= Solution
First apply <momentum conservation> to the rocket and the infinitesimal amount of expelled gas. During $dt$, the exhaust <mass> is $\alpha\,dt$ and its inertial <velocity>, to first order, is $v-u$. With no dust, the momentum balance is
$$
mv=(m-\alpha\,dt)(v+dv)+\alpha\,dt(v-u)+o(dt).
$$
It gives the <rocket equation> \b[$\boxed{m\dot v=\alpha u}$], while $\dot m=-\alpha$.
For a <dust-collecting rocket>, the incoming dust <mass> is $\beta v\,dt$ with zero initial <momentum>. The same balance has rocket <mass> $m+(\beta v-\alpha)dt$ after the interval, so
$$
mv=\bigl(m+(\beta v-\alpha)dt\bigr)(v+dv)+\alpha\,dt(v-u)+o(dt).
$$
Consequently
$$
\boxed{\dot m=\beta v-\alpha,\qquad m\dot v=\alpha u-\beta v^2.}
$$
The second term in the <acceleration> equation accounts for the <momentum> required to bring collected dust up to rocket <speed>.
Set $\lambda=\sqrt{\alpha/(\beta u)}$ and $V=\lambda u$, so $\alpha u=\beta V^2$. As long as the engines operate and $0\leq v<V$, the <velocity> increases and time can be eliminated:
$$
\frac{d\log m}{dv}=\frac{v-\lambda^2u}{V^2-v^2}
=-\frac{\lambda-1}{2(V-v)}-\frac{\lambda+1}{2(V+v)}.
$$
Integrating from $v=0$, $m=m_0$, gives
$$
\log\frac m{m_0}=\frac{\lambda-1}{2}\log\frac{V-v}{V}-\frac{\lambda+1}{2}\log\frac{V+v}{V}.
$$
Thus the required <mass>–<velocity> relation is
$$
\boxed{m=\lambda m_0u\sqrt{\frac{(\lambda u-v)^{\lambda-1}}{(\lambda u+v)^{\lambda+1}}}.}
$$
If $\lambda>1$, this formula would give $m\to0$ as $v\to V^-$. In this interval, $\dot m=\beta(v-\lambda^2u)<0$. A real rocket retains positive body <mass> even after its fuel is gone, and collected dust only increases that residual <mass>. It therefore cannot follow this branch all the way to zero <mass>: \b[its fuel must run out while $v<\lambda u$]. Indeed the hypothetical time integral $dt/dv=m/[\beta(V^2-v^2)]$ is integrable at $V$ when $\lambda>1$, so even the formal zero-mass endpoint would occur at finite time.
If $0<\lambda<1$, the same relation instead gives $m\to\infty$ as $v\to V^-$. The <mass> first decreases, reaches its minimum at $v=\lambda^2u$, and then increases because dust arrives faster than fuel is expelled. The time integral now diverges at $V$: under indefinitely maintained thrust, \b[$v$ approaches $\lambda u$ asymptotically while stored dust grows without bound]. At that limiting <speed>, thrust $\alpha u$ balances the dust-loading term $\beta v^2$, and $\dot m\to\beta u\lambda(1-\lambda)>0$. This is the physical reason the total rocket <mass> no longer signals fuel exhaustion. A rocket with a finite initial fuel supply still runs out at finite time $t=M_{\rm fuel}/\alpha$ and before reaching $V$; increasing total <mass> does not create new fuel.
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