Answer:
The impulse on the ball delivered by the floor is 2.52 kg-m/s.
Explanation:
Given that,
Mass of the ball, m = 0.22 kg
It is dropped from an initial height of 1.80 m. It rebounds back after colliding with the floor to a final height of 1.50 m. Initial velocity and final velocity can be calculated using conservation of energy as :
[tex]u=\sqrt{2gh} \\\\u=\sqrt{2\times 10\times 1.8} \\\\u=6\ m/s[/tex]
Final velocity,
[tex]v=\sqrt{2gh'} \\\\v=\sqrt{2\times 10\times 1.5} \\\\v=5.47\ m/s[/tex]
As the ball rebounds, v = -5.47 m/s
We need to find the impulse on the ball delivered by the floor. We know that impulse is equal to the change in momentum as follows :
[tex]J=m(v-u)\\\\J=0.22\times ((-5.47)-6)\\\\J=-2.52\ kg-m/s[/tex]
So, the impulse on the ball delivered by the floor is 2.52 kg-m/s.
