PHY302K-Sai-HW7-solutions

# PHY302K-Sai-HW7-solutions - Oslund(hno56 Homework 7...

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Oslund (hno56) – Homework 7 – sai – (58015) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Three particles are located in a coordinate system as shown in the fgure below. The acceleration oF gravity is 9 . 8m / s 2 . The mass 7 kg is located at the origin, the mass 5 kg is located on the x coordinate, and the mass 7 kg is located on the y coordinate. x y 7 kg 7 kg 5 kg 7m ±ind the distance oF the center oF gravity From the origin. Correct answer: 3 . 32913 m. Explanation: The center oF gravity oF the x and y compo- nents From the origin respectively are x c = m 0 0+ m 1 x 1 m 0 + m 1 + m 2 = (5 kg) (8 m) (7 kg) + (5 kg) + (7 kg) =2 . 10526 m and y c = m 0 m 2 y 2 m 0 + m 1 + m 2 = (7 kg) (7 m) (7 kg) + (5 kg) + (7 kg) . 57895 m . Then the distance oF the center oF gravity From the origin is d = ± x 2 c + y 2 c = ± (2 . 10526 m) 2 + (2 . 57895 m) 2 =3 . 32913 m . 002 10.0 points Giselle is travelling on her bicycle at a speed oF 15 mph when a car passes her. ±rom her Frame oF reFerence, she estimates that the car is going 45 mph toward her and 45 mph away From her. What is the speed oF the car From the Frame oF reFerence oF someone standing on the ground? Correct answer: 60 mph. Explanation: Since Griselle is traveling in the same direc- tion as the car, it is moving more slowly in his reFerence Frame than the ground’s by 15 mph. The speed oF the car in the reFerence Frame oF the ground is 45 mph. 003 (part 1 oF 4) 10.0 points A 3 kg block is traveling in the negative x direction at 4 . 3m / s, and a 7 . 5 kg block is traveling in the positive x direction at 3 m / s. ±ind the speed oF the center oF mass. Correct answer: 0 . 914286 m / s. Explanation: Let : m 1 = 3 kg , m 2 =7 . 5 kg , v 1 =( - 4 . / s) ˆ ı, and v 2 = (3 m / s) ˆ ı. ±or the center oF mass, ² i m i \$v i = M \$v cm = m 1 1 + m 2 2 m 1 + m 2 ) cm cm = m 1 1 + m 2 2 m 1 + m 2 = (3 kg) ( - 4 . / s) ˆ ı (3 kg) + (7 . 5 kg) + (7 . 5 kg) (3 m / s) ˆ ı (3 kg) + (7 . 5 kg) = (0 . 914286 m / s) ˆ ı . 004 (part 2 oF 4) 10.0 points

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Oslund (hno56) – Homework 7 – sai – (58015) 2 Find the velocity of each block in the center- of-mass reference frame. 1. ± \$u 1 =( - 4 . 3m / s) ˆ ı and 2 = (4 . / s) ˆ ı 2. ± 1 - / s) ˆ ı and 2 = (3 m / s) ˆ ı 3. ± 1 - 5 . 21429 m / s) ˆ ı and 2 = (2 . 08571 m / s) ˆ ı correct 4. ± 1 = (2 . 08571 m / s) ˆ ı and 2 = (2 . 08571 m / s) ˆ ı 5. ± 1 = (4 . / s) ˆ ı and 2 - / s) ˆ ı 6. ± 1 = (2 . 08571 m / s) ˆ ı and 2 - 5 . 21429 m / s) ˆ ı 7. ± 1 - 5 . 21429 m / s) ˆ ı and 2 - 5 . 21429 m / s) ˆ ı 8. ± 1 = (3 m / s) ˆ ı and 2 - 4 . / s) ˆ ı Explanation: In the center of mass reference frame, the velocity of the 3 kg block is 1 = \$v 1 - cm - 4 . / s) ˆ ı - (0 . 914286 m / s) ˆ ı - 5 . 21429 m / s) ˆ ı, and of the 7 . 5 kg block, 2 = 2 - cm = (3 m / s) ˆ ı - (0 . 914286 m / s) ˆ ı = (2 . 08571 m / s) ˆ ı . 005 (part 3 of 4) 10.0 points Find the velocity of each block after the colli- sion in the center-of-mass reference frame. Assume: The collision is head on. 1. ± \$ u ± 1 = (5 . 21429 m / s) ˆ ı and \$ u ± 2 - 2 . 08571 m / s) ˆ ı correct 2. ± \$ u ± 1 = (5 . 21429 m / s) ˆ ı and \$ u ± 2 = (5 . 21429 m / s) ˆ ı 3.
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## This note was uploaded on 02/22/2012 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas.

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PHY302K-Sai-HW7-solutions - Oslund(hno56 Homework 7...

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