PHY 138 - Chapter 6 Solutions

PHY 138 - Chapter 6 Solutions - Problem 6.1 The momentum of...

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Problem 6.1 The momentum of the ball as it hits the floor is i i i i gh m p mgh m p 2 2 2 = = . Similarly, the momentum of the ball when it leaves the floor is f f f f gh m p mgh m p 2 2 2 = = The impulse given to the ball by the floor is then () () () () s N 39 . 1 m 960 . 0 m 25 . 1 s m 81 . 9 2 kg 150 . 0 2 2 2 2 = + = + = = = Δ i f i f i f h h g m gh m gh m p p p . Problem 6.3 (a) The momentum of the proton is () s N 10 35 . 8 s m 10 00 . 5 kg 10 67 . 1 21 6 27 × = × × = = mv p . (b) The momentum of the bullet is () s N 50 . 4 s m 300 kg 0150 . 0 = = = mv p (c) The momentum of the runner is () s N 750 s m 0 . 10 kg 0 . 75 = = = mv p .
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(d) The momentum of the earth is () s N 10 78 . 1 s m 10 98 . 2 kg 10 98 . 5 29 4 24 × = × × = = mv p Problem 6.5 (a) The speed V of the baseball is () () s m 0 . 31 s m 10 50 . 1 kg 145 . 0 kg 003 . 0 3 = × = = = v M m V MV mv . (b) The kinetic energy of the ball is () J 7 . 69 s m 0 . 31 kg 145 . 0 2 1 2 1 2 2 = = = MV K ball The kinetic energy of the bullet is () J 3375 s m 10 50 . 1 kg 003 . 0 2 1 2 1 2 3 2 = × = = mv K bullet Problem 6.6 The average force exerted on the ball is found by using the Momentum-Impulse Theorem
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() () F 1680 s 0.002 s m 61 kg
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PHY 138 - Chapter 6 Solutions - Problem 6.1 The momentum of...

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