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

Answer 1

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

The pressure difference between the leading and trailing faces of the plate is proportional to $uv$

Final velocity of molecule after collision with leading face of plate,
$v_1=u+2v$
Change in velocity for molecules hitting the leading face,

$\Delta V_1=(2u+2v)$
From Newton's second law, force acting on particles hitting leading face,

$F_1=\frac{d{p}_1}{dt}=\rho A(u+v)\times (2u+2v)=2\rho A(u+v{)}^2$Final velocity of molecule after collision with trailing face of plate,

$v_2=(u-2v)$
Change in velocity for molecules hitting the trailing face,

$\Delta v_2=2u-2v$
From Newton's second law force acting on particles hitting trailing plate,

$F_2=\frac{d{p}_2}{dt}=\rho A(u-v)\times (2u-2v)=2\rho A(u-v{)}^2$

From Newton's third law, force $F_1$ and $F_2$ will also act on plate and the net force on plate will be,

$F_{net}=F_1-F_2=2\rho A\left(4uv\right)=8\rho Avu$

$P=\frac{\Delta F}{A}=8\rho uv$
Hence option D that says pressure difference between leading and trailing face of plate is proportional to $uv$
Net resistive force on the plate will be

$F_{net}=(F-\Delta F)=ma$
where $m$ is the mass of the plate.

$F-\left(8\rho Auv\right)=ma=\frac{mdv}{dt}$
Hence the net resistive force is proportional to $v$ and thus option A is correct.
After sufficient time, velocity of plate$\left(v\right)$ keeps on increasing and then external force balances the resistive force. Hence option B is correct.