bukit batok 2020 BBSS O Prelim Sc(Phy) (5076) MS

2020 BBSS SEC 4E5N SCIENCE (PHYSICS) (5076) O PRELIMINARYEXAMMARK SCHEME (FOR TEACHERS ONLY)PAPER 2: (W = working), (C/F = concept / formula),(A & U = answer & unit)● Penalize1 markper questionfor no / wrong unit.● Penalize 1 mark per question for failure to show concept / formula clearly and explicitly at thebeginning of each mathematical working.● Mark for 2 s.f. in Questions4(b), 9(a)(ii) and 9(b)(iii)only.Q Suggested Answer Remarks1a Distance = area under v-t graph = ½(10 × 8.0)=40 m [1]: C/F & W[1]: A1b Deceleration = change in velocity / time = (8.0 –0) / 10=0.80 m/s 2 [1]: C/F & W[1]: A1c Retarding force = ma = (20)(0.80) =16 N● Allow for ecf from part 1b [1]: C/F & W[1]: A1d Work done = F.s = (16)(40)=640 J [1]: C/F & W[1]: A2a ● A to C: decreasing acceleration● C to D: constant deceleration [1][1]2b ● Weight of stone is constant.● Air resistance against stone increases.● Net force acting on stone decreases. [1][1][1]2c No, because v-t graph of stone (between B and C) wasnever straight. [1]3a ● Appropriate scale: minimum of 1.0 cm to 5.0 N (reject 1.0 cmto X Nwhere X >5becausediagramwill betoosmall) andeither 60Nor 80Nforce drawn correctly to stated scale.● Magnitude of resultant force =(100 ± 1.0) N● Angleθ:(37±1)° [1][1][1]3b ● Same magnitude as magnitude of net force found in3(a)● Ball held at equilibrium. Hence tension must be balanced (not equal) byresultant forcein3a(accept“equal inmagnitudeandoppositeindirection”in lieu of “balanced”). [1] [1]4a Point where the weight of a body seems to act (regardless of the body’sorientation) [1]4b(i) [1]Q Suggested Answer Remarks

BBSS / 2020 / O Prelim / Sec 4E5N / Science (Physics)(5076) / Mark Scheme 4b(ii) ● Line of action of lorry’s weight acts outside lorry’sbase.● Clockwisemoment (due to lorry’s weight) acts on lorry. [1][1]4c(i) Perpendicular distance between CG of metre rule andpivot = 20 cmMoment = F × d = (1.0)(20)= 20 Ncm [1][1]: C/F & W[1]: A4c(ii) Apply principle of moments about pivot (70-cm mark)(1.0)(20) = (W)(10)W= 2.0 N (reject “2N”, “2.00N” –mark for 2 s.f.) [1]: C/F & W[1]: AMark for 2 s.f.5a Change in GPE = mgh = (80)(10)(60 – 4.0)=44800 J [1]: C/F & W[1]: A5b ½mv 2 = 23000½(80)v 2 = 23000⇒v =24 m/s(to 2 s.f.) [1]: C/F & W[1]: A 5c Student’s answer should contain the following points:● Loss in gravitational potential energy (GPE) of man● Loss in kinetic energy (KE) of man● Gain in elastic potential energy (EPE) of rope [1][1][1]6a ● Air molecules vibrate parallel to the direction oftravel of sound waves● Forming regions of compression and rarefaction [1][1]6b Sound travels faster through solids than through gases [1]6c(i) Distance = speed×time = 300×0.100=30 m. [1]: C/F & W[1]: A6c(ii) Speed = distance / time = 30 / 0.020=1500 m/s. [1]: C/F, W & A7a Electrons transferred from plastic ball to wool. [1]7b ● Electric field lines originate from positive and pointinto negative.● Electric field lines more closely packed at regions closer to the chargedbodies, and more spaced out