# Physical Sciences Grade 12 P1 Eastern Cape Mock 2023 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape Mock Exam 2023 (English), 16 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/phys-sciences-p1-memo-gr12-sept-2023-afr-english/cb1e457f-61c5-401d-ad52-b79ffaff07b4
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2023
- Exam period: Eastern Cape Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 16

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## About this document

Physical Sciences · Grade 12 · Eastern Cape Mock Exam · 2023 · English. Memorandum, 16 pages. Read online or download the PDF.

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### Page 1

NATIONAL
                     SENIOR CERTIFICATE/
                         NASIONALE
                     SENIORSERTIFIKAAT

                          GRADE/GRAAD 12

                       SEPTEMBER 2023

           PHYSICAL SCIENCES P1/
          FISIESE WETENSKAPPE V1
       MARKING GUIDELINE/NASIENRIGLYN

MARKS/PUNTE:   150

                This marking guideline consists of 16 pages./
                 Hierdie nasienriglyn bestaan uit 16 bladsye.

### Page 2

2               PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                (EC/SEPTEMBER 2023)

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.

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(EC/SEPTEMBER 2023)         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.

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4               PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                   (EC/SEPTEMBER 2023)

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.

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(EC/SEPTEMBER 2023)         PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                     5

QUESTION 1/VRAAG 1:           MULTIPLE-CHOICE QUESTIONS
                              MEERVOUDIGEKEUSE-VRAE

1.1     D                                                                               (2)

1.2     C                                                                               (2)

1.3     B                                                                               (2)

1.4     C                                                                               (2)

1.5     C                                                                               (2)

1.6     A                                                                               (2)

1.7     C                                                                               (2)

1.8     D                                                                               (2)

1.9     A                                                                               (2)

1.10    C                                                                               (2)
                                                                                         [20]

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6               PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                (EC/SEPTEMBER 2023)

QUESTION 2/VRAAG 2

2.1      Newton’s third law 
         Newton se derde wet                                                                (1)

2.2

         Notes
         Mark awarded for arrow and label.
         Do not penalise for length of arrows since drawing is not drawn to scale,
         Any other additional force(s). Max. 4⁄5
         If force(s) do not make contact with body. Max. 4⁄5
         Aantekeninge
         Punt word toegeken vir byskrif en pyltjie.
         Moenie vir die lengte van die pyltjie penaliseer nie
         Enige addisionele krag(te). Maks. 4⁄5
         As krag(te) nie kontak maak met liggaam nie. Maks. 4⁄5                             (5)

2.3.1    Fnet = ma          Any one / Enige een 
         F – T – f = ma
         9 000 – T – 3 500 = 0 
         T = 5 500 N                                                                       (3)

         POSITIVE MARKING FROM / POSITIEWE NASIEN VANAF 2.3.1
2.3.2    Fnet = ma
         T – f = ma       Any one / Enige een 
         T – µkN = ma
         T – µkmg = ma
         5 500 – µk x 950 x 9,8 = 0 
         ∴µk = 0,59 (0,591)                                                                (4)

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(EC/SEPTEMBER 2023)         PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                     7

2.4        Fnet = ma
           -f = ma           Any one / Enige een 
           -(µkN) = ma
           -(µkmg) = ma
           -(0,59 x 950 x 9,8) = 950a 
           ∴a = -5,78 m.s-2
           ∴a = 5,78 m.s-2 (to the right / na regs)                                                   (3)
                                                                                                      [16]

QUESTION 3/VRAAG 4

3.1       9,8 m.s-2                                                                                    (1)

3.2.1     UPWARD POSITIVE/OPWAARTS AS                   UPWARD NEGATIVE/OPWAARTS AS
          POSITIEF                                      NEGATIEF
          Δy = viΔt + ½ aΔt  2                         Δy = viΔt + ½ aΔt2 
                                             2
          -2 = vi x 0,125 + ½ x (-9,8)(0,125)          2 = vi x 0,125 + ½ (9,8)(0,125) Any one/
                           -1                Any one/                   -1               Enige een 
          vi = -15,388 m.s                   Enige een vi = 15,388 m.s
            2     2
          vf = vi + 2aΔy                                  2    2
                                                        vf = vi + 2aΔy
                    2   2
          -15,388 = 0 + 2(-9,8) Δy                     15,3882 = 02 + 2(9,8) Δy 
          ∴Δy = 12,08 m                                 ∴Δy = 12,08 m
          Height / Hoogte = 12,08 + 2                  Height / Hoogte = 12,08 + 2 
          Height / Hoogte = 14,08 m                    Height / Hoogte = 14,08 m             (5)

3.2.2     UPWARD POSITIVE/OPWAARTS AS                   UPWARD NEGATIVE/OPWAARTS AS
          POSITIEF                                      NEGATIEF
          vf = vi + aΔt                                vf = vi + aΔt 
          -15,388 = 0 + (-9,8)(Δt)                     15,388 = 0 + (9,8)(Δt) 
          ∴Δt = 1,57 s                                  ∴Δt = 1,57 s
          Δt = 1,57 + 0,125                            Δt = 1,57 + 0,125 
          Δt = 1,70 s                                  Δt = 1,70 s                                    (4)

                                          OPTION 1/OPSIE 1
3.2.3     UPWARD POSITIVE/OPWAARTS AS              UPWARD NEGATIVE/OPWAARTS AS
          POSITIEF                                 NEGATIEF
            2    2
          vf = vi + 2aΔy                          vf2 = vi2 + 2aΔy 
            2          2
          vf = -15,388 + 2(-9,8)(-2)              vf2 = 15,3882 + 2(9,8)(2) 
          ∴vf = 16,61 m.s-1 downwards / afwaarts  ∴vf = 16,61 m.s-1 downwards/afwaarts 

                                         OPTION 2/OPSIE 2
          UPWARD POSITIVE/OPWAARTS AS             UPWARD NEGATIVE/OPWAARTS AS
          POSITIEF                                NEGATIEF
          vf = vi + aΔt                          vf = vi + aΔt 
          vf = -15,388 + (-9,8)(0,125)           vf = 15,388 + (9,8)(0,125) 
          vf = - 16,61                            ∴vf = 16,61 m.s-1 downwards/afwaarts 
                          -1
          ∴vf = 16,61 m.s downwards / afwaarts 

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8               PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1               (EC/SEPTEMBER 2023)

                                    OPTION 3/OPSIE 3
    UPWARD POSITIVE/OPWAARTS AS              UPWARD NEGATIVE/OPWAARTS AS
    POSITIEF                                 NEGATIEF
          v + vi                                   v + vi
    Δx = f       Δt                         Δx = f       Δt 
             2                                       2
          v + (−15,388)                           v + 15,388
    -2 = ( f            ) x 0,125           2= ( f           ) x 0,125 
                 2                                     2
    ∴vf = 16,61 m.s-1 downwards /afwaarts  ∴vf = 16,61 m.s-1 downwards /afwaarts              (3)

    UPWARD POSITIVE/OPWAARTS AS               UPWARD NEGATIVE/OPWAARTS AS
    POSITIEF                                  NEGATIEF

            CRITERIA FOR MARKING / NASIENKRITERIA
            Correct shape / Korrekte vorm                                       
            Initial position indicated / Aanvanklike posisie aangedui           
            Time when ball hits the ground / Tyd wanneer bal die grond tref                     (3)
                                                                                                [16]

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(EC/SEPTEMBER 2023)         PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                            9

QUESTION 4/VRAAG 4

4.1   A system on which the net external force is zero. 
      ʼn Sisteem waar die netto eksterne krag nul is.                                                (2)

4.2   ∑pi = ∑pf
                                        Any one / Enige een
      mAvi + mBvi = mAvf + mBvf         
      (mA + mB)vi = mAvf + mBvf
      (2 + 1) x 0 = 1 x 5 + 2vf 
      vf = 2,5 m.s-1 left / links (west / wes)                                                     (4)

      OPTION 1 / OPSIE 1                            OPTION 2 / OPSIE 2
4.3   Fnet∆t = ∆p                                   Fnet = ma
      Fnet∆t = m∆v            Any one / Enige een   -80 = 1 x a 
      Fnet∆t = mvf – mvi                           a = -80 m.s-2
      -80 x 0,05  = 1 x vf – 1 x 5                vf = vi + a∆t 
      vf = 1 m.s-1 left / links (west / wes)       vf = 5 + (-80) x 0,05 
                                                    vf = 1 m.s-1 left / links (west / wes)         (4)

4.4   Collision in which kinetic energy is not conserved. 
      Collision where total kinetic energy before the collision is not equal to the total
      kinetic energy after the collision. 
      Botsing waarin die kinetiese energie nie behoue is nie.
      Botsing waar die totale kinetiese energie voor die botsing nie gelyk is aan die
      totale kinetiese energie na die botsing nie.                                                 (2)
                                                                                                  [12]

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10              PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                (EC/SEPTEMBER 2023)

QUESTION 5/VRAAG 5

5.1    The work done on an object by a net force is equal to the change in kinetic
       energy of the object. 
       Die arbeid verrig op ʼn voorwerp deur ʼn netto krag is gelyk aan die
       verandering in die voorwerp se kinetiese energie.                                   (2)

       OPTION 1 / OPSIE 1
5.2    Wnet = ∆Ek
       WFg + Wf = ∆Ek
       -∆Ep + Wf = ∆Ek                               Any one / Enige een 
       [-mg(h2 – h1)] + f.∆xcos θ = ½mvf2 - ½ mvi2

       - 50 x 9,8 (0 – 5) + 72∆xcos180o  = ½ x 50 x 82 – 0 

       ∆x = 11,81 m
                    5
       θ = sin-1(       )
                  11,81
       θ = 25,05° 

       OPTION 2 / OPSIE 2
       Wnc = ∆Ep + ∆Ek
       Wnc = [mg(h2 – h1)] + ½mvf2 – ½mvi2      Any one / Enige een 
       Wf = [mg(h2 – h1)] + ½ mvf2 – ½ mvi2

       72∆xcos180° = 50 x 9,8 (0 – 5) + ½ x 50 x 82 – 0 

       ∆x = 11,81 m
                    5
       θ = sin-1           
                   11,81
       θ = 25,05° 

       OPTION 3 / OPSIE 3
       Wnet = ∆Ek                                Any one / Enige een 
                                 2         2
       (Fg - Ff).Δx.cos θ = ½ mvf – ½ mvi
       (mg sin θ – 72).Δx.(1) = ½ (50)(82 – 02) 
        50(9,8)(5) x
                      – 72Δx = 1 600
             x

       Δx = 11,81 m
                     5
       θ = sin-1           
                   11,81
       θ = 25,05°                                                                         (6)

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(EC/SEPTEMBER 2023)         PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                    11

5.3

                                                                                           (3)

5.4        OPTION 1 / OPSIE 1
           Wnet = ∆Ek
           Wf = ∆Ek                Any one / Enige een 
           Wf = ½mvf2 – ½mvi2
           Wf = 0  – ½ x 50 x 82 
           Wf = 1 600 J 

           OPTION 2 / OPSIE 2
           Wnc = ∆Ep + ∆Ek
                             Any one / Enige een 
           Wf = ∆Ep + ∆Ek
           Wf= 0  – ½ x 50 x 82 
           Wf = 1 600 J 

           OPTION 3 / OPSIE 3
           vf2 = vf2 + 2aΔx
           02 = 82 + 2a(10) 
           ∴ a = -3,2 m·s-1

           Fnett = ma                   (Any ONE / Enige EEN )
                 = 50 x (-3,2) 
                 = -160 N (= Ff)

           Wf = Ff . Δx . cos θ
              = (160)(10)(cos 180°)
              = 1 600 J                                                                   (4)
                                                                                          [15]

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12                PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                    (EC/SEPTEMBER 2023)

QUESTION 6/VRAAG 6

6.1     The (apparent) change in frequency observed by a listener because the
        listener and source of sound have different velocities relative to the medium
        of sound propagation.  
        Die verandering in frekwensie van die klank waargeneem deur ʼn luisteraar
        omdat die klankbron en luisteraar verskillende snelhede relatief tot die
        medium waarin die klank voortgeplant word, het.

        The (apparent) change in frequency observed by a listener due to relative
        motion between the sound source and the listener.  

        Die verandering in die (waargenome) frekwensie waargeneem deur ʼn
        luisteraar omdat daar relatiewe beweging is tussen die luisteraar en die
        klankbron.                                                                                (2)

6.2     Towards observer B.  Frequency detected by observer B is higher than the
        frequency detected by observer A. 

        Na waarnemer B. Die frekwensie wat waargeneem word deur luisteraar B
        is hoër as die waargenome frekwensie deur luisteraar A.                                   (2)

6.3            v ± vL
        fL =          fs 
               v ± vs
                       343
        545  =                  x fs ……….(1)
                     343 + vs
                      343
        615  =                  x fs ……….(2)
                     343 – vs
        vs = 20,70 m.s-1 (20,69827586 m.s-1)

               545(343 + 27,70)                  615(343 – 27,70)
        fs =                       OR/OF fs =
                     343                               343
        fs = 577,91 Hz  (range / gebied 577,89 Hz to 577,91 Hz)                                  (7)

6.4     The star is moving away. 
        The spectral lines show a decrease in frequency (towards red). 

                                                  OR/OF

        The spectral lines show an increase in wavelength (towards red). 

        Die ster beweeg weg.
        Die spektralelyne toon ʼn afname in frekwensie (na rooi).

                                                  OR/OF

        Die spektralelyne toon ʼn toename in golflengte (na rooi).                                (2)
                                                                                                [13]

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### Page 13

(EC/SEPTEMBER 2023)         PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                           13

 QUESTION 7/VRAAG 7

 7.1     The electrostatic force exerted by one point charge on another is directly
         proportional to the product of the charges and inversely proportional to the
         square of the distance between them. 

         Die elektrostatiese krag wat een puntlading op ʼn ander puntlading uitoefen
         is direk eweredig aan die produk van die groottes van die ladings en
         omgekeerd eweredig aan die kwadraat van die afstand tussen hulle.                        (2)

 7.2
                                            Marking criteria / Nasienkriteria
                                         Direction / Rigting
                                         Straight lines / Reguitlyne
                                         Touching and vertically from charge./
                                          Raak en is loodreg vanaf die lading
                                                                                                  (3)

 7.3          kQ1Q2
         F=         
                r2
         F = (9 x 109)(36 x 10-6)(16 x 10-6) 
                       (0,03)2
         F = 5 760 N                                                                             (3)

 7.4     Negative.  For the test charge to experience a net force of zero, the electric
         field due to P and T should be in opposite (directions).  The electric field of
         P is to the right, for the electric field of T to move to the left, (away) from point
         X, it must be a negative charge. 

         Negatief. Vir die puntlading om ʼn netto krag van nul te ervaar moet die
         elektrieseveld a.g.v. P en T in teenoorgestelde (rigtings) wees. Die
         elektrieseveld van P is regs sodat die elektrieseveld van T na links beweeg
         (weg) vanaf punt X. so dit moet ʼn negatiewe lading wees.                                 (3)

 7.5         kQ
         E=       
              r2
         E1 = (9 x 109)(36 x 10-6)  right / regs (Note distance to be correct / Let wel
                   (0,03 + r)2                    die afstand moet korrek wees)
                                         (Any 1 of Top substitutions / Enige van die
                                          vervangings)
         E2 = (9 x 109)(16 x 10-6)  left/links
                       r2

         Enet = E1 + E2
         0 = (9 x 109)(36 x 10-6) + (-(0,03 + r)2)  (Correct addition / Korrekte
                 (0,03 + r)2              r2          addisie)

         r = 0,06 m                                                                              (5)
                                                                                                 [16]

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### Page 14

14                    PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                 (EC/SEPTEMBER 2023)

QUESTION 8/VRAAG 8

8.1     The total energy per coulomb of charge that a battery can supply. 
        Die totale energie per coulomb lading wat ʼn battery kan verskaf.                         (2)

8.2.1   P = I2 R 
        16 = 22 x R 
        R=4Ω                                                                                    (3)

8.2.2   Positive marking from / Positiewe nasien vanaf 8.2.1
                OPTION 1 / OPSIE 1                  OPTION 2 / OPSIE 2
           V                                     12
        R=                               I10 = x 2 
            I                                    10
              V                           I 10 = 2,4 A
        12 = p                           I = I10 + I2
               2                          I = 2,4 + 2 
        V = 24 V                          I = 4,4 A 
                 24
        I10 =         
                 10
        I10 = 2,4 A
        I = I10 + I2
        I = 2,4 + 2 
        I = 4,4 A                                                                               (5)

8.2.3   Positive marking from 8.2.1 and 8.2.2. / Positiewe nasien vanaf 8.2.1
        en 8.2.2
        1        1        1                             R1 R2
             =        +                        Rp =             
        Rp       R1       R2                           R1 + R2
        1        1        1
             =        +                               10 x 12
        Rp       10       12
                                                Rp =             
                                                       10 + 12
        Rp = 5,4545 Ω
                                                Rp = 5,4545 Ω
        Rext = 7,4545 Ω
        Emf / Emk= I(Rext + r) 
        [36 = 4,4 (7,4545 + r)] 
        R = 0,73 Ω                                                                              (6)

8.3     Decrease.  The total external resistance will increase, and the current will
        decrease. 
        Afneem. Die totale eksterne weerstand sal toeneem en die stroom sal
        afneem.                                                                                  (2)
                                                                                                [18]

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### Page 15

(EC/SEPTEMBER 2023)                 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                   15

QUESTION 9/VRAAG 9

9.1      Mechanical energy  to electrical energy. 
         Meganiese energie na elektriese energie.                                                 (2)

9.2

                  EMF/EMK (V)
                                                                               t (s)

         Marking criteria / Nasienkriteria
         Correct shape/ Korrekte vorm                               
         Two complete waves / Twee volledige golwe                                               (2)

9.3      AC can be transmitted over long-distances because it causes minor energy
         loss. 
                                          OR/OF

         The potential difference can be increased or decreased.
         WS kan oor lang afstande met minimale energie verlies oorgedra word.

                                                  OR/OF

         Die potensiaalverskil can verhoog of verlaag word.                                       (1)

9.4      Vrms/ wgk OR Root mean square voltage/wortel gemiddelde kwadraat
         spanning                                                                                 (1)

9.5.1            I
          I rms = max 
                   2

                  6,25
         Irms =                 
                  √2

         Irms = 4,42 A                                                                           (3)

9.5.2    Positive marking from / Positiewe nasien vanaf 9.3.1
         OPTION 1 / OPSIE 1
                                 2
         Paverage / gemiddeld = 𝐼𝑟𝑚𝑠 R
                              = 4,422 x 45 
                              = 879,14 W                                                         (3)
                                                                                                 [12]

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### Page 16

16              PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                   (EC/SEPTEMBER 2023)

QUESTION 10/VRAAG 10

10.1      The process whereby electrons are ejected from a (metal) surface, when
          light of suitable frequency is incident on that surface. 
          Die proses waardeur elektrone uit ʼn (metaal)oppervlakte vrygestel word,
          wanneer lig van geskikte frekwensie invallend op die oppervlak is.                 (2)

10.2      Increasing the power rating of the light bulb increases the intensity of the
          photons.  Increase in intensity does not affect the maximum kinetic
          energy or maximum speed of the photo-electrons. 
          Toename in die drywinggradering van die gloeilampies verhoog die
          intensiteit van die fotone. Toename in die intensiteit affekteer nie die
          maksimum kinetiese energie nie of maksimum spoed van die
          fotoëlektrone nie.                                                                 (2)

10.3      Light B.  It has the higher intensity. More photo-electrons will be ejected
          and therefore more current.  (Ammeter reading will be higher.)

          Lig B. Dit het ʼn hoër intensiteit. Meer foto-elektrone word vrygestel en
          daarom meer stroom. (Ammeterlesing sal hoër wees).                                 (2)

10.4.1        OPTION 1 / OPSIE 1                      OPTION 2 / OPSIE 2
          E = hc                          c = f𝜆
               𝜆                               3x10 8
                                          f=
               6,63x10 -34 x3x10 8             490x10 - 9
          E=                       
                   490x10 - 9             f = 6,12244898 Hz
          E = 4,06 x 10-19 J             E = hf 
                                          E = 6,63 x 10-34 x 6,12244898 x 1014 
                                          E = 4,06 x 10-19 J                                (3)

10.4.2    Positive marking from / Positiewe nasien vanaf 10.4.1.
          E = W0 + Ek(max/ maks.) 
          4,06 x 10-19  = Wo + ½ x 9,11 x 10-31 x (7,5 x 105)2 
          W0 = 1,50 x 10-19 J                                                               (4)
                                                                                            [13]

                                                                  TOTAL / TOTAAL:           150

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