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UNIVERSITYOFCAMBRIDGEINTERNATIONALMARKSCHEMEfortheMay/June2012questionpaperfortheguidanceofteachers97099709PaperumrawmarkThismarkschemeispublishedasanaidtoteachersandcandidates,toindicatetherequirementsoftheexamination.ItshowsthebasisonwhichExaminerswereinstructedtoawardmarks.ItdoesnotindicatethedetailsofthediscussionsthattookceatanExaminers’meetingbeforemarkingbegan,whichwouldhaveconsideredtheacceptabilityofalternativeanswers.MarkschemesmustbereadinconjunctionwiththequestionpapersandthereportontheCambridgeispublishingthemarkschemesfortheMay/June2012questionpapersformostIGCSE,GCEAdvancedLevelandAdvancedSubsidiaryLevelsyllabusesandsomeOrdinaryLevelPagePageMarkScheme:Teachers’GCEAS/ALEVEL–May/June??UniversityofCambridgeInternationalExaminationsMarkSchemeMarksareofthefollowingthreeMMethodmark,awardedforavalidmethodappliedtotheproblem.Methodmarksarenotlostfornumericalerrors,algebraicslipsorerrorsinunits.However,itisnotusuallysufficientforacandidatejusttoindicateanintentionofusingsomemethodorjusttoquoteaformula;theformulaorideamustbeappliedtothespecificprobleminhand,e.g.bysubstitutingtherelevanttiesintotheformula.CorrectapplicationofaformulawithouttheformulabeingquotedobviouslyearnstheMmarkandinsomecasesanMmarkcanbeimpliedfromacorrectanswer.Accuracymark,awardedforacorrectanswerorintermediatestepcorrectlyobtained.Accuracymarkscannotbegivenunlesstheassociatedmethodmarkisearned(orMarkforacorrectresultorstatementindependentofmethodWhenapartofaquestionhastwoormore“method”steps,theMmarksaregenerallyindependentunlesstheschemespecificallysaysotherwise;andsimilarlywhenthereareseveralBmarksallocated.ThenotationDMorDB(ordep*)isusedtoindicatethataparticularMorBmarkisdependentonanearlierMorB(asterisked)markinthescheme.Whentwoormorestepsareruntogetherbythecandidate,theearliermarksareimpliedandfullcreditisgiven.Thesymbol√impliesthattheAorBmarkindicatedisallowedforworkcorrectlyfollowingonfrompreviouslyincorrectresults.Otherwise,AorBmarksaregivenforcorrectworkonly.AandBmarksarenotgivenforfortuitously“correct”answersorresultsobtainedfromincorrectworking. B2orA2meansthatthecandidatecanearn2orB2/1/0meansthatthecandidatecanearnanythingfrom0toThemarksindicatedintheschememaynotbesubdivided.Ifthereisgenuinedoubtwhetheracandidatehasearnedamark,allowthecandidatethebenefitofthedoubt.Unlessotherwiseindicated,marksoncegainedcannotsubsequentlybelost,e.g.wrongworkingfollowingacorrectformofanswerisignored.Wrongormissingunitsinananswershouldnotleadtothelossofamarkunlesstheschemespecificallyindicatesotherwise.Foranumericalanswer,allowtheAorBmarkifavalueisobtainedwhichiscorrectto3s.f.,orwhichwouldbecorrectto3s.f.ifrounded(1d.p.inthecaseofanangle).Asstatedabove,anAorBmarkisnotgivenifacorrectnumericalanswerarisesfortuitouslyfromincorrectworking.ForMechanicsquestions,allowAorBmarksforcorrectanswerswhicharisefromtakinggequalto9.8or9.81insteadof10.Thefollowingabbreviationsmaybeusedinamarkschemeorusedonthescripts: AnyEquivalentForm(ofanswerisequallyacceptable) AnswerGivenonthequestionpaper(soextracheckingisneededtoensurethatthedetailedworkingleadingtotheresultisvalid)BODBenefitofDoubt(allowedwhenthevalidityofasolutionmaynotbeabsoluyCAOCorrectAnswerOnly(emphasisingthatno“followthrough”fromapreviouserrorisallowed) CorrectWorkingOnly–oftenwrittenbya‘fortuitous’answer IgnoreSubsequentWorking PrematureApproximation(resultinginbasicallycorrectworkthatisinsufficiently SeeOtherSolution(thecandidatemakesabetttemptatthesame SpecialRuling(detailingthemarktobegivenforaspecificwrongsolution,oracasewheresomestandardmarkingpracticeistobevariedinthelightofaparticularcircumstance)MR–1ApenaltyofMR–1isdeductedfromAorBmarkswhenthedataofaquestionorpartquestionaregenuinelymisreadandtheobjectanddifficultyofthequestionremainunaltered.InthiscaseallAandBmarksthen e“followthrough√”marks.MRisnotappliedwhenthecandidatemisreadshisownfigures–thisisregardedasanerrorinaccuracy.AnMR–2penaltymaybeappliedinparticularcasesifagreedatthecoordinationmeeting.PA–1ThisisdeductedfromAorBmarksinthecaseofprematureapproximation.ThePA–1penaltyisusuallydiscussedatthemeeting. ForusingWD=WD=6×(0.5× Workdoneis21.9 Tcosθ+Tsinθ=11.2 (or–Tcosθ+Tsinθ=0.16g)–Tcosθ+Tsinθ=(orTcosθ+Tsinθ= Forresolvingshorizontallyor [Tcosθ=4.8andTsinθ=6.4 T=4.8+6.4ortanθ= [4T(cosθ+sinθ) (11.2–1.6)+(11.2+or2Tsinθ÷2Tcosθ(11.2+1.6)÷(11.2–or(Tcosθ+Tsinθ)÷(–Tcosθ+Tsinθ=11.2÷ T=8(orθ= θ=53.1orT= ForfindingTcosθandTsinθandhencefindingTorθ, forfindingthevalue 4T(cosθ+sinθ)or2Tsinθ÷2Tcosθor(Tcosθ+Tsinθ)÷(–Tcosθ+Tsinθ s=0.027(10t/3–t s=0.027[10000/3– Distanceis22.5 Forusings=∫ForfindingthevalueoftatAandusinglimitsorequivalent2 [0.027(20t–3t)=0t= vmax=0.027(4000/9– 1umspeedis4 Forusingdv/dt= ftincorrecttin0.027(10t–t [When4<v< aave=(6–4)/(0.5–when19<vaave=(21–19)/(24.5– Averageaccelerations 4msand0.244 Forusinga≈ DF(5)=P/5andDF(20)=P/20 B1[DF–R= P/5–R=1230×4and P/20–R=1230×0.244P= (orR= R= (orP= ForusingNewton’s2ftincorrectaverageaftP/5–1230a1orP/20–or5(1230a1+R)or20(1230a2+ WD =800× [2800000=PEgain+400 [2400000=16000g×500sinα α= ForusingWDbythedriving=PEgain+WDagainstForusingPEgain= [KEgain=2400000+2400000 8004000000 [?16000(v–20)=4000 1Speedis30 ForusingKEgain=WDbythedriving+PEloss–WDagainstftPE ForKEgain=?m(v–20)attemptingtosolveforSR(max2/4)forcandidateswhoassumeconstantdrivingandconstantwithout UsesNewton’sSecondLawandv=u+2as[4800+16000gsinα–1600=16000a,v=20+2a×500)]1Speedis30 AlternativeMethodforPart =2800000÷ [DF–mgsinα–R=m×0] ForusingNewton’ssecondlaw[16000×10sinα=5600– Forsolvingtheresultantequationfor α= F=5.9–6.1 R=6.1cos [5.9–6.1sinα≤μ(6.1cosα) 4μ> ForresolvingsparalleltotheForusingF≤μR [6.1×(11/61)+5.9–μ6.1×(60/61)> 7μ< ForusingF=μRand‘netdownward>0’ [6.1×(11/61)+5.9–μ6.1×(60/61)0.61× μ= ForusingNewton’s2lawandF= [T–0.12g=0.12a&0.38g–T= 0.38+0.12g 2Accelerationis5.2 ForusingNewton’ssecondforAandBorforusing mM+ [v2=2×5.2×0.65;0.65=? 1SpeedofBis2.6ms–orTB= 1TB=0.5orSpeedofBis Forusingv=2ahors=?ftincorrect [–2.6=2.6–10(T– T= Correctgraphfor0<t<1.02 ftincorrectvaluesofV,TandTBForusing–V=V–g(T–TB)orftincorrectVand/or [0.65+0.5(1.02– Totaldistanceis1.326m(accept Forusing‘total =?(VTB)+2x GCEGCE20125/697099709/43紙張4，最大原始標(biāo)記有表明評(píng)分開始前員會(huì)議上進(jìn)行的討論的細(xì)節(jié)，該討論會(huì)考慮替代答案的可接受性。大學(xué)正在發(fā)布大多數(shù)IGCSE、GCE高級(jí)和高級(jí)附屬級(jí)別課程以及一些普通級(jí)別課程2012年月/6第2頁(yè)評(píng)分方案：教師版教學(xué)大綱文件GCEAS/ALEVEL–2012年5月/6月9709M。然而，對(duì)于候選人來說，僅僅表明使用某種方法的意圖或僅僅一個(gè)通常是不夠的或想法必須應(yīng)用于當(dāng)前的具體問題，例如將相關(guān)量代入即可。正確應(yīng)用而不 可以獲得M分，并且在某些情況下，正確答案可以暗示M分。準(zhǔn)確度分?jǐn)?shù)，正確的答案或正確獲得的中間步驟。除非獲得（或暗示）相關(guān)方法分?jǐn)?shù)，否則無B當(dāng)問題的一部分有兩個(gè)或多個(gè)“方法”步驟時(shí)，M分配了多個(gè)B標(biāo)記時(shí)也是如此。符號(hào)DM或DB（或dep*）用于指示特定的M或B標(biāo)記依賴于方案中較早的M或B（帶星號(hào)）標(biāo)記。當(dāng)考生同時(shí)跑兩個(gè)或多個(gè)步驟時(shí)，將暗示較早的分?jǐn)?shù)并給符號(hào)√表示指示的A或BA或B分?jǐn)?shù)。對(duì)于偶然的“正確”答案或不正確的工作獲得的結(jié)果，不會(huì)給出A和B注：B2或A2表示候選人可以獲得2或0。B2/1/0表示候選人可以獲得0到2o。除非另有說明，否則一旦獲得分?jǐn)?shù)就不能再丟失，例如遵循正確答案形式的錯(cuò)誤工作將被忽略。除非方案另有明確說明，否則答案中錯(cuò)誤或缺失的單位不應(yīng)導(dǎo)致失分。對(duì)于數(shù)字答案，如果獲得的值正確于3s.f.，或者將正確于3s.f.，則允許標(biāo)記A或B是圓形的（角度情況下為1d.p.）。如上所述，如果由于不正確的工作而偶然得到正確的數(shù)字案，則不會(huì)給出A或B分?jǐn)?shù)。對(duì)于力學(xué)問題，如果g等于9.8或9.81而不是10，確案可得A或B分。第3頁(yè)評(píng)分方案：教師版教學(xué)大綱文件GCEAS/ALEVEL–2012年5月/6月9709以下縮寫可用于標(biāo)記方案或中AEF任何等效形式（答案同樣可以接受AG答案在試卷上給出（因此需要額外檢查以確保導(dǎo)致結(jié)果的詳細(xì)工作有效BOD懷疑的好處（當(dāng)解決方案的有效性可能不完全清楚時(shí)允許CAO僅正確答案（強(qiáng)調(diào)不允許從先前的錯(cuò)誤中“跟進(jìn)CWO僅正確工作–通常由“偶然”答案編寫ISW忽略后續(xù)工作MRPA過早近似（導(dǎo)致基SOSSeeOtherSolution（考生在同一問題上做了更好的嘗試SR特別裁決（詳細(xì)說明針對(duì)特定錯(cuò)誤解決方案要給出的標(biāo)記，或者根據(jù)特定情況改變某些標(biāo)準(zhǔn)標(biāo)記實(shí)踐的情況）罰分MR–1如果問題或部分問題的數(shù)據(jù)確實(shí)被，并且問題的目標(biāo)和難度保持不變，則從A或B分中扣除MR–1的罰分。在這種情況下，所有A和B標(biāo)記都會(huì)變成“followthrough√”標(biāo)記。當(dāng)候選人自己的數(shù)字時(shí)，不適用MR——這被視為準(zhǔn)確性錯(cuò)誤。如果協(xié)調(diào)會(huì)議同意，在特殊情況下可適用MR–2處罰。PA–1在過早近的情況下從A或B分?jǐn)?shù)中扣除。PA–1處罰通常在會(huì)議上討論第4頁(yè)評(píng)分方案：教師版教學(xué)大綱文件GCEAS/ALEVEL–2012年5月/6月9709oWD=6×(0.5×8)cos24所做功為21.9JM1A1A1[3]使用WD=(i)(ii)Tcosθ+Tsinθ=11.2（或–Tcosθ+Tsinθ=0.16g）–TcosθTsinθ=0.16g（或Tcosθ+Tsinθ=11.2）[Tcosθ=4.8且Tsinθ=6.4且=4.8+6.4或tanθ=6.4/4.8]222[4T(cosθ+sinθ)=22(11.2–1.6)++1.6)或2Tsinθ2Tcosθ=(11.2+1.6)(11.2–1.6)或(Tcosθ+TsinθTcosθ+Tsinθ)=11.21.6]T=8（或θ=53.1）θ=53.1或T=8M1A1M1A1A1[3][3]TcosθTsinθTθ2224T(cosθsinθ2Tsinθ2Tcosθ(TcosθTsinθTcosθTsinθ)的值(i)(ii)34s=0.027(10t/3–t/4)(+C)s=0.027[10000/3–10000/4]22.5m