Chapter 39457*P39.71Take the two colliding protons as the systemEKmc12=+Emc22=Epcmc1212224p20=.In the final state, EKMcff2:cMc224.By energy conservation, EEEf12+=, soEEEEEf1222++=pcmcK mc mcMcf1222++ejBy conservation of momentum,ppf1=.ThenKmcKm cmc22424=+=+McmcKmc2212.By contrast, for colliding beams we haveIn the original state,c12c22.In the final state,EMcf=2f:c Kmc Mc2McmcKmc22FHGIKJ.FIG. P39.71*P39.72Conservation of momentum γmu:muucmuMvvcmuff13112312 2−+−−=−=−af.Conservation of energy mc2:mcmcMcmcf2222114−+−=−=−.To start solving we can divide:
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .