= Solution
For <constant-area capture of stationary dust>, let $x=0$ at cloud entry, $M(0)=M_p$, $v(0)=v_\infty$, and define $k=\rho A$ with fixed collecting area $A$. Every swept volume $A\,dx$ adds stationary <mass>, so
$$
\frac{dM}{dt}=kv,\qquad M=M_p+kx.
$$
There is no external longitudinal <force>, and the incoming dust has zero <momentum> in the cloud frame. The <variable-mass system> therefore satisfies $d(Mv)/dt=0$, rather than $M\dot v=0$. Using $Mv=M_pv_\infty$ gives
$$
M\dot M=kM_pv_\infty,\qquad M^2=M_p^2+2kM_pv_\infty t.
$$
Hence
$$
\boxed{M(t)=M_p\sqrt{1+\frac{2\rho Av_\infty t}{M_p}},\qquad v(t)=\frac{v_\infty}{\sqrt{1+2\rho Av_\infty t/M_p}},}
$$
$$
\boxed{x(t)=\frac{M_p}{\rho A}\left[\sqrt{1+\frac{2\rho Av_\infty t}{M_p}}-1\right].}
$$
The <kinetic energy> decreases during sticking: $dK/dt=-\tfrac12\dot M v^2$. The lost <energy> becomes internal <energy>. Radiation, gravity, sputtering and changes in area are excluded by the ideal complete-collection model. The zero-density limit recovers $M=M_p$, $v=v_\infty$ and $x=v_\infty t$.
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