# Technical Sciences Grade 12 P1 Eastern Cape Mock 2020 Memo

> Grade 12 Technical Sciences Paper 1 memorandum from the Eastern Cape Mock Exam 2020, 8 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/tech-sciences-p1-gr12-memo-sept2020-ae/916d09e7-a42d-4988-928a-ab6e90ee2fa5
- Updated: 2026-09-27
- Subject: Technical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2020
- Exam period: Eastern Cape Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 8

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Technical Sciences · Grade 12 · Eastern Cape Mock Exam · 2020. Memorandum, 8 pages. Read online or download the PDF.

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### Page 1

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                            NATIONAL
                       SENIOR CERTIFICATE/
                           NASIONALE
                       SENIOR SERTIFIKAAT

                         GRADE/GRAAD 12

                         SEPTEMBER 2020

            TECHNICAL SCIENCES P1/
          TEGNIESE WETENSKAPPE V1
        MARKING GUIDELINE/NASIENRIGLYN

MARKS/PUNTE:   150

                     This marking guideline consists of 8 pages./
                     Hierdie nasienriglyn bestaan uit 8 bladsye.

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2                TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1                (EC/SEPTEMBER 2020)

 QUESTION/VRAAG 1

 1.1    C                                                                                    (2)

 1.2    D                                                                                    (2)

 1.3    C                                                                                    (2)

 1.4    A                                                                                    (2)

 1.5    C                                                                                    (2)

 1.6    A                                                                                    (2)

 1.7    D                                                                                    (2)

 1.8    D                                                                                    (2)

 1.9    B                                                                                    (2)

 1.10 A                                                                                     (2)
                                                                                             [20]

 QUESTION/VRAAG 2

 2.1

                                                                                               (2)

 2.2    An object will continue its state of rest or uniform motion in a straight line
        unless it is acted upon by a net force. 
        ŉ Voorwerp sal is sy toestand van rus bly of aanhou beweeg met
        uniforme beweging in ŉ reguitlyn tensy ŉ netto krag daarop inwerk.                   (2)

 2.3    Block will continue its state of uniform motion (constant velocity). 
        Die blok sal aanhou beweeg met uniforme beweging (konstante
        snelheid).                                                                           (2)
                                                                                               [6]

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(EC/SEPTEMBER 2020)         TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1                      3

 QUESTION/VRAAG 3

 3.1   When a net force acts on an object, the object will accelerate in the direction
       of the force. This acceleration is directly proportional to the net force and
       inversely proportional to the mass of the object. 
       Indien ŉ netto krag op ŉ voorwerp inwerk, sal die voorwerp in die rigting van
       die krag versnel. Die versnelling is direk eweredig aan die netto krag en
       omgekeerd eweredig aan die massa van die voorwerp.                                    (2)

 3.2   No net external force act on the system. 
       Daar is geen eksterne krag wat op die stelsel inwerk nie.                             (2)

 3.3

                                                                                               (5)

       3.4.1 Fnet = ma        Any one / Enige een
             T – f = ma
             50 – (0,2) (20)(9,8) = 20a 
             a = 0,54 m.s-2 left/west / links/ wes
                (Positive marking from QUESTION 3.4.1)(Positiewe merk vanaf VRAAG 3.4.1)       (4)

       3.4.2 OPTION/OPSIE 1
             Fnet = ma
             Fcosθ – T – µkN = ma      Any one / Enige een 
             FH – T – f = ma
             Fcos 30°  – 50 – (0,2)( 30x9,8 – Fsin30) = (30)(0,54) 
             F(cos30° + (0,2) sin30°) = 125
             F = 129,4 N 

               OPTION/OPSIE 2
               Fnet = ma                  Any one / Enige een
               FH – f1 – f2 = ma
               Fcos30° – (0,2)[(30x9,8) – Fsin30]  – (0,2) (20)(9,8) = (50)(0,54)
               F = 129,4 N                                                                    (5)

 3.5   3.5.1 Decreases / Neem af                                                              (1)

       3.5.2 Decreases / Neem af                                                              (1)

 3.6   Fnet decreases, therefore acceleration decreases. 
       Fnet neem af, dus neem die versnelling ook af.                                        (2)
                                                                                              [22]

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4                TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1            (EC/SEPTEMBER 2020)

 QUESTION/VRAAG 4

 4.1    Momentum is the product of mass of the object and its velocity. 
        Momentum is die produk van die massa van die voorwerp en sy
        snelheid.                                                                        (2)

 4.2    Momentum                                                                           (1)

 4.3    ∑pi = ∑pf            East = +ve)(Oos = +)      Any ONE / Enige EEN 
        m1v1i + m2v2i = m1v1f + m2v2f
        (5)(4) + (3)(0)  = (5)( - 0,25) + (3)(v2f)
        v2f = 7,08 m.s-1 east / oos                                                       (4)

 4.4    Law of conservation of momentum / Wet van behoud van momentum 
             (Negative marking)(Negatiewe merk)
        Total linear momentum in a closed system remains constant. 
        Totale lineêre momentum in ŉ geslote sisteem bly konstant.                      (3)
                                                                                         [10]

 QUESTION/VRAAG 5

 5.1    100 000 N  due east / reguit ooswaarts                                           (2)

 5.2     -ve   Newton’s Third law of Motion / Newton se Derde Bewegingswet 
              When object A exerts a force on object B, object B simultaneously
        exerts an oppositely directed force of equal magnitude on object A. 
        Indien voorwerp A ŉ krag op voorwerp B uitoefen, oefen voorwerp B ŉ
        krag uit wat in die teenoorgestelde rigting maar van gelyke grootte op
        voorwerp A uit.                                                                  (3)

 5.3    It is the product of the resultant/net force acting on an object  and the
        time the resultant/net force acts on the object.
        Dit is die produk van die resultante/netto krag wat op ŉ voorwerp inwerk 
        en die tyd wat die resultante / netto krag op die voorwerp inwerk.                (2)

 5.4    OPTION/OPSIE 1
        West is taken as positive / Wes word as positief geneem.
        Fnet ∆t = ∆p 
        Fnet(truck) ∆t = ( mvf – mvi)
        (- 100000)(0,4) = (5000)[(vf) – 20] 
         vf = 12 m.s-1 west / wes 

        OPTION/OPSIE 2
        East is taken as positive / Oos word as positief geneem.
        Fnet ∆t = ∆p 
        Fnet(truck) ∆t = ( mvf – mvi)
        (100000)(0,4) = (5 000)[(vf) – (-20)] 
         vf = - 12 m.s-1
        vf = 12 m.s-1 west / wes                                                         (4)
                                                                                         [11]

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(EC/SEPTEMBER 2020)         TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1                   5

 QUESTION/VRAAG 6

 6.1    6.1.1    Work done is defined as the product of the force applied on an
                 object and the displacement in the direction of the force. 
                 Die arbeid verrig word gedefinieer as die produk van die
                 toegepaste krag op die voorwerp en die verplasing in die rigting
                 van die krag.                                                           (2)

        6.1.2    Tension  Force of gravity 
                 Spanning  Gravitasiekrag                                                (2)

        6.1.3    Fnet = 0 
                 Block of bricks is moving with constant velocity
                 Houer met die bakstene beweeg met konstante snelheid                     (2)

        6.1.4    P = Fv                F = mg = (50)(9,8) = 490 N 
                 P = (490)(5) 
                 P = 2 450 W                                                              (4)

 6.2    6.2.1    Total mechanical energy of a closed system remains constant.
                 Totale meganiese energie van ŉ geslote sisteem bly konstant.            (2)

        6.2.2    ME at a height h above the ground:
                 ME op hoogte h bokant die grond:
                 MEh = ½mv2 + mgh 
                 MEh = ½(0,5)(10)2 + (0,5)(9,8)h  ……..         (1)
                 ME⅔h = ½(0,5)(20,34)2 + (0,5)(9,8)(⅔h )  …… (2)
                    (1) = (2)
                 ½(0,5)(10)2 + (0,5)(9,8)h = ½(0,5)(20,34)2 + (0,5)(9,8)(⅔h )
                 (⅓h)(9,8)(0,5) = 78,43
                 h = 48,02 m 
                 ME = ½(0,5)(10)2 + (0,5)(9,8)(48,02) 
                 ME = 260,3 J                                                             (6)

        6.2.3    (ME)at h = (ME)ground 
                 260,3 = mgh + ½mv2
                 260,3 = 0 + ½(0,5)(v2) 
                 v = 32,27 m.s-1                                                          (3)

 6.3    6.3.1    0J
                 Normal force is perpendicular to the direction of the
                 displacement. 
                 Normaalkrag is loodreg aan die rigting van verplasing.                  (3)

        6.3.2    W = F∆xcosθ 
                 W = (10)(4)(cos0°) 
                 W = 40 J                                                                (3)
                                                                                         [27]

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6                TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1                  (EC/SEPTEMBER 2020)

 QUESTION/VRAAG 7

 7.1    A body which regains its original shape and size completely when the
        deforming force is removed. 
        Dit is ŉ liggaam wat sy oorspronklike grootte en vorm volkome herwin as
        die vervormingskrag verwyder word.                                                       (2)

 7.2    The internal restoring force per unit area is called stress. 
        Die interne herstelkrag per eenheidsoppervlakte van die liggaam word
        druk genoem.                                                                           (2)

 7.3    7.3.1    Stress is directly proportional to strain. 
                 Druk is direk eweredig aan rekking.                                           (2)

        7.3.2    Hooke’s law 
                 Hooke se wet                                                                   (1)

        7.3.3    Tungsten 
                 Graph of tungsten is steeper than steel. 
                 Grafiek van tungsten (wolfram) is steiler as staal.                            (2)

        7.3.4           5,4 ×109 −0
                 m=             
                     3×10−2 −0
                 m = 1,8 × 10  11

                 Modulus of elasticity / Elastisiteitsmodulus = 1,8 × 1011 Pa                   (3)

 7.4    7.4.1    The property of the fluid to oppose relative motion between the
                 two adjacent layers. 
                 Die eienskap van ŉ vloeistof om die relatiewe beweging tussen
                 twee aangrensende vlakke teen te werk.                                        (2)

        7.4.2    Water 
                 The lower the viscosity, the faster the flow. 
                 Hoe laer die viskositeit – hoe vinniger die vloei.                              (2)

        7.4.3    As the temperature of the fluid increases, its viscosity
                 decreases. 
                 Indien die temperatuur van die vloeistof toeneem, verminder sy
                 viskositeit.                                                                  (2)

 7.5    7.5.1    Pascal's law states that in a continuous liquid at equilibrium, the
                 pressure applied at a point is transmitted equally to the other
                 parts of the liquid. 
                 Pascal se wet stel dat in ŉ kontinue vloeistof in ewewig die druk
                 wat by enige punt toegepas word eweredig na die ander dele van
                 die vloeistof versprei word.                                                  (2)

        7.5.2    F1        F2
                       =        
                 A1        A2
                      1765              F2
                                =              
                 284×10−6            507×10−6

                 F2 = 3 150,9 N                                                                 (4)

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(EC/SEPTEMBER 2020)                   TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1                    7

        7.5.3       Increases / Neem toe                                                                   (1)

        7.5.4       Bulldozer’s working systems, dentists’ chairs, hydraulic lifts used to
                    lift heavy loads, car jacks  (ANY TWO)
                    Stootskraper se werkende sisteme, tandarts-stoele, hidrouliese
                    hysers om swaar vragte te lig, motors se domkragte. 
                    (ENIGE TWEE)                                                                        (2)
                                                                                                       [27]

 QUESTION/VRAAG 8

 8.1   8.1.1    The process of adding impurities to intrinsic / inpure
                semiconductors.
                Die proses waardeur onsuiwerhede by intrinsieke / onsuiwer
                halfgeleiers gevoeg word.                                                                 (2)

       8.1.2    5                                                                                          (1)

       8.1.3    n-type / n-tipe                                                                           (2)

       8.1.4    In n-type material current is due to electrons whereas in p-type the
                current is due to holes. 
                In n-tipe materiaal is die stroom as gevolg van elektrone terwyl in p-
                tipe is die stroom as gevolg van holtes.                                                  (2)
                        𝜀 𝐴
 8.2   8.2.1    C= 0             
                         𝑑
                         8,85 ×10−12 ×10×10−4
                C =                   
                          1×10−2
                C = 8,85 × 10 F 
                              -13

                Q = CV 
                Q= (8,85 × 10-13)(12) 
                Q = 1,06 × 10-11 C                                                                         (6)

       8.2.2    Zero / Nul                                                                                 (1)

       8.2.3    Area of the plate / Oppervlak van die plate
                Distance between the plate / Afstand tussen die plate. ( Any TWO)
                Dielectric used / Diëlektrikum gebruik                 (Enige TWEE)                         (2)

 8.3   8.3.1    Energy transferred per second. / Energie oorgedra per sekonde.                            (2)

       8.3.2    1       1        1                                 𝑅1 𝑅2
                    =        +                             Rp =             
                𝑅       𝑅1       𝑅2                                𝑅1 + 𝑅2

                1       1        1                                   26 × 52
                    =        +                    OR/OF       =                 
                𝑅       26       52                                  26 + 52

                1       3
                    =
                𝑅       52
                                                               = 17,33 Ω 
                R = 17,33 Ω                                                                                (3)

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8                TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1          (EC/SEPTEMBER 2020)

      8.3.3    OPTION/OPSIE 1              OPTION/OPSIE 2
                    𝑉2                       𝑉  230
               P=                         I= =     = 8,85 A 
                     𝑅                        𝑅     26
                    (230)2                 P = VI
               P=       
                    26
               P = 2 034,62 W             = (230)(8,85)  = 2 034,62 W               (3)

      8.3.4    OPTION/OPSIE 1              OPTION/OPSIE 2
               W = Pt                     W = VIt 
                 = (2 034,62)(2)(60)        = (230)(8,85)(120) 
                 = 244 154,4 J              = 244 154,4 J                           (3)
                                                                                     [27]

                                                             TOTAL/TOTAAL:            150

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