# Technical Sciences Grade 10 P1 2026 Memo (Eng)

> Grade 10 Technical Sciences Paper 1 memorandum from the November Mock Exam 2 2026 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-10-technical-sciences-2026-november-mock-exam-2-memorandum-paper-1/28bbec3b-2c27-4a61-80da-a281b5da6b1c
- Updated: 2026-10-29
- Subject: Technical Sciences
- Grade: Grade 10
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam 2
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 11

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## About this document

Technical Sciences · Grade 10 · November Mock Exam 2 · 2026 · English. Memorandum, 11 pages. Read online or download the PDF.

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## Matching paper and memo

- [Grade 10 Technical Sciences P1 November 2026 Mock 2](https://www.hlayiso.com/library/grade-10-technical-sciences-2026-november-mock-exam-2-question-paper-paper-1/adf4ceb7-e439-47f5-becb-d075eb68c752) — QP · 2026 · Eng

## Document text

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### Page 1

Mock

                             hlayiso mock exams
                             Practice Series:
                             Grade 10 Exam Preparation
                             WWW.HLAYISO.COM

                                  NATIONAL
                             SENIOR CERTIFICATE

                                     GRADE 10

                             TECHNICAL SCIENCES P1
                     NOVEMBER EXAMINATION 2026 (MOCK)
                               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 and the choice is
     incorrect, no marks are awarded for the explanation.
  6. Definitions: 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, gaps between the arrows and the dot, dotted lines,
     or additional (non-applicable) forces.
  8. Vectors: Where a direction is asked for (displacement, velocity, acceleration, resultant
     force), the final mark is only awarded if the direction is given.

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### Page 3

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

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

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### Page 4

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

QUESTION 2
2.1.1   2,5 × 60 × 60 ✓ = 9 000 s ✓                                                                   (2)
         750
2.1.2         ✓ = 0,75 m ✓
        1 000                                                                                         (2)
2.1.3   4,8 × 107 ✓ m ✓        (1 mark for 4,8; 1 mark for the correct power of ten)                  (2)
2.2.1   Scalar ✓                                                                                      (1)
2.2.2   Vector ✓                                                                                      (1)
2.3.1   1 200 + 1 800 ✓ = 3 000 m ✓        (3 km without conversion: 1/2)                             (2)
2.3.2   North positive: ∆x = 1,2 + (–1,8) ✓
        ∆x = –0,6 km
        ∆x = 0,6 km (600 m) ✓ south ✓                                                                 (3)
2.3.3

                              1,8 km S
         1,2 km N

                                          resultant: 0,6 km S

        Criteria: 1,2 km vector north and 1,8 km vector south, labelled, drawn head-to-tail ✓;
        resultant drawn from tail of first to head of last vector ✓; resultant labelled 0,6 km (600
        m) south ✓.                                                                                   (3)
2.3.4   POSITIVE MARKING FROM QUESTION 2.3.1
                distance
        speed =          ✓
                    ∆t
                 3 000
        speed =         ✓
                15 × 60
        speed = 3,33 m·s–1 ✓
        If 15 is used for ∆t (200 m·s–1 ): max 1/3.                                                   (3)
2.3.5   POSITIVE MARKING FROM QUESTION 2.3.2
            ∆x 600
        v=     =     ✓
            ∆t 900
        v = 0,67 m·s–1 ✓ south ✓
                               0,6
        Accept other units:        = 2,4 km·h–1 south, or 0,04 km·min–1 (40 m·min–1 ) south.
                              0,25                                                                    (3)
2.4.1   YES ✓
        The distance between successive dots increases (in equal time intervals). ✓                   (2)

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### Page 5

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

                  1
2.4.2   ∆t = 5 ×    ✓    (5 spaces between 6 dots)
                 50
        ∆t = 0,1 s ✓                                                                             (2)
2.4.3   POSITIVE MARKING FROM QUESTION 2.4.2
        ∆x = 1 + 2 + 3 + 4 + 5 = 15 cm = 0,15 m ✓
            ∆x 0,15
        v=      =       ✓
            ∆t     0,1
        v = 1,5 m·s–1 ✓ (magnitude only asked: direction not required)                           (3)
            108                           36
2.5     vi =     = 30 m·s–1 and vf =          = 10 m·s–1 ✓
             3,6                          3,6
            ∆v
        a=      ✓
            ∆t
            10 – 30
        a=          ✓
               5
        a = –4 m·s–2 , i.e. a = 4 m·s–2 west ✓
        Accept: 4 m·s–2 opposite to the motion / deceleration of 4 m·s–2 . No conversion (a =
        –14,4): max 2/4.                                                                         (4)
                                                                                                [33]

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### Page 6

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

QUESTION 3
3.1     The single force that has the same effect as all the forces acting on an object
        together. ✓✓
        Accept: the vector sum of all the forces acting on an object.                            (2)
3.2.1   Fg = mg ✓
        Fg = (12)(9,8) ✓
        Fg = 117,6 N (downwards) ✓                                                               (3)
3.2.2   POSITIVE MARKING FROM QUESTION 3.2.1
        117,6 N ✓
        The toolbox is at rest (in equilibrium), so the resultant force is zero / the upward
        tension balances the weight. ✓                                                           (2)
3.3
                     N / FN / normal force ✓

         T2 / 40 N ✓
                                                 T1 / 90 N / applied force ✓
             f / friction ✓

                        Fg / w / weight ✓
        The 40 N pull and the friction both act to the west: two separate westward arrows are
        preferred; accept arrows drawn on top of each other provided BOTH are labelled. Do
        not penalise for relative lengths of arrows. Accept a box drawn instead of a dot.        (5)
3.4     East positive: Fres = 90 + (–40) + (–15) ✓
        Fres = 35 N ✓ east ✓
        Friction omitted (50 N east): max 2/3.                                                   (3)
                                                                                                [15]

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### Page 7

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

QUESTION 4
4.1.1   Fulcrum / pivot ✓                                                                     (1)
4.1.2   Moment (of a force) / torque ✓                                                        (1)
4.1.3   Effort (force) ✓                                                                      (1)
4.2     τ = F × r⊥ ✓
        τ = (300)(0,17) ✓
        τ = 51 N·m ✓                                                                          (3)
4.3.1   Take moments about A:
        Σ clockwise moments = Σ anticlockwise moments ✓
        (1 200)(1,5) + (900)(4) + (600)(6) ✓ = RB (5) ✓
        9 000 = 5RB
        RB = 1 800 N ✓                                                                        (4)
4.3.2   POSITIVE MARKING FROM QUESTION 4.3.1
        Upward forces = downward forces: RA + 1 800 = 1 200 + 900 + 600 ✓✓
        RA = 900 N ✓
        OR moments about B: RA (5) + (600)(1) ✓ = (1 200)(3,5) + (900)(1) ✓ ⇒ RA = 900 N ✓    (3)
4.3.3   POSITIVE MARKING FROM QUESTIONS 4.3.1 AND 4.3.2
        Upward forces: 900 + 1 800 = 2 700 N ✓
        Downward forces: 1 200 + 900 + 600 = 2 700 N ✓
        ∴ upward forces = downward forces                                                     (2)
4.4.1   Second-class lever (class 2) ✓                                                        (1)
4.4.2   Σ clockwise moments = Σ anticlockwise moments (about the fulcrum) ✓
        L(0,02) = (20)(0,10) ✓ (accept cm throughout: L(2) = (20)(10))
        L = 100 N ✓                                                                           (3)
4.4.3   POSITIVE MARKING FROM QUESTION 4.4.2
              load
        MA =        ✓
             effort
             100
        MA =      ✓
              20
        MA = 5 ✓      (no unit)                                                               (3)

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### Page 8

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

4.5
         effort                                load

                            fulcrum
        Criteria: fulcrum between the effort and the load ✓; effort at one end ✓; load at the
        other end, with all three labelled ✓. Accept a labelled sketch of a real object (e.g.
        see-saw, crowbar, scissors, pliers) if the three positions are correctly labelled.       (3)
                                                                                                [25]

QUESTION 5
5.1     Ek = 12 mv2 ✓
        60 000 = 12 (1 200)v2 ✓
        v = 10 m·s–1 ✓                                                                           (3)
5.2     Ep = mgh ✓
        19 110 = (65)(9,8)h ✓
        h = 30 m ✓                                                                               (3)
5.3.1   The sum of the gravitational potential energy and the kinetic energy of an
        object. ✓✓                                                                               (2)
5.3.2   EM = Ep + Ek ✓
        EM = (0,5)(9,8)(2,4) ✓ + 21 (0,5)(8)2 ✓
        EM = 11,76 + 16 = 27,76 J ✓                                                              (4)
5.3.3   0 J (zero) ✓
        The ball is (momentarily) at rest / its velocity is zero at Y. ✓                         (2)
5.3.4   POSITIVE MARKING FROM QUESTION 5.3.2
        EM(Y) = EM(X) : (0,5)(9,8)h + 0 = 27,76 ✓
        h = 5,67 m ✓
                             82
        Accept h = 2,4 +          = 5,67 m.
                           2(9,8)                                                                (2)
                                                                                                [16]

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### Page 9

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

QUESTION 6
6.1.1   GAIN ✓                                                                                (1)
6.1.2   POSITIVE ✓                                                                            (1)
            Q
6.2.1   n=     ✓
            e
             –12 × 10–9
        n=                ✓
            –1,6 × 10–19
        n = 7,5 × 1010 electrons ✓                                                            (3)
6.2.2   To prevent the charge from flowing (leaking) away (to the support / ground /
        earth). ✓✓
        Accept: insulators do not conduct, so the spheres keep their charge.                  (2)
6.2.3   The net (total) charge in an isolated (closed) system remains constant (is
        conserved). ✓✓                                                                        (2)
            Q1 + Q2
6.2.4   Q=          ✓
               2
            (–12 × 10–9 ) + (+4 × 10–9 )
        Q=                               ✓
                         2
        Q = –4 × 10–9 C (on each sphere) ✓                                                    (3)
6.2.5   FROM X TO Y ✓                                                                        (1)
                                                                                            [13]

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### Page 10

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

QUESTION 7
7.1     The rate of flow of charge. ✓✓
        Accept: the amount of charge that passes a point in a conductor per second.                (2)
             Q
7.2.1   I=      ✓
             ∆t
                 Q
        0,25 =        ✓
               2 × 60
        Q = 30 C ✓
        If 2 is used for ∆t (0,5 C): max 1/3.                                                      (3)
7.2.2   POSITIVE MARKING FROM QUESTION 7.2.1
            Q        30
        n= =                 ✓
            e 1,6 × 10–19
        n = 1,88 × 1020 electrons ✓ (accept 1,875 × 1020 )                                         (2)
7.3
                        S (closed)
                                      A
                         2 cells
         V

        Criteria: battery of TWO cells, closed switch and ammeter in series ✓; two bulbs in
        parallel with each other ✓; voltmeter in parallel across the battery ✓; correct symbols
        used and circuit complete ✓.                                                               (4)
7.4.1   Length of the (nichrome) wire ✓                                                            (1)
7.4.2   Any ONE ✓: Thickness (cross-sectional area) of the wire / type of material /
        temperature                                                                                (1)
                                                                                                  [13]

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### Page 11

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

QUESTION 8
8.1     The work done (energy transferred) per unit (coulomb) of charge between two
        points in a circuit. ✓✓                                                        (2)
8.2     R12 = R1 + R2 = 4 + 8 = 12 Ω ✓
         1     1    1
           =      +    ✓
        Rp R12 R3
         1    1 1
           =     + ✓
        Rp 12 6
        Rp = 4 Ω ✓                                                                     (4)
8.3     POSITIVE MARKING FROM QUESTION 8.2
        RT = R4 + Rp = 2 + 4 ✓
        RT = 6 Ω ✓                                                                     (2)
8.4     POSITIVE MARKING FROM QUESTION 8.3
           V
        R= ✓
            I
           18
        6=    ✓
            I
        I=3A✓                                                                          (3)
8.5     POSITIVE MARKING FROM QUESTIONS 8.2 AND 8.4
        V = IRp = (3)(4) ✓
        V = 12 V ✓
        OR: VR4 = (3)(2) = 6 V; V = 18 – 6 = 12 V                                      (2)
8.6     POSITIVE MARKING FROM QUESTION 8.5
          V 12
        I= =     ✓
          R    6
        I=2A✓                                                                          (2)
                                                                                      [15]
                                                                           TOTAL:     150

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