# Physical Sciences Grade 12 P1 Eastern Cape June 2022 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape June 2022 (Afrikaans), 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/phys-sciences-p1-gr12-memo-june-afrikaansenglish/0a20dc30-b20e-41dc-8398-6c09483074a4
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Language: Afrikaans
- Document type: Memorandum
- Year: 2022
- Exam period: Eastern Cape June
- Paper: 1
- Curriculum: CAPS
- Pages: 14

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## About this document

Physical Sciences · Grade 12 · Eastern Cape June · 2022 · Afrikaans. Memorandum, 14 pages. Read online or download the PDF.

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## Matching paper and memo

- [PHYSICAL SCIENCES P1 GR12 QP JUNE2022 Afrikaans](https://www.hlayiso.com/library/physical-sciences-p1-gr12-qp-june2022-afrikaans/7da2a467-33ba-4d42-9de8-a501165ff1e0) — QP · 2022 · Afr

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                           NATIONAL
                      SENIOR CERTIFICATE /
                          NASIONALE
                      SENIOR SERTIFIKAAT

                       GRADE / GRAAD 12

                      JUNE / JUNIE 2022

                 PHYSICAL SCIENCES P1
                   MARKING GUIDELINE
                FISIESE WETENSKAPPE V1
                      NASIENRIGLYN

MARKS/ PUNTE:   150

                 This marking guideline consists of 14 pages./
                  Hierdie nasienriglyn bestaan uit 14 bladsye.

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2                     PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1               (EC/JUNE/JUNIE 2022)

 GENERAL GUIDELINES/ALGEMENE RIGLYNE

 1.    CALCULATIONS/BEREKENINGE

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

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

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

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

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

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

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

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

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(EC/JUNE/JUNIE 2022)        PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                           3

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

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

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

 2.    UNITS/EENHEDE

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

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

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

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

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4                     PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1             (EC/JUNE/JUNIE 2022)

 3. GENERAL/ALGEMEEN

     3.1 If one answer or calculation is required, but two are given by the candidate, only
         the first one will be marked, irrespective of which one is correct. If two answers
         are required, only the first two will be marked, etc.
         Indien een antwoord of berekening verlang word, maar twee word deur die
         kandidaat gegee, sal slegs die eerste een nagesien word, ongeag watter een
         korrek is. Indien twee antwoorde verlang word, sal slegs die eerste twee nagesien
         word, ens.

     3.2 For marking purposes, alternative symbols (s, u, t etc) will also be accepted.
         Vir nasiendoeleindes sal alternatiewe simbole (s, u, t ens.) ook aanvaar word.

     3.3 Separate compound units with a multiplication dot, no a full stop, for example,
         m.s-1.
         For marking purposes, m.s-1 and m/s will also be accepted.
         Skei saamgestelde eenhede met ʼn vermenigvuldigingspunt en nie met ʼn punt
         nie, byvoorbeeld m.s-1. Vir nasiendoeleindes sal m.s-1 en m/s ook aanvaar word.

 4. POSITIVE MARKING/POSITIEWE NASIEN

     Positive marking regarding calculations will be followed in the following cases:
     Positiewe nasien met betrekking tot berekeninge sal in die volgende gevalle geld:

     4.1 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 of 3.3), e.g., if
         the answer for 3.1 is incorrect and is substituted correctly in 3.2 or 3.3, full marks
         are to be awarded for the subsequent 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 vir die daaropvolgende subvraag toegeken.

     4.2 A multistep question in a sub-question: If the candidate has to calculate, for
         example, current in die first step and gets it wrong due to a substitution error, the
         mark for the substitution and the final answer will be forfeited.
         ʼn Vraag met veelvuldige stappe in ʼn subvraag: Indien ʼn kandidaat bv. die
         stroom verkeerd bereken in ʼn eerste stap as gevolg van ʼn substitusiefout,
         verloor die kandidaat die punt vir die substitusie sowel as die finale antwoord.

 5. NEGATIVE MARKING/NEGATIEWE NASIEN

     Normally an incorrect answer cannot be correctly motivated if based on a conceptual
     mistake. If the candidate is therefore required to motivate in QUESTION 3.2 the
     answer given in QUESTION 3.1, and 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 could be considered.
     ʼn Verkeerde antwoord, indien dit op ʼn konsepsuele fout gebaseer is, kan normaalweg
     nie korrek gemotiveer word nie. Indien ʼn kandidaat gevra word om in VRAAG 3.2 die
     antwoord op VRAAG 3.1 te motiveer en 3.1 is verkeerd, kan geen punte vir VRAAG
     3.2 toegeken word nie. Indien die antwoord op bv. 3.1 egter op ʼn berekening
     gebaseer is, kan die motivering vir die verkeerde antwoord in 3.2 oorweeg word.

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(EC/JUNE/JUNIE 2022)          PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                         5

 QUESTION/VRAAG 1

 1.1     C                                                                                     (2)

 1.2     C                                                                                     (2)

 1.3     B                                                                                     (2)

 1.4     D                                                                                     (2)

 1.5     A                                                                                     (2)

 1.6     C                                                                                     (2)

 1.7     B                                                                                     (2)

 1.8     A                                                                                     (2)

 1.9     C                                                                                     (2)

 1.10    A                                                                                    (2)
                                                                                               [20]

 QUESTION/VRAAG 2

 2.1     When a resultant/net force acts on an object, it accelerates in the direction of
         the force. The acceleration is directly proportional to the force and inversely
         proportional to the mass of the object. 
         Wanneer ʼn resulterende/netto krag op ʼn voorwerp inwerk, versnel dit in die
         rigting van die krag. Die versnelling is direk eweredig aan die krag en
         omgekeerd eweredig aan die massa van die voorwerp.                                     (2)

 2.2        OPTION 1 / OPSIE 1                    OPTION 2 / OPSIE 2
                       fk              N               fk                 N

                                             F

                                                                                F// F 
                                                                F⊥
                                W
                                                  F⊥ and / en F// 

         Mark awarded for arrow and label / Punt vir beide pyl en byskrif toegeken.
         Do not penalise for length of arrows since drawing is not drawn to scale
         Moenie vir die lengte van die pyltjie penaliseer nie aangesien skets nie na
         skaal geteken is nie
         Any other additional force(s) / Enige ander addisionele kragte 3⁄4
         If force(s) do not make contact with body. Max. 3⁄4
         Indien krag(te) nie met die voorwerp kontak maak nie. Maks. 3⁄4                         (4)

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6                     PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                        (EC/JUNE/JUNIE 2022)

 2.3    The force that opposes the motion of an object and acts parallel to the
        surface. 
        Die krag wat die beweging van ʼn voorwerp teenstaan en parallel met die
        oppervlak inwerk.                                                                                (2)
 2.4.1 OPTION 1 (Upwards on the plane is positive) / OPSIE 1 (Teen die
       skuinsvlak op as positief)
       Fnet = ma
       Fnet = 0 N            Any one/ Enige een 
       fk – FR – F// = 0
       fk – FR – mg sinθ = 0

        [fk – 240 – (50)(9,8)sin 12˚]  = 0
        fk = 341,88 N                                                                                   (4)
 2.4.2 Positive marking from QUESTION 2.3 / Positiewe nasien vanaf
       VRAAG 2.3
       Marking criteria / Nasienkriteria
       μk = fk
            N             Any one/ Enige een 
               fk
       μk = 𝑚𝑔 𝑐𝑜𝑠 𝜃

                     341,88
        μk = [ (50 𝑥 9,8) 𝑐𝑜𝑠 12˚ ] 

        μk = 341,88
             479,29
        μk = 0,71                                                                                      (4)
                                                                                                       [16]
 QUESTION/VRAAG 3
 3.1    Maximum static force of friction. 
        Maksimum statiese wrywingskrag.                                                                 (1)
 3.2
                   Fapp                       FN 

                                                                        Fstring/ tou 
                                                      fs(max/ maks) 

                                               Fg 
        Mark awarded for arrow and label / Punt vir beide pyl en byskrif.
        Do not penalise for length of arrows since drawing is not drawn to scale.
        Moenie vir die lengte van die pyltjie penaliseer nie aangesien nie na skaal
        geteken nie
        Any other additional force(s) / Enige ander addisionele krag(te) 4⁄5
        If force(s) do not make contact with body. Max. 4⁄5
        Indien krag(te) nie met die voorwerp kontak maak nie. Maks. 4⁄5                                 (5)

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(EC/JUNE/JUNIE 2022)        PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                        7

 3.3.1 OPTION 1 / OPSIE 1
         Fnet = ma
         FT – mg = ma
         Fx – FT – fs(max) = ma Any one / Enige een 
         FT – (5)(9,8) = 0 
                 = 49 N
         Fx = Fapp cos 30˚
             = 65 cos 30˚ 
             = 56,29 N
         56,29 – 49 – fs(max) = 0 
         fs(max)      = 7,29 N 
         OPTION 2/OPSIE 2
         Fnet = ma
         FT - mg = ma             Any one/ Enige een 
         Fx – FT – fs (max) = ma
         [65 cos 30˚ – (5)(9,8)  – fs(max) = 0] 
         56,29 – 49 – fs(max) = 0
         fs(max) = 7,29 N                                                                   (5)
 3.3.2 Vertical forces on metal box are in equilibrium / Vertikale kragte op
       metaalhouer is in ewewig:
       Fg – Fv – FN = 0       Any one / Enige een 
       mg = Fapp sin 30˚ – FN
       mg = [65 sin 30˚ – 36,1] 
       m x 9,8 = 68,6
       m = 7 kg                                                                             (4)
                                                                                            [15]

 QUESTION 4 / VRAAG 4
 4.1     There exists a force of attraction between any two masses in the universe.
         This force is directly proportional to the product of the masses and
         inversely proportional to the square of the distance between their centres. 
         Daar bestaan ʼn aantrekkingskrag tussen enige twee massas in die heelal.
         Die krag is direk eweredig aan die produk van die massas en omgekeerd
         eweredig aan die kwadraat van die afstand tussen hul middelpunte.                    (2)
 4.2     F = Gm1m2 
                d2
           = (6,67 x 10-11)(1,2 x 1018)(3 x 1018) 
                          (2 x 1010)2 
           = 6,0 x 106 N                                                                     (5)
 4.3     LESS THAN 
         Mass of planet P < Mass of planet Q 
         g ∝ Mass
         KLEINER AS
         Massa van planeet P < Massa van planeet Q
         g ∝ Massa                                                                            (2)
                                                                                              [9]

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8                     PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1              (EC/JUNE/JUNIE 2022)

 QUESTION/VRAAG 5

 5.1     Motion of an object upon which the only force acting on it is gravitational
         force. 

         Die beweging waartydens die enigste krag wat op ʼn voorwerp inwerk, die
         gravitasiekrag is.                                                     (2)

 5.2.1         UPWARDS IS POSITIVE /                  DOWNWARDS IS POSITIVE /
               OPWAARTS AS POSITIEF                      AFWAARTS AS POSITIEF
         vf = vi + gΔt                          vf = vi + gΔt 
         0 = 20  + (-9,8)Δt                    0 = -20  + 9,8Δt 
         Δt = 2,04 s                            Δt = 2,04 s                                   (4)

 5.2.2 OPTION 1 / OPSIE 1
       Positive marking from QUESTION 5.2.1 / Positiewe nasien vanaf
       VRAAG 5.2.1
       UPWARDS IS POSITIVE/               DOWNWARDS IS POSITIVE /
       OPWAARTS AS POSITIEF               AFWAARTS AS POSITIEF
       Total time /                       Total time/
       Totale tyd = 2,04 + 1,66  = 3,7 s Totale tyd = 2,04 + 1,66  = 3,7 s
       Δy = viΔt + ½gΔt2 
       Δy = (20)(3,7)  + ½(-9,8)(3,72)  Δy = viΔt + ½gΔt2 
       Δy = 6,92 m                       Δy = (-20)(3,7)  + ½(9,8)(3,72) 
                                          Δy = -6,919 m
                                          Δy = 6,92 m 
                                          (If answer left as negative, then loses
                                          last mark)
                                          (As antwoord as negatief gelaat word,
                                          verloor dan laaste punt)

         OPTION 2 / OPSIE 2
         Positive marking from QUESTION 5.2.1 / Positiewe nasien vanaf
         VRAAG 5.2.1
         UPWARDS IS POSITIVE/             DOWNWARDS IS POSITIVE /
         OPWAARTS AS POSITIEF             AFWAARTS AS POSITIEF
         Time taken from B to ground      Time taken from B to the ground
         Tyd geneem van B na grond        Tyd geneem van B na grond

         Δt= 2,04 - 1,66 = 0,38 s               Δt= 2,04 - 1,66  = 0,38 s
         Δy = viΔt + ½gΔt2                      Δy = viΔt + ½gΔt2 
         Δy = (20)(0,38)  + ½(-9,8)(0,382)     Δy = (-20)(0,38)  + ½(9,8)(0,382) 
         Δy = 6,89 m                            = -6,89 m
                                                 Height / Hoogte = 6,89 m above the
                                                 ground / bo die grond 

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(EC/JUNE/JUNIE 2022)        PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                            9

         OPTION 3 / OPSIE 3               OPTION 3 / OPSIE 3
         Positive marking from QUESTION 5.2.1 / Positiewe nasien vanaf
         VRAAG 5.2.1
         UPWARDS IS POSITIVE/             DOWNWARDS IS POSITIVE /
         OPWAARTS AS POSITIEF             AFWAARTS AS POSITIEF
         Calculate the maximum height:    Calculate the maximum height:

         Bereken die maksimum hoogte:              Bereken die maksimum hoogte:
                                        Any one                                    Any one
         vf 2= vi2 + 2gΔx               formula    vf 2= vi2 + 2gΔx
                                                                                   formula
         02 = 202 + 2(-9,8).Δx          Enige een 0 = (-20) + 2(9,8).Δx 
                                                    2          2
                                                                                   Enige een
         Δx = 20,41 m                   formule  Δx = 20,41 m                     formule 
         Ball falls for 1,66 m / Bal val vir       Ball falls for 1,66 m Bal val vir 1,66 m
         1,66 m                                    Δy = viΔt + ½gΔt2
         Δy = viΔt + ½gΔt2                         Δy = (0)(1,66) + ½(9,8)(1,662) 
         Δy = (0)(1,66) + ½(-9,8)(1,662)          Δy = 13,05 m
         Δy = -13,05 m
                                                   Height of the balcony/
         Height of the balcony /                   Hoogte van die balkon = 20,41 –
         Hoogte van die balkon = 20,41 –           13,51 
         13,51 
                                                   Height of the balcony/
         Height of the balcony /                   Hoogte van die balkon        = 6,91 m 
         Hoogte van die balkon = 6,91 m                                                          (5)

 5.2.3 OPTION 1 / OPSIE 1
       Positive marking from QUESTION 5.2.1 / Positiewe nasien vanaf
        VRAAG 5.2.1
       UPWARDS IS POSITIVE/             DOWNWARDS IS POSITIVE /
       OPWAARTS AS POSITIEF             AFWAARTS AS POSITIEF
       vf = vi + gΔt                   vf = vi + gΔt 
       vf = 0 + (-9,8)(1,66)           vf = 0 + 9,8(1,66) 
       vf = 16,27 m.s -1               vf = 16,27 m.s-1 

         OPTION 2 / OPSIE 2
         Positive marking from QUESTION 5.2.1/ Positiewe nasien vanaf
         VRAAG 5.2.1
         UPWARDS IS POSITIVE/             DOWNWARDS IS POSITIVE /
         OPWAARTS AS POSITIEF             AFWAARTS AS POSITIEF
         vf = vi + gΔt                   vf = vi + gΔt 
         vf = 20 + (-9,8)(3,7)           vf = -20 + 9,8(3,7) 
         vf = -16,26                      vf = 16,26 m.s-1 
         vf = 16,26 m.s -1                                                                       (3)

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10                    PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1         (EC/JUNE/JUNIE 2022)

 5.3    Positive marking from QUESTION 5.2.3 / Positief nasien vanaf
        VRAAG 5.2.3
        UPWARDS IS POSITIVE / OPWAARTS AS POSITIEF
          V (m.s-1)
               20

                  0
                                           2,04                     3,7       Δt (s)

             -16,27

        DOWNWARDS IS POSITIVE / AFWAARTS AS POSITIEF
         V (m.s-1)
             16,27

                  0
                                           2,04                     3,7      Δt (s)

                -20

         Marking criteria / Nasienkriteria
         Shape / Vorm                                                            
         Initial velocity / Begin snelheid                                       
         Time at maximum height / Tyd by maksimum hoogte                         
         Velocity when man catches the ball / Snelheid wanneer man bal vang      
         Time when man catches the ball / Tyd wanneer man bal vang               
                                                                                         (5)
                                                                                        [19]

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(EC/JUNE/JUNIE 2022)        PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                          11

QUESTION/VRAAG 6

6.1     The momentum of an object is the product of its mass and velocity. 
        (2 or 0)
        Die momentum van ʼn voorwerp is die produk van sy massa en snelheid.
        (2 of 0)                                                                                       (2)

6.2.1 p = mv 
      p = 3200 x 18 
      p = 56 700 kg.m.s-1                                                                             (3)

6.2.2 Σpi = Σpf               Any one/Enige een 
      pTi + pCi = pTf + pCf
      57600 + (12 x 1800)  = (10 x 3200) + 1800 vCf 
      vCf = 26,22 m.s-1 to the right / regs (or any indication of direction / of enige
      aanduiding van rigting)                                                                         (5)

6.3     Inelastic. (Only momentum is conserved) Kinetic energy is not conserved. 
        Onelasties (Slegs momentum word behoue) Kinetiese energie is nie behoue
        nie.                                                                                           (3)

6.4     Airbags/seatbelts/Crumble zones (Any two )
        Accept headrests
        Lugsakke/veiligheidsgordel/frommelsone (Enige twee)
        Aanvaar hoofstutte                                                                          (2)
                                                                                                  [15]

QUESTION/VRAAG 7

7.1     UPWARDS IS POSITIVE/                 DOWNWARDS IS POSITIVE /
        OPWAARTS AS POSITIEF                 AFWAARTS AS POSITIEF
        vf = vi + gΔt                       vf = vi + gΔt 
        vf = 0 + -9,8(0,35)                 vf = 0 + -9,8(0,35) 
        vf = -3,43 m.s -1                    vf = 3,43 m.s-1  downwards/afwaarts 
        vf = 3,43 m.s  downwards/afwaarts 
                       -1                                                           (4)

7.2     Impulse is the product of the net force acting on an object and the time for which
        the net force acts on the object. 

        Impuls is die produk van die netto krag wat op ʼn voorwerp inwerk en die tyd wat
        die netto krag op die voorwerp inwerk.                                                     (2)

        Positive marking from QUESTION 7.1 / Positiewe nasien vanaf VRAAG 7.1
7.3     Fnet.∆t = ∆p           Any one / Enige een 
        Fnet.∆t = mvf – mvi
        Fnet x 0,69  = 45 x 0 – 45 x 3,43 
        Fnet = 223,7 N  Upwards / Opwaarts                                                       (5)

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12                    PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1              (EC/JUNE/JUNIE 2022)

 7.4    Fnet = mΔv
                 Δt
        Δv remains the same. 
        If the knees are not bent, the time to come to rest is very small (Δt decreases) 
        Fnet increases and can cause injury. 
        Fnet = mΔv
                 Δt
        Δv bly dieselfde
        As die knieë nie gebuig is nie, is die tyd om tot rus te kom baie klein
        (Δt verminder)
        Fnetto verhoog en kan beserings veroorsaak.                                               (3)
                                                                                                 [14]

 QUESTION/VRAAG 8

 8.1    In an isolated system, the total mechanical energy remains constant. 
        In ʼn geïsoleerde sisteem bly die totale meganiese energie konstant.                           (2)

 8.2    Emec(B) = Emec(C)
        (Ep + Ek)B = (Ep + Ek)C           Any one / Enige een 
        (mgh + ½mv2)B = (mgh + ½mv2)C
        (540)(9,8)(0) + ½(540)vB2 = (540)(9,8) (4,5 sin 7,5˚)  + ½(540)(02)
        vB = 3,39 m.s-1                                                                              (4)

 8.3    fk = μkN 
        fk = (0,8) x (540)(9,8) 
        fk = 4233,6 N                                                                                (3)

 8.4    The net work done on an object is equal to the change in kinetic energy of the
        object. 
        Die netto werk verrig op ʼn voorwerp is gelyk aan die verandering in kinetiese
        energie van die voorwerp.                                                                     (2)

 8.5.1 Positive marking from QUESTION 8.2 / Positiewe nasien van VRAAG 8.2.
       OPTION 1 / OPSIE 1
       Wnet = ΔEk         Any one / Enige een 
       Wf = ½m(vf2 – vi2)
       Wf = ½ x 540)(3,39)2 – ½ x 540(122) 
       Wf = -35 777,13 J 

        OPTION 2 / OPSIE 2
        Wnc = ΔEk + ΔEp        Any one / Enige een 
        Wf = ½m(vf2 – vi2) +0
        Wf = ½ x 540 x 3,392 – ½ x 540 x 122 
        Wf = -35 777,13 J                                                                            (4)

 8.5.2 Positive marking from QUESTION 8.3 and 8.5.1
       Positiewe nasien vanaf VRAAG 8.3 en 8.5.1
       Wf = Ff.Δx.cosθ 
       -35777,13 = (4233,6).Δx.cos 180˚
       Δx = 8,45 m                                                                               (4)
                                                                                                 [19]

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(EC/JUNE/JUNIE 2022)        PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1                       13

 QUESTION/VRAAG 9

 9.1 The change in frequency or pitch detected by a listener because the listener
     and the sound source have different velocity relative to the medium of sound
     propagation. 

       Die verandering in frekwensie of toonhoogte van die klank waargeneem deur ʼn
       luisteraar, omdat die klankbron en die luisteraar verskillende snelhede relatief
       tot die medium waarin die klank voortgeplant word, het.                               (2)

 9.2 fL = v ± vL . fS 
          v ± vS

       875 = (340 + 0) .  680 
              (340 – vs)

       vs = 75,77 m.s-1

       vs = 75,77 x 3,6 
       vs = 272,77 km.h-1                                                                   (6)

 9.3 Decreases 
     As it moves away, wavelength increases  but velocity remains constant. 
     Frequency decreases.

       Afneem
       Soos dit wegbeweeg, verleng die golflengte maar die snelheid bly konstant.
       Frekwensie neem af.                                                                   (3)

 9.4 Measuring foetal heartbeat 
     Monitoring and measuring of blood flow 

       Meet fetus se hartklop
       Monitering en meting van bloedvloei.                                                  (2)
                                                                                            [13]

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14                    PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1         (EC/JUNE/JUNIE 2022)

 QUESTION/VRAAG 10

 10.1
                                             Marking criteria / Nasienkriteria
                                             Direction / Rigting 
                                             Shape / Vorm 
                                             Lines all touching charges and not
                                             crossing / Lyne raak aan ladings en
                                             kruis nie 
                                                                                            (3)

 10.2 F = kQ1Q2 
             r2
      F = (9 x 109)(4 x 10-6)(4 x 10-6) 
                 (0,75)2 
      F = 0,256 N right / regs                                                             (4)

 10.2 Qe = nqe 
      4 x 10-6 = n x (1,6 x 10-19) 
      n = 2,5 x 1013 protons / protone                                                     (3)
                                                                                           [10]

                                                                 TOTAL/ TOTAAL:            150

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