A basketball with a mass of 0.61 kg falls vertically to the floor where it hits with a velocity of −7 m/s. (We take the positive direction to be upward here.) The ball rebounds, leaving the floor with a velocity of 4.6 m/s. (Indicate the direction with the signs of your answers.)

a. What impulse act on the ball during its collision with the floor?

b. If the ball is in contact with the floor for 0.04 s, what is the average force of the ball on he floor?

The answers to the question are

a. The impulse acting on the ball during its collision with the floor is

7.076 kg·m/s

b. The average force of the ball on the floor with contact of 0.04 s is

176.9 N

Explanation:

Mass of the ball = 0.61 kg

Initial velocity = -7 m/s

Final velocity = 4.6 m/s

Impulse = Δp = FΔt = mΔv

Where Δt =change in time

m = mass of Average force of he the ball

Δv = velocity change

Therefore impulse = m×(Final velocity - initial velocity)

= 0.61 × (4.6-(-7)) = 0.61×11.6

= 7.076 kg·m/s

b. Average force of the ball on the floor is given by

×Δt = mΔv = 7.076 kg·m/s

where Δt = change in time = 4.6 m/s

Therefore =mΔv /Δt = 7.076/0.04 = 176.9 kg·m/s²

= 176.9 N

b

explanation: