# Technical Sciences Grade 12 Paper 1 NSC June 2023 Memo

> Grade 12 Technical Sciences Paper 1 memorandum from the NSC June Exam 2023, 16 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/technical-sciences-p1-may-june-2023-mg-afr-eng/0daca110-3e48-4873-a1ba-771fa67fd9d7
- Updated: 2026-10-29
- Subject: Technical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2023
- Exam period: NSC June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 16

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## About this document

Technical Sciences · Grade 12 · NSC June Exam · 2023. Memorandum. Read online or download the PDF.

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### Page 1

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             SENIOR CERTIFICATE EXAMINATIONS/
         NATIONAL SENIOR CERTIFICATE EXAMINATIONS
               SENIORSERTIFIKAAT-EKSAMEN/
          NASIONALE SENIORSERTIFIKAAT-EKSAMEN

                              TECHNICAL SCIENCES P1
                             TEGNIESE WETENSKAPPE V1

                                          2023

                      MARKING GUIDELINES/NASIENRIGLYNE

                                                                 MARKS/PUNTE: 150

                     These marking guidelines consist of 16 pages./
                      Hierdie nasienriglyne bestaan uit 16 bladsye.

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Technical Sciences/P1/Tegniese Wetenskappe/V1        2                                   DBE/2023
                               SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION 1/VRAAG 1

1.1       C                                                                                           (2)

1.2       B                                                                                           (2)

1.3       A                                                                                           (2)

1.4       C                                                                                           (2)

1.5       A                                                                                           (2)

1.6       C                                                                                           (2)

1.7       D                                                                                           (2)

1.8       C                                                                                           (2)

1.9       A/D                                                                                         (2)

1.10      D                                                                                            (2)
                                                                                                        [20]

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QUESTION 2/VRAAG 2
2.1.1     The passengers in the moving bus will continue moving with the same
          velocity as the bus due to inertia and the passengers in a stationary bus will
          remain at rest due to inertia. 
          Die passasiers in die bewegende bus sal aanhou beweeg teen dieselfde
          snelheid as die bus agv traagheid en die passasiers in 'n stilstaande bus sal
          in rus bly agv traagheid.                                                                      (2)
2.1.2     Newton’s first law of motion. 
          An object will remain at rest or continue moving at a constant velocity (or at
          constant speed in a straight line) unless a non-zero external resultant/net
          force/unbalanced force acts on it.. 
          Newton se eerste bewegingswet
          'n Voorwerp sal in sy toestand van rus of uniforme beweging volhard tensy 'n
          nie-nul resulterende/netto krag daarop inwerk.                                                 (3)
2.2.1
                                            N/FN

                              T                               FA

                                            w/Fg
                                            /
           ACCEPTABLE LABELS/                       NOTES/NOTAS:
           AANVAARBARE BYSKRIFTE:

              N/FN/Normal/Normaal/12 740 N        One mark for each force
                                                   represented by an arrow with a
              FA /10 500 N/ Fenjin/Ftoegepas      correct label.
              Fg/w/Force due to gravity/          Een punt vir elke krag voorgestel
               weight/12 740 N/Gravitasiekrag/     deur 'n pyl met korrekte byskrif.
               Gewig
                                                    Penalise once for each of the
              T/Ft/Tension/Spanning                following:/ Penaliseer (een keer)
                                                    vir elk van die volgende:
                                                    No arrows/Geen pyltjies
                                                    There is no dot/Geen kol nie
                                                    Gap between the line and the
                                                       dot/ Spasie tussen lyn en kol
                                                    Dotted lines are
                                                       used/Stippellyne gebruik
                                                    Additional force is included/
                                                       Ekstra krag is ingesluit
                                                    A force diagram is given / 'n
                                                       Kragtediagram word gegee                         (4)

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2.2.2      OPTION 1/ OPSIE 1                         OPTION 2/ OPSIE 2
           To the right is positive/Na regs is       To the right is negative/Na regs
           positief.                                 is negatief.
           For the 1300 kg car/Vir die 1300 kg       For the 1300 kg car/ Vir die
           kar                                       1300 kg kar
           Fnet = ma                                 Fnet = ma           Any one
                              Any one
           FA +(– T) = ma                            -FA +( T) = ma
                                                     -10500 + T = 1300a .......…….(1)
           10500 – T = 1300a ....………….(1)
                                                     For the 900 kg trailer
           For the 900 kg caravan                    T = - 900a ……….……………(2)
           T = 900a..………….……………(2)
                                                     Add (1) with (2)
           Add (1) with (2)                          -10500  = 2200a
           10500 = 2200a                                  ∴ a = - 4,77 m∙s-2
              ∴ a = 4,77 m∙s-2 right  /regs               a = 4,77 m∙s-2 right /regs

           OPTION 3                                  OPTION 4
             To the right is positive/Na regs        To the right is negative/Na regs
           is positief.                              is negatief.
           Fnet = ma 
           10 500  = (1300+900)a                    Fnet = ma 
                   a = 4,77 m∙s-2 right /regs       -10 500 = (1300+900)a 
                                                           a = - 4,77 m∙s-2
                                                           a = 4,77 m∙s-2 right /regs
                                                                                                        (4)

2.2.3     POSITIVE MARKING FROM QUESTION 2.2.2
          POSITIEWE NASIEN VANAF VRAAG 2.2.2
           OPTION 1/ OPSIE 1           OPTION 2/ OPSIE 2
           T = 900(4,77)              Fnet = ma
           T = 4293 N                               10500 - T = 1300(4,77) 
           Accept/Aanvaar T = 4295,45 N                      T = 4299 N 
                                                     Accept/Aanvaar T = 4295,46 N
           Range/ Gebied: 4293 N to 4299 N                                                              (2)

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2.3.1      OPTION 1/ OPSIE 1                        OPTION 2/ OPSIE 2
           w = mg                                   Fnet = fk + mg 
           w = 1600 x 9,8                           Fnet = – 3700 + (1600)(9,8) 
           w = 15680 N                                                            
           Fnet = w – Fupwards                     Fnet = 11980 N downwards/afwaarts
                = 15680 – 3700 
                                          
               = 11980 N downwards/afwaarts                                                            (3)

2.3.2     POSITIVE MARKING FROM QUESTION 2.3.1
          POSITIEWE NASIEN VANAF VRAAG 2.3.1
           OPTION 1/OPSIE 1                 OPTION 2/OPSIE 2
           Downward is positive/Afwaarts is Downward is negative/ Afwaarts is
           positief.                        negatief.
             Fnet = ma                         Fnet = ma 
           11980 = 1600a                    -11980 = 1600a 
              a = 7,49 m∙s-2 downwards           a = -7,49 m∙s-2
                              afwaarts            a = 7,49 m∙s-2downwards
                                                                  afwaarts                             (3)

2.4.1     When object A exerts a force on object B, object B simultaneously exerts an
          oppositely directed force of equal magnitude on object A. 
          Wanneer voorwerp A 'n krag op voorwerp B uitoefen, sal voorwerp B
          tegelykertyd 'n teenoorgesteld gerigte krag met 'n gelyke grootte uitoefen op
          voorwerp A.                                                                                  (2)
2.4.2     Force exerted by the apple on the earth / Fapple on earth
          Force exerted by the earth on the apple / Fearth on apple
          Krag uitgeoefen deur die appel op die aarde / F appel op aarde
          Krag uitgeoefen deur die aarde op die appel /F aarde op appel                                 (2)
                                                                                                       [25]

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QUESTION 3/VRAAG 3

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

3.2.1     OPTION 1/OPSIE 1                                OPTION 2/ OPSIE 2
          Right is positive/ Regs is positief.            Right is negative/Regs is
          Σpi = Σp f                                      negatief.
                                            Any
                                             one/        Σpi = Σp f                Any
          mAv i + mBv i = mAvf + mBvf
                                                          m v +m v =m v +m v         one/
                                            Enige           A i     B i     A f     B f
          mAv i + mBv i = (mA + mB )v f     een                                      Enige
                                                          mAv i + mbv i = (mA + mB )v f
          100((+5) + 75(-4) = (100 + 75) v f                                       een
                                                          100(-5) +75(+4)  = (100 + 75) v f 
                       v f = +1,14
                                                                 v f = -1,14
                       v f = 1,14 m∙s -1 right/regs
                                                                  v f = 1.14 m∙s -1 right/regs    
                                                                                                           (4)

3.2.2     Principle of conservation of linear momentum. Die beginsel van behoud van
          lineêre momentum.
          The total linear momentum of an isolated system remains constant/is
          conserved (in both magnitude and direction).

          Die totale lineêre momentum in 'n geïsoleerde sisteem bly konstant/bly
          behoue in grootte en rigting.
          OR/OF
          In an isolated system, the total linear momentum before a collision is equal to
          the total linear momentum after the collision in both magnitude and direction.

          In 'n geïsoleerde sisteem is die totale lineêre momentum voor die botsing
          gelyk aan die totale lineêre momentum na die botsing in beide grootte en
          rigting.                                                                                         (3)

3.3.1     The net (resultant) force acting on an object is equal to the rate of change of
          its momentum in the direction of the net (resultant) force. 

          Die netto (resulterende) krag wat op 'n voorwerp inwerk is gelyk aan die
          tempo van verandering in momentum, in die rigting van die netto
          (resulterende) krag.                                                                              (2)

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3.3.2                                  1
          Area of /Oppervlakte van  = 2 bh 
                                       1
          Area of /Oppervlakte van  = 2 (2 × 200) 
          Area of /Oppervlakte van  = 200 N∙s  or 200 kg∙m∙s -1                                         (3)

3.3.3     Impulse/Impuls                                                                                 (1)

3.3.4                              Fnet (N) versus time (s) graph

                                        1       2

                         MARKING CRITERIA/NASIEN KRITERIA
           Correct shape of the graph/ 
           Korrekte vorm van die grafiek
           Magnitude of maximum Fnet of graph B is lower than that in graph A 
           Maksimum grootte van Fnet by grafiek B is laer as in grafiek A
           Value of contact time of graph B is greater than that in graph A 
           Waarde van die kontaktyd van grafiek B is groter as in grafiek A
           Accept ONE graph if values of Fnet; less than 200 N; and contact time;
           more than 2 s; are indicated.
           Aanvaar EEN grafiek indien die waarde van Fnet minder is as 200 N
           en die tyd minder is as 2 s, aangedui is op die grafiek.           3/3
           One graph; no values are indicated; shape of the graph.            1/3
           Indien EEN grafiek aangedui is met geen waardes; vorm van grafiek
           If graphs are not labelled A and B; correct shape of the graph.    1/3
           Indien die grafieke nie benoem is nie; korrekte vorm.                                          (3)

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3.4       Crumple zones 
          Airbags 
          Safety belts
          Padded dashboards                                                  Any two

          Krummel sone
          Lugsakke
          Veiligheids gordels
          Opgestopte paneelborde                                           Enige twee
                                                                                                        (2)
3.5       During collision the air bags/crumple zone/padded dash boards increase the
          time taken to reach (the hard surface of ) the dash board/car engine/steering
          wheel , thus decreases the net force/force of impact and the extent of
          injuries. 

          Gedurende botsings sal die lugsakke/ krummelsones en opgestopte
          paneelborde die tyd verleng voor die harde oppervlakte van die
          paneelbord/stuurwiel/enjin getref word, en verlaag dus die netto krag/krag van
          impak en die graad van beserings.                                                              (2)
                                                                                                        [22]

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QUESTION 4/VRAAG 4

4.1.1      OPTION 1/ OPSIE 1                       OPTION 2/ OPSIE 2
          W = F∆xcosθ                              ∆Ep = mg∆h 
                            
          W = mg∆xcosθ                             ∆Ep = (600)(9,8)(25) 
          W = (600)(9,8)(25)(cos0º)               ∆Ep = 1,47×105 J 
          W = 1,47×105 J 
                                                                                                        (3)

4.1.2     POSITIVE MARKING FROM 4.1.1/POSITIEWE NASIEN VAN 4.1.1
          OPTION 1/ OPSIE 1
          PO                           OPTION 2/ OPSIE 2
             W                        Pave = Fv ave
          P=                                                 Any one/
             Δt                                    Δy    Enige een
                                            = mg
             1,47 ×10 5                           Δt
           =
                120                                       25 
                                            = (600)(9,8)(     )
           = 1225 W                                      120 
                                            = 1225 W 
                                                                                                        (4)

4.2      The energy that an object has because of its position above the surface of
         the Earth. 
         Die energie wat n voorwerp besit agv sy posisie bo die aard oppervlakte                        (2)

4.3.1         1
         Ek =   mv 2 
              2        Any one/
             1         Enige een
         K = mv 2
             2
              1
           = (3)(7)2 
              2
            = 73,5 J                                                                                   (3)

4.3.2     POSITIVE MARKING FROM QUESTION 4.3.1
          POSITIEWE NASIEN VANAF VRAAG 4.3.1
           OPTION 1/ OPSIE 1
                        (Ep + Ek)top = (Ep + Ek)bottom 
                        1                              1
           (3)(9,8)h + ( )(3)(0)2  = (3)(9,8)(0) + ( )(3)(7)2 
                        2                              2
                                  h = 2,5 m 
           OPTION 2/ OPSIE 2
           ∆Ep = ∆Ek  = 73,5 J
           73,5  = (3)(9,8)h 
                h = 2,5 m                                                                               (4)
                                                                                                        [16]

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QUESTION 5/VRAAG 5

5.1      In a continuous liquid in equilibrium, the pressure applied at a point is
         transmitted equally to the other parts of the liquid.                 I
         In 'n kontinue vloeistof by ewewig word die druk by enige punt eweredig
         na al die ander dele van die vloeistof oorgedra.                                                (2)

5.2           OPTION 1/OPSIE 1                    OPTION 2/OPSIE 2
                                     πd 2         Area/Oppervlakte = πr 2
              Area / Oppervlakte =
                                      4
                                                        = π(0,02523)2 
                      π(5,046  10 2 ) 2
                                                      = 1,9998 x 10-3 m2
                             4
                    = 1,9998 x 10-3 m2
                                                   F1  F
                                                      2 
                                                   A1 A 2
                  F1  F
                      2                              200              F2
                  A1 A 2                                            =
                                                  1,99 98 ×10 -3        5,25 
                  200              F2
                               =                  F2 = 525 052,51 N 
              1,9998×10 -3         5,25 

              F2 = 525 052,51 N 
              OPTION 3/OPSIE 3                    OPTION 4/ OPSIE 4
                 F1   F                           F1  F
                     2                              2 
                 A1 A 2                           A1 A 2

                   200            F                   200       F
                                 = 2                            = 2
              π(5,046 × 10 )
                          -2 2
                                  5,25            π(0,02523) 
                                                             2
                                                                 5,25 
                               
                    4
                                                  F2 = 525 055,15 N 
              F2 = 525 055,15 N 
              Range/Gebied: 100 480 N - 525 055,15 N                                                     (5)

5.3      Remain the same / Bly dieselfde                                                                (1)

5.4            (Hydraulic) brakes/(Hidrouliese) remme
               (Hydraulic) car lifts/jacks/(Hidrouliese) motor hysers/ domkrag
               (Hydraulic) dentist chairs/(Hidrouliese) tandaarts stoele
               Forklifts/(Hidrouliese) hysers
               (Hydraulic) Steering systems/(Hidrouliese) stuurwiele
               (Hydraulic) Press/(Hidrouliese) drukpers               Any two/Enige twee                (2)

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5.5.1                  σ
                K=       
                       ε
                           σ
          2 × 1011 =
                       0,16 ×10 -2 
                             8
                σ = 3,2 ×10 Pa                                                                           (3)

5.5.2    POSITIVE MARKING FROM QUESTION 5.5.1
         POSITIEWE NASIEN VANAF VRAAG 5.5.1
          OPTION 1/OPSIE 1           OPTION 2/OPSIE 2
                               πd 2  Area/Oppervlakte = πr 2
          Area / Oppervlakte =
                                4
                                           π(0,01)2 
                  π(0,02)2
               =                       A = 3,142×10-4 m2
                     4

              A = 3,142×10-4 m2                                   F
                                                            σ=      
                                                                  A
                             F                                      F
                        σ=                         3,2 ×10 8 =              
                             A                                  3,142×10 - 4
                                 F                          F = 100 544 N 
               3,2 ×10 8 =                
                             3,142×10 - 4
                         F = 100 544 N 

           OPTION 3/OPSIE 3                       OPTION 4/OPSIE 4
                        F                                      F
                   σ=                                    σ= 
                        A                                      A
                            F                                     F           
            3,2 ×10 8 =                            3,2 ×10 8 =
                         π(0,02)2                             π(0,01)2 
                                  
                            4                              F = 100 530,96 N 
                    F = 100 530,96 N 

           Range/Gebied: 100 530,96 N - 100 544 N
                                                                                                          (4)
                                                                                                         [17]

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QUESTION 6/VRAAG 6

6.1.1     The change in direction of a wave upon striking the interface between two
          materials. 
          OR
          Bouncing back of a wave when it strikes an interface between two media.

          Die verandering van rigting van 'n golf `wanneer dit die oorgangsfase tussen
          twee stowwe/materiale tref.
          OF
          Die terugkaatsing van 'n golf wanneer dit die oorgangsfase tussen twee
          mediums tref.                                                                               (2)
6.1.2     Reflected ray 
          Geweerkaatste straal/ weerkaatsingsstraal                                                   (1)

6.1.3     Angle of incidence 
          Invalshoek                                                                                  (1)

6.1.4     Angle of reflection 
          Weerkaatsingshoek                                                                           (1)

6.1.5     Angle 1 (angle of incidence) is equal to angle 2 (angle of reflection) 
          Hoek 1 (invalshoek) is gelyk aan hoek 2 (weerkaatsingshoek)                                 (2)

6.2.1     Total internal reflection 
          Totale interne weerkaatsing                                                                 (1)

6.2.2        Optic fibre in telecommunication 
             Automotive rain sensors 
             Optical fingerprinting devices
             Sparkling brilliance of diamonds
             Binoculars
             Periscopes
             Endoscope                                                           Any two

             Opties viber in telekommunikasie
             Outomatiese reënsensors
             Optiese vingerafdruktoestelle
             Glinstering in diamante
             Verkykers
             Periskoop
             Endoskoop                                                          Enige twee           (2)

6.2.3        The light must be travelling from a more dense medium into a less dense
              medium (i.e. glass to air).
             The angle of incidence must be greater than the critical angle .
              Die lig moet vanaf n opties digter medium na n opties minder digte
              medium beweeg.
              Die invalshoek moet groter as die kritiese hoek vir daardie medium wees.                (2)

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6.3.1     Convex 
          Konveks                                                                                     (1)

6.3.2        Real 
             Inverted 
             Enlarged
             Beyond 2F in the opposite side of the lens                       Any two

              Reëel
              Omgekeerd
              Vergroot
              Verder as 2F aan die teenoorgestelde kant van die lens.         Enige twee              (2)

6.3.3        Magnifying glasses 
             Optical microscopes 
             Projectors 
             Cameras 
             Human eye
             Telescope
             Eye glasses
             Peep holes in doors
             Binoculars                                                      Any two

              Vergroot glase
              Optiese mikroskope
              Projektors
              Kameras
              Menslike oog
              Teleskoop
              Brille
              Loergaatjies in deure
              Verkykers                                                        Enige twee             (2)

6.3.4     The refracted rays (and the rays passing through the optical centre) are
          parallel/do not meet, therefore the image will not be formed. 
          Die gebuigde strale (en die strale wat deur die optiese middelpunt gaan) loop
          parallel/hulle kruis nie, daarom word geen beeld gevorm nie.                                 (2)
                                                                                                      [19]

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                                     SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION 7/VRAAG 7

7.1.1   E = hf 
        E= 6,63 x 10 -34 x 2,4 x 109 
        E= 1,59 x 10 -24 J                                                                           (3)

7.1.2    OPTION 1/OPSIE 1                OPTION 2/ OPSIE 2
               c                         POSITIVE MARKING FROM QUESTION 7.1.1
          λ=     
               f                         POSITIEWE NASIEN VANAF VRAAG 7.1.1
                                                      hc
                 3 ×108                           E=      
          λ =                                          λ
                         
                2,4 ×109                               3 ×108 × 6,63 ×10 -34
                                                  -24
                                          1,59 ×10 =                         
          λ = 0,125 m                                          λ
                                                   λ = 0,125 m 
                                                                                                      (3)

7.2       POSITIVE MARKING FROM QUESTION 7.1.2
          POSITIEWE NASIEN VANAF VRAAG 7.1.2
          Microwaves 
          Mikrogolwe                                                                                  (1)

7.3     POSITIVE MARKING FROM QUESTION 7.1.2 AND QUESTION 7.2
        POSITIEWE NASIEN VANAF VRAAG 7.1.2 EN VRAAG 7.2
         Microwave oven/cooking 
         Cellphones/telecommunicationAny one
          Mikrogolf oond/ kook
          Sellulêre fone/ telekommunikasie           Enige een                                        (1)
                                                                                                      [8]

QUESTION 8/VRAAG 8

8.1       A device that stores electrical charge. 
          'n Toestel wat elektriese lading stoor.                                                     (2)

8.2       The capacitance is directly proportional to the charge on the plates. 
          Die kapasitansie is direk eweredig aan die lading op die plate.

          OR/OF
          CαQ                                                                                         (2)

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8.3.1        Q
        C=     
             V
                 0,3 ×10 -6
        6 ×10 =-12
                            
                     V
             V = 50 000 V or 50 kV                                                                   (3)
8.3.2                  εοΑ
                     C=     
                         d
                       (8,85 × 10 -12 )( Α )
          6 × 10 -12 =                       
                           5 × 10 -2
                  A = 3,39 x 10-2 m2                                                                  (3)
                                                                                                      [10]

QUESTION 9/VRAAG 9

9.1.1    Potential differenceVoltage/ V
         Potensiaalverskil                                                                            (1)

9.1.2    Current/I
         Stroom                                                                                       (1)

9.2      The current passing through the resistor R is directly proportional to the
         potential difference accross its ends when temperature is kept
         constant.
         As the current passing through the conductor increases the potential
         difference across the ends of the conductor also increases.                                  (2)

         Die stroom wat deur resistor R vloei is direk eweredig aan die potensiaal-
         verskil oor sy punte mits/indien die temperatuur konstant bly.

9.3
          OPTION 1                       OPTION 2             OPTION 3
                                                                                                      (3)
          P = VI                              V              I = 0,6 A ; V= 1,5 V
            = 1,5 x 0,6                  R=
                                               I
            = 0,9 W                           1,5                 V2
                                             =                P=       
                                               0,6                 R
                                             = 2,5                 (1,5) 2
                                                                 =         
                                                                     2,5
                                         P = I2R                = 0,9 W
                                           = (0,6)2(2,5)
                                           = 0,9 W

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         Lower than
9.4      Laer as                                                                                (1)

9.5      NEGATIVE MARKING FROM QUESTION 9.4
         The gradient of graph B is lower than that of graph A.The gradients of
         the graphs is the inverse of resistance,  therefore graph B will have
         higher resistance.
         OR
         For the same current the potential difference in graph B is higher than the
         that of graph A. Potential difference is directly proportional to resistance,
         therefore graph B will have higher resistance.
         OR
         For the same current the power output in graph B is higher than that in
         graph A. Power is directly proportional to resistance, therefore graph B will
         have higher resistance.

         Die gradiënt van grafiek B is laer as die van grafiek A. Die gradiënte van
         die grafieke is die inversie/omgekeerde van die weerstand, daarom sal
         grafiek B 'n hoër weerstand het.
         OF
         Met dieselfde stroom sal die potensiaalverskil in grafiek B is hoër wees as
         dit van grafiek A. Potensiaalverskil is direk eweredig aan weerstand, dus
         sal grafiek A 'n hoër weerstand hê.
         OF
         Met dieselfde stroom is die drywing vir grafiek B hoër as vir grafiek A.
         Drywing is direk eweredig aan weerstand, dus sal grafiek B 'n hoër
         weerstand hê.                                                                            (2)
                                                                                                 [10]

QUESTION 10/VRAAG 10

10.1     Decrease
         Verlaag                                                                                 (1)

10.2     Increase
         Verhoog                                                                                 (1)

10.3     Increase
         Verhoog                                                                                 (1)
                                                                                                 [3]
                                                                        TOTAL/TOTAAL:            150

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