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

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Unformatted text preview: 2-2 (a) Scalar equations can be considered in this case because relativistic and classical velocities are in the same direction. p="mv=1.90mv=mv1#v c( )2$%&()1 2*11#v c( )2$%&()1 2=1.90*v=1#11.90+,-./2$%&&())1 2c=0.85c(b) No change, because the masses cancel each other 2-3 As Fis parallel to v, scalar equations are used. Relativistic momentum is given by p="mv=mv1#v c( )2$%&()1 2, and relativistic force is given by F=dpdt=ddtmv1"v c( )2#$%&(1 2)*++,++-.++/++F=dpdt=m1"v2c2( )[ ]3 2dvdt123452-7 E="mc2, p="mu; E2="mc2( )2; p2="mu( )2; E2"p2c2=#mc2( )2"#mu( )2c2=#2mc2( )2"mc( )2u2$%&()=mc2( )21"u2c2*+,-./1"u2c2*+,-./"1=mc2( )2Q.E.D.E2=p2c2+mc2( )22-8 (a) ER=mc2=1.67"10#27kg( )3"108m s( )2=1.503"10#10J=939.4 MeV(Numerical round off gives a slightly larger value for the proton mass) (b) E="mc2=1.503#10$10J1$0.95c c( )2%&(1 2=4.813#10$10J)3.01#103MeV(c) K=E"mc2=4.813#10"10J"1.503#10"10J=3.31#10"10J=2.07#103MeV2-9 (a) When K=1( )Mc2=5Mc2, "=6and E="mc2=6 0.511 0 MeV( ) =3.07 MeV. (b) 1"=1#vc$%&()2*+,,-.//1 2and v=c11$%&()2*+,,-.//1 2=c116$%&()2*+,,-.//1 2=0.986c2-12 (a) When Ke=Kp, mec2"e#1( )=mpc2"p#1( ). In this case mec2=0.511 0 MeVand Mpc2=938 MeV, "e=1#0.75( )2[ ]1 2=1.511 9. Substituting these values into the first equation, we find "p...
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hw2 - 2-2 (a) Scalar equations can be considered in this...

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