A ball with a mass of 0.49 kg is dropped on a table. The initial height of the b

Question

A ball with a mass of 0.49 kg is dropped on a table.
The initial height of the ball is 0.61 meters.
After the ball hits the table it rebounds to a height of 0.43 meters.
The time that the ball is in contact with the table is 0.023 seconds.
Note that this is not a purely elastic or inelastic collision.

A/ Calculate the change in the momentum of the ball during its collision with the table.

B/ Calculate the average force acting on the ball during its collision with the table.

C/ Calculate the change in kinetic energy of the ball during its collision with the table.

Explanation / Answer

Part A)

For the initial velocity at the impact...

vf2 = vo2 + 2ad

vf2 = (0) + 2(9.8)(.61)

v = 3.46 m/s downward

Then for the rebound velocity...

vf2 = (0) + 2(9.8)(.43)

v = 2.90 upward

Change in momemtum = m(delta v)

p = .49(2.9 - (-3.46))

p = 3.12 kg m/s

Part B)

F(t) = p

F(.023) = 3.12

F = 135.5 N

Part C)

Change in KE = .5mv2 - .5mv2

KE = .5(.49)(3.462 - 2.92)

KE = .864 J

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