# Physical Sciences Grade 12 P1 Mpumalanga November 2017 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Mpumalanga November 2017, 16 pages. Read online or download the free PDF.

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- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2017
- Exam period: Mpumalanga November
- Paper: 1
- Curriculum: CAPS
- Pages: 16

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Physical Sciences · Grade 12 · Mpumalanga November · 2017. Memorandum, 16 pages. Read online or download the PDF.

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          NATIONAL SENIOR CERTIFICATE EXAMINATION

                                        GRADE 12

                        PHYSICAL SCIENCES: PHYSICS (P1)
                                        SEPTEMBER 2017
                                         MEMORANDUM

MARKS/PUNTE: 150

                          This memorandum consists of 16 pages

                       Hierdie memorandum bestaan uit 16 bladsye

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Physical Sciences P1                          2                       MDO/September 2017
                                        NSC-Memorandum

GENERAL GUIDELINES
1. CALCULATIONS
1.1   Marks will be awarded for: correct formula, correct substitution, correct
      answer with unit.
      Punte sal toegeken word vir: korrekte formule, korrekte substitusie, korrekte
      antwoord met eenheid.

1.2    No marks will be awarded if an incorrect or inappropriate formula is used,
       even though there may be relevant symbols and applicable substitutions.
       Geen punte sal toegeken word waar 'n verkeerde of ontoepaslike formule
       gebruik word nie, selfs al is daar relevante simbole en relevante substitusies.

1.3    When an error is made during substitution into a correct formula, a mark
       will be awarded for the correct formula and for the correct substitutions, but
       no further marks will be given.
       Wanneer 'n fout gedurende substitusie in 'n korrekte formule begaan word, sal
       'n punt vir die korrekte formule en vir korrekte substitusies toegeken word, maar
       geen verdere punte sal toegeken word nie.

1.4    If no formula is given, but all substitutions are correct, a candidate will
       forfeit one mark.
       Indien geen formule gegee is nie, maar al die substitusies is korrek, verloor die
       kandidaat een punt.

1.5    No penalisation if zero substitutions are omitted in calculations where
       formula/principle is given correctly.
       Geen penalisering indien nulwaardes nie getoon word nie in berekeninge waar
       die formule/beginsel korrek gegee is nie.

1.6    Mathematical manipulations and change of subject of appropriate formulae
       carry no marks, but if a candidate starts off with the correct formula and then
       changes the subject of the formula incorrectly, marks will be awarded for the
       formula and the correct substitutions. The mark for the incorrect numerical
       answer is forfeited.
       Wiskundige manipulasies en verandering van die onderwerp van toepaslike
       formules tel geen punte nie, maar indien 'n kandidaat met die korrekte formule
       begin en dan die onderwerp van die formule verkeerd verander, sal punte vir die
       formule en korrekte substitusies toegeken word. Die punt vir die verkeerde
       numeriese antwoord word verbeur.

1.7    Marks are only awarded for a formula if a calculation had been attempted.
       i.e. substitutions have been made or a numerical answer given.
       Punte word slegs vir 'n formule toegeken indien 'n poging tot 'n berekening
       aangewend is, d.w.s. substitusies is gedoen of 'n numeriese antwoord is gegee.

1.8    Marks can only be allocated for substitutions when values are substituted into
       formulae and not when listed before a calculation starts.
       Punte kan slegs toegeken word vir substitusies wanneer waardes in formule
       ingestel is en nie vir waardes wat voor 'n berekening genoem is nie.

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Physical Sciences P1                          3                      MDO/September 2017
                                        NSC-Memorandum

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

1.10   If a final answer to a calculation is correct, full marks will not automatically be
       awarded. Markers will always ensure that the correct/appropriate formula is
       used and that workings, including substitutions, are correct.
       Indien 'n finale antwoord tot 'n berekening korrek is, sal volpunte nie outomaties
       toegeken word nie. Nasieners sal altyd verseker dat die korrekte/toepaslike
       formule gebruik word en dat bewerkings, insluitende substitusies korrek is.

1.11   Questions where a series of calculations have to be made (e.g. a circuit
       diagram question) do not necessarily always have to follow the same order.
       FULL MARKS will be awarded provided it is a valid solution to the problem.
       However, any calculation that will not bring the candidate closer to the answer
       than the original data, will not count any marks.
       Vrae waar 'n reeks berekenings gedoen moet word (bv. 'n stroomdiagramvraag)
       hoef nie noodwendig dieselfde volgorde te hê nie. VOLPUNTE sal toegeken word
       op voorwaarde dat dit 'n geldige oplossing vir die probleem is. Enige berekening
       wat egter nie die kandidaat nader aan die antwoord as die oorspronklike data bring
       nie, sal geen punte tel nie.

2. UNITS/EENHEDE
2.1   Candidates will only be penalised once for the repeated use of an incorrect
      unit within a question or subquestion.
      'n Kandidate sal slegs een keer gepenaliseer word vir die herhaaldelike gebruik
      van 'n verkeerde eenheid in 'n vraag of subvraag.

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

2.3    Marks are only awarded for an answer, and not for a unit per se. Candidates
       will therefore forfeit the mark allocated for the answer in each of the following
       situations:
       - Correct answer + wrong unit
       - Wrong answer + correct unit
       - Correct answer + no unit
       Punte word slegs vir 'n antwoord en nie vir 'n eenheid per se toegeken nie.
       Kandidate sal derhalwe die punt vir die antwoord in die volgende gevalle verbeur:
       - Korrekte antwoord + verkeerde eenheid
       - Verkeerde antwoord + korrekte eenheid
       - Korrekte antwoord + geen eenheid

2.4    SI units must be used except in certain cases, e.g. V·m -1 instead of N·C-1, and
       cm·s-1 or km·h-1 instead of m·s-1 where the question warrants this.
       SI-eenhede moet gebruik word, behalwe in sekere gevalle, bv. V∙m -1 in plaas van
       N∙C-1, en cm∙s-1 of km∙h-1 in plaas van m∙s-1 waar die vraag dit regverdig.

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Physical Sciences P1                          4                        MDO/September 2017
                                        NSC-Memorandum

3. GENERAL/ALGEMEEN
3.1   If one answer or calculation is required, but two given by the candidate, only
      the first one will be marked, irrespective of which one is correct. If two
      answers are required, only the first two will be marked, etc.
      Indien een antwoord of berekening verlang word, maar twee word deur die
      kandidaat gegee, sal slegs die eerste een nagesien word, ongeag watter een
      korrek is. Indien twee antwoorde verlang word, sal slegs die eerste twee nagesien
      word, ens.
3.2    For marking purposes, alternative symbols (s,u,t, etc.) will also be accepted.
       Vir nasiendoeleindes sal alternatiewe simbole (s, u, t, ens.) ook aanvaar word.
3.3    Separate compound units with a multiplication dot, not a full stop, for
       example, m·s-1. For marking purposes m.s-1 and m/s will also be accepted.
       Skei saamgestelde eenhede met 'n vermenigvuldigpunt en nie met 'n punt nie,
       byvoorbeeld, m·s-1. Vir nasiendoeleindes sal m.s-1 em m/s ook aanvaar word.
4. POSITIVE MARKING/POSITIEWE NASIEN
      Positive marking regarding calculations will be followed in the following cases:
      Positiewe nasien met betrekking tot berekenings sal in die volgende gevalle geld:
4.1    Subquestion to subquestion: When a certain variable is calculated in one
       subquestion (e.g. 3.1) and needs to be substituted in another (3.2 or 3.3), e.g.
       if the answer for 3.1 is incorrect and is substituted correctly in 3.2 or 3.3, full
       marks are to be awarded for the subsequent subquestions.
       Subvraag na subvraag: Wanneer 'n sekere veranderlike in een subvraag (bv.
       3.1) bereken word en dan in 'n ander vervang moet word (3.2 of 3.3), bv. indien die
       antwoord vir 3.1 verkeerd is en korrek in 3.2 of 3.3 vervang word, word volpunte
       aan die daaropvolgende subvraag toegeken.
4.2    A multistep question in a subquestion: If the candidate has to calculate, for
       example, current in the first step and gets it wrong due to a substitution error,
       the mark for the substitution and the final answer will be forfeited.
       'n Vraag met veelvuldige stappe in 'n subvraag: Indien 'n kandidaat
       byvoorbeeld, die stroom verkeerd bereken in 'n eerste stap as gevolg van 'n
       substitusiefout, verbeur die kandidaat die punt vir die substitusie sowel as die finale
       antwoord.
5.     NEGATIVE MARKING/NEGATIEWE NASIEN
       Normally an incorrect answer cannot be correctly motivated if based on a
       conceptual mistake. If the candidate is therefore required to motivate in
       question 3.2 the answer given to question 3.1, and 3.1 is incorrect, no marks
       can be awarded for question 3.2. However, if the answer for e.g. 3.1. is based
       on a calculation, the motivation for the incorrect answer in 3.2 could be
       considered.

       'n Verkeerde antwoord, indien dit op 'n konsepsuele fout gebaseer is, kan
       normaalweg nie korrek gemotiveer word nie. Indien 'n kandidaat gevra word om in
       vraag 3.2 die antwoord op vraag 3.1 te motiveer en 3.1 is verkeerd, kan geen
       punte vir vraag 3.2 toegeken word nie. Indien die antwoord op bv. 3.1 egter op 'n
       berekening gebaseer is, kan die motivering vir die verkeerde antwoord in 3.2
       oorweeg word.

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Physical Sciences P1                          5                        MDO/September 2017
                                        NSC-Memorandum

QUESTION 1 / VRAAG 1

1.1     B 
1.2     A 
1.3     C 
1.4     C 
1.5     D 
1.6     B 
1.7     D 
1.8     D 
1.9     C 
1.10    A                                                                                 [20]

QUESTION 2 / VRAAG 2

2.1    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.                                      (2)

2.2    Accepted labels / Aanvaarde benoemings
       W     Fg / Fw/force of earth on block/weight / 49 N / mg / gravitational force
       T1    Tension in rope 1 / FT1
       T2    Tension in rope 2 / FT2

                        T2 

                  T1      w

                                                                                                    (3)
       Notes/Aantekeninge:
           Any additional forces: deduct 1 mark: max 2⁄3
           No arrows: 0⁄3
           Force(s) not touching object: deduct 1 mark: max 2⁄3

2.3    T2                                                                                          (1)

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Physical Sciences P1                          6                          MDO/September 2017
                                        NSC-Memorandum

2.4    For 5 kg block:                                OPTION 2: Systems approach: Max 4⁄6
       Fnet = ma                                     Fnet = ma 
       T2 + (-Fg) + (-T1) = ma                        120-8(9,8) = 8a
         120 – 5(9,8) – T1 = 5a                      a = 5,2 m·s-2
       71 – T1 = 5a
                                                      T1 -3(9,8) =3a
       For 3 kg block:    Any one: 5a or 3a              T1 = 3(9,8) + 3(5,2) 
       Fnet = ma                                         T1 = 45 N 
       T1 + (-Fg) = ma
        T1 - 3(9,8) = 3a                             String will NOT break 
       T1 – 29,4 = 3a
                  OR
       71-29,4 = 8a            71 – T1 = 5a …….i
       a = 5,2 m·s-2           T1 – 29,4 = 3a …..ii

       T1 = 29,4 + 3(5,2)      i×3: 213-3T1 = 15a
          = 45 N              ii×5: 5T1-147=15a
                                      -8T1 = -360
       String will NOT                 T1 = 45 N 
       break 
                               String will NOT
                               break                                                                 (6)

2.5    Remains the same  / Bly dieselfde                                                              (1)
                                                                                                      [13]

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Physical Sciences P1                          7                       MDO/September 2017
                                        NSC-Memorandum

QUESTION 3
3.1 Motion under the influence of the force of gravity only. 
    Beweging as slegs gravitasiekrag op die voorwerp inwerk.                                       (2)

3.2.1 Fnet = (0,5)(9,8) = 4,9 N                                                                   (1)

3.2.2 Downwards                                                                                   (1)

3.3.1 Upwards positive:                            Downwards positive:

       Δy = viΔt + ½aΔt2                          Δy = viΔt + ½aΔt2 
       -21 = vi(2,88) + ½(-9,8)(2,88)2           21 = vi(2,88) + ½(9,8)(2,88)2
       vi = 6,82 m·s-1                            vi = -6,82
                                                   vi = 6,82 m·s-1                                (4)

3.3.2 POSITIVE MARKING FROM Q3.3.1
      Upwards positive:                            Downwards positive:

       vf2 = vi2 + 2aΔy                           vf2 = vi2 + 2aΔy 
       02 = (6,82)2 + 2(-9,8) Δy                  02 = (-6,82)2 + 2(9,8) Δy 
       Δy = 2,37 m                                 Δy = -2,37 m

       Max height= 2,37 + 21+0,6= 23,97 m        Max height= 2,37 + 21+0,6= 23,97 m
                                                                                                   (4)

3.4    POSITIVE MARKING FROM Q3.3.1
       Upwards positive:                           Downwards positive:

       vf = vi + aΔt                             vf = vi + aΔt
          = 6,82 + (-9,8)(2,88)                    = -6,82 + (9,8)(2,88) 
          = -21,40 m·s-1         Both formulae     = 21,40 m·s-1           Both formulae 

       Fnet·Δt = mvf - mvi                         Fnet·Δt = mvf - mvi
       (N-Fg) (0,1)  = 0,5(18)-0,5(-21,40)       (Fg-N) (0,1)= 0,5(-18) – 0,5(21,40) 
       N-(0,5)(9,8) = 197                          (0,5)(9,8) - N = -197
       N = 201,9 N                                N = 201,9 N 
                                                                                                    (5)
                                                                                                   [17]

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Physical Sciences P1                          8                   MDO/September 2017
                                        NSC-Memorandum

QUESTION 4 / VRAAG 4
4.1 The energy that an object has due to its height above the ground / a reference
    point 
    Die energie wat 'n voorwerp het as gevolg van die hoogte bo die grond / bo 'n
    verwysingspunt.                                                                            (2)

4.2.1 Ep = mgh 
         = (5,05)(9,8)(0,06) 
         = 2,97 J                                                                             (3)

4.2.2 USE ENERGY PRINCIPLES ONLY. ANY OTHER METHOD: 0⁄
                                                             4
      POSITIVE MARKING FROM Q4.2      Wnet = ΔEk 
      (mgh + ½mv2)R = (mgh + ½mv2)S  (5,05×9,8)(0,06)Cos180°=0-½(5,05)v2 
        0 + ½(5,05)v  = 2,97  + 0
                    2                 v = 1,08 m·s-1 
                   v = 1,08 m·s-1                                                             (4)

4.2.3 POSITIVE MARKING FROM Q4.3
                 Σpi = Σpf
                                    Any one 
      (mvi)1 + (mvi)2 = (m1+m2)vf
      (0,05)vi + 0= (5,05)(1,08) 
             vi = 109,08 m·s-1 
                                                                                                (4)
                                                                                               [13]

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Physical Sciences P1                          9                         MDO/September 2017
                                        NSC-Memorandum

QUESTION 5 / VRAAG 5

5.1    The contact force which a surface exerts on an object  and which is perpendicular
       to the surface. 
       Die kontakkrag wat 'n oppervlak op 'n voorwerp uitoefen en wat loodreg op die
       oppervlak is.                                                                                 (2)

5.2

                           N

                        w

                                                                                                     (2)
       Accepted labels / Aanvaarde benoemings
       w    Fg / Fw/force of earth on block/weight / 49 N / mg / gravitational force
       N    Normal force / FN / Force of incline on trolley

5.3    OPTION 1
       (mgh + ½mv2)X = (mgh + ½mv2)Y 
         (76)(9,8)(1,5) + ½(76)(3)2  = 0 + ½(76)v2 
                      v = 6,2 m·s-1 

       OPTION 2
       Wnet = ΔEk 
       W w = ½mvf2 – ½mvi2
       (76×9,8)(1,5)Cos 0° = ½(76)vf2 – ½(76)(3)2
               vf = 6,2 m·s-1 

       OPTION 3
       Wnc = ΔEk + ΔEp 
       0 = (½mvf2 – ½mvi2) + (mghf - mghi)
       0 = [½(76)vf2 – ½(76)(3)2]  + [0 - (76)(9,8)(1,5)] 
               vf = 6,2 m·s-1 
                                                                                                     (4)

5.4    fk = μk·N 
          = μk·mg Cos Θ
          = (0,21)(76×9,8×Cos 10°) 
          = 154,03 N                                                                                (3)

5.5    The normal force is perpendicular to the dispacement / motion. 
       OR
       Θ = 90° → Cos 90° = 0 
       OR
       WN = N∆x Cos90° = 0                                                                          (1)

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Physical Sciences P1                          10                  MDO/September 2017
                                        NSC-Memorandum

5.6    POSITIVE MARKING FROM Q5.3
       Wnet = ΔEk 
       W w// + Wf = ½mvf2 – ½mvi2
       (76×9,8×Sin10°)(∆x) Cos0° + (154)(∆x) Cos180° = 0 - ½(76)(6,2)2
                129,33∆x – 154∆x = -1460,72
        ∆x = 59,22 m

       h = ∆x Sin10°
         = 59,22 Sin10° 
         = 10,28 m                                                                            (6)

5.7    Remains the same                                                                        (1)
                                                                                               [19]

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Physical Sciences P1                                        11                  MDO/September 2017
                                                      NSC-Memorandum

QUESTION 6

6.1    The number of waves that passes a fixed point  per second 
       Die aantal golwe wat per sekonde by ‘n vaste punt verby beweeg                                        (2)

6.2    Wavelength (λ) in front of source decreases 
       More waves per second reaches the listener 
       Frequency will increase                                                                               (2)

6.3                               𝑣±𝑣
                          𝑓𝐿 = 𝑣±𝑣𝐿 𝑓𝑆 
                                     𝑆

                                    340
       980,6  =                            (945) 
                                   340-vs

                                 vs = 12,34 m·s-1 
                                                                                                             (4)

6.4

        Frequency (Hz)
                         980,6
                          945

                                                      Time (s)
       Criteria for graph
       Horizontal line at 980,6 Hz                                     
       Horizontal line below 945 Hz                                    
       945 Hz correctly indicated                                                                            (3)
                                                                                                             [11]

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Physical Sciences P1                               12                         MDO/September 2017
                                             NSC-Memorandum

QUESTION 7 / VRAAG 7
7.1    The magnitude of the electrostatic force exerted by one point charge (Q1) on
       another point charge (Q2) is directly proportional to the product of the magnitudes of
       the charges  and inversely proportional to the square of the distance (r) between
       them 

       Die grootte van die elektrostatiese krag wat een puntlading (Q1) op 'n ander
       puntlading (Q2) uitoefen, is direk eweredig aan die produk van die groottes van die
       ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen
       hulle.                                                                                             (2)

7.2         Q1 +Q2
       Q=
               2
            20+0
          =        
               2
         = 10 μC  (10×10-6 C)
                                                                                                           (2)

7.3    Weight (Fg / w) 
       Electrostatic force                                                                                (2)

7.4.1 POSITIVE MARKING FROM Q7.2
         kQ1 Q2
      F=        
           r2
             (9×109 )(10×10-6 )(10×10-6 )
         =
                       (0,7)2 
       F = 1,84 N                                                                                         (4)

7.4.2 POSITIVE MARKING FROM 7.4.1
      FE = mg tan7°                                       7°    T
                                                      mg
      1,84 = m(9,8) tan7° 
      m = 1,53 kg 
                                                           FE = 1,84 N
                                                                                                            (3)
                                                                                                           [13]

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Physical Sciences P1                            13                       MDO/September 2017
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QUESTION 8

8.1    The potential difference across a conductor is directly proportional to the current in
       the conductor  at constant temperature. 
       Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by
       konstante temperatuur.                                                                         (2)

8.2 - Smaller than  / A2 > A1
      A2 is in the series circuit and reads the total current, while A1 is in a parallel branch
      of the circuit and reads only a fraction of the total current.                                 (2)

8.3.1 P=I2 R 
      64 = (22) R 
      R = 16 Ω                                                                                       (3)

8.3.2 OPTION 1                                   OPTION 2
      P = VI                                     R//1 : R//2
      64 = V(2)                                   18 : 12 
      V = 32 V                                       3:2

       V2Ω = IR                                    I1 : I2
           = (2)(2)                                 2:3
           =4V                                    2A:3A

       V// = 4 + 32 = 36 V                       ITOT = 2+3  = 5 A 

       V12Ω = IR 
       36 = I(12) 
       I12Ω = 3 A

       ITOT = 2+3 = 5 A                                                                              (4)

8.3.3 POSITIVE MARKING FROM Q8.3.1+Q8.3.2
                      OPTION 1                                    OPTION 2
       1    1    1              R R                               V// = IR 
                        Rp = R 1+R2 
                               1    2                                = (5)(7,2) 
      R p R1 R 2               (18)(12)
                             = 18+12                                = 36 V
             1 1
          =    +            = 7,2 Ω
            18 12                                                 V2,5Ω = IR
      Rp = 7,2 Ω
                          Re = 7,2 + 2,5  = 9,7 Ω                    = (5)(2,5) 
                                                                     = 12,5 V
       Re = 7,2 + 2,5  = 9,7 Ω
                                        V1 = Ve = IR                         
                                           = (5)(9,7)            V1 = 12,5 + 36 = 48,5 V 
       V1 = Ve = IR
          = (5)(9,7)                      = 48,5 V 
          = 48,5 V                                                                                   (5)

8.4    Voltmeter 
       A voltmeter has a very high resistance and doesn't allow any flow of current through
       it                                                                                             (2)
                                                                                                      [18]
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### Page 14

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            Physical Sciences P1                          14                       MDO/September 2017
                                                    NSC-Memorandum

            QUESTION 9 / VRAAG 9

            9.1       The resistance against the flow of charge through the battery                             (1)

            9.2

V (V)

 10

                  •                                       Criteria for graph
                                                          3 or more points plotted correctly 
        8                                                 Only 2 points plotted correctly    
                                                          Straight line drawn through points              (3)
                                   •
        6

                                                     •

        4

        2                                                                    •

        0               0,1            0,2          0,3            0,4           0,5             0,6
                                                                                                                 I (A)

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9.3.1    10 V 
         OR
         Value for y-intercept from graph                                                   (1)

9.3.2    Use ANY 2 points ON THE LINE:

                          ∆V
         Gradient = ∆I
                           7,2 - 9
                     = 0,17 - 0,06 
                     = -16,36                                                              (2)

9.3.3    POSITIVE MARKING FROM 9.3.2
         r = 16,36 Ω  (Positive answer of 9.3.2)                                           (1)
                                                                                            [8]

QUESTION 10 / VRAAG 10

10.1     (Split ring) Commutator                                                           (1)

10.2     Replace commutator with (two) slip rings                                          (1)

10.3.1           I        6,43
          I rms  max  =       = 4,55 A
                   2        2                                                               (3)

10..3.2 POSITIVE MARKING FROM Q10.3.1
        OPTION 1
              V
        Irms  rms 
               R
                  Vrms
        ∴4,55 = 48,4 

         Vrms= 220,22 V

                   Vmax
          Vrms           
                     2
                         Vmax
         220,22 =       
                     2
         Vmax = 311,44 V 

         OPTION 2
         Vmax = ImaxR
              = (6,43)  (48,4) 
              = 311,21 V                                                                    (5)
                                                                                            [10]

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### Page 16

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Physical Sciences P1/Fisiese Wetenskappe V1    16                             September 2017
                                              NSC/
                                              NSS

QUESTION 11 / VRAAG 11

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

11.2.1         hc
         E=       
                λ
               (6,63×10-34 )(3×108 )
           =                           
                   (260×10-9 )
           = 7,65×10-19 J                                                                 (3)

11.2.2 POSITIVE MARKING FROM Q11.2.1
       E = W o + Ek(max) 
       7,65×10-19 = hf0 + ½mv2
       7,65×10-19 = (6,63×10-34)f0 + ½(9,11×10-31)(5×105)2 
       f0 = 9,82×1014 Hz                                                                  (3)
                                                                                           [8]

                                                                  TOTAL/TOTAAL: 150

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