A small sphere of mass 1kg is moving with a velocity `(6hati+hatj)m//s` . It hits a fixed smooth wall and rebounds with velocity `(4hati+hatj)m//s`. The coefficient of restitution between the sphere and the wall is

Options

A. `(3)/(2)`
B. `(2)/(3)`
C. `(9)/(16)`
D. `(4)/(9)`

Explanation

To find the coefficient of restitution between the sphere and the wall, we can follow these steps:

### Step 1: Identify Initial and Final Velocities
The initial velocity of the sphere before hitting the wall is given as:
\[ \vec{v_i} = 6\hat{i} + \hat{j} \, \text{m/s} \]
The final velocity of the sphere after rebounding off the wall is:
\[ \vec{v_f} = 4\hat{i} + \hat{j} \, \text{m/s} \]

### Step 2: Determine the Components of Velocity
Since the wall is smooth, it only affects the component of velocity perpendicular to it. We need to identify the components of the initial and final velocities.

- The component of the initial velocity along the direction of the wall (assumed to be along the x-axis) is:
\[ v_{i,x} = 6 \, \text{m/s} \]
- The component of the final velocity along the direction of the wall is:
\[ v_{f,x} = 4 \, \text{m/s} \]

### Step 3: Calculate Speed of Approach and Speed of Separation
- The speed of approach (the speed before hitting the wall) is:
\[ \text{Speed of approach} = v_{i,x} = 6 \, \text{m/s} \]
- The speed of separation (the speed after rebounding from the wall) is:
\[ \text{Speed of separation} = v_{f,x} = 4 \, \text{m/s} \]

### Step 4: Apply the Coefficient of Restitution Formula
The coefficient of restitution \( e \) is defined as the ratio of the speed of separation to the speed of approach:
\[ e = \frac{\text{Speed of separation}}{\text{Speed of approach}} = \frac{v_{f,x}}{v_{i,x}} = \frac{4}{6} = \frac{2}{3} \]

### Step 5: Conclusion
Thus, the coefficient of restitution between the sphere and the wall is:
\[ e = \frac{2}{3} \]