# Physical Sciences Grade 12 P1 Mpumalanga November 2014 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Mpumalanga November 2014, 18 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mpu-phys-sc-p1-sept-2014-memo/42fe566a-4b70-433e-97ef-e69cb2fe938a - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2014 - Exam period: Mpumalanga November - 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 Mpumalanga November 2014 Memo](https://www.hlayiso.com/library/mpu-phys-sc-p1-sept-2014-memo/42fe566a-4b70-433e-97ef-e69cb2fe938a) ## About this document Physical Sciences · Grade 12 · Mpumalanga November · 2014. 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 EXAM PHYSICAL SCIENCES: PHYSICS (P1) GRADE 12 SEPTEMBER 2014 MEMORANDUM MARKS: 150 TIME: 3 HOURS ### Page 2 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/September 2014 CAPS GENERAL GUIDELINES 1. CALCULATIONS 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 may be relevant symbols and applicable substitutions. Geen punte sal toegeken word waar 'n 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 'n fout gedurende substitusie in 'n korrekte formule begaan word, sal 'n 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 formula/principle is given correctly. 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 the 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 'n kandidaat met die korrekte formule begin en dan die onderwerp van die formule verkeerd verander, sal 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 had been attempted. i.e. substitutions have been made or a numerical answer given. Punte word slegs vir 'n formule toegeken indien 'n poging tot 'n berekening aangewend is, d.w.s. substitusies is gedoen of 'n 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 formule ingestel is en nie vir waardes wat voor 'n berekening genoem is nie. ### Page 3 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/September 2014 CAPS 1.9 All calculations, when not specified in the question, must be done to two decimal places. Alle berekenings, wanneer nie in die vraag gespesifiseer word nie, moet tot 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 'n finale antwoord tot 'n 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 'n reeks berekenings gedoen moet word (bv. 'n stroomdiagramvraag) hoef nie noodwendig dieselfde volgorde te hê nie. VOLPUNTE sal toegeken word op voorwaarde dat dit 'n 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 or subquestion. 'n Kandidate sal slegs een keer gepenaliseer word vir die herhaaldelike gebruik van 'n verkeerde eenheid in 'n vraag of subvraag. 2.2 Units are only required in the final answer to a calculation. Eenhede word slegs in die finale antwoord tot 'n 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 word slegs vir 'n antwoord en nie vir 'n eenheid per se toegeken nie. Kandidate sal derhalwe 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. ### Page 4 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/September 2014 CAPS 3. GENERAL/ALGEMEEN 3.1 If one answer or calculation is required, but two 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 'n vermenigvuldigpunt en nie met 'n punt nie, byvoorbeeld, m·s-1. Vir nasiendoeleindes sal m.s-1 em 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 berekenings 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 or 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 'n sekere veranderlike in een subvraag (bv. 3.1) bereken word en dan in 'n ander vervang moet word (3.2 of 3.3), bv. indien die antwoord vir 3.1 verkeerd is en korrek in 3.2 of 3.3 vervang word, word volpunte aan die daaropvolgende subvraag toegeken. 4.2 A multistep question in a subquestion: If the candidate has to calculate, for example, current in the first step and gets it wrong due to a substitution error, the mark for the substitution and the final answer will be forfeited. 'n Vraag met veelvuldige stappe in 'n subvraag: Indien 'n kandidaat byvoorbeeld, die stroom verkeerd bereken in 'n eerste stap as gevolg van 'n substitusiefout, verbeur 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 to 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 in 3.2 could be considered. 'n Verkeerde antwoord, indien dit op 'n konsepsuele fout gebaseer is, kan normaalweg nie korrek gemotiveer word nie. Indien 'n 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 'n berekening gebaseer is, kan die motivering vir die verkeerde antwoord in 3.2 oorweeg word. ### Page 5 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/September 2014 CAPS SECTION A QUESTION 1 / VRAAG 1 1.1 A  1.2 A  1.3 C  1.4 C  1.5 D  1.6 C  1.7 A  1.8 B  1.9 B  1.10 D  [20] TOTAL SECTION A/TOTAAL AFDELING A: 20 SECTION B/AFDELING B QUESTION 2 / VRAAG 2 2.1 Fx = F·Cos Θ = 8,25 Cos 20°  = 7,75 N  (2) 2.2 FN + Fy + Fg = 0 FN + (8,25×Sin20°) + (-19,6) = 0 FN = 16,78 N  (3) 2.3 POSITIVE MARKING FROM QUESTION 2.1 + 2.2 f = µk ·FN  7,75 = µk(16,78 +1,5×9.8)  µk = 0,25 (4) 2.4 Increases  / Toeneem (1) [10] ### Page 6 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/September 2014 CAPS QUESTION 3 / VRAAG 3 3.1 9,8 m·s-2 downwards / afwaarts (2) 3.2 OPTION 1 Notes: upwards positive: Accept / Aanvaar: ∆y = vi∆t + ½a∆t2  g or/of a 21 = vi(3,1) + ½(-9,8)(3,1)2  ∆x = vi∆t + ½a∆t2 vi = 21,96 m·s-1  downwards positive ∆y = vi∆t + ½a∆t2  -21 = vi(3,1) + ½(9,8)(3,1)2  vi = -21,96 m·s-1 vi = 21,96 m·s-1  (4) 3.3 POSITIVE MARKING FROM 3.2 OPTION 1/OPSIE 1 Notes/Aantekeninge: Upwards as positive Accept/Aanvaar: vf2 = vi2 + 2a∆y  g or/of a 02 = 21,962 + 2(-9,8)∆y  vf2 = vi2 + 2a∆x ∆y = 24,6 m v2 = u2 + 2as Max height = 24,6 + 0,6 = 25,2 m  (4) Downwards as positive vf2 = vi2 + 2a∆y  02 = (-21,96)2 + 2(9,8)∆y  ∆y = -24,6 m Max height = 24,6 + 0,6 = 25,2 m  (4) OPTION 2/OPSIE 2 (Ep + Ek)bottom = (Ep + Ek)top (mgh + ½mv2)bottom = (mgh + ½mv2)top } 0 + ½(21,96)2 = 9,8h + 02  h = 24,6 m } Max height above the ground = 24,6 + 0,6 = 25,2 m  (4) ### Page 7 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/September 2014 CAPS OPTION 3/OPSIE 3 Notes/Aantekeninge: Upwards as positive Accept/Aanvaar: vf = vi + a∆t g or/of a 0 = 21,96 + (-9,8)∆t  v = u + at ∆t = 2,24 s s = ut + ½ at2 ∆x = vi∆t + ½a∆t2 ∆y = vi∆t + ½a∆t2 = (21,96)(2,24) + ½(-9,8)(2,24)2  ∆y = 24,6 m Max height = 24,6 + 0,6 = 25,2 m  (4) Downwards as positive vf = vi + a∆t 0 = (-21,96) + (9,8)∆t  ∆t = 2,24 s ∆y = vi∆t + ½a∆t2 = (-21,96)(2,24) + ½(9,8)(2,24)2  ∆y = -24,6 m Max height = 24,6 + 0,6 = 25,2 m  (4) OPTION 4/OPSIE 4 Notes/Aantekeninge: Upwards as positive Accept/Aanvaar: vf = vi + a∆t g or/of a 0 = (21,96) + (-9,8)∆t  v = u + at  ∆t = 2,24 s s=       ∆x =   ∆t ∆y = ∆t    ,  = 2,24   ∆y = 24,6 m Max height = 24,6 + 0,6 = 25,2 m  (4) Downwards as positive vf = vi + a∆t 0 = (-21,96) + (9,8)∆t  ∆t = 2,24 s   ∆y =   ∆t   ,  = 2,24   ∆y = -24,6 m Max height = 24,6 + 0,6 = 25,2 m  (4) ### Page 8 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/September 2014 CAPS 3.4 Inelastic, the ball bounces lower as the max height before the bounce  Onelasties, die bal hop laer as die maksimum hoogte voor die bons (2) OR/OF Inelastic, Max height before bounce > max height after bounce  Onelasties Maksimum hoogte voor bons > maksimum hoogte na bons (2) 3.5 POSITIVE MARKING FROM QUESTION 3.3 OPTION 1/OPSIE 1 Upwards as positive y (m) 3,6 3,1 4,2 t (s) -21 Marking criteria for graph: Nasienriglyne vir grafiek: Graph starts at -21 m at t = 0 s  Grafiek begin by -21 m by t = 0 s Correct shape as shown up to t = 3,1 s  Maximum height = 3,6 m [(Answer of h in Q3.3) -21]  Time of bounce as t = 3,1 s  Correct shape for second part from t = 3,1 s to t = 4,2 s  ### Page 9 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/September 2014 CAPS DOWNWARDS as positive y (m) 21 3,1 4,2 t (s) -3,6 Marking criteria for graph: Nasienriglyne vir grafiek: Graph starts at 21 m at t = 0 s  Grafiek begin by 21 m by t = 0 s Correct shape as shown up to t = 3,1 s  Maximum height = -3,6 m [(Answer of ∆y in Q3.3) +21]  Time of bounce as t = 3,1 s  Correct shape for second part from t = 3,1 s to t = 4,2 s  NOTES/AANTEKENINGE: OTHER POSSIBILITIES [Position from where he kicked, as reference point] Upwards as positive y (m) 24,6 Max: 45 t (s) 0 3,1 4,2 Marking criteria for graph: Nasienriglyne vir grafiek: Graph starts at -21 m at t = 0 s × Grafiek begin by -21 m by t = 0 s Correct shape as shown up to t = 3,1 s  Maximum height = 24,6 m  Time of bounce as t = 3,1 s  Correct shape for second part from t = 3,1 s to t = 4,2 s  ### Page 10 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/September 2014 CAPS [Ground as reference point] Upwards as positive y (m) 25,2 Max: 45 0,6 3,1 4,2 t (s) Marking criteria for graph: Nasienriglyne vir grafiek: Graph starts at -21 m at t = 0 s × Grafiek begin by -21 m by t = 0 s Correct shape as shown up to t = 3,1 s  Maximum height = 25,2 m  Time of bounce as t = 3,1 s  Correct shape for second part from t = 3,1 s to t = 4,2 s  [22] ### Page 11 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/September 2014 CAPS QUESTION 4/VRAAG 4 4.1 The total (linear) momentum remains constant  (in magnitude and direction) in a closed system  (2) Die totale (liniêre) momentum in ‘n geslote sisteem bly konstant (in grootte en rigting) Notes/Aantekeninge: Allocate mark for “closed system” only in conjunction with momentum 4.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Σpi = Σpf  ∆pboy = -∆ptrolley  45(5) + 18(5) = 45(-1) + 18vf  45(-1) – 45(5)  = -(18vf – 18(5))  vf = 20 m·s-1 west vf = 20 m·s-1 west  (4) (4) OR/OF Σpi = Σpf  (45+18)(5) = 45(-1) + 18vf  vf = 20 m·s-1 west  (4) 4.3 Remains the same / Dieselfde  F remains the same (newton III) and ∆t remains the same OR/OF Remains the same / Dieselfde  ∆ptrolley = -∆pboy  and impuls = ∆p: F∆t = ∆p  (3) 4.4 POSITIVE MARKING FROM 4.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Fnet·∆t = mvf – mvi  Fnet·∆t = mvf – mvi  Ftrolley (0,4) = 18(20) – 18(5) Fboy(0,4) = 45(-1) – 45(5) Ftrolley = 675 N west  Fboy = -675 N (4) Fboy = 675 N east Ftrolley = 675 N west  (4) [13] ### Page 12 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/September 2014 CAPS QUESTION 5/VRAAG 5 5.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 N F N F w w//   (3) (3) Take note: DON’T penalise for relative lengths of vectors, although: [N and  should be equal in length] [F should be shorter than w//] Notes/Aantekeninge: • Any additional forces (like friction) max 23 • No arrows: 03 • Force(s) not touching object: max 23 5.2 No, it is not an isolated system Nee, dit is nie ‘n geslote sisteem nie OR/OF No, there is an external force acting on the object Nee, daar is ‘n eksterne krag wat op die voorwerp inwerk (2) 5.3 Gravitational force / weight  Gravitasie (aantrekkings) krag / gewig Note: NOT Fg/w/w///parallel component of weight 01 (1) 5.4 The force is perpendicular to the displacement  OR/OF θ = 90º therefore Cosθ = 0 (1) 5.5 B to A  5.6 OPTION 1/OPSIE 1 Wnet = ∆Ek  Fg//·∆x·Cosθ + F·∆x·Cosθ = ½mvf2 – ½mvi2 (25×9,8×Sin30º)  (3,5)Cos180º + F(3,5)Cos0º = ½(25)(10,8)2 – ½(25)(12)2 (63,41)(3,5)(-1) + F(3,5)(1) = 1458 – 1800 F = 24,79 N  (6) OPTION 2/OPSIE 2 Wnet = ∆Ek  Fg·∆x·Cosθ + F·∆x·Cosθ = ½mvf2 – ½mvi2 (25×9,8)  (3,5)Cos120º+ F(3,5)Cos0º = ½(25)(10,8)2 – ½(25)(12)2 (245)(3,5)(-0,5) + F(3,5)(1) = 1458 – 1800 F = 24,79 N (6) ### Page 13 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/September 2014 CAPS OPTION 3/OPSIE 3 Wnc = ∆Ek + ∆Ep  F·∆x·Cosθ = ½mvf2 – ½mvi2 + mghf - mghi F(3,5)Cos0º= [½(25)(10,8)2 – ½(25)(12)2]  + [(25)(9,8)  (3,5Sin30º)  - 0] F = 24,79 N  (6) [14] QUESTION 6/VRAAG 6 6.1 The change in frequency (or pitch) of the sound detected by a listener because the sound source and the listener have different velocities relative to the medium of sound propagation. Die verandering in frekwensie (of toonhoogte) van die klank waargeneem deur 'n luisteraar omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium waarin die klank voortgeplant word, het. (2) 6.2 Higher / Hoër  (1) 6.3 [Boat = S, Whale = L] ±  = ±! #  " $%&' = $%&  23000 = 23128,223 Hz (4) 6.4 Futher away λ increases ⟹ If c=constant, f will decrease [ ∝ 1* + , = ,-./0.] Verder weg λ neem toe ⟹ As c=konstant dan sal f afneem [ ∝ 1* 0/ , = 2-./0.] OR Futher away A shift towards the red end of the spectrum is observed ⟹ If c=constant, f will decrease Verder weg ‘n Verskuiwing na die rooi end van die spektrum word waargeneem ⟹ As c=konstant dan sal f afneem (4) [11] ### Page 14 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/September 2014 CAPS QUESTION 7/VRAAG 7 7.1 Marking Criteria : • Shape • Arrows pointing towards Q • Fieldlines don’t touch/cross & perpendicular on charge (3) 7.2 45 3 = 67  9 8 : ; $8 ?:  4,44 8 10&  = 67  -3 r = 9×10 ∴d = 9×10-3 - 6×10-3 d = 3×10-3 m (OR 3 mm) (5) 7.3 The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them. Die grootte van die elektrostatiese krag wat een puntlading (Q1) op 'n ander puntlading (Q2) uitoefen, is direk eweredig aan die produk van die groottes van die ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle. (2) 7.4 45> 57 =5 =  67 9 8 : ;9$8 ?: ; $8 ?:  =  , 7 F = 4×10-3 N (west or left) 2@ @% =5% = A 9 8 : ;9$8 ?: ; &8 ?:  =  , 7 -3 F = 1,8 × 10 N (south or down) 2 -3 2 -3 2 4×10-3 N F = (4×10 )  + (1,8 × 10 )  1,8×10-3 N ∴F = 4,39 × 10-3 N Fnet (6) [16] ### Page 15 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/September 2014 CAPS QUESTION 8/VRAAG 8 8.1 Criteria for investigative question/Kriteria vir ondersoekende vraag: Dependent and independent variables correctly identified.  Afhanklike en onafhanklike veranderlikes korrek geïdentifiseer. Question about the relationship between the independent and dependent variables correctly formulated.  Vraag oor die verwantskap tussen die afhanklike en onafhanklike veranderlikes korrek geformuleer. Example/Voorbeeld: What is the influence of the type of material of a conductor on the conductivity of the conductor? / Wat is die verband tussen die tipe materiaal van die geleier op die geleidingsvermoë van die geleier Dependent variable/Afhanklike veranderlike: • Conductivity of the conductor / Geleidingsvermoë van die geleier Independent variable/Onafhanklike veranderlike: • Type of material/metal / Tipe materiaal/metaal Example/Voorbeeld How will the conductivity of a conductor change if different types of metarials/metals are used? Hoe sal die geleidingsvermoë van ‘n geleier verander indien verskillende tipe metale gebruik word? Notes/Aantekeninge: A question that results in a ‘yes’ / ‘no’ answer: max 12 ‘n Vraag wat ‘n ‘ja’ of ‘nee’ as antwoord het: maks 12 Example: Will the type of metal have an influence on the conductivity of the metal? 8.2 Any TWO:  Temperature / Temperatuur Thickness of conductor / Dikte van die geleier Length of conductor / lengte van geleier. (2) 8.3 P Gradient = B  Gradient Q < gradient P ∴RQ >RP  (3) [7] ### Page 16 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/September 2014 CAPS QUESTION 9/VRAAG 9 9.1 Due to internal resistancev energy is transferred inside the battery” As gevolg van interne weerstandv word energie binne in die battery oorgedra. Vv (2) 9.2.1 V=IRY 4,5 =1(3) Vv 1=1,5AVY (3) 9.2.2 POSITIVE MARKING FROM QUESTION 9.2.1 OR/OF: e = |(R4r) ¥ V=IRV = 1,5¥ ((12V+40)+3v) = 1,5¥ (12% +40+3v) € = 82,5 Vv V = 82,5 Vv (5) (5) 9.2.3 | OPTION 1/OPSIE 1 OPTION 2/OPSIE2 V, =IR _ 20 = 1,5(12) P= vay) = 18V I =0,6 AY V=IR 18Y =1(307) 1=0,6AVY (3) 9.3 DecreasesY Total resistance increases V Total current decreases Vv If R remains constant” then P will decrease as well (P « J”) Afneemv Totale weerstand neem toe Vv Totale stroomsterkte neem af ¥ Aangesien R konstant bly” sal P dus afneem (P « J”) (4) [17] ### Page 17 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/September 2014 CAPS QUESTION 10/VRAAG 10 10.1 Mechanical energy to electric energy  Meganiese energie na elektriese energie  OR/OF Kinetic energy to electric energy  Kinetiese energie na elektriese energie  (1) 10.2.1 Brush(es) / Carbon brush(es)  Borsel(s) / Koolstofborsel(s) (1) 10.2.2 Slipring(s) / Sleepring(e)  (1) 10.3.1 A  (1) 10.3.2 OPTION 1/OPSIE 1 VIKL VHIJ = √2 % =  √ Vrms = 212,13 V COPQ IHIJ = √2  =  √ C6OR = 14,85 A Pavg = Vrms ·Irms  = (212,13)(14,85)  = 3150,13 W  (5) OPTION 2/OPSIE 2 Pavg = Vrms ·Irms  S W = TUV 8 TUV √ √ %8  =  = 3150 W  (5) [8] ### Page 18 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/September 2014 CAPS QUESTION 11/VRAAG 11 11.1 Frequency of P < threshold frequency of metal cathode R  Frekwensie van P < drumpelfrekwensie van metaalkatode R OR/OF Energy of light P < Workfunction of metal cathodeR  Energie van P < werksfunksie van metaalkatode R (1) 11.2 R S CRITERIA Criteria for marking: • X-intercept to the right of R • Graph S parallel to graph R (2) 11.3 E = hf  = (6,63×10-34)(1,5×1015)  E = 9,95×10-19 J  (3) 11.4 W = W o + Ek  9,95×10-19 = (6,63×10-34)(5,79×1014)  + Ek Ek = 6,11×10-19 J  (4) [11] QUESTION 12/VRAAG 12 12.1 ∆E = Ef – Ei = -5,44 × 10-19 –(-2,40 × 10-19)  ∆E = 3,04×10-19 J XY 3=  Z , %8 ?[\  %8 ]  3,04×10-19  =  Z λ = 6,54×10-7 m  ∴ RED/ROOI  (6) [6] TOTAL SECTION B/TOTAAL AFDELING B: 130 GRAND TOTAL/GROOTTOTAAL: 150 ## Related documents - [mpu physical sciences p2 sept 2014 memo](https://www.hlayiso.com/library/mpu-physical-sciences-p2-sept-2014-memo/b6800da7-3221-43f2-ba55-a68d6aaa7dab) — Physical Sciences · Memo · 2014 - [PHYS SC P1 MEMO ENGAFR](https://www.hlayiso.com/library/phys-sc-p1-memo-engafr/da9b9ae0-718d-442a-95bc-aa0bfbf308f3) — Physical Sciences · Memo · 2014 - [2014AFR gau physical sciences p1 memo](https://www.hlayiso.com/library/2014afr-gau-physical-sciences-p1-memo/db77042f-0bdc-4e06-8f6b-ce760a7e4148) — Physical Sciences · Memo · 2014 - [2014ENG gau physical sciences p1 memo](https://www.hlayiso.com/library/2014eng-gau-physical-sciences-p1-memo/d8eca56a-f776-4202-84cd-3e2e7e299ae3) — Physical Sciences · Memo · 2014 - [Physical Sciences P1 Nov 2014 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2014-memo-afr-eng/44924678-c33b-42e1-8937-fcfda16ea7b3) — Physical Sciences · Memo · 2014 · Eng - [wkp kwod physc p1 sept 2014 eng memo](https://www.hlayiso.com/library/wkp-kwod-physc-p1-sept-2014-eng-memo/bb06a52a-4a49-454d-ae6e-5128be28cc24) — Physical Sciences · Memo · 2014 · Eng ## 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. When citing, link the page URL above, not a PDF or /api/ URL.