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`

Answer

To solve the problem, we need to determine the velocity of the sphere after it collides with a vertical wall. The initial velocity of the sphere is given as \( \vec{v_i} = 2\hat{i} + 2\hat{j} \), and the coefficient of restitution \( e = \frac{1}{2} \).

### Step-by-Step Solution:

1. **Identify the components of the initial velocity**:
The initial velocity vector can be broken down into its components:
- \( v_{ix} = 2 \) (the component along the x-axis)
- \( v_{iy} = 2 \) (the component along the y-axis)

2. **Understand the effect of the wall**:
The wall is vertical and parallel to the \( \hat{j} \) direction. This means that the collision will affect only the x-component of the velocity, while the y-component will remain unchanged.

3. **Apply the coefficient of restitution**:
The coefficient of restitution \( e \) is defined as:
\[ e = \frac{\text{velocity of separation}}{\text{velocity of approach}} \]
Here, the velocity of approach is the x-component of the initial velocity, which is \( v_{ix} = 2 \).

4. **Determine the velocity of separation**:
Let the final x-component of the velocity after the collision be \( v_{fx} \). According to the definition of the coefficient of restitution:
\[ e = \frac{v_{fx}}{v_{ix}} \]
Substituting the known values:
\[ \frac{1}{2} = \frac{v_{fx}}{2} \]
Solving for \( v_{fx} \):
\[ v_{fx} = 2 \times \frac{1}{2} = 1 \]
However, since the sphere bounces back, the direction of the x-component changes, so:
\[ v_{fx} = -1 \]

5. **Determine the final y-component of the velocity**:
Since the y-component of the velocity does not change during the collision, we have:
\[ v_{fy} = v_{iy} = 2 \]

6. **Combine the final components into a velocity vector**:
The final velocity vector after the collision can be expressed as:
\[ \vec{v_f} = v_{fx}\hat{i} + v_{fy}\hat{j} = -1\hat{i} + 2\hat{j} \]

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