Analytical Mech Homework Solutions 93

# Analytical Mech Homework Solutions 93 - P1 7.18...

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φ P 1 ψ 7.18 Conservation of momentum: P 1 () 11 2 cos cos PP P φ ψφ ′′ =+ − 12 0s i n s i n ψ =− − From Appendix B for ( ) sin α β + and ( ) cos + : P 2 2 cos cos cos sin sin P =+ + i n s i n c o s c o s s i n =− 22 2 2 2 cos cos cos 2cos cos P 2 2 sin sin sin sin + + 2 2 cos cos cos sin sin ++ 2 2 2 0 sin sin cos 2sin cos cos sin cos sin + 2 2s i n s i n c o s c o s s i n −− Adding: 222 11 2 1 2 2c o s P P P =++ Conservation of energy: PPP Q mmm = ( ) 1 2 o s QP P P P P mm = cos Q m = 7.19 2 1 1 2 Tm = 1 1 v 2 1 2 v 1 = let 2 2 Tv r == … ratio of scattered particle to incident particle energy Looking at Figure 7.6.2 … ( 1 1 cm cm vv vv vv ⋅= − ⋅ − ) 2 1 o s cm cm vvv v v 1 =+− hence 2 1 2 cm cm v v v γ =−+ where 1 cos = 2 2 1 1 2 1 2 cm cm v v r v but ′ = …the center of mass speeds of the incident and scattered particle are the same. 121 1 vm vmm …from equation 7.6.12 where
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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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