Physics 6091 Sec 4 Prelim Paper 2 Markscheme 2021

DUNMANSECONDARYSCHOOLSECONDARY4EXPRESSPHYSICS6091PRELIMINARYEXAMINATION2021SUGGESTEDMARKSCHEME PAPER2–SECTIONA1(a) Thedifferencebetweenascalarquantityandavectorquantityisthatascalarquantityhasonlymagnitudewhereasavectorquantityhasbothmagnitudeanddirection. B1 1(b) Thevectorquantityisdisplacement/velocity/acceleration/moment. B11(c) 30°TB =400sin30°=200N (note:todeterminebasedondiagramandcanacceptmathscalculation) TC =400cos30°=346N (note:todeterminebasedondiagramandcanacceptmathscalculation) B1 B1 B1 2(a) Thegradientofthegraphincreaseswhichindicatesthatthevelocityisincreasing. B12(b) Terminalvelocity=(60–20)/(3.5–2.0)=27m/s(2sf) M1A12(c) Asitfallsfromrest,weightactsdownwards,thusresultantforceisactingdownwards.Asspeedincreases,airresistancealsoincreasesResultantforceactingdownwardsdecreasestozero B1B12(d) A1 3(a)(i) Fnet =ma4.0-0.80=2.0aa=1.6m/s 2 1.6= 𝑣 − 𝑢∆𝑡 M1

= 𝑣 − 05.0 v=8.0m/s A13(a)(i) SinceFnet =0whenFbalancesthefrictionalforceof0.8N,BasedonNewton’sSecondLaw,theaccelerationbecomeszero.BasedonNewton’sFirstLaw,theboxcontinuestomoveataconstantspeed(of8.0m/s)inthesamedirection. B1 B1 3(b) SinceFistheactionforceexertedbytheboyonthebox,theotherforce,F1,whichisequaltoFinmagnitude,butactingintheoppositedirectionofFisthereactionforcebytheboxactingontheboy.F1ispartoftheaction-reactionpairwithFbyNewton’sThirdLaw. B1B1 4(a)(i) kineticenergyatB=gravitationalpotentialenergyatA=mgh=50x10x(350/100)=1750J=1800J(2sf) M1A14(a)(ii) (½)mv 2 =1750 OR1800v=8.37m/s=8.4m/s(2sf) ORv=8.49=8.5m/s(2sf) M1A14(b)(i) energylostduetofriction=50x10x[(350–300)/100]=250J M1A14(b)(ii) workdoneagainstfrictionis250Jforcexdistance=250Jforce=250/15=16.7N=17N(2sf)[Note:itisokforthisanswernottohaveunitNandallowECFfrom4(b)(i)]] A1 5(a) Energymust beprovidedbyexternal sourcetoweakentheintermolecular forceofattractionORfortheliquidmoleculestogaininternalpotentialenergyintheliquidinorderforittobecomegas. B1 5(b) Total energylost =energylost bywater duringcooling+latent energylost duringfreezing =mc +mLf ∆θ =500(4.2)(30–0)+500(330)=228000J=230000J(2sf) M1M1A1 5(c) Theheatedwaterneartheheateratthebottomofthetankexpandsandrisesduetodecreaseddensity. Thecoolerwaterfromthetopsinksasithasahigherdensity. Thissetsupaconvectioncurrentuntilallwaterinthetankisheated. B1 B1

6(a) X1 andY1 orX2 andY2 B1 6(b) PA =PB 1.20xρA xg=0.76x13600xgρA =8613=8600kg/m 3 (2sf) M1A16(c) Whenthetapisopened,theliquidlevelsinbothtubeswill dropandachievethesameheight astheliquidsinthecontainers. B1 B1 7(a) Electronsaretransferredfromthepipetothefuelduringtherubbingaction. B17(b) Sparkmightjumpfromthechargedplaneandignitethefuelcausingexplosion. B1B17(c) Metalisanelectricalconductor(orhaslowresistance)whichwill allowelectronstoflowthroughit fromtheaeroplanetotheground. ORaeroplanetobeearthed. B1B1 8(a) Whentheswitchisclosed,currentwillflowthroughthecoilandtheironcorebecomesanelectromagnet.Theelectromagnet will inducemagnetismintheironbolt. Hence, theelectromagnetattractstheironbolt,causingittomovetotheleft,allowingt