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Unformatted text preview: bhakta (vsb255) – HW 10 – coker – (58245) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 2 kg steel ball strikes a wall with a speed of 14 . 9 m / s at an angle of 49 . 5 ◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 1 4 . 9 m / s 2 kg 1 4 . 9 m / s 2 kg 49 . 5 ◦ 49 . 5 ◦ If the ball is in contact with the wall for . 277 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 139 . 737 N. Explanation: Let : M = 2 kg , v = 14 . 9 m / s , and θ = 49 . 5 ◦ . The y component of the momentum is un changed. The x component of the momentum is changed by Δ P x = − 2 M v cos θ . Therefore, using impulse formula, F = Δ P Δ t = − 2 M v cos θ Δ t = − 2 (2 kg) (14 . 9 m / s) cos 49 . 5 ◦ . 277 s bardbl vector F bardbl = 139 . 737 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign. 002 10.0 points A 0.35 kg soccer ball approaches a player horizontally with a velocity of 16 m / s to the north. The player strikes the ball and causes it to move in the opposite direction with a velocity of 24 m / s. What impulse was delivered to the ball by the player? Correct answer: − 14 kg · m / s. Explanation: Let north be positive: Let : m = 0 . 35 kg , v i = 16 m / s , and v f = − 24 m / s . Δ vectorp = mvectorv f − mvectorv i − (0 . 35 kg)(16 m / s) = − 14 kg · m / s , which is directed to the south. 003 (part 1 of 2) 10.0 points A railroad car of mass 42700 kg moving at 3 . 9 m / s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1 . 89 m / s. What is the speed of the three coupled cars after the collision? Correct answer: 2 . 56 m / s. Explanation: Given : m = 42700 kg , v 1 = 3 . 9 m / s , and v 2 = 1 . 89 m / s . By conservation of momentum, mv 1 + (2 m ) v 2 = (3 m ) v f bhakta (vsb255) – HW 10 – coker – (58245) 2 v f = mv 1 + 2 mv 2 3 m = (42700 kg) (3 . 9 m / s) 3 (42700 kg) + 2 (42700 kg) (1 . 89 m / s) 3 (42700 kg) = 2 . 56 m / s . 004 (part 2 of 2) 10.0 points How much kinetic energy is lost in the colli sion? Correct answer: 57504 . 1 J. Explanation: The change in kinetic energy is Δ KE = KE f − KE i = 1 2 (3 m ) v 2 f − bracketleftbigg 1 2 mv 2 1 + 1 2 (2 m ) v 2 2 bracketrightbigg = 3 2 (42700 kg) (2 . 56 m / s) 2 − 1 2 (42700 kg) (3 . 9 m / s) 2 − (42700 kg) (1 . 89 m / s) 2 = − 57504 . 1 J , which represents a loss of 2 . 56 m / s . 005 (part 1 of 2) 10.0 points Given two masses, m 1 = 4 kg and m 2 = 6 kg, m 1 is moving with a velocity v 1 colliding with m 2 , which is suspended by a string of length 8 m....
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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