A flat plane is moving normal to its plane through a gas under the action of a constant force F. The

Q: Why can we neglect collisions between gas molecules in this problem?

The problem states that the gas is kept at a very low pressure. At extremely low pressures, the mean free path of the molecules is much larger than the dimensions of the plate, meaning intermolecular collisions are extremely rare and can be neglected.

Q: How does the elastic collision condition help us find the final velocity?

For a perfectly elastic collision with a massive wall or plate, the relative velocity of separation equals the relative velocity of approach. This allows us to write $v_{\text{separation}} = v_{\text{approach}}$ in the plate's frame and then transform back to the ground frame.

Q: Why is the collision rate different for the leading and trailing faces?

Because the plate is moving, it sweeps through more volume per unit time in the forward direction, increasing the relative speed of approach for molecules in front ($u+v$). Conversely, it moves away from molecules behind, reducing their relative speed of approach ($u-v$).

Q: Does the plate accelerate indefinitely?

No. As the velocity of the plate increases, the resistive drag force also increases. When the drag force equals the constant external force $F$, the net force becomes zero, and the plate reaches a constant terminal velocity.

Q: How does this drag force compare to viscous drag in liquids?

In rarefied gases, drag arises from individual molecular impacts ($F_R \propto v$). In dense fluids, drag is dominated by viscosity and fluid layers sliding past each other, which also yields a linear dependence on velocity for low speeds (Stokes' Law), but through a completely different physical mechanism.

Question

A flat plane 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?

Options:

(A) At a later time the external force $F$ balances the resistive force

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

(C) The resistive force experienced by the plate is proportional to $v$

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

Sigma Percentile · Accepted Answer

### The Microscopic Tug-of-War

Imagine a flat plate of area $A$ slicing through a highly rarefied gas.

At very low pressures, we enter a regime known as **free molecular flow**.

In this regime, the gas molecules are so far apart that they almost never collide with each other.

Instead, their primary interactions are direct, elastic collisions with the moving plate.

As the plate is pulled to the right by a constant external force $F$, it collides with molecules on both its front (leading) and rear (trailing) faces.

This creates a microscopic tug-of-war, resulting in a net resistive force that opposes the motion.

### Kinematics of Elastic Collisions

Let us analyze the collision of a single gas molecule of mass $m$ with the plate.

Since the plate is moving to the right with velocity $v$, and the molecules are moving randomly with an average speed $u$, we must look at the collisions from the perspective of relative motion.

#### 1. The Leading Face

For molecules colliding with the front face of the plate, they approach from the right with a speed $u$ relative to the gas.

Since the plate is moving towards them at speed $v$, the **relative velocity of approach** is:

$$v_{	ext{rel, approach}} = u + v$$

Because the collision is perfectly elastic ($e = 1$), the **relative velocity of separation** must be equal to the relative velocity of approach:

$$v_{	ext{rel, sep}} = u + v$$

In the ground frame, if the molecule rebounds with a velocity $v_1$ to the right, we have:

$$v_1 - v = u + v \implies v_1 = u + 2v$$

The change in momentum of a single molecule during this collision is:

$$\Delta p_1 = m(v_1 - (-u)) = 2m(u + v)$$

To find the total force, we need the rate of collisions.

The number of molecules colliding with the leading face per unit time depends on their relative speed:

$$\frac{dn_1}{dt} = ho A (u + v)$$

where $ho$ is the molecular number density.

Thus, the force exerted on the leading face is:

$$F_1 = \Delta p_1 \frac{dn_1}{dt} = 2ho m A (u + v)^2 = 2ho_0 A (u + v)^2$$

where $ho_0 = ho m$ is the mass density of the gas.

#### 2. The Trailing Face

For the rear face, the plate is moving away from the molecules.

Only molecules with speed $u > v$ can catch up and collide with it.

The relative velocity of approach is:

$$v_{	ext{rel, approach}} = u - v$$

Following the same elastic collision logic, the force exerted on the trailing face is:

$$F_2 = 2ho_0 A (u - v)^2$$

### Calculating the Net Drag

The net resistive force $F_R$ acting on the plate is the difference between the forces on the leading and trailing faces:

$$F_R = F_1 - F_2 = 2ho_0 A \left[ (u + v)^2 - (u - v)^2 ight]$$

Using the algebraic identity $(a+b)^2 - (a-b)^2 = 4ab$, we get:

$$F_R = 2ho_0 A (4uv) = 8ho_0 A uv$$

This elegant result shows that the resistive force is **directly proportional to the velocity $v$** of the plate.

Therefore, **Option (c) is correct**.

### Pressure Difference

The pressure difference $P$ between the leading and trailing faces is simply the net force divided by the area of the plate:

$$P = \frac{F_R}{A} = 8ho_0 uv$$

This clearly shows that the pressure difference is **proportional to $uv$**.

Therefore, **Option (d) is correct**.

### Reaching Terminal Velocity

Now, let us write the equation of motion for the plate of mass $M$:

$$F_{	ext{net}} = F - F_R = Ma \implies F - (8ho_0 A u)v = Ma$$

Initially, when $v = 0$, the resistive force is zero, and the plate accelerates with $a = F/M$.

As the plate speeds up, the resistive force increases linearly with $v$.

Eventually, the resistive force grows to perfectly balance the constant external force $F$.

At this point, the net force becomes zero, the acceleration drops to zero, and the plate continues to move at a constant **terminal velocity**:

$$v_t = \frac{F}{8ho_0 A u}$$

Thus, at a later time, the external force balances the resistive force, and the acceleration becomes zero.

Therefore, **Option (a) is correct**, and **Option (b) is incorrect**.

In conclusion, the correct options are **(a), (c), and (d)**.

Select Answer:

Visualized Solution

Visualizing the Moving Plate

The Collision Mechanism

Kinematics on the Leading Face

Momentum Transfer on the Leading Face

Force on the Leading Face

Force on the Trailing Face

Net Resistive Force

Analyzing Drag and Pressure

Terminal Velocity and Equilibrium

The Way Forward

The Sigma Insight: Kinetic Theory of Gases

The Microscopic Tug-of-War

Kinematics of Elastic Collisions

# 1

# 2

Calculating the Net Drag

Pressure Difference

Reaching Terminal Velocity

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