A super ball of mass 0.1kg is dropped from a height of 3.0m above the floor. It

Question

A super ball of mass 0.1kg is dropped from a height of 3.0m above the floor. It bounces off the table and rises to a height of 2.6m. This is an elastic collision.

a. Calculate the velocity of the ball the instant before it reached ground level, v1.

Hint: You can use the conservation of mechanical energy and the height it is dropped from to get this velocity

b. Now use conservation of energy to calculate the velocity the ball must have after the collision at the instant it leaves the ground to reach a height of 2.6m.

c. These velocities are the initial and final velocities for the collision with the ground. Use them to calculate the change in momentum of the ball. Dont forget the directions are not the same so one will be +, the other -.

Explanation / Answer

Part A)

PE = KE

mgh = .5mv2 (mass cancels)

9.8(3) = .5(v2)

v = -7.67 m/s (Negative since it is going downward)

Part B)

PE = KE

mgh = .5mv2

(9.8)(2.6) = .5(v2)

v = 7.14 m/s

Part C)

Change in momentum = m(delta v)

p = (.1)(7.14 + 7.67)

p = 1.48 kg m/s

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