After a 0.200-kg rubber ball is dropped from a height of 1.95 m, it bounces off a concrete floor and rebounds to a height of 1.45 m.

(a) Determine the magnitude and direction of the impulse delivered to the ball by the floor.

(b) Estimate the time the ball is in contact with the floor to be 0.08 seconds. Calculate the average force the floor exerts on the ball.

To determine the magnitude and direction of the impulse delivered to the ball by the floor, we need to use the impulse-momentum principle, which states that the impulse of a force acting on an object is equal to the change in momentum of the object.

(a) The change in momentum of the ball is given by the equation Δp = mΔv, where m is the mass of the ball and Δv is the change in velocity.
The initial velocity of the ball is v_initial = 0 m/s (since it was dropped), and the final velocity is v_final = 0 m/s (since it rebounds to the same height). Thus, the change in velocity Δv = v_final - v_initial = 0 - 0 = 0 m/s.

The impulse is then given by the equation J = FΔt, where F is the average force applied by the floor and Δt is the time interval in which the force acts.

Since Δv = 0, the impulse J = Δp = mΔv = 0. Hence, the magnitude of the impulse delivered to the ball by the floor is zero.

(b) To calculate the average force the floor exerts on the ball, we can use the equation F = Δp/Δt, where Δp is the change in momentum and Δt is the time interval in which the force acts. In this case, we are given that the time interval is Δt = 0.08 s.

The change in momentum Δp can be calculated using the equation Δp = m(v_final - v_initial). The initial velocity v_initial = 0 m/s, and the final velocity v_final can be determined using the equation v_final = √(2gh), where g is the acceleration due to gravity (approximately 9.8 m/s²) and h is the change in height between the initial and final positions (1.95 m - 1.45 m = 0.5 m).

Plugging in the values, we have v_final = √(2 * 9.8 * 0.5) = √9.8 ≈ 3.13 m/s.

The change in momentum Δp = m(v_final - v_initial) = 0.2 kg * (3.13 - 0) m/s = 0.626 kg·m/s.

Now, we can calculate the average force using F = Δp/Δt = 0.626 kg·m/s / 0.08 s ≈ 7.825 N.

Therefore, the average force the floor exerts on the ball is approximately 7.825 Newtons, in the opposite direction of the velocity.

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