# Physical Sciences Grade 12 P1 Eastern Cape Mock 2020 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape Mock Exam 2020, 18 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sciences-p1-gr12-memo-gr12-sept2020-a-e/6806f8f9-0b21-4e67-a9e7-bd0c32f0e649 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2020 - Exam period: Eastern Cape Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 18 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 P1 Eastern Cape Mock 2020 Memo](https://www.hlayiso.com/library/physical-sciences-p1-gr12-memo-gr12-sept2020-a-e/6806f8f9-0b21-4e67-a9e7-bd0c32f0e649) ## About this document Physical Sciences · Grade 12 · Eastern Cape Mock Exam · 2020. Memorandum, 18 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2020 PHYSICAL SCIENCES P1/ FISIESE WETENSKAPPE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 18 pages./ Hierdie nasienriglyn bestaan uit 18 bladsye. ### Page 2 Downloaded from hlayiso.com 2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) 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. 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. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 3 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 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. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com 4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) 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 Subquestion to subquestion: When a certain variable is calculated in one subquestion (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 subquestions. 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 the a subquestion: 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. 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 ### Page 5 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 5 QUESTION/VRAAG 1: MULTIPLE-CHOICE QUESTIONS/ MEERVOUDIGEKEUSE-VRAE 1.1 C  (2) 1.2 C  (2) 1.3 A  (2) 1.4 D  (2) 1.5 A  (2) 1.6 B  (2) 1.7 D  (2) 1.8 B  (2) 1.9 D  (2) 1.10 C  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com 6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 2 2.1 When a 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 resulterende/netto krag op ʼn voorwerp inwerk, versnel die voorwerp in die rigting van die krag. Die versnelling is direk eweredig aan die krag en omgekeerde eweredig aan die massa van die voorwerp.  (2) 2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 N N T T f f W//  W┴ W (4) Mark awarded for arrow and label. / Punt toegeken vir byskrif en pyltjie. Do not penalise for length of arrows since drawing is not drawn to scale./ Moenie vir die lengte van die pyltjies penaliseer nie, aangesien die skets nie volgens skaal geteken is nie. 3 Any other additional force(s) / Enige addisionele kragte 4 If force(s) do not make contact with body. / Indien die kragte nie kontak maak 3 met die voorwerp nie. Max./Maks. 4 2.3 OPTION 1 (To the right is OPTION 2 (To the right is negative) positive) OPSIE 2 (Na regs is negatief) OPSIE 1 (Na regs is positief) Fnet = ma Fnet = ma Fnet = 0 N Any one / Fnet = 0 N Any one / Fg – T = ma Enige een  T - Fg = ma Enige een  15 x 9,8 - T = 0  T – 15 x 9,8 = 0  T = 147 N T = 147 N T – Wsin θ – f = ma Wsin θ + f – T = ma T – mgsin θ – f = ma mgsin θ + f – T = ma 147– 22 x 9,8sin θ – 43,86 = 0 22 x 9,8 sin θ + 43,86 – 147  = 0 θ = 28,58°  θ = 28,58°  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 7 2.4 2.4.1 The kinetic friction force depends on the angle of the inclined to the horizontal and nature of the surface.  Die kinetiese wrywingskrag is afhanklik van die hoek van die skuinsvlak met die horisontaal en die aard van die oppervlakte.  (2) 2.4.2 Positive marking from Question 2.3 / Marking criteria / Merk kriteria Merk positier vanaf Vraag 2.3 For 15 kg block / Vir 15 kg-blok Fnet = ma Fnet = ma  F - Fg – T = ma F - Fg – T = ma T + Wsin θ – f = ma Any one / 172 – 147 – T = 15a …(1) Enige een  T + mgsin θ – f = ma Calculating Fg// For 22 kg block / Vir 22 kg-blok Berekening van Fg//  T + W sinθ– f = ma Substitution for the 15 kg block T + mg sinθ– f = ma Instelling vir die 15 kg-blok  [T + 22 x 9,8 sin 28,58o – 43,86] = Substitution for the 22 kg block 22a  Instelling vir die 22 kg-blok  [T + 101,14 – 43,86] = 22a (2) Substitution for either 15a or 22a Instelling vir 15a of 22a  ∴ a = 2,278 m.s-2  Answer/Antwoord  (6) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com 8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 3 3.1 UPWARD POSITIVE UPWARD NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + a∆t  vf = vi + a∆t  vf = -3,28 + (-9,8) x 1,2  vf = 3,28 + (9,8) x 1,2  vf = - 15,04 vf = 15,04 m.s-1 downwards/afwaarts  vf = 15,04 m.s-1 downwards/afwaarts  (3) 3.2 OPTION/OPSIE 1 UPWARD POSITIVE UPWARD NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF 1 1 ∆y = vi ∆t + a∆t2  ∆y = vi∆t+ a∆t2  2 2 1 1 ∆𝑦 = -3,28 x 1,2 + (-9,8)1,22  ∆y = 3,28 x 1,2 + (9,8)1,22  2 2 ∆y = -10,99 ∆𝑦 = 10,99 m downwards/afwaarts  ∆y = 10,99 m downwards/afwaarts  OPTION /OPSIE 2 Positive marking from Question 3.1/Positiewe merk vanaf Vraag 3.1 UPWARD POSITIVE UPWARD NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + 2a∆y  2 2 vf2 = vi2 + 2a∆𝑦  -15,04  = -3,28 + 2 (-9,8) ∆y  2 2 15,042 = 3,282 + 2 (9,8) ∆y  ∆y = 10,99 m downwards/afwaarts  ∆y= 10,99 m downwards/afwaarts  OPTION/OPSIE 3 Positive marking from Question 3.1/Positiewe merk vanaf Vraag 3.1 UPWARD POSITIVE UPWARD NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF v  vi v  vi ∆y = f Δt  ∆y = f Δt  2 2 -15,04 + ( 3,28) 15,04 + 3,28  ∆y = x 1,2  ∆y = x 1,2  2 2 ∆y = -10,99 m ∆y = 10,99 m downwards/afwaarts  ∆y = 10,99 m downwards/afwaarts  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 9 3.3 OPTION 1/OPSIE 1 UPWARD POSITIVE UPWARD NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf2= vi2 + 2a∆y  vf2= vi2 + 2a∆y  vf2= -3,282 + 2 (-9,8) (30)  vf2= 3,282 + 2 (9,8) (30)  vf2= - 24,47 m.s-1 vf2= 24,47 m.s-1 downwards/afwaarts  vf2= 24,47 m.s-1 downwards/afwaarts  OPTION 2/OPSIE 2 UPWARD POSITIVE UPWARD NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF 1 1 ∆y = vi∆t+ a∆t2 ∆y = vi∆t+ a∆t2 2 2 1 1 -30 = -3,28∆𝑡 + (-9,8)∆𝑡 2 30 = 3,28∆𝑡 + (9,8)∆𝑡 2 2 2 ∆𝑡 = 2,16 s ∆t = 2,16 s vf = vi + a∆t  vf = vi + a∆t  vf = -3,28 + (-9,8)(2,16)  vf = 3,28 + (9,8)(2,16)  vf= -24,45 m.s-1 vf = 24,45 m.s-1 downwards/afwaarts  vf = 24,45 m.s-1 downwards/afwaarts  (3) 3.4 Positive marking from Question 3.2/Positiewe merk vanaf Vraag 3.2 UPWARD POSITIVE UPWARD NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF Fnet.∆t = m(vf - vi)  Fnet. ∆t = m(vf - vi)  205 x 0,1 = 0,5 [𝑣𝑓 – (- 24,47)]  -205 x 0,1  = 0,5 (vf– 24,47)]  vf= 16,53 m.s upwards/opwaarts  -1 vf= -16,53 m.s-1 vf = 16,53 m.s-1 upwards/opwaarts  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com 10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) 3.5 OPTION 1/OPSIE 1 Positive marking from Question 3.1 and 3.3 Positiewe merk vanaf Vraag 3.1en 3.3 UPWARD POSITIVE UPWARD NEGATIVE OPWAARTS POSITIEF OPWAARTS NEGATIEF vf = vi + 2a∆y  2 2 vf2= vi2 + 2a∆y  0  = 16,53 + 2 (-9,8) ∆y  2 2 02  = -16,532 + 2 (9,8) ∆y  ∆y = 13,94 m ∆y = - 13,94 m width of window / wydte van venster width of window / wydte van venster = 30 – (10,99+13,94)  = 30 – (10,99 +13,94)  = 5,07 m  = 5,07 m  OPTION 2/OPSIE 2 Wnet = ∆Ek Fnet ∆y cos θ = ∆Ek Any one/Enige een  1 1 Fg∆y cos θ = mvf2 - mvi2 2 2 1 4,9 ∆ycos 180o = 0 - x 0,5 x 16,532  2 ∆y = 13,94 m width of window / wydte van venster = 30 – (10,99 + 13,94)  = 5,07 m  (5) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 11 QUESTION/VRAAG 4 4.1 In an isolated system total linear momentum is conserved.  In ʼn geïsoleerde sisteem bly die totale lineêre momentum behoue.  (2) 4.2 4.2.1 MEi = MEf mgh1 + ½ mvi2 = mgh2 + ½ mvf2 0 + ½ x 2,005 x vi2 = 2,005 x 9,8 x 0,06 + 0  vi = 1,08 m.s-1 ∑pi = ∑pf Any one/Enige een  mblockviblock + mbulletvibullet = (mblock + mbullet) vf 0 + 0,005 x vibullet = 2,005 x 1,08  vbullet = 433,08 m.s-1 right/regs  (5) 4.2.2 Positive marking from Question 4.2.1/Positiewe merk vanaf Vraag 4.2.1 ∑pi = ∑pf (mgun + mbullet)vi = mgunvfgun + mbulletvfbullet = Any one/Enige een  0 = 5 x vf(gun)+ 0,005 x 433,08  vf(gun= - 0,43 m.s-1 vf(gun = 0,43 m.s-1 left/links  (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com 12 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 5 5.1 In an isolated system the total mechanical energy is conserved.  In ’n geïsoleerde sisteem bly die totale meganiese energie behoue.  (2) 5.2 MEi = MEf (Ep + Ek)A = (Ep + Ek)B Any one/Enige een  mgh1 + Ek(A) = mgh2 + Ek(B) 55 x 9,8 x 15 + 0 = 0 + Ek(B)  Ek(B) = 8085 J  (3) 5.3 The net work done on an object is equal to the object’s change in kinetic energy.  OR The work done by a net force is equal to the object’s change in kinetic energy.  Die netto arbeid verrig op ʼn voorwerp is gelyk aan die voorwerp se verandering in kinetiese energie.  OF Die arbeid verrig deur ʼn netto krag is gelyk aan die voorwerp se verandering in kinetiese energie.  (2) 5.4 5.4.1 Positive marking from Question 5.2/Positiewe merk vanaf Vraag 5.2 OPTION 1/OPSIE 1 Wner = ∆Ek Wf + WFg// = ∆Ek Any one/Enige een  f x ∆x cos θ + mgsin θ ∆x cos θ = ∆Ek 15 x 10 cos180o  + 55 x 9,8 sin 15o x 10 cos 180o  = 0 – Eki Eki = 1545,035 J  OPTION 2/OPSIE 2 Wnc = ∆Ek + ∆Ep Wf = ∆Ek + ∆Ep Any one/Enige een  f x ∆x cos θ = ∆Ep + ∆Ek 15 x 10 cos180o = [55 x 9,8 x (10 sin15o)] - 0 + 0 - Eki -150 = 1395,035 - Eki Eki = 1545,035 J  OPTION 3/OPSIE 3 Wner = ∆Ek Wf + WFg = ∆Ek Any one/Enige een  f x ∆x cos θ + mg ∆xcos θ = ∆Ek 15 x 10 cos180o  + 55 x 9,8 x 10 cos (90 + 15)  = 0 – Eki Eki = 1545,035 J  (4) 5.4.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Wnet = ∆Ek Wnc = ∆Ek + ∆Ep Wf = ∆Ek Any one/Enige een Wf = ∆Ek + ∆Ep Any one/ f x ∆x cos θ = ∆Ek  f x ∆x cos θ = ∆Ek + ∆Ep Enige een  f x 10 cos180o  = 1545,035  - 8085  f x 10 cos180o = 1545,035  654,00 N  - 8085  f = 654,00 N  (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 13 QUESTION/VRAAG 6 6.1 Doppler effect/Doppler effek  (1) 6.2 AWAY.  The observed frequency is less than the source frequency.  WEG VAN.  Die waargenome frekwensie is minder as die bron se frekwensie.  (2) 6.3 As the listener’s velocity increases, the observed frequency decreases.  As die luisteraar se snelheid toeneem, verminder die waargenome frewensie.  (2) 6.4 The intercept on the vertical axis represents the frequency of the source, fs.  Die afsnit op die vertikale as verteenwoordig die frekwensie van die bron, fs.  (2) 6.5 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 492  470 v ± vL Gradient/Helling =  fL = fs  5  20 v ± vs 22 340  15 =  492 = x fs 15 340 f fs = 499,34 Hz  = - s  v 22 fs  = -  15 340 fs = 498,67 Hz  OPTION 3/OPSIE 3 v ± vL fL = fs  v ± vs 340  15 470 = x fs 340 fs = 499,38 Hz  Range (498,67 Hz – 499,38 Hz) (5) 6.6 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 v = fλ v = fλ v = fλ 340 = 498,67 λ  340 = 499,34 λ  340 = 499,38 λ  𝜆 = 0,68 m  𝜆 = 0,68 m  𝜆 = 0,68 m  The wavelength The wavelength produced The wavelength produced by the tuning by the tuning fork is less produced by the tuning fork is less than the than the required fork is less than the required wavelength. wavelength. required wavelength. ∴ It is not suitable.  ∴ It is not suitable.  ∴ It is not suitable.  Die golflengte wat Die golflengte wat Die golflengte wat voortgeplant word deur voortgeplant word deur die voortgeplant word deur die stemvurk is minder stemvurk is minder as die die stemvurk is minder as as die benodigde benodigde golflengte. die benodigde golflengte. golflengte. ∴ Dit is nie geskik nie.  ∴ Dit is nie geskik nie.  ∴ Dit is nie geskik nie.  (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com 14 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 7 7.1 Q n=  qe 2 x 106 n=  1,6 x1019 n = 1,25 x 1013  (3) 7.2 Criteria for marking/Kriteria vir merk Correct shape / Korrekte vorm  Direction of electric field / Rigting van elektriese veld  Lines not crossing each other / Lyne sny nie mekaar nie.  (3) 7.3 7.3.1 vf = vi + a∆t  6,25 x 103 = 0 + 2 x 10-3 a  Any one /Enige een  a = 3,125 x 106 m.s-2 Fnet = ma Fnet = 5 x 10-6 x 3,125 x 106  Fnet = 15,625 N Fnet Enet =  q 15,625 Enet =  2 x10 6 Enet = 7,81 x 106 N.C-1  (6) 7.3.2 Positive marking from Question 7.3.1/Positiewe merk vanaf Vraag 7.3.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 kQ vf 2 = vi2 + 2a Δx  E= 2  r (6,25 x 103)2 = 02 + 2(3,125 x 106)Δx  9 x109 x 2 x 106 Δx = 6,25 m  7,81 x 106 = 2  r ∴ r = 4,80 x 10-2 m  Using equations of motion / Gebruik van bewegingsvergelykings OPTION 3/OPSIE 3 OPTION/OPSIE 4 Δx = ½(vf + vi) Δt  Δx = vi Δt + ½ a Δt²  = ½(6,25 x 103+ 0)(2 x 10-3)  = 0 + ½(3,125 x 106)(2 x 10-3)²  Δx = 6,25 m  Δx = 6,25 m  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 15 7.3.3 Positive marking from Question 7.3.1 and 7.3.2 Positiewe merk vanaf Vraag 7.3.1en 7.3.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 kQ1Q2 kQ1Q2 FE = 2  FE =  r r2 9 x109 x Q x 2 x 10 6 15,625 = (9 x 109)(Q1)(2 x 10-6) 15,625 =  6,25²   4,80 x 102  2 ∴ Q = 3,39 x 10-2 C  ∴ Q = 2,00 x 10-6 C  (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com 16 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 8 8.1 Resistance/Weerstand  (1) 8.2 8.2.1 4,2   0,2 Gradient/Helling =  24  0 1 = 6 1 = emf 1 1 =  emf 6 ∴ emf = 6 V  (4) 8.2.2 Positive marking from Question 8.2.1/Positiewe merk vanaf Vraag 8.2.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 r 𝜀 = I (R + r )  Intercept on vertical axis = emf 1 6= (24 + r )  r 4,2 0,2 =  emf r = 1,2 Ω  r 0,2 =  6 r = 1,2 Ω  (3) 8.3 8.3.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 P = VI  V2 P= 3 = I x 1,5  R I=2A 1,5 2 3= R R = 0,75 Ω P = I2 R  3 = I2 x 0,75  I=2A (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 17 8.3.2 Positive marking from Question 8.2.1en 8.2.2 Positiewe merk vanaf Vraag 8.2.1en 8.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 V2 V2 P= P= R toy car R toy car 1,52 1,52 3=  3=  R toy car R toy car Rtoy car = 0,75 Ω Rtoy car = 0,75 Ω emf = I (Rext + r )  emf = I (Rext + r)  6 = 2 (Rext + 1,2)  6 = 2 (Rext + 1,2)  Rext = 1,8 Ω Rext = 1,8 Ω Rext = Rtoy car + Rp Rext = Rtoy car + Rp 1,8 = 0,75 + Rp  1,8 = 0,75 + Rp  Rp = 1,05 Ω Rp = 1,05 Ω 1 1 1 R1 R2 = + Rp = Rp R1 R2 R1 +R2 1 1 1 3x R = +  1,05 =  1,05 3 R 3 +R R = 1,62 Ω  R = 1,62 Ω  (6) 8.4 No,  the external resistance will increase. Current will decrease.  P = I2R, P 𝛼 I2 since resistance remain constant, power will decrease.  Nee,  die eksterne weerstand sal toeneem. Stroom sal afneem.  P = I2R, P 𝛼 I2 aangesien die weerstand constant bly sal die drywing afneem.  (3) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com 18 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 9 9.1 (Electric) motor/ (Elektriese) motor  (1) 9.2 Electrical energy to mechanical energy  Elektriese energie  na meganiese energie  (2) 9.3 CLOCKWISE / KLOKSGEWYS  (1) 9.4 Replace the split ring with slip rings.  Replace the cell with a resistor / ammeter / voltmeter.  Vervang die split ring met sleepringe  Vervang die sel met ‘n resistor / ammeter / voltmeter.  (2) 2 9.5 9.5.1 Vrms Paverage   R 2302 1500=  R R = 35,27 Ω  (3) 9.5.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 V Paverage = VrmsIrms R = rms 1 500 = Irms x 230  Irms Irms = 6,52 A 2302 I 35,27 =  Irms Irms = max  2 Irms = 6,52 A 𝐼𝑚𝑎𝑥 6,52 =  Imax √2 Irms =  𝐼𝑚𝑎𝑥 = 9,22 A  2 Imax 6,52 =  2 Imax = 9,22 A  (4) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ## Related documents - [PHYS SCIENCES P2 MEMO GR12 SEPT2020 A E](https://www.hlayiso.com/library/phys-sciences-p2-memo-gr12-sept2020-a-e/91c998b5-2ceb-4a4b-b327-c403d1661716) — Physical Sciences · Memo · 2020 - [WC 2020 Gr 12 SEPT MCED Physics P1 COMMON CLUSTER EXAM MEMORANDUM 2](https://www.hlayiso.com/library/wc-2020-gr-12-sept-mced-physics-p1-common-cluster-exam-memorandum-2/b6743005-6dd1-403e-b52a-d4224e11cb5b) — Physical Sciences · Memo · 2020 - [PHYS SCIENCES P1 GR12 MEMO SEPT2021 Xho English B Copy](https://www.hlayiso.com/library/phys-sciences-p1-gr12-memo-sept2021-xho-english-b-copy/813b1849-8c40-4d71-92b0-1f6a8e44c4f9) — Physical Sciences · Memo · 2021 · Eng - [Physical Sciences P1 Memo](https://www.hlayiso.com/library/physical-sciences-p1-memo/2d2b7939-aa98-49f1-afa7-ca27e7f59489) — Physical Sciences · Memo · 2017 - [PHYS SC P1 MEMO ENGAFR](https://www.hlayiso.com/library/phys-sc-p1-memo-engafr/da9b9ae0-718d-442a-95bc-aa0bfbf308f3) — Physical Sciences · Memo · 2014 - [PHYS SCIENCES P1 QP GR12 SEPT 2023 Xhosa](https://www.hlayiso.com/library/phys-sciences-p1-qp-gr12-sept-2023-xhosa/2bbd2fc3-2b94-4d0c-9012-591942a529e4) — Physical Sciences · QP · 2023 ## More from Grade 12 Physical Sciences - [PHYSICAL SCIENCE P1 QP JCT 2025 AFR D watermark](https://www.hlayiso.com/library/physical-science-p1-qp-jct-2025-afr-d-watermark/b9429ac5-acbb-4fae-989d-af9e1478a96b) — QP · 2025 · Afr - [PHYSICAL SC P2 QP GR12 SEPTEMBER 2025 AFR1 watermark](https://www.hlayiso.com/library/physical-sc-p2-qp-gr12-september-2025-afr1-watermark/37cff2db-7c1a-4ec8-8ac5-07ec5bcf87a9) — QP · 2025 - [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 KZN](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-kzn/6d225dc8-24c6-4c28-a181-df906d5a2c0e) — QP · 2025 - [Physical Sciences Physics Grade 12 PS01 QP March 2025 Limpopo](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-ps01-qp-march-2025-limpopo/482ee9af-e88a-4818-9120-11ab9e045a1c) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-limpopo/b65e9832-a6be-4312-8f5a-57425d647121) — QP · 2025 ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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