To solve the problem, we need to calculate the impulse and the average force exerted between the ball and the ground. Here’s a step-by-step breakdown of the solution:
### Step 1: Calculate the velocity just before the ball hits the ground (V)
The ball falls from a height of 2 m. Using the equation of motion:
\[ V^2 = U^2 + 2gH \]
Where:
- \( U = 0 \) (initial velocity)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( H = 2 \, \text{m} \) (height)
Substituting the values:
\[ V^2 = 0 + 2 \times 9.8 \times 2 \]
\[ V^2 = 39.2 \]
\[ V = \sqrt{39.2} \approx 6.26 \, \text{m/s} \]
### Step 2: Calculate the velocity just after the ball rebounds (V')
The ball rebounds to a height of 0.5 m. Again, using the equation of motion:
\[ V'^2 = U'^2 + 2gH' \]
Where:
- \( U' = 0 \) (initial velocity after rebound)
- \( H' = 0.5 \, \text{m} \)
Substituting the values:
\[ V'^2 = 0 + 2 \times 9.8 \times 0.5 \]
\[ V'^2 = 9.8 \]
\[ V' = \sqrt{9.8} \approx 3.13 \, \text{m/s} \]
### Step 3: Calculate the change in momentum (ΔP)
The mass of the ball is given as 60 g, which we convert to kg:
\[ m = 60 \, \text{g} = 0.06 \, \text{kg} \]
The change in momentum is given by:
\[ \Delta P = m(V' - (-V)) \]
Here, \( V \) is negative because it is directed downwards:
\[ \Delta P = 0.06 \, (3.13 - (-6.26)) \]
\[ \Delta P = 0.06 \, (3.13 + 6.26) \]
\[ \Delta P = 0.06 \times 9.39 \]
\[ \Delta P \approx 0.5634 \, \text{kg m/s} \]
### Step 4: Calculate the impulse (I)
Impulse is equal to the change in momentum:
\[ I = \Delta P \approx 0.5634 \, \text{Ns} \]
### Step 5: Calculate the average force (F)
The average force can be calculated using the formula:
\[ F = \frac{\Delta P}{\Delta t} \]
Where \( \Delta t = 0.2 \, \text{s} \):
\[ F = \frac{0.5634}{0.2} \]
\[ F \approx 2.817 \, \text{N} \]
### Final Answers:
- Impulse \( I \approx 0.563 \, \text{Ns} \)
- Average Force \( F \approx 2.817 \, \text{N} \)
