# Technical Sciences Grade 11 P1 2026 Memo (Eng)

> Grade 11 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-11-technical-sciences-2026-november-mock-exam-2-memorandum-paper-1/a75d0df2-9481-4a66-8480-3170a11f92d2
- Updated: 2026-10-29
- Subject: Technical Sciences
- Grade: Grade 11
- 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 11 · November Mock Exam 2 · 2026 · English. Memorandum, 11 pages. Read online or download the PDF.

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## Matching paper and memo

- [Grade 11 Technical Sciences P1 November 2026 Mock 2](https://www.hlayiso.com/library/grade-11-technical-sciences-2026-november-mock-exam-2-question-paper-paper-1/72fc7f87-da72-42f9-a6fd-e7b7b2ade7e9) — QP · 2026 · Eng

## Document text

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### Page 1

Mock

                             hlayiso mock exams
                             Practice Series:
                             Grade 11 Exam Preparation
                             WWW.HLAYISO.COM

                                  NATIONAL
                             SENIOR CERTIFICATE

                                     GRADE 11

                             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 (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, gaps between the arrows and the dot, dotted lines,
     or additional (non-applicable) forces.
  8. Scale drawings: Accept answers within the stated range. No marks for the answer if
     no scale drawing is shown.
  9. 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     C ✓✓                                                      (2)
1.3     D ✓✓                                                      (2)
1.4     A ✓✓                                                      (2)
1.5     C ✓✓                                                      (2)
1.6     B ✓✓                                                      (2)
1.7     D ✓✓                                                      (2)
1.8     A ✓✓                                                      (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     A single force that, acting together with the other forces, produces equilibrium
        / is equal in magnitude but opposite in direction to the resultant. ✓✓              (2)
2.2.1   The angle between the (two) forces ✓                                                (1)
2.2.2   The magnitude of the resultant (force) ✓
        Accept: reading on the third spring balance / equilibrant.                          (1)
2.3     R2 = F21 + F22 ✓
        R2 = (40)2 + (25)2 ✓
        X = R = 47,17 N ✓                                                                   (3)
2.4

             R
                              25 N (50 mm)
                                 120◦

            40 N (80 mm)
        Criteria:
        Both forces drawn to the correct scale (80 mm and 50 mm) ✓
        Correct angle between the directions of the forces (120◦ , i.e. 60◦ inside the
        triangle) ✓
        Resultant drawn from the tail of the first to the head of the last vector, with
        arrow ✓
        Y = 35 N (range 34 N – 36 N) ✓                                                      (4)
2.5     As the angle between the forces INCREASES, the magnitude of the resultant
        DECREASES. ✓✓
        Accept the inverse statement. Accept: The resultant is a maximum at 0◦ and a
        minimum at 180◦ (1 mark only).                                                      (2)
2.6     65 N ✓                                                                              (1)
                                                                                           [14]

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### Page 5

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

QUESTION 3
3.1     VECTOR ✓                                                                              (1)
3.2     (–5 ; 7) ✓✓      (1 mark for each correct coordinate)
        Accept (–100 ; 140) (in metres).                                                      (2)
3.3     East–west component: 100 m (west) ✓
        North–south component: 140 m (north) ✓
        If grid units (5 and 7) are given without converting to metres: 1/2.                  (2)
3.4     POSITIVE MARKING FROM QUESTION 3.3
        ∆x2 = (100)2 + (140)2 ✓
        ∆x = 172,05 m ✓                                                                       (2)
               100
3.5     tan θ =     ✓
               140
        θ = 35,54◦ (west of north) ✓
        Bearing = 360◦ – 35,54◦ = 324,46◦ ✓
        OR: tan α = 140             ◦                                ◦       ◦        ◦
                    100 ✓; α = 54,46 (north of west) ✓; bearing = 270 + 54,46 = 324,46 ✓.
                        100 or cos θ = 140 .
        Accept sin θ = 172,05         172,05                                                 (3)
                                                                                            [10]

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### Page 6

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

QUESTION 4
4.1     Static frictional force: The frictional force that acts on an object when it
        is stationary (not moving) relative to the surface / opposes the tendency of
        motion. ✓✓
        Kinetic frictional force: The frictional force that acts on an object while it is
        moving (sliding) over the surface. ✓✓                                                    (4)
4.2.1   100 N ✓      to the left (opposite to the push) ✓                                        (2)
4.2.2   N = Fg = mg = (45)(9,8) = 441 N ✓
        fmax
         s   = µs N ✓
         max
        fs = (0,40)(441) ✓
        fmax
         s   = 176,40 N ✓
        Minimum force needed = 176,40 N (horizontally, to the right)
        Accept: (slightly) more than / greater than 176,40 N.                                    (4)
4.3.1
                        N / FN / normal force ✓

                                                  F / FA / 180 N / tension ✓

         fk / friction ✓

                           Fg / w / weight ✓
        Do not penalise for relative lengths of arrows. Accept the components of F instead of
        F only if F itself is not also drawn as an extra force.                                  (4)
4.3.2   Fy = F sin θ = 180 sin 25◦ ✓
        Fy = 76,07 N ✓
        N + Fy = Fg OR N = Fg – Fy ✓
        N = 441 – 76,07 ✓
        N = 364,93 N                                                                             (4)
4.3.3   fk = µk N ✓
        fk = (0,28)(364,93) ✓
        fk = 102,18 N ✓                                                                          (3)
4.4     DECREASES ✓
        (A larger vertical component of F reduces the normal force.)                             (1)
                                                                                                [22]

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### Page 7

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

QUESTION 5
5.1     The distance between two successive (consecutive) points in phase (e.g. two
        successive crests or troughs). ✓✓                                               (2)
5.2.1   Crest ✓                                                                         (1)
5.2.2   Trough ✓                                                                        (1)
5.3     Any ONE pair: P and S / P and U / S and U / Q and T ✓                           (1)
            1,6
5.4     A=      ✓
             2
        A = 0,8 m ✓
        If 1,6 m is given: 0/2.                                                         (2)
           27
5.5     λ=    ✓
            3
        λ =9m✓                                                                          (2)
             50
5.6     T=      =5s✓
             10
            1 1
        f= = ✓
            T 5
        f = 0,2 Hz ✓
        OR: f = numbertime
                        of waves = 10 ✓✓ = 0,2 Hz ✓
                                   50                                                   (3)
5.7     POSITIVE MARKING FROM QUESTIONS 5.5 AND 5.6
        v = fλ ✓
        v = (0,2)(9) ✓
        v = 1,8 m·s–1 ✓                                                                 (3)
5.8     POSITIVE MARKING FROM QUESTION 5.7
            ∆x
        v=     ✓
            ∆t
              150
        1,8 =      ✓
               ∆t
        ∆t = 83,33 s ✓                                                                 (3)
                                                                                      [18]

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### Page 8

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

QUESTION 6
6.1     The highness or lowness of a sound, which is determined by its frequency.
        ✓✓
        Accept: The effect produced in the ear by a sound of a particular frequency.                (2)
6.2.1   Q✓
        Q shows MORE complete waves in the same time interval ✓ / shorter period,
        so it has a HIGHER frequency (and pitch increases with frequency). ✓                        (3)
6.2.2   P✓
        P has the LARGER amplitude (loudness increases with amplitude). ✓                           (2)
6.3.1   NO ✓
        40 kHz is ABOVE the upper limit of the human hearing range (20 kHz) / it is
        ultrasound. ✓                                                                               (2)
                                                       4,0 × 10–3
6.3.2   Time for the pulse to reach the wall: ∆t =                = 2,0 × 10–3 s ✓
                                                            2
             ∆x
        v=      ✓
             ∆t
                  ∆x
        340 =            ✓
              2,0 × 10–3
        ∆x = 0,68 m ✓
        OR: 2∆x = v∆t = (340)(4,0 × 10–3 ) = 1,36 m ✓✓✓; distance = 0,68 m ✓.     If 1,36 m is
        given as the distance: max 2/4.                                                             (4)
6.3.3   v = fλ ✓
        340 = (40 × 103 )λ ✓
        λ = 8,5 × 10–3 m (0,0085 m) ✓
        If 40 is used instead of 40 000 Hz: max 2/3.                                                (3)
6.4     Any ONE: Scanning a foetus (pre-natal scans) / imaging internal organs
        / breaking up kidney stones / physiotherapy (treating injured muscles) /
        measuring blood flow ✓                                                                     (1)
                                                                                                  [17]

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### Page 9

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

QUESTION 7
7.1     The region (in space) around a magnet in which another magnet or a
        ferromagnetic material will experience a (magnetic) force. ✓✓                         (2)
7.2.1   P: NORTH ✓         Q: SOUTH ✓
        (Field lines point away from a north pole and towards a south pole.)                  (2)
7.2.2   ATTRACT ✓                                                                             (1)
7.2.3   X✓
        The field lines are CLOSER TOGETHER (denser) at X. ✓                                  (2)
7.3     Charged particles (solar wind) from the Sun ✓ are deflected by the Earth’s
        magnetic field and guided towards the (magnetic) poles. ✓
        There they collide with gas atoms/molecules in the upper atmosphere, which
        give off (emit) light. ✓                                                              (3)
7.4     Any TWO: Electric motors / generators (dynamos) / loudspeakers and
        microphones / MRI scanners / maglev trains / compasses / magnetic door
        catches / hard drives and bank cards / electromagnets on scrap-yard cranes /
        relays ✓✓                                                                            (2)
                                                                                            [12]

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### Page 10

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

QUESTION 8
8.1     The magnitude of the electrostatic force exerted by one point charge on
        another is directly proportional to the product of their charges and inversely
        proportional to the square of the distance between them. ✓✓                                (2)
8.2

             X                   Y
             +                   –

        Criteria: Correct shape (lines from X curve towards Y) ✓; Direction of arrows
        from X (positive) to Y (negative) ✓; Field lines start/end at the surfaces of the
        spheres and do not cross ✓.
        The outer lines may be omitted.                                                            (3)
             kQ1 Q2
8.3     F=           ✓
               r2
                    (9 × 109 )(4 × 10–9 )(6 × 10–9 )
        6 × 10–5 =                                   ✓✓
                                   r2
        r = 0,06 m ✓
        1 mark for substitution of the charges and k; 1 mark for the force. Ignore the negative
        sign if it is used.                                                                        (4)
8.4     The (electrostatic) force experienced per unit positive charge placed at that
        point. ✓✓                                                                                  (2)
            V
8.5.1   E=     ✓
            d
              3 000
        E=             ✓
            20 × 10–3
        E = 1,5 × 105 N·C–1 (V·m–1 ) ✓
        If 20 m is used (no conversion): max 2/3.                                                  (3)
8.5.2   A✓     (the positive plate)                                                                (1)
                                                                                                  [15]

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### Page 11

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

QUESTION 9
9.1     The potential difference across a conductor is directly proportional to the
        current in the conductor at constant temperature. ✓✓                                  (2)
9.2     The maximum energy (work done) provided by a battery per coulomb (unit
        charge) passing through it. ✓✓
        Accept: The potential difference across the terminals of a battery when no current
        flows (open circuit).                                                                 (2)
            V
9.3.1   R=     ✓
            I
              24
        RT =     ✓
               3
        RT = 8 Ω ✓                                                                            (3)
            V
9.3.2   R=    ✓
            I
           V
        4= 1 ✓
            3
        V1 = 12 V ✓                                                                           (3)
9.3.3   POSITIVE MARKING FROM QUESTION 9.3.1
        Rp = RT – R1 = 8 – 4 = 4 Ω ✓
         1     1     1
            =     +    ✓
        Rp R2 Rx
        1    1     1
          =     +     ✓
        4 12 Rx
        Rx = 6 Ω ✓
                                        12 = 1 A ✓; I = 3 – 1 = 2 A ✓; R = 12 = 6 Ω ✓
        OR: Vp = 24 – 12 = 12 V ✓; I2 = 12           x                  x   2                 (4)
9.3.4   POSITIVE MARKING FROM QUESTION 9.3.2
        V2 = ε – V1 = 24 – 12 ✓
        V2 = 12 V ✓
        OR: V2 = IRp = (3)(4) ✓ = 12 V ✓                                                      (2)
9.4     DECREASES ✓                                                                           (1)
9.5     The total (equivalent) resistance of the circuit INCREASES (only Rx is left in
        the parallel section). ✓
        The emf stays the same, so the current decreases (I = RV ). ✓
                                                                                              (2)
9.6     POSITIVE MARKING FROM QUESTION 9.3.3
        RT = R1 + Rx = 4 + 6 = 10 Ω ✓
            V 24
        I= =        ✓
            R 10
        I = 2,4 A ✓                                                                           (3)
                                                                                             [22]
                                                                                   TOTAL:    150

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