Question

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?

magnitude?

direction? upward or downward

Answer 1

Solution:

         From newton's second law of motion, Force acting for a given amount of time changes the momentum of an object is called impulse.

from equation 3 one needs initial and final velocity of the ball to find out impulse. Lets say V_f is final velocity of the ball which is pointing upwards after hitting the floor and positive beause it is upwards and V_i is initial velocity and negative because it is pointing downwards.

The initial and final velocites can be find out from potential and kinetic energies of the ball.

The potential energy of the ball at 1.25m is:

P.E = mgh = 0.120*9.8*1.25 = 1.47kg m^2/s^2. This is the kinetice energy while hits down the floor. Therefore

V_i = -4.95 m/s^2

Applying same equation for final velocity as above for The ball at 0.6m, we got

V_f = 3.43 m/s^2

So the impulse of the ball as it rebounds from the floor is =

Impulse = 1 Kgm/s

Direction of Impulse in this case is upward direction.