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Unformatted text preview: roofner (bar784) – Homework 12 – Weathers – (17104) 1 This printout should have 12 questions. Multiplechoice 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 . 4 kg can of soup is thrown upward with a velocity of v 2 = 6 m / s. It is immediately struck from the side by an m 1 = 0 . 48 kg rock traveling at v 1 = 8 . 3 m / s. The rock ricochets off at an angle of α = 51 ◦ with a velocity of v 3 = 5 . 8 m / s. What is the angle of the can’s motion after the collision? Correct answer: 70 . 3081 ◦ . 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 . 48 kg) (8 . 3 m / s) (0 . 48 kg) (5 . 8 m / s) cos51 ◦ = 2 . 23197 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 . 4 kg) (6 m / s) (0 . 48 kg) (5 . 8 m / s) sin51 ◦ = 6 . 23642 kg m / s . Thus tan β = m 2 v 4 sin β m 2 v 4 cos β = (6 . 23642 kg m / s) (2 . 23197 kg m / s) = 2 . 79413 , and β = arctan(2 . 79413) = 70 . 3081 ◦ . 002 (part 2 of 2) 10.0 points With what speed does the can move immedi ately after the collision? Correct answer: 4 . 7313 m / s. Explanation: Using equation (1) above, v 4 = m 1 v 1 m 1 v 3 cos α m 2 cos β = (0 . 48 kg) (8 . 3 m / s) (1 . 4 kg) cos70 . 3081 ◦ (0 . 48 kg) (5 . 8 m / s) cos(51 ◦ ) (1 . 4 kg) cos(70 . 3081 ◦ ) = 4 . 7313 m / s or using equation (2) above, v 4 = m 2 v 2 m 1 v 3 sin α m 2 sin β = (6 m / s) sin(70 . 3081 ◦ ) (0 . 48 kg) (5 . 8 m / s) sin(51 ◦ ) (1 . 4 kg) sin(70 . 3081 ◦ ) = 4 . 7313 m / s . 003 10.0 points Two ice skaters approach each other at right angles. Skater A has a mass of 74 . 4 kg and travels in the + x direction at 2 . 6 m / s. Skater B has a mass of 101 kg and is moving in the + y direction at 0 . 853 m / s. They collide and cling together. Find the final speed of the couple. Correct answer: 1 . 20729 m / s. roofner (bar784) – Homework 12 – Weathers – (17104) 2 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 (193 . 44 kg m / s) 2 + (86 . 153 kg m / s) 2 74 . 4 kg + 101 kg = 1 . 20729 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 document was uploaded on 10/26/2011 for the course PHYS 1710 at North Texas.
 Fall '08
 Weathers
 mechanics, Work

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