= Solution
For a fixed set of bodies in an <inertial frame>, total <linear momentum> is constant if the resultant external <force> vanishes. Summing <Newton's second law> cancels internal <force> pairs by <Newton's third law>. For an open subsystem, <momentum> carried across its boundary must also be included.
For an <equal-intake-fuel mass-loading rocket>, enclose the original craft and the stationary material collected during $dt$. Its initial <momentum> is $mu$. At the end, the craft has <mass> $m-\mu\,dt$ and the exhaust has <mass> $2\mu\,dt$. To first order,
$$
mu=(m-\mu\,dt)(u+du)+2\mu\,dt(u-u_0),
\qquad
\boxed{m\frac{du}{dt}=2\mu u_0-\mu u.}
$$
The first term is <rocket thrust>; the second is the <momentum> cost of collecting stationary matter. It vanishes in the conventional <rocket equation>, which therefore has no analogous fixed exhaust-speed ceiling in <Newtonian mechanics>. Here the <acceleration> vanishes at \b[$u_{\max}=2u_0$]; from rest it stays positive below that value while fuel remains. Relativistic limits are outside this model.
For constant $\mu$, <mass conservation> gives $m=m_i-\mu t$. Separating variables,
$$
\int_0^u\frac{dw}{2u_0-w}=\int_0^t\frac{\mu\,d\tau}{m_i-\mu\tau}
\quad\Longrightarrow\quad
\frac{2u_0-u}{2u_0}=\frac{m_i-\mu t}{m_i}.
$$
Therefore
$$
\boxed{u(t)=\frac{2\mu u_0}{m_i}t,\qquad \dot u=\frac{2\mu u_0}{m_i}.}
$$
The formal limiting <speed> would require $t=m_i/\mu$ and zero remaining craft <mass>. A positive <rocket dry mass> and finite carried fuel terminate the burn earlier. Maintaining nonzero collection rate even at rest is an idealization explicitly prescribed by the problem.
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