hw10 - karna (pk4534) – HW 10 – coker – (58245) 1...

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Unformatted text preview: karna (pk4534) – HW 10 – coker – (58245) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 5 kg steel ball strikes a wall with a speed of 10 . 5 m / s at an angle of 50 . 2 ◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 1 . 5 m / s 5 kg 1 . 5 m / s 5 kg 50 . 2 ◦ 50 . 2 ◦ If the ball is in contact with the wall for . 184 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 365 . 28 N. Explanation: Let : M = 5 kg , v = 10 . 5 m / s , and θ = 50 . 2 ◦ . 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 (5 kg) (10 . 5 m / s) cos 50 . 2 ◦ . 184 s bardbl vector F bardbl = 365 . 28 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign. 002 10.0 points A 0.41 kg soccer ball approaches a player horizontally with a velocity of 17 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: − 16 . 81 kg · m / s. Explanation: Let north be positive: Let : m = 0 . 41 kg , v i = 17 m / s , and v f = − 24 m / s . Δ vectorp = mvectorv f − mvectorv i − (0 . 41 kg)(17 m / s) = − 16 . 81 kg · m / s , which is directed to the south. 003 (part 1 of 2) 10.0 points A railroad car of mass 12800 kg moving at 3 . 96 m / s collides and couples with two cou- pled railroad cars, each of the same mass as the single car and moving in the same direc- tion at 2 . 23 m / s. What is the speed of the three coupled cars after the collision? Correct answer: 2 . 80667 m / s. Explanation: Given : m = 12800 kg , v 1 = 3 . 96 m / s , and v 2 = 2 . 23 m / s . By conservation of momentum, mv 1 + (2 m ) v 2 = (3 m ) v f karna (pk4534) – HW 10 – coker – (58245) 2 v f = mv 1 + 2 mv 2 3 m = (12800 kg) (3 . 96 m / s) 3 (12800 kg) + 2 (12800 kg) (2 . 23 m / s) 3 (12800 kg) = 2 . 80667 m / s . 004 (part 2 of 2) 10.0 points How much kinetic energy is lost in the colli- sion? Correct answer: 12769 . 7 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 (12800 kg) (2 . 80667 m / s) 2 − 1 2 (12800 kg) (3 . 96 m / s) 2 − (12800 kg) (2 . 23 m / s) 2 = − 12769 . 7 J , which represents a loss of 2 . 80667 m / s . 005 (part 1 of 2) 10.0 points Given two masses, m 1 = 18 kg and m 2 = 12 kg, m 1 is moving with a velocity v 1 collid- ing with m 2 , which is suspended by a string of length 7 m....
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This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw10 - karna (pk4534) – HW 10 – coker – (58245) 1...

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