Question 317: Work, Energy and Power & Circular Motion

Question

A flat plate is moving normal to its plane through a gas under the reaction 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?

Accepted Answer

It is given that the average speed of gas molecules is u; the speed of the plate is v; the force on the plate if F.
$$\bullet$$ Just before the collision : The gas molecules are approaching the leading and trailing faces of the plate with speed u.
$$\bullet$$ Just after the collision : The gas molecules bounce back with speed u
1
and u
2
.
$$\bullet$$ At trailing face
The speed before collision is u $$-$$ v.
The speed after collision is u
2
+ v.
From law of conservation of linear momentum, we have
u
2
+ v = u $$-$$ v
That is,
u
2
= u $$-$$ 2v ....... (1)
and
$$\Delta$$u
2
= 2u $$-$$ 2v ....... (2)
$$\bullet$$ At leading face
The speed before collision is u + v.
The speed after collision is u
1
$$-$$ v.
From law of conservation of linear momentum, we have
u
1
$$-$$ v = u + v
That is,
u
1
= u + 2v ..... (3)
and
$$\Delta$$u
1
= 2u + 2v ...... (4)
Now, from Newton's second law of motion, we have
$$F = {{dp} \over {dt}}$$
$$\bullet$$ For leading face : $${F_1} = {{d{p_1}} \over {dt}}$$.
The volume swept by the plate is A(u + v). Therefore,
$${F_1} = {{d{p_1}} \over {dt}} = \rho A(u + v)\Delta {u_1}$$
$${F_1} = \rho A(u + v)(2u + 2v)$$ ....... (5)
$$\bullet$$ For trailing face : $${F_2} = {{d{p_2}} \over {dt}}$$
The volume swept by the plate is A(u $$-$$ v). Therefore,
$${F_2} = {{d{p_2}} \over {dt}} = \rho A(u - v)\Delta {u_2}$$
$${F_2} = \rho A(u - v)(2u - 2v)$$ ....... (6)
The net force on the plate is
$${F_{net}} = {F_1} - {F_2}$$
$$ = \rho A(u + v)(2u + 2v) - \rho A(u - v)(2u - 2v)$$
$$ = 2\rho A(u + v)(u + v) - 2\rho A(u - v)(u - v)$$
$$ = 2\rho A{(u + v)^2} - 2\rho A{(u - v)^2}$$
$$ = 2\rho A[{(u + v)^2} - {(u - v)^2}]$$
$$ = 2\rho A[{u^2} + {v^2} + 2uv - ({u^2} + {v^2} - 2uv)]$$
$$ = 2\rho A({u^2} + {v^2} + 2uv - {u^2} - {v^2} + 2uv)$$
$$ = 2\rho A(4uv)$$
$$ = 8\rho Auv$$
Therefore, the pressure difference is
$${{{F_{net}}} \over A} = {{8\rho Auv} \over A} = 8\rho uv$$
Thus, the pressure difference between the leading and trailing faces of the plate is proportional to uv. Hence, option (A) is correct.
Now,
$${F_{net}} = 8\rho Auv$$
The net resistive force is
$$F - (8\rho Au)v = {{mdv} \over {dt}}$$
where F is the constant force applied by the gas molecules on the plate.
The resistive force experienced by the plate is proportional to v.
Hence, option (B) is also correct.
At a later time, the velocity v becomes sufficient and the external force $$F( = 8\rho Auv)$$ balances the resistive force.
Hence, option (D) is also correct.

Answer

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

Answer

The resistive force experimenced by the plate is proportional to $$v$$

Answer

The plate will continue to move with constant non-zero acceleration, at all times

Answer

At a later time the external force $$F$$ balances the resistive force

Work, Energy and Power & Circular Motion

A flat plate is moving normal to its plane through a gas under the reaction 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?

Choose the correct answer: