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# 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 Δ 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 Δ r = 17 . 5 cm = 0 . 175 m . r v F c m The kinetic energy is KE = 1 2 I ω 2 = 1 2 ( mr 2 ) parenleftBig v r parenrightBig 2 = 1 2 mv 2 Angular momentum is conserved, so I i ω i = I f ω 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 = Δ 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 mℓ 2 . Correct answer: 1 . 0225 rad / s....
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TradOldMidterm3_2 - Salamah(jms5723 – Practice Midterm 03...

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