# Physical Sciences Grade 12 P1 2017 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Northern Cape November 2017, 23 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/2017-physics-p1-memo-nc-fs/dc8bc0b3-17c4-4042-96d7-6356caabc15d - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2017 - Exam period: Northern Cape November - Paper: 1 - Curriculum: CAPS - Pages: 23 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 2017 Memo](https://www.hlayiso.com/library/2017-physics-p1-memo-nc-fs/dc8bc0b3-17c4-4042-96d7-6356caabc15d) ## About this document Physical Sciences · Grade 12 · Northern Cape November · 2017. Memorandum, 23 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 Physical Sciences/ Gr 12/Memorandum/P1 1 NCDOE/ SEPTEMBER 2017 PCE NORTHERN CAPE DEPARTMENT OF EDUCATION NOORDKAAP DEPARTEMENT VAN ONDERWYS PROVINCIAL COMMON QUESTION PAPER PROVINSIALE GEMEENSKAPLIKE VRAESTEL GRADE 12 GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) SEPTEMBER 2017 MEMORANDUM MARKS/ PUNTE: 150 This memorandum consists of 22pages Hierdie memorandum bestaan uit 22 bladsye 1 ### Page 2 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 2 NCDOE/ SEPTEMBER 2017 PCE 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 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 correct 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 has 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. 2 ### Page 3 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 3 NCDOE/ SEPTEMBER 2017 PCE 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 gelys is nie. 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 'n minimum van twee desimale plekke gedoen word. 2. UNITS/ EENHEDE 2.1 Candidates will only be penalised once for the repeated use of an incorrect unit within a question or sub-question. Kandidate sal slegs een keer gepenaliseer word vir die herhaaldelike gebruik van 'n verkeerde eenheid in 'n vraag. 2.2 Units are only required in the final answer to a calculation. Eenhede word slegs in die finale antwoord op '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. -1 SI-eenhede moet gebruik word, behalwe in sekere gevalle, bv. V∙m in plaas van -1 -1 -1 -1 N∙C , en cm∙s of km∙h in plaas van m∙s waar die vraag dit regverdig. 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 3 ### Page 4 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 4 NCDOE/ SEPTEMBER 2017 PCE 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, -1 -1 byvoorbeeld, m·s . Vir nasiendoeleindes sal m.s 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 berekenings sal in die volgende gevalle geld: 4.1 Sub-question to sub-question: When a certain variable is calculated in one sub- question (e.g. 3.1) and needs to be substituted in another (3.2 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 sub-questions. 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 4.2 die antwoord vir 3.1 verkeerd is en word korrek in 3.2 of 3.3 vervang, word volpunte vir die daaropvolgende subvraag toegeken. A multi-step question in a sub-question: 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, verloor die kandidaat die punt vir die substitusie sowel as die finale antwoord. 4.5 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. 4 ### Page 5 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 5 NCDOE/ SEPTEMBER 2017 PCE QUESTION 1 / VRAAG 1 1.1 D ✓✓ (2) 1.2 C ✓✓ (2) 1.3 D ✓✓ (2) 1.4 B ✓✓ (2) 1.5 B ✓✓ (2) 1.6 B ✓✓ (2) 1.7 A ✓✓ (2) 1.8 C ✓✓ (2) 1.9 C ✓✓ (2) 1.10 D ✓✓ (2) [20] QUESTION 2 / VRAAG 2 2.1 (gravitational) acceleration ✓ (Gravitasie) versnelling (1) 2.2 Option 1/Opsie1 Option 2/Opsie2 Up as positive / Op as positief Down as positive / Af as positief 2.5 s – 5 s 2.5 s – 5 s 𝑣𝑣𝑓𝑓 = 𝑣𝑣𝑖𝑖 + 𝑔𝑔∆𝑡𝑡✓ 𝑣𝑣𝑓𝑓 = 𝑣𝑣𝑖𝑖 + 𝑔𝑔∆𝑡𝑡✓ 0 = 𝑣𝑣𝑖𝑖 + (−9.8)(2.5) ✓ 0 = 𝑣𝑣𝑖𝑖 + (9.8)(2.5) ✓ 𝑣𝑣𝑓𝑓 = 24,5𝑚𝑚 ∙ 𝑠𝑠 −1 ✓ 𝑣𝑣𝑓𝑓 = −24.5 𝑚𝑚. 𝑠𝑠 −1 𝑣𝑣𝑓𝑓 = 24,5 𝑚𝑚 ∙ 𝑠𝑠 −1 ✓ Option 3/Opsie3 Option4/Opsie4 Up as positive / Op as positief Down as positive / Af as positief 0 - 2.5s 0 – 2.5 s 𝑣𝑣𝑓𝑓 = 𝑣𝑣𝑖𝑖 + 𝑔𝑔∆𝑡𝑡✓ 𝑣𝑣𝑓𝑓 = 𝑣𝑣𝑖𝑖 + 𝑔𝑔∆𝑡𝑡✓ 𝑣𝑣𝑓𝑓 = 49 + (−9.8)(2.5) ✓ 𝑣𝑣𝑓𝑓 = −49 + (9.8)(2.5) ✓ 𝑣𝑣𝑓𝑓 = 24,5𝑚𝑚 ∙ 𝑠𝑠 −1 ✓ 𝑣𝑣𝑓𝑓 = − 24.5𝑚𝑚. 𝑠𝑠 −1 𝑣𝑣𝑓𝑓 = 24,5𝑚𝑚 ∙ 𝑠𝑠 −1✓ NB: Do not penalise if learners used a instead of g./ LW: Moenie penaliseer as leerders (a) in plaas van (g) gebruik het nie. 5 ### Page 6 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 6 NCDOE/ SEPTEMBER 2017 PCE 2.3.1 Displacement = area underneath the graph Verplasing = area onder grafiek OR/OF 1 ∆𝑥𝑥 = 𝑏𝑏ℎ 2 1 = 2 (5)(49) ✓ = 122.5 𝑚𝑚✓ (2) 2.3.2 POSITIVE MARKING FROM 2.2 / POSITIEWE NASIEN VANAF 2.2 Displacement = area underneath the graph between graph and t-axis. Verplasing = area onder grafiek OR/OF 1 ∆𝑥𝑥 = 𝑏𝑏ℎ 2 1 = 2 (2.5)(24.5) ✓ =30.63 𝑚𝑚✓ (2) 2.4 Y(m) 122,5 30,63 0 5 10 12,5 t(s) Criteria/ Marks/Punte Correct shape from 0 s to 10 s.  Correct shape from 10 s to 12,5 s.  The position of the projectile at 5 s.  The time the projectile reaches the projection point.  Position of the projectile at 12,5 s.  (5 [13] 6 ### Page 7 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 7 NCDOE/ SEPTEMBER 2017 PCE QUESTION 3 3.1 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 'n resulterende/netto krag op 'n voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel teen 'n versnelling direk eweredig aan die krag en omgekeerd eweredig aan die massa van (2) die voorwerp. 3.2   NB: Do not penalise if arrows  omitted (vectors notation). LW: Moenie penaliseer inden  pyle uitgelaat is nie (vektornotasie) (4) 3.3 Option 1/Opsie 1 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗ OR/OF ∑ 𝐹𝐹⃗ = 𝑚𝑚𝑎𝑎⃗ Block A/Blok A (In x direction/ In x rigting) F gx ( A) + f f ( A) + F BA = ma Anyone/ enigeen Fgx ( A) − f f ( A) − FBA = ma mg sin α − µ A N A − FBA = ma Block A OR B /Blok A OF B (In y direction/ In y rigting) 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = �⃗ 0 OR/OF ∑ 𝐹𝐹⃗ = �⃗ 0 F gy + N = 0 Anyone/ enigeen − Fgy + N = 0 N = Fgy = mgcoα mg sin α − µ Amg cos α − FBA = ma 7 ### Page 8 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 8 NCDOE/ SEPTEMBER 2017 PCE Block B(In x direction/ In x rigting) F gx ( B ) + f f ( B ) + F AB = ma Fgx ( B ) − f f ( B ) + FAB = ma Anyone/ enigeen mg sin α − µ B N B + FAB = ma mg sin α − µ B mg cos α + FAB = ma Solving/ oplossing mg sin α − µ Amg cos α − FBA + mg sin α − µ B mg cos α + FAB = ma + ma 1 a = g sin 45o − ( µ A + µ B ) g cos 450 2 1 a = (9,8) sin 45o  − (0,01 + 1,0)(9,8) cos 45 0  2 a = +  3,43 m ⋅ s −2  OR/OF a = 3,43 m ⋅ s −2 in the direction of motion/in die rigting van beweging Option 2 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗ OR/OF ∑ 𝐹𝐹⃗ = 𝑚𝑚𝑎𝑎⃗ Block A/Blok A (In x direction/ In x rigting) F gx ( A) + f f ( A) + F BA = ma Anyone/ enigeen Fgx ( A) − f f ( A) − FBA = ma mg sin α − µ A N A − FBA = ma Block A OR B /Blok A OF B (In y direction/ In y rigting) 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = �0⃗ OR/OF ∑ 𝐹𝐹⃗ = �0⃗ F gy + N = 0 Anyone/ enigeen − Fgy + N = 0 N = Fgy = mgcoα 8 ### Page 9 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 9 NCDOE/ SEPTEMBER 2017 PCE mg sin α − µ Amg cos α − FBA = ma (0,2)(9,8) sin 45o − (0,01)(0,2)(9,8)co450 − FBA = (0,2)a Anyone/ enigeen 1,386 - 0,01386 - FBA =(0,2)a OR/OF 1,372 - FBA =(0,2)a Block B(In x direction/In x rigting) F gx ( B ) + f f ( B ) + F AB = ma Fgx ( B ) − f f ( B ) + FAB = ma mg sin α − µ B N B + FAB = ma Anyone/ enigeen mg sin α − µ B mg cos α + FAB = ma (0,2)(9,8) sin 45 − (1,0) (0,2)(9,8) cos 450 + FAB = (0,2)a 1,39 -1,39+ FBA=(0,2)a OR/OF 0+ FBA=(0,2)a Anyone/ enigeen Solving/ oplossing 1,372 - FBA + 0+ FBA =(0,2)a+(0,2)a 1,372=(0,4)a a = +  3,43 m ⋅ s −2  OR/OF a = 3,43 m ⋅ s −2 in the direction of motion/in die rigting van beweging Option 3/Opsie 3 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗ OR/OF ∑ 𝐹𝐹⃗ = 𝑚𝑚𝑎𝑎⃗ F gx + f f 1 + f f 2 = (m A + m B )a Fgx − f f 1 − f f 2 = (m A + m B )a Anyone/ enigeen (m A + m B ) g sin α − ( µ A + µ B ) N = (m A + m B )a 9 ### Page 10 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 10 NCDOE/ SEPTEMBER 2017 PCE In y direction/ In y rigting 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = �0⃗ OR/OF ∑ 𝐹𝐹⃗ = �0⃗ F gy + N = 0 Anyone/ enigeen − Fgy + N = 0 N = Fgy = mg cos α m A = mB m A + m B = 2m 2mg sin α − µ A mg cos α − µ B mg cos α = 2ma 1 1 g sin α − µ A mg cos α − µ B mg cos α = a 2 2 Anyone/ enigeen 1 a = g sin 45o − ( µ A + µ B ) g cos 450 2 1 a = (9,8) sin 45o  − (0,01 + 1,0)(9,8) cos 45 0  2 a = +  3,43 m ⋅ s −2  OR/OF (8) a = 3,43 m ⋅ s −2  in the direction of motion/in die rigting van beweging 10 ### Page 11 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 11 NCDOE/ SEPTEMBER 2017 PCE 3.4 POSITIVE MARKING FROM 3.3/POSITIEWE NASIEN VANAF 3.3 Option 1/Opsie 1 mg sin α − µ Amg cos α − FBA = ma  (0,2)(9,8) sin 45o − (0,01) (9,8) cos 45 − FBA = (0,2)(3,43)  FBA= 0,69 N Option 2/Opsie 2 mg sin α − µ B mg cos α + FAB = ma  (0,2)(9,8) sin 45 − (1,0) (0,2)(9,8) cos 450 + FAB = (0,2)(3,43)  FAB=FBA= 0,69 N Option 3/Opsie 3 1 FBA = mg ( µ B − µ A ) cos α  2 1 FBA = (0,2)(9,8)(1,0 − 0,01) cos 45  2 FBA= 0,69 N (3) (3. Decreases 5 Acceleration of the blocks is directly proportional to gravitational acceleration (g) Acceleration due to gravity on the Moon is less that on the Earth Verminder Versnelling van die blokke is direk eweredig aan gravitasieversnelling (g) gravitasieversnelling op die Maan is kleiner as op die Earth (3) [20] 11 ### Page 12 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 12 NCDOE/ SEPTEMBER 2017 PCE QUESTION 4 / VRAAG 4 4.1 Newtons First law✓ / Newton se Eerste wet✓ OR/OF Law of inertia /Wet van Traagheid (1) 4.2 The total linear momentum in an isolated/closed system✓ remains constant (is conserved). Die totale lineere momentum in ‘n geisoleerde/geslote sisteem ✓bly (2) constant. ✓ 4.3 𝑝𝑝⃗𝑇𝑇(𝑏𝑏 𝑏𝑏𝑏𝑏 )/𝑣𝑣 ) = 𝑝𝑝⃗𝑇𝑇(𝑎𝑎 𝑏𝑏𝑏𝑏 𝑏𝑏 )/𝑛𝑛 ) 𝑣𝑣𝑣𝑣𝑣𝑣 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑛𝑛  Anyone/ enigeen (𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠 )�𝑣𝑣⃗𝑠𝑠𝑠𝑠𝑠𝑠 � = 𝑚𝑚𝐽𝐽 𝑣𝑣⃗𝐽𝐽 + 𝑚𝑚𝑃𝑃/𝑇𝑇 𝑣𝑣𝑃𝑃/𝑇𝑇 𝑠𝑠𝑠𝑠 𝑠𝑠 (160)(2)✓ = [(60)(2.5) + 100𝑣𝑣⃗]✓ 𝑣𝑣 = 1.7 𝑚𝑚. 𝑠𝑠 −1  𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑟𝑟𝑟𝑟 ℎ𝑡𝑡 /𝑛𝑛 𝑟𝑟 ��⃗ 𝑟𝑟 𝑛𝑛 𝑟𝑟𝑟𝑟𝑟𝑟 ✓ (5) [8] NB: Do not penalise if vector notation is not used. LET WEL: Moenie penaliseer as vectornotasie nie gebruik is nie. Notes/Aantekeninge: The mark for 'closed/isolated system' is only awarded if used in conjunction momentum. Die punt vir 'geslote/geïsoleerde sisteem' word slegs toegeken indien saam met momentum gebruik QUESTION 5 / VRAAG 5 5.1 The total mechanical energy remains constant/is conserved ✓in an isolated/closed system. ✓ Die totale meganiese energie bly konstant/bly behoue✓ in 'n geïsoleerde/ geslote sisteem✓ OR/OF The sum of the potential and kinetic energy remains constant ✓in an isolated/closed system.✓ Die som van die potensiële en kinetiese energie bly konstant✓in 'n geïsoleerde / geslote sisteem ✓ OR/OF 12 ### Page 13 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 13 NCDOE/ SEPTEMBER 2017 PCE When the work done by the non-conservatives forces is zero the total mechanical energy of the system of bodies is conserved. Wanneer die werk wat gedoen is deur die nie-konserwatiewe kragte nul is, is die totale meganiese energie van die stelsel van liggame behoue. (2) Notes/Aantekeninge: The mark for 'closed/isolated system' is only awarded if used in conjunction with enery Die punt vir 'geslote/geïsoleerde sisteem' word slegs toegeken indien saam met energie gebruik 5.2 𝐾𝐾𝑖𝑖 + 𝑈𝑈𝑖𝑖 = 𝐾𝐾𝑓𝑓 + 𝑈𝑈𝑓𝑓 ✓ 1 1 Notes/ Aantekeninge 𝑚𝑚𝑣𝑣 2 + 𝑚𝑚 ℎ = 𝑚𝑚𝑣𝑣 2 + 𝑚𝑚 ℎ Accept/ Aanvaar 2 2 𝑚𝑚 𝑚𝑚 Ep and EK 1 0 + (40)(9.8)(1.5)✓ = (40)𝑣𝑣 2 + 0✓ 2 𝑣𝑣 = 5.42 𝑚𝑚 · 𝑠𝑠 −1 ✓ (4) 5.3 Option 1/Opsie 1 Accept/ Aanvaar 𝑤𝑤𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐾𝐾✓ 1 1 ∆y / ∆x 𝑚𝑚 ∆𝑥𝑥𝑥𝑥 𝑥𝑥 + 𝑓𝑓∆𝑥𝑥𝑥𝑥 𝑥𝑥 = 𝑚𝑚𝑣𝑣𝑓𝑓2 − 𝑚𝑚𝑣𝑣𝑖𝑖2 2 2 𝑚𝑚 𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥 1 (22)(9.8)(3) cos 600 ✓ + (1.9)(3)cos1800 ✓ = (22)(𝑣𝑣𝑖𝑖2 − 00 )✓ 2 𝑣𝑣𝑓𝑓 = 5,37𝑚𝑚 · 𝑠𝑠 −1 ✓ Option 2/Opsie 2 Accept/ Aanvaar 13 ### Page 14 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 14 NCDOE/ SEPTEMBER 2017 PCE 𝑤𝑤𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐾𝐾✓ ∆y / ∆x 1 1 𝑚𝑚 ∆𝑥𝑥𝑥𝑥 𝑥𝑥 + 𝑓𝑓∆𝑥𝑥𝑥𝑥 𝑥𝑥 = 𝑚𝑚𝑣𝑣𝑓𝑓2 − 𝑚𝑚𝑣𝑣𝑖𝑖2 2 2 𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 1 (22)(9.8)𝑠𝑠 300 (3) cos 00 ✓ + (1.9)(3)cos1800 ✓ = (22)(𝑣𝑣𝑖𝑖2 − 00 )✓ 2 𝑠𝑠𝑠𝑠 𝑣𝑣𝑓𝑓 = 5,37𝑚𝑚 · 𝑠𝑠 −1 ✓ Option 3/Opsie 3 Accept/ Aanvaar 𝑤𝑤𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐾𝐾✓ h / ∆y / 1 1 ∆x 𝑚𝑚 ℎ + 𝑓𝑓∆𝑥𝑥𝑥𝑥 𝑥𝑥 = 𝑚𝑚𝑣𝑣𝑓𝑓2 − 𝑚𝑚𝑣𝑣𝑖𝑖2 2 2 𝑥𝑥𝑥𝑥 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑚𝑚 1 (22)(9.8)(1.5) cos 00 ✓ + (1.9)(3)cos1800 ✓ = (22)(𝑣𝑣𝑖𝑖2 − 00 )✓ 2 𝑣𝑣𝑓𝑓 = 5,37𝑚𝑚 · 𝑠𝑠 −1 ✓ 𝑤𝑤𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐾𝐾✓ Accept/ Aanvaar 1 1 −∆𝑈𝑈 + 𝑓𝑓∆𝑥𝑥𝑥𝑥 𝑥𝑥 = 𝑚𝑚𝑣𝑣 2 − 𝑚𝑚𝑣𝑣 2 2 2 h / ∆y / 𝑥𝑥𝑥𝑥 ∆x 1 −(0 − (22)(9,8)(1.5))✓ + (1.9)(3)cos1800 ✓ = (22)(𝑣𝑣𝑖𝑖2 − 00 )✓ 2 𝑣𝑣𝑓𝑓 = 5,37𝑚𝑚 · 𝑠𝑠 −1 ✓ (5) 5.4 Equal to / Gelyk aan ✓ (1) [12] QUESTION 6 / VRAAG 6 6.1 Doppler Effect ✓/ Doppler Effek ✓ Doppler Effect is 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. Doppler Effek is 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. (3) 14 ### Page 15 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 15 NCDOE/ SEPTEMBER 2017 PCE OR/OF The change in the observe frequency when there is relative motion between the source and the observer. Die verandering in die waargenomefrekwensie as daar relatiewe beweging tussen die klankbron en die luisteraar. 6.2 𝑣𝑣±𝑣𝑣 𝑣𝑣 𝑓𝑓𝑙𝑙 = 𝑣𝑣±𝑣𝑣𝐿𝐿 𝑓𝑓𝑠𝑠 ✓ OR/OF 𝑓𝑓𝑙𝑙 = 𝑣𝑣±𝑣𝑣 𝑓𝑓𝑠𝑠 𝑠𝑠 𝑠𝑠 340 𝑓𝑓𝐿𝐿 = (420)✓ (340 − 16) 𝑓𝑓𝐿𝐿 = 440.74 𝐻𝐻𝐻𝐻✓ (5) 6.3.1 Smaller than ✓/ Kleiner as ✓ (1) 6.3.2 Increases ✓/ Toeneem✓ (1) 6.4 Sun and stars emit light. When a star moves away from the Earth, its spectrum shifts to longer wavelength ✓ (lower frequency) in other words, the red shift ✓ [ NB - Red has a longer wavelength then blue). The stars then appears red. The Universe is expanding  Sun en sterre straal lig uit. As ‘n ster weg van die aarde beweeg, verskuif sy spectrum na ‘n langer golflengte ✓ (laer frekwensie) in ander woorde, rooi verskuiwing ✓ (NB Rooi het ‘n langer golflengte as blou). Die ster kom dan voor as rooi. Die heelal brei uit (3) [13] QUESTION 7 / VRAAG 7 7.1 + Q1 - Q2 15 ### Page 16 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 16 NCDOE/ SEPTEMBER 2017 PCE Criteria / Kriteria Marks / Punte Shape of field line between charges 1 Vorm van veldlyne tussen ladings Shape of field lines outside charges 1 Vorm van veldlyne buite ladings Direction of field lines 1 Rigting van die veldlyne (3) 7.2 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. OR/OF The electrostatic force between two point charges is directly proportional to product of the charges ✓and inversely proportional to the square of the distance between them. ✓ Die elektrostatiese krag tussen twee punt ladings is direk eweredig aan die produk van die ladings✓ en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. ✓ (2) 7.3 Option 1/Opsie 1 𝑘𝑘𝑄𝑄1 𝑄𝑄2 𝐹𝐹 = 𝑟𝑟 2 ✓ (9𝑥𝑥109 )(5𝑥𝑥10−6 )(5𝑥𝑥10−6 )✓ 𝐹𝐹 = (10 × 10−2 )2 ✓ 𝐹𝐹 = 22.5 𝑁𝑁 ✓ Option 2/Opsie 2 𝑘𝑘𝑄𝑄1 𝑄𝑄2 𝑘𝑘(𝑄𝑄)2 𝐹𝐹 = 2 ✓ OR/OF 𝐹𝐹 = 2 𝑟𝑟 𝑟𝑟 (9𝑥𝑥109 )(5𝑥𝑥10−6 )2 ✓ 𝐹𝐹 = (10 × 10−2 )2 ✓ 𝐹𝐹 = 22.5 𝑁𝑁 ✓ (4) 16 ### Page 17 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 17 NCDOE/ SEPTEMBER 2017 PCE 7.4 Same magnitude ✓but opposite in direction✓ Dieselfde grootte maar teenoorgestelde rigting (2) 7.5 F will become 9 times smaller✓ F sal 9 keer kleiner word. (1) 7.6 𝐸𝐸�⃗𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 = 𝐸𝐸�⃗1 + 𝐸𝐸�⃗2 ✓ OR 𝐸𝐸𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 = 𝐸𝐸1 + 𝐸𝐸2 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝐸𝐸𝑛𝑛𝑛𝑛𝑛𝑛 = + ✓ (Mark for/Punt vir ) 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝑟𝑟 2 𝑟𝑟 2 𝑟𝑟 2 (9𝑥𝑥109 )(5𝑥𝑥10−6 ) (9𝑥𝑥109 )(5𝑥𝑥10−6 ) 𝐸𝐸𝑛𝑛𝑛𝑛𝑛𝑛 = ✓ + ✓ (2 × 10−2 )2 (8 × 10−2 )2 ���⃗ 𝐸𝐸𝑛𝑛𝑛𝑛𝑛𝑛 = 2,81 × 108 N ✓towards charge 𝑄𝑄2  (6) [18] NB: Do not penalise if vector notation is not used. LW: Moenie penaliseer as vektornotasie nie gebruik is nie. QUESTION 8 / VRAAG 8 8.1.1 Criteria for hypothesis /Kriteria vir hipotese: Mark/Punt The dependent and independent variables are stated. ✓ Die afhanklike en onafhanklike veranderlikes is genoem. Makes a statement about the relationship between dependent and independent variables. ✓ Maak ’n stelling oor die verwantskap tussen die afhanklike en onafhanklike veranderlikes. Notes/ Independent variable/Onafhanklike veranderlike: Aantekeninge: Amount of potato cells / Aantal aartappelselle A statement that does not contain a relationship: Max ½ Dependent variable/Afhanklike veranderlike: ’n Stelling wat geen verwantskap Bulb glows or not /Gloeilampie gloei of nie bevat nie: Maks. ½ Example/ Voorbeeld: OR/OF Three potato cells are needed to make penlight bulb glow. ½ Current (strength) /Stroom(sterkte) Drie aartappelselle is nodig om die gloei-lampie te laat gloei. ½ 17 ### Page 18 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 18 NCDOE/ SEPTEMBER 2017 PCE Examples/Voorbeelde: More potato cells will increase the brightness of the bulb/ Meer aartappelselle sal die helderheid van die gloeilampie laat toeneem. More potato cells will increase the current through the bulb/ Meer aartappelselle sal die grootte van die stroom deur die gloeilampie laat toeneem. (2) 8.1.2 Dependent variable/Afhanklike veranderlike: Bulb glows or not /Gloeilampie gloei of nie  OR/OF Current (strength) /Stroom(sterkte)  (1) 8.2 1,6 V ✓ (1) 8.3 The battery has an internal resistance /  Die battery het ’n interne weerstand ✓ (1) 8.4 Option 1/Opsie 1 𝑉𝑉 2 𝑃𝑃 = 𝑅𝑅 ✓ (0.02)2 𝑃𝑃 = 2 𝑃𝑃 = 2 × 10−4 𝑊𝑊 ✓ Option 2/Opsie 2 𝑉𝑉 𝐼𝐼 = 𝑅𝑅 0.02 𝐼𝐼 = 2 𝐼𝐼 = 0.01 𝐴𝐴 𝑃𝑃 = 𝐼𝐼  𝐼𝐼 𝑃𝑃 = 0.01 × 0.02 𝑃𝑃 = 2 × 10−4 𝑊𝑊  Option 3/Opsie 3 𝑉𝑉 𝐼𝐼 = 𝑅𝑅 0.02 𝐼𝐼 = 2 𝐼𝐼 = 0.01 𝐴𝐴 𝑃𝑃 = 𝐼𝐼 2 𝑅𝑅 𝑃𝑃 = (0.01)2 (2) 𝑃𝑃 = 2 × 10−4 𝑊𝑊  (3) [8] 18 ### Page 19 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 19 NCDOE/ SEPTEMBER 2017 PCE QUESTION 9 / VRAAG 9 9.1 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. ✓ Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur. (2) 9.2 𝑉𝑉 𝑅𝑅 = 𝐼𝐼 ✓ 12.9 𝑅𝑅 = 1.5 ✓ 𝑅𝑅 = 8.6 Ω ✓ (3) 9.3 POSITIVE MARKING FROM QUESTION 9.2 POSITIEWE NASIEN VAN VRAAG 9.2 1 1 1 𝑅𝑅𝑇𝑇 = 𝑅𝑅 + 𝑅𝑅 ✓ 1 𝑠𝑠 1 1 1 𝑅𝑅𝑇𝑇 = 8.6 + (4+5) ✓ 1 88 = 𝑅𝑅𝑇𝑇 387 𝑅𝑅𝑇𝑇 = 4.4 Ω ✓ (3) 9.4 POSITIVE MARKING FROM QUESTION 9.3 POSITIEWE NASIEN VAN VRAAG 9.3 Option 1/Opsie1 Option 2/Opsie 2 𝑉𝑉 𝑉𝑉 𝑅𝑅 = 𝐼𝐼 ✓ 𝑅𝑅 = 𝐼𝐼 ✓ 12.9 12.9 4.4 = 𝐼𝐼 ✓ 4.4 = 𝐼𝐼 ✓ 𝐼𝐼 = 2,93 𝐴𝐴 𝐼𝐼 = 2,93 𝐴𝐴 19 ### Page 20 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 20 NCDOE/ SEPTEMBER 2017 PCE 𝑉𝑉 𝜀𝜀 = 𝐼𝐼 (𝑅𝑅 + 𝑟𝑟 ) 𝑟𝑟 = 𝐼𝐼 15 = 2.93 (4.4 + 𝑟𝑟) ✓ 15−12.9 𝑟𝑟 = 2.93 ✓ 𝑟𝑟 = 0,72 Ω ✓ 𝑟𝑟 = 0,72 Ω ✓ Option 3/Opsie 3 Option 4/Opsie 4 𝑉𝑉 𝑉𝑉 𝑅𝑅 = 𝐼𝐼 ✓ 𝑅𝑅 = 𝐼𝐼 ✓ 12.9 12.9 9= 𝐼𝐼 ✓ 9= 𝐼𝐼 ✓ 𝐼𝐼 = 1.43 𝐴𝐴 𝐼𝐼 = 1.43 𝐴𝐴 𝐼𝐼 = 1.5 + 1.43 𝐼𝐼 = 1.5 + 1.43 𝐼𝐼 = 2.93 𝐴𝐴 𝐼𝐼 = 2.93 𝐴𝐴 𝑉𝑉 𝑟𝑟 = 𝐼𝐼 𝜀𝜀 = 𝐼𝐼 (𝑅𝑅 + 𝑟𝑟 ) 15−12.9 𝑟𝑟 = ✓ 15 = 2.93 (4.4 + 𝑟𝑟) ✓ 2.93 𝑟𝑟 = 0,72 Ω ✓ 𝑟𝑟 = 0,72 Ω ✓ (4) [12] QUESTIONS 10 / VRAAG 10 10.1 Direct Current Generator ✓/ Gelykstroom-generator ✓ OR/OF DC generator/ GS generator (1) 10.2 Alternating Current Generator ✓ / Wisselstroom-generator ✓ OR/OF AC generator/WS generator (1) 10.3 DC (Direct Current) generator – has a splitring commutator ✓ GS (Gelykstroom) generator – bevat ‘n splitring – kommutator ✓ AC (Alternating Current) generator – has slip rings ✓ WS (Wisselstroom) generator – bevat sleepringe ✓ (2) 𝑉𝑉 10.4 𝑉𝑉 𝑟𝑟 = 𝑚𝑚𝑚𝑚𝑚𝑚 ✓ 𝑟𝑟 𝑟𝑟 √2 20 ### Page 21 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 21 NCDOE/ SEPTEMBER 2017 PCE 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 220 = ✓ √2 𝑉𝑉 𝑚𝑚𝑚𝑚 𝑚𝑚 = 311.13 𝑉𝑉✓ (3) 10.5 Vmax 311.13 Potential Difference (V) Vrms 220 0 Potensiaalverskil (V) Time (s) Vrms - 220 Vmax -311.13 Criteria Marks Shape / Vorm ✓ One cycle / Een siklus ✓ Both rms – values / Beide wgk-waardes ✓ Both peak – values / Beide piek-waardes ✓ (4) 10.6 ANY TWO CORRECT APLICATIONS  • Car alternator • Bicycle dynamo • Hybrid electric vehicle (HEV) drive systems, • Aircraft auxiliary power generation, • Wind generators, • High speed gas turbine generators ENIGE TWEE KORREKTE TOEPASSINGS • Motor alternator / • Fiets dinamo • Hibriede elektriese voertuig (HEV) dryfstelsels, 21 ### Page 22 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 22 NCDOE/ SEPTEMBER 2017 PCE • Vliegtuig hulp kragopwekkers, • Windgenerators, • Hoë spoed gas turbine kragopwekkers.  (2) [13] QUESTION 11 / VRAAG 11 11.1 Photoelectric effect ✓ Fotoëlektriese effek ✓ (1) 11.2 What effect will light of different frequencies✓ have on the emission of electrons from a metal. ✓ Wat is die effek van verskillende frekwensie✓ van lig op die vrystelling van elektrone uit ‘n metaal.✓ (2) 11.3.1 light of different frequencies ✓ Lig van verskillende frekwensies ✓ (1) 11.3.2 emission of electrons✓ Vrystelling van elektrone✓ (1) 11.3.3 Type of metal/work function /threshold (cut off) frequency (1) 11.4.1 work function is the minimum energy that an electron in the metal needs to be emitted from the metal surface.  Werkfunksie is die minimum energie benodig om 'n elektron uit die oppervlak van 'n metaal vry te stel.  OR/OF Minimum energy needed to eject electrons from the surface of a certain metal. ✓✓ Minimum energie benodig om elektrone uit ‘n metaaloppervlak vry te stel. ✓✓ (2) 1 11.4.2 𝐸𝐸 = 𝑊𝑊 + 2 𝑚𝑚𝑣𝑣 2 ✓ 𝑜𝑜 ℎ𝑐𝑐 𝜆𝜆 = 𝑊𝑊 + 𝐸𝐸𝑘𝑘 𝑜𝑜 22 ### Page 23 Downloaded from hlayiso.com Physical Sciences/ Gr 12/Memorandum/P1 23 NCDOE/ SEPTEMBER 2017 PCE (6,63𝑥𝑥10−34 )(3𝑥𝑥108 )✓ = 1,8 × 10−19 + 𝐸𝐸𝑘𝑘 ✓ 7.8 ×× 10−7 ✓ 𝐸𝐸𝑘𝑘 = 7.5 × 10−20 𝐽𝐽✓ (5) [13] TOTAL/TOTAAL: 150 23 ## Related documents - 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