A ball falls from a height such that it strikes the floor of lift at 10 m/s. If lift is moving in the upward direction with a velocity 1 m/s, then velocity with which the ball rebounds after elastic collision will be

Options

A. 11 m/s
B. 12 m/s
C. 13 m/s
D. 9 m/s

Answer

To solve the problem, we need to analyze the situation involving the ball and the lift, especially focusing on the velocities involved before and after the collision.

### Step-by-Step Solution:

1. **Identify the Initial Velocities**:
- The ball is falling downwards with a velocity of \( v_b = 10 \, \text{m/s} \).
- The lift is moving upwards with a velocity of \( v_l = 1 \, \text{m/s} \).

2. **Determine the Relative Velocity of Approach**:
- Since the ball is moving downwards and the lift is moving upwards, the relative velocity of approach \( v_{\text{approach}} \) can be calculated as:
\[ v_{\text{approach}} = v_b + v_l = 10 \, \text{m/s} + 1 \, \text{m/s} = 11 \, \text{m/s} \]

3. **Apply the Coefficient of Restitution**:
- For an elastic collision, the coefficient of restitution \( E \) is equal to 1. The formula for the coefficient of restitution is:
\[ E = \frac{v_{\text{separation}}}{v_{\text{approach}}} \]
- Here, \( v_{\text{separation}} \) is the velocity of the ball after it rebounds, which we will denote as \( v' \).

4. **Set Up the Equation**:
- From the definition of \( E \):
\[ 1 = \frac{v' - v_l}{v_{\text{approach}}} \]
- Substituting the values we have:
\[ 1 = \frac{v' - 1}{11} \]

5. **Solve for \( v' \)**:
- Rearranging the equation gives:
\[ v' - 1 = 11 \]
- Thus, solving for \( v' \):
\[ v' = 11 + 1 = 12 \, \text{m/s} \]

6. **Conclusion**:
- The velocity with which the ball rebounds after the elastic collision is \( v' = 12 \, \text{m/s} \).

### Final Answer:
The velocity with which the ball rebounds after the elastic collision is **12 m/s**.