A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before and just after impact with the floor is 10 m/s. Determine the magnitude of the impulse delivered to the floor by the steel ball.

Question

Quizplus · Accepted Answer

The impulse delivered to the floor can be calculated using the impulse-momentum theorem:

Impulse = Change in momentum

Before the impact, the momentum of the ball is:

p1 = m*v1 = 0.2 kg * 10 m/s = 2 kg m/s

After the impact, the ball bounces straight up, so its velocity reverses direction. The magnitude of its velocity remains the same (10 m/s), so the momentum of the ball is:

p2 = m*v2 = 0.2 kg * (-10 m/s) = -2 kg m/s

The change in momentum is:

Δp = p2 - p1 = -4 kg m/s

Since the impulse is equal to the change in momentum, the impulse delivered to the floor by the ball is:

Impulse = Δp = -4 kg m/s

Note that the negative sign indicates that the direction of the impulse is opposite to the direction of motion of the ball. The floor exerts an upward force on the ball during the collision, which causes the ball to bounce back up. In accordance with Newton's third law, the ball exerts an equal and opposite force on the floor, which is the impulse delivered to the floor.

A 02-Kg Steel Ball Is Dropped Straight Down onto a Hard