Question

A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change in the ball's momentum occurs when the ball hits the ground

Answer 1

Answer:

$\Delta p=-1.56\ kg-m/s$

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

$u=\sqrt{2gh}$

h = 1.8 m  

$u=\sqrt{2\times 9.8\times 1.8}$

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

$v=\sqrt{2gh'}$

h' = 1.4 m  

$v=-\sqrt{2\times 9.8\times 1.4}$

v = -5.23 m/s

The change in the momentum of the ball is given by :

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

$\Delta p=0.14(-5.23-5.93)$

$\Delta p=-1.56\ kg-m/s$

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.