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Unformatted text preview: homework 06 KIM, JI Due: Oct 10 2007, 4:00 am 1 Question 1, chap 10, sect 1. part 1 of 2 10 points A 3006 caliber hunting rifle fires a bullet of mass 0 . 00601 kg with a velocity of 737 m / s to the right. The rifle has a mass of 6 . 09 kg. What is the recoil speed of the rifle as the bullet leaves the rifle? Correct answer: 0 . 727319 m / s (tolerance 1 %). Explanation: From conservation of momentum p = 0 0 = p b + p r or 0 = m b v b + m r v r . Hence V = m b v b m r = (0 . 00601 kg) (737 m / s) (6 . 09 kg) = 0 . 727319 m / s . Question 2, chap 10, sect 1. part 2 of 2 10 points If the rifle is stopped by the hunters shoul der in a distance of 1 . 59 cm, what is the magnitude of the average force exerted on the shoulder by the rifle? Correct answer: 101 . 307 N (tolerance 1 %). Explanation: The average force is F av = p r t . The time of contact is t = 2 s v r = 0 . 0437222 s , since the average velocity from the guns initial velocity ( v = v r ) to when it stops ( v = 0) is v = v r 2 assuming constant deacceleration. Then the average force is F av = m r v r t = (6 . 09 kg) (0 . 727319 m / s) (0 . 0437222 s) = 101 . 307 N . Question 3, chap 10, sect 2. part 1 of 2 10 points Three spherical masses are located in a plane at the positions shown in the figure be low. A has mass 13 . 8 kg, B has mass 9 . 26 kg, and C has mass 19 . 7 kg. 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 A B C y Distance(m) x Distance (m) Three Masses in a Plane Figure: Drawn to scale. Calculate the xcoordinate of the center of mass. Correct answer: 3 . 85875 m (tolerance 3 %). Explanation: Basic Concepts: x cm = i m i x i i m i (1) y cm = i m i y i i m i (2) There is no acceleration, so a x = a y = 0 . homework 06 KIM, JI Due: Oct 10 2007, 4:00 am 2 Solution: Let: x a = 0 . 5 m , y a = 3 . 5 m , m a = 13 . 8 kg , x b = 7 . 5 m , y b = 4 . 5 m , m b = 9 . 26 kg , x c = 4 . 5 m , y c = 9 . 5 m , m c = 19 . 7 kg . x cm = i m i x i i m i (1) = m a x a + m b x b + m c x c m a + m b + m c = (13 . 8 kg) (0 . 5 m) + (9 . 26 kg) (7 . 5 m) (13 . 8 kg) + (9 . 26 kg) + (19 . 7 kg) + (19 . 7 kg) (4 . 5 m) (13 . 8 kg) + (9 . 26 kg) + (19 . 7 kg) = (165 m kg) (42 . 76 kg) = 3 . 85875 m . Question 4, chap 10, sect 2. part 2 of 2 10 points Calculate the ycoordinate of the center of mass. Correct answer: 6 . 48082 m (tolerance 3 %). Explanation: y cm = i m i y i i m i (1) = m a y a + m b y b + m c y c m a + m b + m c = (13 . 8 kg) (3 . 5 m) + (9 . 26 kg) (4 . 5 m) (13 . 8 kg) + (9 . 26 kg) + (19 . 7 kg) + (19 . 7 kg) (9 . 5 m) (13 . 8 kg) + (9 . 26 kg) + (19 . 7 kg) = (277 . 12 m kg) (42 . 76 kg) = 6 . 48082 m ....
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This note was uploaded on 05/05/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics, Mass, Work

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