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Unformatted text preview: Phys 131 Midterm, Oct 31 st , 2008 PHYS 131 MIDTERM October 31 st , 2008 The exam comprises two parts: 8 short-answer questions, and 4 problems. Calculators are allowed, as well as a formula sheet (one-side of an 8 x 11 sheet) of your own making. Answer all the short-answer questions with a few words or a phrase, but be concise, please! For the problems, your grade will be calculated with the best three problems . Show your work. The short answer problems are worth two points each, and the problems are worth 10 points each. Put all answers in the red and white answer booklets provided; you may keep this exam. Good luck ! Short answer questions (answer all): you should not need to do any calculations for these questions. Answer in a few words, a short phrase, or a simple sketch. 1) [2 pts] a) a system of particles is known to have zero kinetic energy. What can you say about the momentum of the system? b) A system of particles is known to have zero momentum. What can you say about the kinetic energy of the system? Solution: a) If the KE is zero, then all speeds are zero, so the momentum is zero. b) if the momentum is zero, this could be (but is not necessarily) the result of zero speeds. So in case b) there is no general statement that can be made. 2) [2 pts] You drop a ball from a high balcony and it falls freely (and theres no air resistance). Does the balls kinetic energy increase by equal amounts in equal times, or by equal amounts in equal distances? Explain. Solution: during the fall, the PE is converted to KE. PE is equal to mgh where h is the height, so the amount of energy going into KE is increasing by equal amounts for equal heights. This is not the same as equal amounts in equal times, because the fall distance d is given by d= gt 2 . 3) [2 pts] Two drivers are on a highway at the same speed, side by side. At the same instant, they each see an obstacle on the highway and start to brake (ignore reaction times). Driver 1 hits the brakes, locks the wheels, and skids to a stop. Driver 2 hits the brakes and applies the brakes to the verge of locking, but so that the wheels never lock up. Which driver stops in the shorter distance? Solution: Driver 1 is using kinetic friction to stop (his wheels are skidding). Driver 2 is using static friction, and s > k , so driver 2 will stop in a shorter distance....
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