# Physical Sciences Grade 12 P1 Western Cape Mock 2015 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Western Cape Mock Exam 2015, 9 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2015-sept-winelands-physics-p-1-memo/69181b1d-95f6-41d8-8658-ef7da76875ba
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2015
- Exam period: Western Cape Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 9

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## About this document

Physical Sciences · Grade 12 · Western Cape Mock Exam · 2015. Memorandum, 9 pages. Read online or download the PDF.

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## Document text

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### Page 1

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PHYSICAL SCIENCES/ P1            1
                                CAPS

                   CAPE WINELANDS EDUCATION DISTRICT

                         CURRICULUM AND
                        ASSESSMENT POLICY
                            STATEMENT

                             GRADE 12

                 PHYSICAL SCIENCES: PHYSICS (P1)

                  SEPTEMBER 2015 MEMORANDUM

MARKS 150

TIME 3 hours

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PHYSICAL SCIENCES/ P1                           2                       SEPTEMBER 2015
                                               CAPS

QUESTION 1

1.1    A✓✓                                                                                 (2)

1.2    A✓✓                                                                                 (2)

1.3    D ✓✓                                                                                (2)

1.4    A ✓✓                                                                                (2)

1.5    D ✓✓                                                                                (2)

1.6    C ✓✓                                                                                (2)

1.7    A ✓✓.                                                                               (2)

1.8    B ✓✓                                                                                (2)

1.9    B ✓✓                                                                                (2)

1.10   C ✓✓                                                                                (2)

                                                                                           [20]

QUESTION 2

2.1    When a resultant/net force acts on an object, the object will accelerate in the
       direction of the force ✓the acceleration is directly proportional to the force
       and inversely proportional to the mass of the object.✓                              (2)

2.2

                                                                                           (3)

2.3    downward = positive (clockwise)
       Ftension + Fgravitation = Fresultant✓
       Ftension + (2)(9,8) = 2(3)✓                                                         (3)

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PHYSICAL SCIENCES/ P1                       3                           SEPTEMBER 2015
                                           CAPS

       Ftension = -13,6 N
       Ftension = 13,6 N ✓ upward

2.4    Right = positive (clockwise)
       Frope + Ffriction = Fresultant✓
       13,6 N + Ffriction =4(3)✓                                                           (3)
       Ffriction = -1,6 N
       Magnitude of friction force: 1,6 N ✓

2.5    The force that the rope exerts on the box✓ and the force that the box exerts
       on the rope.✓     OR the force that the Earth exerts on the box and the force       (2)
       that the box exerts on the Earth.

                                                                                           [13]

QUESTION 3

3.1    The product of the resultant/net force acting on an object and the time the
       resultant/net force acts on the object.✓✓                                           (2)

3.2    Right = positive
       F𝛥t = m𝛥v =m(vf - vi)✓ (or other correct form of the equation)
       F(0,1) = 0,4[ 1,49 - (-6)]✓                                                         (3)
       F = 29,96 N ✓

3.3    The total linear momentum of a closed system✓ remains constant (is
       conserved)✓                                                                         (2)

3.4    m1v1i + m2v2i = (m1 + m2) v✓(Right = positive)
       50v1i + (0,4)(-6) = 50,4(1,49)✓
       v1i = 1,55 m∙s-1✓                                                                   (3)

3.5    Total kinetic energy before collision:
       ½ m1v1i2 +½ m2v2i2
       = (0,5)(50)(1,55)2 + (0,5)(0,4)(-6)2✓
       =67,26 J✓
       Total kinetic energy after collision:
       ½ (m1 + m2)v2
       = (0,5)(50,4)(1,49)2✓
       = 55,95 J✓
       Ek before≠Ek after✓                                                                 (5)
        inelastic collision.

                                                                                           [15]

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PHYSICAL SCIENCES/ P1                      4                     SEPTEMBER 2015
                                          CAPS

QUESTION 4

4.1    vf2 = v2i +2 g𝛥y✓ (Upwards = positive)
       vf2 =(8)2 + 2(-9,8)(2)✓
        vf = 4,98 m∙s-1✓                                                            (3)

4.2    Time it took to reach the ceiling:
       vf = vi + gt✓
       4,98 = 8 + (-9,8)t✓
       t = 0,31 s.
       Therefore: time it took for ball to bounce back:
       0,65 - 0,31 = 0,34 s✓
       initial velocity of the ball when it bounce back:
       ∆y = vit + ½ gt2✓
       2 = vi(0,34) + (0,5)(9,8)(0,34)2✓                                            (6)
       v = 4,22 m∙s-1✓.

4.3

                                                                                    (5)

                                                                                    [14]

QUESTION 5

5.1    A force for which the work done in moving an object between two points
       depends on the path taken/is not independent of the path taken✓✓             (2)

5.2    0J✓                                                                          (1)

5.3    Fg// - ( f + F) = 0✓ (Accept other correct symbols)
       OR/OF
       F = mg sinθ – fk
       OR/OF
       F = mgsinθ – 266
       F = [100(9,8) sin 25 o]✓ – 266✓

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PHYSICAL SCIENCES/ P1                     5          SEPTEMBER 2015
                                         CAPS

        414,167- 266
       F = 148,17 N✓                                                    (4)
       NOTE/LET WEL
       No mark for diagram
       1 mark for use of any of the three formulae

5.4                                                                     (6)

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PHYSICAL SCIENCES/ P1                      6                             SEPTEMBER 2015
                                          CAPS

                                                                                            [13]

QUESTION 6

6.1.1   The 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)

6.1.2   increase ✓                                                                          (1)

6.1.3   As the police officer move closer to the alarm, he would observe a sound with
        a shorter wavelength✓ than was originally omitted. Since the wavelength is
        inversely proportional to the frequency of the wave, the frequency will
        increase (become more / higher).✓                                                   (2)

6.1.4
             𝑣 + 𝑣𝐿
        𝑓𝐿 =   𝑣
                    𝑓𝑠 ✓ (OR Formula as on data sheet)
            340 + 40
          =          ✓(1200)✓
              340
          = 1 341, 18 Hz✓                                                                   (4)

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PHYSICAL SCIENCES/ P1                               7                   SEPTEMBER 2015
                                                   CAPS

6.2.1   An atomic absorption spectrum is formed when certain frequencies of
        electromagnetic radiation that passes through a medium✓, e.g. a cold gas, is       (4)
        absorbed.✓
        An atomic emission spectrum is formed when certain frequencies of
        electromagnetic radiation are emitted✓ due to an atom's electrons making a
        transition from a high-energy state to a lower energy state.✓

6.2.2   The absorbed electromagnetic radiation for the light from Andromeda appear
        at higher frequencies than the absorbed electromagnetic radiation for light
        from the Sun.✓                                                                     (1)

6.2.3   Blue shift ✓                                                                       (1)

                                                                                           [15]

QUESTION 7

7.1     The magnitude of the electrostatic force exerted by one point charge (Q 1) on
        another point charge (Q2) is directly proportional to the product of the
        magnitudes of the charges and inversely proportional to the square of the
        distance (r) between them:✓✓                                                       (2)

7.2          𝑘𝑄1 𝑄2
        F=          ✓
              𝑟2
                        9 ×109 × 𝑄 × 16 𝑥 10−9
        7,2 x 10-6 =            (0,4)2
                                               ✓

        QA = -8 nC.✓                                                                       (3)

7.3     Electric field at P due to A
              𝑘𝑄
        𝐸 = 2✓
              𝑟
            9 × 109 × 8 × 10−9
          =                    ✓
                  (0,3)2
          = 800 N∙C-1 l✓
        Electric field at P due to B
             9 × 109 × 16 × 10−9
        E=          (0,7)2
                                 ✓
          = 293,88 N∙C-1✓
        800 + 293,88 = 1 093,88 N∙C-1✓                                                     (6)

7.4     B TO A✓                                                                            (1)

7.5     8𝑛𝐶 + 16𝑛𝐶
            2
                   = 12 nC✓
        4 nC electrons were transferred from B to A✓
                                                                                           (4)
          4×10−9
        1,6 ×10−19
                   ✓= 2,5 x 1010 ✓electrons

                                                                                           [16]

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PHYSICAL SCIENCES/ P1                    8                            SEPTEMBER 2015
                                        CAPS

QUESTION 8

8.1     The potential difference across a conductor is directly proportional to the
        current in the conductor✓ at constant temperature.✓                              (2)

8.2     Negative✓                                                                        (1)

8.3     1     1   1                                                                      (3)
        𝑅
          = 𝑅 + 𝑅✓
              1    2
        1   1   1
          = + ✓
        𝑅   4   2
        R = 1,33𝛺✓

8.4     V = IR✓
        3,9 = I x 1,33✓
        I = 2,925 A✓
        𝜀 = 𝐼𝑅 + 𝐼𝑟✓
        6 = 3,9 + 2,925 r✓                                                               (6)
        r = 0,72𝛺✓

8.5     2,925
          3
              ✓= 0,975 A.✓
        or
        V = IR
        3,9 = I (4)✓
                                                                                         (2)
        I = 0,975 A✓

8.6.1   Increase✓

8.6.2   Stays the same ✓                                                                 (1)

8.7     𝜀 = 𝐼( 𝑅 + 𝑟)✓
        6 = I ( 4 + 0,72)✓
        I = 1,27 A✓                                                                      (3)

                                                                                         [19]

QUESTION 9

9.1.1   Generator ✓                                                                      (1)

9.1.2   Kinetic/mechanical energy✓ → electrical energy ✓                                 (2)

9.1.3   B to A ✓                                                                         (1)

9.1.4   DC ✓                                                                             (1)

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PHYSICAL SCIENCES/ P1                         9                                SEPTEMBER 2015
                                             CAPS

9.1.5    The split ring commutator ensures✓ that the current that passes through to
         the external circuit is always in the same direction.✓                                   (2)

9.1.6    Use a coil that consist of more windings ✓
         Increase the strengths of the magnets.✓                                                  (2)

9.2.1    The rms value of AC is the DC potential difference which dissipates the same
         amount of energy as AC✓✓                                                                 (2)
                 𝑉
         Vrms = 𝑚𝑎𝑥
                 2
                    ✓
9.2.2              √
                39,45
            =        ✓
                 √2
         = 27,9 V ✓
                                                                                                  (3)

9.3      It can be stepped up or stepped down / is easier to transmit ✓                           (1)

                                                                                                  [15]

QUESTION 10

10.1.1   Photoelectric effect ✓.                                                                  (1)

10.1.2   Wo = hfo✓
           = 6,63 x 10-34 x 4,389 x 1014 ✓
          = 2,91 x 10-19 J✓                                                                       (3)

10.1.3   E = hf
           = 6,63 x 10-34 x 4,83 x 1014
          = 3,2 x 10-19J

         E = hfo + ½ mv2 ✓
         3,2 x 10-19✓ = 6,63 x 10-34 x 4,39 x 1014 + (0,5)(9,11 x 10-31) v2✓                      (4)
         v = 2,5 x 105 m∙s-1✓

10.2.1   Increase ✓                                                                               (1)

10.2.2   increase✓                                                                                (1)

                                                                                                  [10]

                                                                                    TOTAL         [150]

                                                                               Please turn over

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