# Physical Sciences Grade 12 Paper 1 Limpopo Mock 2022 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Limpopo Mock Exam 2022, 35 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/limpopo-physical-science-grade-12-sept-2022-p1-and-memo/048e1ab3-3292-48b5-8630-f379589f4c76
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: Limpopo Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 35

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## About this document

Physical Sciences · Grade 12 · Limpopo Mock Exam · 2022. Memorandum, 35 pages. Read online or download the PDF.

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‘

                           NATIONAL SENIOR CERTIFICATE

                                    GRADE 12

                       PHYSICAL SCIENCES: PHYSICS (P1)

                                 SEPTEMBER 2022

                              MARKING GUIDELINES

MARKS: 150

                     This marking guidelines consist of 15 pages.

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                               Marking Guidelines

 QUESTION 1

 1.1   B ✓✓                                                                         (2)

 1.2   A ✓✓                                                                         (2)

 1.3   C ✓✓                                                                         (2)

 1.4   A ✓✓                                                                         (2)

 1.5   B ✓✓                                                                         (2)

 1.6   D ✓✓                                                                         (2)

 1.7   D ✓✓                                                                         (2)

 1.8   A ✓✓                                                                         (2)

 1.9   C ✓✓                                                                         (2)

 1.10 A ✓✓                                                                          (2)

                                                                                    [20]

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                               Marking Guidelines

 QUESTION 2

 2.1.1 The force that opposes the tendency of motion of a stationary object relative to
       a surface. ✓✓                                                                    (2)
 2.1.2
             N

                                   fs
                     ●
                                               Accept the following symbols:
                                        N✓     FN/Normal force/Fsurface on crate
                         w              fs ✓   f/Ff/frictional force/static frictional force
                                        w✓     Fg/mg/weight/gravitational force/FEarth on crate
         Take uphill as POSITIVE
          Notes:
          • Mark is awarded for label and arrow.
          • Do not penalize for length of arrows
          • Deduct 1 mark for any additional force.
          • If force(s) do not make contact with dot/body: 2/3
          • If arrows missing: 2/3                                                                 (3)

 2.1.3     Fnet = ma
             max                           Any one ✓
          ∴fs + (-mgsin θ) = mamax
          µs฀ FN - mgsin θ = mamax
          µs฀ mgcos θ - mgsinθ = mamax
            max
          ∴fs     - m(9,8)(sin 10°) = mamax …..①
                 max
          But fs = µ𝑠 N
          ∴fsmax = µs (mgcos θ)….②
          Subst. ② into ① :
          ∴µs (mgcos θ) - m(9,8)(sin 10°) = m(amax)

          ∴(0,35) m(9,8)( cos 10°)✓-m(9,8)( sin 10°) = m(amax)

          (÷m) : ∴(0,35)(9,8)( cos 10°) ✓– (9,8)( sin 10°) ✓= amax

          ∴the maximum acceleration is 1,676 m฀s-2 ✓
                                                                                                   (5)

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                               Marking Guidelines

 2.2.1    Marking criteria
          -1 mark for each key word/phrase omitted in the correct context.

         Each body in the universe attracts every other body with a (gravitational) force
         that is directly proportional to the product of their masses and inversely
         proportional to the square of the distance between their centers. ✓✓
         OR:
         Every particle in the universe attracts every other particle with a (gravitational)
         force that is directly proportional to the product of their masses and inversely
         proportional to the square of the distance between them. ✓✓                         (2)

 2.2.2           GME
          gE =     2    = 9,8 m฀s-2
                  RE
                  GMM
          gM =     2    ✓
                  RM
                     M
                  G ( E)
                     153
              =             ✓
                   𝑅 2
                  ( 5𝐸 )
                  25     GME
              =             2
                  153     RE
                  25
             =          (9,8)
                 153
           =     1,60           m฀s-2   ✓                                                       (3)
        (downwards)
 2.2.3 Equal to ✓                                                                               (1)
                                                                                                [16]

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                               Marking Guidelines

 QUESTION 3

 3.1 Motion during which the only force acting on an object is the gravitational force.✓✓ (2)

 3.2
       OPTION 1:
       UPWARDS POSITIVE:

       For stone X:                            For stone Y:
                      1       2                                 1           2
       ∆y = vi ∆t + a(∆t) ✓                    ∆y = vi ∆t + a(∆t)
                      2                                         2
                          1       2                 49              1
       0 = (3v)(10) + (-9,8) (10) ✓            0 = ( )∆t + (-9,8) (∆t)2 ✓
                          2                          3              2
       0 = 30v – 490                                      49
                                               0 = ∆t (        -4,9∆t)
       490 = 30v                                          3
            49                                 ∆t = 0 s or ∆t = 3,333 s
       v = ( 3 ) m฀s-1
                                                ∆t = 3,33 s (3,333 s) ✓
                                                                                                   (4)

       DOWNWARDS POSITIVE:

       For stone X:                            For stone Y:
                      1       2                                 1           2
       ∆y = vi ∆t + a(∆t) ✓                    ∆y = vi ∆t + 2a(∆t)
                      2
                          1                              49             1
       0 = (3v)(10) + (9,8) (10)2 ✓            0 = (−       )∆t + 2(9,8) (∆t)2 ✓
                          2                               3
       0 = 30฀v + 490                                          49
                                               0 = ∆t (−            +4,9฀∆t)
       -490 = 30฀v                                             3
                 49                            ∆t = 0 s or ∆t = 3,333 s
       v = (− 3 ) m฀s-1
                                                ∆t = 3,33 s (3,333 s) ✓

       OPTION 2:
       UPWARDS POSITIVE:
                                                 For stone Y:
                                                         49
       For stone X:                              vBi = 3 m฀s-1 or 16.333 m฀s-1
       Consider upward motion:
                                                 Consider upward motion:
       vf = vi + a∆t✓
                                                 vf = vi + a∆t
       0 = vi + (-9,8)(5)✓
                                                 0 = 16,333 + (9,8)∆t✓
       vi = 49 m฀s-1
                                                 ∆t = 1,666(1,67 s)
                                                 ∆t(total) = 2 x 1,666 = 3,33 s

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                               Marking Guidelines

      DOWNWARDS POSITIVE:

       For stone X:                                    For stone Y:
                                                                     49
       Consider upward motion:                         vBi = - 3 m฀s-1 or -16.333 m฀s-1
       vf = vi + a∆t✓
                                     Consider upward motion:
       0 = vi + (9,8)(5)✓
                                     vf = vi + a∆t
       vi = -49 m฀s-1
                                     0 = -16,333 + (9,8)∆t✓
                                     ∆t = 1,666(1,67 s)
                                     ∆t(total) = 2 x 1,666 = 3,33 s
 3.3 POSITIVE MARKING FROM QUESTION 3.2:

      OPTION 1:
      UPWARDS POSITIVE:
       For stone Y                                  For stone X
                       1       2                                          1            2
       ∆y = vi ∆t + a(∆t) ✓                         ∆y = vi ∆t + a(∆t)
                       2                                                  2
                                        5 2
             49        5                                                           1              2
                               1                       = (49)(5) ✓+ 2(-9,8)(5) ✓
       H=(        ) (3) + 2(-9,8)(3) ✓
             3                                             245            245
             245                                       =            = 9(           )✓ = 9H
       H=(      )m                                          2                 18
             18

      DOWNWARDS POSITIVE:
       For stone Y                                  For stone X
                       1       2                                          1            2
       ∆y = vi ∆t + 2a(∆t) ✓                        ∆y = vi ∆t + a(∆t)
                                                                          2
                 49        5       1     5 2                                       1              2
       H = (−       ) (3) + 2(9,8)(3) ✓                = (-49)(5) ✓+ 2(9,8)(5) ✓
                  3
                 245                                        245                    245
       H = (−          )m                              =-            = 9(−                 )✓ = 9H
                  18                                            2                  18

       OPTION 2:                                 For stone X
       UPWARDS POSITIVE:                                         1
                                                 ∆y = vi∆t +         a∆t2
                                                                 2
                                                                                           1
        For stone Y                                 = (16.333)(1,666) +                        (9,8)(1,666)2✓
                                                                                           2
                       1                            = 13,61 m = H
        ∆y = vi∆t + a∆t2✓
                       2                         13,61 X 9 = 122,49 m = 9H(122,5 m)✓
                               1
           = (49)(5) ✓+            (-9,8)(5)2✓
                               2
           = 122,5 m

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                               Marking Guidelines

        DOWNWARDS POSITIVE:                            For stone X
                                                                                1
                                                       ∆y = vi∆t +                  a∆t2
          For stone Y                                                           2
                                                                                              1
                            1                                = (-16.333)(1,666) +                 (9,8)(1,666)2✓
          ∆y = vi∆t +       a∆t2✓                                                             2
                            2                             = - 13,61 m = H
                                1
                = (-49)(5) ✓+       (9,8)(5)2✓         -13,61 X 9 = -122,49 m = 9H(122,5 m)✓
                                2
                = -122,5 m                                                                                              (5)

 3.4 POSITIVE MARKING FROM QUESTION 3.2:

         49                                                    49

      Velocity (m฀s-1 )                                     Velocity (m฀s-1 )

            0               5            10                       0                    5            10 Time (s)
                                                 Time (s)

        -49                                                   -49

        Marking criteria:
          • Correct shape (Should intersect t-axis) ✓
          • Final velocity and initial velocity shown ✓
          • 5 s shown for maximum height ✓                                                                              (3)
                                                                                                                        [14]

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                               Marking Guidelines

 QUESTION 4

 4.1   The product of an object’s mass and its velocity. ✓✓            (2 or 0)                      (2)

 4.2   The total (linear) momentum of an isolated system remains constant (is
       conserved). ✓✓                                                         (2)

 4.3
        OPTION 1                                     OPTION 2

        ∑ pi = ∑ pf                      Any one✓ ∆pA = - ∆pB                     Any one✓
        mA vAi +mB vBi = mA vAf + mB vBf          mA (vAf − vAi ) = mB (vBf - vBi )
        mA (4) + mB (-1) ✓= mA (1) + mB (3)✓ mA (1-4) ✓ = -mA (3-(−1)) ✓
        4mA - mA = 3mB + mB                       mA (-3) = −mB (4)
        3mA = 4mB ✓                               mA (-3) = mB (4) ✓
        mA       4                                   mA       -4   4
        mB
             =
                 3
                                                          =    =
                                                     mB       -3   3
        mA : mB = 4:3                                mA : mB = 4:3

        Marking criteria:

        •        Formula

        •        Right hand substitution into formula

        •        Left hand substitution into formula

        •        This step: mA (-3) = mB (4)                                                         (4)

                                                 ✓
 4.4                         pi
                           ✓ pf                                    ∆p ✓                              (3)

        Criteria                                                                        mark
            • Large initial momentum in the same direction as final momentum            ✓
            • Small final momentum in the same direction as initial momentum
                                                                                        ✓
            • Change in momentum in the opposite direction                              ✓
                                                                                                     [11]

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                               Marking Guidelines

 QUESTION 5

 5.1   The net work done on an object is equal to the change in the object’s kinetic
       energy. ✓✓
       OR:
       The work done on an object by the net force is equal to the change in the object’s
       kinetic energy. ✓✓                                                                 (2)
       OPTION 1:
 5.2    Wnet = ∆EK OR Wf + Ww + WN = ∆EK             Any one✓
                                         1
        f∆x฀cos θ + mg∆x(cos θ) + 0 = 2m(vf2 − v2i )
                                                                 1      2     2
        (45)(∆x)( cos 180°) + (10)(9,8)(∆x)( cos 125°) ✓+ 0 = 2(10)(0 - 8,84 )✓
        -45฀∆x – 56,2105฀∆x = -390,728
        ∆x = 3,86 m
        x = 3,86 m✓                                                                             (4)

        NB: The work done by the gravitational force W w can also be calculated
        as follows:
                                         OR: Ww = -∆Ep
        Ww = mgsinθ∆xcos฀
             = (10)(9,8)(sin35°)xcos180°           = mg(hi – hf)
             = (-56,2105x)J                        =(10)(9,8)(0 – xsin35°)
                                                      = (-56,2105x)J
        OPTION 2:
        Wnc = ∆Ep + ∆Ek
                             1
             = mg(hf – hi) + 2m(vf2 − v2i )          Any one✓
                             1
         Wf = mg(hf – hi) + 2m(vf2 − v2i )
                                 1
        f∆xcos θ = mg(hf – hi) + 2m(vf2 − v2i )
                                                      1
        (45)xcos 180° = (10)(9,8)(xsin35° - 0)✓ +2 (10)(02 - 8,842 )✓
        -45x = 56,2105x – 390,728
           x = 3,86 m✓

 5.3            N

                                     f
                     ●                            Accepted labels
                                                  N✓     FN /Normal force/Fnormal
                                                  f✓     f/Ff / frictional force/ fs
                                                  w✓     Fg/mg/weight/gravitational force
                       w
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                               Marking Guidelines

         Notes:
         • Mark is awarded for label and arrow.
         • Do not penalize for length of arrows
         • Deduct 1 mark for any additional force.
         • If force(s) do not make contact with dot/body: 2/3
         • If arrows missing: 2/3                                                            (3)

 5.4     OPTION 1:                   OPTION 2:
         fs = wⅡ = mgsin θ✓          µs = tanθ = tan35° =0,7002
            =(10)(9,8)(sin 35°)✓     fs = µsN = µsmgcosθ (ANY ONE)✓
         fs = 56,21 N ✓                 =(0,7002)(10)(9.8)(cos35°)✓
                                        =56,21 N✓                                            (3)
                                                                                             [12]

 QUESTION 6

 6.1     The apparent change in frequency (or pitch) of the sound detected by a listener,
         because the sound source and the listener have different velocities relative to
         the medium of sound propagation. ✓✓                                              (2)

 6.2      v = fs λ ✓
          340 = (300)λ✓
          λ = 1,13 m ✓                                                                       (3)

 6.3     ANY ONE:
         • To monitor the heartbeat of a foetus (unborn baby).✓
         • To measure the rate of blood flow.✓                                               (1)
 6.4.1    OPTION 1:                        OPTION 2:
                   v±vL                           v- vS
          fL = (   ) fs ✓                  λB =           ✓
               v±v    S
                                                    fs
                    340                            340-30 ✓
            = (340-30) (300) ✓                    =
                                                      300✓
            = 329,032 Hz                          = 1,03 m✓

          v = fL λ
          340 = (329,032)λ ✓
                                                                                             (4)
          λ = 1,033 m ✓

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                               Marking Guidelines

 6.4.2
         OPTION 1:                                             OPTION 2:
                   v±vL                                                  V- VS
         fL = (        ) fs                                    λA = (            )✓
                   v±v   S                                                 fs
                  340    ✓                                                340+30
              = (340+30) (300)✓                                        =(
                                                                            300
                                                                                   )✓
           = 275,676 HZ                                                = 1,23 m ✓
         v=fλ
         (340) = (275,676) λ
                   340
         λ=                                                                                                (3)
               275,676
           = 1,23 m ✓

 6.5     Less than✓                                                                                        (1)
                                                                                                           [14]

 QUESTION 7

 7.1.1 Distance (between the point charges)/medium/air✓                                                    (1)

 7.1.2 The electrostatic force is directly proportional to the product of charges.✓                        (1)

 7.1.3                       ∆F                 NOTE: accept any
         gradient =              2
                        ∆Q                      value from the graph
                   (4-3)X 1012
               =             2         ✓
                      ∆Q
                   1 X1012
               =
                         1✓
                                           -2
               = 1 X 1012 N฀C ✓                                                                            (3)

 7.1.4         KQ1 Q2
         F=                  ✓
                    r2
          F        k
             2 = r2
         Q
                                 9 x 109
         1 X 1012 ✓=                       ✓
                                     r2
         r2 = 9 x 10-3
         r = 0,09487 m (0,095 m)✓
                                 KQ2                               3
         NOTE: If F =                     is used, then maximum:
                                  r2                               4                                       (4)

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                               Marking Guidelines

 7.2.1 A region of space in which an electric charge experiences a force. ✓✓                                             (2)

 7.2.2           kQ
         E=            ✓
                 r2
         Enet,p = 0
         kQ1            kQ2
                 ✓=
          r2               r2
         (9 X 10 )(8 X 10-6 )
                   9
                                              (9 X 109 )(2 X 10-6 )
                                          =                           ✓
                 (0,4-d)2                              d2

          ACCEPT: If 10-6 is omitted since it appears on both sides.

           d2              (2 X 10-6 )
         (0,4-d)2
                       =             -6
                           (8 X 10 )
         = 0,25
           d
                 = 0,5
         0,4-d
         d = 0,1333 m

         ∴The distance is 0,1333 m ✓                                                                                     (4)

 7.2.3   OPTION 1:                                                        OPTION 2:
                       Q1 + Q2                                                       Q1 + Q2
         Qnew =                                                           Qnew =
                           2                                                            2
                       8 X 10-6 +2 X 10-6                                            8 X 10-6 +2 X 10-6
                   =                             ✓                               =                        ✓
                                 2                                                             2
                   = 5 𝑋 10−6 C                                                  = 5 𝑋 10−6 C

               Q                                                             Q
         n= ✓                                                             n= ✓
               e                                                             e
               5 𝑋 10−6 − 8 𝑋 10−6                                              5 X 10-6 - 2 X 10-6
         n=                                     ✓                          n=                         ✓
                      −1,6 X 10−19                                                   -1,6 X 10-19

          n = 1,875 x 1013 electrons ✓                                    n = 1,875 x 1013 electrons ✓                   (4)
                                                                                                                         [19]

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                               Marking Guidelines

 QUESTION 8

 8.1      Emf✓                                                                          (1)

 8.2.1     Vlost = Ir
           24 – 21,6✓ = I(2)
           I = 1,2 A

           Itotal = 1,2 A
           V12 Ω = IR = (1,2)(12)✓ = 14,4 V
           V10 Ω = 21,6 – 14,4 = 7,2 V
                      V     7,2
           I10 Ω = R ✓=           ✓= 0,72 A✓
                            10                                                          (5)
           IA = 0,72 A

 8.2.2     IX = 1,2✓ – 0,72
             = 0,48 A
                  V        7,2
           RX =       ✓=          ✓
                  I        0,48
              = 1,50 Ω✓                                                                 (4)

 8.3       POSITIVE MARKING FROM QUESTION 8.2.1
           P = I2R✓
             = (1,2)2✓ (12)✓
             = 17,28 W✓
                                                                                        (4)
 8.4      Decreases ✓
         • Total resistance in the circuit increases. ✓
                                                I
         • Current in the circuit decreases (I R). ✓
         • P = I 2R; when R is constant, P decreases ✓

                                                                                        (4)

                                                                                        [18]

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                               Marking Guidelines

 QUESTION 9

 9.1.1    From electrical energy to mechanical energy. ✓                                         (1)

 9.1.2    Clockwise. ✓                                                                           (2)

 9.1.3    (carbon) brush✓✓                                                                       (1)

 9.1.4    It reverses the direction of the current in the coil after each half-cycle. ✓          (1)

 9.1.5    Increases. ✓
          The current increases✓                                                                 (2)

 9.2.1    The rms potential difference is the AC potential difference which
          dissipates/produces the same amount of energy as an equivalent DC potential
          difference✓✓
          ACCEPT:
          The rms voltage is the DC potential difference which dissipates/produces the (2)
          same amount of energy as the equivalent AC potential difference✓✓

 9.2.2             Vrms
          Vrms =          ✓
                    √2
                   Vrms
          200 =           ✓
                   √2
          Vrms = (200)(√2)
          Vrms = 282,8427 V✓                                                                     (3)
                                                                                                 [12]

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### Page 35

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Physical           f rom St anmorephysics.
         Sciences/P1                   15       com                 LimpopoDoE/September 2022
                               Marking Guidelines

 QUESTION 10

 10.1.1 The process whereby electrons are ejected from a metal surface when light of
        suitable frequency is incident on that surface. ✓✓                           (2)

 10.1.2   The minimum energy that an electron in the metal needs to be emitted from
          the metal surface.✓✓                                                      (2)

 10.1.3   (a)        Frequency (of the incident light). ✓                                       (1)
          (b)        Frequency (of the incident light).✓                                        (1)

 10.1.4   E = hf
                hc
          E =λ ✓
                     (6,63 X 10-34 )(3 X 108 )
                                                        ✓
          ∴E =                                    -9
                           (450 X 10 ) ✓
          E = 4,42 X 10−19 J ✓
          Since photon energy is less than the work function of the metal, so, no
          emission occurs. ✓                                                                    (5)

 10.2.

                            Increasing energy
                                                                    1880 nm ✓

                                                486 nm ✓ 656 nm ✓

                                                                                                [14]
                                                                                    TOTAL:      150

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