Physical Sciences Grade 12 Paper 1 NSC June 2013 Memo

Downloaded from hlayiso.com basic education Department: Basic Education REPUBLIC OF SOUTH AFRICA SENIOR CERTIFICATE EXAMINATION SENIORSERTIFIKAAT-EKSAMEN PHYSICAL SCIENCE P1 NATUUR- EN SKEIKUNDE V1 HIGHER GRADE/HOER GRAAD 2013 POSSIBLE ANSWERS Copyright reserved/Kopiereg voorbehou Please turn over/Biaai om asseblief

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1_ 2 DBE/2013 SCE/SSE — Memorandum QUESTION 1/ VRAAG 1 11. =D 12 A 13° A 14°C 15 °C 16 A 17, A 18 8B 19 C 110 C 111. B 412 A 113 B 114 C 115 B (15x 4) (60) QUESTION 2/ VRAAG 2 24 OR/OF Vwina VplaneMiegtuig OR/OF O marks if not a complete vector diagram / geen punte as die vektordiagram nie volledig is nie. Y Vres horizontal and direction to left / V res horisontaal en na links YY plane in correct direction / V viiegtuig in korrekte rigting Y V wind in correct direction /V wing in korrekte rigting ¥ 135 ° between V wing aNd V res / Hoek tussen V wing ANC V es . (4) 2.2.1 Positive marking from 2.1 Calc: vv sini35° _ sind ¥ 200% 50” sin6 = 0,177 @ =10,19° north of west “ OR 280,2° (Or 10,18 °) (5) Copyright reserved/Kopiereg voorbehou Please tum over/Blaai om asseblief R |

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1_ 3 DBE/2013 SCE/SSE ~ Memorandum 2.2.2 Positive marking from 2.1 other angle of triangle = 180°-135°-10,2° = 34,8° sin135 ¥ sin34,8” 200 Vres . v Vres = 161, 42 km.h* OF /OR Vres 2 = Vwind” + Vpiane 2 — (2x Vwind XVplanex COS 34,8) ¥ = (50)* + (200) ? — (2 x 50 x 200 x 0,8211) v = 26077,016 Ves = 161,48 km.h ¥ (3) Positive marking to 2.2.3 OR: 2.2 tana 2.22 v Measure 2 cm /meet 2 cm Y must be 45° (135°) / moet 45° v wees 8 cm (200 km.h”) v Measure 8 cm from Vwina; Meet 8 cm vanaf Vwing ¥ Complete triangle with v 2 cm (60 km.h") ar) Horizontal / voitooi driehoek met — . 7 horisontale lyn ¥ completion of: R: 6,4cm+0,2cm Measure angle 6 = 10,2° Vv triangle (156 - 166 km.h" )v North of west” /meethoek 0 = 10,2° noord van wes (range 10,2° + 2°) vv measure and convert R = 161,5 km.h'/ meet en skakel om R = 161,5 km.h" (range 161,5 km.h't 5 km.h") Note ! If no arrow for R, loose 1 mark . Arrows for other 2 vectors not necessary/indien geen py! vir R, verloor 1 punt. Pyle_nie nodig vir ander 2 vektore nie Incorrect scale /verkeerde skaal max 6/8 (8) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief A f

Physical Science/Natuur- en Skeikunde HG/P1/V1 2.2.3 2.3 3.1 3.2 3.3 Downloaded from hlayiso.com SCE/SSE - 4 DBE/2013 Memorandum Positive marking from 2.2.1 and 2.2.2 Positive marking from 2.2.1 and 2.2.2 Construction in 2.2 Equations and symbols used in NCS and CAPS are accepted for all responses Vergelykings en simbole gebruik in NCS en CAPS word aanvaar vir alle antwoorde (calculations) AS A ht -AS v “a 151 =850" 161 42 850 ¥ t=5,28 hours /ure” ty Range: 5,12hrs 5,45hrs (4) t=5,27 hours/ure oS ye Also accept: Vee vere vey As s Ax Veg =—— =—: =—.: == res = Ay Also accept v t Vv At Vv Y. v 180 = 850 t v t= 5,67hours/ure a 0,4 hours longer/0,4 ure langer (4) [20] QUESTION 3/ VRAAG 3 Asv¥ 0,014” “a5 s =— = =0,35 m. Allow v =— for 3.1 and 3.2 Ya “Vac At 2x0,02v""S oe (4) if mm not converted to m, ~y., =A .9 0,034" 5% 85 loose 2 marks in 3.1, but Yo = YEH Ar = 9 0av’ ms" do not penalise again in 3.2 (3) and 3.3 / Indien mm nie omskakel na m nie, verloor 2 punte in 3.1 maar verloor 7 nie weer punte in 3.2 en AVY 0,85- %, 35 =5"n 5? 3.3 nie “At 5x0,02v : (6) [12] Copyright reserved/Kopiereg voorbehou LA f Please turn over/Biaai om asseblief

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1_ 5 DBE/2013 SCE/SSE —- Memorandum QUESTION 4/ VRAAG 4 Candidates may use the symbols v;, v;, Ax, Ay, At, and g for gravitational acceleration. 4. Using g = 9,8 m.s? ; v?=u2+2as ¥ af ae v =u? + 2as = 10° + 2(-10)(1,8) a Qo = 64 = 10? + 2(-9,8)(1,8) v=8ms" = 64,72 v=8,04m.s" OR/OF Using g = 9,8 m.s? v’ = wt 2as 4 v2 =u2 +2as ¥ 8? = 107+ 2(-10)xs 8? = 107 + 2(-9,8)xs s=1,8m s=1,84m (4) 42 v=u+ atv Using g = 9,8 m.s* 8210 +(-40)t , t=02s¥ eet 8=10 + (-9,8)t v=utrat % t=02sy O=u+ (-10)0,5 7 7 ) v=utat 7 u=5m.s v Y O=u + (-9,8).0,5 (7) u=4,9m.s" 43 Ap=mv-mu yv v v =0,5x5-05x8 = -1,5 kg.m.s™ (4) v -1,5 kg.m.s* downwards / afwaarts . 44 1,5 Kg.ms* upwards / opwaarts (Direction marked independently / rigting word onafhankllik gemerk) (2) Rk voorbehou Please turn over/Biaai om asseblief

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1__ 6 DBE/2013 SCE/SSE - Memorandum A Ue W = AE, = pm’ - gmu A 1 x Accept positive or negative = 5(0.8).5° - 5(0,5).8 oa =-9,75J or 9,75) v (5) 4.6 No./Neev v v vv v v Work done by bail in breaking glass / Not isolated system/ work done by glass on ball/ force exerted by glass on ball i! Arbeid verrig deur bal om die glas te breek / nie geisoleerde sisteem nie / krag deur glas op bal toegepas If Yes 0 marks / as ja dan geen punte (3) [25] QUESTION 5/ VRAAG 5 5.1 i 4 v trailer T (cable) TT (cable) ( 1000kg ) 3800N > car ae >TrEeEs (5) friction: 300 N ¥ friction: 500 N ¥ Accept free body diagrams, boxes, Accept: Feapie, cable, T Accept F;, f, Fw (Afrikaans) or 300N and 500 N 6.2 Frese =mMa ¥ v v vYoev v 500.a ={- 30 (equation 1) (3800 — 800) = 1500a as 2 v 1000.a = <0 — T = SB> (equation 2) 3000=1500.a 1500.a = 3000 a=2ms? v a=2ms?v (5) Fres=mxa ¥ vv v OR | 3000 = 1500xa a= 2ms? v Copyright reserved/Kopiereg voorbehou Please turn over/Biaai om asseblief A ik

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1_—7 DBE/2013 SCE/SSE — Memorandum 5.3 DO NOT MARK THIS QUESTION: OMMITED FROM QUESTION PAPER (TOTAL OF PAPER ADJUSTED - 195) MOENIE NASIEN/MERK NIE: VRAAG UITGELAAT IN ENGELSE VRAESTEL (TOTAAL VIR VRAESTEL AANGEPAS - 195) 5.4 5.5 Fr=Fot Fv Fres = Frow—(T + Fy) ¥ v v 500.2 = Fo +(— 300) ¥ 1000.2 = 3800 — (T +500) ¥ Fo = 1300 NY T=1300NV¥ (4) OR: from 5.2 equation 2 v 1000.a = 3800 — T - 500 v 1000.2 = 3800 ~-T — 500 ¥ T = 1300 NV vv Slows down (uniformly) (until it comes to a stop) / negative acceleration // Beweeg (konstant) stadiger (totdat dit tot rus kom, negatiewe versnelling (2) v Increase. v v vv ; . F Fengine the same, Mtotai Smaller / a inversely proportional to mass (aa— ) v v / F ves is larger and Fres directly proportional to a Toeneem Fenjin dieselfde, Miotaa Kleiner / a omgekeerd eweredig aan m(: aon), /F res iS groter en Fres direk eweredig aan a (3) [19] QUESTION 6/VRAAG 6 6.1 Fr=0=Fecae + Fe Fecate = 50.10 =500NVY (2) 6.2 7 Fr= ma (52) Fr = Fecale + Fo 100 = Fecate — 500 V (4) Fecaie = 100 + 500 = 600 N¥ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief KK i

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1_ 8 DBE/2013 SCE/SSE — Memorandum 6.3 DO NOT MARK THIS QUESTION: OMMITED FROM QUESTION PAPER (TOTAL OF PAPER ADJUSTED -195) MOENIE NASIEN/MERK NIE: VRAAG UITGELAAT IN ENGELSE VRAESTEL (TOTAAL VIR VRAESTEL AANGEPAS - 195) zero (Newton) v (0 N) (1) 7] QUESTION 7/ VRAAG 7 7.1 To left as positive / na links as positief To the right as positive / na regs as positief Pbeforecollision = Pafter collision Pbeforecollision = Pafter collision (ZP before collision = Patter collision ) } ~ (2Ppefore collision = 2Patter collision ) } we mu =mMyVv4 + M2V2 y mur MmyV, + M2V2 (3% 2) = (1*8) + 2v (3%-2) = (1*-8) + 2v v= -imstv v=etms? v totheright/naregs ” v v=1ms" tothe right /naregs (6) 7.2 v v v Law of conservation of momentum: The total momentum in a closed system remains constant/ is conserved V (in magnitude and direction) v v OR: Law of conservation of momentum: The total momentum before a v v collision equals the total momentum after the collision in a closed system v v v Wet van behoud van momentum: Die totale momentum in ‘n geslote sisteem is konstant/ bly behoue “ OF: Wet van behoud van momentum: die totale momentum voor ‘n botsing is gelyk aan die totale momentum na die botsing in ‘n geslote sisteem. (4) 7.3 Some Ex converted into heat/work sound done by the spring/deformation of spring /other forms of energy” “” Van die Ex omgesit in hitte/ klank werk verrig deur die veer/ vervorming van veer/ ander vorme van energiew” (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief KK

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1V1_ 9 DBE/2013 SCE/SSE — Memorandum QUESTION 8/ VRAAG 8 8.1 v The force of attraction or repulsion which one charge exert on the other is v directly proportional to the product of the charges and is inversely v proportional to the square of the distance between them. Die aantrekkende of afstotende krag wat een lading op ‘n ander uitoefen is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. (3) 8.2 Curved electric field lines important on outside Geboé elektriese veld en lyne belangrik op buitekant If opposite charges 1 mark: last mark of rubric As teenoorgestelde ladings 1 punt: laaste punt in rubriek Y field lines between charges / veldlyne tussen die ladings Y field lines outside the charges / veldlyne buite die ladings Y direction / rigting Y field lines not going into spheres but touching surface and without touching each other / veldlyne nie binne sfere nie, maar raak opperviak en sonder om aan mekaar te raak (4) 8.3 = w= 0,1NYY (2 or zero) (2) 8.4 Positive marking from Q 8.3 k W=Fe= a2 v r ¥_(9x10°)q? " (0,03)? ¥ q=1x10 oC v Each ball carries half the original charge /elke bal dra helfte van oorspronklike lading 5) “twice of 1x10°’C is 2x10°7C ®) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief EK i

Physical Science/Natuur- en Skeikunde HG/P1/V1 8.5 8.5.1 8.5.2 Downloaded from hlayiso.com 10 DBE/2013 SCE/SSE - Memorandum new charge = 2x107 +(-1x107) 1 QV 7 gx107C ¥ or5,0x10*C (2 marks for answer if no calculation shown / twee punte vir antwoord indien geen berekening getoon word nie) \ charged Vv altwee + gelaai. After contact, each has a charge of + 0,5x107 C -. repel as both + Na kontak het elk ‘n lading van + 0,5x1 07 C - stoot mekaar af omdat 8.5.3 Closer”. Q smallerv’, Fweign the samev / rs S thus r smaller vv kQq Nader aan mekaar Q kleiner, Fyewig dieselfde/ r? = - dus r kleiner QUESTION 9/VRAAG 9 9.1 9.2 ele ly Rparalle! Ry Ro viv = + 6 12 Rparallel = 40 ¥ 1 gow Roarallel Ry Rparallel = 13.0 x R sora 7 Rea +Riza 6x12 v ~64+12% R paraltel =40Ov ¥ = Rea xRio'd Acceptable / aanvaarbaar: 1 14 R =(—- +=) parallel Ry Ro 1 If Roarailel =>-+=>— : Omarks Ri Re i v Rrot 74+2= 60 “Vy SIR =3x6=18VY Copyright reserved/Kopiereg voorbehou LS v VpsiR = (3x4) = 12 Vig =IR = (3x2) =6V¥ Viti = 1246 =18VV (2) (2) (3) [21] (4) (4) Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1 1 DBE/2013 SCE/SSE - Memorandum Positive marking from 9.2 Positive marking from 9.2 (6 Q) 93 | 6a) Emf =IR + Ir ¥ “6 A oA = Mes”. 8 Log v 24 =3(6+r) | 3v 6 r=3 (4) =20V% 9.4 = : Positive marking from 9.2 (18 V) Positive marking from 9.1(4Q) ORIOF | Vog=3x2 ¥ md VV iv =6VY V2 =RI=3x4=12V v bd Vi20 = 18-6 (4) =12VVY a Decrease Y , Rrot larger ¥ Afneem , Rrot groter (2) ee Increase/ Toeneem “ v } smaller, lost volts less, ~ 1 minder, veriore volts minder “ @) 9.5.3 Increase/Toeneem. “ I smaller Vv, V across 2Q smaller, ¥ 1 minder, V oor 2Q minder (3) [24] RN Copyright reserved/Kopiereg voorbehou TOTALITOTAAL: Ra 19 s-