CW9 - Earth (mass = 5 . 98 10 24 kg) moving with an orbital...

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boland (kbb544) – CW9 – mackie – (10611) 1 This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 4) 10.0 points Calculate the magnitude oF the linear momen- tum For each oF the Following cases: A proton with mass 1 . 67 × 10 - 27 kg moving with a speed oF 5 . 23 × 10 6 m / s. Correct answer: 8 . 7341 × 10 - 21 kg · m / s. Explanation: Given : m = 1 . 67 × 10 - 27 kg and v = 5 . 23 × 10 6 m / s . Momentum is Vp = mVv , so p = ( 1 . 67 × 10 - 27 kg ) ( 5 . 23 × 10 6 m / s ) = 8 . 7341 × 10 - 21 kg · m / s . 002 (part 2 of 4) 10.0 points A 14 . 2 g bullet moving with a speed oF 322 m / s. Correct answer: 4 . 5724 kg · m / s. Explanation: Given : m = 14 . 2 g and v = 322 m / s . p = (14 . 2 g) (322 m / s) · 1 kg 1000 g = 4 . 5724 kg · m / s . 003 (part 3 of 4) 10.0 points A 76 . 6 kg sprinter running with a speed oF 12 . 9 m / s . Correct answer: 988 . 14 kg · m / s. Explanation: Given : m = 76 . 6 kg and v = 12 . 9 m / s . p = (76 . 6 kg) (12 . 9 m / s) = 988 . 14 kg · m / s . 004 (part 4 of 4) 10.0 points
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Unformatted text preview: Earth (mass = 5 . 98 10 24 kg) moving with an orbital speed equal to 26208 . 9 m / s . Correct answer: 1 . 56729 10 29 kg m / s. Explanation: Given : m = 5 . 98 10 24 kg and v = 26208 . 9 m / s . p = ( 5 . 98 10 24 kg ) (26208 . 9 m / s) = 1 . 56729 10 29 kg m / s . 005 10.0 points A billiard ball traveling at 4.2 m/s has an elastic head-on collision with a billiard ball oF equal mass that is initially at rest. The frst ball is at rest aFter the collision. What is the speed oF the second ball aFter the collision? Correct answer: 4 . 2 m / s. Explanation: Basic Concept: m 1 Vv 1 ,i = m 2 Vv 2 ,f since v 2 ,i = 0 m/s and v 1 ,f = 0 m/s. Given: v 1 ,i = 4 . 2 m / s m 1 = m 2 Solution: v 2 ,f = m 1 v 1 ,i m 2 = v 1 ,i = 4 . 2 m / s...
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