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# solution19_pdf - arellano (aa39398) – Homework 13 –...

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Unformatted text preview: arellano (aa39398) – Homework 13 – weathers – (17101) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points m 1 1 m 2 before after v v v 2 v m 1 m 2 3 4 α β An m 2 = 1 . 6 kg can of soup is thrown upward with a velocity of v 2 = 6 . 7 m / s. It is immediately struck from the side by an m 1 = 0 . 49 kg rock traveling at v 1 = 8 . 4 m / s. The rock ricochets off at an angle of α = 59 ◦ with a velocity of v 3 = 4 . 7 m / s. What is the angle of the can’s motion after the collision? Correct answer: 71 . 4793 ◦ . 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 α + m 2 v 4 cos β , so that m 2 v 4 cos β = m 1 v 1- m 1 v 3 cos α (1) = (0 . 49 kg) (8 . 4 m / s)- (0 . 49 kg) (4 . 7 m / s) cos59 ◦ = 2 . 92987 kg m / s . Vertically m 1 (0) + m 2 v 2 = m 1 v 3 sin α + m 2 v 4 sin β , so that m 2 v 4 sin β = m 2 v 2- m 1 v 3 sin α (2) = (1 . 6 kg) (6 . 7 m / s)- (0 . 49 kg) (4 . 7 m / s) sin59 ◦ = 8 . 74594 kg m / s . Thus tan β = m 2 v 4 sin β m 2 v 4 cos β = (8 . 74594 kg m / s) (2 . 92987 kg m / s) = 2 . 9851 , and β = arctan(2 . 9851) = 71 . 4793 ◦ . 002 (part 2 of 2) 10.0 points With what speed does the can move immedi- ately after the collision? Correct answer: 5 . 7648 m / s. Explanation: Using equation (1) above, v 4 = m 1 v 1- m 1 v 3 cos α m 2 cos β = (0 . 49 kg) (8 . 4 m / s) (1 . 6 kg) cos71 . 4793 ◦- (0 . 49 kg) (4 . 7 m / s) cos(59 ◦ ) (1 . 6 kg) cos(71 . 4793 ◦ ) = 5 . 7648 m / s or using equation (2) above, v 4 = m 2 v 2- m 1 v 3 sin α m 2 sin β = (6 . 7 m / s) sin(71 . 4793 ◦ )- (0 . 49 kg) (4 . 7 m / s) sin(59 ◦ ) (1 . 6 kg) sin(71 . 4793 ◦ ) = 5 . 7648 m / s . 003 10.0 points Two ice skaters approach each other at right angles. Skater A has a mass of 66 . 4 kg and travels in the + x direction at 1 . 02 m / s. Skater B has a mass of 94 kg and is moving in the + y direction at 2 . 3 m / s. They collide and cling together. Find the final speed of the couple. arellano (aa39398) – Homework 13 – weathers – (17101) 2 Correct answer: 1 . 41247 m / s. Explanation: From conservation of momentum ∆ p = 0 m A v A ˆ ı + m B v B ˆ = ( m A + m B ) v f Therefore v f = radicalbig ( m A v A ) 2 + ( m B v B ) 2 m A + m B = radicalbig (67 . 728 kg m / s) 2 + (216 . 2 kg m / s) 2 66 . 4 kg + 94 kg = 1 . 41247 m / s 004 10.0 points A uniform flat plate of metal with a circular hole is situated in the reference frame shown in the figure below....
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## This note was uploaded on 10/31/2010 for the course PHYSICS 1710 taught by Professor Weathers during the Fall '10 term at North Texas.

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solution19_pdf - arellano (aa39398) – Homework 13 –...

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