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Unformatted text preview: kaplan (hmk378) – Homework 13 – 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 . 6 kg can of soup is thrown upward with a velocity of v 2 = 4 . 5 m / s. It is immediately struck from the side by an m 1 = 0 . 48 kg rock traveling at v 1 = 7 . 1 m / s. The rock ricochets off at an angle of α = 52 ◦ with a velocity of v 3 = 6 . 4 m / s. What is the angle of the can’s motion after the collision? Correct answer: 72 . 3932 ◦ . 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) (7 . 1 m / s) (0 . 48 kg) (6 . 4 m / s) cos52 ◦ = 1 . 51669 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) (4 . 5 m / s) (0 . 48 kg) (6 . 4 m / s) sin52 ◦ = 4 . 77923 kg m / s . Thus tan β = m 2 v 4 sin β m 2 v 4 cos β = (4 . 77923 kg m / s) (1 . 51669 kg m / s) = 3 . 15109 , and β = arctan(3 . 15109) = 72 . 3932 ◦ . 002 (part 2 of 2) 10.0 points With what speed does the can move immedi ately after the collision? Correct answer: 3 . 13383 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) (7 . 1 m / s) (1 . 6 kg) cos72 . 3932 ◦ (0 . 48 kg) (6 . 4 m / s) cos(52 ◦ ) (1 . 6 kg) cos(72 . 3932 ◦ ) = 3 . 13383 m / s or using equation (2) above, v 4 = m 2 v 2 m 1 v 3 sin α m 2 sin β = (4 . 5 m / s) sin(72 . 3932 ◦ ) (0 . 48 kg) (6 . 4 m / s) sin(52 ◦ ) (1 . 6 kg) sin(72 . 3932 ◦ ) = 3 . 13383 m / s . 003 10.0 points Two ice skaters approach each other at right angles. Skater A has a mass of 46 . 4 kg and travels in the + x direction at 2 . 38 m / s. Skater B has a mass of 99 . 7 kg and is moving in the + y direction at 1 . 88 m / s. They collide and cling together. Find the final speed of the couple. kaplan (hmk378) – Homework 13 – Weathers – (17104) 2 Correct answer: 1 . 48904 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 (110 . 432 kg m / s) 2 + (187 . 436 kg m / s) 2 46 . 4 kg + 99 . 7 kg = 1 . 48904 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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