Analytical Mech Homework Solutions 91

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

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() 22 6 0 26 2 10 10 p vvv v v ±+ == DDD D ± 0 , so the positive square root is used. p v ′ > 0.9288 p vv ′ = D 0.657 2 p px py v ′′ == = D v 11 2 1 1 .9288 p v v α =−=− D D 0.1853 ′ = D .9288 2 tan 2 2 .9288 2 p p p p v v v v φ = −− D D 1 tan 1.9134 62.41 D cos 0.086 x αα v D sin 0.164 y v =− D 7.15 Conservation of energy: 2 11 1 1 1 4 22 2 4 2 pp p mv  =+ +   DD 2 2 16 3 4 p D From the conservation of momentum eqn of Prob. 7.14: 2 v + 2 16 Subtracting: 2 5 v 02 v + 4 0 24 2 10 10 p v v D ± Using the positive square root, since 0 p v ′ > : 0.7895 p ′ = D 0.558 2 p px py v D v 1 1 2 2 2 31 .75 .7895 16 4 2 p v v =− =− D D 0.1780 ′ = D
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