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# hw2 - 2-2(a Scalar equations can be considered in this case...

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2-2 (a) Scalar equations can be considered in this case because relativistic and classical velocities are in the same direction. p = " mv = 1.90 mv = mv 1 # v c ( ) 2 \$ % & ( ) 1 2 * 1 1 # v c ( ) 2 \$ % & ( ) 1 2 = 1.90 * v = 1 # 1 1.90 + , - . / 0 2 \$ % & & ( ) ) 1 2 c = 0.85 c (b) No change, because the masses cancel each other 2-3 As F is parallel to v , scalar equations are used. Relativistic momentum is given by p = " mv = mv 1 # v c ( ) 2 \$ % & ( ) 1 2 , and relativistic force is given by F = dp dt = d dt mv 1 " v c ( ) 2 # \$ % & ( 1 2 ) * + + , + + - . + + / + + F = dp dt = m 1 " v 2 c 2 ( ) [ ] 3 2 dv dt 0 1 2 3 4 5 2-7 E = " mc 2 , p = " mu ; E 2 = " mc 2 ( ) 2 ; p 2 = " mu ( ) 2 ; E 2 " p 2 c 2 = # mc 2 ( ) 2 " # mu ( ) 2 c 2 = # 2 mc 2 ( ) 2 " mc ( ) 2 u 2 \$ % & ( ) = mc 2 ( ) 2 1 " u 2 c 2 * + , - . / 1 " u 2 c 2 * + , - . / " 1 = mc 2 ( ) 2 Q.E.D. E 2 = p 2 c 2 + mc 2 ( ) 2 2-8 (a) E R = mc 2 = 1.67 " 10 # 27 kg ( ) 3 " 10 8 m s ( ) 2 = 1.503 " 10 # 10 J = 939.4 MeV (Numerical round off gives a slightly larger value for the proton mass) (b) E = " mc 2 = 1.503 # 10 \$ 10 J 1 \$ 0.95 c c ( ) 2 % & ( 1 2 = 4.813 # 10 \$ 10 J ) 3.01 # 10 3 MeV (c) K = E " mc 2 = 4.813 # 10 " 10 J " 1.503 # 10 " 10 J = 3.31 # 10 " 10 J = 2.07 # 10 3 MeV 2-9 (a) When K = " # 1 ( ) mc 2 = 5 mc 2 , " = 6 and E = " mc 2 = 6 0.511 0 MeV ( ) = 3.07 MeV . (b) 1 " = 1 # v c \$ % & ( ) 2 * + , , - .

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