A superball with a mass m = 61.6 g is dropped from a height h = 1.56 m. It hits

Question

                 A superball with a mass m = 61.6 g is dropped from a height h = 1.56 m. 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 (kg m/s)  of the ball during its collision with the floor?                 
                (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                  A superball with a mass m = 61.6 g is dropped from a height h = 1.56 m. 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 (kg m/s)  of the ball during its collision with the floor?                 
                (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

Explanation / Answer

mass = 0.0616 (convert) to kg)
height attended after the bounce = 0.885 * 1.56 = 1.3806 m
The velocity of the ball at the bottom when dropped from 1.56m can be found with laws of conservation of energy.

PE = KE
mgh = 1/2mv^2
gh = 1/2v^2
square root 2gh = v = 2 * 9.8 * 1.56
v = -5.53 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 1.38m.

PE = KE
mgh = 1/2mv^2
square root 2gh = v
v = 5.2 m/s

Change in momentum = m*change in velocity
= 0.0616(5.2-(-5.53)) =0.66.96 kg x m/s

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