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**Unformatted text preview: **PHYS 3202 Classical Mechanics II - Quiz #1 Solutions 1. BD (correction: problem 1 has 5 points; 45 points in total for quiz 1) 2. No 3. AC 4. BCE 5. C 6. BC Problem 7: The potential energy is U ( r ) =- Z r - Gm 2 r 2 dr =- Gm 2 r . In the center of mass frame the problem is easiest, in which the two particles initially have velocities v , and eventually have velocities v after going to infinite distance. By energy conservation, T i + U i = T f + U f , namely 1 2 mv 2 + 1 2 mv 2- Gm 2 r = 1 2 mv 2 + 1 2 mv 2 + 0 . To minimize v , we need v = 0. So v = q Gm r . Some folks thought that this was the final answer for dr dt . It is not. The two particles move in opposite directions with equal speed. So dr dt = 2 v = 2 r Gm r . If one works in a different reference frame, in which particle 1 has ZERO initial speed, and particle 2 has speed r , then the systems final kinetic energy is NONZERO. By momentum conservation, and because the particles have equal mass, the final velocities satisfy v 1 + v 2 = 0 + r . By energy conservation, we also have 1 2 m 2 + 1 2 m r 2- Gm 2 r = 1 2 mv 2 1 + 1 2 mv 2 2 +0 = m 4 ( v 1 + v 2 ) 2 + m 4 ( v 1- v 2 ) 2 = m 4 r 2 + m 4 ( v 1- v 2 ) 2 ....

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