Question

A 0.14 kg ball is dropped from a height and hits the ground at a velocity of -9 m/s (the negative sign shows the downward direction of the ball's motion). After

touching the ground for a short period of time, the ball bounces back at a velocity of +8.5 m/s. What is the impulse on the ball due to the floor?

Answer 1

Impulse = Δp = 2.45 kg m/s

Step-by-step explanation:

Impulse = change in momentum = final momentum - initial momentum

The initial momentum of the ball before hitting the ground is given by:

m * v_i = 0.14 kg * -9 m/s = -1.26 kg m/s

The final momentum of the ball after bouncing back is given by:

m * v_f = 0.14 kg * 8.5 m/s = 1.19 kg m/s

So the change in momentum is:

Δp = v_f - v_i = 1.19 kg m/s - (-1.26 kg m/s) = 2.45 kg m/s

Impulse = Δp = 2.45 kg m/s