A smooth sphere is moving on a horizontal surface 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?

Explanation

To solve the problem, we need to analyze the collision of the sphere with the wall and apply the concept of the coefficient of restitution.

### Step-by-Step Solution:

1. **Identify the Initial Velocity**:
The initial velocity vector of the sphere is given as:
\[ \vec{U} = 2\hat{i} + 2\hat{j} \]
Here, \(2\hat{i}\) is the component along the x-axis and \(2\hat{j}\) is the component along the y-axis.

2. **Understand the Collision**:
The wall is vertical and parallel to the y-axis. This means that the collision will affect only the x-component of the velocity, while the y-component will remain unchanged.

3. **Coefficient of Restitution**:
The coefficient of restitution \(e\) is given as \( \frac{1}{2} \). The formula for the coefficient of restitution in terms of velocities is:
\[ e = -\frac{V_{separation}}{V_{approach}} \]
where \(V_{separation}\) is the velocity of separation after the collision and \(V_{approach}\) is the velocity of approach before the collision.

4. **Determine the Velocity of Approach**:
The velocity of approach in the x-direction before the collision is:
\[ V_{approach} = U_x = 2 \text{ (since the sphere is moving towards the wall)} \]

5. **Set Up the Equation**:
Let \(V_x\) be the x-component of the velocity after the collision. Since the wall is stationary, we can write:
\[ e = -\frac{V_x}{V_{approach}} = -\frac{V_x}{2} \]
Substituting \(e = \frac{1}{2}\):
\[ \frac{1}{2} = -\frac{V_x}{2} \]

6. **Solve for \(V_x\)**:
Rearranging the equation gives:
\[ V_x = -1 \]
This indicates that the sphere will move in the negative x-direction after the collision.

7. **Determine the y-component of the Velocity**:
Since the collision does not affect the y-component of the velocity, we have:
\[ V_y = U_y = 2 \]

8. **Final Velocity Vector**:
Combining both components, the final velocity vector after the collision is:
\[ \vec{V} = V_x \hat{i} + V_y \hat{j} = -1\hat{i} + 2\hat{j} \]

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