A molecule of mass m moving at a velocity v impinges elastically on the wall at an angle a 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 need to analyze the situation using the principles of momentum and elastic collisions.

### Step-by-Step Solution:

1. **Identify the Components of Velocity:**
- Before the collision, the velocity \( v \) can be broken down into two components:
- The horizontal component (parallel to the wall):
\[ v_x = v \sin \alpha \]
- The vertical component (perpendicular to the wall):
\[ v_y = v \cos \alpha \]

2. **Analyze the Collision:**
- Since the collision is elastic, the horizontal component of the velocity will reverse its direction, while the vertical component will remain unchanged.
- After the collision:
- The horizontal component becomes:
\[ v_x' = -v \sin \alpha \]
- The vertical component remains:
\[ v_y' = v \cos \alpha \]

3. **Calculate Change in Momentum:**
- The change in momentum in the horizontal direction (x-direction) can be calculated as:
\[ \Delta p_x = p_x' - p_x = m v_x' - m v_x = m (-v \sin \alpha) - m (v \sin \alpha) = -2m v \sin \alpha \]
- The change in momentum in the vertical direction (y-direction) is:
\[ \Delta p_y = p_y' - p_y = m v_y' - m v_y = m (v \cos \alpha) - m (v \cos \alpha) = 0 \]

4. **Impulsive Reaction of the Wall:**
- The impulsive reaction from the wall is equal to the change in momentum of the molecule. Since there is no change in the vertical momentum, we only consider the horizontal change:
\[ \text{Impulsive Reaction} = \Delta p_x = -2m v \sin \alpha \]
- The negative sign indicates the direction of the impulse is opposite to the initial direction of the molecule's horizontal momentum.

5. **Final Result:**
- The magnitude of the impulsive reaction exerted by the wall on the molecule is:
\[ |\text{Impulsive Reaction}| = 2m v \sin \alpha \]

### Conclusion:
The impulsive reaction of the wall when a molecule of mass \( m \) collides elastically at an angle \( \alpha \) is given by:
\[ \text{Impulsive Reaction} = 2m v \sin \alpha \]