Analytical Mech Homework Solutions 92

# Analytical Mech Homework Solutions 92 - = tan 1 1.2638 =...

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1 tan 1.2638 51.65 φ == D cos 0.110 x vv αα v ′′ D sin 0.140 y v =− D 7.16 From eqn. 7.6.14, 1 sin tan cos θ γ = + 1 and are the scattering angles in the Lab and C.M. frames respectively. From eqn. 7.6.16, for Q = 0, 1 2 m m = sin n 45 1 1 cos 4 ta + D 1 cos sin 4 += and squaring … 22 11 cos cos 1 cos 16 2 θθ ++= 2 5 os cos 0 21 6 +− 2c = 1 5 242 cos .125 .696 4 −± + ± Since 0 2 π << , 1 cos .571 55.2 =≈ D 7.17 From eqn. 7.6.14, 1 sin tan cos = + From eqn. 7.6.18, 1 2 mQm mTm  =−+   1 2 111 1 1 0.3015 444  =   sin n 45 .3015 cos ta = + D .3015 cos sin ( s i n c
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## This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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