A ball after falling a distance of 5 meter rest hits elastically the floor of a lift and rebounds. At the time of impact the lift was moving up with a velocity of 1m/sec. The velocity with which the ball rebounds just after impact is : `(g=10m//sec^(2))`

Options

A. 10 m/sec
B. 11m/sec
C. 12m/sec
D. 13m/sec

Answer

To solve the problem, we need to find the velocity with which the ball rebounds just after hitting the floor of the lift. We will use the principles of conservation of momentum and the concept of elastic collisions.

### Step-by-step Solution:

1. **Determine the velocity of the ball just before impact**:
The ball falls from a height of 5 meters. We can use the equation of motion to find the velocity just before it hits the ground:
\[ v^2 = u^2 + 2gh \]
where:
- \( u = 0 \) (initial velocity when dropped)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 5 \, \text{m} \) (height fallen)

Substituting the values:
\[ v^2 = 0 + 2 \times 10 \times 5 = 100 \]
\[ v = \sqrt{100} = 10 \, \text{m/s} \]

2. **Consider the velocity of the lift**:
At the moment of impact, the lift is moving upwards with a velocity of \( 1 \, \text{m/s} \).

3. **Determine the relative velocities**:
The velocity of the ball just before impact is \( 10 \, \text{m/s} \) downwards, and the lift is moving upwards at \( 1 \, \text{m/s} \). Therefore, the relative velocity of approach (the velocity of the ball towards the lift) is:
\[ v_{\text{approach}} = v + v_{\text{lift}} = 10 + 1 = 11 \, \text{m/s} \]

4. **Apply the coefficient of restitution**:
Since the collision is elastic, the coefficient of restitution \( e = 1 \). The velocity of separation after the collision can be expressed as:
\[ v_{\text{separation}} = e \times v_{\text{approach}} = 1 \times 11 = 11 \, \text{m/s} \]

5. **Determine the velocity of the ball after impact**:
The ball rebounds upwards after the collision. The velocity of the ball after impact \( v' \) can be calculated as:
\[ v' = v_{\text{lift}} + v_{\text{separation}} = 1 + 11 = 12 \, \text{m/s} \]

Thus, the velocity with which the ball rebounds just after impact is **12 m/s** upwards.