A 0.35kg ball is dropped from rest at a height of 8.30m. Every time it hits the

Question

A 0.35kg ball is dropped from rest at a height of 8.30m. Every time it hits the floor, it loses 11% of its energy. a.) How fast is the ball going Just before it hits the ground the first time? b.) How fast does it rebound the first time? c.) What is the impulse delivered to the ball on the first bounce? d.) If the duration of this first bounce is 3.6ms, what is the average force on the ball during the first bounce? e.) How high does the ball get (max) after the third bounce? f.) Let?s call the force that robs the ball of its energy at each bounce the unelastic force. How much work has the unelastic force done after 9 bounces?

Explanation / Answer

Part A)

From conservation of energy

PE = KE

mgh = .5mv2 (mass cancels)

(9.8)(8.3) = .5(v)2

12.8 m/s

Part B)

Since it loses 11% of its energy...

KE = .89(.5)(.35)(12.8)2 = 25.3 J

25.3 = .5mv2

25.3 = .5(.35)(v2)

v = 12.0 m/s

Part C)

Impulse = m(delta v)

Impulse = (.35)(12 + 12.5)

Impulse = 8.59 Ns

Part D)

Impulse = Ft

8.59 = F(.0036)

F = 2385 N

Part E)

We lose another 11%, so

KE = .89(25.3) = 22.5 J

22.5 = mgh

22.5 = (.35)(9.8)(h)

h = 6.56 m

Part F

Work = change in KE

Our initial KE = 28.5 J

29.5(.89) = 25.3(.89) = 22.6(.89) = 20.1(.89) = 17.9(.89) = 15.9(.89) = 14.1(.89) = 12.6(.89) = 11.2(.89) = 9.97 J

The change is 9.97 - 28.5 = -18.5 J (Its negative since its taking energy away - Ignore the negative if your answer key wants you to)

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