# Physical Sciences Grade 12 P1 Eastern Cape Mock 2012 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape Mock Exam 2012, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/phsc-p1-s12-memo-eng-afr/081b067a-dc7e-45b8-bcba-b7556f428acc
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2012
- Exam period: Eastern Cape Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 14

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Physical Sciences · Grade 12 · Eastern Cape Mock Exam · 2012. Memorandum, 14 pages. Read online or download the PDF.

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                 Province of the
                 EASTERN CAPE
                 EDUCATION

                            NATIONAL
                       SENIOR CERTIFICATE

                       GRADE/GRAAD 12

                       SEPTEMBER 2012

                PHYSICAL SCIENCES P1
               FISIESE WETENSKAPPE V1

                        MEMORANDUM

MARKS/PUNTE:     150

                  This memorandum consists of 14 pages.
                Hierdie memorandum bestaan uit 14 bladsye.

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2                                    PHYSICAL SCIENCES P1 (Memo)                          (SEPTEMBER 2012)

               LEARNING OUTCOMES AND ASSESSMENT STANDARDS
                 LEERUITKOMSTE EN ASSESSERINGSTANDAARDE
            LO/LU1               LO/LU2               LO/LU3
AS 12.1.1                             AS 12.2.1                         AS 12.3.1
Plan and conduct a scientific         Define and discuss basic          Recognise, discuss and compare
investigation to collect data         prescribed and scientific         scientific and indigenous knowledge
systematically with regard to         knowledge.                        systems and knowledge claims by
accuracy, reliability and the need                                      indicating the correlation among
to control variables.                                                   them, and explain the acceptance of
                                                                        different claims.
                                                                        Herken, bespreek en vergelyk
Beplan en voer ŉ wetenskaplike        Definieer en bespreek basiese     wetenskaplike en inheemse
ondersoek uit om data sistematies     voorgeskrewe wetenskaplike        kennissisteme en kennisaansprake
te versamel ten opsigte van           kennis.                           deur die ooreenkoms aan te dui en
akkuraatheid, betroubaarheid en                                         verduidelik die aanvaarding van
die kontroleer van veranderlikes.                                       verskillende aansprake.
AS 12.1.2                             AS 12.2.2                         AS 12.3.2
Seek pattern and trends,              Express and explain prescribed    Identify ethical and moral issues
represent them in different forms     scientific theories, models and   related to the development of
to draw conclusions, and              laws by indicating the            science and technology and evaluate
formulate simple generalisations.     relationship between different    the impact (pros and cons) of the
                                      facts and concepts in own         relationship from a personal
                                      words.                            viewpoint.
Soek patrone en tendense, stel dit    Verduidelik en druk               Identifiseer etiese en morele
in verskillende vorms voor om         voorgeskrewe wetenskaplike        uitkomste in verband met die
gevolgtrekkings te maak en om         beginsels, teorieë, modelle en    ontwikkeling van die wetenskap en
eenvoudige veralgemenings te          wette uit deur die verwantskap    tegnologie en evalueer die impak
formuleer.                            tussen verskillende feite         (voordele en nadele) van die
                                      konsepte in eie woorde aan te     verhouding van ŉ persoonlike
                                      dui.                              oogpunt.

AS 12.1.3                             AS 12.2.3                         AS 12.3.3
Apply known problem-solving           Apply scientific knowledge in     Evaluate the impact of scientific and
strategies to solve multi-step        everyday life contexts.           technological knowledge on
problems.                                                               sustainable development of
                                                                        resources and suggest long-term
                                                                        and short-term strategies to improve
                                                                        the management of resources in the
                                                                        environment.
                                                                        Evalueer die impak van
Pas bekende                           Pas wetenskaplike kennis in       wetenskaplike en tegnologiese
probleemoplossingstrategieë toe       kontekste van die alledaagse      kennis op volhoubare ontwikkeling
om veelvuldige-stapprobleme op        lewe toe.                         van bronne en om kort-termyn en
te los.                                                                 lang-termyn strategieë voor te stel
                                                                        om die bestuur van bronne in die
                                                                        omgewing te verbeter.
AS 12.1.4
Communicate and present
scientific arguments with clarity
and precision.
Kommunikeer en verdedig
wetenskaplike argument duidelik
en presies.

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(SEPTEMBER 2012)               FISIESE WETENSKAPPE V1 (Memo)                                      3
 GUIDELINES FOR MARKING/RIGLYNE VIR NASIEN
 This section provides guidelines for the way in which marks will be allocated. The
 broad principles must be adhered to in the marking of Physical Sciences tests and
 examinations.
 Hierdie afdeling verskaf riglyne vir die manier waarop punte toegeken sal word.
 Die breë beginsels moet tydens die nasien van Fisiese Wetenskappe toetse en
 eksamens gevolg word.
 1.1    MARK ALLOCATION/PUNTE TOEKENNING
        1.1.1      Calculations/Berekeninge:
                   o    Marks will awarded for: correct formula, correct substitution, correct
                        answer with unit.
                   o    Punte sal toegeken word vir: korrekte formule, korrekte substitusie,
                        korrekte antwoord met eenheid.
                   o    No marks will be awarded if an incorrect or inappropriate formula is
                        used, even though there may be relevant symbols and applicable
                        substitutions.
                   o    Geen punte sal toegeken word waar ŉ verkeerde of ontoepaslike
                        formule gebruik word nie, selfs al is daar relevante simbole en
                        relevante substitusies.
        1.1.2      Explanations and interpretations/Verduidelikings en interpretasie:
                   Allocation of marks to questions requiring interpretation or explanation
                   e.g. AS 1.4, 2.2, 2.3, 3.1, 3.2 and 3.3, will differ and may include the use
                   of rubrics, checklists, memoranda, etc. In all such answers emphasis
                   must be placed on scientific concepts relating to the question.
                   Toekenning van punte by vrae wat interpretasie of verduideliking vereis
                   bv. AS 1.4, 2.2, 2.3, 3.1, 3.2 en 3.3, sal verskil en mag die gebruik van
                   rubrieke, kontrolelyste, memoranda, ens. insluit. By al hierdie antwoorde
                   moet die beklemtoning op die wetenskaplike konsepte, met betrekking tot
                   die vraag, val.

 1.2    FORMULAE AND SUBSTITUTIONS/FORMULES EN SUBSTITUSIE
        1.2.1      Mathematical manipulations and change of subjects of appropriate
                   formulae carry no marks, but if a candidate starts 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 manipulering en verandering van die voorwerp van
                   toepaslike formules dra geen punte nie, maar as ŉ kandidaat begin met
                   die korrekte formule en dan die voorwerp van die formule verkeerd
                   uitwerk, sal punte vir die formule en korrekte substitusie toegeken word.

        1.2.2      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 substitusie
                   toegeken word, maar geen verdere punte sal toegeken word nie.

        1.2.3      Marks are only awarded for a formula if a calculation had been
                   attempted, i.e. substitutions have been made or a numerical answer
                   given.
                   Punte sal slegs toegeken word vir ŉ formule as ŉ poging aangewend
                   was om ŉ berekening te doen d.w.s. substitusie was gedoen of ŉ
                   numeriese antwoord word verskaf.

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4                            PHYSICAL SCIENCES P1 (Memo)                          (SEPTEMBER 2012)

      1.2.4   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 is en nie vir waardes wat voor ŉ berekening gelys is
              nie.

      1.2.5   All calculations, when not specified in the question, must be done to two
              decimal places.
              Alle berekenings, wanneer nie in die vraag gespesifiseer word nie, moet
              tot twee desimale plekke gedoen word.

1.3   UNITS/EENHEDE

      1.3.1   Candidates will only be penalised once for the repeated use of an
              incorrect unit within a question or sub-question.
              ŉ Kandidaat sal slegs een keer gepenaliseer word vir die herhaaldelike
              gebruik van ŉ verkeerde eenheid in ŉ vraag of subvraag.

      1.3.2   Units are only required in the final answer to a calculation.
              Eenhede word slegs in die finale antwoord tot ŉ vraag verlang.

      1.3.3   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 ŉ antwoord en vir ŉ 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

      1.3.4   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.
              (This instruction only applies to Paper 1).
              SI-eenhede moet gebruik word behalwe in sekere gevalle, bv. V.m-1 in
              plaas van of N.C-1, en cm•s-1 of km.h-1 in plaas van m•s-1 waar die vraag
              dit verlang. (Hierdie instruksie geld slegs by Vraestel 1).

1.4   POSTIVE 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:

      1.4.1   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 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 aan die daaropvolgende subvraag
              toegeken.

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(SEPTEMBER 2012)              FISIESE WETENSKAPPE V1 (Memo)                                  5

        1.4.2      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.
                   ŉ Vraag met veelvuldige stappe in ŉ subvraag: Indien ŉ
                   kandidaat byvoorbeeld, die aantal mol verkeerd bereken in ŉ
                   eerste stap as gevolg van ŉ substitusiefout, verloor die kandidaat
                   die punt vir die substitusie sowel as die finale antwoord.

        1.4.3      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 tot 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.

        1.4.4      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 diagram vraag) hoef nie noodwendig altyd dieselfde
                   orde te volg nie. VOLPUNTE sal toegeken word mits dit ŉ geldige
                   oplossing tot die probleem is. Maar, enige berekening wat nie die
                   kandidaat nader aan die antwoord bring as die oorspronklike data,
                   sal geen punte tel nie.

        1.4.5      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.

        1.4.6      Normally, if based on a conceptual mistake, an incorrect answer
                   cannot be correctly motivated. 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 for 3.2 could be considered.
                   Normaalweg, as dit gebaseer is op ŉ voorstellingsfout, kan ŉ
                   verkeerde antwoord nie korrek gemotiveer word nie. As die
                   kandidaat derhalwe gevra word met ŉ vraag in 3.2 om die antwoord
                   in vraag 3.1 te motiveer, en 3.1 is verkeerd, sal geen punte vir
                   vraag 3.2 toegeken word nie. Maar, as die antwoord in bv. 3.1
                   gebaseer is op ŉ berekening, kan die motivering vir die verkeerde
                   antwoord oorweeg word.

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6                            PHYSICAL SCIENCES P1 (Memo)                   (SEPTEMBER 2012)

       1.4.7    If instructions regarding method of answering are not followed, e.g.
                the candidate does a calculation when the instruction was to solve
                by construction and measurement, a candidate may forfeit all
                the marks for the specific question.
                Indien instruksies aangaande metode van beantwoording nie
                gevolg word nie, bv. die kandidaat doen ŉ berekening wanneer die
                instruksie los op deur konstruksie en meting was, mag die
                kandidaat al die punte vir die spesifieke vraag verbeur.

       1.4.8    For an error of principle, no marks are awarded (Rule 1) e.g. If
                the potential difference is 200 V and resistance is 25 Ω, calculate
                the current.
                Vir ŉ foutdraendebeginsel, sal geen punte toegeken word nie
                (Reël 1) bv. as die potensiaalverskil 200 V en die weerstand 25 Ω
                is, bereken die stroom.

                CORRECT/     ANSWER (1)     POSSIBLE     ANSWER (2)     POSSIBLE
                KORREK       ANTW (1)       MOONTLIK     ANTW (2)       MOONTLIK

                                                                               

                      

SECTION/AFDELING A

QUESTION/VRAAG 1:           ONE WORD ITEMS/EENWOORD-ITEMS

1.1    Isolated system/Geïsoleerde sisteem                               [12.2.1]     (1)
1.2    Power/Drywing                                                     [12.2.1]     (1)
1.3    Alternator                                                        [12.2.1]     (1)
1.4    Destructive interference/Destruktiewe interferensie               [12.2.1]     (1)
1.5    Ultraviolet light/Ultravioletlig                                  [12.2.1]     (1)
                                                                                       [5]
QUESTION/VRAAG 2:            MULTIPLE CHOICE QUESTIONS/
                             MEERVOUDIGEKEUSE-VRAE
2.1    C                                                                [12.2.3]      (2)
2.2    C                                                                [12.2.3]      (2)
2.3    D                                                                [12.2.3]      (2)
2.4    C                                                                [12.2.3]      (2)
2.5    B                                                                [12.2.3]      (2)
2.6    A                                                                [12.2.3]      (2)
2.7    B                                                                [12.2.3]      (2)
2.8    D                                                                [12.2.3]      (2)
2.9    C                                                                [12.2.3]      (2)
2.10   A                                                                [12.2.3]      (2)
                                                                                       [20]

                                       TOTAL SECTION/TOTAAL AFDELING A:                25

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(SEPTEMBER 2012)            FISIESE WETENSKAPPE V1 (Memo)                                        7

 SECTION B/AFDELING B

 QUESTION 3/VRAAG 3

 3.1    8m                                                                    [12.2.1]    (1)

 3.2    0,8 s  – the time taken by the ball to rise is equal to the time it takes to
        fall to the same position where it was projected                       [12.2.3]
        tyd op = tyd af vanaf ŉ gegewe punt                                                (2)

 3.3    take upward motion as positive [FOR QUESTION 3.3 to 3.7]
        neem opwaarts as positief [VIR VRAE 3.3 tot 3.7]
        vf = vi + g ∆t 
        0 = vi + (-9,8) (0,4) 
           = vi – 3,92
        vi = 3,92 m•s-1 up/op                                                 [12.2.2]    (3)

 3.4    POSITIVE MARKING FROM 3.3
        POSITIEWE NASIEN VAN 3.3
        ∆y   = vi ∆t + ½ g∆t2 
             = (3,92) (0,4) + ½ (-9,8) ( 0,4)2 
             = 0,784 m 
        The maximum height/maksimum hoogte = 8 m + 0,78
                                                 = 8,78 m                     [12.2.1]
        OR/OF

        vf2 = vi2 + 2g ∆ y 
        (0)2 = (3,92)2 + 2 (-9,8) ∆ y 
        ∆ y = 0,784 m 
        Maximum height/maksimum hoogte = 8 m + 0,78 =8,78 m 

        OR/OF

        ∆ y = (vi + vf) t 
                   2
            = (3,92 + 0) 0,4 
                       2
            = 0,78 m 
        Maximum height/maksimum hoogte = 8 m + 0,78 m
                                         = 8,78 m                                         (4)

 3.5    POSITIVE MARKING FROM 3.3 and 3.4
        POSITIEWE NASIEN VAN 3.3 en 3.4
        from maximum height downwards/van maksimum hoogte afwaarts
         vf2 = vi2 + 2g ∆y 
             = (0)2 + 2 (-9,8) (-8,78) 
          vf = 13,12 m•s-1                                       [12.1.2]

        OR/OF
        From the balcony upwards/van die balkon afwaarts
        vf2 = vi2 + 2a∆y 
             = 3,92 + 2 (- 9,8) (-8) 
         vf = 13,12 m•s-1                                                                 (4)

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8                           PHYSICAL SCIENCES P1 (Memo)                         (SEPTEMBER 2012)

3.6   POSITIVE MARKING FROM 3.3 - 3.5
      POSITIEWE NASIEN VAN 3.3 - 3.5
      From maximum height/van maksimum hoogte
      ∆ y = vi ∆t +1/2 g ∆t2
         - 8,78 = 0 x ∆t + ½ (-9,8) ∆ t2 
           t = 1,34 s
      therefore total time/totale tyd = 0,4 + 1,34 = 1,74 s                   [12.1.2]   (3)

      OR/OF                              OR/OF                   OR/OF
      From maximum height/               From balcony/           From balcony/
       van maksimum hoogte               Van balkon              Van balkon
      vf = vi + a ∆t                    ∆y =       t           vf = vi + a ∆t 
      13,12 = 0 + 9,8 t                                         -13,12 = +3,92 +
           t = 1,34 s                    -8 =              t    (-9,8)t 
      total t = 0,4 + 1,34 = 1,74 s     t = 1,74 s                  t = 1,74 s 

3.7   The graph of acceleration vs. time/Die grafiek van versnelling vs. tyd
      One mark for labelling both axes/een punt vir byskrifte van beide asse
      One mark for the shape of the graph/een punt vir die vorm van grafiek

                                                                    [12.1.2]               (2)
                                                                                          [19]
QUESTION 4/VRAAG 4

4.1   mechanical energy/meganiese energie                                     [12.1.2]   (1)

4.2   4.2.1    The speed of the ball at A & B is the same because there is no
               external force acting on the ball, the surface is frictionless  or
               Newton’s first law of motion.                                 [12.3.1]
               Die spoed van die bal by A & B is gelyk  want daar is geen
               eksterne krag wat op die bal inwerk nie, die oppervlak is
               wrywingloos  of Newton se eerste wet                                      (2)

      4.2.2    The ball is moving at a lower speed at B than at A  because
               the rough surface has an external force which is frictional force
                acting on the ball                                       [12.3.2]
               Die bal se spoed is laer by B as by A want die oppervlak is grof
               en ŉ eksterne krag, wrywing,  werk op die bal in                          (2)

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(SEPTEMBER 2012)           FISIESE WETENSKAPPE V1 (Memo)                                     9

 4.3    Investigation 1/Ondersoek 1
        Since/Omdat vA = vB
        which implies that/daarom EkA =EkB and since/en omdat EPA = EPB 
        Mechanical energy at A = Mechanical energy at B 
        Meganiese energie by A = Meganiese energie by B
        Investigation 2/ondersoek 2
        vA > vB
        therefore/daarom EKA ≠ EKB  but EPA = EPB
        Mechanical energy at A is not equal to mechanical at B because there is
        friction and mechanical energy is not conserved.               [12.3.2]
        Meganiese energie by B is nie gelyk aan meganiese energie by A want
        daar is wrywing                                                               (4)

 4.4    The principle of conservation of mechanical  energy states that in an
        isolated system total mechanical energy is conserved              [12.2.1]
        Beginsel van behoud van meganiese energie stel dat in ŉ geslote sisteem
        totale meganiese energie behoue sal bly                                        (2)
                                                                                      [11]
 QUESTION 5/VRAAG 5

 5.1    The free body diagram/Vryeliggaamdiagram
        One mark for each force indicated/Een punt vir elke krag

        Fa- force applied/toegepaste krag
        f- frictional force/wrywing
        N – normal force/normaalkrag
        fg┴ _ ┴ component of gravitational force          Fg - gravitational
        fg//- mgsin Ɵ║component of gravitational force OR force/gravitasiekrag
                                                                          [12.2.3]    (4)

 5.2    Wnet = ∆ Ek 
             = ½ m (22 – 22)
             =0J                                                          [12.2.3]   (2)

 5.3    FNET = ma = 0 
        Fnet = Fa + mg sin Ɵ + f = 0
         Fa = 400 x 9,8 sin 15° + 1400 
             = 2414,57 N                                                  [12.1.3]   (4)

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10                             PHYSICAL SCIENCES P1 (Memo)                      (SEPTEMBER 2012)

5.4   P=Fv
       = 2 414,57 X 2 
       = 4 829,14 W                                                          [12.1.3]    (3)
5.5   sin 15° =h/6 
            h = 1,55 m                                                       [12.2.3]    (2)
5.6   5.6.1    Wclimber = Whiker /Wklimmer = W stapper
      -        The magnitude of their mass is the same, the vertical height of
               the mountain they are climbing is the same 
               Die grootte van hul massas is gelyk, die vertikale hoogte van die
               berg is dieselfde
               E = mgh = W                                               [12.3.2]         (2)
      5.6.2    The hiker has more power/Die stapper het meer drywing 
               The time taken by the hiker is less than the time taken by the
      -
               climber and/die stapper neem korter tyd as die klimmer en
               P =W/t
               Therefore power is inversely proportional to time. 
               Daarom is drywing omgekeerd eweredig aan tyd              [12.3.2]          (2)
                                                                                          [19]
QUESTION 6/VRAAG 6
6.1   moving away/beweeg weg                                                 [12.2.2]    (1)
6.2   Doppler effect                                                         [12.2.3]    (1)
6.3   fL = fS ( v + vL/v + vS) 
           700  = 900 (343/343 + vs) 
      vs = 98 m•s-1                                                          [12.2.3]    (4)
6.4   6.4.1   Controlled variables (any two)/
              Beheerbare veranderlikes (enige twee)
              - direction the sound source moves/rigting waarin klankbron
              beweeg
              - frequency of the sound source/frekwensie van die klankbron
              - speed of the listener (stationary)/spoed van die luisteraar
              (stilstaande)
              - speed of sound or air temperature/spoed van klank of lug temp.
                                                                         [12.1.2]         (2)
      6.4.2                                                                  [12.1.2]
                                         one mark for f (Hz) on y-axis/
                      f (Hz)
                                         een punt vir f (Hz)op die y-as

                                         one mark for v (m.s-1) on x-axis/
                                         een punt vir v (m.s-1)op die x-as

                                           v (m.s-1)
                                                                                          (2)
      6.4.3   Reject the hypothesis, the higher the frequency the stationery
              observer hears  because the source moves towards the
              stationary observe                                       [12.3.2]
              Verwerp die hipotese, hoe hoër die frekwensie wat ŉ stilstaande
              luisteraar hoor want die klankbron beweeg na die luisteraar                  (1)
                                                                                          [11]

### Page 11

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(SEPTEMBER 2012)            FISIESE WETENSKAPPE V1 (Memo)                                   11

 QUESTION 7/VRAAG 7
 7.1    Huygen’s principle states that every point in a wave front acts as a
        secondary wavelet that spreads out in all directions with the same speed
        as the wave.                                                     [12.2.1]
        Huygens se beginsel verklaar dat elke punt op ŉ golffront reageer soos
        ŉ bron van sekondêre golfies wat in alle rigtings met dieselfde spoed as
        die golf uitsprei.                                                                (2)
 7.2    Diffraction pattern B/Diffraksiepatroon B 
        Because yellow light has a longer wavelength therefore the longer the
-
        wavelength the wider the diffraction pattern.  Diffraction pattern B is
        more spread out than diffraction pattern A.                         [12.2.3]
        Geel lig het ŉ langer golflengte en daarom ŉ breër diffraksie patroon.
        Diffraksiepatroon B is meer uitgesprei as diffraksiepatroon A

        OR diffraction is directly proportional to wavelength
        OF diffraksie is direk eweredig aan golflengte                                    (3)
 7.3    Sin Ɵ =m
               = 1 x 460 x 10-9 
                    5 x 10-6
             Ɵ = 5,28°                                                       [12.1.3]    (3)
 7.4    The smaller the width of the slit, the wider the diffraction pattern /the
        dark bands will be further from the centre of the screen
-       Smaller spleet, breër die diffraksiepatroon/die donker bande sal verder
        van die uitmekaar wees
        because diffraction is inversely proportional to slit width. 
        want diffraksie is omgekeerd eweredig aan spleetwydte                  [12.2.2]    (2)
                                                                                          [10]
 QUESTION 8/VRAAG 8
 8.1    –By preventing the harmful UV-rays of the sun from penetrating the skin 
        Deur te voorkom dat die skadelike UV-strale van die son die vel
        binnedring
        –To prevent sunburn /Voorkom sonbrand
        and permanent damage to the skin/skin cancer en permanente
        velskade/velkanker                                              [12.3.2]         (3)
 8.2    Gamma rays/Gammastrale                                               [12.2.1]    (1)
 8.3    X-rays have a higher penetrating ability than radio waves because of their
        higher frequency. (E α f)/
        X-strale het ŉ hoë indringingsvermoë as radiogolwe as gevolg van hul hoë
        frekwensie (E α f)
        X-rays do not penetrate through lead  so the radiologist will be safe from
        harmful effects of x-rays that cause cancer.
        X-strale kan nie deur lood beweeg nie daarom is die radioloog veilig van
        die skadelike effek van x-strale wat kanker kan veroorsaak          [12.3.2]      (2)
 8.4          c=f    
              8    8
        3 x 10 = 10 
                =3m                                                          [12.2.3]    (3)
                                                                                          [9]

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12                            PHYSICAL SCIENCES P1 (Memo)                      (SEPTEMBER 2012)

QUESTION 9/VRAAG 9

9.1    Coulomb’s law of electrostatics states that the electrostatic force
       between two charges is directly proportional to the product of the charges
        and inversely proportional to the square of the distance between the
       charges. 
       Die elektrostatiese krag tussen twee ladings is direk eweredig aan die
       produk van die ladings en omgekeerd eweredig aan die kwadraat van die
       afstand tussen die ladings                                          [12.2.1]      (2)

9.2    one mark for the pattern/een punt vir die patroon
       one mark for the direction of field lines/een punt vir die rigting van die
       veldlyne

                                                                              [12.1.2]   (2)

9.3    F1 = k q1 q 
                  r2
              = 9 x 109 x 5 x 10-9 x q 
                            (0,8)2
       F2 = k q q2
               r2
         = 9 x 109 x q x 3 x 10-9 
                     (0,2)2
       Fnet = F1 + F2                        
       12 x10-4 = 9 x 109 x 5 x 10-9 x q + 9 x 109 x q x 3 x 10-9
                               (0,8)2                 (0,2)2
       12 x10-4 = (45 q) (0,2)2 + (27 q) (0,8)2
                                  0,0256
       12 x10-4 =1,8 q + 17,28 q
                          0,0256
       30,72 x 10-6 = 19,08 q
              q = 1,61 x 10-6 C  (positive/positief)                         [12.1.3]    (6)
                                                                                         [10]
QUESTION 10/VRAAG 10

10.1   one mark for the direction of field lines/een punt vir die
       rigting van die veldlyne
       one mark for the uniform field between the plates/een
       punt vir die uniforme veld tussen die plate
       one mark for non-uniform field near the ends/een punt
       vir die nie-uniforme veld naby die kante                                          (3)

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(SEPTEMBER 2012)             FISIESE WETENSKAPPE V1 (Memo)                             13

 10.2   10.2.1      C = Q/V 
                      = 3 x 10-6 
                            1,5
                      = 2 x 10-6 F                                       [12.2.3]   (3)

        10.2.2      [POSITIVE MARKING FROM 10.2.1]
                    [POSITIEWE NASIEN VAN 10.2.1]
                    C= 0A
                          d
                     2 x 10-6 = 8,85 x 10-12 (0,2 x 3) 
                                           d
                            d = 2,66 x 10-6 m                            [12.2.3]   (3)
                                                                                     [9]
 QUESTION 11/VRAAG 11

 11.1        1/R║ = 1/r1 + 1/(r2 + r3) 

                    = ¼ + 1/(10 +6) 
                   R = 3,2 Ω                                              [12.1.3]   (2)

 11.2   11.2.1      V1 = V ║ =IR║           OR/OF Current through 6 Ω resistor/
                             = 6 x 3,2            Stroom deur 6 Ω resistor
                             = 19,2 V             = 1/5 of/van 6 A 
                     I = /R = 19,2/16 
                        V
                                                  = 1,2 A
                             = 1,2 A

                    Which implies that/Wat impliseer dat V2 =I6Ω x R6Ω
                                                            = 1,2 x 6 
                                                            = 7,2 V      [12.1.3]   (4)

        11.2.2      Vlost = 24 – 19,2 = 4,8 V 
                    r = Vlost/Itot 
                      = 4,8/6
                       = 0,8 Ω 

                    OR/OF

                    emf = I (R + r) 
                    24 = 6 ( 3,2 + r ) 
                      r = 0,8Ω                                           [12.1.3]   (3)
                                                                                     [9]

### Page 14

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14                           PHYSICAL SCIENCES P1 (Memo)                           (SEPTEMBER 2012)

QUESTION 12/VRAAG 12
12.1   vrms = vmax/                                 12.2     Vmax = I max R 
       172 = vmax/                                         243,24 = 20 I 
       Vmax = 243,24 V          [12.2.3]      (3)            I max = 12,16 A    [12.1.3]   (3)
12.3   Pave = Vrms2/R                               OR/OF Pave = Vrms Irms
                  = (172)2                                     = 172 x 8,60
                      20                                        =1479,2 W
                  = 1479,2 W       [12.1.3]
                                                     Irms = Imax/
       OR/OF
       Pave = (Irms)2 R                                   = 12,16/
            = (8,60)2 x 20                                = 8,60 A
            = 1479,2 W                                                                       (3)
                                                                                             [9]
QUESTION 13/VRAAG 13
13.1   Photoelectric effect/fotoëlektriese effek                                 [12.2.1]   (1)
13.2   Intensity/intensiteit                                                     [12.2.2]   (1)
13.3   Increase/toeneem 
-      because when intensity is increased more electrons are ejected per
       second  causing the flow of current to increase as well as the reading on
       the ammeter increases. (I = Q/t)
       wanneer intensiteit verhoog word meer elektrone per sekonde vrygestel,
       wat veroorsaak dat die stroom vergroot en ook die lesing op die ammeter
       (I = Q/t)                                                          [12.2.2]           (2)
13.4   E = W0 + Ek 
           h c/ = W0 + Ek
        6,63 x 10-34 x 3 x 108  = 6,63 x 10-34 x f0 + 2,8 x 10-18 
                       60 x 10-9
                     f0 = 7,77 x 1014 Hz 
       OR/OF
       E = W0 + Ek 
       6,63 x 10-34 x 3 x 108 = W0+ 2,8 x 10-18 
                       60 x 10-9
       W0      = 5,15 x 10-19 J
       But/Maar W0 = h f0
         5,15 x 10-19 = 6,63 x 10-34 f0
                    f0 = 7,7 7 x 1014 Hz                                         [12.2.3]   (4)
13.5   Learners should use solar                The photon of light from the sun will
       cells/photo cells to make                knock off electrons on the surface of
       solar  panels./                         silicon metal in the solar cell and
       Leerders kan                   OR/       electricity will be produced./
       sonselle/fotoselle gebruik     OF        Fotone lig van die son sal elektrone
       om sonpanele te maak                     van die oppervlak van silikon in die
                                                sonsel vrystel en elektrisiteit word
                                                opgewek                          [12.3.2]    (1)
                                                                                             [9]
                                     TOTAL SECTION B/TOTAAL AFDELING B:                      125
                                            GRAND TOTAL/GROOTTOTAAL:                         150

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