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PHYS-1100 PHYSICS I QUIZ 2 MARCH 23, 2000 ANSWERS PART A. 1. C 2. E 3. C 4. B 5. B 6. D 7. D 8. D PART B. 1. The center of mass must be directly below the pivot, and it also must lie on the axis of symmetry of the racket. Therefore, it lies at the intersection of the two lines drawn on the diagram. 2. a) Yes, more particles will emerge. The momentum before the collision is 10 kg·m/s to the right, and only 6 kg·m/s to the right after the collision. b) and c) Based on (a), 4 kg·m/s to the right is yet to come. See the diagram below. d) No. There is not enough information to determine if kinetic energy is consesrved e) The masses of all the particles involved in the collision are needed. PART C. 1. a) v = R = (2.5 rad/s)(0.20 m) v = 0.50 m/s b) J = pf – pi = mv – mv0 = m(v – v0) = (0.10 kg)(0.50 m/s – 1.5 m/s) = – 0.10 kg·m/s Magnitude of J = 0.10 kg·m/s c) TO THE LEFT d) INTO THE PAGE e) I(total) = I + mR² = 0.0080 kg·m² +(0.10 kg)(0.20 m)² I(total) = 0.012 kg·m² f) L = I(total) 0.012 kg·m²)(2.5 rad/s) L = 0.030 kg·m²/s Or, L = mv0R = (0.10 kg)(1.5 m/s)(0.20 m) gives the same answer. g) K = ½ I(total) ² = ½ (0.012 kg·m²)(2.5 rad/s)² K = 0.0375 J h) Angular momentum is conserved. Lout = Lin (I + mR²) I2 2(I + mR²) /I = L/I = (0.030 kg·m²/s)/(0.008 kg·m²) 2 = 3.75 rad/s