A rubber ball (mass 0.21 kg) is dropped from a height of 1.7 m onto the floor. Just after bouncing from the floor, the ball has a speed of 3.6 m/s.

(a) What is the magnitude and direction of the impulse imparted by the floor to the ball?
Magnitude?
Direction?

(b) If the average force of the floor on the ball is 19 N, how long is the ball in contact with the floor?

banana

To find the magnitude and direction of the impulse imparted by the floor to the ball, we can use the principle of conservation of momentum. The impulse is equal to the change in momentum of the ball.

(a) To find the magnitude of the impulse, we need to calculate the change in momentum of the ball.

The change in momentum (Δp) can be calculated using the formula:

Δp = m * Δv

where m is the mass of the ball and Δv is the change in velocity of the ball.

In this case, the mass of the ball is 0.21 kg, and the change in velocity is from 0 m/s (when it's dropped) to 3.6 m/s (after bouncing). So,

Δv = 3.6 m/s - 0 m/s = 3.6 m/s

Δp = (0.21 kg) * (3.6 m/s) = 0.756 kg·m/s

Therefore, the magnitude of the impulse imparted by the floor to the ball is 0.756 kg·m/s.

To determine the direction of the impulse, we consider the direction of the change in velocity. In this case, the ball was dropped from a height and it bounced back upward, so the change in velocity is in the opposite direction to the initial velocity.

Therefore, the direction of the impulse imparted by the floor to the ball is downward.

(b) To find the time the ball is in contact with the floor, we can use the formula:

Average force (F) = Δp / Δt

where Δt represents the time of contact.

Rearranging the formula, we get:

Δt = Δp / F

Substituting the given values, we have:

Δt = (0.756 kg·m/s) / (19 N) = 0.04 seconds

Therefore, the ball is in contact with the floor for 0.04 seconds.