Question

A ball falls from a height of 19.0 m, hits the floor, and rebounds vertically upward to a height of 15.5 m. Assume that m ball=0.380 kg.

What is the impulse (in kg⋅m/s) delivered to the ball by the floor?

Answer 1

The impulse delivered to the ball by the floor is -5.89 kg⋅m/s.

Step-by-step explanation:

The impulse delivered to the ball by the floor can be calculated using the principle of conservation of momentum and the equation:

impulse = change in momentum = mass of the ball * change in velocity

Since the ball rebounds vertically, the change in velocity is equal to the negative of the initial velocity.

Therefore, the magnitude of the impulse is:

impulse = (mass of the ball) * (final velocity - initial velocity)

Substituting the given values, the impulse delivered to the ball by the floor is:

impulse = (0.380 kg) * (-15.5 m/s - (0 m/s))

= -5.89 kg⋅m/s