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 3.30 m/s and rebounds with a speed of 1.60 m/s, determine the following. (a) magnitude of the change in the ball's momentum in kg · m/s (Let up be in the positive direction.)

Answer 1

`\Delta p=1.3475\ kg-m/s`

Step-by-step explanation:

The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-

We represent

The ball mass = m = 275 g = 0.275 kg

Thus it goes to the floor and resurfaces upward.

The ball hits the ground at 3.30 m/s speed that is

u = -3.30 m/s which represents the Negative since the ball hits the ground)

It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)

`\Delta p=p_f-p_i`

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

`\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))`

`\Delta p=1.3475\ kg-m/s`