problem08_98

University Physics with Modern Physics with Mastering Physics (11th Edition)

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8.98: The two fragments are 3.00 kg and 9.00 kg. Time to reach maximum height = time to fall back to the ground. s. 5 . 12 s m 8 . 9 0 . 55 sin ) s m 150 ( 0 sin 2 0 9 = - = - θ = ο t t gt v v The heavier fragment travels back to its starting point, so it reversed its velocity. 86.0 55 cos ) s m (150 cos 0 = = θ = ο v v x s m to the left after the explosion; this is . 9 v Now get 3 v using momentum conversation. ( 29 ( 29 ( 29 ( 29 ( 29 s m 602 s m 0 . 86 kg 00 . 9 kg 00 . 3 s m 0 . 86 kg 12 3 3 9 9 3 3 0 = - + = + = v v v m v m Mv ) 5 . 12 )( s m 602 ( ) s 5 . 12 )( s m 0 . 86 ( explosion After explosion
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Unformatted text preview: Before 3 + = + = x x x launched it was where from m 8600 3 = x Energy released = Energy after explosion Energy before explosion J 10 33 . 5 ) s m . 86 )( kg . 12 ( 2 1 ) s m . 86 )( kg 00 . 9 ( 2 1 ) s m 602 )( kg 00 . 3 ( 2 1 ) ( 2 1 m 2 1 m 2 1 5 2 2 2 2 9 9 3 2 9 9 2 3 3 =-+ = +-+ = v m m v v...
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This document was uploaded on 02/04/2008.

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