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

Downloaded from hlayiso.com NTT basic education Department: Basic Education REPUBLIC OF SOUTH AFRICA SENIOR CERTIFICATE EXAMINATION SENIORSERTIFIKAAT-EKSAMEN PHYSICAL SCIENCE P1/NATUUR- EN SKEIKUNDE V1 oe __PHYSICSIFISIKA STANDARD GRADE/STANDAARDGRAAD pee EDUCATION MARKS: 150 pote -05- 25 PUNTE: 150 TE BAG X (10 PRETORIA 0001 _ PUBLIC EXAMINATION This memorandum consists of 9 pages. Hierdie memorandum bestaan uit 9 bladsye. Vs Ga ya — aS a MALLS (9-9 S-VAL Please turn over/Blaai om asseblief 0 Copyright reserved/Kopiereg voorbehou Aetl | | |

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde/SG/P1/V1 2 DBE/2012 SCE/SSE ~ Memorandum QUESTION 1/VRAAG 1 11 +A 12 B 13° 4A 14 A ik © is) (3) eh 18 C 19 A 10 B 1.11 B 1.12 C 1.13 D 1.14 A 1.15 C [15x3 = 45] [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 > Equivalent wording in all cases to be accepted QUESTION 2/VRAAG 2 2.1. * the resultant force on the block is zero/resulterende krag op blok is nul OR/OF * all the forces acting on the block are balanced/al die kragte wat op die blok inwerk, is gebalanseerd. ORIOF e Itis at rest/dit is in rus. e All forces will form a closed triangle / a/ die kragte vorm ‘n geslote driehoek vv (2) 43.cm 40° 10 cm 10cm (8) v Measure 10 cm (50 N) — must be ‘horizontal’ Meet die 10 cm (50 N) horisontaal ¥ Measure 140° (or 40° from horizontal) Meet die 140° (of die 40° vanaf die horisontaal) v Measure 13 cm (force in XY)/Meet 13 cm (krag in XY) v¥ Complete parallelogram or triangle/Voltooi parallelogram of driehoek v¥ Measure R/Meet R (8.1 cm to 8.5 cm) ¥ Convert R to N/Skakel R om na N_: Range: 40,5 N to Y Direction/Rigting O0°%lrigting 0°/due north/reg noort PARTMENT OF BASIC 42 g2.BH EDUCATION = 41,5 Nin a direction of 0°/rigting 0°/due north/reg ngord 2012 -05- 25 PRIVATE BAG X 110 F=mg v F=mg Y iH PRETORIA 0001 Y v ¥ PUBLIC EXAMINATIONS 41,5=mx10 at 5= mx9,8 a m = 4,15 kg¥ m = 4,23 kg¥ (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde/SG/P1/V1 3 DBE/2012 SCE/SSE — Memorandum QUESTION 3/VRAAG 3 : 34 =u + atv v=utatyv VY Av _6Y v=at a=——=— v _ 0 marks 6 = a(4) =15ms a=15ms? ¥ a=1,5ms* (4) 3.2 | For1 secondat constant velocity/ 112 = = v s ut + pat : a = (6)(1) + gO =6mY OR/OF s = vt= 6)(1)=6 mY OR/OF v + s= (Ite CS =6m¥ Vir 1 sekond teen konstante snelheid If the displacement to come to restis 12 m/As die verplasing om tot rus te kom, 12 m is y2 =u +2as % v (0% es 2(-1)s s=18mv Cyclist need more than 18 mto stop/Fietsryer benodig meer as vv 18m omte stop OR/OF Total displacement/ Totale verplasing =6+18 =24mv _ For1 second at constant velocity/ Vir 1 sekond teen konstante snelheid : 1 =ut+—at? Vv s=ut 3% = 61) + ZO)? =6m/’ ORIOF s=vt=(6)(1)=6m”% OR / OF vv utv 6+6 $= (1 = me =6m Copyright reserved/Kopiereg voorbBO8 0S- 25 PRIVATE BAG * 10 PRETORIA 0001 me Cyclist need more than 18 mto stop/Fietsryer benodig meer as 18m om te stop If the displacement to come to restis12 m/As die verplasing om tot rus te kom, 12 m is y2=u2 +2as % v Cores 2(a)12 =-15 ms? ¥ Cyclist has a constant deceleration of -1 ms"? he must have a deceleration of - 1,5 m.s"2to stop in time /He does not stopintime vv Fietsryer het 'n konstante negatiewe versnelling van-1 m.s”?, hy moet 'n negatiewe versnelling -2 : | .van-1,5m.s"“ he om betyds te stop : Hy stop dus + Gie betyds nie. Please turn over/Blaai om asseblief ane ELAMINATONS PUBLIC EX — K

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde/SG/P1/V1 4 DBE/2012 SCE/SSE - Memorandum For 1 second at constant velocity/ Vir 1 sekond teen konstante snelheid Le =ut+ cat" v s=ut+ sal - 2 = (6)(1) + 30) =6m Vv OR/OF s= v= (6)(1)=6 mY OR / OF wa utv 6+6 s= (t= (SM 6m % 3.3 v If the displacement to come to restis 12 m/As die verplasing om tot rus te kom, 12m is a s=utt+ 2at v v v 12 =6t+ see t=255s ¥ v=utat 0=6+(-1)t t=6s¥% Cyclist must complete final 12 min6 s but has only 2,55 s v Fietsryer moet laaste 12 min6 s voltooi, het net 2,55 s Cyclist will not stop in time / stop nie betyds nie (8) v v Cyclist continues to move forward at same speed/velocity due to inertia/Newton4 Fietsryer beweeg steeds vorentoe teen selfde spoed/snelheid a.g.v. traagheid v v An object will stay at rest or carry on moving at a_ constant velocity in a straight line unless acted on by an external resultant / net force (Newton1) ” n Voorwerp sal in rus bly of teen n konstante spoed in n reqguit lyn volhard tensy daar in eksterne resulterende / netto krag daarop inwerk (Newton1) (3) Copyright reserved/Kopiereg voorbehou [15] DEPARTMENT OF BASIC EDUCATION 202 -05- 25 PHaVATE BAG x 110 PRETORIA 0001 PUBLIC EXAMINATIONS err L Please turn over/Blaai om asseblief Go? I

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde/SG/P /V1 5 DBE/2012 SCE/SSE — Memorandum QUESTION 4/VRAAG 4 If 15 m.s” used without being calculated — full marks as 15 m.s' gebruik word sonder ‘n berekening — volpunte ifg=9,8m. g2 is used " = As g=9,8m.s" 1 v=u* gt gebruik: s=ut+ zat'v = 0 + (10)(1,5) Answer / antwoord: v 2 =15m st ae = 0+ 3001.5) — $= 11,03m 2 ¥5+0”% s=11,03m = 11,25 mv Beuk + 298” aes ns \ 4 @ <0) (toys) |} Y= 14,28 m. (4) s =17,25m v 4.2 on if g = 9,8 ms? is used _ g = 9,8 m.s~ is use = 0 + (10)(18) ae 7 As g=9,8m.s? =15ms7¥ 2 gebruik: OR/OF VA 25 = “5 Ye Answer / antwoord: v? =u? 4 298 ee v=147 ms" = (0) + 2(10)(11,25) DEPARTMENT OFLBASIC v =15m.st% EDUCATION 2012 -05- 25 tt PRIVATE BAG X 110 QUESTION S/VRAAG 5 PRETORIA 0004 PUBLIC EXAMINATIONS 5.1 Aresultant force atting on a mass proauces an Acceleration in the direction of the y nice. The acceleration is directly propoftional to the resultant force and is ¥ inversely propotional to the mass. nResulterende krag wat op n massa inwkrk. sal die massa laat versnel in die rigting van die resulterende krag. Die versnelling is direk eweredig aan die krag en v 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.” Die tempo van momentumverandering van ‘n voorwerp is direk eweredig aan die resulterende/netto krag en die momentumverandering is in die rigting van die resulterende / netto krag. F (3) aa— 1mark only m Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde/SG/P1/V1 6 DBE/2012 SCE/SSE — Memorandum =ma” v 5.2 Fres = ane ov Fres = Ma = Fappliea — Friction 7 4 v v = 110N 55x 2= 115 — Friction Fres = Friction + Fapptied wee aaa, 110 = Friction + 115 ¥ Friction = 115-110 Friction = 5N ¥ on = v > : Ee Feton N a ne Friction = 5 N in the opposite a friction = 9 Nin ; direction of being pulled v I opposite direction of being pulled QUESTION 6/VRAAG 6 v v ¥ 6.1 In an isolated system the total (linear) momentum remains a constant in magnitude and direction, or v v v In an isolated system the total momentum before a collision is equal to the total momentum after the collision. (3) v v v Inn geisoleerde stelsel bly die totale (liniére) momentum konstant in grootte en tigting v v of Inn geisoleerde stelsel is die totale momentum voor fn botsing gelyk aan die totale momentum na die botsing. 6.2 | (total momentuM)pefore = (total MomentuM)arer (mp + mMs)U = Mp_V + MeV v v v (50 410)2 = 50v + 10(-3) v = 3 ms" Vto the right vina regs (6) Pbefore = Pafter “DEPARTMENT OF BASIC EDUCATION wl momentuMpbefore = MOoMEntuUMatter 12 -05 25 19) | PRIVATE BAG A 109 j PRETORIA 0001 PUBLIC EXAMINATION Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde/SG/P 1/V1 if DBE/2012 SCE/SSE — Memorandum QUESTION 7/VRAAG 7 7.1. The sum “of the gravitational potential and kinetic energy of the object. v (2) Die som v van die gravitasie potensiéle energie en die kinetiese energie van die voorwerp ¥ Law of conservation of mechanical energy will also be accepted Wet van behoud van meganiese enerige sal ook aanvaar word. a2 j total mechanical Energy soomm = total Mechanical ENerY vest polht of motion _ v (E, + Ey )atsoomm = (E, + Ey atiowest point of motion /laagste punt van beweging (mgh +5 coon = (mgh +3mv?) at the lowest point of motion laagste punt van beweging YoY A (0,15)(10)(0,1) + 0 = (0,15)(40)(0) + 3(015)v") =1,41 mst” Viowest point of moiion/iaagste punt van beweging If 9,8 m.s” used then the answer is : 14ms7 (6) Note: E,(top) = E,(bottom) will lose only 1 mark can still get 5 E,(bo) = E,(onder) verloor 1 put maar kan nog 5 punte verdien v=./2gh zero o 6 7.3. Equations of motion only apply to motion in a straight line. ~v” OR The bob does not move at uniform acceleration. OR To apply equations of motion the bob must move under uniform acceleration. (2) Bewegingsvergelykings is slegs van toepassing op liniére beweging. VY OF Die massastuk beweeg nie teen uniforme versnelling nie OF Om van toepassing te wees, moet die massastuk uniform versnel. [10] DEPARTMENT OF BASIC EDUCATION 2012 -05- 25 PRIVATE BAG & 110 PRETORIA 0001 L_PUBLIC EXAMINATIONS | Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Bo

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde/SG/P 1/V1 8 DBE/2012 SCE/SSE — Memorandum QUESTION 8/VRAAG 8 v v v v 8.1 Phas an excess of electrons /P het in oormaat elektrone/ (2) v 8.2 The force (electrostatic / force of attraction or repulsion) which one charge exert v on the other is directly proportional to the product of the charges and is inversely v proportional to the square of the distance between them. v Die elektrostatiese (aantrekkende of afstotende) krag wat een lading op n ander v v uitoefen, is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat_van die afstand tussen hulle. (3) Curved electric field line impt. on outside Geboé elektriese veld en lyne belangrik op buitekant ¥ field lines between charges/veldiyne tussen die ladings Y field lines to left and right of the charges/veldlyne links en regs (buite die) van ladings Y direction: from Q to P/rigting: van Q na P toe 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 If dots are used for spheres award full marks for last bullet (4) 8.4 kqyqo y DEPARTMENT OF BASIC fa EDUCATION Jo v _ (9x10°)(6x1079)(3x1079) ma, 2012 -05- 25 F 7 PRIVATE BAG X 110 =18x10-5N % PRETORIA 0001 PUBLIC EXAMINATIONS | 8.5 . 9% 49-9 gy (3) new charge = 2% 107 +310” =-15x1079 C [18] 2v Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief f

Downloaded from hlayiso.com Physical Science/Natuur- en Skeikunde/SG/P1/V1 9 SCE/SSE — Memorandum DBE/2012 QUESTION 9/VRAAG 9 9.1 f Sv ae —— Rtotal series = 14 2=30 R Rea Rio Rat tt iy R Rea Reon a rey a 60 series [ 3 1) “2 “ Ihe =20 = v R effective) parallel = = 2 Q voivy v 92.1 emf=V=IR=6x2=12V OR /O) amis Ry = (20) = 13 9.2.2 | = Vert ThA rs Sled 10+20 R +R ov ~~ Gof Vacross"2 2 aay, “4A x42 8V =x12= V, Flea * Roo ¥ =4x2” =8VVY 9.3.1 Remains the same/bly dieselfde 9.3.2 Decreases/neem af vv 9.3.3 Decreases/neem af, Copyright reserved/Kopiereg voorbehou A Rtotal series = 14 2=30 Rproduct v Reffec = R sum _ (6x3) ¥ “6 +3) Vv R =2Q v (8) (4) = PARTMENT OF BASI DEP EDUCATION - ani? -05- 25°) PRIVATE BAG X 110 PRETORIA 0001 PUBLIC EXAMINATION: ee (2) (2) (2) [21] Question 1: 45 Question 2-9: 105 Grand Total: 150 Ke