# answer - 1-5This is a case of dilation γ′=TT in this...

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Unformatted text preview: 1-5This is a case of dilation. γ′=TT in this problem with the proper time ′ =TT( 29-=-⇒=-1 21 22211TvvTTccT;in this case =2TT, ( 29=-=-1 221 221114LvL therefore =0.866vc.1-6This is a case of length contraction. γ′=LL in this problem the proper length ′ =LL, -=-⇒=-1 21 222211vLLLvcLc; in this case =2LL, ( 29=-=-1 221 221114LvL therefore =0.866vc.1-8γ′=LL( 29(29γ==-′=-=-=′1 2221 21 2221175110.661100LvLcLvcccL1-9γ′=earthLL′=-1 22earth21vLLc, ′L, the proper length so (29==-=1 22earth10.90.436LLLL.1-20(29 (29′++===′++220.900.700.98110.900.70vuccucvu ccc c1-21(29 (29--′=== ---220.500.800.50110.500.80XXXuvccucu v ccc c2-1(29=-1 2221mvpvc(a)(29(29(29--×==×⋅-27211 221.67 10 kg0.015.01 10 kg m s10.01cpc c(b)(29(29(29--×==×⋅-27191 221.67 10 kg0.52.89 10 kg m s10.5cpc c(c)(29(29(29--×==×⋅-27181 221.67 10 kg0.91.03 10 kg m s10.9cpc c(d)--×==×⋅×132281.00 MeV1.602 10 J5.34 10 kg m s2.998 10 m sc so for (a) (29(29--×⋅==×⋅21225.01 10 kg m s100 MeV9.38 MeV5.34 10 kg m scpcSimilarly, for (b) =540 MeVpc and for (c) =1 930 MeVpc.2-2(a)Scalar equations can be considered in this case because relativistic and classical velocities are in the same direction.(29(29(29γ===⇒=⇒=---=1 221 21 222111.901.9011.90110.85mvpmvmvvcv cv cc(b)No change, because the masses cancel each other....
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## This note was uploaded on 11/09/2010 for the course PHYSICS PHASM/G 44 taught by Professor Marklancaster during the Spring '09 term at UCL.

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answer - 1-5This is a case of dilation γ′=TT in this...

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