Sure, let's walk through the detailed solution to find the impulse given to the ball by the floor.
Given Data:
- Mass of the ball, \( m = 500 \, \text{g} = 0.5 \, \text{kg} \) (since mass should be in kilograms for consistency in SI units).
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \).
- Initial height from which the ball is dropped, \( h_{\text{drop}} = 1.5 \, \text{m} \).
- Height to which the ball rebounds, \( h_{\text{rebound}} = 1.2 \, \text{m} \).
### Step-by-Step Solution:
1. Calculate the velocity just before hitting the floor:
We will use the equation of motion for free fall:
[tex]\[
v^2 = u^2 + 2gh
\][/tex]
where \( u \) is the initial velocity (which is 0, since the ball starts from rest), \( g \) is the acceleration due to gravity, and \( h \) is the height.
For the fall:
[tex]\[
v_{\text{before}}^2 = 0 + 2gh_{\text{drop}}
\][/tex]
[tex]\[
v_{\text{before}} = \sqrt{2 \cdot 9.81 \cdot 1.5}
\][/tex]
[tex]\[
v_{\text{before}} \approx 5.424942396007538 \, \text{m/s}
\][/tex]
2. Calculate the velocity just after rebounding:
Similarly, using the conservation of energy principles when the ball rebounds:
[tex]\[
v^2 = u^2 + 2gh
\][/tex]
where \( u \) is the initial velocity just after rebounding.
For the rebound:
[tex]\[
v_{\text{after}}^2 = 0 + 2gh_{\text{rebound}}
\][/tex]
[tex]\[
v_{\text{after}} = \sqrt{2 \cdot 9.81 \cdot 1.2}
\][/tex]
[tex]\[
v_{\text{after}} \approx 4.852215988597375 \, \text{m/s}
\][/tex]
3. Calculate the impulse imparted to the ball by the floor:
Impulse is defined as the change in momentum. Note that upon rebounding, the direction of the ball's velocity changes, so the velocities add up when considering the impulse.
[tex]\[
\text{Impulse} = \Delta p = m \cdot (v_{\text{after}} + v_{\text{before}})
\][/tex]
[tex]\[
\text{Impulse} = 0.5 \cdot (5.424942396007538 + 4.852215988597375)
\][/tex]
[tex]\[
\text{Impulse} \approx 0.5 \cdot 10.277158384604913
\][/tex]
[tex]\[
\text{Impulse} \approx 5.138579192302457 \, \text{Ns}
\][/tex]
Hence, the impulse given to the ball by the floor is approximately [tex]\( 5.14 \, \text{Ns} \)[/tex].
