equation with stationary mass accretion covers increasing or decreasing rocket mass, as long as $M>0$. For equal-rate rocket burning and accretion, in the case $\alpha =\beta >0$, the mass is constant and the linear equation instead givesIt is also the continuous equal-rate limit of the preceding expression when written in terms of $t$.

For the no-accretion limit, expand the power as an exponential:Substitution givesrecovering (a). At a fixed time with $M_0-\beta t>0$, the limiting mass is $M=M_0-\beta t$, so the same calculation recovers the time-dependent no-dust solution.