A ball of mass 1kg is dropped from a height of h = 5m. It rebounds losing 50% of its total mechanical energy during collision. Then choose the correct statement(s) from the following .

Options

A. After collision speed of ball is `5m//s`
B. After collision speed of ball is `5 sqrt(2) m//s`
C. The value of impulse is `( 10 - 5 sqrt( 2)) Ns`.
D. The value of impulse is `( 10 + 5 sqrt(2)) Ns`

Answer

To solve the problem step by step, we will analyze the motion of the ball, calculate its speed just before and after the collision, and determine the impulse experienced by the ball during the collision.

### Step 1: Calculate the velocity just before the collision
The ball is dropped from a height \( h = 5 \, \text{m} \). We can use the principle of conservation of energy to find the velocity just before the ball hits the ground.

The potential energy (PE) at height \( h \) is given by:
\[ \text{PE} = mgh \]
where:
- \( m = 1 \, \text{kg} \) (mass of the ball),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 5 \, \text{m} \).

Calculating the potential energy:
\[ \text{PE} = 1 \times 10 \times 5 = 50 \, \text{J} \]

At the moment just before the collision, all this potential energy will have converted into kinetic energy (KE):
\[ \text{KE} = \frac{1}{2} mv^2 \]
Setting the potential energy equal to the kinetic energy:
\[ 50 = \frac{1}{2} \times 1 \times v^2 \]
Solving for \( v \):
\[ 50 = \frac{1}{2} v^2 \implies v^2 = 100 \implies v = 10 \, \text{m/s} \]

### Step 2: Calculate the speed after the collision
The problem states that the ball loses 50% of its total mechanical energy during the collision. Therefore, the kinetic energy after the collision \( \text{KE}' \) is:
\[ \text{KE}' = \frac{1}{2} \text{KE} = \frac{1}{2} \times 50 = 25 \, \text{J} \]

Using the kinetic energy formula again to find the new speed \( v' \):
\[ \text{KE}' = \frac{1}{2} mv'^2 \]
Substituting the values:
\[ 25 = \frac{1}{2} \times 1 \times v'^2 \]
Solving for \( v' \):
\[ 25 = \frac{1}{2} v'^2 \implies v'^2 = 50 \implies v' = \sqrt{50} = 5\sqrt{2} \, \text{m/s} \]

### Step 3: Calculate the impulse
Impulse is defined as the change in momentum. The initial momentum \( p_i \) just before the collision is:
\[ p_i = mv = 1 \times 10 = 10 \, \text{kg m/s} \]
The final momentum \( p_f \) just after the collision is:
\[ p_f = mv' = 1 \times 5\sqrt{2} = 5\sqrt{2} \, \text{kg m/s} \]

Now, the change in momentum (impulse \( J \)) is:
\[ J = p_f - p_i = 5\sqrt{2} - 10 \]

### Summary of Results
1. The speed of the ball after the collision is \( 5\sqrt{2} \, \text{m/s} \).
2. The impulse experienced by the ball is \( 5\sqrt{2} - 10 \, \text{Ns} \).