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Analytical Mech Homework Solutions 91

# Analytical Mech Homework Solutions 91 - 2v 2v 2 60v 2 v 2...

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( ) 2 2 2 2 60 2 62 10 10 p v v v v v ± + ′ = = D D D D ± 0 , so the positive square root is used. p v ′ > 0.9288 p v v ′ = D 0.657 2 p px py v v v = = = D v ( ) ( ) 1 1 2 2 2 2 2 1 1 .9288 2 2 p v v v v α = = D D 0.1853 v v α ′ = D .9288 2 tan 2 2 .9288 2 p p p p v v v v v v φ = = = D D 1 tan 1.9134 62.41 φ = = D cos 0.086 x v v α α v φ = = D sin 0.164 y v v α α v φ = − = − D 7.15 Conservation of energy: 2 2 2 1 1 1 1 1 4 2 2 2 4 2 p p p p p m v m v m v m v α = + + D D 2 2 2 2 16 3 4 p v v v α = D From the conservation of momentum eqn of Prob. 7.14: 2 2 2 p p v v v v v α = + D D 2 16 Subtracting: 2 2 2 5 p p v v v 0 2 v = − + D D ( ) 2 2 2 2 40 2 42 10 10 p v v v v v ± + ′ = = D D D D ± Using the positive square root, since
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