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8%) Problem 5: A golf ball m=0.185 kg strikes a vertical concrete wall elastically with a horizontal velocity of y = 14 m/s.
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Answer #1

a.

Impulse, I = mv = 0.185 * 14 = 2.59kgm/s

b.

Force, F = \frac{I}{t} = \frac{ 2.59 }{0.1} = 25.9 N

c.

Impulse, I = m(v+u) = 0.185 * (14+7) = 3.88 kgm/s

d.

Kinetic Energyn Lost, KE_{lost} = \frac{1}{2}\left ( v_1^2 - v_2^2 \right ) = \frac{1}{2} * 0.185 ( 14^2 - 7^2) = 13.6 J

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