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Unformatted text preview: madrid (tmm2353) – HW 102 – Antoniewicz – (56445) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 5) 4.0 points In outer space, a rock whose mass is 5 kg and whose velocity was ( 3950 , − 2850 , 3200 ) m / s struck a rock of mass 13 kg and velocity ( 240 , − 285 , 280 ) m / s . After the collision, the 5 kg rock’s velocity is ( 3550 , − 2300 , 3500 ) m / s . What is the final velocity of the 13 kg rock? Start by finding the x component. Correct answer: 393 . 846 m / s. Explanation: Let’s go ahead and do the vector algebra for all three components of this velocity. We use conservation of momentum for this problem. If the first rock is A and the second is B , we can write vectorp sys ,i = vectorp sys ,f vectorp A,i + vectorp B,i = vectorp A,f + vectorp B,f ⇒ vectorv B,f = m A vectorv A,i + m B vectorv B,i − m A vectorv A,f m B = ( 393 . 846 , − 496 . 538 , 164 . 615 ) m / s . 002 (part 2 of 5) 4.0 points Find the y component of the same velocity. Correct answer: − 496 . 538 m / s. Explanation: See the solution to part 1. 003 (part 3 of 5) 4.0 points Find the z component of the same velocity. Correct answer: 164 . 615 m / s. Explanation: See the solution to part 1. 004 (part 4 of 5) 4.0 points What is the change in internal energy of the rocks? Correct answer: 8 . 18125 × 10 6 J. Explanation: To find the change in internal energy, we need to know both the initial kinetic energy and the final kinetic energy and then take the difference K f − K i . Let’s start by finding the initial kinetic energy. Since K = 1 2 m  vectorv  2 , we need to know the magnitude of the velocity of each rock first. We find this with the Pythagorean theorem. vextendsingle vextendsingle vectorv A,i vextendsingle vextendsingle = radicalBig (3950) 2 + ( − 2850) 2 + (3200) 2 m / s = 5827 . 95 m / s . The magnitude of rock B ’s velocity is found the same way. So the initial kinetic energy is K i = K A,i + K B,i = 1 2 m A vextendsingle vextendsingle vectorv A,i vextendsingle vextendsingle 2 + 1 2 m B vextendsingle vextendsingle vectorv B,i vextendsingle vextendsingle 2 = 1 2 (5 kg)(5827 . 95 m / s) 2 + 1 2 (13 kg)(466 . 074 m / s) 2 = 8 . 63245 × 10 7 J ....
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This note was uploaded on 02/05/2012 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner

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