TradOldMidterm3_2 - Salamah (jms5723) Practice Midterm 03...

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Unformatted text preview: Salamah (jms5723) Practice Midterm 03 Yao (58180) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A(n) 74 g particle moving with an initial speed of 24 m / s in the positive x direction strikes and sticks to a(n) 110 g particle moving with 110 m / s in the positive y direction. How much kinetic energy is lost in this collision? Correct answer: 280 . 388 J. Explanation: This is an inelastic collision. Momentum is conserved. m 1 v 1 = ( m 1 + m 2 ) v x m 2 v 2 = ( m 1 + m 2 ) v y Final velocity of both particles is v = radicalBig v 2 x + v 2 y . Energy before collision is K i = 1 2 m 1 v 2 1 + 1 2 m 2 v 2 2 . Energy after collision is K f = 1 2 ( m 1 + m 2 ) v 2 . Energy lost is K = K i- K f . 002 10.0 points The puck in the figure has a mass of 0 . 178 kg. Its original distance from the center of rota- tion is 38 . 7 cm, and the puck is moving with a speed of 84 . 1 cm / s. The string is pulled downward 17 . 5 cm through the hole in the frictionless table. 38 . 7 cm 84 . 1 cm / s . 178 kg What is the magnitude of the work was done on the puck? Treat the hockey puck as a point mass. Correct answer: 0 . 146817 J. Explanation: Given : m p = 0 . 178 kg , r i = 38 . 7 cm = 0 . 387 m , v i = 84 . 1 cm / s = 0 . 841 m / s , and r = 17 . 5 cm = 0 . 175 m . r v F c m The kinetic energy is KE = 1 2 I 2 = 1 2 ( mr 2 ) parenleftBig v r parenrightBig 2 = 1 2 mv 2 Angular momentum is conserved, so I i i = I f f ( mr i 2 ) v i r i = ( mr f 2 ) v f r f v f = v i r i r f Thus the work done on the puck is W = KE = 1 2 mv 2 f- 1 2 mv 2 i = 1 2 m bracketleftbigg v i r i r f bracketrightbigg 2- 1 2 mv 2 i = 1 2 m bracketleftBigg r 2 i r 2 f- 1 bracketrightBigg v 2 i = 1 2 (0 . 178 kg) bracketleftbigg (0 . 387 m) 2 (0 . 212 m) 2- 1 bracketrightbigg (0 . 841 m / s) 2 = 0 . 146817 N Salamah (jms5723) Practice Midterm 03 Yao (58180) 2 which has magnitude of . 146817 N . 003 10.0 points A uniform rod of mass 3 . 4 kg is 14 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 14 m from the center of mass of the rod. Initially the rod makes an angle of 54 with the horizontal. The rod is released from rest at an angle of 54 with the horizontal, as shown in the figure. 14 m 14 m 3 . 4 kg O 54 What is the angular speed of the rod at the instant the rod is in a horizontal position? The acceleration of gravity is 9 . 8 m / s 2 and the moment of inertia of the rod about its center of mass is I cm = 1 12 m 2 . Correct answer: 1 . 0225 rad / s....
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This note was uploaded on 05/02/2011 for the course GE 208K taught by Professor Dutt during the Spring '11 term at University of Texas at Austin.

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TradOldMidterm3_2 - Salamah (jms5723) Practice Midterm 03...

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