Question

A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.40 m/s and rebounds with a speed of 1.70 m/s, determine the following. (a) magnitude of the change in the ball's momentum (Let up be in the positive direction.)

Answer 1

Answer:

Change in momentum is 1.1275 kg-m/s

Explanation:

It is given that,

Mass of the ball, m = 274 g = 0.274 kg

It hits the floor and rebounds upwards.

The ball hits the floor with a speed of 2.40 m/s i.e. u = -2.40 m/s  (-ve because the ball hits the ground)

It rebounds with a speed of 1.7 m/s i.e. v = 1.7 m/s (+ve because the ball rebounds in upward direction)

We have to find the change in the ball's momentum. It is given by :

$\Delta p=p_f-p_i$

$\Delta p=m(v-u)$

$\Delta p=0.275\ kg(1.7\ m/s-(-2.4\ m/s))$

$\Delta p=1.1275\ kg-m/s$

So, the change in the momentum is 1.1275 kg-m/s