Physical Sciences Grade 12 P1 NSC June 2012 Memo (Afrikaans)

basic educatio Downloaded from hlayiso.com n Department: Basic Education SENIOR CERTIFICATE EXAMINATION SENIORSERTIFIKAAT-EKSAMEN REPUBLIC OF SOUTH AFRICA PHYSICAL SCIENCE P1__ __ NATUUR- EN SKEIKUNDE V1 _HIGHER GRADE/HOER GRAAD BOR es MEMORANDUM DEPARTMENT OF BASIC EDUCATION 2012 -05- 25 PRIVATE BAG X 110 PRETORIA 0001 PUBLIC EXAMINATIONS This memorandum consists of 10 pages. Hierdie memorandum bestaan uit 10 bladsye. WR STR eG hp WALL Copyright reserved/Kopiereg voorbehou hastens ceeencecess iis piiiaaisssains Sesissss MARKS: 200 PUNTE: 200 Vio s| 10} 1. Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1 2 DBE/2012 SCE/SSE —- Memorandum QUESTION 1/VRAAG 1 B 14 iw A 13°C 14. 8B 15° 4A 16 8B 17 A 18 D 19 A 110 C 111A 112 B 113 D 114 A 115 D [15x 4=60] [General remarks > Accept all equations / formulae as currently used in NCS / CAPS > Where applicable use g = 9,8 m.s* also acceptable > Rounding off to whole numbers or 1 decimal also accepted >Equivilant wording in all cases to be accepted QUESTION 2/VRAAG 2 v force BC/krag BC/Tac 2.4 force AB/krag AB/ Tas 60° Y Or any other correct angle/ enige a ander korrekte hoek Man’s weight/Gravitational force of earth Fe/Fearn/ W (4) Man se gewig/Aantrekkingskrag van die aarde/Fo/F ade! W OR/ OF v force BC/krag BC/Tac Man’s weight/Gravitational force of earth Fe/Featn/ W ¥ Man se gewig/Aantrekkingskrag van die aardelF G/F aarde! W force AB/krag AB/Tas ¥ OR / OF. y-component of force BC/krag BC/Tac ¥ for both components / beide komponente x-component of force BC/krag BC/Tac 4 force AB/krag AB/Tas v Man’s weight/Gravitational force of earth Fe/Feann/ W ‘Man se gewig/Aantrekkingskrag van die aarde/Fg/Faardel W Note: *No arrows no marks / geen pylpunte geen punte + Vectors/ arrows not starting at same point / vektore begin nie in een punt (- 1) «If any extra forces OR BC and components ( — 1} / as enige ander kragte OF BC en komponente (- 1) 2.2. Horizontal / x component/horisontaal / x komponent / Fac cos 60°V ¥ (2) ee DEPARTMENT OF BASIC : EDUCATION Copyright nine voorbehou 2012-08-25 ' PRIVATE BAGX 10 PRETORIA 0004 PUBLIC EXAMINATIONS Please turn over/Blaai om asseblief h

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1 3 DBE/2012 SCE/SSE— Memorandum v 2.3. When three forces acting at the same point are in equilibrium they can be represented in magnitude and direction v by the sides of a triangle taken “ in order./ Wanneer drie kragte op dieselfde punt inwerk wat in ewewig is, kan hulle voorgestel word in grootte en rigting deur die sye van fn driehoek (3) 2.4 Facey = 400 = Faccos60° vor 400 Fes | Fecy | Fee= 800 ¥ 80 Fac Fac y = 800c0s30° v : a = 692,82 Nv Fac = 800 N 5 400? + Facyy)* = 8007 ee) Feoyy) = 69282N ¥ v cos60°= a = y oa o_ BCIY) (4) Fac = 800N tan60°=— oy y singo° = 200) . Fecty) “ Facty) = 400 x tan 60° = 692,82N Fac B00 v | Facty) = 800 x sin60°= 692,82N 2.5 Fg = Faciy yeuilibriunt/ ewewig Fe = Faciy) equilibriun’/ ewewig F=mx10= 692,82N Fg =mx98 = 692,82N 3) m = 69,28 kg y m= 70,70 kg v [16] DEPARTMENT OF BASIC EDUCATION QUESTION 3/VRAAG 3 a1 AB vv 2012 ~05- 25 (2) PRIVATE BAG X 110 32 va p-200" va PRETORIA 0001 v(velocity) = gradient = -00= 2007 e 0,67m.8" ‘PUBLIC EXAMINATIONS) v=0,67m.s~" opposite to direction of original motion he ee of motion | AS v(velocity) = gFadient= 200-9 4 o6Tms 1 rahe ahaa aa v=0,67ms 1 opposite to direction of original motion 2, zero Because the graph shifted a value of 205 could have been used — answer: 0,68 m.s" Omdat die grafiek geskuif het kan ‘n waarde van 205 gebruik word — antwoord: 0,68 ms Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Science/Natuur- en Skeikunde HG/P1/V1 3.3 eoceoe 3.4 3.5 3.6 3.7 3.8 Downloaded from hlayiso.com DEVY v EF vv 4 DBE/2012 SCE/SSE ~ Memorandum Because the graph shifted the following answers will be acceptable: Omdat die grafiek geskuif het sal die volgende antwoorde aanvaar word: Only / slegs BC VVVY (4) or BC vv and DEVY (4) or (2) or BC and any other set of points / BC en enige ander stel puntev ¥ (2) v v 200 + 200 = 400m OR/OF 205 +205 =410m (3) (2) 1.2 vtu $= —at«v = ut + 3 a s=( 5 yt v v. v 200 = (0x 200) + “ (ay(2002) 200 = rs = 2 500 2 2 (5) a=001ms-2v v=2ms" = v4 “ a A og v2 Fu2 +28” 2 =0+ (a)(200) 2 _ , 2* =0 + 2(a)(200) = 2 ee a=001 ms? v v=urat v? =u? + 2asV re v i v = 0 + (0,01)(200) = 0 + 2(0,01)(200) v=2ms' v v=2ms'v¥ ay 2 | v - 4 (m.s") vv 0,67 0 v “DEPARTMENT OF BASIC EDUCATION | 2012 -05- 25 | PrReesTE BAG A 110 PRETORIA 0001 “PUBLIC EXAMINATIONS 100 200 300 400 500 > 6 te)” (6) Note! If 205 is used the answers in 3.6 and 3.7 will still be the same Copyright reserved/Kopiereg voorbehou Ple: As 205 gebruik word sal die antwoorde van 3.6 en 3.7 dieselfde bly [31] ase turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V14 5 DBE/2012 SCE/SSE -— Memorandum QUESTION 4/VRAAG 4 y 4.1 Aresultant/net force acting on a mass produces an acceleration in the direction of the force. The acceleration is directly proportional to the (resultant/net) force and is inversely proportional to the mass./'n Resulterende/netto krag wat inwerk op 'n massa veroorsaak 'n versnelling in die rigting van die krag. Die versnelling is direk eweredig aan die (resulterende/netto) krag en is omgekeerd eweredig aan die massa : OR / OF v v The rate of change in momentum of an object is directly proportional to the resultant / net force acting on the object and the rate of change of momentum is in the direction of the resultant / net force v Die tempo van momentumverandering van ‘n voorwerp is direk eweredig aan die resulterende/netto krag en die tempo van momentumverandering is in die rigting van die resulterende / netto krag. (3) 4.2 v v v T T v P — Q oo 7 v én 3N < 6) Note: ¢ Any additional horizontal forces(— 1) /enige ekstra horisontale kragte (- 1) ¢ If NOT separate diagrams - linked system (— 1) ! as NIE aparte diagramme nie ~ gekoppelde stelse! (- 1) ¢ If no arrows then (-1) / As daar nie pylpunte is nie (-1) 43 v v P and Qa system: (4) : Fres = T -2 = m.a Fes = 55-2-3 =ma=20.1+mi1iv T-2 =201v” m= 30 kg T=22NV yv A For Q: Fyes = 55-T-3 = 30.1 T=22N v Ad 7, / 7, PandQa system: (4) Fres = 55 ~T — 3y= ma Fres = 55-2-3 = Meal” 55-22 -3 =m1v Mp+ay = 50 kg ¥ m=30kg~ | | mass of Q: 50 — 20 = 30 kg “ [16] DEPARTMENT OF BASIC EDUCATION | Copyright reserved|Kopiereg voorbehou Please turn over/Blaai om asseblief | 2012 -05- 25 Fiylyat Ei BAG X + RRETORIA 0001 Ge = PUBLIC EXAMINATIONS |

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1 6 DBE/2012 SCE/SSE — Memorandum QUESTION 5/VRAAG 5 5.1 Fem) v ae ow) _ 80(0-20) v _ 0-20 _— 0,2 = -8000N “ o = -100ms? Vv F = 8000N/ in the opposite direction F on = (80)(-100) = -8000N of the motion of the car / backwards F=8000N in the opposite direction of the motion of the car / backwards (5) v a v 5.2 The time taken to bring the car to a stop increases and this decreases the “force on the driver because the force is inversely proportional to the time. m Av Ap ee ee t orlof 2 At orfof Fe F At= mAv Die tyd wat dit neem om die motor tot stilstand te bring neem toe en dit verminder die krag op die bestuurder want die krag is omgekeerd eweredig aan die tyd. mAv _ Ap Bate ooh aN orlct F At=mAv (4) 19] QUESTION 6/VRAAG 6 61 Any two particles in the universe exert a force of attraction on each other which is vdirectly proportional to the product of the masses of the particles and inversely” proportional to the square of the distance between the two particles. (Objects / bodies and mass / masses will also be acceptable) Enige twee deeitjies in die heelal oefen ‘n aantrekkingskrag op mekaar uit en die krag is direk eweredig aan die produk van die massas van die deeltjies en omgekeerd eweredig aan die kwadraat van die afstand tussen die deeltjies. (3) DEPARTMENT OF BASIC 1 EDUCATION | 2012 -05- 25 ; : i, PRIVATE BAG X 110 Copyright reserved/Kopiereg voorbehou PRETORIA 0001 Please turn over/Blaai om asseblief k “PUBLIC EXAMINATIONS By i

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1 7 DBE/2012 SCE/SSE- Memorandum Mm 6.2 37 and Fy = w =mg r GMmv_ ev g =o, 4) r2 6.3 GM =e Vv G8OMpianet Qplanet 2 Qearth = — U y yt = Gao 16S od _ BAMpIanet _ 80 ae (1 a 80 ye * 5 Searttiaarde Taey 4 5Qptanefplanee val (3) [10] QUESTION 7/VRAAG 7 a (FE, +E av =(E, + Ew If equations of Motion are used 1 1 zero marks are obtained, (mgh +5? ay = (mgh +5 MV" Dany E,(top) = Ex(bottom) 64)(10)(42) +0 4 Vw = 49 mS" (to the right) 4) DEPARTMENT OF BASIC Whet = AEx : EDUCATION (2 i F,.Ax.cos@ = =m(v4* - Vv; ii 9 i i P 2 i 2012 -05- 25 (0,4 *9,8)(1,2)cos0° = 500 anu? -0) | PRIVATE BAG X 110 . PRETORIA 0001 Vv; = 4,85 m.s"to the right v PUBLIC EXAMINATIONS 7.2 total Ppefore collision/ voorbot sing = totalPaftercollision /nabotsing (mv + MV)before/voor = (M+ M)Vatter/na (mv y ae ee (0,449) +0 = (0,4+05)v v=22mst y (my +M2 Watter/na v If before and after not in equation lose 1 mark for formula As voor en na nie in vergelyking nie - verloor 1 punt Copyright reserved/Kopiereg voorbehou 6) Please turn over/Blaai om asseblief A

Downloaded from hlayiso.com ; Physical Science/Natuur- en Skeikunde HG/P1/V1 8 DBE/2012 SCE/SSE — Memorandum 7A 7.2 7.3 (6) Wv tv ip a E, (before) = yim? = 3 (0.4)(497) 0s v ied A 3 v E, (after) = mv? = 5(0,4+05)(22°) = 2.2J AE, =48-2,2=26) v 115) QUESTION 8/VRAAG 8 v The force (electrostatic / force of attraction or repulsion) which one charge exerts on the other is v 8.1 directly proportional to the product of the charges and is inversely v proportional to the square of the distance between them. (3) Die krag (elektrostatiese / 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. 8.2 negative, “v field lines directed towards the charge. positive test charge will be attracted, the direction in which a positive test charge will experience a force in a field, electric field lines point towards negative charge (any one)” Negatief, veldiyne is gerig na die lading toe, positiewe toetsiading sal aangetrek word, rigting waarin ‘n positiewe toetslading ‘n krag sal ondervind in ‘n veld, elektriese veldiyne wys in rigting van ‘n negatiewe lading (enige een van) (3) 8.3 a Fg +4x107> + (-2x1075) v6 =+1x107°C 2v new charge/nuwelading = (3) 84 | 8.4.1 Sphere A, YY Ais positively charged / electrons moving from negative to positive / A is now less positively chargedv Sfeer A, A is positief gelaai / elektrone beweeg van negatief na positief/ A is nou minder positief gelaai (3) 8.4.2 3x 10°C of charge was transferred/3 x 10° C van die lading is oorgedra: Ignore the signs / ignoreer die tekens v x10°° number of electrons / aantal elektrone =>, 1,6x10° a =1,875 10" electrons a [15] DEPARTMENT OF BASIC : : Copyright reserved/Kopiereg VoopbeHgE ATION Please turn over/Blaai om asseblief 2012 -05- 25 fo PRIVATE BAG X 110 PRETORIA 0001 C BRIO OY ARAIMATIORNC

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde HG/P1/V1 9 QUESTION 9/VRAAG 9 v SCE/SSE ~ Memorandum v 9.1 The current in a resistor is directly proportional to the potential difference across the resistor if the temperature of the resistor remains constant.” Die stroom deur ‘n resistor is direk eweredig aan die potensiaalverskil oor die resistor as die temperatuur van die resistor konstant bly. 9.2 The resistance / opposition / difficulty experienced by charge when moving through the cell or battery vv Die weerstand / opposisie / ondervind deur lading as dit deur die se! of battery beweeg. 9.3 1 i . Jy R : R(product) / Roarallel Ry Ro parallel “R(sum) tv, tv _ (10x15) 40° 15 (10+15)v Rparetlel =8Q v Roarellel =6QY DBE/2012 (3) (2) (4) 576 DEPARTMENT OF BASIC 60. EDUCATION 60 =9,6 Watt ead 2012 -05- 25 R PRIVATE BAG X 110 vv v2 PRETORIA 0001 (5) 86=—E, PUBLIC EXAMINATIONS | V=12V¥ p-W Wert ¥ t =2 _ 876 576 =! 15.60 (4) “ony v "se “to . v =9,6 Watt hoa = 5 = 12K =/2p =|2 Yv P=I?R=1*.15 i, =08+12=2A lysq = 08 AY 12 v =—=I12A loa = 79 = 12 v Iq =08+12=2A Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from h Physical Science/Natuur- en Skeikunde HG/P1/V1 10 SCE/SSE — Memorandum layiso.com DBE/2012 fs 96 | emPaIRtinYy Vostornatokstem = Emf — (2 x 1,5) ¥ 24 V=2(R+6)+(2x15) -%4_av 24 =2R+12+3 = 21V (6) 2R =9 14 06¥ R=45Q0v Rexternal/ekstern = “3” ee R = Rosternaviekstem ~ 6 = 10,5-6 = 4,50 V 9.7 R increase, | decreases, less “lostvolts’, (4) V, increases v [28] QUESTION/VRAAG 1: 60 QUESTION/VRAAG 2-9: 140 GRAND TOTAL/GROOTTOTAAL: 200 | eerie Ba aNATO EXAMINATIONS aot -05° 25 BAG X10 RIA 0001 ee PRIVATE PUBLIC | | eenanornennooet Copyright reserved/Kopiereg voorbehou ff