A smooth sphere of mass m is moving on a horizontal plane with a velocity `(3hati +hat j)` . It collides with smooth a vertical wall which is parallel to the vector `hatj` . If coefficient of restitution `e = 1/2` then impulse that acts on the sphere is

Options

A. `-9/2mhati`
B. `(-3/2hati+hatj)`
C. `3/2mhatj`
D. `(3/2mhatj+1/2mhati)`

Answer

To solve the problem, we need to analyze the motion of the sphere before and after the collision with the wall. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
The sphere is moving with an initial velocity given by the vector:
\[ \vec{v}_i = 3\hat{i} + 1\hat{j} \]
This means that the sphere has a velocity component of 3 in the x-direction and 1 in the y-direction.

### Step 2: Identify the Direction of the Wall
The wall is vertical and parallel to the y-axis (along the \(\hat{j}\) direction). Therefore, it will only affect the x-component of the sphere's velocity during the collision.

### Step 3: Determine the Coefficient of Restitution
The coefficient of restitution \(e\) is given as \( \frac{1}{2} \). The coefficient of restitution relates the relative velocities of separation and approach along the line of impact. It is defined as:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]

### Step 4: Calculate the Velocity of Approach
Since the wall is vertical, the velocity of approach is only due to the x-component of the velocity:
\[ \text{Velocity of approach} = 3 \text{ (the x-component)} \]

### Step 5: Calculate the Velocity of Separation
Using the coefficient of restitution:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \implies \frac{1}{2} = \frac{v_f}{3} \]
Solving for \(v_f\) (the velocity of separation):
\[ v_f = \frac{1}{2} \times 3 = \frac{3}{2} \]

### Step 6: Determine the Final Velocity Components
After the collision, the x-component of the velocity will change direction and become negative:
\[ v_{fx} = -\frac{3}{2} \]
The y-component of the velocity remains unchanged:
\[ v_{fy} = 1 \]

### Step 7: Calculate the Change in Momentum
The change in momentum in the x-direction (since the wall only affects this direction) is given by:
\[ \Delta p_x = m(v_{fx} - v_{ix}) = m\left(-\frac{3}{2} - 3\right) \]
Calculating this:
\[ \Delta p_x = m\left(-\frac{3}{2} - \frac{6}{2}\right) = m\left(-\frac{9}{2}\right) = -\frac{9}{2}m \]

### Step 8: Determine the Impulse
The impulse \(J\) is equal to the change in momentum:
\[ J = \Delta p_x = -\frac{9}{2}m \hat{i} \]

### Final Answer
The impulse that acts on the sphere is:
\[ \boxed{-\frac{9}{2}m \hat{i}} \]