bedok south 4E5N Prelim 2020 Sc(Phy) Marking Scheme

Bedok South Secondary SchoolSec 4Exp/ 5NA Science (Physics) Prelim Exam 2020 Markingscheme Paper 2 Section ANo. Answers Marks1 Markers’ comments:● Most candidates are able to give an appropriate scale.● A number constructed parallelogram without indicatingthe direction of forces.● Some drew the resultant force incorrectly and manywere unable to state thedirection. 2a 1

Correct time period and line drawn (B1) for each line. Markers’ comments:● This question is well attempted.2b Distance travelled = m 12 9 + 17( ) 2.5( )= 32.5 Average speed = 32.5 / 17 = 1.91 m/s[Allowerror carry forwardfrom the diagram, as longas student displayunderstanding in how to calculate the total distancebased on the graph drawn.] Markers’ comments:● Generally well attempted.● Weaker ones still continue to assume that distance= speed x time. B1B1 2c Acceleration = = m/s 2 0−2.55 − 0.50 Deceleration = 0.50 m/s 2 Markers’ comments:● Some left answer as m/s 2 which shows thatcandidates were unable to− 0.50 the differenc between acceleration & deceleration. B1B1 3a 10 m/s 2 or 9.8 m/s 2 or 9.81 m/s 2 Markers’ comments:● Candidates were unable to distinguish between velocity& acceleration as manystated 0. B1 3b(i) 2000 – 800 = 1200 N B13b(ii) 1200 / 80 [C1]= 15 m/s 2 [A1] C1A14a [C1]𝑃 = 𝐹𝐴 = 14 000 𝑁500 𝑐𝑚 2 = 28 N/cm 2 [A1] C1A1 4b Plarger = Psmaller [B1] 𝐹𝐿𝑎𝑟𝑔𝑒𝑟 𝐴𝐿𝑎𝑟𝑔𝑒𝑟 = 𝐹𝑆𝑚𝑎𝑙𝑙𝑒𝑟 𝐴𝑆𝑚𝑎𝑙𝑙𝑒𝑟 = 1 𝐹𝐿𝑎𝑟𝑔𝑒𝑟 𝐹𝑆𝑚𝑎𝑙𝑙𝑒𝑟 = 𝐴𝐿𝑎𝑟𝑔𝑒𝑟 𝐴𝑆𝑚𝑎𝑙𝑙𝑒𝑟 1440 𝑐𝑚 2 500 𝑐𝑚 2 >Therefore Flarger > Fsmaller . [B1] B1 B1 2

OR Flarger = 28 N/cm 2 × 1440 cm 2 = 40 230 N [B1]Since Fsmaller = 14 000 N, Flarger > Fsmaller [B1] Markers’ comments:● Generally well attempted and for those who can recall the Pressure formula. B1B1 5a Elastic Potential Energy B1 5b(i) KE = ½ x m x v 2 1900 = ½ x 45 x v 2 [C1]v = 9.19 m/s [A1] C1A15b(ii) Loss in KE = Gain in GPE1900 = m x g x h1900 = 45 x 10 x h [C1]h = 4.22 m [A1] Markers’ comments:● Some candidates wrote KE = GPE which is not acceptableconceptually. Youare supposed to write loss in KE = Gain in GPE. C1A1 6a(i) Foam-filled wall: Ittraps airwhich is a poor conductorof heat.Itreduces heat loss through conduction Markers’ comments:● There is a need to identify air is trapped in thefoam filled walls as a significantnumber of candidates just stated foam is a poor conductorof heat. B1B1 6a(ii) Shiny outer casing: Ashiny surface is a poor emitterof heat.Itreduces heat loss by radiation. Markers’ comments:● Some candidates stated that shiny surface is a poorabsorber of heat which isnot applicable in this situation.● Some phrases their answer without clarity e.g. Shinyouter casing is a pooremitter of heat. This can be better with just “Shinysurface is a poor emitter ofheat.” B1B1 6b During evaporation,the more energetic particlesatthe surface of the liquidbreakthe forces of attraction andleft the liquid.Leaving behind particles withlower kinetic energyand thus,lowering thetemperatureof the liquid. Markers’ comments:● This question is badly attempted. Most answers didnot use the ideas ofmolecules to explain evaporation. B1 B1 3

7a 1 m for correct ray drawn1m for image and labeling of Image (I)Max penalty of 1m for missing arrows or image is notdotted 27b Real, inverted and same size as objectAny 2 correct1mAll 3 correct2m 2 8a Thefree moving negative chargesin the metal ball will be repelled by thenegatively charged rod andmove to the metal thread. The metal threads containing the excess negative chargeswill repel each other aslike charges repel. Markers’ comments:● Not well attempted as many candidates assumed thatthe threads arenegatively charged initially despite the questionstated that the threads has nooverall charge.● Many were unable to explain the movement of free electronsdue to the rod andthem being evenly distributed to each threads. B1 B1 8b(i) Q = I x t= (0.3 x 10 -3 ) x (50 x 10 -6 ) [C1]= 1.5 x 10 -8 C [A1] C1A18b(ii) E = V x Q= 120000 x 1.5 x 10 -8 = 1.8 x 10 -3 J Alternative method(E = P x t)P = V x I= 120000 x 0.3 x 10 -3 = 36 W E = P x t= 36 x 50 x 10 -6 = 1.8 x 10 -3 J Markers’ comments:● Both parts of Q8b were badly attempted due to thewrong prefixes recalled. C1A1 C1A1 9a V = RI6 = 8 I [C1]I = 0.75 A [A1] C1A1 4

Markers’ comments:● Many candidates were unable to determine that theeffective resistance us 8ohms.9b Same brightness [B1]They are connected in series, current through boththe bulb is the same.Resistance of the bulbs are also the same. [B1] Markers’ comments:Many candidates has the misconceptions of currentdecreasing after passingthrough a component. Hence, mistakenly assumed thatbrightness of L2 will bedecreased as compared to L1. B1B1 9c V = RI6 = 6 I [M1] (Short circuit)I = 1 A [A1] Markers’ comments:Many candidates were unable to apply the effect ofshort circuit. M1A1 Section BNo. Answers Marks10a Centre of gravity isa point onan object wheretheentire weight appears to acton B1 10b W = mg= 50 kg × 10 N/kg=500 N [B1] B110c clockwise moment = 500 N × 0.90 m [C1] [allow e.c.f.from (b)]= 450 Nm [A1] C1A110d F × 1.4 m = 450 Nm [C1] [allow e.c.f. from (c)]F =321 N [A1] C1A110e(i) The additional weights positioned at the shouldersof the woman causes thecentre of gravity of the woman to shift further awayfrom the pivot O.Since moment = force x distance, as the distance increases,the clockwisemoment increases. So, students need to actually talk about what theadditional weights do,-Shifts the centre of gravity further from pivot O-Increase the weight of the setup. B1 B1 10e(ii) Sincethe total clockwise moments is increased, thetotal anti-clockwisemoments increased too. [B1]HenceF increases.[B1] [ECF] (If the students mentioned in 10e(i) that theclockwise moment decreases,)Sincethe total clockwise moments is decreased, thetotal anti-clockwisemoments decreased too. [B1]HenceF decreases.[B1] Another methodBy the principle of moments, the sum of clockwisemoment about a pivot is equalto the sum of anti-clockwise moment about the samepivot. [B1]Hence F increases. [B1] B1B1 Q10 Markers’ comments:● Only Q10b is well attempted. 5

● Q10c to 10d were badly done. Many were unable to explainclearly by usingconcepts of moments. E.g. When weights are strapped,weight increaseshence F increases. 11a(i) Period = 2 ms = 0.002 sFrequency = 1 / 0.002 = 500 Hz B1B111a(ii) Same period but withtwice the amplitude B1B111a(iii) The higher pressure is the compression region whilethe lower pressure region isthe rarefaction region. B1 11b(i) Ultra-violetAlright to accept ultraviolet radiation, ultravioletrays B1 11b(ii) Speed = 1.48 x 10 15 x 2 x 10 -7 = 2.96 x 10 8 m/s C1A111b(iii) Electromagneticwaves aretransversewaves whilesoundwavesarelongitudinalwaves.Electromagnetic wavescan travel through vacuumbutsound wavescannot travelthrough vacuum. B1 B1 Markers’ comments:● Q11a(i) were done badly as many missed out the prefixmilli- for the x axis.● Rest of the question were generally well attempted. 12 a(i) It means that the appliance will function witha powerof 2000 Wwhen connectedto a240 V electrical source(normal operating conditions). B1 12a(ii) I = P/V= 2000/240= 8.33 A M1A112a(iii) The wire with diameter = 1mmIt can accommodate up to 10 A safely and the operating current (8.33 A) is lessthan 10 A B1B1 12a(iv) As diameter increases, cross sectional area increases.Resistance decreases (R inversely proportional toA)P = I 2 R. Withlower resistance, power generated islower with the same currentof 8.33 A flowing through the wire, less heat produced. Hence overheating willnot occur with the operational current. B1 B1 12b(i) P: live wireQ: neutral wireR: Earth wireNot all correct, cannot be full marks.If it is not all wrong, we can give one mark.- 1 or 2 out of 3 correct, we give one marks 2 m 12b(ii) A fuse is a safety device used to limit the current flow in a circuit to preventoverheating. B1 12b(iii) If the appliance does not have a metal casing, theEarth wire is no