Solutions_Quiz2_PHYS2D_2010

Solutions_Quiz2_PHYS2D_2010 - (1) (a) Applying conservation...

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(1) (a) Applying conservation of relativistic momentum we have γ 1 m 1 (0.4 c ) = g 2 m 2 (0.6 c ) where 1 = 1 1- (0.4) 2 , g 2 = 1 1- (0.6) 2 , so m 1 = 1- (0.429 2 1- (0.629 2 0.6 0.4 μ 2 = 1.72 2 (1) Conservation of total energy gives 1 m 1 c 2 + g 2 m 2 c 2 = Mc 2 or 1 1 - (0.429 2 1 + 1 1- (0.629 2 2 = Μ so 1.09 m 1 + 1.25 2 = Using Eq. (1) we get 3.12 m 2 = so m 2 = 0.32 and m 1 = 1.72(0.32 29 = 0.55 (b) Fraction of mass converted to K.E. = ( 1 – 0.55 – 0.32) = 0.13 (c) K.E. of m 1 = ( 1 -129 1 χ 2 = (1.09 -129 (0.5529 Μχ 2 = 0.05 2 K.E. of m 2 = ( 2 -129 2 2 = (1.25 -129 (0.3229 2 = 0.08 2 Total K.E. = 0.13 Mc 2 as expected.
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(2) We have for an electron moving in a circular orbit perpendicular to a magnetic field, p = θΒρ = 1.6.10 -19 .10 -3 .2 κγ . μ . σ -1 = 3.2.10 -22 5.344.10 -22 Μες / χ = 0.6 / Total energy of electron is p 2 c 2 + 2 4 = (0.629 2 +
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This note was uploaded on 02/16/2010 for the course PHYS 2D 2D taught by Professor Sinha during the Winter '10 term at UCSD.

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Solutions_Quiz2_PHYS2D_2010 - (1) (a) Applying conservation...

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