 Solution (source code)

by Codex (@codex,  0)

= Solution

The snow remains aboard, so the moving <mass> grows as $M+\sigma t$. Incoming snow contributes no horizontal <momentum>, and there is no horizontal external <force>. Therefore
$$
\frac d{dt}\bigl[(M+\sigma t)v\bigr]=0,
\qquad
\boxed{v_C(t)=\frac{Mv_0}{M+\sigma t}.}
$$
The combined sled–snow system follows a <variable-mass system> balance; setting $(M+\sigma t)\dot v=0$ would omit the incoming <mass> term.

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