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Physical Sciences Grade 12 P1 Eastern Cape Mock 2021 Memo

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Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape Mock Exam 2021 (English), 17 pages. Read online or download the free PDF.

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Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2021 PHYSICAL SCIENCES P1/ FISIESE WETENSKAPPE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 17 pages./ Hierdie nasienriglyn bestaan uit 17 bladsye.
Downloaded from hlayiso.com 2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021) GENERAL GUIDELINES/ALGEMENE RIGLYNE 1. CALCULATIONS/BEREKENINGE 1.1 Marks will be awarded for: correct formula, correct substitution, correct answer with unit. Punte sal toegeken word vir: korrekte formule, korrekte substitusie, korrekte antwoord met eenheid. 1.2 No marks will be awarded if an incorrect or inappropriate formula is used, even though there are many relevant symbols and applicable substitutions. Geen punte sal toegeken word waar ʼn verkeerde of ontoepaslike formule gebruik word nie, selfs al is daar relevante simbole en relevante substitusies. 1.3 When an error is made during substitution into a correct formula, a mark will be awarded for the correct formula and for the correct substitutions, but no further marks will be given. Wanneer ʼn fout gedurende substitusie in ʼn korrekte formule begaan word, sal ʼn punt vir die korrekte formule en vir korrekte substitusies toegeken word, maar geen verdere punte sal toegeken word nie. 1.4 If no formula is given, but all substitutions are correct, a candidate will forfeit one mark. Indien geen formule gegee is nie, maar al die substitusies is korrek, verloor die kandidaat een punt. 1.5 No penalisation if zero substitutions are omitted in calculations where correct formula/principle is correctly given. Geen penalisering indien nulwaardes nie getoon word nie in berekeninge waar die formule/beginsel korrek gegee is nie. 1.6 Mathematical manipulations and change of subject of appropriate formulae carry no marks, but if a candidate starts off with the correct formula and then changes the subject of the formula incorrectly, marks will be awarded for the formula and correct substitutions. The mark for the incorrect numerical answer is forfeited. Wiskundige manipulasies en verandering van die onderwerp van toepaslike formules tel geen punte nie, maar indien ʼn kandidaat met die korrekte formule begin en dan die onderwerp van die formule verkeerd verander, sal die punte vir die formule en korrekte substitusies toegeken word. Die punt vir die verkeerde numeriese antwoord word verbeur. 1.7 Marks are only awarded for a formula if a calculation has been attempted, i.e. substitutions have been made or a numerical answer given. Punte word slegs vir ʼn formule toegeken indien ʼn poging tot berekening aangewend is, d.w.s. substitusies is gedoen of ʼn numeriese antwoord is gegee. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 3 1.8 Marks can only be allocated for substitutions when values are substituted into formulae and not when listed before a calculation starts. Punte kan slegs toegeken word vir substitusies wanneer waardes in formules ingestel word en nie vir waardes wat voor ʼn berekening gelys is nie. 1.9 All calculations, when not specified in the question, must be done to a minimum of two decimal places. Alle berekenings, wanneer nie in die vraag gespesifiseer word nie, moet tot ʼn minimum van twee desimale plekke gedoen word. 1.10 If a final answer to a calculation is correct, full marks will not automatically be awarded. Markers will always ensure that the correct/appropriate formula is used and that workings, including substitutions, are correct. Indien ʼn finale antwoord van ʼn berekening korrek is, sal volpunte nie outomaties toegeken word nie. Nasieners sal altyd verseker dat die korrekte/toepaslike formule gebruik word en dat bewerkings, insluitende substitusies korrek is. 1.11 Questions where a series of calculations have to be made (e.g. a circuit diagram question) do not necessarily always have to follow the same order. FULL MARKS will be awarded provided it is a valid solution to the problem. However, any calculation that will not bring the candidate closer to the answer than the original data, will not count any marks. Vrae waar ʼn reeks berekeninge gedoen moet word (bv. ʼn stroombaan- diagramvraag) hoef nie noodwendig dieselfde volgorde te hê nie. VOLPUNTE sal toegeken word op voorwaarde dat dit ʼn geldige oplossing vir die probleem is. Enige berekening wat egter nie die kandidaat nader aan die antwoord as die oorspronklike data bring nie, sal geen punte tel nie. 2. UNITS/EENHEDE 2.1 Candidates will only be penalised once for the repeated use of an incorrect unit within a question. Kandidate sal slegs een keer gepenaliseer word vir die herhaaldelike gebruik van ʼn verkeerde eenheid in ʼn vraag. 2.2 Units are only required in the final answer to a calculation. Eenhede word slegs in die finale antwoord op ʼn vraag verlang. 2.3 Marks are only awarded for an answer, and not for a unit per se. Candidates will therefore forfeit the mark allocated for the answer in each of the following situations:  Correct answer + wrong unit  Wrong answer + correct unit  Correct answer + no unit Punte sal slegs vir ʼn antwoord en nie vir ʼn eenheid per se toegeken word nie. Kandidate sal die punt vir die antwoord in die volgende gevalle verbeur:  Korrekte antwoord + verkeerde eenheid  Verkeerde antwoord + korrekte eenheid  Korrekte antwoord + geen eenheid Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com 4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021) 2.4 SI units must be used except in certain cases, e.g. V.m-1 instead of N.C-1, and cm•s-1 or km•h-1 instead of m•s-1 where the question warrants this. SI-eenhede moet gebruik word, behalwe in sekere gevalle, bv. V•m-1 in plaas van N•C-1, en cm•s-1 of km•h-1 in plaas van m•s-1 waar die vraag dit regverdig. 3. GENERAL/ALGEMEEN 3.1 If one answer or calculation is required, but two are given by the candidate, only the first one will be marked, irrespective of which one is correct. If two answers are required, only the first two will be marked, etc. Indien een antwoord of berekening verlang word, maar twee word deur die kandidaat gegee, sal slegs die eerste een nagesien word, ongeag watter een korrek is. Indien twee antwoorde verlang word, sal slegs die eerste twee nagesien word, ens. 3.2 For marking purposes, alternative symbols (s, u, t, etc.) will also be accepted. Vir nasiendoeleindes sal alternatiewe simbole (s, u, t, ens.) ook aanvaar word. 3.3 Separate compound units with a multiplication dot, not a full stop, for example, m•s-1. For marking purposes, m•s-1 and m/s will also be accepted. Skei saamgestelde eenhede met ʼn vermenigvuldigingspunt en nie met ʼn punt nie, byvoorbeeld m•s-1. Vir nasiendoeleindes sal m•s-1 en m/s ook aanvaar word. 4. POSITIVE MARKING/POSITIEWE NASIEN Positive marking regarding calculations will be followed in the following cases: Positiewe nasien met betrekking tot berekeninge sal in die volgende gevalle geld: 4.1 Sub-question to sub-question: When a certain variable is calculated in one sub-question (e.g. 3.1) and needs to be substituted in another (3.2 of 3.3), e.g. if the answer for 3.1 is incorrect and is substituted correctly in 3.2 or 3.3, full marks are to be awarded for the subsequent sub-questions. Subvraag na subvraag: Wanneer ʼn sekere veranderlike in een subvraag (bv. 3.1) bereken word en dan in ʼn ander vervang moet word (3.2 of 3.3), bv. indien die antwoord vir 3.1 verkeerd is en word korrek in 3.2 of 3.3 vervang, word volpunte vir die daaropvolgende subvraag toegeken. 4.2 A multistep question of a sub-question: If the candidate has to calculate, for example, current in die first step and gets it wrong due to a substitution error, the mark for the substitution and the final answer will be forfeited. ʼn Vraag met veelvuldige stappe in ʼn subvraag: Indien ʼn kandidaat bv. die stroom verkeerd bereken in ʼn eerste stap as gevolg van ʼn substitusiefout, verloor die kandidaat die punt vir die substitusie sowel as die finale antwoord. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 5 5. NEGATIVE MARKING/NEGATIEWE NASIEN Normally an incorrect answer cannot be correctly motivated if based on a conceptual mistake. If the candidate is therefore required to motivate in QUESTION 3.2 the answer given in QUESTION 3.1, and QUESTION 3.1 is incorrect, no marks can be awarded for QUESTION 3.2. However, if the answer for e.g. QUESTION 3.1 is based on a calculation, the motivation for the incorrect answer could be considered. ʼn Verkeerde antwoord, indien dit op ʼn konsepsuele fout gebaseer is, kan normaalweg nie korrek gemotiveer word nie. Indien ʼn kandidaat gevra word om in VRAAG 3.2 die antwoord op VRAAG 3.1 te motiveer en VRAAG 3.1 is verkeerd, kan geen punte vir VRAAG 3.2 toegeken word nie. Indien die antwoord op bv. VRAAG 3.1 egter op ʼn berekening gebaseer is, kan die motivering vir die verkeerde antwoord in VRAAG 3.2 oorweeg word. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com 6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021) QUESTION/VRAAG 1: MULTIPLE-CHOICE QUESTIONS/ MEERVOUDIGEKEUSE-VRAE 1.1 C  (2) 1.2 A  (2) 1.3 B  (2) 1.4 D  (2) 1.5 D  (2) 1.6 C  (2) 1.7 A  (2) 1.8 C  (2) 1.9 D  (2) 1.10 B  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 7 QUESTION/VRAAG 2 2.1.1 When a (non-zero) resultant/net force acts on an object, it accelerates in the direction of the force. The acceleration is directly proportional to the force and inversely proportional to the mass of the object.  Indien ʼn (nie-nul) resultante/netto krag op ʼn voorwerp inwerk, versnel dit in die rigting van die krag. Die versnelling is direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp.  (2) 2.1.2 OPTION/OPSIE 1 OPTION/OPSIE 2 N N F Fy cc  f f T Fx  T w w (5) Accept the following symbols./Aanvaar die volgende simbole: N FN/Normal/Normaal/Normal force/Normaalkrag Ff / fk / fr / frictional force/wrywingskrag/kinetic frictional force / f kinetiese wrywingskrag Fg,/mg/weight/FEarth on block/49 N/gravitational force/gewig/Faarde op blok/ w gravitasiekrag T Tension / Spanning / FT / Fs Fapplied/toegepas  F / FA / Applied force / Toegepaste krag Marks awarded for arrow and label/Punte toegeken vir pyltjie en byskrif. Do not penalise for length of arrows since drawing is not drawn to scale. Moenie penaliseer vir pyltjie-lengtes nie (diagram is nie volgens skaal nie). 3 Any other additional force(s)/Enige addisionele krag(te) 4 3 If force(s) do not make contact with body. Max./Maks. 4 Indien krag(te) nie kontak maak met voorwerp nie. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com 8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021) 2.1.3 OPTION/OPSIE 1 OPTION/OPSIE 2 (To the right is positive) (To the right is negative) (Na regs positief) (Na regs negatief) Fnet = ma Fnet = ma Any one  Any one T – f = ma f – T = -ma Enige een  Enige 1 Fcosθ - T – f = ma T + f – Fcosθ = -ma T – 10 = 2 (2) 10 – T = 2(-2)  T = 14 N T = 14 N Fcosθ - T – f = ma T + f – Fcosθ = -ma ° 14 + 15 -Fcos20 = 5(-2)  Fcos20o – 14 – 15 = 5(2)  F = 41,50 N  F = 41,50 N  (5) 2.2 Gm 1m 2 F=  d2 (6,67  10-11 )( 5,98  1024 )(200) 1 842,50  =  d2 d = 6 579 982,80 m distance above earth surface/afstand bo die aardoppervlak = 6 579 982,80 – 6,38 x 106  = 199 982,80 m (1,9998280 x 105 m / 2,00 x 105 m)  (5) [17] QUESTION/VRAAG 3 3.1.1 OPTION/OPSIE 1 UPWARDS POSITIVE UPWARDS NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf2 = vi2 + 2aΔy  vf2 = vi2 + 2aΔy  (-25)2 = (-20)2 + 2(-9,8) Δy  (25)2 = (20)2 + 2(9,8) Δy  Δy = -11,48 Δy = 11,48 m  Δy = 11,48 m  OPTION/OPSIE 2 UPWARDS POSITIVE UPWARDS NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + aΔt vf = vi + aΔt -25 = -20 + (-9,8)(Δt) 25 = 20 + (9,8)(Δt) Δt = 0,51 s Δt = 0,51 s Δy = viΔt + ½gΔt2  Δy = viΔt + ½gΔt2  Δy = (-20)(0,51) + ½(-9,8)(0,51)2  Δy = (20)(0,51) + ½(9,8)(0,51)2  Δy = -11,47 Δy = 11,47 m  Δy = 11,47 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 9 OPTION/OPSIE 3 UPWARDS POSITIVE UPWARDS NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + aΔt vf = vi + aΔt -25 = -20 + (-9,8)(Δt) 25 = 20 + (9,8)(Δt) Δt = 0,51 s Δt = 0,51 s v  vi v  vi Δy = f Δt  Δy = f Δt  2 2 - 25  (-20) 25  20 Δy = x 0,51 Δy = x 0,51 2 2 Δy = -11,48 m Δy = 11,48 m  Δy = 11,48 m  OPTION/OPSIE 4 (ACCEPT/AANVAAR) UPWARDS POSITIVE UPWARDS NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + aΔt vf = vi + aΔt -25 = -20 + (-9,8)(Δt) 25 = 20 + (-9,8)(Δt) Δt = 0,51 s Δt = 0,51 s Δy = lb + ½bh  Δy = lb + ½bh  Δy = 20 x 0,51 + ½(5)0,51  Δy = 20 x 0,51 + ½(5)0,51  Δy = 11,48 m  Δy = 11,48 m  (3) 3.1.2 UPWARDS POSITIVE UPWARDS NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + aΔt  vf = vi + aΔt  -25 = -20 + (-9,8)(Δt)  25 = 20 + (-9,8)(Δt)  Δt = 0,51 s (time to reach to ground) Δt = 0,51 s (time to reach to ground) (tyd om die grond te bereik) (tyd om die grond te bereik) vf = vi + aΔt 0 = 12 + (-9,8)(Δt)  vf = vi + aΔt Δt = 1,22 s 0 = -12 + (9,8)(Δt)  (time to reach maximum height) Δt = 1,22 s (tyd om maks hoogte te bereik) (time to reach maximum height) t = 1,22 + 0,51  (tyd om maks hoogte te bereik) t = 1,73 s  t = 1,22 + 0,51  t = 1,73 s  (5) 3.1.3 OPTION/OPSIE 1 Positive marking from 3.1.1/Positiewe merk vanaf 3.1.1 UPWARDS POSITIVE UPWARDS NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf2 = vi2 + 2aΔy  vf2 = vi2 + 2a  2 2 0 = 12 + 2(-9,8) Δy  02 = -122 + 2(9,8) Δy  Δy = 7,35 m Δy = -7,35 m Displacement = - 11,48 + 7,35  Displacement = 11,48 + (- 7,35)  Verplasing = - 4,13 Verplasing = 4,13 m = 4,13 m (downwards/afwaarts)  (downwards/afwaarts)  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com 10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021) OPTION/OPSIE 2 Positive marking from 3.1.1/Positiewe merk vanaf 3.1.1 UPWARDS POSITIVE UPWARDS NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + aΔt vf = vi + aΔt 0 = 12 + (-9,8)(Δt) 0 = -12 + (9,8)(Δt) Δt = 1,22 s Δt = 1,22 s 2 Δy = viΔt + ½aΔt  Δy = viΔt + ½aΔt2  Δy = 12 x1,22 + ½ (-9,8)(1,22)2  Δy = (-12)(1,22) + ½ (9,8)(1,22)2  Δy = 7,35 m Δy = -7,35 Displacement = - 11,48 + 7,35  Displacement = 11,48 + (- 7,35)  Verplasing = - 4,13 Verplasing = 4,13 m = 4,13 m (downwards/afwaarts)  (downwards/afwaarts)  OPTION/OPSIE 3 Positive marking from 3.1.1 / Positiewe merk vanaf 3.1.1 UPWARDS POSITIVE UPWARDS NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + aΔt vf = vi + aΔt 0 = 12 + (-9,8)(Δt) 0 = -12 + (9,8)(Δt) Δt = 1,22 s Δt = 1,22 s v  vi v  vi Δy = f Δt  Δy = f Δt  2 2 0  (12) 0  (-12) Δy = x 1,22  Δy = x 1,22  2 2 Δy = 7,32 m Δy = - 7,32 m Displacement = - 11,48 + 7,32  Displacement = 11,48 + (- 7,32)  Verplasing = - 4,16 Verplasing = 4,16 m = 4,16 m (downwards/afwaarts)  (downwards/afwaarts)  OPTION/OPSIE 4 Positive marking from 3.1.1 / Positiewe merk vanaf 3.1.1 UPWARDS POSITIVE UPWARDS NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + aΔt vf = vi + aΔt 0 = 12 + (-9,8)Δt 0 = -12 + (9,8)Δt Δt = 1,22 s Δt = 1,22 s Area/oppervlakte = ½bh  Area/oppervlakte = ½bh  = ½ (1,22)(12)  = ½ (1,22)(-12)  = 7,32 m = -7,32 m Displacement = - 11,48 + 7,32  Displacement = 11,48 + (- 7,32)  Verplasing = - 4,16 Verplasing = 4,16 m = 4,16 m (downwards/afwaarts)  (downwards/afwaarts)  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 11 3.2 Positive marking from 3.1.1 and 3.1.2 / Positiewe merk vanaf 3.1.1 en 3.1.2 Position/Posisie(m) 11,48 m Time/Tyd(s) 1,73 m CRITERIA FOR MARKING/MERK KRITERIA Correct shape/Korrekte vorm  Height indicated/Hoogte aangedui (11,48 m)  Time t indicated/Tyd t aangedui (1,73 s)  (3) [15] QUESTION/VRAAG 4 4.1 In an isolated system total linear momentum is conserved.  Die totale lineêre momentum bly behoue in ʼn geïsoleerde/geslote sisteem. (2) 4.2.1 ∑pi = ∑pf Any one/Enige een  mAviA + mBviB = (mA + mB)vf (2 x viA) + (4 x -5) = (2+4)(-1,67)  viA = 4,99 m.s-1 (East/Ooswaarts )  (4) 4.2.2 POSITIVE MARKING FROM POSITIVE MARKING FROM 4.2.1/Positiewe merk vanaf 4.2.1 4.2.1/Positiewe merk vanaf 4.2.1 OPTION/OPSIE 1 OPTION/OPSIE 2 Fnet.∆t = ∆p Fnet.∆t = ∆p Any one Any one Fnet.∆t = m(vf – vi) /Enige een  Fnet.∆t = m(vf – vi) /Enige een  Fnet x 0,01 = 2 (-1,67- 4,99)  Fnet x 0,01 = 2 [1,67- (-4,99)]  Fnet = - 1 332 N Fnet = - 1 332 N Fnet = 1 332 N west/left/Wes/links  Fnet = 1 332 N west/left/Wes/links  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com 12 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021) OPTION/OPSIE 3 OPTION/OPSIE 4 Fnet = ma Any one Fnet = ma Any one v f - vi /Enige een  v - vi Fnet = m ( ) Fnet = m ( f ) /Enige een  t t - 1,67 - 4,99  1,67 - (- 4,99)  Fnet = 2 x ( ) Fnet = 2 x ( ) 0,01  0,01 Fnet = -1 332 Fnet = 1 332 N west/left/Wes/links  Fnet = 1 332 N west/left/Wes/links  OPTION 5/OPSIE 5 OPTION 6/OPSIE 5 Fnet.∆t = ∆p Fnet.∆t = ∆p Any one /Enige Any one /Enige Fnet.∆t = m(vf – vi) een Fnet.∆t = m(vf – vi) een Fnet(0,01)  = 4(-1,67 – -5)  Fnet(0,01)  = 4(1,67 – 5)  Fnet = 1 332 Fnet = -1 332 FAB = - FBA FAB = - FBA Fnet(BA) = 1 332 N west/left  Fnet(BA) = 1 332 N west/left  (4) [10] QUESTION/VRAAG 5 5.1 Gravitational force / Gravitasiekrag  (1) 5.2 12 ∆x =  = 24 m sin30o Wf = f.∆x cos θ  Wf = 35,5 x 24 cos 180o  Wf = - 852 J  (4) 5.3 Zero/0 J  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 13 5.4 Positive marking from 5.2/Positiewe merk vanaf 5.2 OPTION/OPSIE 1 Wnet = Ek Any one/Enige een  Wf + W F + W Fg = Ek f x ∆x cos θ + F xcos θ + mg(h2 – h1) = Ek -852  + (62,5 x 24 cos180o)  + m(9,8)(12-0)  = 0 m = 20 kg  OPTION/OPSIE 2 Wnc = Ep + Ek W f + W F = Ep + Ek Any one/Enige een  f∆xcos θ + F xcos = mg(h2 – h1) + Ek -852 + (62,5)(24)cos180o = m(9,8)(0 – 12) + 0 m = 20 kg  OPTION 3/OPSIE 3 Wnet = Ek W f + W F + W w = Ek Any one/Enige een  f∆xcos θ + F xcos θ + mgΔxcos θ = Ek -852  + (62,5)(24)cos180o  + m(9,8)(24)cos60o  = 0 m = 20 kg  (5) [11] QUESTION/VRAAG 6 6.1.1 520 Hz / 520 waves per second (waves.s-1)/golwe per sekondes (golwe.s-1)  (1) 6.1.2 The change in frequency (or pitch) observed/detected by a listener because the listener and the sound source have different velocities relative to the medium of sound propagation.  Die verandering in frekwensie  (of toonhoogte) waargeneem deur ʼn luisteraar want die luisteraar en die klankbron het verskillende snelhede relatief tot die medium wat klank voortplant.  OR/OF The (apparent) changed in observed/detected frequency (pitch) as a result of relative motion between the sound source and the listener.   Die waarskynlike verandering in die waargenome frekwensie (toonhoogte) as gevolg van die relatiewe beweging tussen die klankbron en die luisteraar.  (2) 6.1.3 TOWARDS  Detected frequency is higher than the source frequency  NA.  Waargenome frekwensie is hoër as die bron se frekwensie  (2) 6.1.4 v ± vL fL = fs  v ± vs 343 520  =  (480)  343 - v s vs = 26,38 m·s-1  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com 14 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021) 6.1.5 Decreases/Neem af  For a constant velocity/speed of sound, if the frequency increases, λ decreases.  Vir ʼn konstante snelheid/spoed van klank, indien die frekwensie toeneem, neem λ af. OR/OF λα at constant velocity/speed/by konstante snelheid/spoed  OR/OF f α at constant velocity/speed/by konstante snelheid/spoed  (2) 6.2 Light from the star is shifted towards longer wavelength (towards the red end of the spectrum) which indicated that the star is moving away from the earth.  Lig van die ster skuif na ʼn langer golflengte (na die rooi kant van die spektrum)  wat aandui dat die ster wegbeweeg van die aarde af.  (2) 6.3 Used to measure the direction and speed of blood flow in arteries and veins.  Word gebruik om die rigting en spoed waarteen bloed in are en slagare vloei, te meet. OR/OF Used to measure the heartbeat of a foetus in the womb. Word gebruik om die hartklop van ʼn fetus in die baarmoeder te meet. (1) [15] QUESTION/VRAAG 7 7.1.1 GAIN/BYGEKRY  (1) 7.1.2 Q n=  qe 5 x 10 -6 n= 19  1,6 x 10 - n = 3,125 x 1013 (electrons/elektrone)  (3) 7.1.3 kQ E=  r2 9 x 109 x 5 x 10 -5 E=  0,12  E = 4,5 x 106 N.C-1  left/links  (5) 7.2.1 Negative/Negatief  Like charges repel each other/Gelyksoortige ladings stoot mekaar af.  OR/OF The charges repel each other. If sphere A is negative, then sphere B must also be negative.  Die ladings stoot mekaar af. Indien sfeer A negatief is, dan moet sfeer B ook negatief wees. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 15 7.2.2 FE = Tsin30o FE = 25sin30o  FE = 12,5 N kQ 1Q 2 FE =  r2 (9  109 )( 5  10 -6 ) Q 12,5 =  0,05 2  Q = 6,94 x 10-7 C  (6) [17] QUESTION/VRAAG 8 8.1.1 Temperature/temperatuur  Length of the conductors/Lengte van die geleiers  (Any two/Enige twee) Thickness of the conductors/Dikte van die geleiers. ACCEPT: Type of material AANVAAR: Tipe materiaal (2) 8.1.2 Gradient is the inverse of the resistance. / Gradiënt is die inverse van die weerstand  OR/OF Gradient =  (1) 8.1.3 Conductor C./Geleier C.  It has the highest resistance. The higher the resistance of a conductor, the more heat is produced in the conductor if the current is constant.  Dit het die hoogste weerstand. Hoe hoër die weerstand, hoe meer hitte geproduseer word in die geleier indien die stroom konstant is. (2) 8.2.1 R =  12 R=  1,5 R=8Ω (3) 8.2.2 OPTION/OPSIE 1 OPTION/OPSIE 2 Rtotal = R + r  = I(R+ r)  [8 = (4 + 3) + r ] 12 = 1,5 [(4 + 3) + r]  r=1Ω r=1Ω (4) 8.2.3 W = I2R∆t  W = (1,5)2(3)(180)  W = 1 215 J  (3) 8.3.1 Decrease/Neem af.  (1) 8.3.2 Increase/Neem toe.  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com 16 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021) 8.4 Increase/Neem toe.  Rext decreases. Current through battery increases.  W = I2r∆t / Energy transfer to the battery/work done by battery increases.  Reks neem af. Stroom deur die battery neem toe.  W = I2rΔt / Energie oorgedra aan die battery/arbeid verrig deur die battery neem toe.  (3) [20] QUESTION/VRAAG 9 9.1 Mechanical energy to electrical energy. Meganiese energie na elektriese energie  (2) 9.2 AC generator has slip rings and DC generator has a split ring / commutator  WS-generator het sleepringe en GS-generator het ’n splitring / kommutator (1) 9.3 Potential Difference (emf) (V) Potensiaalverskil (emk) CRITERIA FOR MARKING/MERK KRITERIA Correct shape/Korrekte vorm  Axes labelled correct/Asse korrek  Vmax indicated on graph/  Vmaks aangedui op grafiek (2) 9.4 Pave = VrmsIrms  Pgem = VwgkIwgk  2000 = Irms x 230  2000 = Iwgk x 230  Irms = 8,70 A Iwgk = 8,70 A I I Irms = max Iwgk = max 2 2 Imax I 8,70 =  8,70 = max  2 2 Imax = 12,30 A  Imaks = 12,30 A  (4) 9.5 Vmax Vrms =  2 Vmax 230 =  2 Vmax/Vmaks = 325,27 V  (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 17 QUESTION/VRAAG 10 10.1 Work function (of the metal) / Werksfunksie (van die metaal)  hc Ek(max) = – W0   The intercept on the vertical axis = W 0. Die afsnit op die vertikale-as = Wo  OR/OF hc = W 0 + Ek(max)   The intercept on the vertical axis is equal to the W o Die afsnit op die vertikale-as is gelyk aan die Wo  (3) 10.2 E = W 0 + Ek(max) hf = W 0 + Ek(max) Any one/Enige een  hf = hf0 + Ek(max) 6,63 x 10-34 x 6,16 x 1014 = 6,63 x 10-34 f0  + 5,6 x 10-20  f0 = 5,32 x 1014 Hz  (5) 10.3.1 Remain the same/Bly dieselfde  The gradient is equal to the product of Planck’s constant and the speed of light in vacuum which are constants.  Die gradient is gelyk aan die produk van Planck se konstante en die spoed van lig in ʼn vakuum wat konstantes is. OR/OF Gradient = hc, which are constants/Gradiënt = hc, wat konstantes is. (2) 10.3.2 Remains the same/Bly dieselfde  Ek(max) (J) Intensity/Intensiteit CRITERIA FOR MARKING/MERK KRITERIA Axes labelled/Asse benoem  Correct shape/Korrekte vorm  (3) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

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