A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?

Before we find impulse, we need to find the initial and final momentum of the ball.

To find the momentum of the ball before it hit the floor, we need to figure out its final velocity using kinematics.

Values we know:

acceleration (a) - 9.81m/s^2 [down]

initial velocity (vi) - 0m/s

distance (d) - 1.25m [down]

This equation can be used to find final velocity:

Vf^2 = Vi^2 + 2ad

Vf^2 = (0) ^2 + (2) (-9.81) (-1.25)

Vf^2 = 24.525

Vf = 4.95m/s [down]

Now we need to find the velocity the ball leaves the floor at using the same kinematics concept.

What we know:

a = 9.81m/s^2 [down]

d = 0.600m [up]

vf = 0m/s

Vf^2 = Vi^2 + 2ad

0^2 = Vi^2 + 2 (-9.81) (0.6)

0 = Vi^2 + - 11.772

Vi^2 = 11.772

Vi = 3.43m/s [up]

Now to find impulse given to the ball by the floor we find the change in momentum.

Impulse = Momentum final - momentum initial

Impulse = (0.120) (3.43) - (0.120) (-4.95)

Impulse = 1.01kgm/s [up]

Here are the given values:

mass (m) = 0.120kginitial velocity (Vo) = 0distance traveled (s) = 1.25 m

We first calculate for the final velocity of the ball:

Vf^2 = 2gs + V o^2

V f^2 = 2 (9.8m / s2) (1.25m)

Vf = 4.95 m/s

Impulse = m (Vf-Vo)

Impulse = 0.120 (4.95)

Impulse = 0.59 Ns