# Physical Sciences Grade 12 P1 Eastern Cape Prelim 2024 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape Prelim Exam 2024 (English), 18 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/phys-sciences-p1-memo-gr12-sept-2024-afr-english/f774af61-24a2-4e5e-b3d2-d1b299c2cf53 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Language: English - Document type: Memorandum - Year: 2024 - Exam period: Eastern Cape Prelim 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 Prelim 2024 Memo](https://www.hlayiso.com/library/phys-sciences-p1-memo-gr12-sept-2024-afr-english/f774af61-24a2-4e5e-b3d2-d1b299c2cf53) ## About this document Physical Sciences · Grade 12 · Eastern Cape Prelim Exam · 2024 · English. 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 SENIORSERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2024 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. Hosted by www.ecexams.co.za. ### Page 2 Downloaded from hlayiso.com 2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) 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 Hosted by www.ecexams.co.za. ### Page 3 Downloaded from hlayiso.com (EC/SEPTEMBER 2024) 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 Hosted by www.ecexams.co.za. ### Page 4 Downloaded from hlayiso.com 4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) 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 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. 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 Hosted by www.ecexams.co.za. ### Page 5 Downloaded from hlayiso.com (EC/SEPTEMBER 2024) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 5 QUESTION 1/VRAAG 1: MULTIPLE-CHOICE QUESTIONS/ MEERVOUDIGEKEUSE-VRAE 1.1 B  (2) 1.2 C  (2) 1.3 C  (2) 1.4 A  (2) 1.5 B  (2) 1.6 A  (2) 1.7 D  (2) 1.8 D  (2) 1.9 B  (2) 1.10 A  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 6 Downloaded from hlayiso.com 6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION 2/VRAAG 2 2.1 When a resultant/net force acts on an object, the object will accelerate in the direction of the force at an acceleration directly proportional to the force and inversely proportional to the mass of the object.  Wanneer ʼn resulterende / netto krag op ʼn voorwerp inwerk, versnel die voorwerp in die rigting van die krag teen ʼn versnelling direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp. OR/OF The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force.  Die resulterende/netto krag wat op ʼn voorwerp inwerk, is gelyk aan die tempo van verandering van momentum van die voorwerp in die rigting van die resulterende/netto krag. (2) 2.2 Accepted labels / Aanvaarde benoemings w Fg/Fw/weight/gravitational force /  Gewig/gravitasiekrag T FT/Tension/Force in string/Spanning/  Spanning in tou N FN/N/Normal Force/Normaalkrag  Fk fk/Frictional force/Wrywingskrag  (4) 2.3 2.3.1 fk = μkN  fk = μkFg fk = (0,25)(4)(9,8)  fk = 9,8 N  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 7 Downloaded from hlayiso.com (EC/SEPTEMBER 2024) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 7 2.3.2 POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 2.3.1 Fnet = ma Fnet = T – f Any one / Enige een ✓ Fnet = Fg – T 4 kg block / 4-kg blok T – 9,8 = 4a T = 4a + 9,8 Any ONE / Enige EEN (4a or/of 2a) ✓ 2 kg block / 2-kg blok (2)(9,8) – T ✓ =2a T = 19,6 – 2a 19,6 – 2a = 4a + 9,8 a = 1,63 m∙s-2 vf2 = vi2 + 2aΔy  vf2 = 02 + 2(1,63)(1)  vf = 1,81 m∙s-1  (7) 2.4 The gravitational force (weight) is not the only force acting on the 2 kg block.  Die gravitasiekrag (gewig) is nie die enigste krag wat op die 2 kg-blok inwerk nie. OR/OF The acceleration is not 9,8 m.s-2.  Die versnelling is nie 9,8 m.s-2 nie. (2) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 8 Downloaded from hlayiso.com 8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION 3/VRAAG 3 3.1 Downwards/Afwaarts  (1) 3.2 Upwards positive/Opwaarts as Downwards positive/Afwaarts as positief positief v f = vi + a t  v f = vi + a t  = 30  + (-9,8)(2,135) = - 30  + (9,8)(2,135) = 9,08 m·s-1, upwards / opwaarts  = - 9,078 m·s-1 = 9,08 m·s-1, upwards/ opwaarts  (3) POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 3.2 3.3 OPTION 1/OPSIE 1 Upwards positive / Opwaarts as Downwards positive / Afwaarts as positief positief Ball A: Δy = v i Δt + 21 at 2  Δy = v i Δt + 21 at 2  ∆y = 12(2,5) + ½(9,8)(2,5)2  ∆y = -12(2,5) + ½(-9,8)(2,5)2  ∆y = 60,625 m ∆y = - 60,625 m (Height / Hoogte = 29,375 m) (Height / Hoogte = 29,375 m) Ball / Bal B: Ball / Bal B: Δy = v i Δt + 21 at 2 Δy = v i Δt + 21 at 2 = -30(2,5) + ½(9,8)(2,5)2  = 30(2,5) + ½(-9,8)(2,5)2  = - 44,375 m = 44,375 m Distance/afstand = 44,375 – 29,375  Distance/afstand = 44,375 – 29,375  = 15 m  = 15 m  OPTION 2/OPSIE 2 Ball/Bal A v f = vi + a t vf = -12 + (-9,8)(2,135) vf = 32,923 m·s-1, downwards/afwaarts Δy = v i Δt + 21 at 2  ∆y = -32(2,5 – 2,135) + ½(-9,8)(2,5 – 2,135)2  ∆y = - 12,6696 m Ball/Bal B Δy = v i Δt + 21 at 2 ∆y = 9,08(2,5 – 2,135) + ½(-9,8)(2,5 – 2,135)2 ∆y = 2,66 m Distance/afstand = 2,66 + 12,6696  15,33 m  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 9 Downloaded from hlayiso.com (EC/SEPTEMBER 2024) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 9 3.4 OPTION 1/OPSIE 1 Upward positive/Opwaarts as positief OPTION 2/OPSIE 2 Upward negative/Opwaarts negatief Criteria for graph / Nasienkriteria vir Marks / grafiek Punte Shape for ball A up till zero position.  Vorm vir bal A tot die nulposisie. Shape for ball B up till intersection of lines time. 2,135 s/  Vorm vir bal B tot die snypunt van lyne. 2,135 s. Indication of time 2,135 s/  Aandui van tyd 2,135 s Ground not zero position (provided everything else is correct): 2/3/ Grond nie as nulpunt nie (mits alles korrek is) 2/3 (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 10 Downloaded from hlayiso.com 10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION 4/VRAAG 4 4.1 The total linear momentum ✓ of an isolated system ✓ remains constant / is conserved. Die totale lineêre momentum in ʼn geїsoleerde sisteem bly konstant/ behoue. (2) 4.2 Right as positive/Reg as positief Σpi = Σpf Any one / Enige een ✓ (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 (5 000)(15) + (2 000)(-20 ✓ = (5 000)vf + (2 000)(5) ✓ vi = 5 m·s-1 ✓ Left as positive / Links as positief Σpi = Σpf Any ONE / Enige EEN ✓ (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 (5 000)(-15) + (2 000)(20) ✓ = (5000)vf + ( 2000)(-5) ✓ vi = -5 m·s-1 Magnitude of velocity / Grootte van snelheid = 5 m∙s-1 ✓ (4) POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 4.2/ 4.3 OPTION 1/OPSIE 1 Fnet∆t = ∆p Any one / Enige een ✓ Fnet Δt = mvf – mvi Fnet (0,4) ✓ = (5 000)(5) – (5 000)(15) ✓ Fnet = -125 000 N Fnet = 125 000 N to the right/na regs ✓ OPTION 2/OPSIE 2 Fnet∆t = ∆p Any one / Enige een ✓ Fnet Δt = mvf – mvi Fnet (0,4) ✓ = (2000)(-5) – (2000)(20) ✓ Fnet = -125 000 N Fnet = 125 000 N to the right/na regs ✓ (4) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 11 Downloaded from hlayiso.com (EC/SEPTEMBER 2024) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 11 QUESTION 5/VRAAG 5 5.1 Backwards/behind him / Terugwaarts/agter hom  (1) 5.2 Newton’s third Law of motion/Newton se derde wet van beweging When one body exerts a force on a second body, the second body exerts a force of equal magnitude  in the opposite direction on the first body.  Wanneer voorwerp A ʼn krag uitoefen op voorwerp B, sal voorwerp B gelyktydig ʼn krag van gelyke grootte in in die teenoorgestelde op voorwerp A uitoefen. (3) 5.3 OPTION 1/OPSIE 1 Wnet = ∆K Any one/Enige een ✓ Wg + Wf = ∆K Fg∆xcosӨ + f∆xcosӨ = ∆K (57)(9,8)(4)cos180°  + 40∆xcos180°  = 0 – ½(57)(62)  ∆x = - 30,21 m 4 Sin Ө = 30,21 Ө = 7,61°  OPTION 2/OPSIE 2 Wnc = ΔU + ΔK / Wnc = ΔEp + ΔEk  40∆xcos180o  = (57)(9,8)(4) – (57)(9,8)(0)  + ½ (57)(0)2 – ½ (57)(6)2  ∆x = -30,21 m 4 Sin Ө = 30,21 Ө = 7,61°  (5) 5.4 OPTION 1/OPSIE 1 Wnet = ∆K Any ONE/Enige EEN ✓ WT + Wg + Wf = ∆K (80)(5)(4)cos0° + (4)(9,8)sin30° •(5)cos180° + (15)(5)cos180°  = ½(4)vf2 – ½(4)(32)  vf = 11,07 m·s-1  OPTION 2/OPSIE 2 Wnc = ΔU + ΔK Any one/Enige een ✓ WT + Wf = ΔU + ΔK (80)(5)(4)cos0° + (15)(5)cos180°  = (4)(9,8)(sin30°)(5) – (4)(9,8)(0)  + ½ (4)vf2 – ½ (4)(3)2  vf = 11,07 m·s-1  (6) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 12 Downloaded from hlayiso.com 12 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION 6/VRAAG 6 6.1 The (apparent) change in frequency/pitch  of the sound detected by a listener, because the listener and the sound source have different velocities relative to the medium of sound propagation.  Die (skynbare) verandering in frekwensie/toonhoogte van die klank wat deur ʼn luisteraar waargeneem word, omdat die luisteraar en die klankbron verskillende snelhede het relatief tot die medium van klankvoortplanting. OR/OF The (apparent) change in frequency/pitch  of the (sound) wave heard by a listener when there is relative motion between the listener and the source producing the sound waves. / Die (waarskynlike) verandering in frekwensie / toonhoogte van (klank) soos gehoor deur ʼn luisteraar wanneer daar relatiewe beweging tussen die luisteraar en bron wat die klankgolwe produseer. (2) 6.2 6.2.1 INCREASE/TOENEEM  (1) 6.2.2 DECREASE/AFNEEM  (1) 6.3 v  vL v + vL fL = f s  or/of fL = 𝑓𝑆  v  vs v 330 + 30 fL=  x 1 800  330 − 0 fL = 1 963,6 Hz  (5) 6.4 Marking Criteria/Nasienkriteria Labels of axes/Benoeming van asse  Higher frequency initially/Hoër  aanvanklike frekwensie Lower frequency as time increases/  Laer frekwensie soos tyd toeneem (3) 6.5 Bloodflow meter  / Doppler flow meter  Bloedvloeimeter / Doppler-vloeimeter (1) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 13 Downloaded from hlayiso.com (EC/SEPTEMBER 2024) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 13 QUESTION 7/ VRAAG 7 7.1 The electrostatic force between two point charges is directly proportional to the product of the magnitude of the charges,  and inversely proportional to the square of the distance between them.  Die elektrostatiese krag tussen twee puntladings is direk eweredig aan die produk van die grootte van die ladings, en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. (2) 7.2 Since/Aangesien FK on M = FG on M  (Fnet)2 = (FK on M)2 + (FG on M)2 (2,864 x 10-6)2  = F2 + F2 F = 2,025 x 10-6 N  (4) F = kQ1Q2  r2 2,025 x 10-6  = 9 x 109 x 6 x 10-9 x 6 x 10-9  (X)2 X = 0,4 m  (8) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 14 Downloaded from hlayiso.com 14 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION 8/VRAAG 8 8.1 Criteria for marking / Nasienkriteria Correct shape / Korrekte vorm  Correct direction / Korrekte rigting  Lines touching charge and not crossing/  Lyne raak lading en kruis nie (3) 8.2 It is the electrostatic force exerted per unit positive charge placed at that point. / Dit is die elektrostatiese krag wat per eenheidspositiewe-lading wat by daardie punt geplaas is. (2) 8.3 EP = kQ1  r2 EP = 9 x 109 x 200 x 10-9  (0,2)2 EP = 45 000 N.C-1 to the right / na regs EQ = kQ2 r2 EQ = 9 x 109 x 200 x 10-9  (0,4)2 EQ = 11 250 N.C-1 to the left / na links Enet = 45 000 + (-11 250)  Enet = 33 750 N.C-1 to the right / na regs  (5) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 15 Downloaded from hlayiso.com (EC/SEPTEMBER 2024) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 15 QUESTION 9/VRAAG 9 9.1 The potential difference is directly proportional to current,  provided the temperature remains constant.  Die potensiaalverskil is direk ewerewig aan die stroom, mits die temperatuur konstant bly. OR/OF The ratio of potential difference to current  at constant temperature  is a constant./ Die verwantskap tussen potensiaalverskil tot stroom by konstante temperatuur is konstant. (2) 9.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 V2 P = VI  P =  IR = P/V = 13,5/18  = 0,75 A R (18)2 V = IR 13,5 =  R = V/IR = 18/0,75 = 24 Ω  R R = 24 Ω  (3) POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 9.2 9.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 V 1 1 1 R =  = +  I Rp 𝑅1 R2 1 1 1  = +  24 = (18) RP 12 24 RP = 8 Ω  I V = IRP 18 = I (8)  I = 0,75 A I = 2,25 A  Vp R= I12 18 12 = I12 I12 = 1,5 A  ITotal = 1,5 + 0,75 = 2,25 A  (5) 9.4 Internal resistance is the opposition to the flow of charge within a battery / Interne weerstand is die teenstanding van die vloei van lading binne die battery. (2) POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 9.3 9.5 V10 = IR10  V10 = 2,25 x 10  V10 = 22,5 V  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 16 Downloaded from hlayiso.com 16 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) 9.6 OPTION 1/OPSIE 1 V1 = Vext V1 = VP + V10 V1 = 18 + 22,5  V1 = 40,5 V When the switch is open the Emf / Wanneer die skakelaar oop is, is die emk = 45,9 V The ‘lost volts’ is / Die ‘verlore spanning’ is: Vlost = emf/emk – Vext = 45,9 – 40,5  = 5,4 V V r= L  I 5,4 r=  2,25 r = 2,4 Ω  OPTION 2/OPSIE 2 1 1 1 Emf = I (R + r)  = + Rp r1 r2 1 1 1 45,9 = (2,25) (8 + 10 + r)  = +  RP 12 24 r = 2,4 Ω  RP = 8 Ω (5) 9.7 Remain the same/Bly dieselfde (1) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 17 Downloaded from hlayiso.com (EC/SEPTEMBER 2024) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 17 QUESTION 10/VRAAG 10 10.1 To ensure continuous rotation of the coil. / Om te verseker dat die spoel aanhoudend roteer. (1) 10.2 (i) Replace the source of potential difference with load/resistor. / Vervang die bron van die potensiaalverskil met ʼn resistor (ii) Replace the split ring (commutator) with (two) slip rings / Vervang die splitring (kommutator) met (twee) sleepringe. (2) POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 10.2 10.3 I I rms = max  2 0,54 Irms =  √2 Irms = 0,38 A Pave = VrmsIrms  60 = Vrms x 0,38  Vrms = 157,89 V  (5) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za. ### Page 18 Downloaded from hlayiso.com 18 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION 11/VRAAG 11 11.1 c = f x λ Accept/Aanvaar v = f λ 3 x 108 = f x 229 x 10-9  f = 1,31 x 1015 Hz  (3) 11.2 Threshold frequency / Drumpelfrekwensie  (1) 11.3 Work function is the minimum amount of energy required to eject an electron out of a surface of a given solid, usually a metal. / Werkfunksie is die minimum energie benodig om ʼn elektron uit die oppervlak van ʼn metaal vry te stel. (2) 11.4 E = Wo + Ek ( max/maks )  hf = Wo + ½ mv2 6,63 x 10-34 x f  = (6,63 x 10-34 x 1,31 x 1015) + ½ (9,11 x 10-31) (1,57 x 106)2  f = 3 x 1015 Hz  (4) 11.5 INCREASES / TOENEEM  A decrease in wavelength results in an increase in the frequency.  Work function increases.  ʼn Afname in die golflengte lei na ʼn toename in die frekwensie. Arbeidsfunksie neem toe. 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