中南大学大学物理双语版答案Problem-1-22

Problem1.Answers:1.216vij;8aj;7.13.(cosavav)2.1/3(3/)ftvk3.a-e,b-d,c-f.4.[d]:222xyL,0dxdyxydtdtdxvdt,Bdyvdt,0Bxvyv,cotBxvvvy5.(a)32(102)3trittj,(Answer)(b)912rij,(3)(0)343avgrrvij,(Answer)(3)(0)343avgvvaij(Answer)(c)92vij2tan9yxvv,12.5(Answer)6.Solution:Fromthedefinitionofaccelerationforastraightlinemotiondvadt,andthegivenconditionakv,wehavedvkvdt.Applychainruletodv/dt,theequationcanberewrittenasdvdxdvkvvdxdtdxSeparatingthevariablesgivesvdvkdxTakedefiniteintegrationforbothsidesoftheequationwithinitialconditions,wehave000xvvdvkdx,or3/2023xvk(Answer)Problem2.Answers:1.13.0m/s2,5.7m/s,7.5m/s2,2.gR,(21)R.Solution:Atinitialmomentwhentheballisjustkickedout:2irvag,2ivg.Inordertheballnottohittherock,2ivRg,ivgR;(Answer)Fortheverticalmotion:212gtR,2/tRg,ivtR(21)R(Answer)3.d.4.d.5.Solution:atanymomentthespeedoftheprojectileis220()vvgtTangentialacceleration:22220tdvgtadtvgt,Radialacceleration:2202220rtgvagavgt,From2rva,wehavetheradiusofcurvatureisgivenby2223/2200()rvgtvagv(Answer)6.86.7Solution:sdsggdvvvsgdgvv3510/3600tan17.360.8dgsgvv86.78642Problem3.Answers:1.Solution:212coscosvTTmR12sinsinTTmg(a)21108N2cos2sinmvmgTR,(b)2156NsinmgTT2.(a)Mg,(b)/Mgrm3.d.4.e5.Solution:fromthegivenconditionctivveandwith10.0m/siv,at20s,5.0m/sv,wehave(a)10.0347s,With40.0s,wecanget(b)2.50m/s,Solution:(c)Fromctivve,wehavectidvcvedtThereforeacv,(Answer)whichmeanstheaccelerationoftheboatisproportionaltothespeedatanytime.6.Solution:FromNewton’ssecondlaw,wehave2kmvmaBythedefinitionofacceleration,thisequationcanberewrittenas2dvkmvmdtSeparatingthevariablesobtains2dvkdtvTakethedefinitionintegralwiththeinitialconditions,wehve020vtvdvkdtv011ktvvThen00/(1)vvktv.(Answer)Fig.3-1Problem4.Answers:1.(a)21J,(b)17cos513,=19.42.23(79)3Fxyixj.3.c.4.d5.(a)002()xxUFdrAxBxdx=2323ABxx,(b)51923ABU(c)0UK,19532BAKU6.Solution:(a)Intheprocessofthependulumswingingdown,onlythegravitydoeswork,mechanicenergyisconserved:21(1cos),2mgLmvWhenthesphereisreleasedfromacertainheight,inordertomaketheballwillreturntothisheightafterthestringstrikesthepeg:21()2mvmgLd,(1cos)()mgLmgLdcosLd.(Answer)Solution:(b)Ifthependulumistoswinginacompletecirclecenteredonthepeg,atthetopofthepath,thetensionTonthecordmustnotbezero.2vTmgmLd,Because0TSo2()vgLdFromtheconservationofmechanicalenergy,wehave212()2mgLmvmgLdInserting2()vgLdintoaboveequation,obtains1()2()2mgLmgLdmgLd35Ld(Answer)Problem5.Answers:1.(a)52104ij32()pFdt2.(a)0.284,(b)114.6fJ,45.4fJ.Solution:(a)Momentumandkineticenergyareconserved:0ncppp(1)2220222ncpppmmM,12MmSo,22201212ncppp(2)FromEq.1,0ncppp,insertEq.2222001212()ccpppp22200122412cccppppp202413ccppp,02413cpp222001242()0.28412132ccpKMpKm.(Answer)0000.284ncKKKKK15114.610J114.6fJ(Answer)1500.28445.410J45.4fJcKK(Answer)3.[a].12FdKK,22121222ppmm,1122pmpm4.[a]Solution:Form:intheelasticcollisionprocess,00()Ftmvmvmvv,0()mvvFtintheinelasticcollision:00()Ftmvmvmvv,0()mvvFt00,()()ttvvvvFF5.4MvglmSolution:/2mvMVmv,2mVvM2122MVMgl,4Mvglm6.proofproblemSolution:()mvMmV,mVvmM212hgt,2mvhdVtMmg()2MmdgvmhProblem6.Answers:1.144rad.Solution:106t,ddt，20(106)103ttdttt2103dttdt,4223400(103)(5)144radttdttt(Answer)2.(a)22853IMdmd,Solution:2221(2)2(2)3ImdmdMd(b)212KI22245()32Mdmd3.c.rF(4i5j)(2i3j)2k4.a.Solution:Theinitialangularaccelerationoftherodandtheinitialtranslationalaccelerationoftherightendoftherod:I,2123LMgML,32gL32tgaL5.21.5NSolution:Fortheflywheel:()uRRTTrI212RIMR,21.67rad/sR,135NuT21.5NT(Answer)6.222sinkdmgdImRSolution:fromtheconservationofmechanicalenergy:222111sin222mgdkdmvIvR,222sinkdmgdImRFig.6-2Fig.6-4Fig.6-5Fig.6-6Problem7.Answers:1.(a)230Ltk,60dLtkdtSolution:(a)Lrp,2(3i5j)mrtt,2.0kgm,(b)dLdt.2.(a)Mvd,(b)2vv,(c)23Mv.Solution:(a)Angularmomentumisconserved:22fiddLLMvMvMvd;(b),fiLL24dMvMvd2vv(c)2222211222243WKMvMvMvMvMv3.d.4.a5.(a)2202233MLmlvvMLml,022603mlvMLml,(b)212IMghSolution:(a)Intheprocessofcollision,angularmomentumandmechanicalenergyareconserved:201()3mvlmvlML(1)222201111()2223mvmvMLOr222201()3mvmvML222201()()3mvvML22001()()()3mvvvvML(2)FormEq.1,wehave201()3vvmlML(3)Eq.(2)Eq.(3):0vvl(4)Fig.7-2Fig.7-5FromEq.(3),wehave2013vvMLml(5)Eq.(4)+Eq.(5)20123vlMLml022603mlvMLml(Answer)InsertingintoEq.(4),obtains2202233MLmlvvMLml(Answer)Solution:(b)212IMgh,213IML22IhMg220226()3mlvLgMLml6.Solution:(a)angularmomentumisconserved221()2imvdmRMR22202imvdMRmR(Answer)(b)Mechanicalenergyisnotconservedbecausetheclayandthesolidcylinderundergoaninelasticcollision.Inthisprocess,somekineticenergymustbelost.:2222212(2)ifmvdEIMmR22220112111/22fiIEdERMmmv，0fEEProblem9.Answers:1.(a)3.07MeV（Or:6mc2,4.921013J）,(b)0.986c.Solution:(a)22(1)5mcmc