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

The impulse on the ball delivered by the floor is zero.

Step-by-step explanation:

To determine the impulse on the ball delivered by the floor, we can use the principle of conservation of momentum. Impulse is defined as the change in momentum, and momentum is defined as mass multiplied by velocity. Since the ball is initially at rest, its momentum before the collision is zero. After the collision, the ball rebounds and moves upwards, so its final momentum is also zero. We can write the equation:

Impulse = (final momentum) - (initial momentum)

Since momentum is zero before and after the collision, the impulse on the ball delivered by the floor is also zero.

Answer 2

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

Step-by-step 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=√(2gh) \\\\u=√(2* 10* 1.8) \\\\u=6\ m/s`

Final velocity,

`v=√(2gh') \\\\v=√(2* 10* 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* ((-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.