Question

Consider a ball of 0.22 kg, initially at rest, 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. Determine the impulse on the ball delivered by the floor.

Answer 1

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 :

$u=\sqrt{2gh} \\\\u=\sqrt{2\times 10\times 1.8} \\\\u=6\ m/s$

Final velocity,

$v=\sqrt{2gh'} \\\\v=\sqrt{2\times 10\times 1.5} \\\\v=5.47\ m/s$

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 :

$J=m(v-u)\\\\J=0.22\times ((-5.47)-6)\\\\J=-2.52\ kg-m/s$

So, the impulse on the ball delivered by the floor is 2.52 kg-m/s.