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prelim2006 - Table of Contents Complete Partially Complete...

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Table of Contents Complete Partially Complete Not Complete Preliminary Examination 2006 2 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Problem 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Problem 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Problem 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Problem 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Problem 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Problem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1
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Preliminary Examination September , 2006 1. Two identical rods of mass M and length R are lying on their sides, motionless, on a frictionless table. At time t = 0 , each is hit by a small clay ball of mass m and velocity v directed perpendicular to the rods. Ball A hits in the center of its rod, while ball B hits near an end. Both clay balls stick to the rods after the collisions. (a) What quantity (or quantities) are conserved in each collision? After the collision, for which ball and rod setup (A or B) does the center of mass travel a distance D first, or can you not determine this from the information given? (b) Now imagine that two such rods, each of mass M and length L , are connected at each end by a spring of equilibrium length d and spring constant k , as shown below. As in part a), a small clay ball of mass m is moving perpendicular to the rods at speed v , and strikes the center of left-hand rod at time t = 0 . If the ball sticks to the rod after the collision, find the subsequent motion of the left-hand rod. 2
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Preliminary Examination September , 2006 2. Consider this integral ˆ −∞ d x e i tx x 2 + 4 = F ( t ) . Carefully work out the explicitly form for F ( t ) , for −∞ <t< . Solution Let f ( z ) e i tz z 2 +4 . Then f ( z ) has singular points at z = ± 2i . We must choose a closed contour in the complex plane that i) includes as some part of the contour the entire real axis and ii) converges. When t < 0 , f ( z ) converges in the lower half plane. To check convergence, take z = i | z | . Then e i tz = e i t ∗− i | z | = e t | z | , which goes to zero along the negative imaginary axis in the lower half plane. So we choose the contour shown: ˛ C 1 e i tz z 2 + 4 d z = 2 π i Res z = 2i ( e i tz z 2 + 4 ) = 2 π i e i t ( 2i) ( 2i) 2i = 2 π i e 2 t 4i = π 2 e 2 t Notice that we have a negative sign because we’re going in a clockwise contour. Also, ¸ C 1 f ( z ) d z = ´ −∞ f ( z ) d z since f ( z ) goes to zero faster than 1 | z | along the bottom semicircle. When t> 0 , f ( z ) converges in the upper half plane. Again, we can check this by taking z out to i along the positive imaginary axis z = | z | i . e i tz = e i t (i | z | ) = e t | z | . So we choose a contour in the upper half plane: ˛ C 2 e i tz z 2 + 4 d z = 2 π i Res z =2i ( e i tz z 2 + 4 ) = 2 π i e i t (2i) (2i) + 2i = 2 π i e 2 t 4i = π 2 e 2 t Again, ¸ C 2 f ( z ) d z = ´ −∞ f ( z ) d z since ´ x f ( z ) d z = 0 . So our final solution is ˆ −∞ e i tx x 2 + 4 d x = π 2 e 2 | t | 3
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Preliminary Examination September , 2006 3. Consider a sphere of radius R characterized by a charge density ρ ( r,θ ) . The potential outside the sphere is found to be ϕ ( r,θ,φ ) = Q 2 r 3 Y 2 , 0 ( θ,φ ) where Q is an experimentally determined constant and Y 2 , 0 is the spherical harmonic Y l,m with l = 2 ,m = 0 : Y 2 , 0 ( θ,φ ) = 1 2 5
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