S08 AM261 MT S

S08 AM261 MT S - Phys 263/Amath 261 Midterm SOLUTIONS 1....

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Unformatted text preview: Phys 263/Amath 261 Midterm SOLUTIONS 1. (a) To find the CM coordinates use R = 1 M X k m k r k (1) = 1 m 1 + m 2 ( m 1 r 1 + m 2 r 2 ) (2) = 1 m 1 + m 2 2 t 2 m 1 + (1 + t 2 ) m 2 , 3 tm 1 , 4 m 1 + 4 t 2 m 2 (3) (b) V = R = 1 m 1 + m 2 [4 tm 1 + 2 tm 2 , 3 m 1 , 8 tm 2 ] (4) and A = V = 1 m 1 + m 2 [4 m 1 + 2 m 2 , , 8 m 2 ] (5) (c) The linear momentum is P = M V = [8 , 3 , 16] (6) 2. To calculate the velocity and position requires two integrations, and solving of the two integration constants via initial conditions. (a) To get the velocity e- t = m v = m dv dt (7) dv dt = 1 m e- t (8) v ( t ) = 1 m- 1 e- t t (9) v ( t ) =- 1 m ( e- t- 1 ) (10) (b) Integrate again to get the position v = dx dt (11) x ( t ) =- 1 m- 1 e- t- t t (12) x ( t ) = 1 m 1 e t + t- 1 (13) 1 (c) As t approaches infinity: v ( t ) 1 ( e- + 1 ) = 1 (14) (d) The velocity we are looking for is v ( t ) = 1- e- 1 (15) 1- e t/ 2 = 1- e- 1 (16) which implies t = 2 (17)...
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This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.

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S08 AM261 MT S - Phys 263/Amath 261 Midterm SOLUTIONS 1....

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