hw 21 - homework 21 HETTIGER, CHRISTOF Due: Mar 20 2008,...

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Unformatted text preview: homework 21 HETTIGER, CHRISTOF 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 1250 kg have in order to have the same mo- mentum as a 2270 kg pickup truck traveling at 28 m / s to the east? Correct answer: 50 . 848 m / s (tolerance 1 %). Explanation: Let : m 1 = 1250 kg , m 2 = 2270 kg , and v 2 = 28 m / s to the east . vectorp = m 1 vectorv 1 = m 2 vectorv 2 v 1 = m 2 v 2 m 1 = (2270 kg) (28 m / s) 1250 kg = 50 . 848 m / s to the east. Question 2, chap 10, sect 1. part 1 of 1 10 points A(n) 88 . 7 kg fisherman jumps from a dock into a 123 . 4 kg rowboat at rest on the east side of the dock. Note: Assume the boats interaction with the water is frictionless. If the velocity of the fisherman is 4 . 57 m / s to the east as he leaves the dock, what is the final speed of the fisherman and the boat? Correct answer: 1 . 91117 m / s (tolerance 1 %). Explanation: Let east be positive: Let : m 1 = 88 . 7 kg , m 2 = 123 . 4 kg , and v i, 1 = 4 . 57 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 = (88 . 7 kg) (4 . 57 m / s) 88 . 7 kg + 123 . 4 kg = 1 . 91117 m / s , which is 1 . 91117 m / s to the east. Question 3, chap 10, sect 1. part 1 of 1 10 points A(n) 67 kg student holding a 2 . 1 kg physics textbook stands in the center of a frozen pond of radius 16 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 7 . 6 m / s towards the northern shore of the pond.. How long does it take him to reach the southern shore? Correct answer: 67 . 1679 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 books velocity, we can find the students 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 = 2 . 1 kg 67 kg 7 . 6 m / s = 0 . 238209 m / s . homework 21 HETTIGER, CHRISTOF Due: Mar 20 2008, 4:00 am 2 The distance from the ponds center to the shore is the radius R = 16 m. So the time it takes the student to slide through this dis-...
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hw 21 - homework 21 HETTIGER, CHRISTOF Due: Mar 20 2008,...

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