hw21_s - homework 21 – YOO, HEE – Due: Mar 20 2008,...

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Unformatted text preview: homework 21 – YOO, HEE – Due: Mar 20 2008, 4:00 am 1 Question 1, chap 10, sect 1. part 1 of 1 10 points What velocity must a car with a mass of 1110 kg have in order to have the same mo- mentum as a 2300 kg pickup truck traveling at 26 m / s to the east? Correct answer: 53 . 8739 m / s (tolerance ± 1 %). Explanation: Let : m 1 = 1110 kg , m 2 = 2300 kg , and v 2 = 26 m / s to the east . vectorp = m 1 vectorv 1 = m 2 vectorv 2 v 1 = m 2 v 2 m 1 = (2300 kg) (26 m / s) 1110 kg = 53 . 8739 m / s to the east. Question 2, chap 10, sect 1. part 1 of 1 10 points A(n) 89 kg fisherman jumps from a dock into a 136 . 2 kg rowboat at rest on the east side of the dock. Note: Assume the boat’s interaction with the water is frictionless. If the velocity of the fisherman is 5 . 45 m / s to the east as he leaves the dock, what is the final speed of the fisherman and the boat? Correct answer: 2 . 15386 m / s (tolerance ± 1 %). Explanation: Let east be positive: Let : m 1 = 89 kg , m 2 = 136 . 2 kg , and v i, 1 = 5 . 45 m / s . The boat and fisherman have the same final speed, and v i, 2 = 0 m/s, so m 1 vectorv i, 1 + m 2 vectorv i, 2 = ( m 1 + m 2 ) vectorv f m 1 vectorv i, 1 = ( m 1 + m 2 ) vectorv f v f = m 1 v i m 1 + m 2 = (89 kg) (5 . 45 m / s) 89 kg + 136 . 2 kg = 2 . 15386 m / s , which is 2 . 15386 m / s to the east. Question 3, chap 10, sect 1. part 1 of 1 10 points A(n) 82 kg student holding a 1 . 5 kg physics textbook stands in the center of a frozen pond of radius 25 m. He is unable to walk to the other side because there is no friction between his shoes and the ice. To overcome this difficulty, he uses recoil: he throws the textbook with a horizontal velocity 5 . 8 m / s towards the northern shore of the pond.. How long does it take him to reach the southern shore? Correct answer: 235 . 632 s (tolerance ± 1 %). Explanation: The recoil works by conservation of mo- mentum: Since the ice is frictionless, there are no external horizontal force, hence the net horizontal momentum is conserved, vector P net = const . Initially, there is no motion and hence vector P net = 0. Consequently, after the student throws the book, their net momentum is still zero, P net = M s vectorv s + M b vectorv b = vector . Hence, given the book’s velocity, we can find the student’s velocity as vectorv s = − M b M s vectorv b . Thus, he moves South (in the opposite direc- tion to the book) with speed v s = M b M s × v b = 1 . 5 kg 82 kg × 5 . 8 m / s = 0 . 106098 m / s . homework 21 – YOO, HEE – Due: Mar 20 2008, 4:00 am 2 The distance from the pond’s center to the shore is the radius R = 25 m. So the time it takes the student to slide through this dis-...
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This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw21_s - homework 21 – YOO, HEE – Due: Mar 20 2008,...

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