# Physical Sciences Grade 12 P1 Western Cape Mock 2018 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Western Cape Mock Exam 2018, 18 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/wp-2018-paper-1-memorandum/624f3cc3-b897-459d-9d31-876f0074f311
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: Western Cape Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 18

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## About this document

Physical Sciences · Grade 12 · Western Cape Mock Exam · 2018. Memorandum, 18 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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### Page 1

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                                                                1

                            MEMORANDUM

PHYSICAL SCIENCE/FISIESE WETENSKAPPE           SEPTEMBER 2018

                       CW PLC COMMON PAPER 1

QUESTION 1

1.1    A                                                        (2)

1.2    A                                                        (2)

1.3    C                                                        (2)

1.4    D                                                        (2)

1.5    D                                                        (2)

1.6    A                                                        (2)

1.7    B                                                        (2)

1.8    B                                                        (2)

1.9    D                                                        (2)

1.10   D                                                        (2)

                                                                [20]

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                                                                                                   2

QUESTION 2

21.   2.1.1   When a resultant/net force acts on an object, the object accelerates in the
              direction of the (net) force at an acceleration directly proportional to the force
              √ and inversely proportional to the mass of the object. √
              Wanneer ’n resultante /netto krag op ’n voorwerp uitgeoefen word, versnel
              die voorwerp in die rigting van die (netto) krag met ’n versnelling direk
              eweredig aan die grootte van die krag √ en omgekeerd eweredig aan die
              massa van die voorwerp. √                                                            (2)

      2.1.2

                                                                                                   (3)

      2.1.3   (a)
                                                                                                         Commented [1]: If a candidate add the masses (12 kg)
                                                                                                         directly into Fnet = ma, 2 marks maximum!

                                                                                                         -1 mark for formula
                                                                                                         - 1 mark for correct calculation of combined weight
                                                                                                         (117,6 N)
                                                                                                         Commented [2]: Accepted

              Option 2 for resolving two equations:
              140,8 - T = 4a………(i)
              -78,4 + T = 8a ……..(ii)
                  62,4 = 12 a
                    a = 5,2 ms-2
              T = 8 a + 78,4                                                                             Commented [3]: Candidate receives full marks for (b) if
              = 120 N                                                                              (5)
                                                                                                         Option 2 is used in (a). Candidate must show the
                                                                                                         answer in (b)
                                                                                                         Commented [4]: Yes...answer MUST be shown in (b).

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                                                                           3

      2.1.3   (b)

                                                                           (2)

2.2

      2.2.1   2 x 103 N. √                                                 (1)

      2.2.2   Option 1:

              Distance above Earth’s surface/ Afstand bo Aardoppervlak =

              Option 2:                                                           Commented [5]: I added Option 2
                                                                                  Commented [6]: Thanks

                                                                           (5)    Commented [7]: I suggest that the candidate receives
                                                                                  full marks for a final answer in meter.
                                                                                  Commented [8]: Accepted
                                                                           [18]

QUESTION 3

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                                             4

3.1   g = 9,8 m.s-1 ⩗ downwards/afwaarts ⩗   (2)

3.2   3.2.1

                                             (3)

      3.2.2

                                             (3)

      3.2.3

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                           5

                           (4)

3.3

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                                                                               6

   Correct shape/Korrekte vorm ⩗
(NOTE: The graphs are not drawn to scale. The length on the y-axis between 0
 and 17 should be longer than the length between 0 and 14,51)

                                                                               (4)

(NOTE: The graphs are not drawn to scale. The length on the y-axis between 0
 and 17 should be longer than the length between 0 and 14,51)

                                                                               [15]

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                                                                                            7

QUESTION 4

4.1           The total linear momentum in a closed system remains constant./is conserved /
              Die totale lineêre momentum in 'n geslote stelsel _bly konstant/bly behoue. √√

              OR/OF

      In a closed/isolated system, the total momentum before a collision is equal to the total
      momentum after the collision./In 'n geslote/geïsoleerde stelsel is die totale momentum
      voor 'n botsing gelyk aan die totale momentum na die botsing.
                                                                                            (2)

4.2   4.2.1

                                                                                            (4)

      4.2.2

                                                                                            (4)

4.3                                                                                         (1)
      Greater than/groter as ⩗

                                                                                            [11]

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                                                                                          8

QUESTION 5

5.1

                                                                                          (3)

      Option 3:                                                                                 Commented [9]: I added option 3
      Wnc = ∆Ep + ∆Ek √                                                                         Commented [10]: Thank you

      0√ = mg(hf - hi) +    ½m(vf2 - vi2)
      0 = 3 x 9,8 (0 - 0,15)√ + ½ x 3 (vf2 - 02)√
        vf = 5,42 m.s-1 √

5.2

           0 J. Nul resulterende krag/ konstante snelheid/kinetiese energie is die
      selfde/slegs konserwatiewe kragte is teenwoordig                                    (2)

5.3

      Die netto/totale arbeid op ‘n voorwerp verrig is gelyk aan die verandering in die
      kinetiese energie van die voorwerp
      OF
      Die arbeid verrig op ‘n voorwerp deur ‘n resulterende/netto krag is gelyk aan die
      voorwerp se verandering in kinetiese energie.
                                                                                          (2)

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                                                     9

5.4

                                                     (3)
                                   OR

5.5

      Option 3:
      Wnc = ∆Ep + ∆Ek √
      -30√ = mg(hf - hi) + ½m(vf2 - vi2)
      -30 = 3 x 9,8 (0 - h)√ + ½ x 3 (02 - 5,422)√
        h = 0,48 m √                                 (5)

                                                     [15]

QUESTION 6

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                                                                                            10

6.1

                                OR

                                                                                            (2)

6.2   6.2.1     170 Hz                                                                      (1)

      6.2.2     130 Hz                                                                      (1)
6.3

                                                                                                   Commented [11]: 6.3: 3/6 if candidate take fs = 150
                                                                                                   Hz. (fs is in fact 147,33 Hz)
                                                                                                   give the first three marks

                                                                                            (6)

6.4   ANY ONE/ENIGE EEN
      Doppler flow meter/Doppler-vloeimeter
      Measuring foetal heartbeat/Meet van fetale hartslag
      Measure speed of blood flow/Meet spoed van bloedvloei
      Ultra sound/Ultraklank
      Sonar
      Radar (for speeding/vir jaag)                                                         (1)

6.5   The red shift occurs when the spectrum of a distant star moving away from the earth
      is shifted toward the red end of the spectrum. √√

      Rooi verskuiwings vind plaas wanneer die spektrum van 'n vêr afgeleë ster wat vanaf   (2)
      die aarde wegbeweeg na die rooi ent van die spektrum skuif.

                                                                                            [13]

QUESTION 7

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                                                                                          11

7.1   The force of attraction that two charges at rest exert on each other is directly
      proportional to the product of the two charges✓ and inversely proportional to the
      square of the distance between their centres.✓

      Die aantrekkingskrag wat twee ladings in rus op mekaar uitoefen is direk eweredig
                                                                                          (2)
      aan die produk van die twee ladings en✓ omgekeerd eweredig aan die vierkant van
      die afstand tussen die middelpunte.✓

7.2

                                                                                          (3)

7.3

                                                                                          (5)

                                                                                          [10]

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                                                                   12

QUESTION 8

8.1   The electric force ✓ per unit charge ✓ at that point.        (2)

8.2

      Mark allocation:
      Number of electric field lines from both spheres the same✓
                                                                   (3)
      Correct direction of electric field lines ✓
      Correct pattern✓

8.3   OPTION 1:

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                                 13

OPTION 2:

                                 (6)

                                 [11]

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                                                                                   14

QUESTION 9

9.1        I = V/R ✓     = 4/4 ✓         =1 A ✓                                    (3)

9.2   Opsie 1:
      ( I parallel) :
      One branch ( 4 + 4 = 8 Ω ) current is 1 A.
      The 16 Ω branch current is 1 A/2 = 0,5 A✓ (dubbel resistance    means half
      the current )
              I tot = I2 = 1 + 0,5 ✓ = 1,5 A ✓

      Opsie 2:
      Vp = 2 x 4 = 8 V therefore V(R1) = 20 - 8 = 12 V
      I(R1) = V/R ✓
                                                                                   (3)
             = 12/8 ✓
             = 1,5 A ✓

9.3   1/Rp = 1/R1 + 1/R2
               = 1/(4 +4) + 1/16 ✓
                 = 3/16 ✓
                 = 5,33 Ω ✓
      or
             Rp = R1 x R2 / R1 + R2 ✓
                                                                                   (3)
                   =       8 x 16 / 8 + 16 ✓
                  = 5,33 Ω ✓

9.4   emf = I ( R + r) ✓
      24    = 1,5 [ (5,33 + 8) + r ] ✓
      24    = 19,995 + 1,5 r
       r     = 2,67 Ω ✓
                          OR
                                                                                   (3)
      V(int) = Ir. ✓
      4 = 1,5 x r. ✓
      r = 2,67 ohm ✓

9.5   With S open the total resistance increases and the current decreases. ✓

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                                                                                      15

        emf = IR + Ir , emf and r is constant thus Ir will decrease. Therefore V1 will
        increase ✓
        Wanneer S oop is neem die totale weerstand van die stroombaan toe, dus sal
        die stroom afneem
        Emk = IR + Ir, emk en r bly konstant daarom sal Ir afneem, dus sal V1          (2)
        toeneem
                                                                                       [14]

QUESTION 10

10.1.   Move the bar magnet very quickly up✓ and down inside the coil ✓
        Beweeg die staafmagneet baie vinnig op                                         (2)
        en af binne in die spoel.

10.2.   10.2.1   Electromagnetic induction/Elektromagnetiese induksie ✓                (1)

        10.2.2   Commutator/kommutator / split ring/splitring ✓                        (1)

10.3    10.3.1

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                              16

                              (3)

10.3.2

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                                                                                              17

                                                                                               (4)

                                                                                               [11]

QUESTION 11

11.1   11.1.1   Photoelectric effect/foto-elektriese effek✓                                    (1)

       11.1.2   Increase/neem toe✓
                An increase in intensity, will result in more electrons being emitted per
                second/time unit / ‘n toename in intensiteit lei tot meer elektrone wat per
                                                                                               (2)
                sekonde/tydseenheid vrygestel word✓

       11.1.3   Increase in Kinetic Energy/ toename in kinetiese energie✓
                Blue light has higher frequency than red light/blou lig het ‘n hoër frekwensie
                as rooi lig✓
                Therefore the energy of blue is higher than red because E = hf/daarom is       (3)
                die energie van blou lig hoër omdat E = hf✓

11.2

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                                                                  18

            E = W0 + Kmax ✓
                                                                  (5)
        -20
2,95 x 10     ✓= 1 x 10-20 ✓ + 0,5 (9,11 x 10-31) v2 ✓
            v = 2,07 x 105 m.s-1 ✓

                                                                  [11]

                                                         TOTAAL: 150

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