HW6 Solution

# HW6 Solution - choi(hc5878 – hw6 – Demkov –(59910 1...

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Unformatted text preview: choi (hc5878) – hw6 – Demkov – (59910) 1 This print-out should have 54 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider an inertial reference frame. In this coordinate system 1. there is no inertia. 2. nothing can accelerate. 3. inertial mass depends on your elevation. 4. objects with no forces acting on them have no acceleration. correct Explanation: Solution: Inertial reference frames are coordinate sys- tems in which the law of inertia holds; i.e., objects will not accelerate unless acted upon by a net force. 002 10.0 points Who first proposed the concept of inertia? 1. They came up with the concept of inertia about the same time. 2. Galileo a few years earlier than Newton 3. Newton a few years earlier than Galileo 4. Galileo after Newton was born 5. Galileo before Newton was even born cor- rect Explanation: Isaac Newton was born in 1642, several months after Galileo died. 003 10.0 points Margie (of mass 44 kg) and Bill (of mass 63 kg), both with brand new roller blades, are at rest facing each other in the parking lot. They push off each other and move in opposite directions, with Margie moving at a constant speed of 14 ft / s . At what speed is Bill moving? Correct answer: 9 . 77778 ft / s. Explanation: Let : m M = 44 kg , m B = 63 kg , and Δ v M = 14 ft / s . By Newton’s third law, Margie (M) and Bill (B) experience the same force. We can write F M = m M a M = m M Δ v M t M and F B = m B a B = m B Δ v B t B . Since F M = F B , m M Δ v M t M = m B Δ v B t B The times are the same, so m M Δ v M = Δ v B m B Δ v B = m M m B Δ v M = 44 kg 63 kg (14 ft / s) = 9 . 77778 ft / s . 004 (part 1 of 4) 10.0 points Tracy (of mass 46 kg) and Tom (of mass 73 kg) are standing at rest in the center of the roller rink, facing each other, free to move. Tracy pushes off Tom with her hands and re- mains in contact with Tom’s hands, applying a constant force for 0 . 7 s. Tracy moves 0 . 45 m during this time. When she stops pushing off Tom, she moves at a constant speed. What is Tracy’s constant acceleration dur- ing her time of contact with Tom? Correct answer: 1 . 83673 m / s 2 . Explanation: Let : d = 0 . 45 m and t = 0 . 7 s . choi (hc5878) – hw6 – Demkov – (59910) 2 Since d = 1 2 at 2 , a = 2 d t 2 = 2 (0 . 45 m) (0 . 7 s) 2 = 1 . 83673 m / s 2 . 005 (part 2 of 4) 10.0 points What is Tracy’s final speed after this con- tact? Correct answer: 1 . 28571 m / s. Explanation: Let : a = 1 . 83673 m / s 2 and t = 0 . 7 s . Δ v = at = (1 . 83673 m / s 2 ) (0 . 7 s) = 1 . 28571 m / s . 006 (part 3 of 4) 10.0 points What force was applied to Tracy during this time?...
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## This note was uploaded on 11/01/2008 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas.

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HW6 Solution - choi(hc5878 – hw6 – Demkov –(59910 1...

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