# Physical Sciences Grade 10 P1 November Mock 2026 Memo (Eng)

> Grade 10 Physical Sciences Paper 1 memorandum from the November Mock Exam 2026 (English), 8 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-10-physical-sciences-2026-november-mock-exam-memorandum-paper-1/34ed52fc-13b0-4309-90e5-14c85c1516da
- Updated: 2026-10-29
- Subject: Physical Sciences
- Grade: Grade 10
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 8

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## About this document

Physical Sciences · Grade 10 · November Mock Exam · 2026 · English. Memorandum, 8 pages. Read online or download the PDF.

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## Document text

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### Page 1

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

                                  NATIONAL
                             SENIOR CERTIFICATE

                                    GRADE 10

                       PHYSICAL SCIENCES: PHYSICS (P1)
                     NOVEMBER EXAMINATION 2026 (MOCK)
                              MARKING GUIDELINES

  MARKS: 100

                     These marking guidelines consist of 8 pages.

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### Page 2

Physical 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 (✓). A formula is not
     awarded a mark if it 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 to a previous question is used correctly
     in a later calculation, the later calculation is marked positively.
  5. Negative marking: If a candidate must choose (e.g. INCREASES/DECREASES) and
     then explain, 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 or used in an incorrect context. Accept equivalent wording.
  7. Vectors: A direction must be given wherever a vector quantity (displacement, velocity,
     acceleration) is asked for; candidates may choose either direction as positive if applied
     consistently.

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### Page 3

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

QUESTION 1
 1.1         C ✓✓              (2)
 1.2         B ✓✓              (2)
 1.3         D ✓✓              (2)
 1.4         A ✓✓              (2)
 1.5         B ✓✓              (2)
 1.6         D ✓✓              (2)
 1.7         C ✓✓              (2)
 1.8         A ✓✓              (2)
                                                                                          [16]

QUESTION 2
2.1     A wave in which the particles of the medium vibrate perpendicular (at right
        angles) to the direction of motion of the wave. ✓✓                                  (2)
2.2     0,15 m ✓                                                                            (1)
2.3     A and C ✓      (OR B and D)                                                         (1)
2.4     2,5 wavelengths fit into 3,0 m ✓
            3,0
        λ=      = 1,2 m ✓
            2,5                                                                             (2)
           10
2.5     f=    = 5 Hz ✓
            2
            1      1
        T = ✓ = = 0,2 s ✓
            f      5
                     2s
        OR: T =            ✓✓ = 0,2 s ✓
                  10 waves                                                                  (3)
2.6     v = fλ ✓
        v = (5)(1,2) ✓
        v = 6 m·s–1 ✓
        OR v = λ /T = 1,2/0,2 = 6 m·s–1 . POSITIVE MARKING FROM 2.4 AND 2.5.               (3)
                                                                                          [12]

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### Page 4

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

QUESTION 3
3.1.1   Longitudinal ✓                                                                         (1)
3.1.2   Total distance travelled by the pulse:
            ∆x              ∆x
        v=      ✓ 340 =          ✓
            ∆t            0,025
        ∆x = 8,5 m (to the moth and back) ✓
                            8,5
        Distance to moth =       = 4,25 m ✓
                              2
        OR: ∆t one way = 0,0125 s ✓; ∆x = v∆t ✓ = (340)(0,0125) ✓ = 4,25 m ✓                   (4)
3.2.1   A packet (quantum) of energy ✓ of light/electromagnetic radiation ✓.                   (2)
              c
3.2.2   E=h     ✓
             λ
            (6,63 × 10–34 )(3 × 108 )
        E=                            ✓✓
                  250 × 10–9
        E = 7,96 × 10–19 J ✓
        One mark for h and c substituted correctly, one mark for converting 250 nm to 250 ×
        10–9 m.                                                                                (4)
                                                                                              [11]

QUESTION 4
4.1     The net charge of an isolated system remains constant during any physical
        process (e.g. when two charges make contact and are separated). ✓✓                     (2)
            Q1 + Q2     (+6 × 10–9 ) + (–14 × 10–9 )
4.2     Q=            =                              ✓
               2                     2
        Q = –4 × 10–9 C (–4 nC) on EACH sphere ✓                                               (2)
4.3     Change in charge on A: ∆Q = –4 – (+6) = –10 nC ✓
            Q      –10 × 10–9
        n=      =               ✓
            e     –1,6 × 10–19
        n = 6,25 × 1010 electrons ✓
        Electrons are transferred from B TO A. ✓
        Accept ∆Q for B: –4 – (–14) = +10 nC. POSITIVE MARKING FROM 4.2.                       (4)
                                                                                               [8]

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### Page 5

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

QUESTION 5
5.1     The total energy supplied per coulomb of charge by the battery (OR the
        potential difference across the terminals of a battery when no current flows).
        ✓✓                                                                                (2)
         1      1    1      1 1
5.2         =     +       = +     ✓
        Rp R1 R2            6 12
        Rp = 4 Ω ✓
        Rtotal = Rs = 4 + 4 ✓ = 8 Ω ✓                                                     (4)
             V       12
5.3     R=      ✓ 8=    ✓
              I       I
        I = 1,5 A ✓
        POSITIVE MARKING FROM 5.2.                                                        (3)
5.4     Q = I∆t ✓
        Q = (1,5)(2 × 60) ✓
        Q = 180 C ✓
        POSITIVE MARKING FROM 5.3.                                                        (3)
5.5     DECREASES ✓
        The total (external) resistance INCREASES (from 8 Ω to 10 Ω), so for the same
        emf the current decreases. ✓                                                      (2)
                                                                                         [14]

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### Page 6

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

QUESTION 6
6.1     The rate of change of velocity. ✓✓                                               (2)
               1 000
6.2     90 ×         = 25 m·s–1 ✓
               3 600                                                                     (1)
6.3     ∆x = v∆t (OR ∆x = vi ∆t + 12 a∆t2 with a = 0)
        ∆x = (25)(0,7) ✓
        ∆x = 17,5 m ✓                                                                    (2)
6.4     Take the direction of motion as positive.
        v2f = v2i + 2a∆x ✓
        02 = (25)2 + 2(–5,5)∆x ✓✓
        ∆x = 56,82 m ✓
        One mark for vf = 0 and vi = 25; one mark for a = –5,5. POSITIVE MARKING FROM
        6.2.                                                                             (4)
6.5     YES
        Total stopping distance = 17,5 + 56,82 = 74,32 m ✓
        74,32 m < 75 m, so the taxi stops (0,68 m) before reaching the cow. ✓
        POSITIVE MARKING FROM 6.3 AND 6.4.                                               (2)
6.6     vf = vi + a∆t ✓
        0 = 25 + (–5,5)∆t ✓
        ∆t = 4,55 s ✓                                                                    (3)
                                                                                        [14]

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### Page 7

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

QUESTION 7
7.1     Constant velocity (of 8 m·s–1 east) OR uniform motion. ✓                  (1)
            ∆v       8–0
7.2     a=     ✓ =        ✓
            ∆t       4–0
        a = 2 m·s–2 east ✓                                                        (3)
7.3     ∆x = area under the graph ✓
        ∆x = 12 (4)(8) + (6)(8) + 21 (4)(8) ✓✓
        ∆x = 80 m east ✓
        OR: area of trapezium = 12 (14 + 6)(8) ✓✓✓ = 80 m east ✓                  (4)
               ∆x       80
7.4     vave =    ✓ =      ✓
               ∆t       14
        vave = 5,71 m·s–1 east ✓
        POSITIVE MARKING FROM 7.3.                                                (3)
7.5
         a (m·s–2 )

           2

                                        10          14
                                                         t (s)
                        4

          –2

        Marking criteria
        Horizontal line at +2 from 0 to 4 s ✓
        Zero acceleration from 4 s to 10 s ✓
        Horizontal line at –2 from 10 s to 14 s ✓                                (3)
                                                                                [14]

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### Page 8

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

QUESTION 8
8.1     The energy an object possesses due to its position (height) relative to a
        reference point (in the gravitational field). ✓✓                                               (2)
8.2     EP = mgh ✓
        EP = (55)(9,8)(3,2) ✓
        EP = 1 724,8 J ✓                                                                               (3)
8.3     The total mechanical energy of an isolated system remains constant. ✓✓
        Accept: In the absence of friction/air resistance, the sum of the kinetic and gravitational
        potential energy of an object remains constant.                                                (2)
8.4     EM(A) = EM(C)
        (EP + Ek )A = (EP + Ek )C ✓
        1 724,8 + 0 ✓ = (55)(9,8)(1,5) + 12 (55)v2 ✓
        v = 5,77 m·s–1 ✓
        OR: mghA + 0 = mghC + 12 mv2 ⇒ v2 = 2(9,8)(3,2 – 1,5). POSITIVE MARKING FROM
        8.2.                                                                                           (4)
                                                                                                      [11]
                                                                                            TOTAL: 100

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