Homework_13 - Nguyen, Thanh Homework 13 Due: Oct 19 2007,...

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Unformatted text preview: Nguyen, Thanh Homework 13 Due: Oct 19 2007, 7:00 pm Inst: D Weathers 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Two ice skaters approach each other at right angles. Skater A has a mass of 32 . 6 kg and travels in the + x direction at 1 . 48 m / s. Skater B has a mass of 46 kg and is moving in the + y direction at 1 . 25 m / s. They collide and cling together. Find the final speed of the couple. Correct answer: 0 . 954972 m / s. Explanation: From conservation of momentum p = 0 m A v A + m B v B = ( m A + m B ) v f Therefore v f = p ( m A v A ) 2 + ( m B v B ) 2 m A + m B = p (48 . 248 kg m / s) 2 + (57 . 5 kg m / s) 2 32 . 6 kg + 46 kg = 0 . 954972 m / s keywords: 002 (part 1 of 2) 10 points m 1 1 m 2 before after v v v 2 v m 1 m 2 3 4 An m 2 = 1 . 7 kg can of soup is thrown upward with a velocity of v 2 = 5 . 6 m / s. It is immediately struck from the side by an m 1 = 0 . 72 kg rock traveling at v 1 = 8 . 6 m / s. The rock ricochets off at an angle of = 46 with a velocity of v 3 = 4 . 1 m / s. What is the angle of the cans motion after the collision? Correct answer: 60 . 7552 . Explanation: Basic Concepts: Conservation of Mo- mentum p before = p after . Solution: Horizontally m 1 v 1 + m 2 (0) = m 1 v 3 cos + m 2 v 4 cos , so that m 2 v 4 cos = m 1 v 1- m 1 v 3 cos (1) = (0 . 72 kg)(8 . 6 m / s)- (0 . 72 kg)(4 . 1 m / s)cos46 = 4 . 14137 kg m / s . Vertically m 1 (0) + m 2 v 2 = m 1 v 3 sin + m 2 v 4 sin , so that m 2 v 4 sin = m 2 v 2- m 1 v 3 sin (2) = (1 . 7 kg)(5 . 6 m / s)- (0 . 72 kg)(4 . 1 m / s)sin46 = 7 . 39651 kg m / s . Thus tan = m 2 v 4 sin m 2 v 4 cos = (7 . 39651 kg m / s) (4 . 14137 kg m / s) = 1 . 78601 , and = arctan(1 . 78601) = 60 . 7552 . 003 (part 2 of 2) 10 points With what speed does the can move immedi- ately after the collision? Correct answer: 4 . 98647 m / s. Explanation: Using equation (1) above, v 4 = m 1 v 1- m 1 v 3 cos m 2 cos = (0 . 72 kg)(8 . 6 m / s) (1 . 7 kg)cos60 . 7552 - (0 . 72 kg)(4 . 1 m / s)cos(46 ) (1 . 7 kg)cos(60 . 7552 ) = 4 . 98647 m / s Nguyen, Thanh Homework 13 Due: Oct 19 2007, 7:00 pm Inst: D Weathers 2 or using equation (2) above, v 4 = m 2 v 2- m 1 v 3 sin m 2 sin = (5 . 6 m / s) sin(60 . 7552 )- (0 . 72 kg)(4 . 1 m / s)sin(46 ) (1 . 7 kg)sin(60 . 7552 ) = 4 . 98647 m / s . keywords: 004 (part 1 of 1) 10 points Assume an elastic collision (ignoring friction and rotational motion)....
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Homework_13 - Nguyen, Thanh Homework 13 Due: Oct 19 2007,...

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