# Physical Sciences Grade 12 P1 Free State November 2021 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Free State November 2021, 35 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/fs-ps-trial-2021-p1-memo/a2aebf71-c1f7-4580-b78e-398d5d0a015f
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2021
- Exam period: Free State November
- Paper: 1
- Curriculum: CAPS
- Pages: 35

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## About this document

Physical Sciences · Grade 12 · Free State November · 2021. Memorandum, 35 pages. Read online or download the PDF.

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                                                                     FS/September 2021
Grade 12 Prep.Exam.

INSTRUCTIONS AND INFORMATION

1.     Write your name and other information in the appropriate spaces on the
       ANSWER BOOK.

2.     This question paper consists of 10 questions. Answer ALL questions in the
       ANSWER BOOK.

3.     Start EACH question on a NEW page in the ANSWER BOOK.

4.     Number the answers correctly according to the numbering system used in this
       question paper.

5.     Leave one line between two sub-questions, for example between r&sTION  .._
       2.1 and QUESTION 2.2.

6.     You may use a non-programmable pocket calculator.

7.     You may use appropriate mathematical instrumen
                                                       �
8.     You are advised to use the attached DATA SH�    i-

9.     Show ALL formulae and substitutions in �\culations.

10.    Round off your FINAL numerical ans1 ,..
                                          k�to a minimum of TWO decimal
       places where applicable.

11.    Give brief motivations, discu��s, et cetera where required.
                                 Q
12.    Write neatly and legibly \\

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QUESTION 1

Various options are provided as possible answers to the following questions. Choose
the answer and write down only the letter ( A - D) next to the question number (1.1-
1.10) in your ANSWER BOOK, e.g. 1.11 E.

1.1    A car is moving at a CONSTANT velocity on an incline.

       Which ONE of the following statements about the forces acting on the car is
       CORRECT?

       A       There are no forces acting on the car.
       B       The net force acting on the car is zero.
       C
       D                                                      �f
               There is a non-zero net force acting on the car.
               The weight of the car is equal to the normal                           (2)

1. 2   A ball of mass m strikes a wall perpendicularly q.)�peed 2v. Immediately
       after the collision the ball moves in the opposi ��frection at a speed v, as
       shown in the diagram below.

       The magnitude of the impulse of the t,a.� ...

                                  m
                                                 -,
       A
                                      111    \

                                                 ,_.,
                                                 m      I

       B       :v
       C       2mt\O
       D       �◊                                                                     (2)

1.3    A Ot�ne is dropped from a certain height above the ground.
       Which ONE of the following combinations is TRUE while the stone is moving
       downwards? Ignore the effects of air resistance.

                 NET FORCE ACTING ON              MECHANICAL ENERGY OF
                          THE STONE                        THE STONE
           A     Remain the same                  Increases
           B     Decreases                        Remains the same
           C     Remain the same                  Remain the same
           D                                                                          (2)
                 Increases                        Decreases

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1.4    Which ONE of the following forces is an example for conservative force?

       A       Gravitational force.
       B       Tension in a chord.
       C       Frictional force.
       D       Normal force.                                                             (2)

1.5    Which ONE of the following combinations is TRUE for an elastic
       collision?
                TOTAL LINEAR MOMENTUM           TOTAL KINETIC ENERGY
           A    Conserved                       Conserved
           B    Not conserved                   Not conserved
           C    Conserved                       Not conserved
           D    Not conserved                   Conserved
                                                                                     (2)

1.6    SPECTRUM A represents the visible light spectrum recorded on earth.
       SPECTRA B and C are recorded from two stars Y and Z respectively

                        '-------'---II -----'-'I_------'----_
                    Blue                                         Red

       SPECTRUM A

                                Blue                                    Red
                            I
       SPEC TRUM B
                        �         ____...
                                        ____._l.,_I ______

                        �II �I��
                                __                            _
                                                          ..__ _
                                Blue                                    Red

       SPECTRUMC

       Which ONE of the following combinations most correctly describes the
       movement of the stars relative to earth?

       A       Star Y is moving away from the earth and star Z is moving towards
               the earth.

       B       Star Y is moving towards the earth and star Z is moving away from
               the earth.

       C       Both Star Y and star Z are moving towards the earth.

       D       Both Star Y and star Z are moving away from the earth.                (2)

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1.7    A stone starting from rest and rolls down a frictionless track, reaches the
       bottom of the track with a speed v. What change must be effected on the
       track for the stone to reach a speed of 2v at the bottom of the track?
                                                 Stone

       A       Double the height of the track.
       B       Increase the height by a factor   ✓2 .
       C       Increase the height by four times.
       D       keep the height the same, but make the track twice steeper.         (2)

1.8    Two identical charged spheres kept at a distance d apart. The force exerted on
       one charge by th other is F. What is the new force, when the charge on one
       sphere i OOUBLED?

                                                                                       (2)

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1.9    5 0 and 10 0 resistors are connected in parallel as shown in the diagram.

                     Ii
                                5n

                     12         10.0

                          A   : I Ii
The relationship between current Ii and current 12 can be expressed as:

       A

       B

                      1
       C       12 = - Ii

                      3
       D       12 = -11                                                                (2)
                      2

1.10 When an electron moves from a higher energy level to a lower energy level a
     specific ...

       A       emission line in an emission spectrum is observed.
       B       absorption line in an emission spectrum is observed.
       C       emission line in an absorption spectrum is observed.
       D       absorption line in an absorption spectrum is observed.                  (2)
                                                                                       [20]

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QUESTION 2

A block of mass 100 kg is sliding down a rough inclined surface at a CONSTANT
speed when a force F is applied parallel to the inclined plane as indicated in the
diagram below.

The coefficient of kinetic friction for the surfaces is 0,29.

2.1    Define the term normal force in words.                                           (2)

2.2    Draw a labelled free-body diagram for the block.                                 (4)

2.3    Calculate the magnitude of the applied force, F.                                 (5)

2.4    Name the force that will form an action-reaction pair with the normal force
       in this case.                                                               (1 )
                                                                                   [12]

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QUESTION 3

The position - time graph below, not according to scale, represents the motion of a
ball thrown vertically downwards with a speed of 3,28 m-s- 1 from the edge of the roof
of a 35 m high building. The ball hits the ground and is in contact with the ground for
0,1 s. The ball then bounces off to a maximum height x above the ground.
Ignore the effects of air resistance.

                          35

                               '-       -'-                   time (s)
                           Q    ---t 1--12---'---
                                            3,86

3.1    Explain the term projectile.                                                     (2)

3.2    Calculate the:

       3.2.1 Speed of the ball at time ti                                               (3)

       3.2.2 Time, ti, it takes the ball to hit the ground                                  (3)

       3.2.3 Height, x, reached by the ball after it bounces off the ground                 (7)

3.3    Draw the velocity-time graph for the ball from the moment it is thrown until it
       reaches the maximum height x after the first bounce.

       Indicate:

       (i)     The values of time h and 12
       (ii)    The velocities of the ball at time ti and 12                             (3)
                                                                                        [18]

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QUESTION 4

Car A of mass 1 085 kg travelling east at a speed of 33 m-s-1, collides head on with
car B of mass 3 450 kg travelling west at a speed of 28 m-s·1. Immediately after the
collision, car A moves west at a speed of 5 m-s·1. Ignore friction.

                33 m-s·1                          28 m-s·1
                     )

Car A                                                          CarB

                mass - 1085 kg              mass - 3450 kg
                                                                                s

4.1     State Newton's second law of motion in terms of momentum in words.                (2)

4.2     Calculate the:

        4.2.1 Velocity of car B after collision                                           (5)

        4.2.2 Magnitude of the impulse on car B during the collision                      (3)

4.3     Airbags serves as a safety feature during collision.

        4.3.1 Use the relevant laws of physics to explain the purpose of air bags in
              cars.                                                                ( 3)

        4.3.2 Which driver sustains more injuries during above collision?
              Choose from CAR A or CAR B                                                  (1)
                                                                                      [14]

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QUESTION 5

A 70 kg crate, starts from rest at point A (top of slope) and slides down a slope
inclined at 25° to the horizontal. The distance from point A to point B is 135 m.

                     135 m

             B

The total kinetic friction between the crate and the surface is 150 N.

5.1    Define the term non-conservative force in words.                                     (2)

5.2    Calculate the net work done on the crate, as it slides from point A to
       point B.                                                                         (4)

5.3    Calculate the time it takes the crate to slide from point A to point B.              (6)
                                                                                        (12]

QUESTION 6

An ambulance moving at a constant speed along a straight horizontal road, emits
sound with a constant frequency. When the ambulance approaches a detector
mounted next to the road a frequency of f is recorded. After passing the detector, a
frequency of 0,55 f is recorded.

6.1    State the Doppler effect in words.                                               (2)

6. 2   Calculate the speed of the ambulance. Take the speed of sound in air as
       340 m-s· 1.                                                             (6)

6.3    State TWO applications of the Doppler effect.                                        (2)
                                                                                        (10]

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QUESTION 7

x is a point 0,2 m from point charge, Qi, carrying charge -3x10-5 C.

                              0,2 m       >I
                     •---------------------•
                     Qi                     X

7 .1   Define electric field at a point in words.                                         (2)

7.2    Draw the electric field pattern due to the charge Qi.                          (2)

Now another point charge Q2, carrying a charge of +12x10-5 C is placed at a distance
of 0,35 m from Qi as shown below.                                      N
                      ----- 0,35 m-----

                     ,-----------------------------------.
                      I<     0,2 m        >I            Q2
                     Qi                    X

7.3    Calculate the net electric field at point x due to the two charges.            (6)

7.4    State Coloumb 's law in words.                                                     (2)

7.5    Calculate the magnitude of the electrostatic force exerted by charge Q2 on
       charge Qi.                                                                 (3)

7.6    Charge Q2 is moved further west. How will this change affect the net
       electrostatic force at this new distance?
       Choose from INCREASE, DECREASE or REMAIN THE SAME.                             (1 )
                                                                                      [16]

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QUESTION 8

A battery has an emf(£) of 12 V and an internal resistance of 0,2 0. The resistances
of the connecting wires are negligible, as shown in the circuit below.

                                           10 0

                     R                 15 0

       s

                                  E= 12V

                                      0,2 0

8.1    What is the meaning of the underlined words?                                   (2)

When switch S is CLOSED, the voltmeter reading decreases from 12 V and the
ammeter reading is 0,9 A.

8.2    Explain why the voltmeter reading decreases.                                   (2)

Calculate the:

8.3    Total current in the circuit                                               (4)

8.4    Resistance of resistor R                                                   (5)

8.5    The total power dissipated by the battery                                  (3)

8.6    Will the ammeter reading INCREASE, DECREASE or REMAIN THE SAME if
       the 15 n resistor is removed? Give a reason for the answer.   (2 )
                                                                     [18]

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QUESTION 9

9.1    The diagram below represents a simplified electric machine. The coil abed
       rotates at 20 cycles per second.
                                                                    "T             •
       The direction of the current in the coil is from a to b.

                                                                             com

                        N

       9.1.1 What type of electric machine is represented by the above diagram? (1)

       9.1.2 Name the principle on which the machine operates.                          (1)

       9.1.3 In which direction is the coil rotating?
             Choose from CLOCKWISE or ANTICLOCKWISE.                                   (1)

       9.1.4 How long does it take for the coil abed to complete one rotation?          (3)

       9.1.5 State ONE way in which this machine can produce a lower output
             voltage without changing its construction.                                 (1)

9.2    An electrical device is rated 220 V, 1850 W.

       9.2.1 Define the term rms voltage in words.                                     (2)

       Calculate the:

       9.2.2    Rms current passing through the appliance when it is operating.         (3)

       9.2.3 Resistance of the device.                                                 (3)
                                                                                       [15]

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QUESTION 10

An experiment is conducted to investigate the relationship between the frequency of
light incident on a metal cathode and the maximum kinetic energy of the emitted
electrons from the surface of a metal cathode of a photocell.
The graph represents the results obtained.

            Graph of maximum kinetic energy versus frequency of light

               >
               0
               Cl)
               C
               Cl)
               u
              ·.;::::;
                Cl)
                C

                                 0, 5    10         >5'   2,0   f (x1015 Hz)
                                          ,       _-�·

                         -X ,

10.1   For this investigation, write down the following:

       10.1.1               Controlled variable                                               (1)

       10.1.2               Dependent variable                                                (1)
10.2   Name the phenomenon on which this experiment is based.                                 (1)

10.3   Define the term work function in words.                                                    (2)

10.4   Calculate the:

       10.4.1               Value of x as shown on the graph                                      (3)

       10.4.2               Frequency of the incident light that will emit electrons from
                            the surface with a maximum speed of 5,23x105 m•s- 1                   (4)

10.5   The intensity of the light is now increased. State how this change affects the
       current in the photocell .
       Choose from INCREASES, DECREASES or REMAINS THE SAME.

       Explain your answer.                                                                   (3)
                                                                                              [15]

                                                                         GRAND TOTAL: 150

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                                                          FS/SEPTEMBER/ 2021

                        DATA FOR PHYSICAL SCIENCES GRADE 12
                                  PAPER 1 (PHYSICS)

                     GEGEWENS VIR FIS/ESE WETENSKAPPE GRAAD 12
                                 VRAESTEL 1 (FIS/KA)

TABLE 1: PHYSICAL CONSTANTS/ TABEL 1: FIS/ESE KONSTANTES

NAME/ NAAM                         SYMBOL/ S/MBOOL   VALUE/ WAARDE
Acceleration due to gravity
                                          g                 9,8 m-s-1
Swaartekragversnelling
Universal gravitational constant
                                          G           6,67 x 10-11 N-m 2-kg·2
Universe/e gravitasiekonstant
Speed of light in a vacuum
                                          C              3,0 x 108 m-s· 1
Spoed van Jig in 'n vakuum
Planck's constant
                                          h              6,63 X 10-34 J·S
Planck se konstante
Charge on electron
                                          e               -1,6 X 10-19 C
Lading op elektron
Electron mass
                                         me              9,11 X 10-31 kg
Elektronmassa
Mass of the Earth
                                          M               5,98 X 1024 kg
Massa van die Aarde
Radius of the Earth
                                         RE               6,38 X 106 m
Radius van die Aarde

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TABLE 2: FORMULAE/TABEL 2: FORMULES

MOTION/BEWEG/NG

                                                   2
V1       =V;       +2MX Oro
                          If V 1             =V;       +2My
     2         2                         2

FORCE/KRAG

Fn et = ma                                                          P= ITW
f max
 s
         =µ s N                                                     fk       µk N
                                                                         =
Fnet l'.lt =
             L'.lp
                                                                    W=l'TY:J
Llp = mv 1                mv i
                                                                               Mf
 F=G-2 -                                     F=G-2 -                                                       g =G-
          m1m                                          m1m
                                                                    g =G
       2                                           2
                             or/of                                                       or/of
                                                                               c
                                                                                                               2
               d                                        r                                                       r

WORK, ENERGY AND POWER/ARBE/D, ENERGIE EN DRYWING

W=F�xcose                                                           U=mgh                    or/of         Ep = mgh
                                                                    Wnet = �K                or/of         wnet = b.Ek
          mv 2                                     1
K=_!
                                                            2
                            or/of        Ek =-rnv
  2                                          2
                                                                    �K = K, -K;              or/of         �Ek = Ekf -Eki
                                                                         W
Wnc = �K + �U or/of wnc =�Ek + �Ep                                  P=
                                                                         b.t
Pav = FVav               / Pgemid = FVgemid

WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG

V=fA                                                                T=�
                                                                      f

         V±VL                              v v
f=                   f               f =    ± L f
                                                 b                  E= hf           or /of    E=h�
 L       v+
          - vs       s               L     v v
                                            ± b                                                       "-
E = Wo + E k(max) or/of E =WO + K max where/waar
                                                                1                                -1
E =hf          and/en Wo = hfo and/en Ek (max) = rnv �ax or/of K max - rnv 2max
                                                2                     2

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ELECTROSTATICS/ELEKTROSTA TIKA

  k                                                            kQ
F= Q l2Q 2                                                E=        2
    r                                                           r

  W                                                         F
V=                                                        E=
   q                                                         q

 n=
      Q
                    or/of       n=   _2_
                                     qe

ELECTRIC CIRCUITS/ELEKTR/ESE STROOMBANE

                                                          emf ( £)= l(R + r)
   V
R =-
                                                          emk ( £)= l( R + r)

R s =R1 +R 2 + ...
1 1 1                                                     q=I fit
-=-+-+ ...
RP     Rl          R2

W=Vq                                                         W
                                                          P=
                                                              Lit
W =VI fit
                                                          P =VI
W= 1 2Rfit
                                                                2
                                                          P =l R
            2
w----
   V Lit
    R
                                                            v
                                                          P=-
                                                                    2

                                                            R

ALTERNATING CURRENT/W/SSELSTROOM

                                                          Pa1.e =Vrms l rms       I   pgemiddeld = Vwgkl wgk
                                               _
l r ms =
           ! max
           ✓2               I          I wgk      ✓
                                               - lma2ks

                                                          pa1.e = l �sR       I       pgemiddeld = l�kR

            vmax                                 vmaks
v,ms =        2 ✓           I        Vwgk -
                                          -       ✓2                vr�s                              V1gk
                                                          pa1.e =
                                                                     R
                                                                              I       pgem iddeld =   R

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                       �1-
               �� �..�   .          \•",

           l.                           �
           '                                '   Department of
                                                Education
                                                FREE STATE PROVI CE

                PREPARATORY EXAMINATION
                VOORBEREIDENDE EKSAMEN

                            GRADE/GRAAD 12

  PHYSICAL SCIENCES: PHYSICS (Pl)
  FIS/ESE WETENSKAPPE: FIS/KA (V1)

                             SEPTEMBER 2021

                           MARKS/PUNTE: 150

                        MARKING GUIDELINES
                              NASIENRIGL YNE

                    This marking guidelines consists of 17 pages.
                    Hierdie nasienriglyne bestaan uit 17 bladsye.

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                                Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 1

1.1    B ✓✓                                                                                  (2)
1.2    o ✓✓                                                                                  (2)
1.3    c ✓✓                                                                                  (2)
1.4    A✓✓                                                                                   (2)
1.5    A✓✓                                                                                   (2)
1.6    A✓✓                                                                                   (2)
1.7    C✓✓                                                                                   (2)
1.8    A✓✓                                                                                   (2)
1.9    B ✓✓                                                                                  (2)
1.10   A✓✓                                                                                   (2)
                                                                                             [20]

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                                  Marking Guidelines/Nasienriglyne

QUESTIONNRAAG 2

      Marking criteria/Nasienkriteria:
      If any of the underlined key words/phrases in the correct context are omitted:
      - 1 mark per word/phrase
      As een van die onderstreepte s/eutelwoorde/frases in die regte konteks
      weggelaat word:
      - 1 punt per woordlfrase

2.1      The force or the component of a force which a surface exerts on an object
         with which it is in contact, and which is perpendicular to the surface. ✓ ✓
         Die kraq of die komponent van 'n kraq wat 'n oppervlak op 'n voorwerp
         uitoefen waarmee dit in kontak is, en wat /oodreq op die oppervlak is.      (2)

2.2
               N                                          ACCEPTIAANVAAR
                                      F

                                                             Fg11

             Accept the following symbols/Aanvaar die volgende simbole
               F    Fapplied / Force applied / FA                                 ✓
                    Froeaeoas I Krag toegepas I FA
              Fg    w / Fw / weight/ gewig                                        ✓
               N    FN / Normal / Normal force                                    ✓
                    FN I Normaal I Normaalkrag
               f    FtI fk / friction / frictional force/kinetic frictional force ✓
                    Ft/ fk / wrywing / wrywingskrag I kinetiese
                    wrywingskrag

         NOTE/LET WEL:
         •     Mark awarded for label and arrow/Punt toegeken vir byskrif en pyl
         •     Do not penalise for length of arrow since drawing is not to scale. Moenie
               penaliseer vir die lengte van die pyl nie, want die tekening
               is nie volgens skaal nie.                                3
         •     Any additional force(s)!Enige bykomende krag(te)         4
         •     If force(s) do not make contact with body/As krag(te) nie kontak
               met liggaam maak nie        �

         •     No labels/Geen byskrifte
                                              4

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                                Marking Guidelines/Nasienriglyne

2.3
              Marking criteria / Nasienkriteria:
              • Showing Fnet/netto = O ✓
              •
              •    Substitution of first term0 ✓
              •    Substitution of second term
              •    Final answer of/ Finale antwoord van 157,48 N✓
                   Range (152,98N to/tot 157,45N)
              OPTION / OPSIE 1:
              Downwards as positive/Afwaarts positief:
              Fnet = ma or Fnet = O
              -F - fk + F911 = 0        } Any ✓
              -F - µkmgcos0 + mgsin0 = 0

              -F - O,289x(1OOx9,8cos25 °) ✓ + (1OOx9,8)sin25° ✓ = O ✓
              F = 157,48 N ✓
              Downwards as negative / Afwaarts positief

              Fnet = ma or Fnet = 0
              F + fk - Fgtt = 0
              F + µkmgcos0 - mgsin0 = 0
                                              }
              F + O,289x(1OOx9,8cos25 ) ✓ - (1OOx9,8)sin25° ✓ = 0 ✓
                                          °

              F = 157,48 N✓                                                                  (5)

2.4    Perpendicular force tha the water tank exerts on the surface. ✓
       Loodreg krag wat die watertenk op die oppervlak uitgeoefen.

       ACCEPT/AANVAAR
       mgcose                                                                                (1)
                                                                                             [12]

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                                            Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 3

3.1      An object upon which the only force acting is the force of gravity (2 or 0) ✓ ✓
        'n Voorwerp waarop die eniqste kraq wat daarop inwerk, swaartekraq is.        (2)

3.2.1
         Marking criteria / Nasinkriteria:
        Appropriate formula/ Toepas/ike formu/e ✓
        Substitution of/ Vervanging van 3,28 & 35 ✓
        Final answer of/ Finale antwoord van 26,396 m-·1 ✓
        Range (26,397 to 26,41)
           Downwards as positive/                                    Upwards as positive/
           Afwaarts positief                                         Opwaarts posiitief

                 OPTION 1 / OPSIE 1

            v 21 = v 2i + 2a�y ✓                         v: = v� + 2a�y✓
                 = ( ,28 )2 + 2 (9,8 )( 5) ✓
                    3                          3             = ( ,28 )2 + 2(9,8 )( 5) ✓
                                                                 3                     3

                 = 26, 96 m • s- 1 ✓                    v t = 26, 96 m · s- ✓
                                                                           1
                            3                                        3
           Vr

                OPTION 2/ OPSIE 2

           v 1 = v i + a�t ✓                            v, = v i + a�t ✓
           = ,28 + (9,8 X2, 59) ✓
             3                          3               = - ,2s+ (-9,sX2, 59)✓
                                                             3                     3

           v 1 = 26, 98 m · s-1✓
                        3                               v 1 = 26, 98 m · s-1 ✓
                                                                     3

           OPTION 3/ OPSIE 3

                    Vr+V                                      Vt +V-
           �y = (                   )�t✓
                                    I                   �y = (     I ) �t ✓
                            2                                            2

                    Vf + 3,28                                = (Vf \ •28
           35 = (
                                                                             3
                                        )( 2,359)✓      35                       )(2,359) ✓
                                2

           v 1 = 26, 98 m · s-1 ✓
                        3                               v 1 = 26, 98 m · s-1 ✓
                                                                     3
                                                                                                               ( )
                                                                                                               3

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                                     Marking Guidelines/Nasienriglyne

3 .2.2    POSITIVE MARKING FROM QUESTION 3.2.1 / POSIT/EWE NASIEN
          ANAF 3.2.1

                Downwards as positive/                    Upwards as positive/
                Afwaarts positief                          Opwaarts posiitief

              OPTION 1/0PS/E 1

              V 1 = V; + aL'it ✓                     V 1 = V; + aL'it ✓
              26 ,396 = 3,28 + ( 9, 8 )L'i t ✓       -26,396 = 3,28 + (-9, 8 )L'it ✓
              L'i t = 2359 s ✓
                     '                               L'it = 2,359 s ✓
              OPTION 2/0PS/E 2

              L'iy = V;L'it + ½a!it2 ✓               L'iy = V;L'it + ½a!it2 ✓
                                                                         1
                                         2
              3 5=3,28L'it+ !( 9, 8 )L'it ✓          - 35=- 328L'i'    t+-(-9 8 )L'it2 ✓
                           2                                             2   '
              L'i t = 2'359 s ✓
                                                                     ✓
                                                     L'i t = 2,359 s

              OPTION 3/OPS/E 3                       OPTION 3/OPS/E 3

                         v1    v;                          V V;
              L'iy=(          ; ) L'it ✓             liy =( 1 ; )lit ✓

                   = 26,39 + 3,28                         = -26,396 (- 3,28 )
              35    (     �       )L'it ✓            35    (       ;         )L'it ✓

              L'it = 2,359 s ✓                       L'it = 2,359 s ✓
                                                                                                   (3)

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3.2.3     Calculate velocity at tz/ Bereken sne/heid by ti

          12 = 2,359 + 0.1✓
             = 2,459 s

          Time to max(x) = 3,86 - 2.459 ✓
                         = 1,40 s
                     V r =V;+ai"1t ✓
                     0 = V; +9,8(1,40) ✓
                              V =-1373m•S-1
                               I              ,

                    Downward as positive/                            Upward as positive/
                    Afwaarts positief                                Opwaarts positief

                    OPTION 1/ OPSIE 1
              v2 -   2
                 - v 1 +2a6..y
                               ✓                                   v1 = v� +2a6..y ✓
                f
                                                                                         2
                    0=(-13, 73) +2(9,8)Lly ✓
                                             2
                                                                       0 = (13,73) + 2(- 9,8)Lly ✓
              6..y = -9,62 m                                       6,.y = 9,62 m
                   :. Height/ Hoogte = 9,62 m ✓                        :. Height/ Hoogte = 9,62 m ✓

                    OPTION 2/ OPSIE 2
              6..y = V 6..t + 2.la6..t2 ✓
                         I
                                                                   6..y =V;6..t + -½a6..t2✓
                                                 2                                                         2
                     =(-1 3 ,73 X1,4)+ 1 (9,sX1A) ✓        = (1 3 ,73 X1,4 )+ 1 (- 9,8 X1 ,4 )✓
                                       2                                      2
              6,.y = -9,62 m                        6,.y = 9,62 m
                   :. Height/ Hoogte = 9,62 m ✓          :. Height/ Hoogte = 9,62 m ✓

                    OPTION 3/ OPSIE 3
                         V;       V                                           V;    V
              6,.y = (        ;       f   )6,.t✓                   6,.y = (        ; f )6,.t✓

                                                                                    +(- 3 , 73 )
                    = ( (o
                              + - 3 , 73
                               ( �       ))}1,4 ✓                       = ((o          �      ) )1.4) ✓
                                               )
              6..y = -9,62 m                                       6..y = -9,62 m
                   :. Height/ Hoogte = 9,62 m ✓                        :. Height/ Hoogte = 9,62 m ✓
                                                                                                                     (7)

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                                                 Marking Guidelines/Nasienriglyne

3.3    POSITIVE MARKING OF QUESTION 3.2.1 AND 3.3.2 I POSIT/EWE NASIEN
       VAN 3.2.1 EN 3.2.2

       OPTION 1: Downward Positive
       OPSIE 1: Afwaarts Positief
                                                                  Marking criteria/ Nasienkriteria:
                                                                  • Velocity striking the ground
                                                                    and leaving the ground at hand t2. ✓
                      26 ,39 ----------------------                 Snelheid waarteen grand getref en
                                                                    grand verlaat word by t1 en t2
                                                                  • Contact time hand t2 I Kontaktyd t1 en
                                                                      t2 ✓
         :"S;!
          (l)
                          3,28                                    • Lines parallel / Paralelle lyne ✓
         :S
         (l)
                                                                 2,4 6             3, 86
         >,
         :!::                                         2,36
          u                                                                            Time(s)/Tyd
          0

                 -13, 76 ---------------------------------

OPTION 2: Upwards positive
OPS/E 2: Opwaarts positief

                 .............
                                  13, 76 -- --------------------------------- ---------------,
                                                                              I              I

                 :"S;!
                 :S
                 (l)

                                                               2 36
                 >,                                                      2,46              3,86 Time(s)/Tyd
                 ·g               -3,28
                  >
                    Q)

                                 -26 , 39 ------------------------                                                       (3)
                                                                                                                         [18]

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QUESTION/VRAAG 4

4.1     The resultant/net force acting on an object is equal to the rate of change
        of momentum of the object in the direction of the net/resultant force. ✓ ✓
        (2 or 0)
        Die resulterendelnetto kraq wat op 'n voorwerp inwerk is qelyk aan die
        tempo van veranderinq van momentum van die voorwerp in die rigting
        van die nettolresulterende krag. (2 of 0)                                                     (2)

4.2.1
          OPTION 1 / OPSIE 1

          Motion to the east(+)/ Beweging na o s(+)
                                              r Any!Enige ✓
                            "     _ L..P
                            L-P t - " j
                    m 1 v 1 i + m 2 v 2i,/'° m 1v 11 + m 2v 21 ✓
          1085 (+33) + 3450 (-28) = 1 085 (-5) + 3450v 21
                                 v 21 = - 15, 05       ✓ west / forward ✓
                                                             1
                                        = 16,05 m · s-   wes I voorwaarts

          OPTION 2/ OPSIE 2
          Motion to the east(-)/ Beweging na oos(-)

                                LP, - LP,
                     m,v,i + m 2 v 2i = m 1 v 11 + m 2 V 21 J
                                                                 l
                                                              Any/Enige ✓

          1 085 (-33) + 34 50 ( +28)✓-= 1 085 ( +5 ) + 3450v 21✓
                                 V 21 =+16,05

                                        = 16,05 m · s-           ✓ west / forward ✓
                                                         1

                                                                   wes I voorwaarts
                                                                                                      (5)

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4.2.2 POSITIVE MARKING FROM QUESTION 4.2.1/ POSIT/EWE NASIEN VANAF
      4,2,1

            OPTION 1: CAR B                       OPTION 2: CAR B
            Motion to the west(+)                 Motion to the west(-)
            OPSIE 1: KAR B                        OPSIE 2: KAR B
            Beweging na wes (+)                   Beweging na wes (-)

            Impulse= �P = m�v✓                     Impulse= �P = m�v✓
                   = 3450(16,05 - 28)✓                    = 3450(-16,05 - ( - 28))✓
                   = - 41227,5                            = 41227,5 N·s ✓
                   = 41227,5 N-s✓
            OPTION 3: CAR A                       OPTION 4: CAR A
            Motion to the west(+)                 Motion to the west(-)
            OPSIE 3: KAR A                        OPSIE 4: KAR A
            Beweging na wes (+)                   Beweging na wes (-)

            Impulse= �P = m�v ✓                    Impulse= �P = m�v✓
                   = 1085(5 -( - 3 3))✓                   = 3450(-5 - ( + 3 3)) ✓
                   = 412 30 N-s✓                          = - 412 30
                                                          = 412 30 N-s ✓
                                                                                              (3)

4.3.1 According to Newton's second law ✓ I Fnec     �p
                                                     M
       Air bag causes an increase in collision time ✓ which results in a lesser /
       smaller force ✓ ( for the driver/passenger)                                           (3)

4.3.2 Car A / Kar A ✓                                                                        (1)
                                                                                             [14]

QUESTION/VRAAG 5

5.1    A force for which the work done in moving an object between two points
       is dependent of the path taken.✓✓
       'n Krag waarvan die werk wat verriq word om 'n voorwerp tussen twee
       punte te skuif, afhanklik is van die pad wat qevolq word.                              (2)

5.2
        Wnet = W1 + WFg//
              = Mxcose + mgsin8�xcos8             ✓
                                              }
              = 150(135)C0Sl80 ✓+ 70(9,8)sin25 °(135)COS0° ✓
                                   °

              = -20 250 + 39 138, 677
              = 18 888,677 J ✓                                                               (4)

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5.3    POSITIVE MARKING FROM QUESTION 5.2/ POSIT/EWE NASIEN VANAF
       5.2

       OPTION 1/0PS/E 1

                Wnet   = L'iK
                         1    2   2   }Anv ✓
                                v
                       = m(v t - i  )
                         2

       18 888,677✓= !(70)(v; -0) ✓
                          2

                  v 1 = 23 ,?3 m · s-1
                               ''
                                  '
                  L'iy = (V f +''V\) L'it ✓
                              2
                                      23 , 2            0)L'i
                       135 = (              �+                t   ✓

                  L'it = 11,625 s ✓

       OPTION 2/ OPSIE 2
                                                                                  -
                         Wnc        = L'iK + L'iU
                           1                                                       - Any ✓
                   faycos0=-m(v; -v;}+mg(h 1 -h i )
                                        2

       150( 135)cos180 ° ✓ = !(70 Xv; -o) +(70X9,8Xo -sin 25° x 135) ✓
                             2
                             v 1 = 23,2� m · s-1
                                                ''
                                                    I

                             L'iy
                                    =   (v + � )L'it ✓
                                            1

                                                2       \
                                                     23, 2        + o)L'it
                                      135 = (               �                ✓

                              L'it = 11,625 s✓

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        OPTION 3/ OPSIE 3

          Wnet = Fnet 6.xcos0
                              ✓

          18 888,68 = Fnei (135)cosO ° ✓

           Fnet = 139,92 N

          Fnet = ma
          139,92 = 7Oa ✓

          a= 2 m-s-2

                          1
           6.y = V;6.t + -a6.t
                                 2
                                     ✓
                          2
          135 = 0 + !(2)6.t ✓
                              2

                      2
          6.t = 1 1,625 s ✓
                                                                                                  (6)
                                                                                                  [12]

QUESTION/VRAAG 6
6.1    It is 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. ✓
       Dit is die (skynbare) verandering in frekwensie (of toonhoogte) van die
       klank (waargeneem deur 'n /uisteraar) omdat die klankbron en die
       luisteraar verskillende snelhede het in verhouding tot die medium van
       klankverspreiding.

                                                    OR/OF

       An (apparent) change in (observed/detected) frequency (pitch),
       (wavelength) ✓ as a result of the relative motion between a source and
       an observer ✓ (listener).
       'n (Skynbare) verandering in (waargenome) frekwensie (toonhoogte),
       (golflengte) as gevolg van die relatiewe beweging tussen 'n bron en
       'n waarnemer (luisteraar).                                                                 (2)

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                                  Marking Guidelines/Nasienriglyne

6.2

                Towards: fl    = ( v-vv Jt         s

                        340
                tV: ( 340-  Jf ·•·equation 1
                             Vs
                                      ✓
                                      5

                Away : fl   = ( +v Jt
                                  V       V5
                                               s

                0,55 f   = ( 340340 Jt5✓ ... equation 2
                                  +Vs

                Equating-eqn 1 & eqn 2
                153f = 1,55v J
                V 5 =98,71m•S- 1 ✓                                                                 (6)

6.3    ANY TWO ✓✓

       Doppler /Blood flow meter
       Measuring the heart beat of a foetus
       Radar
       Sonar
       Use to determine whether stars are receding or approaching earth                            (2)
                                                                                                   (10]

QUESTION/VRAAG 7

7 .1   The electric field at a point is the electrostatic force ✓ experienced per
       unit positive charge (placed at that point). ✓
       Die elektrieseveld by 'n punt is die elektrostatiese krag ervaar per
       eenheid positiewe lading (op daardie punt geplaas).                                          (2)

       Note: If electric field defined (       ½) /    Let we/: As elektriese veld gedefinieer    (½)

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7.2
                                             Markin criteria/Nasienkriteria:
                                              Correct direction towards the sphere          ✓
                                              Korrekte ri tin na die steer
                                              Field lines not crossing and radial           ✓
                                              Veld/ ne wat nie die radial kruis nie                   (2)

7.3
               EXQl    = kQ
                         r2
                            ✓

                        9 X 109 (3 X 10-5)
                       =-----           -✓
                              (0,2)2
                       =6,75xl0 6 N-c-1 west

               EXQ2 = kQ
                           r2
                       _ 9 X 109 (12 X 10- 5) ✓
                       -
                                      2
                               (0,15)
                       = 4' 8 x10 7 N • c-1 west

                             ✓
                E net = EXQl + EXQ2
                       = 6,75xl0 6 + 4,8xl0 7
                       = 5,475 x10 7 N-c-1✓ west✓
                                                                                                     (6)

7.4    The electrostatic force(of attraction/repulsion) between two point charges is
       directly proportional to the product of the charges and inversely proportional
       to the squareof the distance between them. ✓✓                                (2)

7.5            = k0 , 0 2
           F                   ✓
                      r2
                           9            5            5
                 9 X 1 0 ( 3 X 1 0 - 0( 12 X 1 0 -
               =---------
                                                         )
                     2
                          ✓
                                   (0,35 )
               = 264,49 N ✓                                                                          (3)

7.6    Increase / Neem aan ✓                                                                         (1)
                                                                                                     [16]

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