problem44_58

# problem44_58 - γ m mM M m m M γ mM M m M m γ = = from...

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44.58: a) For mass m , in Eq. . 1 so and , , ) 23 . 37 ( 2 cm 0 cm 0 0 cm c v v v v v v v v u m - - = = - = For mass . so , 0 , , cm cm v v v v u M M - = = - = b) The condition for no net momentum in the center of mass frame is M m M M m m γ and γ v v where , 0 = + correspond to the velocities found in part (a). The algebra reduces to , ) ( 0 0 M m m γ γ γ β - = where , , cm 0 0 c v c v = = = and the condition for no net momentum becomes = - M m γ γ ) ( 0 0 . ) / ( 1 and , 1 1 , 2 0 0 cm 2 0 0 0 0 c v M m mv v M m m M or γ M M - + = - + = + = c) Substitution of the above expression into the expressions for the velocities found in part (a) gives the relatively simple forms . M m γ v ,v M m M γ v v M γ m + - = + = 0 0 0 0 0 0 After some more algebra, 0 0 0 0 2 2 2 2 2 2 γ γ M γ
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Unformatted text preview: γ m mM M m m M γ , mM M m M m γ + + + = + + + = , from which . γ 2 γ 2 2 mM M m M m M m + + = + γ This last expression, multiplied by , 2 c is the available energy a E in the center of mass frame, so that , 2 ) ( ) ( 2 2 ( 2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 a m E Mc Mc mc ) c )(mγ Mc ( ) (Mc ) (mc )c mMγ M m E + + = + + = + + = which is Eq. (44.9)....
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## This note was uploaded on 04/03/2008 for the course PHYS 2500 taught by Professor Kil during the Spring '08 term at Eastern Michigan University.

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