A molecule of mass m moving at a velocity v impinges elastically on the wall at an angle `alpha` with the wall. Then

Options

A. the impulsive reaction of the wall is `2mv cos alpha`
B. the impulsive reaction of the wall is `2mvsinalpha`
C. the impulsive reaction of the wall is nonzero
D. given data is insufficient to calculate impulsive reaction of the wall

Answer

To solve the problem of a molecule of mass \( m \) moving at a velocity \( v \) impinging elastically on a wall at an angle \( \alpha \), we can follow these steps:

### Step 1: Resolve the Velocity Components
The velocity \( v \) of the molecule can be resolved into two components:
- The component perpendicular to the wall: \( v_{\perp} = v \sin \alpha \)
- The component parallel to the wall: \( v_{\parallel} = v \cos \alpha \)

### Step 2: Determine Initial Momentum
The initial momentum of the molecule before the collision can be calculated using the perpendicular component since only this component will change direction upon collision:
\[ p_{\text{initial}} = m v_{\perp} = m (v \sin \alpha) \]

### Step 3: Determine Final Momentum
Since the collision is elastic, the component of velocity perpendicular to the wall will reverse direction while the parallel component remains unchanged. Therefore, the final momentum after the collision is:
\[ p_{\text{final}} = -m v_{\perp} + m v_{\parallel} = -m (v \sin \alpha) + m (v \cos \alpha) \]
However, since we are interested in the change in momentum only in the perpendicular direction:
\[ p_{\text{final}} = -m (v \sin \alpha) \]

### Step 4: Calculate Change in Momentum
The change in momentum (\( \Delta p \)) is given by:
\[ \Delta p = p_{\text{final}} - p_{\text{initial}} = -m (v \sin \alpha) - m (v \sin \alpha) = -2m (v \sin \alpha) \]

### Step 5: Calculate the Impulsive Reaction Force
The impulsive reaction force \( F \) exerted by the wall on the molecule can be calculated using the formula for force as the rate of change of momentum:
\[ F = \frac{\Delta p}{\Delta t} \]
Assuming the time duration of the collision \( \Delta t \) is very small, we can express the impulsive force as:
\[ F = -2m (v \sin \alpha) / \Delta t \]
This force acts in the direction opposite to the initial momentum of the molecule.

### Conclusion
The impulsive reaction force exerted by the wall on the molecule is:
\[ F = -2m (v \sin \alpha) / \Delta t \]
In terms of magnitude, we can express it as:
\[ |F| = \frac{2m (v \sin \alpha)}{\Delta t} \]