第二泛珠三角物理林匹克

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1、PanPearlRiverDeltaPhysicsOlympiad2006Notetograders:•Ifthefinalansweriscorrectandtherearesufficientstepstoshowtheprocess,givefullmarks.•Ifthefinalansweriswrong,findthestep(s)wheremistakesaremadeanddeductpointsthereonly.Norepeatdeductionofpointsforwronganswersbeingus

2、edforfurthersteps.Thisistheprincipleoferrornon-progressivenesswhichwefollowhere.QI(8points)Ek=一E.(m2+)(m3+m2)KineticenergyofmesonInternalenergyofmesonSuggestedsteps:(1)Useenergyandmomentumconservationtoobtainthespeedofm2aftercollisionwithm】.2m{m2.v2=[―—E(2points)(加

3、]+m2)Theproblemthenbecomesapairofweightsconnectedbyaspring,withonemasssuddenlyacquiresvelocity.(2)Findthecenter-of-massspeedVcofm2・rri3system(m3isstillatrestrightaftercollision).ThekineticenergyEkofthem2+m3systemisthen(1112+013)/2timesthesquareofVc-(3points)(3)Thei

4、nternalenergy冊川]Mt+2M2+2M3Q3(10points)(a)Pressureisproportionaltodepth,andthetotalforceisfoundbyintegratingtheentiredepth・TheforceactingonthestripisP=pgh(1point)Ecanthenbefoundusingenergyconservation,countingboththemesonandtheelectronaftercollision,orjustcountingt

5、heenergym2hasaftercollision.(3points)Q2(8points)Whenthemomentofinertiaofthepulleyisthelargest,whichmeansputtingmassalongtheedge.Thenuseenergyconservationthattheinitialpotentialenergyisequaltothekineticenergyofmassm3andthatofthepulley.(2points)MomentofinertiaI+M2R^(

6、2points)v=Rco(1points)(2points)(1points)(2points)M3gh=—Ico=JPdA=d(pghdh=^pgdH+—A/3v2(b)Step-1Findtheshapeoftheliquidsurface(2points)Method・1:Intherotatingreferenceframetheinertiaforceisco^r.andtheassociatedpotentialis122U=—(/)r.Theliquidsurfaceisanequal-potentialsu

7、rface,andthegravitationpotentialisgh.Sotheshapeisarotatingparaboladeterminedbygh+—a)2r2=constant.Method-2:Theslopeofthesurface—shouldbesuchthatnormalforceshouldbalancethegravityintheverticaldirection,andgivetheconcentricforceforinthehorizontaldirection.Thatleadstot

8、hesameanswerasMethod-1.Method-3F{)=pA^>AW/?=-PAVwhereAV=2^rArA/?Fc=mco2r=>=mco2rrwherem=17iphrr△叱=工甲=0()in(2兀师临门0)rAr=pgh(27irrh)dhA/?co2=>——