Homework 13

# Homework 13 - Gozick Brandon Homework 13 Due 5:00 pm Inst D...

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Gozick, Brandon – Homework 13 – Due: Oct 27 2006, 5:00 pm – Inst: D Weathers 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd 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 26 kg and travels in the + x direction at 1 . 04 m / s. Skater B has a mass oF 108 kg and is moving in the + y direction at 1 . 86 m / s. They collide and cling together. ±ind the fnal speed oF the couple. Correct answer: 1 . 51262 m / s. Explanation: ±rom conservation oF momentum Δ p = 0 m A v A ˆ ı + m B v B ˆ = ( 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 (27 . 04 kg m / s) 2 + (200 . 88 kg m / s) 2 26 kg + 108 kg = 1 . 51262 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 α β An m 2 = 1 . 6 kg can oF soup is thrown upward with a velocity oF v 2 = 4 . 9 m / s. It is immediately struck From the side by an m 1 = 0 . 66 kg rock traveling at v 1 = 7 . 2 m / s. The rock ricochets o² at an angle oF α = 60 with a velocity oF v 3 = 4 . 5 m / s. What is the angle oF the can’s motion aFter the collision? Correct answer: 58 . 194 . 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 . 66 kg) (7 . 2 m / s) - (0 . 66 kg) (4 . 5 m / s) cos 60 = 3 . 267 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 . 9 m / s) - (0 . 66 kg) (4 . 5 m / s) sin 60 = 5 . 2679 kg m / s . Thus tan β = m 2 v 4 sin β m 2 v 4 cos β = (5 . 2679 kg m / s) (3 . 267 kg m / s) = 1 . 61246 , and β = arctan(1 . 61246) = 58 . 194 . 003 (part 2 oF 2) 10 points With what speed does the can move immedi- ately aFter the collision? Correct answer: 3 . 8742 m / s. Explanation: Using equation (1) above, v 4 = m 1 v 1 - m 1 v 3 cos α m 2 cos β = (0 . 66 kg) (7 . 2 m / s) (1 . 6 kg) cos 58 . 194 - (0 . 66 kg) (4 . 5 m / s) cos(60 ) (1 . 6 kg) cos(58 . 194 ) = 3 . 8742 m / s

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Gozick, Brandon – Homework 13 – Due: Oct 27 2006, 5:00 pm – Inst: D Weathers 2 or using equation (2) above, v 4 = m 2 v 2 - m 1 v 3 sin α m 2 sin β = (4 . 9 m / s) sin(58 . 194 ) - (0 . 66 kg) (4 . 5 m / s) sin(60 ) (1 . 6 kg) sin(58 . 194 ) = 3 . 8742 m / s . keywords: 004 (part 1 of 1) 10 points Assume an elastic collision (ignoring friction and rotational motion). A queue ball initially moving at 4 . 1 m / s strikes a stationary eight ball of the same size and mass. After the collision, the queue ball’s Fnal speed is 1 . 8 m / s at an angle of θ with respect to its original line of motion.
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Homework 13 - Gozick Brandon Homework 13 Due 5:00 pm Inst D...

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