# Technical Sciences Grade 12 P1 NSC November Mock 2026 Memo

> Grade 12 Technical Sciences Paper 1 memorandum from the NSC November Mock Exam 2026 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-technical-sciences-2026-nsc-november-mock-exam-memorandum-paper-1/5543b9af-5bb3-4e43-95e3-ad6028ca21e8
- Updated: 2026-10-29
- Subject: Technical Sciences
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Mock Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 11

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## About this document

Technical Sciences · Grade 12 · NSC November Mock Exam · 2026 · English. Memorandum, 11 pages. Read online or download the PDF.

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### Page 1

hlayiso mock exams
                             Practice Series:
                             Grade 12 NSC Preparation
                             WWW.HLAYISO.COM

                                  NATIONAL
                             SENIOR CERTIFICATE

                                     GRADE 12

                             TECHNICAL SCIENCES P1
                             MOCK EXAMINATION 2026
                               MARKING GUIDELINES

  MARKS: 150

                     These marking guidelines consist of 11 pages.

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### Page 2

Technical Sciences/P1                       2                              Hlayiso/Mock 2026
                                 NSC – Marking Guidelines

GENERAL MARKING GUIDELINES

  1. Calculations: Marks are awarded for the correct formula, correct substitution and the
     correct final answer WITH the correct unit, as indicated by the ticks (✓). No mark is
     awarded for a formula that is not followed by a substitution. If an incorrect formula is
     used, no further marks are awarded for that calculation.
  2. Units: The final-answer mark is only awarded if the correct unit is given. Penalise an
     omitted or incorrect unit only ONCE per question.
  3. Rounding: Accept final answers rounded to a minimum of TWO decimal places. Small
     differences caused by rounding of intermediate values are accepted.
  4. Positive marking: Where an incorrect answer from a previous question is used correctly
     in a later calculation, full marks are awarded for the later calculation.
  5. Negative marking: If a candidate must explain or motivate a choice (e.g. INCREASES
     or DECREASES) and the choice is incorrect, no marks are awarded for the explanation.
  6. Definitions/laws: Where key words are underlined, deduct 1 mark for EACH key word
     or phrase that is omitted. Accept equivalent wording that conveys the same meaning.
  7. Free-body diagrams: ONE mark for each force represented by an arrow with a correct
     label. Penalise ONCE for: no arrows, no dot, gaps between the arrows and the dot,
     dotted lines, or additional (non-applicable) forces.
  8. Sign convention: Candidates may choose either direction as positive, provided it is
     applied consistently.

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### Page 3

Technical Sciences/P1              3               Hlayiso/Mock 2026
                        NSC – Marking Guidelines

QUESTION 1
1.1     B ✓✓                                                      (2)
1.2     D ✓✓                                                      (2)
1.3     A ✓✓                                                      (2)
1.4     C ✓✓                                                      (2)
1.5     D ✓✓                                                      (2)
1.6     A ✓✓                                                      (2)
1.7     B ✓✓                                                      (2)
1.8     C ✓✓                                                      (2)
1.9     B ✓✓                                                      (2)
1.10    A ✓✓                                                     (2)
                                                                [20]

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### Page 4

Technical Sciences/P1                              4                                      Hlayiso/Mock 2026
                                        NSC – Marking Guidelines

QUESTION 2
2.1     When a net force acts on an object, the object will accelerate in the direction
        of the net force. The acceleration is directly proportional to the net force and
        inversely proportional to the mass of the object. ✓✓
        OR The net force acting on an object is equal to the rate of change of
        momentum of the object. ✓✓                                                                          (2)
2.2
                                 N / FN ✓

                                                     FA / 300 N ✓

         fk / friction ✓

                            Fg / w / weight ✓
        Accept: FA : applied force / FT / tension; fk : f / Ff / frictional force; N: normal force; Fg :
        w / 784 N.
        Do not penalise for relative lengths of arrows.                                                     (4)
2.3     Fx = 300 cos 25◦ = 271,89 N ✓
        Fnet = ma ✓
        Fx – fk = ma
        271,89 – fk ✓ = (80)(0,6) ✓
        fk = 223,89 N ✓                                                                                     (5)
2.4     N + F sin θ = Fg OR N = mg – F sin θ ✓
        N = (80)(9,8) – 300 sin 25◦ ✓
        N = 784 – 126,79 = 657,21 N ✓                                                                       (3)
2.5     POSITIVE MARKING FROM QUESTIONS 2.3 AND 2.4
        fk = µk N ✓
        223,89 = µk (657,21) ✓
        µk = 0,34 ✓     (Range: 0,34 – 0,35. No unit.)                                                      (3)
2.6     INCREASES ✓                                                                                         (1)
2.7     The vertical (upward) component of the applied force is now zero, so the
        normal force increases. ✓
        Since fk = µk N / friction is directly proportional to the normal force, the frictional
        force increases. ✓                                                                                  (2)
                                                                                                           [20]

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### Page 5

Technical Sciences/P1                         5                               Hlayiso/Mock 2026
                                   NSC – Marking Guidelines

QUESTION 3
3.1     The product of the mass of an object and its velocity. ✓✓                            (2)
3.2     p = mv ✓
        p = (1 200)(20) ✓
        p = 24 000 kg·m·s–1 east ✓                                                           (3)
3.3     East is positive
        ∑ pi = ∑ pf OR m1 vi1 + m2 vi2 = (m1 + m2 )vf ✓
        (1 200)(20) + (1 800)(0) ✓ = (1 200 + 1 800)vf ✓
        vf = 8 m·s–1 east ✓                                                                  (4)
3.4     POSITIVE MARKING FROM QUESTION 3.3
        Ek,before = 12 mv2 = 12 (1 200)(20)2 = 240 000 J ✓
        Ek,after = 12 (3 000)(8)2 = 96 000 J ✓
        Ek,before ̸= Ek,after (kinetic energy is not conserved) ✓
        INELASTIC ✓
        Any one of the formula marks may be awarded for Ek = 12 mv2 .                        (4)
3.5     POSITIVE MARKING FROM QUESTION 3.3
        Fnet ∆t = ∆p = m(vf – vi ) ✓
        Fnet (0,15) = 1 200(8 – 20) ✓
        Fnet = –96 000 N
        Magnitude: 96 000 N ✓                                                                (3)
3.6     Any TWO:
        Airbags ✓ Seat belts ✓ Crumple zones                 Collapsible steering column
        Padded dashboard Head rests                                                         (2)
                                                                                           [18]

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### Page 6

Technical Sciences/P1                         6                              Hlayiso/Mock 2026
                                   NSC – Marking Guidelines

QUESTION 4
4.1.1   The product of the force (component) in the direction of the displacement and
        the magnitude of the displacement. ✓✓
        OR Work is done when a force causes an object to move in the direction of the
        force.                                                                              (2)
4.1.2   W = F∆x cos θ ✓
        W = (120)(15) cos 40◦ ✓
        W = 1 378,88 J ✓                                                                    (3)
4.1.3   Zero / 0 J ✓
        The lawnmower moves at constant velocity, so the net force on it is zero / there
        is no change in its kinetic energy. ✓                                               (2)
4.2.1   The total mechanical energy in an isolated (closed) system remains constant
        / is conserved. ✓✓                                                                  (2)
4.2.2   Ep = mgh ✓
        Ep = (15)(9,8)(3,5) ✓
        Ep = 514,50 J ✓                                                                     (3)
4.2.3   POSITIVE MARKING FROM QUESTION 4.2.2
        (Ep + Ek )top = (Ep + Ek )bottom ✓
        514,50 + 0 ✓ = 0 + 21 (15)v2 ✓
        v = 8,28 m·s–1 ✓                                                                    (4)
4.3     Pave = Fvave ✓
        Pave = (500)(9,8)(0,8) ✓     (F = mg since velocity is constant)
        Pave = 3 920 W ✓                                                                    (3)
4.4     POSITIVE MARKING FROM QUESTION 4.3
        3 920
              = 5,25 hp ✓
         746                                                                                (1)
                                                                                           [20]

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### Page 7

Technical Sciences/P1                         7                                Hlayiso/Mock 2026
                                   NSC – Marking Guidelines

QUESTION 5
5.1     The (internal restoring) force per unit (cross-sectional) area. ✓✓                     (2)
            F
5.2     σ=     ✓
            A
              20 000
        σ=             ✓
            5,0 × 10–5
        σ = 4,0 × 108 Pa ✓                                                                     (3)
            ∆L
5.3     ε=      ✓
             L
            8,0 × 10–3
        ε=             ✓
                4,0
        ε = 2,0 × 10–3 (0,002; no unit) ✓                                                      (3)
5.4     POSITIVE MARKING FROM QUESTIONS 5.2 AND 5.3
            σ
        K= ✓
            ε
             4,0 × 108
        K=             ✓
            2,0 × 10–3
        K = 2,0 × 1011 Pa ✓                                                                    (3)
5.5     B✓                                                                                     (1)
5.6     The resistance of a fluid to flow. ✓✓
        OR A measure of the internal friction in a fluid.                                      (2)
5.7     DECREASES ✓                                                                            (1)
                                                                                              [15]

QUESTION 6
6.1     The pressure applied to an enclosed fluid is transmitted undiminished to every
        part of the fluid, as well as to the walls of the container. ✓✓                        (2)
6.2     F2 = mg = (1 500)(9,8) = 14 700 N ✓
        A1 = π r2 = π (0,02)2 = 1,26 × 10–3 m2 and A2 = π (0,16)2 = 8,04 × 10–2 m2 ✓
        F1 F2
           =      ✓
        A1 A2
             F1            14 700
                       =              ✓
        1,26 × 10  –3    8,04 × 10–2
        F1 = 229,69 N ✓         (Range: 229,69 – 230,38 N)
        Areas in cm2 (12,57 cm2 and 804,25 cm2 ) are also accepted, since the units cancel.    (5)
6.3     P = ρ gh ✓
        P = (1 025)(9,8)(12) ✓
        P = 120 540 Pa (1,21 × 105 Pa) ✓                                                       (3)
                                                                                              [10]

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### Page 8

Technical Sciences/P1                            8                       Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

QUESTION 7
7.1     The change in direction of light (a wave) as it passes from one medium to
        another (of different optical density), due to a change in speed. ✓✓             (2)
7.2     Any TWO:
        The speed of light decreases. ✓
        The light bends towards the normal / changes direction. ✓
        The wavelength decreases.                                                        (2)
7.3     The angle of incidence in the optically denser medium for which the angle of
        refraction is 90◦ . ✓✓                                                           (2)
7.4     Total internal reflection ✓                                                      (1)
7.5     Light must travel from an optically denser medium to an optically less dense
        medium. ✓
        The angle of incidence must be greater than the critical angle. ✓                (2)
7.6     Any ONE: Optical fibres (endoscopes / telecommunication) ✓       Binoculars
        Cat’s eye road reflectors Sparkle of diamonds                                   (1)
                                                                                       [10]

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### Page 9

Technical Sciences/P1                         9                              Hlayiso/Mock 2026
                                   NSC – Marking Guidelines

QUESTION 8
8.1.1   c = fλ ✓
        3,0 × 108 = (2,45 × 109 )λ ✓
        λ = 0,12 m (1,22 × 10–1 m) ✓                                                           (3)
8.1.2   E = hf ✓
        E = (6,63 × 10–34 )(2,45 × 109 ) ✓
        E = 1,62 × 10–24 J ✓                                                                   (3)
8.1.3   Any ONE: Infrared / visible light / ultraviolet / X-rays / gamma rays ✓                (1)
8.2.1   Convex / converging lens ✓                                                             (1)
8.2.2   Real ✓
        Inverted ✓
        Diminished / smaller than the object ✓
        Also accept: formed between F and 2F on the opposite side of the lens (only if a
        property above is missing).                                                           (3)
                                                                                             [11]

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### Page 10

Technical Sciences/P1                           10                          Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

QUESTION 9
9.1.1   The ratio of the charge on one plate of a capacitor to the potential difference
        across the plates. ✓✓
        OR The charge stored per volt.                                                     (2)
            ε A
9.1.2   C= 0 ✓
             d
            (8,85 × 10–12 )(0,02)
        C=                        ✓
                  5 × 10–4
        C = 3,54 × 10–10 F ✓                                                               (3)
9.1.3   POSITIVE MARKING FROM QUESTION 9.1.2
            Q
        C= ✓
            V
                       Q
        3,54 × 10–10 =    ✓
                       12
        Q = 4,25 × 10–9 C ✓                                                                (3)
         1      1    1    1 1
9.2.1       =     +    = +        ✓      Rp = 4 Ω
        Rp R2 R3 6 12
        Rtotal = R1 + Rp = 4 + 4 ✓
        Rtotal = 8 Ω ✓                                                                     (3)
9.2.2   POSITIVE MARKING FROM QUESTION 9.2.1
            V
        I= ✓
            R
            12
        I=      ✓
             8
        I = 1,5 A ✓                                                                        (3)
9.2.3   POSITIVE MARKING FROM QUESTION 9.2.2
        P = I2 R ✓
        P = (1,5)2 (4) ✓
        P=9W✓
        OR: V1 = IR1 = 6 V; P = VI = (6)(1,5) = 9 W                                        (3)
                                                                                          [17]

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### Page 11

Technical Sciences/P1                       11                             Hlayiso/Mock 2026
                                  NSC – Marking Guidelines

QUESTION 10
10.1.1 Slip rings ✓                                                                      (1)
10.1.2 The magnitude of the induced emf is directly proportional to the rate of change
       of magnetic flux (linkage). ✓✓                                                    (2)
10.1.3 Mechanical (kinetic) energy to electrical energy ✓                                (1)
        Vs Ns
10.2        =    ✓
        Vp Np
         12     Ns
             =       ✓
        230 1 150
        Ns = 60 turns ✓                                                                  (3)
10.3    POSITIVE MARKING FROM QUESTION 10.2
        The secondary coil has fewer turns (60) than the primary coil (1 150) ✓,
        therefore the output (secondary) voltage is lower than the input (primary)
        voltage. ✓                                                                       (2)
                                                                                         [9]
                                                                             TOTAL:      150

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