PHYS310_homework_02_solutions

PHYS310_homework_02_solutions - 1 2 c v dv dx m-= -max 1 2...

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Homework set 2, due Wednesday, September 7, 11:30 am 1. 2.2 (20 points) (a) dx dx dx dx x x dt dx dt dx = = = ( 29 1 dx x F cx dx m = + o ( 29 1 xdx F cx dx m = + o 2 2 1 1 2 2 cx x F x m = + o ( 29 1 2 2 x x F cx m = + o (b) 1 cx dx x x F e dx m - = = o 1 cx xdx F e dx m - = o ( 29 ( 29 2 1 1 1 2 cx cx F F x e e cm cm - - = - - = - o o ( 29 1 2 2 1 cx F x e cm - = - o (c) ( 29 1 cos dx x x F cx dx m = = o cos F xdx cxdx m = o 2 1 sin 2 F x cx cm = o 1 2 2 sin F x cx cm = o &
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2. 2.5 (30 points). Skip the computer plot for (e). (a) ( 29 3 2 kx F x kx A - + so ( 29 3 4 2 2 2 0 1 1 2 4 x kx kx V x kx dx kx A A = - = - (b) ( 29 ( 29 4 2 2 1 1 2 4 kx T x T V x T kx A = - = - + o o (c) E T = o (d) ( 29 V x has maximum at ( 29 0 m F x 3 2 0 m m kx kx A - = m x A = ± ( 29 4 2 2 2 1 1 1 2 4 4 m kA V x kA kA A = - = If ( 29 m E V x < turning points exist. Turning points @ ( 29 1 0 T x let 2 1 u x = 2 2 1 1 0 2 4 ku E ku A - + = solving for u , we obtain 1 2 2 2 4 1 1 E u A kA = ± - or 1 2 1 2 4 1 1 E x A kA = ± - - 3. 2.6 (10 points) ( 29 x v x x α = = 2 2 3 x x x x = - = - ( 29 2 3 m F x mx x = = - &&
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4. 2.11 (20 points) 3 2 dv dv dx dv c a v v dt dx dt dx m = = = ⋅ = -
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Unformatted text preview: 1 2 c v dv dx m-= -max 1 2 v x v c v dv dx m-=- o 1 2 max 2 c v x m-= -o 1 2 max 2 mv x c = o 5. 2.12 (20 points) Going up: 2 2 x F mg c v = --2 dv a v g kv dx = = - -, 2 2 c k m = 2 v x v vdv dx g kv =- - o (Letting x = starting from ground) ( 29 2 1 ln 2 v v g kv x k-- -= o 2 2 2 kx g kv e g kv-+ = + o 2 2 2 kx g g v v e k k- = +- o Going down: 2 2 x F mg c v = -+ 2 dv v g kv dx = - + 2 v x vdv dx g kv =- + ( 29 2 1 ln 2 v g kv x x k- + =-o 2 2 2 1 kx kx k v e e g--= o 2 2 2 kx kx g g v e e k k- =- o...
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This note was uploaded on 12/07/2011 for the course PHYS 310 taught by Professor Jones,m during the Fall '08 term at Purdue University-West Lafayette.

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PHYS310_homework_02_solutions - 1 2 c v dv dx m-= -max 1 2...

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