Final answer:
The impulse acting on a 1.1 kg ball that hits the floor and rebounds is 51.7 kg·m/s. The magnitude of the average force exerted on the floor by the ball, when it is in contact for 0.0131 s, is approximately 3946.6 N.
Step-by-step explanation:
The student is asking about two aspects related to the motion of a ball in contact with a floor: the impulse experienced by the ball and the average force exerted on the floor by the ball. To answer these questions, we will apply principles from physics, specifically the concepts of impulse and force.
Impulse:
(a) Impulse is defined as the change in momentum of an object when a force is applied over a period of time. It is given by the formula:
Impulse (J) = Change in Momentum (Δp)
Since momentum is the product of the mass (m) and the velocity (v), we can write the change in momentum as:
Δp = m × (v_final - v_initial)
In this case, the impulse acting on the ball during the contact with the floor is:
Δp = 1.1 kg × (14 m/s - (-33 m/s)) = 1.1 kg × 47 m/s = 51.7 kg·m/s
Average Force:
(b) The average force (F_avg) can be calculated using the impulse and the time (t) for which the force is applied:
F_avg = Impulse (J) / Time (t)
The magnitude of the average force on the floor from the ball is:
F_avg = 51.7 kg·m/s / 0.0131 s ≈ 3946.6 N
The direction of this force is upward, opposing the direction of the ball's impact.