# Physical Sciences Grade 12 Northern Cape Mock 2021 Memo

> Grade 12 Physical Sciences memorandum from the Northern Cape Mock Exam 2021, 26 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/trial-exam-2021-physics-nc-memo/b1008360-918f-48bd-92ee-dca83e598f5f
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2021
- Exam period: Northern Cape Mock Exam
- Curriculum: CAPS
- Pages: 26

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## About this document

Physical Sciences · Grade 12 · Northern Cape Mock Exam · 2021. Memorandum, 26 pages. Read online or download the PDF.

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### Page 1

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Physical Sciences P1/Grade 12                             1                        NC/ September 2021
Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

                                              —————————————————————
                                              DEPARTMENT OF EDUCATION
                                              —————————————————————
                                              DEPARTEMENT VAN ONDERWYS
                                              —————————————————————
                                              LEFAPHA LA THUTO
                                              —————————————————————
                                              ISEBE LEZEMFUNDO
                                              —————————————————————

                  PROVINCIAL PREPARATORY EXAMINATION/
                  PROVINSIALE VOORBEREIDINGSEKSAMEN

                                        GRADE/GRAAD 12

                       PHYSICAL SCIENCES P1: PHYSICS/
                       FISIESE WETENSKAPPE V1: FISIKA
                   MARKING GUIDELINES/NASIENRIGLYNE

                                   SEPTEMBER 2021

MARKS/PUNTE: 150

TIME/TYD: 3 hours/uur

                         This memorandum consist of 26 pages
                      Hierdie memorandum bestaan uit 26 bladsye.

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Physical Sciences P1/Grade 12                             2                        NC/ September 2021
Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

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.

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.
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Physical Sciences P1/Grade 12                             3                        NC/ September 2021
Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

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
                -1         -1       -1                -1
       van 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
       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.

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Physical Sciences P1/Grade 12                             4                          NC/ September 2021
Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

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 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 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.

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Physical Sciences P1/Grade 12                             5                        NC/ September 2021
Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

QUESTION 1/VRAAG 1
1.1    D                                                                                                 (2)
1.2    B                                                                                                 (2)
1.3    A                                                                                                 (2)
1.4    B                                                                                                 (2)
1.5    C                                                                                                 (2)
1.6    D                                                                                                 (2)
1.7    D                                                                                                 (2)
1.8    A                                                                                                 (2)
1.9    B                                                                                                 (2)
1.10   C                                                                                                 (2)
                                                                                                           [20]

QUESTION 2/VRAAG 2
2.1
      NOTE: -1 mark for each key word/phrase omitted in the correct context.
      LET WEL:-1 punt vir elke sleutel word/frase in die korrekte konteks weggelaat.

        A body will remain in its state of rest or motion at constant velocity unless a
        non-zero resultant/net force acts on it. 
       'n Liggaam sal in sy toestand van rus of beweging teen konstante snelheid
       volhard, tensy 'n nie-nul resulterende/netto krag daarop inwerk.

       OR/OF
       A body will remain in its state of motion (at rest or moving at constant velocity)
       until a net force acts on it. 
       'n Liggaam sal in sy toestand van beweging (in rus of beweging teen konstante
       snelheid) volhard, totdat 'n netto krag daarop inwerk.
                                                                                                           (2)
2.2    OPTION 1/OPSIE 1
       Block A/Blok A
       ⃗Fnet =0
                        Any one/Enige een
        𝑓s +T ⃗ =0
        -fs +T=0
        Block B/Blok B
        ⃗Fg +T ⃗ =0
                        Any one/Enige een
         Fg -T=0
         mB g-T=0
         Solving
         -fs +T+mB g-T=0
         fs =mB g
         fs =5×9,8
         𝑓𝑠 = 49,00 𝑁 
         Note/Aantekening: Do not penalise if vectors notation is not used./Moenie                           (4)
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Physical Sciences P1/Grade 12                             6                        NC/ September 2021
Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

        penaliseer indien vektornotasie nie gebruik word nie.
2.3.1
                 ⃗T                               ACCEPT/AANVAAR                    ⃗T

          𝐹                                                                 𝐹
                 𝐹𝑔                                                                 𝐹𝑔

         Accepted labels/Aanvaarde benoemings
         F     Fapp / FA/applied force                                     
               Ftoegepas / FT/toegepaste krag
         Fg    w/Fw /weight/mg/gravitational force
               w/Fw /gewig/mg/gravitasiekrag                               
         T     FT/Tension/Spanning                                         
        Notes/Aantekeninge
             Marks awarded for label and arrow./Punt toegeken vir benoeming en pyltjie.
             Do not penalise for length of arrows since drawing is not to scale./Moenie vir
              die lengte van die pyltjies penaliseer nie aangesien die tekening nie volgens
              skaal is nie.
                                                                                   2
             Any other additional force(s)./Enige ander ekstra krag(te). Max/Maks 3
              If force(s) do not make contact with body./Indien krag(te) nie met die
                                                    2
               voorwerp kontak maak nie: Max/Maks 3
              Do not penalise if vector notation is not used./Moenie penaliseer indien
               vektor notasie nie gebruik is nie.                                       (3)

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Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

2.3.2   OPTION 1/OPSIE 1
        Direction of motion as positive/Rigting van beweging as positief
        Block A/Blok A
        ∑ ⃗F=ma ⃗ OR/OF ⃗Fnet =ma ⃗ OR/OF Fnet =ma
        ⃗T+fk = mA ⃗a OR/OF T+fk = mA a
        T-fk = mA a
                                                                           Any one/Enige een
        T − 46 = 4a  (1)
        Block B/Blok B:
        ⃗ +F
        T  ⃗ g +F
                ⃗ = mB ⃗a OR/OF T+Fg +F= mB a
        -T+Fg(B) +F= mB a
        -T+mB g+F= mB a

        -T+(5)(9,8)+15= 5a        4a OR/OF 5a 
        -T+49+15= 5a
        -T+64 = 5a (2)
        Solving (1) and (2)/Los (1) en (2) op:
        T − 46 − T+(5)(9,8)+15 = 4a+5a
        T − 46 − T+49+15 = 9a
        a= 2,00 m∙s-2  Accept/Aanvaar a= 2 m∙s-2
        OPTION 2/OPSIE 2
        Direction of motion as negative/Rigting van beweging as negatief
        Block A/Blok A
        ∑ ⃗F=ma ⃗ OR/OF ⃗Fnet =ma ⃗ OR/OF Fnet =ma
        ⃗T+fk = mA ⃗a OR/OF T+fk = mA a
        −T+fk = mA a

        −T + 46 = 4a  (1)                                                Any one/Enige een
        Block B/Blok B:
        ⃗T+F
           ⃗ g +F
                ⃗ = mB ⃗a OR/OF T+Fg +F= mB a
        +T − Fg(B) − F= mB a
        +T − mB g − F= mB a

        +T − (5)(9,8) − 15= 5a         4a OR/OF 5a 
        +T − 49 − 15 = 5a
        +T − 64 = 5a (2)
        Solving (1) and (2)/Los (1) en (2) op:
        −T+46+T − (5)(9,8) − 15 = 4a+5a
        −T+46+T − 49 − 15 = 9a
        a=2,00 m∙s-2  Accept/Aanvaar a= 2 m∙s-2
        Note/Aantekening: Do not penalise if vector notation is not used./Moenie
        penaliseer indien vektornotasie nie gebruik is nie.                                                (5)

2.4 - Smaller than /Kleiner as                                                                             (2)
      -
     - Fnet decreases./Fnet Neem af.
                                                                                                           [16]

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 Physical Sciences P1/Grade 12                             8                        NC/ September 2021
 Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

QUESTION 3/VRAAG 3
3.1   A projectile is an object which has been given an initial velocity and then it
      moves under the influence of the gravitational force only. 
      'n Projektiel is 'n voorwerp waaraan 'n beginsnelheid gegee is en wat dan slegs
      onder die invloed van die gravitasiekrag beweeg.
                                                                           (2 or/of 0)
3.2   ∆y= 0 (m)/Zero                                                                                       (1)
3.3.1 Marking criteria/Nasienriglyne
            Any vorrect formula/Enige kprrekte formule
            Whole substitution/Hele vervanging
            Final answer/Finale antwoord

        OPTION 1/OPSIE 1
        Positive downwards/Positief afwaarts:
                   1                     1
            ⃗ i ∆t+ ⃗a∆t2 OR/OF ∆y=vi ∆t+ a∆t2 
         ⃗ =v
        ∆y         2                     2
                             1
        +19,6 =(0)∆t+ 2 (+9,8)∆t2 
        ∆t = 2s
        ΔtTOTAL = 4s/ total t = 4 s
        OPTION 2/OPSIE 2
        Positive upwards/ Positief opwaarts:
                   1                     1
            ⃗ i ∆t+ ⃗a∆t2 OR/OF ∆y=vi ∆t+ a∆t2 
         ⃗ =v
        ∆y         2                     2
                             1
        −19,6 =(0)∆t+ 2 (-9,8)∆t2 
        ∆t = 2s
        ΔtTOTAL = 4s/ total t = 4 s

        OPTION 3/OPSIE 3
        Positive downwards/Positief afwaarts:
         2     2
       ⃗vf = ⃗vi +2a      ⃗ OR/OF v2f = v2i +2a∆y
                       ⃗ ∆y
        v2 = 02 +2(9,8)(+19,6)
        vf =+ 19,6 m∙𝑠 −1
                ⃗v +v⃗                       v +v
        ∆y⃗ = ( i f ) ∆t OR/OF ∆y= ( i f ) ∆t
               2                           2
                    0+19,6
        +19,6= ( 2 ) ∆t
        ∆t = 2s
        ΔtTOTAL = 4s/ total t = 4 s

        OPTION 4/OPSIE 4
        Positive upwards/ Positief opwaarts:
         2     2
       ⃗vf = ⃗vi +2a      ⃗ OR/OF v2f = v2i +2a∆y
                       ⃗ ∆y
        v2 = 02 +2(9,8)( − 19,6)
        vf =- 19,6 m∙𝑠 −1
                ⃗v +v⃗                       v +v
        ∆y⃗ = ( i f ) ∆t OR/OF ∆y= ( i f ) ∆t
               2                           2
                   0+(−19,6)
        -19,6= [ 2 ] ∆t
        ∆t = 2s
        ΔtTOTAL = 4s/ total t = 4 s

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 Physical Sciences P1/Grade 12                             9                        NC/ September 2021
 Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

         OPTION 5/OPSIE 5
         Positive downwards/Positief afwaarts:
          2      2
        ⃗vf = ⃗vi +2a   ⃗ OR/OF v2f = v2i +2a∆y
                     ⃗ ∆y
         v2 = 02 +2(9,8)(+19,6)
         vf =+ 19,6 m∙𝑠 −1
        ⃗vf =v    ⃗ ∆t OR/OF vf =vi +a∆t
             ⃗ i +a
         +19,6=0+(+9,8)∆t
         ∆t = 2s
         ΔtTOTAL = 4s/ total t = 4 s

         OPTION 6/OPSIE 6                                                                                   (3)
         Positive upwards/ Positief opwaarts:
          2      2
        ⃗vf = ⃗vi +2a   ⃗ OR/OF v2f = v2i +2a∆y
                     ⃗ ∆y
         v2 = 02 +2(9,8)( − 19,6)
         vf =- 19,6 m∙𝑠 −1
        ⃗vf =v    ⃗ ∆t OR/OF vf =vi +a∆t
             ⃗ i +a
         -19,6=0+(-9,8)∆t
         ∆t = 2s
         ΔtTOTAL = 4s/ total t = 4 s
         Note/Aantekening: Do not penalise if vector notation is not used./Moenie
         penaliseer indien vektornotasie nie gebruik is nie.

3.3.2   Marking criteria/Nasienriglyne
            Any correct formula/Enige kprrekte formule
            Whole substitution/Hele vervanging
            Final answer/Finale antwoord

        POSITIVE MARKING FROM 3.3.1/POSITIEWE NASIEN VANAF 3.3.1
        OPTION 1/OPSIE 1
        From bottom to top/Van onder na bo
        Positive upwards/ Positief opwaarts:
             ⃗v +v⃗                      v +v
         ⃗ = ( i f ) ∆t OR/OF ∆y= ( i f ) ∆t
        ∆y      2                     2
                      v +0
        +19,6= ( i2 ) 2
        vi = + 19,6 m∙𝑠 −1
        OR/OF
        vi = 19,6 m∙𝑠 −1upwards/afwaarts. Accept
        OPTION 2/OPSIE 2
        From bottom to top/Van onder na bo
        Positive downwards/Positief afwaarts:
             ⃗v +v⃗                      v +vf
         ⃗ = ( i f ) ∆t OR/OF ∆y= ( i
        ∆y                                       ) ∆t
               2                           2
                   v +0
        -19,6= ( i2 ) 2
        vi = − 19,6 m∙𝑠 −1
        OR/OF
        vi = 19,6 m∙𝑠 −1upwards/afwaarts.
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Physical Sciences P1/Grade 12                             10                       NC/ September 2021
Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

       OPTION 3/OPSIE 3
       From bottom to top/Van onder na bo
       Positive upwards/ Positief opwaarts:
      ⃗vf =v    ⃗ ∆t OR/OF vf =vi +a∆t
           ⃗ i +a
       0=vi +( − 9,8)2
       vi = + 19,6 m∙𝑠 −1
       OR/OF
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.
       OPTION 4/OPSIE 4
       From bottom to top/Van onder na bo
       Positive downwards/Positief afwaarts:
      ⃗vf =v    ⃗ ∆t OR/OF vf =vi +a∆t
           ⃗ i +a
       0=vi +(+9,8)2
       vi = − 19,6 m∙𝑠 −1
       OR/OF
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.

       OPTION 5/OPSIE 5
       From bottom to top/Van onder na bo
       Positive upwards/ Positief opwaarts:
        2     2
                     ⃗  OR/OF v2f = v2i +2a∆y
                  ⃗ ∆y
      ⃗vf = ⃗vi +2a
        2
       0 = v2i +2(-9,8)( + 19,6)
       vi = + 19,6 m∙𝑠 −1
       OR/OF
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.
       OPTION 6/OPSIE 6
       From bottom to top/Van onder na bo
       Positive downwards/Positief afwaarts:
        2     2
                     ⃗  OR/OF v2f = v2i +2a∆y
                  ⃗ ∆y
      ⃗vf = ⃗vi +2a
               2
       02 = vi +2(+9,8)( − 19,6)
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.
       OR/OF
       vi = + 19,6 m∙𝑠 −1
       OPTION 7/OPSIE 7
       From top to bottom/ Van bo na onder
       Positive downwards/Positief afwaarts:
            ⃗v +v⃗                      v +v
        ⃗ = ( i f ) ∆t OR/OF ∆y= ( i f ) ∆t
       ∆y      2                     2
                     0+v
       +19,6= ( 2 f ) 2
       vf = + 19,6 m∙𝑠 −1
       vi = − 19,6 m∙𝑠 −1
       OR/OF
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.

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       OPTION 8/OPSIE 8
       From top to bottom/ Van bo na onder
       Positive upwards/ Positief opwaarts:
            ⃗v +v⃗                      v +v
        ⃗ = ( i f ) ∆t OR/OF ∆y= ( i f ) ∆t
       ∆y      2                     2
                 0+v
       -19,6= ( 2 f) 2
       vf = − 19,6 m∙𝑠 −1
       vi = + 19,6 m∙𝑠 −1
       OR/OF
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.
       OPTION 9/OPSIE 9
       From top to bottom/ Van bo na onder
       Positive downwards/Positief afwaarts:
      ⃗vf =v    ⃗ ∆t OR/OF vf =vi +a∆t
           ⃗ i +a
       vf =0+(+9,8)2
       vf =+ 19,6 m∙𝑠 −1
       vi =- 19,6 m∙𝑠 −1
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.

       OPTION 10/OPSIE 10
       From top to bottom/ Van bo na onder
       Positive upwards/ Positief opwaarts:
      ⃗vf =v     ⃗ ∆t OR/OF vf =vi +a∆t
            ⃗ i +a
       vf =0+( − 9,8)2
       vf = − 19,6 m∙𝑠 −1
       vi = + 19,6 m∙𝑠 −1
       OR/OF
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.
       OPTION 11/OPSIE 11
       From top to bottom/ Van bo na onder
       Positive downwards/Positief afwaarts:
        2       2
      ⃗vf = ⃗vi +2a    ⃗  OR/OF v2f = v2i +2a∆y
                    ⃗ ∆y
       v2f = 02 +2(+9,8)( + 19,6)
       vf = + 19,6 m∙𝑠 −1
       vi = -19,6 m∙𝑠 −1                                                                            (3)
       OR/OF
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.
       OPTION 12/OPSIE 12
       From top to bottom/ Van bo na onder
       Positive upwards/ Positief opwaarts:
        2       2
      ⃗vf = ⃗vi +2a    ⃗  OR/OF v2f = v2i +2a∆y
                    ⃗ ∆y
       v2f = 02 +2(-9,8)( − 19,6)
       vf =19,6 m∙𝑠 −1
       vi = 19,6 m∙𝑠 −1upwards/afwaarts.
       OR/OF
       vi = + 19,6 m∙𝑠 −1
       Note/Aantekening: Do not penalise if vector notation is not used./Moenie
       penaliseer indien vektorenotasie nie gebruik is nie.
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3.4     POSITIVE MARKING FROM 3.3.1/POSITIEWE NASIEN VANAF 3.3.1
        OPTION 1/ OPSIE 1
        Positive upwards/Positief opwaarts:
            y(m)
           19,6

                0           2             4    t (s)

        OPTION 2/ OPSIE 2
        Positive upwards/Positief opwaarts:

                                y(m)
                                    0                  2          4 t (s)

                                -19,6

        OPTION 3/OPSIE 3
        Positive downwards/Positief afwaarts:
                y (m)

                        0            2
                                                   4
                                                           t

                -19,6

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       OPTION 4/OPSIE 4
       Positive downwards/Positief afwaarts:

                 y (m)

                   19,6

                      0
                                    2             4 t (s)

        Criteria for marking the graph/                                                   Marks/
        Kriteria vir die nasien van die grafiek.                                          Punte
        Initial and final position (19,6 m) /Aanvanklike en finale posisie
                                                                                           
        (19,6m).
        The time taken by the ball to reach the ground and the time taken by
        the ball to reach the starting point./ Die tyd wat die bal neem om die
                                                                                           
        grond te bereik en die tyd wat die bal neem om die beginpunt te
        bereik.
        Correct shape/Korrekte vorm.                                                                  (3)
                                                                                                      [12]

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QUESTION 4/VRAAG 4

4.1     NOTE: -1 Mark for each key word /phrase omitted in the correct context.
        LET WEL: - 1 punt vir elke sleutelwoord/frase weggelaat        in die korrekte
        konteks.
                                                                                                        (2)
      The total mechanical energy (sum of gravitational potential energy and kinetic
      energy) in an isolated system remains constant.
      Die totale meganiese energie (som van gravitasie- potensiële energie en
      kinetiese energie) in 'n geslote sisteem bly konstant.

4.2   OPTION 1/OPSIE 1
      ∆EM =0
      EMf -EMi =0
      EMi =EMf                                                    Any One/Enige een
      EKi + EPi = EKf + EPf
      1                             1
        (mP +mQ )v2i +(mP +mQ )ghi = (mP +mQ )v2f +(mP +mQ )ghf
      2                             2
       1             2                  1
      [ (4+6)(4,5) +(4+6)(9,8)(6)] = [ (4+6)v2f +(4+6)(9,8)(0)] 
       2                                2
      OR/OF
       1                              1
      [ (10)(4,5)2 +(10)(9,8)(6)] = [ (10)v2f +(10)(9,8)(0)] 
       2                              2
      v=11,74 m∙s  -1

      OPTION 2/OPSIE 2
      ∆EM =0
      ∆EK +∆Ep =0
                                                                                    Any One/Enige een
      (EKf -EKi )+(EPf -EPi )= 0
       1                1
      [ (mP +mQ )v2f - (mP +mQ )v2i ] +[(mP +mQ )ghf -(mP +mQ )ghi ]=0
       2                2
       1            1
      [ (4+6)v2f - (4+6)(4,5)2 ] +[(4+6)(9,8)(0) − (4+6)(9,8)(6)]=0
       2            2
      OR/OF
       1          1
      [ (10)v2f - (10)(4,5)2 ] +[(10)(9,8)(0) − (4+6)(9,8)(6)]=0
       2          2
      v=11,74 m∙s-1 

      OPTION 3/OPSIE 3
      ∆EM =0
      ∆EK +∆Ep =0                                                        Any One/Enige een
      (EKf -EKi )= -(EPf -EPi )
       1               1
      [ (mP +mQ )v2f - (mP +mQ )v2i ] = − [(mP +mQ )ghf − (mP +mQ )ghi ]
       2               2
       1           2
                       1
      [ (4 + 6)𝑣𝑓 − (4 + 6)(4,5)2 ] +[(4+6)(9,8)(0) − (4+6)(9,8)(6)]
       2               2
      OR/OF
       1             1
      [ (10)𝑣𝑓2 − (10)(4,5)2 ] = − [(10)(9,8)(0) − (4+6)(9,8)(6)]
       2             2
      v=11,74 m∙s-1 
                                                                                        (4)
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4.3   OPTION 1/OPSIE 1                                                                                        (1)
      ∑ ⃗pafter/na - ∑ ⃗p(before/voor) =0
                                                                               Any one / Enige een
       ∑ ⃗p(before/voor) = ∑ ⃗p(after/na)
      mP⃗vP(bef/voor) +mQ⃗vQ(bef/voor) =(mP )v
                                             ⃗ P(after/na) +mQ⃗vQ(aft/na)
                                                                                   NOTE/LET WEL:
       positive to the right/na regs is positief:                                  11,74 - 100%
       (4)(+11,74)+(6)(+11,74) = (4)v      ⃗ P(after/na) +(6)(+12,914)           X - 10 %
                                                                                        11,74 × 10
      ⃗vP(after/na) = +9,979 m·s  OR/OF ⃗vP(after/na) = +9,98 m·s-1
                                  -1
                                                                                   𝑋=              = 1,174
                                                                                           100
       OR/OF                                                                       vQ= 11,74 + 1,174
      ⃗vP(after/na) = 9,98 m·s-1 to the right/na regs
                                                                                   vQ= 12,914 m∙s-1
       OPTION 2/OPSIE 2
      ∑ ⃗pafter/na - ∑ ⃗p(before/voor) =0
       ∑ ⃗p(before/voor) = ∑ ⃗p(after/na)                                      Any one / Enige een
      mP⃗vP(bef/voor) +mQ⃗vQ(bef/voor) =(mP )v
                                             ⃗ P(after/na) +mQ⃗vQ(aft/na)

       positive to the left/na links is positief:
       (4)( − 11,74)+(6)( − 11,74) = (4)v     ⃗ P(after/na) +(6)(−12,914)
      ⃗vP(after/na) = −9,979 m·s  OR/OF ⃗vP(after/na) = −9,98 m·s-1
                                  -1

       OR/OF
      ⃗vP(after/na) = 9,98 m·s-1 to the right/na regs
       Note/Aantekening: Do not penalise if vector notation is not used./Moenie
       penaliseer indien vektornotasie nie gebruik is nie.

4.4   OPTION 1/OPSIE 1
      ⃗F∆t=∆p            ⃗ OR/OF FΔt = Δp
       ⃗J=∆p          ⃗ OR/OF I=∆p     ⃗ OR/OF Impulse/Impuls=∆p      Any one/Enige een
        ⃗J=𝑝𝑓 − 𝑝𝑖
         ⃗J=m(v             ⃗ fi ) OR/OF ⃗J=mv
                       ⃗ f -v                ⃗ f -mv
                                                   ⃗ fi
          positive to the right/na regs is positief:
          ⃗J=4[(+9,98-(+11,74)]  OR/OF ⃗J= [(4)(+9,98)]-[(4)(+11,74)]
           ⃗J= - 7,00 N∙s
            J= 7,00 N∙s Accept/Aanvaar J/I/Impulse/Impuls= 7 N∙s
            OPTION 2/OPSIE 2
            ⃗F∆t=∆p      ⃗ OR/OF FΔt = Δp
             ⃗J=∆p    ⃗ OR/OF I=∆p     ⃗ OR/OF Impulse/Impuls=∆p      Any one/Enige een
              ⃗J=𝑝𝑓 − 𝑝𝑖
               ⃗J=m(v       ⃗ fi ) OR/OF ⃗J=mv
                       ⃗ f -v                ⃗ f -mv
                                                   ⃗ fi
                positive to the left/na links is positief:
                ⃗J=4[(-9,98-(-11,74)]  OR/OF ⃗J= [(4)(−9,98)]-[(4)(−11,74)]
                 ⃗J= +7,00 N∙s
                  J= 7,00 N∙s Accept/Aanvaar J/I/Impulse/Impuls= 7 N∙s
                  Note/Aantekening: Do not penalise if vector notation is not used./Moenie
                  penaliseer indien vektornotasie nie gebruik is nie.                                         (4)
                                                                                                              [13]

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QUESTION 5/VRAAG 5

5.1
        NOTE: -1 Mark for each key word /phrase omitted in the correct context.
        LET WEL: - 1 punt vir elke sleutelwoord/frase weggelaat in die korrekte
        konteks.
                                                                                                     (2)
       The net work done on an object is equal to the change in the object's kinetic
       energy.
       Die netto arbeid verrig op 'n voorwerp is gelyk aan die verandering in kinetiese
       energie van die voorwerp
        OR/OF
       The work done on an object by a net force is equal to the change in the object's
       kinetic energy.
       Die arbeid verrig op die voorwerp deur 'n netto krag is gelyk aan die
       verandering in kinetiese energie van die voorwerp.

5.2    OPTION 1/OPSIE 1
       Wnet =∆EK
       (Fg -ff )∆xcos θ = EKf − EKi                 Any One /Enige een 
                          1     1
       (Fg -ff )∆xcos θ = mv2f - mv2i
                          2     2
                                            1         1
       [(80×9,8)-172 ]×4×(cos0o )= 2 (80)v2f - 2 (80)(0)2 )
       v = 7,82 m∙s-1 

       OPTION 2/OPSIE 2
       Wnet =∆EK
       𝑊𝑓 + 𝑊𝐹𝑔 = EKf − EKi
        𝑓𝑓 ∆xcos1800 + Fg∆xcos0o =EKf − EKi                   Any One /Enige een 
                                  1     1
       𝑓𝑓 ∆xcos1800 +mg∆xcos0o = mv2f - mv2i
                                  2     2
                                                1         1
       172 ×4×(-1)+(80)(9,8)(4)(1)= 2 (80)v2f - 2 (80)(0)2 )
       v = 7,82 m∙s-1 
       OPTION 3/OPSIE 3
       Wnet =∆EK
       𝑊𝐹𝑔 + 𝑊𝑓 = EKf − EKi
       −∆𝐸𝑝 + 𝑊𝑓 = EKf − EKi                                     Any One /Enige een 
                                     1  0   1
       -(mghf -mghi ) + 𝑓𝑓 ∆xcos180 = mv2f - mv2i
                                     2      2
                                                                1          1
       −[(80)(9,8)(0) − (80)(9,8)(4)] + 172 ×4×(-1)= 2 (80)v2f - 2 (80)(0)2 )
       v = 7,82 m∙s-1                                                                               (4)

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5.3    No/Nee 
       The net work done by the non-conservative forces is not zero./Die netto arbeid
       deur die nie-konservatiewe kragte is nie nul nie.

       OR/OF
       There is friction/non-conservative forces doing work./Daar is wrywing/nie-
       konservatiewe kragte wat arbeid verrig.
       OR/OF
       It is not an isolated system.
       Dit is nie ‘n geïsoleerde sisteem nie.
                                                                                                     (2)

5.4    OPTION 1/OPSIE 1
       𝐸𝑀𝑓 = 𝐸𝑀𝑖 + 𝑊𝑛𝑐
        𝑚𝑣𝑓2              𝑚𝑣𝑖2
       (     + 𝑚𝑔ℎ𝑓 ) = (      + 𝑚𝑔ℎ𝑖 ) + 𝑊𝑛𝑐
         2                 2                          Any one/Enige een
        𝑚𝑣𝑓2                𝑚𝑣𝑖2
       (     + 𝑚𝑔(−𝑑)) = (       + 𝑚𝑔ℎ) + 𝑊𝑛𝑐
         2                     2
        80 × 02                        80 × 02
       (        + 80 × 9,8(−𝑑))  = [(         + 80 × 9,8 × 4) + (−6240)] 
           2                              2
       d=3,96 m

       OPTION 2/OPSIE 2

       Wnc =∆EK +∆Ep 
       Wnc =(EKf -EKf )+(EPf -EPi )
               mv2    mv2
       Wnc = ( 2 f − 2 i ) − (mghf -mghi )
                                                             Any one/Enige een
               mv2f mv2i
       Wnc = (     -     ) +(mg(-d)-mgh)
                2    2
                     80 × 02 80 × 02
       −6240 = (           −        ) + [80 × 9,8 × (−𝑑) − ((80 × 9,8 × 4))]
                        2       2
       d=3,96 m

       OPTION 3/OPSIE 3

       Wnet =∆EK
       𝑊𝑐 + 𝑊𝑛𝑐 = 𝐸𝐾𝑓 − 𝐸𝐾𝑖
                        1         1
                                                                     Any one/Enige een
       −∆𝐸𝑃 + 𝑊𝑛𝑐 = 2 𝑚𝑣𝑓2 − 2 𝑚𝑣𝑖2
                            1        1
       −(𝐸𝑃𝑓 − 𝐸𝑝𝑖 ) + 𝑊𝑛𝑐 = 𝑚𝑣𝑓2 − 𝑚𝑣𝑖2
                            2        2
                                    1    1
       −[(𝑚𝑔(−𝑑) − 𝑚𝑔ℎ)] + 𝑊𝑛𝑐 = 𝑚𝑣𝑓2 − 𝑚𝑣𝑖2
                                    2    2
                                                               1         1
       −[(80 × 9,8(−𝑑) − (80)(9,8)(4)] + (−6240) =              80(0)2 − 80(0)2 
                                                               2         2
       d=3,96 m

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       OPTION 4/OPSIE 4

       Wnc =∆EM
       Wnc = EMf − EMi 
       Wnc =(EKf -EKf )+(EPf -EPi )
               mv2            mv2
                                                        Any one/Enige een
       Wnc = ( 2 f +mghf ) - ( 2 i +mghi )
               mv2f            mv2i
       Wnc = (      +mg(-d) - (     +mgh)
                2               2
                 80×02                    80×02
       −6240= [       +80×9,8 ×(-d)] - [       +(80×9,8×4)] 
                   2                        2

       d=3,96 m

       OPTION 5/OPSIE 5

       Wnc =∆EK +∆Ep 
       Wnc =(EKf -EKf )+(EPf -EPi )                   Any one/Enige een
               mv2f   mv2i
       Wnc = ( 2 − 2 ) − (mghf -mghi )
                  80 × 02 80 × 02
       −6240 = (        −        ) + [80 × 9,8 × (0) − 80 × 9,8 × (4+d)]
                     2       2

       d=3,96 m
                                                                                                     (4)
                                                                                                     [12]

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QUESTION 6/VRAAG 6

6.1 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. 
    OR/OF
    The change in the observed frequency when there is relative motion between
    the source and the observer.
    Die verandering in die waargenome frekwensieas daar relatiewe beweging
    tussen die klankbron en die luisteraar is.                                                         (2)

6.2 Longer/Langer

     For the same (constant) speed of sound the frequency of sound is inversely
     proportional to the wave length and as the fire truck moves away from the listener
     the listener hears a lower frequency..
     Vir dieselfde (konstante) spoed van klank is die frekwensie van klank
     omgekeerd eweredig aan die golflengte en as die brandweerwa beweeg weg van
     die luisteraar die luisteraar hoor 'n laer frekwensie.
     OR/OF
     Speed of sound is constant, frequency detected (by the observer) is lower
     wevelenghts is inversely proportinal to frequncy.
     Klankspoed is konstant, frekwensie waargeneem deur die waarnemer is laer,                          (3)
     golflengtes is omgekeerd eweredig aan frekwensie

6.3 f = (v±vL ) f  OR/OF f = ( v ) f
     L   v±v     𝑠         L   v±v    s
            s                        s
     Approaching
            340
     460= (340-v ) fs 
                  s
     Moving away
             340
     410 = (340+v ) fs
                      s

     Solving
                340       410
     460 = (          ) ( 340 )
               340-vs
                           340+vs
     vs=19,54 m∙s-1

                  340                                 340
     460 = (340-19,54) fs OR/OF 410 = (340+19,54) fs                                                (6)

     fs= 433,56 Hz

6.4 The star is movin away from the Erath  because the spectrum shows a shift
    towards the red end which is a lower frequnecy 
    Die ster beweeg van die aarde af omdat die spektrum 'n verskuiwing na die rooi
    einde toon wat 'n laer frekwensie is.                                                             (3)

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6.5 ANY ONE
        To measure rate of blood flow.
         Detect heartbeat of foetus.
        It is used (in flow meters) in medical science to measure:
         - the speed and direction (velocity) of blood flow.
         - movement of the heart of a foetus.
        To find the rate of blood flow (Doppler scanning)
        To see the unborn child (Ultra sound scanning)
        To hear the heart of a foetus (Ultra sound scanning)
        It is used in medical sonography to generate images (and sounds) of
         flowing blood.
        To detect blood clotting (Doppler ultrasound test)

      ENIGE EEN
          Om die bloedvloeitempo te meet.
          Ontdek die hartklop van die fetus.
          Dit word (in vloeimeters) gebruik in mediese wetenskap om die
           volgende te meet:
           - Die spoed en rigting (snelheid) van bloedvloei.
           - Beweging van 'n fetus se hart.
          Om die tempo van bloedvloei te meet (Doppler skandering)
          Om 'n ongebore baba te sien (Ultraklankskandering)
          Om die hart van 'n fetus te hoor (Ultraklankskandering)
          Dit word gebruik in mediese sonografie om beelde (en klanke) te vorm van
           vloeiende bloed.
          Om vorming van bloedklonte op te spoor. (Doppler- ultraklamktoets)
                                                                                                        (1)
                                                                                                       [15]

QUESTION 7/VRAAG 7

7.1    Positive/Positief.                                                                              (1)
                                 q
7.2     Q = ne  OR/OF n=
                            e
        Q = 938(1,6 × 10−19 C) OR/OF Q = 938(1,6 × 10−19 )
        Q =1,50×10−16 C 
                                                                                                      (3)

7.3      NOTE: -1 Mark for each key word /phrase omitted in the correct context.
         LET WEL: - 1 punt vir elke sleutelwoord/frase weggelaat in die korrekte
         konteks.

        The electric field at a point is the electrostatic force experienced per unit
        positive charge placed at that point.
        Die elektriese veld by 'n punt is die elektrostatiese krag wat per
        eenheidspositiewe-lading wat by daardie punt geplaas is, ondervind word.                      (2)

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 Fisiese Wetenskappe V1/Graad 12          Marking Guidelines/Nasienriglyne

7.4
                                          Marking criteria/ Nasienkriteria
                                          Shape (radial)/Vorm (radiaal)           

                                          Correct direction/Korrekte rigting      

                                                                                                             (3)

7.5 At A.
    The distance from A to the charged sphere X is smaller than the distance from B
    to the charged sphere X and the electric field at a point due to a point charge is
    inversely proportional to the square distance between the point and the charge
         1
    (E∝ 2) 
          d
      By A
      Die afstand vanaf A na die gelaaide sfeer X is kleiner as die afstand van B na die
      gelaaide sfeer X en die elektriese veld op 'n punt as gevolg van 'n puntlading
      omgekeerd eweredig is aan die vierkante afstand tussen die punt en die lading
          1
      (E∝ 2).                                                                                                (3)
          d

 7.6.1 The magnitude of the electrostatic force exerted by one point charge on another
       point charge is directly proportional to the product (of the magnitudes) of the
       charges and inversely proportional to the square of the distance between
       them.
       Die grootte van die elektrostatiese krag wat een puntlading op 'n ander
       puntlading 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)
 7.6.2    OPTION 1/OPSIE 1
            KQ Q             KQ Q
          F= r12 2  OR/OF F= rX2 Y
                    9×109 ×(1,5×10-16 ×(2,8×10-16 )
          FYonX =                                     
                                (0,03)2
          FYonX =4,2 ×10-19 N .to the right/na regs
             KQX QZ
          F=
                 r2
                  9×109 ×(1,5×10-16 ×(3,2×10-16 )
          FZonX =                                 
                             (0,01)2
          FZonX =43,2 ×10-19 N . to the right/na regs
          Positive to the right/Positief na regs
          𝐹𝑛𝑒𝑡 = 𝐹YonX + 𝐹ZonX OR/OF 𝐹𝑛𝑒𝑡 = FYonX + FZonX
          𝐹𝑛𝑒𝑡 = 4,2 ×10-19 + 43,2 ×10-19 
          𝐹𝑛𝑒𝑡 = 47,4×10-19 N.to the right/na regs
          OR/OF
          𝐹𝑛𝑒𝑡 = 4,74×10-18 N to the right/na regs

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Fisiese Wetenskappe V1/Graad 12            Marking Guidelines/Nasienriglyne

        OPTION 2/OPSIE 2
          KQ Q             KQ Q
        F= r12 2  OR/OF F= rX2 Y
                  9×109 ×(1,5×10-16 ×(2,8×10-16 )
        FYonX =                                        
                           (0,03)2
        FYonX =4,2 ×10-19 N .to the right/na regs
           KQX QZ
        F=
              r2

                  9×109 ×(1,5×10-16 ×(3,2×10-16 )
        FZonX =                        2
                                                      
                             (0,01)
                         -19
        FZonX =43,2 ×10 N . to the right/na regs
        Positive to the left/Positief na links
        𝐹𝑛𝑒𝑡 = 𝐹YonX + 𝐹ZonX OR/OF 𝐹𝑛𝑒𝑡 = −FYonX − FZonX
        𝐹𝑛𝑒𝑡 = −4,20 ×10-23 − 43,2 ×10-23 
        𝐹𝑛𝑒𝑡 = 47,4×10-19 N.to the right/na regs
        OR/OF
        𝐹𝑛𝑒𝑡 = 4,74×10-18 N to the right/na regs

        OPTION 3/OPSIE 3
        Positive to the right/ Positief na regs
               𝑄
        E=𝐾 2
               𝑟
                         (3,2×10−16 )
        E𝑥 = 9 × 109 × (0,01)2 
        E𝐴 = 28,8 × 10−3 𝑁 ∙ 𝐶 −1 to the right/na regs
                           (2,8×10−16 )
        E𝑌 = 9 × 109 ×                 
                             (0,03)2
        ⃗E𝐵 = 2,8 × 10−3 𝑁 ∙ 𝐶 to the right/na regs
                                  −1

         ⃗Enet = ⃗EX + ⃗EY
          Enet = EX + 𝐸B
          Enet = 28,8 × 10−3 + 2,8 × 10−3
          Enet = 31,6 × 10−3 𝑁 ∙ 𝐶 −1
          ⃗Enet = 31,60 × 10−3 𝑁 ∙ 𝐶 −1to the right/na regs

        𝐹 = 𝑞𝐸 
        𝐹 = 1,5 × 10−16 × 31,60 × 10−3 
        𝐹 = 47,4×10-19 𝑁 to the right/na regs
        OR/OF
        𝐹 = 4,74×10-18 𝑁 to the right/na regs
        OPTION 4/OPSIE 4
        Positive to the left/ Positief na links
               𝑄
        E=𝐾 2
               𝑟
                         (3,2×10−16 )
        E𝑥 = 9 × 109 × (0,01)2 
        E𝐴 = 28,8 × 10−3 𝑁 ∙ 𝐶 −1 to the right/na regs

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 Fisiese Wetenskappe V1/Graad 12                Marking Guidelines/Nasienriglyne

                             (2,8×10-16 )
         E𝑌 = 9 × 109 ×                     
                           (0,03)2
         ⃗E𝐵 = 2,8 × 10 𝑁 ∙ 𝐶 to the right/na regs
                        −3       −1

          ⃗Enet = ⃗EX + ⃗EY
           Enet = −EX − 𝐸B
           Enet = −28,8 × 10−3 − 2,8 × 10−3
           Enet = −31,60 × 10−3 𝑁 ∙ 𝐶 −1
           ⃗Enet = 31,60 × 10−3 𝑁 ∙ 𝐶 −1to the right/na regs

         𝐹 = 𝑞𝐸 
         𝐹 = 1,5 × 10−16 × 31,60 × 10−3 
                       -19
         𝐹 = 47,4×10         𝑁 to the right/na regs
         OR/OF
         𝐹 = 4,74 × 10-18 𝑁 to the right/na regs

         Note/Aantekening: Do not penalise if vector notation is not used./Moenie (5)
         penaliseer indien vektornotasie nie gebruik is nie.
                                                                                   [18]

 QUESTION 8/VRAAG 8

8.1     Power is the rate at which work is done 

        OR/OF
        Power is the rate at which energy is supplied to the device.                                         (2)
                                                                                       (2 or/of 0)

8.2.1   OPTION 1/OPSIE 1
        Rs =RA +RB
        Rs =8+8 
        Rs = 16 Ω
             𝑅𝑠 × 𝑅𝐶
        RP =
             𝑅𝑠 + 𝑅𝐶
             16×8
        RP = 16+8
        RP = 5,333 Ω
        V=IR
        V=2×5,333 
        V=10,67 V 
        OPTION 2/OPSIE 2
        Rs =RA +RB
        Rs =8+8 
        Rs = 16 Ω
             𝑅𝐴𝐵 × 𝑅𝐶
        RP =
             𝑅𝐴𝐵 + 𝑅𝐶
             [(8+8)×8]              (16×8)
        RP = [(8+8)+8]  OR/OF RP = (16+8) 
        RP = 5,333 Ω
        V=IR
        V=2×5,333 
        V=10,67 V 
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 Physical Sciences P1/Grade 12                             24                       NC/ September 2021
 Fisiese Wetenskappe V1/Graad 12         Marking Guidelines/Nasienriglyne

        OPTION 3/OPSIE 3
        Rs =RA +RB
        Rs =8+8 
        Rs = 16 Ω
         1     1     1
            =     +
        R 𝑃 R1 R 2
         1    1    1
           =     +  
        R 𝑃   16   8
        RP = 5,333 Ω
        V=IR
        V=2×5,333 
        V=10,67 V                                                                                  (5)

8.2.2   POSITIVE MARKING FROM 8.2.1/POSITIEWE NASIEN VANAF 8.2.1.
        OPTION 1/OPSIE 1
        ε = Vext +Vint    Any one/Enige een
        ε = IR+Ir

        12 =10,67 +2r
        r=0,67 Ω
        PTION 2/OPSIE 2
        Vint = ε − Vext             Any one/Enige een
        Ir = ε − Vext

        2r = 12 − 10,67
        r=0,67 Ω
        PTION 3/OPSIE 3
            ε
        I= R+r
                12
        2I= 5,33+r
        r=0,67 Ω
        PTION 4/OPSIE 4
            ε
        I= R+r
                   12                          12
        2= [(8+8)×8]       OR/OF      2= (16×8)         
              ([(8+8)+8])+r                ((16+8))+ r

        r=0,67 Ω                                                                                   (3)
8.3     Bulbs A and B are connected in series across the battery, therefore they
        glow equally brightly, whereas bulb C is connected itself across the
        battery, then the voltage drop across C has the same magnitude as the
        terminal potential difference,     whereas the same terminal potential
        difference splits between bulbs A and B, as a result , bulb C glows more
        brightly than either of bulb A and B. 
        Gloeilampe A en B word in serie oor die battery gekoppel, daarom gloei
        hulle ewe helder, terwyl gloeilamp C self oor die battery gekoppel is, dan is
        die spanningsval oor C dieselfde grootte as die potensiaalverskil van die
        terminale, terwyl dieselfde potensiaalverskil dieselfde is verdeel tussen
        gloeilampe A en B, gevolglik gloei gloeilamp C helderder as een van
        gloeilamp A en B.                                                                             (4)

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