To solve the problem step by step, we will analyze the motion of the ball, calculate its mechanical energy, and determine the impulse after the collision.
### Step 1: Calculate the initial potential energy (PE) of the ball
The potential energy when the ball is at height \( h = 5 \, \text{m} \) is given by the formula:
\[
PE = mgh
\]
Where:
- \( m = 1 \, \text{kg} \) (mass of the ball)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 5 \, \text{m} \) (height)
Substituting the values:
\[
PE = 1 \times 10 \times 5 = 50 \, \text{J}
\]
### Step 2: Determine the kinetic energy (KE) just before the collision
Just before the ball hits the ground, all its potential energy is converted into kinetic energy. Therefore:
\[
KE = PE = 50 \, \text{J}
\]
### Step 3: Calculate the velocity of the ball just before the collision
Using the kinetic energy formula:
\[
KE = \frac{1}{2} mv^2
\]
Setting \( KE = 50 \, \text{J} \):
\[
50 = \frac{1}{2} \times 1 \times v^2
\]
Solving for \( v^2 \):
\[
v^2 = 100 \implies v = 10 \, \text{m/s}
\]
### Step 4: Calculate the mechanical energy after the collision
The problem states that the ball loses 50% of its total mechanical energy during the collision. Therefore, the mechanical energy after the collision is:
\[
\text{Mechanical Energy after collision} = 50 \, \text{J} \times 0.5 = 25 \, \text{J}
\]
### Step 5: Determine the velocity after the collision
After the collision, the potential energy is again zero (as it is at ground level), so the kinetic energy after the collision is:
\[
KE' = 25 \, \text{J}
\]
Using the kinetic energy formula:
\[
KE' = \frac{1}{2} mu'^2
\]
Setting \( KE' = 25 \, \text{J} \):
\[
25 = \frac{1}{2} \times 1 \times u'^2
\]
Solving for \( u'^2 \):
\[
u'^2 = 50 \implies u' = 5\sqrt{2} \, \text{m/s}
\]
### Step 6: Calculate the impulse
Impulse is defined as the change in momentum. The initial momentum just before the collision is:
\[
p_{\text{initial}} = mv = 1 \times 10 = 10 \, \text{kg m/s}
\]
The final momentum after the collision is:
\[
p_{\text{final}} = mu' = 1 \times 5\sqrt{2} \, \text{kg m/s}
\]
Calculating the impulse:
\[
\text{Impulse} = p_{\text{final}} - p_{\text{initial}} = (5\sqrt{2} - 10) \, \text{kg m/s}
\]
### Conclusion
The ball rebounds with a speed of \( 5\sqrt{2} \, \text{m/s} \) and the impulse experienced by the ball is \( 5\sqrt{2} - 10 \, \text{kg m/s} \).
