Physics Exam Problems and Formula Sheet - Prof. Masashi Yamaguchi, Exams of Physics

Download Physics Exam Problems and Formula Sheet - Prof. Masashi Yamaguchi and more Exams Physics in PDF only on Docsity! Name _______________________________________________________________________ 1 Exam #2 Physics I Spring 2005 If you would like to get credit for having taken this exam, we need your name (printed clearly) at the top and section number below. Your name should be at the top of every page. Section # _____ 1 M/R 8-10 (Bedrosian) _____ 2 M/R 10-12 (Washington) _____ 3 M/R 10-12 (Shannon) _____ 4 M/R 12-2 (Bedrosian) _____ 5 M/R 2-4 (Bedrosian) _____ 7 M/R 4-6 (LaGraff) _____ 9 T/F 10-12 (Yamaguchi) _____ 10 T/F 10-12 (Wilke) _____ 11 T/F 12-2 (Korniss) _____ 12 T/F 2-4 (Wilke) _____ 14 M/R 12-2 (Shannon) _____ 15 T/F 12-2 (Yamaguchi) You may not unstaple this exam. Only work written on the same page as the question will be graded. Cheating on this exam will result in an F in the course. Questions Value Score Part A 32 B-1 20 C-1 24 C-2 24 Total 100 Name _______________________________________________________________________ 2 On this exam, please neglect any relativistic and/or quantum mechanical effects. If you don’t know what those are, don’t worry, we are neglecting them! On all multiple-choice questions, choose the best answer in the context of what we have learned in Physics I. On graphing and numerical questions, show all work to receive credit. Part A – Multiple Choice – 32 Points Total (8 at 4 Points Each) For questions 1-4, please refer to the figure below. Two trains, A and B, start at rest at t = 0 in a rail yard and move in one direction following parallel straight tracks for 20 seconds. Both trains experience the same net force as shown in the graph below. Train B has more mass than train A. t (sec)0 F (N) 20 Fmax 0 10 _______1. Which train has the greater displacement from t = 0 to t = 20 seconds? A) Train A. B) Train B. C) Both have the same displacement. D) There is not enough information to decide which one. _______2. Which train has the greater magnitude of (linear) momentum at t = 20 seconds? A) Train A. B) Train B. C) Both have the same magnitude of linear momentum. D) There is not enough information to decide which one. _______3. Which train has the greater kinetic energy at t = 20 seconds? A) Train A. B) Train B. C) Both have the same kinetic energy. D) There is not enough information to decide which one. _______4. Which train has the greater magnitude of angular momentum at t = 20 seconds? A) Train A. B) Train B. C) Both have the same magnitude of angular momentum. D) There is not enough information to decide which one. Name _______________________________________________________________________ 5 Problem C-1 (24 Points) (Put your work and answers on the next page.) A block with a mass of 10. kg is released from rest at a position 0.48 m above the equilibrium position of a spring at time t = 0. It falls straight down and contacts the spring at time A. The block continues moving downward as the spring slows it, until it temporarily comes to rest at time B after compressing the spring 0.020 m from equilibrium. The system is the block and the spring. The spring is massless and obeys Hooke’s Law. Ignore air resistance. Use g = 9.8 N/kg. Take the position of the block at time B as the zero of gravitational potential energy. time A time B V = ? time t = 0 x = 0.02 m Vf = 0 Vi = 0 h = 0.48 m The questions to answer about this problem are on the next page. Please put your work and answers on the next page. Name _______________________________________________________________________ 6 Problem C-1 (24 Points) (Put your work and answers below.) A. Total mechanical energy at time 0 = _________________________________ units ________ B. Total mechanical energy at time A = ________________________________ units ________ C. Speed of the block at time A = _____________________________________ units ________ D. Total mechanical energy at time B = ________________________________ units ________ E. Spring constant = ________________________________________________ units ________ Name _______________________________________________________________________ 7 Problem C-2 (24 Points) Consider the system of two point masses moving in the XY plane as shown below. The mass of particle 1 is m1 = 0.001 kg and its velocity is 20 m/s in the +X direction. The mass of particle 2 is m2 = 0.002 kg and its velocity is 10 m/s in the –X direction. The origin of the coordinate system is at the center of mass of the two-particle system and the coordinates are in meters. Find all three components (X,Y,Z) of the angular momentum of the system about the origin. origin (0.0,0.0) m1 X Y (+0.3,-0.4) (Z out of page) (-0.6,+0.8) v1 v2 m2 Angular Momentum X Component: ____________________________ units ________ Angular Momentum Y Component: ____________________________ units ________ Angular Momentum Z Component: ____________________________ units ________ Physics I – Exam 2 – Spring 2005 Answer Key Part A – 1: A, 2: C, 3: A, 4: D, 5: E, 6: F (2 pts for E), 7: C, 8: D. 4 points each, possible partial credit for 6. B-1 20 Points There is no external torque on the system., so angular momentum is conserved. KE is not conserved. Idisk ωinitial = (Idisk + Iring) ωfinal ωinitial = (Idisk + Iring) ωfinal / Idisk = 30 rad/s Looking For: Trying to use conservation of angular momentum in some form. ω is constant for first 0.1 sec. Correct initial value of ω (30 rad/s). Straight line going down from 0.1 to 0.2 sec . ω is constant from 0.2 to 0.3 sec. Final value of ω = .10 rad/s disk ω (rad/s) t (sec) 10 0.1 0.30.2 0 20 30 Part C Must show work to receive credit. C-1 24 points The only forces are conservative, and so we can use conservation of energy, kinetic plus potential. A. U = m g h = (10) (9.8) (0.5) = 49 J. K = 0. E = U+K = 49 J. B. Same as A; energy is conserved. C. U = m g x = (10) (9.8) (0.02) = 1.96 J. → K = 49–U = 47.04 J. ½ m v 2 = 47.04 J. → v = 3.07 m/s D. Same as A; energy is conserved. E. ½ k x 2 = 49 J. k = 245,000 N/m. C-2 24 points The velocity of each particle has only an X component, and so in taking pr rr × only the Y component of r counts. By right-hand rule, both angular momenta are into the page (–Z). k̂016.0)20)(001.0)(8.0(1 −=−=l r kg m 2 /s k̂008.0)10)(002.0)(4.0(2 −=−=l r kg m 2 /s k̂024.0L 21 −=+= ll rrr kg m 2 /s