# Technical Sciences Grade 12 P1 Gauteng Prelim 2024 Memo

> Grade 12 Technical Sciences Paper 1 memorandum from the Gauteng Prelim Exam 2024 (English), 9 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr-12-technical-sciences-p1-english-2024-possible-answers/59537e86-1667-495e-9909-26971e5c41db
- Updated: 2026-09-27
- Subject: Technical Sciences
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2024
- Exam period: Gauteng Prelim Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 9

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## About this document

Technical Sciences · Grade 12 · Gauteng Prelim Exam · 2024 · English. Memorandum, 9 pages. Read online or download the PDF.

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- [Gr 12 Technical Sciences P1 (English) 2024 Question Paper](https://www.hlayiso.com/library/gr-12-technical-sciences-p1-english-2024-question-paper/ab8131c1-99a0-450f-885d-25eaf13497ec) — QP · 2024 · Eng

## Document text

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      PREPARATORY EXAMINATION

                        2024

           MARKING GUIDELINES

          TECHNICAL SCIENCE (PAPER 1) (11101)

9 pages

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                                           11101/24

QUESTION 1: MULTIPLE-CHOICE QUESTIONS

1.1   A                                      (2)

1.2   REMOVED                                  (2)

1.3   B                                      (2)

1.4   D                                      (2)

1.5   A                                      (2)

1.6   C                                      (2)

1.7   A                                      (2)

1.8   B                                      (2)

1.9   D                                      (2)

1.10 D                                       (2)
                                              [18]

QUESTION 2: MATCHING ITEMS

2.1   D                                       (1)

2.2   H                                       (1)

2.3   A                                       (1)

2.4   F                                       (1)

2.5   B                                       (1)

2.6   G                                       (1)

2.7   C                                       (1)

2.8   E                                       (1)
                                               [8]

                                  2

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QUESTION 3

3.1 3.1.1 The body will remain at rest or continue to move in a straight line at a
          constant velocity unless it is acted by a net force.                         (2)

     3.1.2

                                             OR

                                                                                         (3)

             CRITERIA                   MARKS
             Tension van A (T)          
             Tension of B (Fb)          
             Gravitational force (Fg)   

     3.1.3 Option 1
           W = mg 

             9000 = m x 9,8

                  9000
             𝑚𝑚 = 9,8

               = 918,37 kg                                                              (3)

     3.1.4 Option 1                                   Option 2

             Fnet = 0                                 Fnet = 0
             FAx + TBx = 0                           Fay + Fby + Fg = 0
             FAcosƟ + (- TBcosα) = 0                 FAsinƟ + (- Fbsinα) = 0
             6 500 cos30o – Tcos45o  = 0           6500sin300 + Fbsin450 – 9000 = 0

             T = 6 500 cos30o                         Fbsin450 = 5 750
                   cos45o
               = 7 960,84 N                                     = 8 131,73 N            (4)

                                                3

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3.2 3.2.1 When the net force is exerted on an object, the object will accelerate in the
          direction of the net force with the acceleration that is directly proportional
          to the net force and inversely proportional to the mass of the object.            (2)

     3.2.2 Fnet = ma
             FA + Fbx + Ff = ma              
             FA + FBcosƟ + Ff = ma
             FA + 70cos30o  – 25 = 200 x 2 
             FA = 364,38 N to the right                                                      (4)

     3.2.3 Increases                                                                         (1)

     3.2.4 Increasing the angle will increase the y component of FB that is acting
           downward which will add to the gravitational force. Adding the component
           of FB to gravitational force will increase the normal force.                     (2)
                                                                                             [21]

QUESTION 4

4.1 4.1.1    The total linear momentum of an isolated system remains constant in
             magnitude and direction.                                                       (2)

     4.1.2   Σp before = Σp after
                 or
             (m1 + m2)vi = m1v1f + m2v2f 
             (2 100 + 1 200)7  = 2 100vf + 1 200 x 10 
             23 100 = 2 100vf + 12 000
             Vf = 5,29 m.s-1                                                                 (4)

4.2 4.2.1    The product of the net force acting on an object and the time the net force
             acts on the object.                                                            (2)

     4.2.2   Fnet Δt = ΔP
             Increasing time decreases the net force./The relationship is inversely
             proportional.                                                                  (2)

     4.2.3   Equal                                                                           (1)

     4.2.4   During the collision the car is still moving at the same speed as it was. The
             airbags will be deployed and the body will hit the airbag with the same
             speed. 
             The airbags will slow down the body by increasing the time of impact and
             decreasing the force of the impact. 
             Time is inversely proportional to the force.                                    (3)

                                                 4

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     4.2.5    OPTION 1                                    OPTION 2
              Fnet Δt = Δp                               Fnet Δt = Δp
              Fnet Δt = m(vf – vi)                        Fnet Δt = m(vf – vi)
              -55000 x Δt = 1000(0 – 12)              55000 x Δt = 1000(0 – (-12))
              Δt = 0,22 s                                           Δt = 0,22 s                 (4)
                                                                                                [18]

QUESTION 5

5.1 The total mechanical energy in an isolated system remains constant.                        (2)

5.2 Ep = mgh 
     Ep = 250 x 9,8 x 30 
         = 73 500 J                                                                             (3)

5.3 Ek = ½mv2 
     Ek = ½ x 250 x 52 
        = 3 125 J                                                                               (3)

5.4 The sum of kinetic energy and potential energy.                                            (2)

5.5 EM = mgh + ½mv2 
         = 2 x 9,8 x 2 + 0             (If 0 is omitted then 2/3)
         = 39,2 J                                                                               (3)

5.6 Positive marking from 5.5
     MEtop = MEbottom
     EP + EK = EP + EK                      
     mgh + ½mv2 = mgh + ½mv2
     2 x 9,8 x 2 + ½ x 2 x 02  = 2 x 9,8 x 0 + ½ x 2 x v2 
     39,2 + 0 = 0 + v2
             v = 6,26 m.s-1                                                                     (4)
                                                                                                [17]

QUESTION 6

6.1 Within the elastic limit, the stress is directly proportional to the strain.               (2)

6.2 •     Stress ball.
    •     Spring.                                                                (any two) 
    •     Kicking a rugby or soccer ball.                                                        (2)

                                                 5

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6.3 •   Power steering.
    •   Shock absorbers.
    •   Windshields.                                                       (any two) 
    •   Brake.
    •   Dentists' chair.                                                                   (2)

6.4 6.4.1    σ=F
               A
               =        F          
                       πr2
               =       6 000   
                   π x 0,025 
                             2

               = 3 055 744,91 Nm-2       
               = 3,06 x 106 Nm-2                                                           (4)

     6.4.2   ε = Δl 
                   L
               = 0,003 
                  5
               = 0,0006 
               = 6 x 10-4                                                                  (3)

     6.4.3   POSITIVE MARKING FROM 6.4.1 and 6.4.2

             OPTION 1                             OPTION 2

             K=σ                                 K=σ
                   ε                                ε
               = 3,06 x 106                        = 3 055 744,91
                 6 x 10-4                            6 x 10-4
               = 5,1 x 109 N.m-2                  = 5,09 x 109 N.m-2                      (2)
                                                                                          [15]
QUESTION 7

7.1 Distance between the plates, area of the plates and dielectric of material used.
    (any two).                                                                           (2)

7.2 7.2.1    C = εo A 
                  d
               = 8,85 x 10-12 x 0,04 
                   0,002
               = 1,77 x 10-10 CV-1          (Or the unit F can be used)                   (3)

                                              6

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     7.2.2   POSITIVE MARKING FROM 7.2.1

             C=Q
               V
             1,77 x 10-10 = Q 
                            12
             Q = 2,12 x 10-9 C 
                                                                                            (3)
                                                                                            [8]
QUESTION 8

8.1 The electric current is proportional to the voltage and inversely proportional to the
    resistance when the temperature is constant.
    OR the potential difference across the conductor is directly proportional to the
    current in the conductor at constant temperature                                      (2)

8.2 8.2.1 Rp = R5 x R3                            OR
               R5 + R3
                                                    1   1   1
                                                      =   +   
                = 5x3                              RP R5 R3
                  5+3
                                                       1 1
             Req = Rp + R2                         =    + 
                                                       5 3
                 = 1,88 + 2 
                 = 3,88 Ω                         Rp = 1,88 Ω

                                                   Req = Rp + R2
                                                        = 1,88 + 2 
                                                        = 3,88 Ω                           (4)

     8.2.2 R = V 
               I

             3,88 = 3 
                    I

             V = IR

             I = 0,77 A
               = 0,77 x 2 
              = 1,54 V                                                                     (4)

8.3 8.3.1 The rate at which energy is transferred. /The rate at which work is
          done.                                                                           (2)

                                               7

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      8.3.2 P = V2 
                R

               = 2202 
                 18

               = 2 688,89 W
                                  
               = 2,67 kW                                                     (3)

      8.3.3 • Stove
            • Heater                                         (any two) 
            • Kettle
            (Any other relevant appliances are correct)                      (2)
                                                                            [17]

QUESTION 9

9.1

        CRITERIA                                   MARKS
        Object (at 2F)                             
        Inverted Image                             
        Image position (2F)                        
        Light ray passing through the centre       
        Light ray passing through F2.              
        Correct Lens (Convex lens)                                          (6)

9.2    9.2.1   Red                                                          (1)

       9.2.2   Violet                                                       (1)

       9.2.3   Dispersion                                                   (1)

       9.2.4   The phenomenon whereby light breaks up into its component
               colours.                                                    (2)
                                                                            [11]

                                               8

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QUESTION 10

10.1 • They originate from accelerated electric charges.
     • Propagate as changing electric field and magnetic fields that are perpendicular
       to each other.
     • They are transverse waves.
     • They transfer energy.
     • They can travel through the vacuum and a medium.
     • They have a speed of 3 x 108 m·s-1                                    (any two)    (2)

10.2 10.2.1 Infrared                                                                     (1)

       10.2.2 Infrared                                                                   (1)

10.3 10.3.1 E = hf 
                = 6,63 × 10-34 × 5,56 × 1014 
                = 3,69 × 10-19 J                                                         (3)

       10.3.2 E = hc 
                   ʎ
                = 6,63 × 10-34 × 3 × 108 
                          390 x 10-9 
                = 5,1 × 10-19 J                                                          (4)
                                                                                         [11]

QUESTION 11

11.1   Commutator (split ring)                                                           (1)

11.2 Anticlockwise                                                                       (1)

11.3 Electrical energy to mechanical energy                                             (2)

11.4 DC generator has commutator (split rings) and an AC generator has slip
     rings.                                                                             (2)
                                                                                          [6]

                                                                              TOTAL:     150

                                             9

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