solution-HWVI

# solution-HWVI - Version 001 Homework VI belay (20480) 1...

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Unformatted text preview: Version 001 Homework VI belay (20480) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Serway CP 04 02 001 10.0 points A football punter accelerates a 0 . 7 kg football from rest to a speed of 11 m / s in 0 . 24 s. What constant force does the punter exert on the ball? Correct answer: 32 . 0833 N. Explanation: Given : m = 0 . 7 kg , v f = 11 m / s , and t = 0 . 24 s . Applying impulse, vector F t = mvectorv f mvectorv i = mvectorv f since v i = 0 m/s. F = mv f t = (0 . 7 kg) (11 m / s) . 24 s = 32 . 0833 N . Stopping a Car 002 10.0 points A force of 2570 N is needed to bring a car moving at 20 . 7 m / s to a halt in 16 . 4 s. What is the mass of the car? Correct answer: 2036 . 14 kg. Explanation: Basic Concepts: Impulse-momentum re- lationship m v = F t . Solution: The car is brought to a halt, so v = 0 v = v The force is negative, so m ( v ) = ( F ) t m = F t v = F t v . Tipler PSE5 08 63 003 10.0 points A 7 g handball moving with a speed of 10 . 5 m / s strikes a wall at an angle of 51 with the horizontal and then bounces off with the same speed at the same angle. It is in contact with the wall for 4 . 7 ms. What is the magnitude of average force exerted by the ball on the wall? Correct answer: 19 . 683 N. Explanation: Let : m = 7 g , v = 10 . 5 m / s , and = 51 . x y vectorv f m vectorv i m Perpendicular to the wall, vectorv i = ( v cos ) vectorv f = vectorv i Applying the impulse relationship, vector F av on ball t = vectorp = m vectorv vector F av on ball = m vectorv t = m ( vectorv f vectorv i ) t = 2 mv cos t . Version 001 Homework VI belay (20480) 2 The magnitude is F av on ball = 2 mv cos t = 2(0 . 007 kg)(10 . 5 m / s)(cos 51 ) 4 . 7 ms = 19 . 683 N , so the average force exerted by the ball on the wall is also 19 . 683 N by Newtons third law. Collision 01 004 (part 1 of 2) 10.0 points a 20 . 6 g object moving to the right at 29 . 5 cm / s overtakes and collides elastically with a 13 . 3 g object moving in the same di- rection at 7 . 32 cm / s. Find the velocity of the 20 . 6 g object after the collision. Correct answer: 12 . 0962 cm / s. Explanation: Let : m 1 = 20 . 6 g , v 1 = 29 . 5 cm / s , m 2 = 13 . 3 g , and v 2 = 7 . 32 cm / s . For head-on elastic collisions, v 1 i v 2 i = ( v 1 f v 2 f ) and for the relative velocities, v 1 v 2 = v 2 f v 1 f v 2 f = v 1 v 2 + v 1 f . Momentum is conserved, so m 1 v 1 + m 2 v 2 = m 1 v 1 f + m 2 v 2 f m 1 v 1 + m 2 v 2 = m 1 v 1 f + m 2 ( v 1 v 2 + v 1 f ) m 1 v 1 + m 2 v 2 m 2 ( v 1 v 2 ) = v 1 f ( m 1 + m 2 ) v 1 f = m 1 v 1 + m 2 v 2 m 2 ( v 1 v 2 ) m 1 + m 2 = (20 . 6 g)(29 . 5 cm / s) 20 . 6 g + 13 . 3 g + (13 . 3 g)(7 . 32 cm / s) 20 . 6 g + 13 . 3 g (13 . 3 g)(29 . 5 cm / s 7 . 32 cm / s) 20...
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## This note was uploaded on 11/18/2010 for the course PHY 2048 taught by Professor Belay during the Fall '10 term at Florida A&M.

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solution-HWVI - Version 001 Homework VI belay (20480) 1...

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