Past exam of the physics 1 paper of the University of Oxford / 2015 / 3

Take $M$ as the constant mass of the sled and rider when snow is immediately removed. Assume the falling snow has zero horizontal velocity in the ice frame, and its arrival rate $\sigma$ is the specified constant mass flux onto the sled. In a snow-collecting sled, horizontal momentum flux, including the departing snow, determines the change in sled momentum.

Snow leaves perpendicular to the motion in the rider's frame, so its outgoing horizontal velocity in the ice frame is $v$. During $dt$, incoming snow brings zero horizontal momentum and outgoing snow carries $\sigma vdt$. ThusAlthough the sled does not retain snow, it must continually supply the snow's departing horizontal momentum.

Snow now leaves with zero horizontal velocity in the ice frame, the same as its incoming horizontal velocity. The net horizontal momentum flux is zero:Any forward momentum given to the snow on capture is returned to the sled while sweeping it off. “Perpendicular” specifies a different outgoing velocity here than in case A; changing the reference frame without transforming that velocity would miss this difference.

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. ThereforeThe combined sled–snow system follows a variable-mass system balance; setting $(M+\sigma t)\dot{v}=0$ would omit the incoming mass term.