A rubber ball, mass m = 0.83 kg, is dropped from a height of 2.0 meters above th

Question

A rubber ball, mass m = 0.83 kg, is dropped from a height of 2.0 meters above the ground. The initial speed of the ball is zero. Assume no air resistance. Obtain the kinetic energy of the ball just before it hits the ground. The ball bounces back up to just 1.8 meters. How much mechanical energy is lost during the collision of the ball and the ground? What is the speed of the ball just after it leaves the ground? Obtain the change in momentum of the ball. (Clearly define your coordinate system!) If we assume the force of the ground on the ball is ~ constant, and the impact time was 0.025 seconds, obtain the acceleration of the ball during impact. (Clearly define your coordinate system!)

Explanation / Answer

A) KE_ball = change in Potential energy = m*g*h = 0.83*9.81*2 = 16.3 J

B) lost in mechanical energy is m*g*(h1-h2) = 0.83*9.81*(2-1.8) = 1.63 J

C) v = sqrt(2*g*h) = sqrt(2*9.81*1.8) = 5.94 m/s

D) change in momentum of the ball is m*v = 0.83*5.94 = 4.93 kg*m/s

E) Favg*dt = change in momentum = 4.93

Favg= 4.93/0.025 = 197.2 N

Favg = m*a = 197.2

a = 197.2/0.83 = 237.6 m/s^2

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