Question

A flat plate is moving normal to its plane through a gas under the action of a constant force $F$. The gas is kept at a very low pressure. The speed of the plate $v$ is much less than the average speed $u$ of the gas molecules. Which of the following options is/are true?

• A. The pressure difference between the leading and trailing faces of the plate is proportional to $uv$
• B. The resistive force experienced by the plate is proportional to $v$
• C. The plate will continue to move with constant non-zero acceleration, at all times
• D. At a later time the external force $F$ balances the resistive force

Answer 1

Answer: (A), (B), (D)

Consider area $A$ of the plate moving towards right (see figure). In next dt time, only those particles of the gas can collide with the plate (on its right face) as are within $(v+u)dt$ distance from it. Number of these particles $=n(u+y)$ dt $A$, where $n$ is number of gas particles per unit volume. Only half of these particles will collide with the plate, the other half will be moving away from it.
$\therefore$ Number of particles, colliding with the plate from right, in dt time $=\frac{1}{2}n(u+v)$ dt $A$ Upon collision each particle (of mass $m$ ) imparts momentum of $2m(u+v)$ to the plate towards left, as they are moving with a velocity $u+v$ relative to the plate.
The force applied on the right face of the plate by the collisions $=\frac{\text{ momentum imparted }}{\text{ time }}$
$=\frac{\left[\frac{1}{2}n(u+v)dtA\right][2m(u+v)]}{dt}=mnA(u+v{)}^2$
$\therefore$ Pressure on the right face of plate $=\frac{\text{ Force }}{\text{ Area }}=\frac{mnA(u+v{)}^2}{A}=mn(u+v{)}^2$
Similarly, pressure on its left plate $=mn(u-v{)}^2$ [The particles on the left, move in with a velocity u-v relative to the plate].
And, the difference of pressure $=mn\left[(u+v{)}^2-(u-v{)}^2\right]=4mnuv\propto$ uv With passage of time, $v$ will increase (or decrease) such that the resistive force (due to this pressure difference) balances $F$ : Also, the resistive force is proportional to v