A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m (v-u) ... Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = - 1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5 - (-1.5) ]

I = 2.4 (2.5+1.5)

I = 2.4 (4)

I = 9.6 Ns