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Resource Quiz 7

# Resource Quiz 7 - schroeder(cs39963 Resource 7 Momentum...

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schroeder (cs39963) – Resource 7: Momentum, Work and Energy – balasubramanya – (1401310) 1 This print-out should have 40 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A 10 metric ton train moves toward the south at 40 m / s. At what speed must it travel to have five times its original momentum? A metric ton is 1000 kg. Correct answer: 200 m / s. Explanation: Let : v = 40 m / s and p final = 5 p initial . Since momentum is proportional to speed, its speed must be increased to v f = 5(40 m / s) = 200 m / s . 002 (part 2 of 3) 10.0 points At what speed must it travel to have a momentum of 4 × 10 5 kg · m / s? Correct answer: 40 m / s. Explanation: Let : m = 10 metric ton and p = 4 × 10 5 kg · m / s . Since a metric ton is 1000 kg, p = m v v = p m = 4 × 10 5 kg · m / s 10 metric ton · 1 metric ton 1000 kg = 40 m / s . 003 (part 3 of 3) 10.0 points If there were a speed limit for this train as it traveled through a city, but not a weight limit, what mass must be added to the train to slow it down to 30 m / s while at the same time keeping the momentum the same as in the second part? Correct answer: 3333 . 33 kg. Explanation: Let : p = 4 × 10 5 kg · m / s , v = 30 m / s , and m = 10000 kg . p = m v m = p v = 4 × 10 5 kg · m / s 30 m / s = 13333 . 3 kg is the required mass, so we must add m = 13333 . 3 kg 10000 kg = 3333 . 33 kg . 004 (part 1 of 4) 10.0 points Calculate the magnitude of the linear momen- tum for each of the following cases a) a proton with mass 1 . 67 × 10 27 kg mov- ing with a velocity of 6 × 10 6 m / s. Correct answer: 1 . 002 × 10 20 kg · m / s. Explanation: Let : m = 1 . 67 × 10 27 kg and v = 6 × 10 6 m / s . Momentum is vectorp = mvectorv , so p = ( 1 . 67 × 10 27 kg ) ( 6 × 10 6 m / s ) = 1 . 002 × 10 20 kg · m / s . 005 (part 2 of 4) 10.0 points b) a 1 . 6 g bullet moving with a speed of 312 m / s to the right. Correct answer: 0 . 4992 kg · m / s.

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schroeder (cs39963) – Resource 7: Momentum, Work and Energy – balasubramanya – (1401310) 2 Explanation: Let : m = 1 . 6 g and v = 312 m / s . p = (1 . 6 g) (312 m / s) · 1 kg 1000 g = 0 . 4992 kg · m / s . 006 (part 3 of 4) 10.0 points c) a 8 kg sprinter running with a velocity of 13 . 3 m / s. Correct answer: 106 . 4 kg · m / s. Explanation: Let : m = 8 kg and v = 13 . 3 m / s . p = (8 kg) (13 . 3 m / s) = 106 . 4 kg · m / s . 007 (part 4 of 4) 10.0 points d) Earth ( m = 5 . 98 × 10 24 kg) moving with an orbital speed equal to 29400 m / s. Correct answer: 1 . 75812 × 10 29 kg · m / s. Explanation: Let : m = 5 . 98 × 10 24 kg and v = 29400 m / s . p = ( 5 . 98 × 10 24 kg ) (29400 m / s) = 1 . 75812 × 10 29 kg · m / s . 008 10.0 points Two blocks of masses m and M [ M = 3 . 35 m ] are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord holding them together is burned, after which the block of mass M moves to the right with a speed of 2 . 23 m / s. m M m M Before After (a) (b) v What is the speed of the block of mass m ? Correct answer: 7 . 4705 m / s. Explanation: From conservation of momentum Δ p = 0, in our case we obtain 0 = M v M m v m .
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Resource Quiz 7 - schroeder(cs39963 Resource 7 Momentum...

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