# Technical Sciences Grade 12 P1 GP Prelim 2023 Memo (English)

> Grade 12 Technical Sciences Paper 1 memorandum from the GP Prelim 2023 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/technical-sciences-p1-english-memo/036cde07-8a86-4b6d-a9ba-2a4da6e21b9e
- Updated: 2026-09-27
- Subject: Technical Sciences
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2023
- Exam period: GP Prelim
- Paper: 1
- Curriculum: CAPS
- Pages: 11

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## About this document

Technical Sciences · Grade 12 · GP Prelim · 2023 · English. Memorandum, 11 pages. Read online or download the PDF.

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## Matching paper and memo

- [Technical Sciences P1 (English) QP](https://www.hlayiso.com/library/technical-sciences-p1-english-qp/2c25951e-f501-41d6-bf5d-eac3e246f09a) — QP · 2023 · Eng

## Document text

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                                                 :

                          AMENDMENT TO MARKING GUIDELINES

                              PREPARATORY EXAMINATION /
                             VOORBEREIDENDE EKSAMEN 2023

 FOR ATTENTION / VIR AANDAG:
 THE CHIEF INVIGILATOR / DIE HOOF TOESIGHOUER
 SUBJECT / VAK                                        TECHNICAL SCIENCES/TEGNIESE WETENSKAPPE
 SUBJECT CODE / VAKKODE                               11101
 PAPER / VRAESTEL                                     1
 DATE OF EXAMINATION / DATUM VAN EKSAMEN              22 SEPTEMBER 2023

The TECHNICAL SCIENCES/TEGNIESE WETENSKAPPE (Paper 1 / Vraestel 1) written on 22
September 2023 has reference. It has come to our attention that there was an anomaly on the
data sheet which impacted on questions 6.1.3, 6.1.4 and 6.1.6.

To ensure that your candidates are not disadvantaged and prejudiced in way, you are advised to
please ask Educators to ignore the above-mentioned questions.

In other words, the paper must be marked out of a total of 140 instead of 150 and then the learners’
marks must be converted back to a mark out of 150. E.g., Should a learner have attained 80/140,
then that mark is recalculated as 86/150.

              Use the        a                                  b
              formula:     140
                                  x 100 =       b.    Then,
                                                               100
                                                                     x 150 =       c
                   c is the mark that is entered into SASAMS out of 150.

  DIRECTOR: EXAMINATIONS MANAGEMENT
  22 September 2023

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                                                 11101/23

      PREPARATORY EXAMINATION

                        2023

           MARKING GUIDELINES

          TECHNICAL SCIENCES (PAPER 1) (11101)

9 pages

                           3

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QUESTION 1: MULTIPLE-CHOICE QUESTIONS

1.1   C ✓✓                                      (2)

1.2   A ✓✓                                      (2)

1.3   B ✓✓                                      (2)

1.4   C ✓✓                                      (2)

1.5   A ✓✓                                      (2)

1.6   D ✓✓                                      (2)

1.7   C ✓✓                                      (2)

1.8   B ✓✓                                      (2)

1.9   D ✓✓                                      (2)

1.10 B ✓✓                                       (2)
                                               [20]

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 force parallel to the surface that opposes the motion of an object and
          acts in the direction opposite to the motion the object. ✓✓                      (2)

     3.1.2

                                    FA must not be equal or same length with Ff (-1)
             Symbols                Name                      Mark allocation
             Ff, f, FK              Force of friction         ✓
             Fg, W,                 Gravitational force       ✓
             FA, F                  Applied force             ✓
             FN, N                  Normal force              ✓                            (4)

     3.1.3 Fnet = ma                    ✓
           FA – Ff – = ma
           FA – 750 ✓ = 960 x 2 ✓
           FA = 1 920 + 750
              = 2670 N to the right ✓                                                      (4)

     3.1.4 Fg = 960 x 9,8 ✓
              = 9408 N ✓                                                                   (2)

                                             3

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3.2 3.2.1 When a net/resultant force is applied to an object of mass m, it accelerates
          the object in the direction of the net force. The acceleration is directly
          proportional to the net/resultant force and inversely proportional to the
          mass of the object. ✓✓                                                              (2)

     3.2.2 5 forces ✓ or 5 ✓                                                                  (1)

     3.2.3 Fnet = ma
           FH + FDx – Ff = ma                        ✓
           FH + FD cos Ɵ – Ff = ma
           120 + 180 cos 15 – 100 ✓= 70a ✓
           193,87 = 70a
           a = 2,77 m·s–2 to the right ✓                                                      (4)

     3.2.4 DECREASE ✓                                                                         (1)
                                                                                             [20]

QUESTION 4

4.1 4.1.1    It is the product of the net force and the time interval for which the net
             force acts on the body. ✓✓                                                       (2)

     4.1.2   1 kg = 1 000 g                   1 kg : 103 g
             1 kg = 1 000 g                     x : 150 g
             1000 1 000                    OR        150
             1 g = 0,001 kg                     x= 3
                                                     10
             150 g = 0,15 kg ✓✓
                                                  = 150  10-3 kg ✓✓                          (2)

     4.1.3   DECREASE ✓                                                                       (1)

     4.1.4   Option 1                               Option 2

              Fnet Δt = mvf – mvi = Δp ✓                  Fnet Δt = mvf – mvi = Δp ✓
                       = 0,15(32 – (–50)) ✓                        = 0,15(-32 – (+50)) ✓
                            0,012 ✓                                     0,012 ✓
                       = 1 025 N towards the wall ✓                = -1 025 N away from
                                                          the wall ✓

                                                                                              (4)

4.2 4.2.1    From Fnet = Δp
                         Δt

             The air bag causes the delay of impact by increasing the time of impact
             and decreasing the net force. ✓✓                                                 (2)

     4.2.2   Increase in velocity will increase the impulse since the impulse is directly
             proportional to the velocity. ✓✓                                                 (2)
                                                                                             [13]
                                                4

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QUESTION 5

5.1 Work done is the product of the applied force on an object and the displacement in
    the direction of the force. ✓✓                                                        (2)

5.2 W = F Δx cos Ɵ ✓
      = 5 x 3 cos 0° ✓
      = 15 J ✓                                                                            (3)

5.3 5.3.1    The principle of conservation of mechanical energy: The total mechanical
             energy (sum of gravitational potential energy and kinetic energy) in an
             isolated system remains constant. ✓✓                                         (2)

     5.3.2   EK = ½mv2 ✓
                = ½ x 2 x 1,52 ✓
                = 2,25 J ✓                                                                (3)

     5.3.3   EP = mgh ✓
                = 2 x 9,8 x 0,5 ✓
                = 9,8 J ✓                                                                 (3)

     5.3.4   positive marking from 5.3.2 and 5.3.3
             ME top = ME bottom                            ✓
             (½mv2 + mgh)top = (½mv2 + mgh)bottom
             ½ x 2 x 1,52 + 2 x 9,8 x 0,5 ✓= ½ x 2 x v2 ✓+ 2 x 9,8 x 0 ✓
             2,25 + 9,8 = v2 + 0
             v2 = 12,05
             v = 3,47 m·s–1 ✓                                                             (5)
                                                                                         [18]

QUESTION 6

6.1 6.1.1    Deforming force is a force that changes the shape and size of a body. ✓✓    (2)

     6.1.2   Area = πr2 ✓
                  = 3,14 x 0,032 ✓
                  = 0,0028 m2 ✓ (2,83 x 10-3 m2)                                         (3)

     6.1.3    σ = F ✓ (do not mark) Due to Data sheet
                  A
                = 6 000 ✓
                  0,0028
                = 2 142 857,14
                = 2,14 x 106 N.m–2 ✓                                                     (3)

     6.1.4   ε = ΔL ✓ (do not mark) Due to Data sheet
                  l
               = 0,01 ✓
                    3
               = 0,0033✓ (3,33 x 10-3)                                                   (3)
                                              5

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     6.1.5   Hooke’s law states that, within the limit of elasticity, stress is directly
             proportional to the strain. ✓✓                                                    (2)

     6.1.6   K = σ ✓ (do not mark) Due to Data sheet
                 ε
               = 2,14 x 106 ✓
                  0,0033 ✓
               = 648 484 848.5
               = 6,48 x 108 Pa or N.m–2 ✓                                                      (4)

6.2 6.2.1    Viscosity is the property of the fluid to oppose relative motion between
             the two adjacent layers. ✓✓                                                       (2)

     6.2.2   5W-40 ✓                                                                           (1)

     6.2.3   As the temperature decreases, the viscosity increase. ✓                           (1)

     6.2.4   5W-20 ✓                                                                           (1)

     6.2.5   • To reduce wear on the moving parts
             • Cleans the moving parts
             • Reduces corrosion
             • Cools the engine
             • Improves sealing
             (Any ONE) ✓                                                                       (1)
                                       Mark this question out of 13 marks                     [13]

QUESTION 7

7.1 Capacitance of a capacitor is the amount of charge it can store per volt. ✓✓               (2)

7.2 C = Q
        V                         Any two pairs of coordinates can be used to
      = 10 – 5 ✓               OR calculate capacitance (gradient may be used/ any
        4 – 2✓                    value may be used from the graph)
      = 2,5 C.V–1 (F)✓                                                                         (3)

7.3 The charge is directly proportional to potential difference. OR
    As the charge increases, the potential difference increases. ✓✓                            (2)

7.4 • Filter circuits in power supplies
    • Separation of frequencies between the woofer (base) speaker and tweeter
      (high pitch) speaker/electronic noise filtering
    • Power factor correction/improvement in electrical transmission systems
    (Any TWO) ✓✓                                                                               (2)
                                                                                               [9]

                                                 6

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QUESTION 8

8.1 Ohm's law states that: The potential difference across a conductor is directly
    proportional to the current in the conductor at constant temperature. ✓✓             (2)

8.2 8.2.1   1 = 1+ 1 ✓
            Req    R1 R2
                 = 1 + 1✓
                    4    2
                 =3
                   4
            Req = 1,33 Ω

            Rtotal = Req + Rs ✓
                   = 1,33 + 3
                   = 4,33 Ω ✓
                                                                                         (4)

     8.2.2 RT = VT ✓
                 IT
           4,33 = 12 ✓
                    IT
           IT = 12
                4,33
              = 2,77 A ✓                                                                 (3)

     8.2.3 Req = V                           V = IR
                  I                            = (2,77) (3)
           1,33 ✓ = V ✓                        = 8,31
                  2,77                       12 – 8,31
           V = 1,33 x 2,77                     = 3,69 V
             = 3,68 V ✓                                                                  (3)

     8.2.4 Option 1
           12 V – 3,68 V = 8, 32 V ✓ (positive marking from previous questions)

            P = VI ✓
              = 8,32 x 2,77 ✓
              = 23,05 W ✓
            Option 2
            P = V2 / R
              =8,322 / 3
              = 23,07 W
            Option 3
            P = I2 x R                                                                   (4)
              =2,772 x 3
              = 23,02 W

8.3 INCREASE ✓
    NEGATIVE MARKING                                                                     (2)

                                              7

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The effective resistance decreases✓
                                              [18]

                                      8

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                                                                                        11101/23

QUESTION 9

9.1

                                          Normal
         Incident ray                                       Reflection ray

                                      ɵi ɵr

                                    Mirror

      Angle of Incident                                 ✓
      Reflected beam                                    ✓
                                                              (If the arrow is not
      Normal                                            ✓
                                                              shown – 1 mark)
      Incident beam                                     ✓
      Angle of reflection                               ✓                                (5)

9.2 Critical angle: The critical angle is measured when the incident ray passes
    from an optically more dense to an optically less dense medium, such that
    the refracted ray just passes through the surface of the two media. ✓✓

                                               OR

      The critical angle is the angle incidence where the angle of refraction is 90°.
      The refracted beam then travels along the boundary between the two
      media.✓✓                                                                           (2)

9.3 Dispersion: The phenomenon whereby (white) light breaks up into its
    component colours. ✓✓ (Award the marks in Afrikaans)                                 (2)

9.4 When light travels from air to water, it moves from a less dense to an
    optically ✓ denser medium. The colour of the light is determined by
    frequency ✓ and the colour of light does not change, but the speed of the
    light and the wavelength decrease. ✓                                                 (3)
                                                                                        [12]

                                                    9

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QUESTION 10

10.1   Electromagnetic wave: A change of a magnetic and an electric field
       perpendicular to each other and to the direction of propagation of the
       wave ✓✓                                                                           (2)

10.2   • EM waves move at a constant speed.
       • EM waves do not need a medium.
       • They have properties of waves.
       • EM waves have particle properties.
       • EM waves are transverse waves.
       (Any 4) ✓✓✓✓                                                                      (4)

10.3   X-rays/Gamma rays ✓
       Infrared for Physiotherapy.
       UV rays to sterilize equipment.
       Marker may use own discretion on equipment or places within context.
                                                                                         (1)

10.4   X-rays are used to ascertain if there are bone fractures.
       Gamma rays are used to treat cancer.
       Marker may use own discretion on equipment context.

       (Look at 10.3’s answer) ✓✓                                                        (2)
                                                                                         [9]

                                                                                TOTAL:   140

                                            10

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