A ball falls from a height of 5 m and strikes a lift which is moving in the upward direction with a velocity of `1 m s^(-1)`, then the velocity with which the ball rebounds after collision will be

Options

A. `11 ms^(-1)` downwards
B. `12 ms^(-1)` upwards
C. `13 ms^(-1) ` upwards
D. `12 ms ^(-1)` downwards

Answer

To solve the problem of a ball falling from a height of 5 meters and colliding with a lift moving upwards at a velocity of 1 m/s, we can follow these steps:

### Step 1: Calculate the velocity of the ball just before the collision
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 lift.

Using the formula:
\[ v = \sqrt{2gh} \]
where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 5 \, \text{m} \)

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

### Step 2: Determine the velocities during the collision
At the moment of collision:
- The velocity of the ball (downward) is \( 10 \, \text{m/s} \).
- The velocity of the lift (upward) is \( 1 \, \text{m/s} \).

### Step 3: Apply the principle of conservation of momentum and the coefficient of restitution
Since the lift has a much larger mass compared to the ball, we can assume that the lift's velocity remains approximately unchanged during the collision.

Using the coefficient of restitution \( e \), which is assumed to be 1 for an elastic collision, we can write:
\[ v_2 - v_1 = e(u_1 - u_2) \]
where:
- \( v_1 \) = final velocity of the ball after collision (upward)
- \( v_2 \) = final velocity of the lift after collision (upward)
- \( u_1 \) = initial velocity of the ball before collision (downward)
- \( u_2 \) = initial velocity of the lift before collision (upward)

Substituting the known values:
- \( u_1 = -10 \, \text{m/s} \) (downward is negative)
- \( u_2 = 1 \, \text{m/s} \) (upward is positive)

The equation becomes:
\[ v_2 - v_1 = 1 \times (-10 - 1) \]
\[ v_2 - v_1 = -11 \]

### Step 4: Solve for the final velocities
Since the lift's velocity after the collision is approximately unchanged (we can assume it remains at \( 1 \, \text{m/s} \)):
\[ v_2 = 1 \, \text{m/s} \]

Now substituting back into the equation:
\[ 1 - v_1 = -11 \]
\[ v_1 = 1 + 11 \]
\[ v_1 = 12 \, \text{m/s} \]

### Conclusion
The velocity with which the ball rebounds after the collision is \( 12 \, \text{m/s} \) upward.