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Unformatted text preview: Muraj, Hamza Homework 20 Due: Mar 20 2006, 4:00 am Inst: Florin 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points An object having an initial momentum that may be represented by the vector below strikes an object that is initially at rest. Which of the following sets of vectors may represent the momenta of the two objects af- ter the collision? Note carefully: The original vector above and the following vectors are all drawn to the same length scale. 1. 2. 3. 4. 5. correct 6. 7. Explanation: There is no external force for the two-object system, so the total momentum is a constant. From the choices, applying the vector sum- mation for the momenta of the two objects, we can easily identify the correct choice. The figure below shows the sum of the x- and y-components of the vectors which repre- sent the correct answer. initial momentum The horizontal vectors add to be the same length as the vector presented in the question. The vertical vectors cancel, as expected, since there is no vertical momentum. 002 (part 1 of 1) 10 points The bird perched on the swing in the figure has a mass of 44 g, and the base of the swing has a mass of 127 . 6 g. Assume that the swing and the bird are originally at rest and that the bird then takes off horizontally at 2 . 76 m / s. The acceleration of gravity is 9 . 8 m / s 2 . Muraj, Hamza Homework 20 Due: Mar 20 2006, 4:00 am Inst: Florin 2 If the base can swing freely (without fric- tion) around the pivot, how high will the base of the swing rise above its original level? Correct answer: 0 . 0462132 m. Explanation: Let : m b = 44 g , m s = 127 . 6 g , and v b = 2 . 76 m / s . By momentum conservation we can find the speed of the swing when the bird takes off v s = m b m s v b = 44 g 127 . 6 g 2 . 76 m / s = 0 . 951724 m / s . Then, because of energy conservation, the height the swing can reach is h = v 2 s 2 g = (0 . 951724 m / s) 2 2 (9 . 8 m / s 2 ) = . 0462132 m . 003 (part 1 of 1) 10 points A uranium nucleus 238 U may stay in one piece for billions of years, but sooner or later it de- cays into an particle of mass 6 . 64 10- 27 kg and 234 Th nucleus of mass 3 . 88 10- 25 kg, and the decay process itself is extremely fast (it takes about 10- 20 s). Suppose the uranium nucleus was at rest just before the decay....
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.
- Spring '09