# Physical Sciences Grade 12 P1 Eastern Cape Mock 2017 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape Mock Exam 2017, 15 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sciences-p1-memo/2d2b7939-aa98-49f1-afa7-ca27e7f59489 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2017 - Exam period: Eastern Cape Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 15 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 2017 Memo](https://www.hlayiso.com/library/physical-sciences-p1-memo/2d2b7939-aa98-49f1-afa7-ca27e7f59489) ## About this document Physical Sciences · Grade 12 · Eastern Cape Mock Exam · 2017. Memorandum, 15 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/NASIONALE SENIOR CERTIFICATE/SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2017 PHYSICAL SCIENCES P1/ FISIESE WETENSKAPPE V1 MARKING GUIDELINE/NASIENRIGLYNE MARKS/PUNTE: 150 This marking guideline consists of 15 pages./ Hierdie nasienriglyne bestaan uit 15 bladsye. ### Page 2 Downloaded from hlayiso.com 2 PHYSICAL SCIENCES P1 (EC/SEPTEMBER 2017) 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 verkeerde 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 ʼn berekening aangewend is, d.w.s. substitusies is gedoen of ʼn numeriese antwoord is gegee. Kopiereg voorbehou Please turn over ### Page 3 Downloaded from hlayiso.com (EC/SEPTEMBER 2017) 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 formule 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 no 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 Blaai om asseblief ### Page 4 Downloaded from hlayiso.com 4 PHYSICAL SCIENCES P1 (EC/SEPTEMBER 2017) 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, no 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 in 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. Kopiereg voorbehou Please turn over ### Page 5 Downloaded from hlayiso.com (EC/SEPTEMBER 2017) 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 3.1 is incorrect, no marks can be awarded for QUESTION 3.2. However, if the answer for e.g. 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 3.1 is verkeerd, kan geen punte vir VRAAG 3.2 toegeken word nie. Indien die antwoord op bv. 3.1 egter op ʼn berekening gebaseer is, kan die motivering vir die verkeerde antwoord in 3.2 oorweeg word. Copyright reserved Blaai om asseblief ### Page 6 Downloaded from hlayiso.com 6 PHYSICAL SCIENCES P1 (EC/SEPTEMBER 2017) QUESTION/VRAAG 1 MULTIPLE-CHOICE QUESTIONS/MEERVOUDIGEKEUSE-VRAE 1.1 D ✓✓ (2) 1.2 D ✓✓ (2) 1.3 C ✓✓ (2) 1.4 B ✓✓ (2) 1.5 B ✓✓ (2) 1.6 D✓✓ (2) 1.7 A ✓✓ (2) 1.8 C ✓✓ (2) 1.9 C ✓✓ (2) 1.10 A ✓✓ (2) [20] Kopiereg voorbehou Please turn over ### Page 7 Downloaded from hlayiso.com (EC/SEPTEMBER 2017) FISIESE WETENSKAPPE V1 7 QUESTION/VRAAG 2 2.1 If the resultant/net force acts on an object, the object will accelerate in the direction of the resultant/net force with an acceleration that is directly proportional to the resultant/net force ✓ and inversely proportional to the mass ✓ of the object. Wanneer ʼn resultante/netto krag op ʼn voorwerp inwerk, versnel die voorwerp in die rigting van die resultante/netto krag met ʼn versnelling wat direk eweredig is aan die resultante/netto krag ✓ en omgekeerd eweredig aan die massa van die voorwerp. ✓ (2) 2.2 ✓ for both components / vir albei komponente (5) 2.3 2.3.1 (For the vertical motion) FAY is the vertical component of (Vir die vertikale beweging) FA. Fnet = 0 ✓ any one FAY is die vertikale komponent FN + FAY +(- w) = 0 (enige een) van FA. FN + FAY = w FA = 5 N FN + FA.sin30° = 5 × 9,8 ✓ FN = 49 – (5 × 9,8 × sin 30°) FN = 24,5 N ✓ (3) 2.3.2 fk = μ k N ✓ any one fk = μk mg enige een fk = 0,2 × 24,5 ✓ fk = 4,9 N ✓ (2) 2.3.3 Positive marking from 2.3.1: At 12 kg block A / By 12 kg blok (Positiewe merk van 2.3.1): W + (-T) = ma At 5 kg block B: (By 5 kg blok) mg – T = 12 × a 12 × 9,8 – T = 12 × a Fnet = ma T = 117,6 – 12a ✓........ (2) T + (-f) + (-FAX) = ma (1) = (2): T =5×a✓ 117,6 –12a = 5a + 4,9 + 5cos30° T = 5a + 4,9 + 5 cos 30°✓.....(1) 117,6 – (4,9+5cos30°)=5a + 12a 108,369873 = 17a a = 6,37 m•s-2 ✓ (4) [16] Copyright reserved Blaai om asseblief ### Page 8 Downloaded from hlayiso.com 8 PHYSICAL SCIENCES P1 (EC/SEPTEMBER 2017) QUESTION/VRAAG 3 3.1 Free-fall is the motion of an object when the only force acting on it is gravitational force ✓✓ Vryval is die beweging van ʼn voorwerp indien die enigste krag wat daarop inwerk, gravitasiekrag is. ✓✓ (2) 3.2 3.2.1 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 (downwards positive) (upwards positive) (afwaarts positief) (opwaarts positief) vf = vi + a∆t ✓ vf = vi + a∆t ✓ 0 = (-15) + (9,8)x ∆t ✓ 0 = (15) + (-9,8)∆t ✓ ∆t = 1,53 s✓ ∆t = 1,53 s✓ (3) 3.2.2 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 (downwards positive) (upwards positive) (afwaarts positief) (opwaarts positief) vf = vi + a∆t ✓ vf = vi + a∆t ✓ = (-15) + (9,8)(2,4) ✓ = (15) + (-9,8)(2,4) ✓ = 8,52 = -8,52 vf = 8,52 m•s ✓-1 vf = 8,52 m•s-1 ✓ (downwards / afwaarts) ✓ (downwards / afwaarts) ✓ OPTION 3 / OPSIE 3 OPTION 2 / OPSIE 2 (downwards positive) (upwards positive) (afwaarts positief) (opwaarts positief) Descent (Afwaarts): Y – X: Descent (Afwaarts): Y – X: ∆t = 2,4 – 1,53 = 0,87 s ∆t = 2,4 – 1,53 = 0,87 s vf = vi + a∆t ✓ vf = vi + a∆t ✓ = 0 + (9,8)(0,87) ✓ = 0 + (-9,8)(0,87) ✓ = 8,53 m•s ✓ -1 = -8,53 m•s-1 (downwards / afwaarts) ✓ = 8,53 m•s-1 ✓ (downwards / afwaarts) ✓ (4) 3.3 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 (upwards positive/opwaarts positief) (downwards positive/afwaarts positief) A – Y: A – Y: ∆y = vi∆t + ½ a∆t ✓ 2 ∆y = vi∆t + ½ a∆t2 ✓ 1 1 = 15 × 1,53 + 2 (-9,8)(1,53)2✓ = -15 × 1,53 + 2 (9,8)(1,53)2 ✓ = 11,48 m ✓ = -11,48 m ✓ Y – X: = 11,48 m upwards/opwaarts ∆t = 2,4 – 1,53 = 0,87 s Y – X: 1 ∆y = vi∆t + 2a∆t2 ∆t = 2,4 – 1,53 = 0,87 s 1 1 ∆y = vi∆t + 2 a∆t2 = 0 × 0,87 + 2 (-9,8)(0,87)2✓ 1 = -3,71 ✓ = 0 × 0,87 + 2 (9,8)(0,87)2✓ = 3,71 m downwards/afwaarts = 3,71 m ✓ Height of the building is given by: Height of the building is given by: Hoogte van die gebou gegee as: Hoogte van die gebou gegee as: h = 11,48 + 1,8 – 3,71 h = -11,48 – 1,8 +3,71  h = 9,57 m ✓  h = -9,57 m  h = 9,57 m ✓ Kopiereg voorbehou Please turn over ### Page 9 Downloaded from hlayiso.com (EC/SEPTEMBER 2017) FISIESE WETENSKAPPE V1 9 OPTION 3 / OPSIE 3 OPTION 4 / OPSIE 4 (upwards positive/opwaarts positief) (upwards positive/opwaarts positief) (Y-X): (Y-X): vf = vi + a∆t vf = vi + a∆t -8,52 = (0) + (-9,8) ∆t -8,52 = (0) + (-9,8) ∆t ∆t = 0,87 s ✓ ∆t = 0,87 s ✓ (A-Roof): (A-Roof): 1,53 = ∆t + 0,87 1,53 = ∆t + 0,87 ∆t = 0,66 s ✓ ∆t = 0,66 s ✓ (A-X): (A-X): vf2 = vi2 + 2a∆y ✓ 1 ∆y = vi . ∆t +2 𝑔∆t 2✓ (8,52)2 = (15)2 + 2(-9,8)∆y ✓ 1 = 15 × 0,66 + 2 𝑥 – 9,8 × 0,662 ✓ ∆y = 7,78 m ✓ Height/Hoogte: h = 7,78 + 1,8 = 7,77 m ✓ Height/Hoogte: h = 7,78 + 1,8 = 9,58 m ✓ = 9,58 m ✓ (6) 3.4 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 (downwards positive) (upwards positive) (afwaarts positief) (opwaarts positief) Δt (s) Δt (s) Criteria for graph / Kriteria vir grafiek Marks/Punte Initial velocity / Beginsnelheid ✓ Final velocity / Eindsnelheid ✓ Time at Y, maximum height / Tyd by Y, maksimum hoogte ✓ (3) [18] Copyright reserved Blaai om asseblief ### Page 10 Downloaded from hlayiso.com 10 PHYSICAL SCIENCES P1 (EC/SEPTEMBER 2017) QUESTION/VRAAG 4 4.1 In an isolated system ✓the total mechanical energy remains constant. ✓ In ʼn geïsoleerde sisteem ✓ bly die totale meganiese energie konstant. ✓ (2) 4.2 OPTION 1 / OPSIE 1 Emech at/by A = Emech at/by B ✓ Any one 1 2 1 2 (mgh + 2mv ) at/by A = (mgh + 2mv ) at/by B Enige een 1 1 4 × 9,8 × (7 sin 60°) ✓+ 2 × 4 × 0 = 4 x 9,8 × (4 sin 60°)✓+ 2 × 4 × v2 v = 7,14 m•s-1 ✓ OPTION 2 / OPSIE 2 Emech at/by A = Emech at/by B ✓ 1 1 (mgh + 2mv2) at A = (mgh + 2mv2) at B 1 1 4 × 9,8 × (3 sin 60°) ✓+ × 4 × 0 = 4 × 9,8 × (0)✓+ × 4 × v2 2 2 v = 7,14 m•s-1 ✓ (4) 4.3 4.3.1 Work done by a net force is equal✓ to the change in kinetic energy✓ of an object. OR The net (total) work done ✓is equal to the change in kinetic energy of the object. ✓ Arbeid verrig deur ʼn netto krag is gelyk ✓ aan die verandering in kinetiese energie van ʼn voorwerp ✓ OF Die netto (totale) arbeid op ʼn voorwerp is gelyk ✓ aan die verandering in kinetiese energie van die voorwerp ✓ (2) 4.3.2 ✓ for both components of weight ✓ vir albei gewigskomponente (3) 4.3.3 OPTION 1 / OPSIE 1 W net = ∆ EK ✓ any one 1 Wf + W g// = 2 m(v2f – v2i) (enige een) 1 f ∙ ∆x ∙ cos𝜃 + mgsin 60 ∙ ∆x ∙ cos𝜃 = 2 × 4(32 – 7,142) ✓ 2 2 f × cos 60° ✓ × -1 + 4 × 9,8 sin 60 ✓ × cos 60° × 1 = -83,9592 -4f + 135,7927833 = -83,9592 f = 54,94 N ✓ OPTION 2 / OPSIE 2 W nc = ∆ Ep + ∆ EK 1 1 W f = mghf -mghi + 2 mv2f – 2 m v2i ✓ any one / enige een 1 f ∙ cos𝜃 ∙ ∆x = mg (hf –hi) + 2 m (v2f – v2i) 1 f ∙ cos 0 × 4 ✓ = 4 × 9,8 (0 – 4 sin 60°)✓ + 2 × 4 (32 – 7,142)✓ f = 54,94 N ✓ (5) 4.4 Remains the same / Bly dieselfde ✓ (1) [17] Kopiereg voorbehou Please turn over ### Page 11 Downloaded from hlayiso.com (EC/SEPTEMBER 2017) FISIESE WETENSKAPPE V1 11 QUESTION/VRAAG 5 5.1 The total linear momentum of an isolated system remains constant (is conserved) ✓✓ OR Total linear momentum before a collision equals total linear momentum after a collision in an isolated system. ✓✓ Die totale lineêre momentum van ʼn geïsoleerde sisteem bly konstant (bly behoue) ✓✓ OF Totale lineêre momentum voor ʼn botsing is gelyk aan die totale lineêre momentum na ʼn botsing in ʼn geïsoleerde sisteem. ✓✓ (2) 5.2 5.2.1 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 1 ∆y = v i ∆t + 2 a ∆t ✓ 2 vi  v f 2 ∆y = ∆t ✓ 0,32 = v i × 0,33 + 0 ✓ 2 2𝑣 Vi = 0,97 m•s-1 ✓ 0,32 = 2 × 0,33 ✓ V = 0,97 m•s-1 ✓ (3) 5.2.2 Consider LEFT as positive / Beskou LINKS as positief ∑pi = ∑pf (mc + mm)vi = mcvcf + mmvmf ✓ 0 ✓ = 2 × -0,97 + 0,005 × vmf ✓ vfm = 392 m•s-1 east/right/forward / oos/regs/vorentoe ✓ (4) 5.3 Impulse = Fnet .∆t = ∆p = m(vf –vi) ✓ any one / enige een = 2 (-0,4 – 0,97) ✓ = - 2,74 N•s = 2,97 N•s ✓ (3) [12] Copyright reserved Blaai om asseblief ### Page 12 Downloaded from hlayiso.com 12 PHYSICAL SCIENCES P1 (EC/SEPTEMBER 2017) QUESTION/VRAAG 6 6.1 Doppler effect / Doppler-effek ✓ (1) 6.2 The number of waves reaching the detector per unit ✓ time decreases ✓. OR As the ambulance is moving away from the scene /detector, the wavelengths become longer resulting in less waves reaching the detector ✓ per unit time ✓ hence the frequency decreases. Die aantal golwe wat die detektor per eenheid ✓ tyd neem af. ✓ OF Soos die ambulans wegbeweeg vanaf die toneel / detektor, word die golflengte langer wat veroorsaak dat minder golwe die detektor ✓ per eenheidstyd ✓ bereik en dus neem die frekwensie af. (2) 6.3 v ± vL fL = fs ✓ any one / enige een v ± vs 𝑣 fL = v+𝑣 fs s 340 90% × 890 ✓ = 340+𝑣 ✓ × 890 ✓ s 340 × 890 801 = 340+𝑣s 340 × 890 340 +vs = 801 vs = 37,78 m•s-1 ✓ (5) 6.4 Doppler flow meter is used to determine whether arteries are clogged / narrowed ✓ OR to determine the rate of flow of blood ✓ Die Doppler-vloeimeter word gebruik om te bepaal of die are vernou/verstop is ✓ OF om die tempo van bloedvloei te bepaal. ✓ (1) 6.5 6.5.1 The shift is to a longer wavelength, lower frequency, as the star is moving away from the Earth. ✓ Die skuif is na ʼn langer golflengte, laer frekwensie, so die ster beweeg weg van die Aarde. ✓ (1) 6.5.2 A greater shift, therefore it shows that the distant star is moving away at a greater speed than a nearby star. ✓✓ ʼn Groter skuif, so dit toon dat die verafgeleë ster beweeg weg teen ʼn groter spoed as die nabygeleë ster. ✓✓ (2) [12] Kopiereg voorbehou Please turn over ### Page 13 Downloaded from hlayiso.com (EC/SEPTEMBER 2017) FISIESE WETENSKAPPE V1 13 QUESTION/VRAAG 7 7.1 The electrostatics force between the two charges is directly proportional the product of the two charges and inversely proportional to the square of the distance between them. ✓✓ Die elektrostatiese krag tussen die twee ladings is direk eweredig aan die produk van die twee ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. ✓✓ (2) 𝑘𝑄 𝑄 7.2 𝐹 = 𝐽 𝐿✓ 𝐽 𝑟2 (9 × 109 )(3 × 10−6 )(2 × 10−6 ) ✓ 𝐹𝐽 = (0,2)2✓ FJ = 1,35 N right / regs ✓ (4) 7.3 7.3.1 QL = -3 × 10-6 ✓ (1) 7.3.2 OPTION 1/OPSIE 1 OPTION 2 /OPSIE 2 n𝑒 = Q𝑓 −Q𝑖 ✓ Q𝑓 − Q𝑖 q𝑒 n𝑒 = ✓ −3×10−6 −2×10−6 q𝑒 n𝑒 = ✓ −3 × 10−6 + 8 × 10−6 1,6×10−19 n𝑒 = ✓ n𝑒 = 3,125 × 10 ✓ 13 1,6 × 10−19 n𝑒 = 3,125 × 1013 ✓ (3) 7.3.3 Criteria for sketch / Kriteria vir skets Marks/Punte Correct shape / Korrekte vorm ✓ Correct direction / Korrekte rigting ✓ Field lines not crossing each other ✓ Veldlyne kruis nie (3) 7.4 OPTION 1 / OPSIE 1 k𝑄 9 × 109 × 3 × 10−6 E𝐽 = 𝑟 2𝐽 ✓ = ✓= 1 875 000 N•C-1 West/Left / (Wes / Links) (0,12)2 k𝑄𝑀 9 × 10 × 3 × 10−6 9 E𝑀 = = ✓ = 4 218 750 N•C-1 West/Left / (Wes / Links) 𝑟2 (0,08)2 Left positive / Links positief: E𝑛𝑒𝑡 = E𝐽 + E𝑀 ✓ E𝑛𝑒𝑡 = (+1 875 000) + (+4 218 750) ✓(both substitutions)(albei instellings) E𝑛𝑒𝑡 = 6 093 750 N•C-1 West/Left / (Wes / Links) ✓ OPTION 2 /OPSIE 2 Enet = EJ + EM ✓ k𝑄 k𝑄𝑀 = 𝑟 2𝐽 + ✓ ✓ one of the two/enige een van die twee 𝑟2 (9×109 )× (3×10−6 ) (9×109 )× (3×10−6 ) = ✓ + (0,12)2 (0,08)2 = 6 093 750 N•C-1 West/Left (Wes / Links) ✓ (6) [19] Copyright reserved Blaai om asseblief ### Page 14 Downloaded from hlayiso.com 14 PHYSICAL SCIENCES P1 (EC/SEPTEMBER 2017) QUESTION/VRAAG 8 8.1 8.1.1 V max ✓ OR/OF maximum voltage (maksimum spanning) ✓ (1) 8.1.2 P𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = V𝑟𝑚𝑠 . I𝑟𝑚𝑠 ✓ (Pgem = Vwgk . Iwgk ✓) 94,3 3 94,3 =( ) ✓ . ( 100 × ) ✓ = 133,39 W ✓ (4) √2 √2 8.1.3 Replacing the slip rings with a split-ring commutator ✓✓ Vervang die sleepringe met ʼn split-ring kommutator ✓✓ (2) 8.2 8.2.1 (V𝑟𝑚𝑠 )2 𝑃= ✓ (P = Vwgk²/R ✓) R 902 56✓ = 𝑅 ✓ 𝑅 = 144,64 Ω ✓ (4) 8.2.2 Too bright ✓ The power of the generator is greater than the power of the light bulb. ✓ OR The power of the light bulb is smaller than the power of the generator. ✓ Te helder ✓ Die drywing van die generator is groter as die drywing van die gloeilamp ✓ OF Die drywing van die gloeilamp is kleiner as die drywing van die generator. ✓ (2) [13] QUESTION/VRAAG 9 9.1 Emf / Emk ✓ (1) 9.2 9.2.1 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 Vlost/verlore =Ir ✓ Emf (Emk) = I (R + r) ✓ 0,9 ✓= 4,5 × r ✓ 22,14 =Vext/ekst + Ir r = 0,2 Ω ✓ 22,14✓ = 21,24 + 4,5 × r ✓ r = 0,2 Ω ✓ (4) 9.2.2 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 V4Ω = IR4Ω V4Ω = IR4Ω = 4,5 × 4 ✓ = 4,5 × 4 ✓ = 18 V = 18 V Vext = Vp + V4Ω ✓ Vext = Vp + V4Ω 21,24 = Vp + 18 ✓ 21,24 = Vp + 18 ✓ Vp = 3,24 V ✓ Vp = 3,24 V ✓ 3,24 𝑉 3,24 𝑅𝑒𝑓𝑓 = 4,5 ✓ I4Ω = R = 4 = 0,81 A 𝑉 3,24 𝑅𝑒𝑓𝑓 = 0,72 Ω ✓ I3Ω = R = 3 = 1,08 A I𝑅1 = I𝑝 − (I4Ω + I3Ω ) I𝑅1 = 4,5 – (0,81 + 1,08) = 2,61 A ✓ 3,24 R1 = 2,61 = 1,24 Ω ✓ 1 1 1 1 = 4 + 3 + 1,24 = 1,39 𝑅𝑒𝑓𝑓 , :. 𝑅𝑒𝑓𝑓 = 0,72 Ω ✓ (6) 9.3 Temperature / Temperatuur ✓ (1) 9.5 Increase / Neem toe ✓ (1) [13] Kopiereg voorbehou Please turn over ### Page 15 Downloaded from hlayiso.com (EC/SEPTEMBER 2017) FISIESE WETENSKAPPE V1 15 QUESTION/VRAAG 10 10.1 Work function of a metal is a minimum energy needed by a metal in order to release/ejects electrons from its surface. ✓✓ Werksfunksie van ʼn metaal is die minimum energie wat ʼn metaal benodig om elektrone van sy oppervlak vry te laat ✓✓ (2) 10.2 W0 = hf0 ✓ = (6,63 × 10-34) × (1,4 × 1015) ✓ = 9,28 × 10-19J ✓ (3) 10.3 10.3.1 Remains the same / Bly dieselfde ✓ (1) 10.3.2 Increase / Neem toe ✓ (1) 1 10.4 Ek = 2 𝑚𝑣 2 ✓ 1 0,7 × 10-18 = 2 × 9,11 × 10−31 × 𝑣 2 ✓ 0,7 × 10−18 × 2 v = √ 9,11 ×10−31 v = 1,24 × 106 m•s-1 ✓ (3) [10] TOTAL/TOTAAL: 150 Copyright reserved Blaai om asseblief ## Related documents - [2017 Physics P1 Memo GP](https://www.hlayiso.com/library/2017-physics-p1-memo-gp/3c8a7aaa-341e-4611-bd94-085d68de1cb6) — Physical Sciences · Memo · 2017 - [Physical Sciences P1 May June 2017 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-may-june-2017-memo-afr-eng/ef24ecaa-5af8-4262-bea3-b96eddcae564) — Physical Sciences · Memo · 2017 · Eng - [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 - [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 P2 SEPT2017 ERRATA](https://www.hlayiso.com/library/phys-sciences-p2-sept2017-errata/14ef3406-a298-492e-92e0-6cd6a067d1c0) — Physical Sciences · Errata · 2017 - [Phsc P1 Jun2017 Qp Afr](https://www.hlayiso.com/library/phsc-p1-jun2017-qp-afr/b5663af8-09c3-43af-85ec-615d384497f5) — Physical Sciences · QP · 2017 · Afr ## 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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