A sphere is moving with velocity vector `2hati+2hatj` immediately before it hits a vertical wall. The wall is parallel to `hatj` and the coefficient of restitution of the sphere and the wall is `e=1/2`. Find the velocity of the sphere after it hits the wall?

Options

A. `hati - hatj`
B. `-hati + 2hatj`
C. `-hati - hatj`
D. `2hati - hatj`

Explanation

To solve the problem, we need to analyze the motion of the sphere before and after it hits the vertical wall. The wall is parallel to the y-axis, which means it will only affect the x-component of the sphere's velocity due to the collision.

### Step-by-Step Solution:

1. **Identify the Initial Velocity Components:**
The initial velocity vector of the sphere is given as:
\[ \vec{u} = 2\hat{i} + 2\hat{j} \]
From this, we can identify the components:
\[ u_x = 2 \, \text{m/s} \quad \text{(velocity in the x-direction)} \]
\[ u_y = 2 \, \text{m/s} \quad \text{(velocity in the y-direction)} \]

2. **Understand the Effect of the Wall:**
The wall is vertical (parallel to the y-axis), so it will only affect the x-component of the velocity. The y-component of the velocity will remain unchanged.

3. **Apply the Coefficient of Restitution:**
The coefficient of restitution \( e \) is given as \( \frac{1}{2} \). The formula to find the final velocity in the x-direction after the collision is:
\[ v_x = -e \cdot u_x \]
Substituting the values:
\[ v_x = -\frac{1}{2} \cdot 2 = -1 \, \text{m/s} \]

4. **Determine the Final Velocity Components:**
Since the y-component remains unchanged:
\[ v_y = u_y = 2 \, \text{m/s} \]

5. **Combine the Final Velocity Components:**
The final velocity vector after the collision can be written as:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} = -1\hat{i} + 2\hat{j} \, \text{m/s} \]

### Final Answer:
The velocity of the sphere after it hits the wall is:
\[ \vec{v} = -1\hat{i} + 2\hat{j} \, \text{m/s} \]