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Quiz1_solution_2D_2010

Quiz1_solution_2D_2010 - c 2 c 2 = 1.512 so γ p – 1...

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1. (a) Using formula for relativistic addition of velocities, we have (measuring velocities to the right as positive) Velocity of spaceship B relative to A = - 0.6 c - (0.6 c ) 1 - ( - 0.6 c )(0.6 c ) / c 2 = - 1.2 c/(1+ 0.36) = - 0.88 c (moving to left) = ν say (b) Let proper length of spaceship B in its own rest frame be L B Then L B as seen by spaceship A = ( 1 - v 2 / c 2 ) L B =40 m Where v is relative velocity of A and B = -0.88 c from part (a) So L B = 40/ 1 - v 2 / c 2 m = 40 / 1 - (0.88) 2 m = 84.2 m (c) L A as seen by spaceship B = ( 1 - v 2 / c 2 ) L A = 150.0 1 - (0.88) 2 = 71.2 m 2. (a) K.E. of electron = γ e m e c 2 - m e c 2 = ( e - 1) m e c 2 K.E. of proton = p m p c 2 - m p c 2 = ( p - 1) m p c 2 Equating these, we get p - 1 = ( m e m p )( e - 1) Now e = 1 1 - (0.75
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Unformatted text preview: c ) 2 / c 2 = 1.512 so γ p – 1 = (0.511/938.3) 0.512 = 2.288 . 10 -4 This is a very small number so γ p ~ 1, so u p << c Expanding, γ p = 1 1-( u p / c ) 2 = 1 + 1 2 .2.( u p / c ) 2 + ....... So, ( u p / c ) 2 = 2.288.10-4 or u p = 1.512. 10 -2 c (b) Momentum of electron = γ e m e u e Momentum of proton = γ p m p u p Equating these , we get p u p = ( m e m p ) e u e = ( 0.511 938.3 )(1.512)(0.75 c ) = 6.175.10-4 c ( u p / c ) 1-( u p / c ) 2 = 6.175.10-4 This is satisfied to a good approximation if u p = 6.175. 10 -4 c...
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Quiz1_solution_2D_2010 - c 2 c 2 = 1.512 so γ p – 1...

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