# Physical Sciences Grade 12 P1 Gauteng June 2024 Memo (Afr)

> Grade 12 Physical Sciences Paper 1 memorandum from the Gauteng June Exam 2024 (Afrikaans), 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr-12-physical-sciences-p1-english-and-afrikaans-june-2024-possible-answers/f78ca708-f7f0-4bc5-8ff9-36594b1b1052
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Language: Afrikaans
- Document type: Memorandum
- Year: 2024
- Exam period: Gauteng June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 14

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## About this document

Physical Sciences · Grade 12 · Gauteng June Exam · 2024 · Afrikaans. Memorandum, 14 pages. Read online or download the PDF.

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### Page 1

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                    JUNE EXAMINATION
                     JUNIE EKSAMEN

                     GRADE/GRAAD 12

                             2024
               MARKING GUIDELINES/
                 NASIENRIGLYNE

                   PHYSICAL SCIENCES: PHYSICS/
                   FISIESE WETENSKAPPE: FISIKA

                       (PAPER/VRAESTEL 1)

13 pages/bladsye

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                                                         PHYSICAL SCIENCES: PHYSICS
                            MARKING GUIDELINES
                                                         FISIESE WETENSKAPPE: FISIKA
                            NASIENRIGLYNE
                                                         (PAPER/VRAESTEL 1)     GR12 0624

QUESTION/VRAAG 1

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

QUESTION/VRAAG 2

2.1    The force or the component of a force which a surface exerts on an object with
       which it is in contact, and which is perpendicular to the surface. ✓✓

       Die krag of komponent van ŉ krag wat ŉ oppervlak op ŉ voorwerp waarmee dit
       in kontak is, uitoefen en wat loodreg op die oppervlak is.                             (2)
       Marking criteria/Nasienkriteria
       If any of the underlined key words/phrases in the correct context is omitted deduct
       1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte
       konteks uitgelaat is, trek 1 punt af.

2.2             

                

                              Accepted labels/Aanvaarde byskrifte
               Fg/Fw/weight/mg/gravitational force do not accept gravity
       w
               Fg/Fw/gewig/gravitasiekrag moet nie gravitasie aanvaar nie.
               FT/tension
       T
               FT/spanning/spankrag
               Deduct 1 mark for any additional force. Mark is given for both arrow
               and label. If everything is correct, but no arrows, deduct a mark              (2)

                                                2

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                                                               PHYSICAL SCIENCES: PHYSICS
                                 MARKING GUIDELINES
                                                               FISIESE WETENSKAPPE: FISIKA
                                 NASIENRIGLYNE
                                                               (PAPER/VRAESTEL 1)     GR12 0624

              Trek een punt af vir enige addisionele kragte. Punte word gegee vir
              beide pylpunt en byskrif. Indien alles korrek is, maar geen pyle nie, trek
              ŉ punt af.

2.3   In x-direction for mass M:/In x-rigting vir massa M:

      Fnet = ma                 
                                              Any one/Enige een
      T + f – wII = 0
      2(9,8)  + f – 10(9,8)sin30°  = 0            OR      2(9,8)  - f – 10(9,8)sin30°  = 0 
        f     = –2(9,8) + 10(9,8)sin30°
              = –19,6 + 49
              = 29,4 N (up the slope/teen die helling op)                                               (5)

2.4   2.4.1                                                         Accepted labels
                                                                       Aanvaarde byskrifte
                                                              Fg/Fw/weight/mg/gravitational force
                                                         w
                                                               Fg/Fw/gewig/mg/gravitasiekrag
                                                               Ffriction/Ff/friction/fk
                                                         f
                                                               Fwrywing/Ff/wrywing/fk
                                                         N     FN/Normal force/Normaalkrag
                                                               Deduct 1 mark for any additional force.
                                                               Mark is given for both arrow and label
                                                               DO NOT ACCEPT COMPONENTS IN
                                                               FREE BODY DIAGRAM
                                                              Trek 1 punt af vir addisionele kragte.
                                                               Punte word gegee vir beide pyl en
                                                               byskrif.
                                                               MOET NIE KOMPONENTE AANVAAR
                                                               IN DIE VRYLIGGAAMDIAGRAM NIE              (3)

      2.4.2     Fnet = ma                
                wII – f = ma
                10(9,8)sin30° – 25  = 10 a 
                     a       = 2,4 m·s –2 
                                                                                                         (4)

2.5   2.5.1     vf       =     vi + a∆t 
                         =     0 + (9,8)(0,5)                            A learner can use
                                                                          the option of -9,8
                vf       =     4,9 m·s –1 (down / afwaarts)              The same answer
                                                                          will be obtained.

                OR
                Fnet Δt = Δp     ✓
                2(9,8)(0,5) = 2vf – 2vi ✓
                vf = 4,9 m.s-1 (down / afwaarts) ✓
                                                                                                         (3)

                                                     3

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                                                        PHYSICAL SCIENCES: PHYSICS
                          MARKING GUIDELINES
                                                        FISIESE WETENSKAPPE: FISIKA
                          NASIENRIGLYNE
                                                        (PAPER/VRAESTEL 1)     GR12 0624

      2.5.2    REMAIN THE SAME/BLY DIESELFDE ✓

               The object is moving under the force of gravity only ✓and through the
               same height.✓ That will have no effect on the final velocity.
               OR
               When using equations of motion and falling at gravitational acceleration
               only, the mass has no influence on the final velocity. The height will
               though.

               Die voorwerp beweeg slegs onder die grativitasiekrag en deur dieselfde
               hoogte. Dit sal geen effek op die finale eindsnelheid hê nie.
               OF
               Wanneer bewegingsvergelykings gebruik word en slegs teen gravitasie-
               versnellings val, het die massa geen invloed op die finale eindsnelheid
               nie. Die hoogte sal wel.                                                      (3)
                                                                                            [22]

QUESTION/VRAAG 3

3.1   Each body in the universe attracts every other body with a force that is directly
      proportional to the product of their masses ✓and inversely proportional to the
      square of the distance between their centres. ✓

      Elke liggaam in die heelal trek elke ander liggaam aan met ŉ krag direk eweredig
      aan die produk van hul massas en omgekeerd eweredig aan die kwadraat van die
      afstand tussen hul middelpunte.
                                                                                            (2)
      Marking criteria/Nasienkriteria
      If any of the underlined key words/phrases in the correct context is omitted deduct
      1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte
      konteks uitgelaat is, trek 1 punt af.

3.2   Weight/Gewig w = ma
                     = 899 (9,8) ✓
                     = 8 810,2 N ✓                                                          (2)

3.3           GM
      g =        ✓
              r2
           6,67  10-11(6,42  1023 ) ✓
         =
            (0,5325  6,38  106 )2 ✓

         = 3,71 m·s-2 ✓

      The gravitational acceleration on Earth is 9,8 m·s-2 while on
      Mars it is 3,71 m·s-2. (range of 3,70 – 3,72)
      Die grativitasieversnelling op Aarde is 9,8 m·s-2 terwyl dit op Mars 3,71 m·s-2 is.
      (gebied van 3,70 – 3,72)                                                              (4)

                                               4

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                                                           PHYSICAL SCIENCES: PHYSICS
                          MARKING GUIDELINES
                                                           FISIESE WETENSKAPPE: FISIKA
                          NASIENRIGLYNE
                                                           (PAPER/VRAESTEL 1)     GR12 0624

3.4   No/Nee ✓

      Mass represents the matter an object is made up of and will not differ, no matter
      where in the universe you are. ✓
      OR
      The mass will be the same on both planets.

      Massa verteenwoordig die materie waaruit ŉ voorwerp bestaan en wat nie sal
      verskil nie, ongeag waar in die heelal jy is.
      OF
      Die massa sal dieselfde wees op beide planete.                                             (2)
                                                                                                [10]

4.1   Motion of an object is free fall. ✓ Beweging van ŉ voorwerp.

      Object given an initial velocity is projectile. ✓ ŉ Voorwerp wat begin met ŉ
      beginsnelheid.

      OR
      Accept if definitions are given. / Aanvaar indien definisies gegee word
      OR
      Free fall and projectile motion is the same. Vryval en projektielbeweging is
      dieselfde.                                                                                 (2)

4.2   4.2.1   OPTION/OPSIE 1 (down as              OPTION/OPSIE 2 (up as positive/op as
              positive/af as positief)             positief)

              vf    =   vi + a∆t                  vf     =      vi + a∆t 
              0     =   (–11) + (9,8)∆t           0      =      (11) + (–9,8)∆t 
               Δt =     1,12 s                         Δt =     1,12 s                         (3)

      4.2.2   Positive marking from QUESTION               OPTION 2 down as negative/OPSIE
              4.2.1                                        2 af as negatief
              OPTION 1 down as positive/
              Positiewe nasien vanaf VRAAG
              4.2.1
              OPSIE 1 af as positief

              vf    =   vi + a∆t                          vf = vi + a∆t 
              0     =   vi +(9,8)(1,12 – 0,72)                0 = vi + (–9,8)(1,12 – 0,72) 
               0 =      vi + (9,8)(0,40)                   vi = 3,92 m·s-1 ✓
                                                           upwards/opwaarts ✓
              vi = 3,92 m·s-1 ✓
              upwards/opwaarts                                                                  (4)

                                              5

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                                                  PHYSICAL SCIENCES: PHYSICS
                             MARKING GUIDELINES
                                                  FISIESE WETENSKAPPE: FISIKA
                             NASIENRIGLYNE
                                                  (PAPER/VRAESTEL 1)     GR12 0624

4.2.3   Positive marking from                     Positive marking from 4.2.1 and
        QUESTION 4.2.1 and                        4.2.2/Positiewe nasien vanaf
        4.2.2./Positiewe nasien vanaf             VRAAG 4.2.1 en 4.2.2.
        VRAAG 4.2.1 en 4.2.2.
                                                  OPTION 2 (upwards positive)
        OPTION 1 (down as positive)               Ball A height above the ground/
        Ball A height above the ground:           OPSIE 2 (op as positief)
        OPSIE 1 (af as positief)                  Bal A hoogte bo die grond
        Bal A hoogte bo die grond
                                                  𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑦 ✓
        𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑦 ✓                          0 = (11)2 + 2(−9,8)∆𝑦 ✓
          0 = (–11)2 +2(9,8)∆y ✓                  ∆y = 6,17 m
        ∆y = 6,17 m ✓

        Ball B height above the building/         Ball B height above the building/
        Bal B hoogte bo die gebou:                Bal B hoogte bo die gebou:
                                                  𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑦
        𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑦
                                                   0 = (3,92)2 +2(–9,8)∆y ✓
          0 = (–3,92)2 +2(9,8)∆y ✓
                                                  ∆y = 0,784 m
        ∆y = 0,784 m

        Height of building/Hoogte van die Height of building/Hoogte van die
        gebou:                             gebou:
                             ✓                                 ✓
        Height/Hoogte = 6,17 – 0,784       Height/Hoogte = 6,17 – 0,784
                      = 5,39 m (5,388 m) ✓               = 5,39 m ✓

        OPTION/OPSIE 3                            OPTION/OPSIE 4
        Ball A above the ground/                  Ball A above the ground/
        Bal A bo die grond:                       Bal A bo die grond:

                 v f + vi                                     1
        ∆x = (       2      ) t ✓                ∆𝑥 = 𝑣𝑖 ∆𝑡 + 2 𝑎∆𝑡 2 ✓
                  0 + 11
        ∆x A = ( 2 ) x (1,12) ✓                   ∆xA = 11 × 1,12
                                                          1
                                                        + 2(–9,8) (1,12)2 ✓
        ∆x = 6,16 m
                                                        ∆xA = 6,17 m
        Ball B above the building/                Ball B above the building/
        Bal B bo die gebou:                       Bal B bo die gebou:
                 v f + vi                                        1
        ∆x = (       2      ) t ✓                ∆𝑥 = 𝑣𝑖 ∆𝑡 + 2 𝑎∆𝑡 2 ✓
                   0 + 3, 92
        ∆xB = ( 2 ) (0,4) ✓                       ∆xB = 3,92 × 0,4 + ✓
                                                  1
                                                   (–9,8) (0,4)2
        ∆x = 0,784 m                              2
                                                  ∆xB = 0,784 m
        Height of building/Hoogte van die         Height of building/Hoogte van die
        gebou:                                    gebou:
                              ✓                                          ✓
        Height/Hoogte = 6,16 – 0,784              Height/Hoogte = 6,17 – 0,784
           = 5,38 m (5,376 m) ✓                       = 5,39 m (5,386 m) ✓
                                                                                      (6)

                                           6

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                                                         PHYSICAL SCIENCES: PHYSICS
                          MARKING GUIDELINES
                                                         FISIESE WETENSKAPPE: FISIKA
                          NASIENRIGLYNE
                                                         (PAPER/VRAESTEL 1)     GR12 0624

               Positive marking from QUESTION 4.2 for answers of QUESTION 4.3/
               Positiewe nasien van VRAAG 4.2 vir antwoorde van VRAAG 4.3

4.3    4.3.1   6,16 m ✓                                                                            (1)

       4.3.2   5,38 m ✓                                                                            (1)

       4.3.3   0,72 s ✓                                                                            (1)

       4.3.4   1,12 s ✓                                                                            (1)
                                                                                                  [19]

QUESTION/VRAAG 5
 5.1   5.1.1   The total mechanical energy (sum of gravitational potential energy and
               kinetic energy) in an isolated system remains constant. ✓✓

               Die totale meganiese energie (som van gravitasie-potensiële energie en
               kinetiese energie) in ŉ geïsolleerde sisteem bly konstant.                         (2)
               Marking criteria/Nasienkriteria
               If any of the underlined key words/phrases in the correct context is omitted
               deduct /If momentum is mentioned 0/2

               1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte
               konteks uitgelaat is, trek 1 punt af.
               Indien momentum genoem word 0/2

       5.1.2   OPTION/OPSIE 1:

                           Emech i = Emech f
               1                        1                   Any one/Enige een
                 m X v 2Xi + m X gh X = m X v 2Xf + m Xghh
               2                        2
                                          1
               0 + 950(9,8)(3,5) = (950)(v 2 ) + 950(9,8)(1)
                                          2
                                      v = 7 m  s−1

               OPTION/OPSIE 2:

                               Emech i = Emech f
                                                              Any one/Enige een
                   1       2               1       2
                     m X v Xi + m X gh X = m X v Xf + m Xghh
                   2                       2
                                           1
               0 + 950(9,8)(3,5-1) = (950)(v 2 ) + 950(9,8)(0)
                                           2
                                     v = 7 m  s−1

                                                   7

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                                                 PHYSICAL SCIENCES: PHYSICS
                      MARKING GUIDELINES
                                                 FISIESE WETENSKAPPE: FISIKA
                      NASIENRIGLYNE
                                                 (PAPER/VRAESTEL 1)     GR12 0624

        OPTION 3/OPSIE 3
          Wnc   = Ek + Ep
                     1         1                                  Any one/Enige een
                            2         2
               0 = ( m X v Xf - m X v Xi ) + (m Xghh - m Xgh X )
                     2         2
                     1
               0 = ( (950)(v 2 ) − 0) + (950(9,8)(1) - 950(9,8)(3,5))
                     2
                v = 7 m  s −1                                                        (4)

5.1.3   Positive marking from QUESTION 5.1.2/Positiewe nasien vanaf VRAAG
        5.1.2

        Fnet∆t = m∆v ✓
        Fnet(0,1)         = (950)(0 – 7) ✓
        Fnet      =       – 66 500 N
        Fnet      =       66 500 N ✓

        OR

          vf = vi + aΔt                            Fnet = ma    ✓
           0 = 7 + a(0,1)                              = 950(-70) ✓
            a = - 70 m.s-2                             = - 66 500 N
                                                       = 66 500 N ✓                   (3)

5.1.4   Positive marking from 5.1.3/positiewe nasien vanaf 5.1.3

        66 500 N to the left/west/na links/wes ✓ OR Fball on wall

        Newtons third law. ✓ When object A exerts a force on object B, object B
        SIMULTANEOUSLY exerts an oppositely directed force of equal
        magnitude on object A. ✓
        Newton se derde wet. Wanneer voorwerp A ŉ krag op voorwerp B
        uitoefen, sal voorwerp B GELYKTYDIG ŉ krag van gelyke grootte en in die
        teenoorgestelde rigting op voorwerp A uitoefen.                               (3)

                                             8

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                                                                         PHYSICAL SCIENCES: PHYSICS
                                  MARKING GUIDELINES
                                                                         FISIESE WETENSKAPPE: FISIKA
                                  NASIENRIGLYNE
                                                                         (PAPER/VRAESTEL 1)     GR12 0624

 5.2                1
        Ek after =    mv 2
                    2
                    1
             146 =    (75 + 2 + 0,5)v 2
                    2
                v = 1,94 m  s−1
                        pbefore = pafter
                                                                                Any one/Enige een
       mlearner v leaerner i + mball v ball i = (mlearner + mball )v f
             (75+2)(2) + 0,5(v)            = 77,5(1,94)                  OR   m1v1 + m2v2 = (m1+m2)vf
                             v             = -7,3
                                     v ball i = 7,3 m  s−1
       Range 7,13 – 7,3                                                                                  (5)
                                                                                                        [17]

QUESTION/VRAAG 6
 6.1                

                    

                               Accepted labels/Aanvaarde byskrifte
                Fg/Fw/weight/mg/gravitational force
        w
                Fg/Fw/gewig/mg/gravitasiekrag
                FT/tension
        T
                FT/spanning/spankrag
                Deduct 1 mark for any additional force. Mark is given for both arrow
                and label. If everything is correct, but no arrows, deduct a mark
                Trek een punt af vir enige addisionele kragte. Punte word gegee vir
                beide pylpunt en byskrif. Indien alles korrek is, maar geen pyle nie, trek
                ŉ punt af.                                                                                (2)

 6.2    Tension/Spanning ✓                                                                                (1)

                                                               9

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                                                   PHYSICAL SCIENCES: PHYSICS
                       MARKING GUIDELINES
                                                   FISIESE WETENSKAPPE: FISIKA
                       NASIENRIGLYNE
                                                   (PAPER/VRAESTEL 1)     GR12 0624

6.3   OPTION/OPSIE 1:

      WF = Fxcos
                                         Any one/Enige een
          = mgxcos
          = (85)(9,8)(11)cos180o
          = - 9 163 J

      OPTION/OPSIE 2:

      Ww = - EP
                                          Any one/Enige een
          = - (mgh - 0)
          = - (85)(9,8)(11)
          = - 9 163 J                                                                    (3)

6.4   The net work done on an object is equal to the change in the object's kinetic
      energy. ✓✓ (2 or 0)
      OR
      The work done on an object by a net force is equal to the change in the object's
      kinetic energy.

      Die netto arbeid verrig op ŉ voorwerp is gelyk aan die verandering in kinetiese
      energie van die voorwerp.
      OF
      Die arbeid verrig op die voorwerp deur ŉ netto krag is gelyk aan die verandering
      in kinetiese energie van die voorwerp.                                             (2)

6.5                      Wnet = K
                               1           1                  Any one/Enige een
                  Fnet xcos   =mv 2f - mv i2
                               2           2
                               1
                    maxcos = (85)v 2f - 0
                               2
                          o
      (85)(0,75)(11) cos0   = 42,5v 2f
                          vf    = 4,06 m  s−1                                           (5)

                                           10

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                                                         PHYSICAL SCIENCES: PHYSICS
                              MARKING GUIDELINES
                                                         FISIESE WETENSKAPPE: FISIKA
                              NASIENRIGLYNE
                                                         (PAPER/VRAESTEL 1)     GR12 0624

 6.6    OPTION 1/OPSIE 1                                   OPTION 2/OPSIE 2
         vf = vi + a∆t
                                                                    Fnet + w = Fmotor
        4,06 = 0 + (0,75)∆t ✓                              85(0,75) + 85(9,8) ✓ = Fmotor
                                                                       Fmotor = 896,75 N
          t = 5,41 s
                                                           Pave = Fvave ✓
        Positive marking from QUESTION 6.3 for                               0+4,06
        Work done by gravitational force./Positiewe              = (896,75) ✓ 2 ✓
        nasien vanaf VRAAG 6.3 vir arbeid verrig                 = 1 820,4 W
        deur gravitasiekrag.
                                                                           1 820,4
                                                           % efficiency =        𝑥 100 ✓
                                                                          30 000
                Wnet = ΔEk
                                                                        = 6,07% ✓
        -9163 + Wmotor = ½ (85)( 4,062 – 02) ✓
            Wmotor = 9 863,55 J
                       𝑊
                 P= 𝑡             ✓
                       9 863,55
                   =                  ✓
                           5,41
                   = 1 823,21 W
                           1 823,21
        % efficiency = 30 000 𝑥 100 ✓
                     = 6,08% ✓                                                                 (6)
                                                                                              [19]
QUESTION/VRAAG 7
 7.1   7.1.1   The change in frequency (or pitch) of the sound detected by a listener,
               because the sound source and the listener have different velocities
               relative to the medium of sound propagation. ✓✓

               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.

               OR/OF
               An (apparent) change in (observed/detected) frequency/pitch as a result
               of the relative motion between a source and an observer (listener).

               'n (Skynbare) verandering in (waargenome) frekwensie/toonhoogte as
               gevolg van die relatiewe beweging tussen die bron en 'n
               waarnemer/luisteraar.
               Marking criteria/Nasienkriteria
               If any of the underlined key words/phrases in the correct context is omitted
               deduct
               1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die
               korrekte konteks uitgelaat is, trek 1 punt af.                                 (2)

                                               11

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                                                              PHYSICAL SCIENCES: PHYSICS
                                MARKING GUIDELINES
                                                              FISIESE WETENSKAPPE: FISIKA
                                NASIENRIGLYNE
                                                              (PAPER/VRAESTEL 1)     GR12 0624

        7.1.2    Towards/Na ✓

                      •     The apparent frequency was higher/bigger ✓
                      •     The wave fronts were closer together ✓ and
                      •     The wavelength was shorter/smaller. ✓

                      •     Die skynbare frekwensie was hoër,
                      •     Die golffronte was nader aan mekaar en
                      •     Die golflengte was korter.                                             (4)

        7.1.3                v ± vL
                     fL =           fs
                             v ± vs
                                 340
                 405        =             360
                                340 - v s
                     v s = 37,78 m  s−1                                                           (4)

 7.2    7.2.1    360 (Hz) ✓                                                                        (1)

        7.2.2    Less than/Minder as 360 (Hz) ✓ OR < 360 (Hz)
                 Any value less than 360 given is accepted. / Enige waarde minder as
                 360 word aanvaar.                                                                 (1)

        7.2.3    340 (m·s-1 ) ✓                                                                    (1)
                                                                                                  [13]

QUESTION/VRAAG 8

  8.1    Positive/Positief ✓                                                                        (1)

  8.2                Q                                   Q    NOTE: do not penalize if –1,6 is
                n=     ✓                        or n =        substituted, but answer must be
                     e                                   qe
                                                              positive.
                          Q         ✓                         NOTA: moenie penaliseer indien
         1238 =
                     1,6 x 10−19                               –1,6 vervang, maar antwoord moet
                                                              positief wees.
            Q = 1,981 x 10−16 C ✓                                                                   (3)

  8.3    An electric field is a region of space in which an electric charge experiences a
         force. ✓✓ (2 or 0)

         ŉ Elektriese veld is ŉ area of gebied waarin ŉ elektriese lading ŉ krag sal
         ondervind. (2 of 0)                                                                        (2)

                                                     12

### Page 13

Downloaded from hlayiso.com
                                                       PHYSICAL SCIENCES: PHYSICS
                         MARKING GUIDELINES
                                                       FISIESE WETENSKAPPE: FISIKA
                         NASIENRIGLYNE
                                                       (PAPER/VRAESTEL 1)     GR12 0624

8.4

       Criteria for sketch/Kriteria vir skets                               Marks/
                                                                            Punte
       Correct direction of electric field – away from positive and         ✓
       towards negative. Do not penalize if + and – are not indicated.
       /Korrekte rigting van elektriese veld –weg van positief en na
       negatief. Moenie penaliseer indien + en – nie aangedui word
       nie.
       Correct shape of the electric field lines between, above and on      ✓
       the outside of the charges. Lines curve in the correct direction./
       Korrekte vorm van die elektriese veldlyne, tussen, bo en aan
       die buitekant van die ladings. Lyne buig in die regte rigting.
       No field lines crossing each other. Field lines touch the charge,    ✓
       start at an angle perpendicular to the charge, do not go inside
       the charge./Geen veldlyne wat mekaar kruis nie. Veldlyne raak
       aan die ladings, begin by ŉ hoek loodreg op die lading, gaan
       nie binne die lading nie.

8.5   A✓

      The field is strongest closest ✓to the charged sphere. Point A is closer to
      sphere X than point B. The electric field is inversely proportional to the
                                        1
      distance squared. OR E                ✓/
                                        r2
      Die veld is die sterkste, naaste aan die gelaaide sfeer. Punt A is nader aan
      sfeer X as punt B. Die elektriese veld is omgekeerd eweredig aan die afstand
                             1
      kwadraat. OF E 
                             r2                                                      (3)

8.6       kQ X
      E=
           r2
           9x109 (2x10−9 )
        =
               (0,03)2
        = 20 000 N  C−1 left / links                                                (3)

                                                  13

### Page 14

Downloaded from hlayiso.com
                                                         PHYSICAL SCIENCES: PHYSICS
                          MARKING GUIDELINES
                                                         FISIESE WETENSKAPPE: FISIKA
                          NASIENRIGLYNE
                                                         (PAPER/VRAESTEL 1)     GR12 0624

  8.7   8.7.1              FYX

                           FZX

                  Marking criteria/Nasienkriteria:                        Marks/Punte
                  Line and arrow with label//Lyn en pyl met byskrif       ✓
                  Line and arrow with label/Lyn en pyl met byskrif        ✓                      (2)

        8.7.2   Marking criteria/Nasienriglyne
                 • Correct formula for Coulombs law/Korrekte formule vir Coulomb wet
                 • Correct substitution for FYX/Korrekte vervanging vir FYX
                 • Correct substitution for Fzx/Korrekte vervanging vir FZX
                 • Addition of forces/Optel van kragte
                 • Correct final answer/Korrekte finale antwoord
                       kQY Q X    kQZQ X
                Fnet =     2
                                +
                         r           r2
                        9x109 (2,8x10 −9 )(2x10 −9 )      (9x109 )(3,2x10 −9 )(2x10 −9 )
                     =                               +
                                 (0,03)2                            (0,01)2
                     = 6,32 x 10−4 N                                                             (5)
                                                                                                [22]

QUESTION/VRAAG 9

  9.1    Marking criteria/Nasienkriteria
         If any of the underlined key words/phrases in the correct context is omitted, deduct
         1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte
         konteks uitgelaat is, trek 1 punt af.
         The potential difference across a conductor is directly proportional to the current
         in the conductor at constant temperature. ✓✓

         Die potensiaalverskil oor ŉ geleier is direk eweredig aan die stroom in die geleier
         by konstante temperatuur.                                                              (2)

  9.2    Lamp ✓ or bulb/gloeilamp of lamp                                                       (1)

  9.3    12 V ✓                                                                                 (1)

  9.4    12 V ✓                                                                                 (1)

  9.5    0V✓                                                                                    (1)

  9.6    Stays the same/Bly dieselfde ✓                                                         (1)

  9.7    Stays the same/Bly dieselfde ✓                                                          (1)
                                                                                                 [8]
                                                                            TOTAL/TOTAAL:       150

                                               14

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