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ph1a_07_finals - Ph1a Final Exam Solution Chefung Chan Fall...

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Ph1a Final Exam Solution Chefung Chan, Fall 2007 Problem 1: Stay and Sway (10 points) (a) (3 points) Notice the diagram uses capital M while the question uses m . Here uses m for the whole question. By conservation of momentum: m 1 (0) - m 2 v ˆ x = ( m 1 + m 2 ) ~v 0 ~v 0 = m 1 m 1 + m 2 v ( - ˆ x ) (b) (3 points) By Hooke’s law, F = - kx m ¨ x = - kx ¨ x = - k m 1 + m 2 x Above equation is simple harmonic oscillation in x with ω 2 0 = k m 1 + m 2 . Therefore, m 2 = k ω 2 0 - m 1 (c) (4 points) We know x = A sin ( ω 0 t )+ B cos ( ω 0 t ). We are going to match with the initial conditions to find A and B . x ( t = 0) = 0 B = 0 ˙ x ( t = 0) = v 0 = - m 1 m 1 + m 2 v ω 0 A = - m 1 m 1 + m 2 v A = - m 1 m 1 + m 2 v ω 0 Problem 2: Misdirected Meteor (8 points) (a) (2 points) First, we want to find the initial speed of the meteor. Since the meteor is in a circular orbit that the gravitational force on the meteor from the planet is equal to the centripetal force. mv 2 i r = GMm r 2 v 2 i = GM r = GM R + 3 R v i = r GM 4 R 1
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After the 50% kinetic energy loss, K f = 1 2 K i 1 2 mv 2 f = 1 4 mv 2 i v 2 f = 1 2 v 2 i = GM 8 R v f = r GM 8 R K f = 1 2 mv 2 f = GMm 16 R U f = U i = - GMm 4 R L f = m (4 R ) v f = m 2 GMR (b) (2 points) Total energy E = U f + K f = - 3 GMm 16 R < 0. The new orbit is either circular or elliptical. Then with
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