HW #10-2-solutions

# HW #10-2-solutions - pasha(sep635 – HW 10-2 –...

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Unformatted text preview: pasha (sep635) – HW 10-2 – Antoniewicz – (56445) 1 This print-out should have 12 questions. Multiple-choice 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 h 3650 ,- 2950 , 3050 i m / s struck a rock of mass 14 kg and velocity h 250 ,- 245 , 300 i m / s . After the collision, the 5 kg rock’s velocity is h 3400 ,- 2250 , 3550 i m / s . What is the final velocity of the 14 kg rock? Start by finding the x component. Correct answer: 339 . 286 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 ~ p sys ,i = ~ p sys ,f ~ p A,i + ~ p B,i = ~ p A,f + ~ p B,f ⇒ ~ v B,f = m A ~ v A,i + m B ~ v B,i- m A ~ v A,f m B = h 339 . 286 ,- 495 , 121 . 429 i m / s . 002 (part 2 of 5) 4.0 points Find the y component of the same velocity. Correct answer:- 495 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: 121 . 429 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: 4 . 11973 × 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 di ff erence K f- K i . Let’s start by finding the initial kinetic energy. Since K = 1 2 m | ~ v | 2 , we need to know the magnitude of the velocity of each rock first. We find this with the Pythagorean theorem. ~ v A,i = q (3650) 2 + (- 2950) 2 + (3050) 2 m / s = 5597 . 1 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 ~ v A,i 2 + 1 2 m B ~ v B,i 2 = 1 2 (5 kg)(5597 . 1 m / s) 2 + 1 2 (14 kg)(461 . 004 m / s) 2 = 7 . 98064 × 10 7 J ....
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## This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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HW #10-2-solutions - pasha(sep635 – HW 10-2 –...

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