A superball with a mass m = 61:6 g is dropped from a height h = 1.76m. It hits t

Question

A superball with a mass m = 61:6 g is dropped from a height h = 1.76m. It hits the floor and then rebounds to a new height that is 88.5% of the initial height. Use upward as the positive direction.

(a) What is the momentum change of the ball during its collision with the floor? (kg m/s)

(b) Suppose you drop a ball of putty of the same mass from the same height and the putty sticks to the floor. What then would be the second ball's momentum change during its collision with the floor? (kg m/s)

The correct answers are (a) 0.702, (b) 0.362

Although i have the answers I would like to understand how to do the problem.

Explanation / Answer

mass = 0.0616 (convert) to kg)

The velocity of the ball at the bottom when dropped from 1.764m can be found with laws of conservation of energy.

PE = KE
mgh = 1/2mv^2
gh = 1/2v^2
square root 2gh = v
v = -5.88 m/s

(negative because it is falling down)

Now do the same to find the velocity of the ball as it leaves the floor and rebounds to a height of

1.764m*0.885=.1.56114

PE = KE
mgh = 1/2mv^2
square root 2gh = v
v = 5.53439850390266m/s

Change in momentum = m*change in velocity

= 0.0616(5.53-(-5.88))

=0.702856kg x m/s

lly the 2nd one

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