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Physics I – Exam 2 – Spring 2004 Answer Key Part A – 1. B, 2. C, 3. 41, 4. A 4 points each, no partial credit. B-1 8 Points Potential energy at the top = potential energy at the bottom: m g h = ½ k (-h)2 k = 2 m g / h = 98. N/m Net Force (N) +4.9 B-2 22 Points Straight line for force as shown. Parabolas for energy as shown. Max/min energy at 5 cm Features Force: straight line slope = -98 N/m F = 0 at y = 5 cm Total PE: parabola 0.49 J at y = 0 or y = 10. min value at 5 cm = 0.3675 J same value at y = 10 and y = 0. KE: inverted shape of PE curve 0 J at y = 0 or y = 10. max value at 5 cm = PEmax – PEmin same value at y = 10 and y = 0. 0 y (cm) 5 10 5 10 5 10 -4.9 Total PE (J) 0.49 0.3675 y (cm) 0 KE (J) 0.1225 y (cm) 0 0 Part C Must show work to receive credit. C-1 16 points Can solve this problem in angle units of rev, rev/s, and rev/s2. Conversion from rev to rad is OK but unnecessary. t 2 4 t1 Equation 31: 1 12 1 t 1 0.4 t 1 2 2 Equation 33: 2 12 2 t 2 0.1 t 2 1.6 t 1 2 2 2 1 2 2.0 rev 2.0 t 1 t 1 1s t1 t 2 5 t1 5s 2 C-2 18 points Equation 26: (0 12 m v 2 ) (m g h 0) F d 1 (80)(8) 2 (66) 2352.655 J 2 2352.655 h 3.0 m (80)(9.8) mgh C-3 20 points Use conservation of angular momentum. It is in the +Z direction. The bug: l r p (0.1)(0.003)( 2.0) k 6.0 10 4 k kg m 2 / s (y directed velocity does not contribute to the cross product.) OK to consider instantaneous 0 = vtang/r of bug before landing and then L0 = Ibug 0. I total I turntable I bug 5.97 10 3 (0.003)(0.1) 2 6.00 10 3 kg m 2 L 6.0 10 4 L I total 0.1 rad / s I 6.0 10 3