A small sphere of mass 1 kg is moving with a velocity `(6hat(i)+hat(j))ms^(-1)`. It hits a fixed smooth wall and rebound with velocity `(4hat(i)+hat(j))ms^(-1)`. The coefficient of restitution between the sphere and the wall is

Options

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

Answer

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

### Step 1: Identify the initial and final velocities
The initial velocity \( \mathbf{V_i} \) of the sphere is given as:
\[ \mathbf{V_i} = 6 \hat{i} + 1 \hat{j} \, \text{m/s} \]
The final velocity \( \mathbf{V_f} \) after rebounding is:
\[ \mathbf{V_f} = 4 \hat{i} + 1 \hat{j} \, \text{m/s} \]

### Step 2: Determine the components of the velocities along the normal direction
Since the wall is fixed and smooth, we need to find the normal direction. For a wall aligned with the y-axis, the normal direction is along the x-axis (i.e., \(\hat{i}\)).

- The component of the initial velocity along the normal (x-direction) is:
\[ V_{i,n} = 6 \, \text{m/s} \]
- The component of the final velocity along the normal (x-direction) is:
\[ V_{f,n} = 4 \, \text{m/s} \]

### Step 3: Calculate the velocity of approach and velocity of separation
- The velocity of approach (before collision) is the component of the initial velocity along the normal:
\[ V_{approach} = V_{i,n} = 6 \, \text{m/s} \]
- The velocity of separation (after collision) is the component of the final velocity along the normal:
\[ V_{separation} = V_{f,n} = 4 \, \text{m/s} \]

### Step 4: Apply the formula for the coefficient of restitution
The coefficient of restitution \( e \) is defined as the ratio of the velocity of separation to the velocity of approach:
\[ e = \frac{V_{separation}}{V_{approach}} = \frac{4}{6} = \frac{2}{3} \]

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