exam2b_solns - TAM 212, Spring, 2004, Exam 2B solutions 1....

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Unformatted text preview: TAM 212, Spring, 2004, Exam 2B solutions 1. Just before to just after the collision, e is defined by ( ) ( ) 1 1 2 2 R P P R v v v v e = . Using work energy leading up to the collision, ( ) gH v gH m v m R R R R 2 2 / 1 1 2 1 = = , and after the collision, back to when the ram is at rest, ( ) ( ) gH v gH m v m R R R R 5 / 2 2 / 1 2 2 2 = = (where the positive and negative signs are chosen for the velocities according to down, positive). Conservation of linear momentum during the collision gives: 2 2 1 4 P R R mv mv mv + = ( 1 = P v ). So, 2 4 5 / 2 4 5 / 2 2 P mv gH m gH m gH m = = 5 / 2 2 gH v P = , and ( ) ( ) ( ) ( ) 5 / 2 2 5 / 2 5 / 2 1 1 2 2 = = = gH gH gH v v v v e R P P R 2. Using R O B F O B R F R ' / ' / / r v v v + + = v , with i v v F R = / ; ( ) ( ) j i v + = 2 2 / / B R B R v ; / ' = F O v ; k B B = v , and ( ) j i r + = d P O ' , ( ) ( ) ( ) ( )( ) ( ) i j j i j i k j i i + + = + + + =...
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This note was uploaded on 04/09/2008 for the course TAM 212 taught by Professor Keane during the Spring '08 term at University of Illinois at Urbana–Champaign.

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