# Physical Sciences Grade 10 P1 Gauteng June 2022 Memo (Eng)

> Grade 10 Physical Sciences Paper 1 memorandum from the Gauteng June 2022 (English), 6 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr10-ps-p1-eng-june-2022-possible-answers/40efa26e-8296-44ef-b393-3eb33ed9a902
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 10
- Document type: Memorandum
- Year: 2022
- Exam period: Gauteng June
- Paper: 1
- Curriculum: CAPS
- Pages: 6

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## About this document

Physical Sciences · Grade 10 · Gauteng June · 2022. Memorandum, 6 pages. Read online or download the PDF.

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## Document text

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### Page 1

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          PROVINCIAL EXAMINATION
                   JUNE 2022
                   GRADE 10
           MARKING GUIDELINES
             PHYSICAL SCIENCES (PAPER 1)

6 pages

### Page 2

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                                                                PHYSICAL SCIENCES
                                     MARKING GUIDELINES
                                                                (PAPER 1)      GRADE 10

QUESTION 1

1.1   B ✓✓

1.2   C ✓✓

1.3   B ✓✓

1.4   B ✓✓

1.5   C ✓✓

1.6   A ✓✓

1.7   D ✓✓

1.8   D ✓✓
                                                                            (2 x 8)   [16]

QUESTION 2

2.1   2.1.1   Scalar is physical quantity with magnitude only. ✓✓                      (2)

      2.1.2                              4,5 cm ✓

                                                    6,75 cm ✓
                           R = 8,11 cm ✓

                                                                                       (3)

      2.1.3   Fnet 2 = X 2 + Y 2 ✓

              Fnet = (4,5) 2 + (6,75) 2 ✓✓

              Fnet = 8,11 cm ✓                                                         (4)

      2.1.4   Yes ✓, the solution must be the same irrespective of method used.✓       (2)

2.2   2.2.1   Distance = 7,5 km ✓                                                      (1)

      2.2.2   Displacement = 4,5 – 3 = 1,5 km ✓, North ✓                               (2)
                                                                                      [14]

                                             2

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                                                               PHYSICAL SCIENCES
                                     MARKING GUIDELINES
                                                               (PAPER 1)      GRADE 10

QUESTION 3

3.1   The total mechanical energy remains constant in an isolated system. ✓✓

      OR

      The sum of gravitational potential energy and kinetic energy at the top equals
      the sum of gravitation potential energy and kinetic energy at the bottom in the
      absence of friction. ✓✓                                                            (2)

3.2   Ep = m.g.h ✓

           = 112 x 9,8 x 10 ✓

           = 10 976 J ✓                                                                  (3)

3.3   EMtop = EMbottom ✓

      m.g.h + .m.v2 = m.g.h + .m.V2

      (112 x 9,8 x 10) + 0 ✓ = (112 x 9,8 x 8) +   x 112 x V2 ✓

      10 976 = 8 780,8 + 56.V2

       V = 6,26 m·s-1 ✓                                                                  (4)

3.4   EMtop = EMbottom ✓

      m.g.h + .m.v2 = m.g.h + .m.V2

      (112 x 9,8 x 10) + 0 ✓ = 0 +     x 112 x V2 ✓

      10976 = 56·V2

       V = 14 m·s-1 ✓                                                                    (4)

3.5   Remains the same. ✓ Total mechanical energy is conserved. No friction. ✓           (2)
                                                                                        [15]

                                              3

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                                                                 PHYSICAL SCIENCES
                                           MARKING GUIDELINES
                                                                 (PAPER 1)      GRADE 10

QUESTION 4

4.1   Every charge in the universe consists of integer multiple of the electron
      charge. ✓✓                                                                         (2)

4.2
      n=      ✓

      n=                      ✓

        = 5,125 x 1010 electrons ✓                                                       (3)

4.3   Attraction force ✓                                                                 (1)

      4.4.1   S to T ✓                                                                   (1)

      4.4.2   The net charge of an isolated system remains constant during any
              physical process. ✓✓                                                       (2)

      4.4.3               QT + QS
              Q new =             ✓
                             2

                  ( +5,4  10 −9 ) + ( −8,2  10 −9 )
              =                                       ✓
                                   2

              = − 1,4  10 −9 C ✓                                                        (3)

4.5   Static electricity is formed much better when the air is dry or the humidity is
      low. When the air is humid, water molecules can collect on the surface of
      various materials. This can prevent the build-up of electrical charges. ✓✓         (2)
                                                                                        [14]

                                                          4

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                                                                PHYSICAL SCIENCES
                                         MARKING GUIDELINES
                                                                (PAPER 1)      GRADE 10

QUESTION 5

5.1   Maximum energy given to each coulomb of charge passing through battery. ✓✓             (2)

5.2   Parallel ✓                                                                             (1)

5.3   5.3.1
                                    ✓

                  1    1   1
                     =   +   ✓
                  Rp   24 16

                  Rp = 9,6  ✓                                                              (3)

      5.3.2            V
                  I=     ✓
                       R

                       12
                  I=       ✓
                       9,6

                  I = 1,25 A✓                                                                (3)

5.4   5.4.1       Rs = R1 + R2

                  Rs = 24 + 16 ✓

                               Rs = 40  ✓                                                   (2)

      5.4.2            V
                  I=
                       R

                      12
                  =      ✓
                      40

                  = 0,3 A ✓                                                                  (2)

5.5        Q
      I=      ✓
           Δt

               Q
      0,3 =       ✓
              120

      Q = 36 C ✓                                                                            (3)

5.6   Circuit 1 ✓, connecting resistors in parallel decreases the total resistance of the
      circuit ✓, thus increasing the total current passing through the circuit/
      increasing the brightness of the bulb. ✓                                               (3)
                                                                                            [19]

                                                5

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                                                                PHYSICAL SCIENCES
                                           MARKING GUIDELINES
                                                                (PAPER 1)      GRADE 10

QUESTION 6

6.1   Rate of change in velocity. OR Change in velocity per second. ✓✓               (2)

6.2   77 km .h −1
                   1 000 ✓ = 21,39 m·s-1 ✓
        3 600                                                                        (2)

6.3   V f = Vi + at ✓

      V f = 0 + (1,5)(6,5) ✓✓

      Vf = 9,75 m·s-1 ✓                                                              (4)

6.4   X = Vi t + 1 2 at 2 ✓

      X = (0)(6,5) + 1 2 (1,5)(6,5)2 ✓✓

      X = 31,69 m ✓                                                                 (4)

6.5         V f + Vi 
      X =          t ✓
             2 

            21,39 + 0 
      X =            (6,5) ✓✓
                2     

      X = 69,52 m ✓                                                                 (4)

6.6   No, ✓ CAR A is moving with 2 times the speed of CAR B. ✓                       (2)

6.7   X = Vi t + 1 2 at 2 ✓

      25 = (21,39 )(2,5) + 1 2 a(2,5) ✓✓
                                  2

      a = −9,112 m.s −2 ✓ (Decelerating)                                             (4)
                                                                                    [22]

                                                                          TOTAL:    100

                                                  6

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