hw4 solution

hw4 solution - NR rel NR NR K mc mv mc K mv K K mc K K K β...

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Homework 4 solution 1a) b) E = 2+4 = 6GeV p = 4*2^0.5 GeV/c v/c = pc/E v = 2*2^0.5 *c/3 = 0.943c 2a) b) MeV E E E p pi 4 . 1122 = + = λ 3) 4a) Since F is perpendicular to dr as body moves. So, the velocity is constant in magnitude and changes only in direction. So, 0 = dt d γ , so a m F = b) for F parallel to dr as body moves, a c v dt d 2 3 = 2 2 2 2 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ( )( ) ( )( ) ( )( ) 1 m v c m c v c mc v c c mc mc v c v c E = + = + = + = - - = MeV E c MeV p pc pc pc pc pc pc E E E pi p pi 4 . 184 / 120 ) ( 140 ) 938 2 938 140 1116 ( ) ( 140 ) ( 140 938 2 938 ) ( 1116 ) ( 140 938 ) ( 1116 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 = = = - - - + + + + = + + + = + + = 2 4 2 4 2 2 2 4 2 2 4 2 2 3 1 ..... 2 8 3 1 3 (1 ..... ) ...... 2 8 2 8 1 2 3 8 3 4 3 0.01 0.115 4 rel NR rel
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Unformatted text preview: NR rel NR NR K mc mv mc K mv K K mc K K K β = + + + = + + + = + + =-=-= = = = So, 2 2 3 2 3 2 2 (1 ) mv v F ma a ma ma c c γ = + = + = ur r r r r 5 (a) Lab Frame: Moving Frame: Consider the Velocity Addition Formula: (b) In the Earth Frame, the observer that: , Apply velocity addition formula...
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This note was uploaded on 03/31/2011 for the course PHY 126 taught by Professor Hobunchan during the Spring '11 term at HKUST.

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hw4 solution - NR rel NR NR K mc mv mc K mv K K mc K K K β...

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