# Physical Sciences Grade 12 FS March 2022 Memo

> Grade 12 Physical Sciences memorandum from the FS March 2022, 20 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/fs-ps-march-2022-gr12-qp-memo/f97c530d-e810-40ed-87b4-481a9109da2d
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: FS March
- Curriculum: CAPS
- Pages: 20

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Physical Sciences · Grade 12 · FS March · 2022. Memorandum, 20 pages. Read online or download the PDF.

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### Page 1

education
                                       Department of
                                       Education
                                       FREE STATE PROVINCE

                                       OL TEST

                              MARKS: 100

                            TIME: 2 HOURS

            This paper consists of 11 pages and one information sheet.

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### Page 2

t Grade 12
                                                                       FS/March 2022
INSTRUCTIONS AND INFORMATION

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

2.     This question paper consists of SIX 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 �STION
                                                                      L-
       2.1 and QUESTION 2.2.

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

7.     You may use appropriate mathematical instrument     �0
                                                   �
8.     You are advised to use the attached DATA SH��-

9.     Show ALL formulae and substitutions in �culations.

10.    Round off your FINAL numerical ans��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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### Page 3

FS/March 2022
Grade 12

QUESTION 1: MULTIPLE-CHOICE QUESTIONS

Four options are provided as possible answers to the following questions. Each
question has only ONE correct answer. Choose the answer and write down only the
letter A, B, C or D next to the question number (1.1-1.10) in your ANSWER BOOK.

1.1    When two objects collide during an ELASTIC COLLISION,

       A           both momentum and kinetic energy are conserved.

           B       both impulse and momentum are conserved.

       C           only kinetic energy is conserved.

       D           only momentum is conserved.                                        (2)

1.2    A vector quantity with the same DIRECTION as the veloo  ;n object
                                                             �
       is the ... of the object.
                                                              ...._\C,
       A           rate of the change in momentum

               B   momentum

       C           impulse

       D           inertia                                                            (2)

1.3    An airbag can protect the dr��f a vehicle from serious injuries during
       a collision. Which one of t (\JPtlowing best describes how that is possible?

                   Net force on
                   the driv

                                     Remain the same
                                                                                      (2)

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FS/March 2022

1.4    Two forces of 200 N and 100 N are simultaneously applied to a stationary
       box that has been placed on a flat surface. One of the forces is horizontal and
       the other one is applied at an angle of 30° to the horizontal as shown below.
                                                                F1 = 200 N

                                                      30°
                                  50 kg

       The normal force acting on the box is DECREASED by ...

       A       increasing the angle at which the 200 N force is acting.

       B       decreasing the angle at which the 200 N force is acting.

       C       decreasing the magnitude of F1.

       D       increasing the magnitude of F2.                                         (2)

1.5    A ball is thrown vertically upward and returns to the thrower's hand.
       Taking upward as POSITIVE, which one of the following combinations
       best describes the velocity and acceleration of the ball when it is moving
       DOWNWARDS towards the thrower's hand? Ignore air resistance.

              Velocity Acceleration
        A        +          +
        B        +          -
        C        -          +
        D        -          -
                                                                                      (2)

1.6    Consider the velocity versus time
       graph for an object moving
       VERTICALLY. Upward is taken as
       positive.

       Which one of the following
       statements is correct?

       A       The object's speed is decreasing throughout the motion.

       B       The object is travelling downwards throughout the motion.

       C       The object is travelling with a constant velocity throughout the motion.

       D       The object is travelling with a constant acceleration throughout
               the motion.                                                            (2)

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FS/March 2022

1. 7   An object is dropped from a hot air balloon moving upward at a constant
       velocity. Which one of the following position versus time graphs best
       represents the motion of the object UNTIL IT HITS THE GROUND?

       A   --                                      B
                                                           --
                                                           E
            E                                              C:
                                                           0
            C:                                             :2
            0                                              (J)
           :;:::;
           ·;:;;
                                                            0             time (s)
            0                                              0...
           0...

                                time (s)

           --
            E
            C:
                                                           -
                                                           E
       C    0                   time (s)           D       C:
                                                           0
           ·;:;;
           :;:::;
                                                           :E
                                                            (J)
            0                                               0
           0...                                            0...           time (s)
                                                                                         (2)

1.8    The condensed structural formula off an organic compound is shown below.
                                                   CH3

                                  CH3 --CH--CH
                                                       I
                                             I
                                             Br
                                                       I
                                                       CH3
       Which one of the following is the correct IUPAC name of this compound?

       A            2-methyl-3-bromobutane

       B            2-bromo-3-methylbutane

       C            2-bromo-1, 1-dimethylpropane

       D            2-bromo-3,3-dimethylpropane                                          (2)

1.9    Consider the functional group on the right. For which          0
       one of the following homologous series is this the
       functional group?                                              11  I
                                                                      C-0-C-
       A            Aldehydes
                                                                  /                  I

       B            Alcohols

       C            Ketones

       D            Esters                                                               (2)

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FS/March 2022
Grade 12

1.10   The MELTING point of a compound is the ...

       A       minimum temperature at which it boils.

       B       maximum temperature at which it boils.

       C       temperature at which its vapour pressure equals atmospheric
               pressure.

       D       temperature at which the solid and liquid phases of a substance
               are at equilibrium.                                                  (2)
                                                                                    [20]

QUESTION 2

Two crates, A and 8, with masses of 8 kg and 5 kg respectively, are stationary
on a rough, horizontal surface. The crates are connected by a light, inextensible
string. When force F, with a magnitude of 80 N and making an angle of 30 ° with
the horizontal, is applied to the 8 kg block, both blocks move to the right.
                                                                        F
                         8                         A

                          5 kg                   8 kg

2.1    State Newton's second law of motion in words.                                (2)

2.2    Draw a free-body diagram of ALL the forces acting on the 5 kg block          (4)

2.3    The magnitudes of the frictional foroes acting on crates A and 8 are
       7,68 N and 4,9 N respectively. Calculate the magnitude of the
       acceleration of block 8.                                                     (5)
                                                                                    [11]

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FS/March 2022
Grade 12

QUESTION 3

A truck of mass 2 000 kg is moving      -        Vi
                                                      truck

                                                                           ___
eastward and collides with a car of
mass 900 kg moving at a speed of
4,5 m-s- 1. After the collision, the
                                        -....E
                                         >,
                                                               I
                                                               I

truck and the car are entangled         ·u0 3,2 -----------�--------,,._...,
                                                               I

into a wreck which moves as ONE                0
                                                           2 / 3,5                time (s)
UNIT with a speed of 3,2 m·s- 1.                            :/
The graph (not drawn to scale)
                                           - 4,5  ___      y
represents the motion of the                         car
vehicles just before and after the
collision.

3.1    Are the vehicles moving in the SAME or OPPOSITE directions BEFORE
       the collision?                                                                     (1)

3.2    State the principle of conservation of momentum in words.                          (2)

3.3    What do you understand by the term isolated system as used in physics? (2)

Use the information in the graph to answer the following questions.

3.4    How long is the collision between the car and the truck?                           (1)

3.5    Use a calculation to show that the speed of the truck before the collision,
       v. in the graph, is equal to 6,665 m-s- 1.                                         (4 )

3.6    Calculate the average net force acting on the car during the collision.            (4 )

3.7    Determine, by means of calculations, what type of collision this is
       (elastic or inelastic) and give a reason as well for your choice.                  (5)

3.8    How does the magnitude of the average net force exerted by the truck on
       the car compare with the magnitude of the average net force exerted by
       the car on the truck? Choose from greater than, smaller than or equal to.
       Name a physics law or a principle to support your answer.                 (2)
                                                                                 [21]

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FS/March 2022
Grade 12

QUESTION 4

An object is projected vertically upward at a velocity of 10 m-s- 1 from the top of
a building, which is 40 m high. Ignore air resistance.

                                    • 10 m-s- 1

                                                    40 m

                                                  , Ground
4.1    Define the term projectile in words.                                            (2)

4.2    Calculate the:

       4.2.1 Maximum height the object reaches above the ground.                       (4)

       4.2.2 Time it takes the object to hit the ground (from the instant it has
             been projected).                                                          (4)

4.3    Draw the following set of axes in your answer book.

                         a

       Use it to draw a sketch graph of acceleration versus time to represent
       the motion of the object from the moment it is projected from the top of
       the building (at t = 0) until it strikes the GROUND.

       t1 and t2 represent the times when the object is at its HIGHEST position
       and when it strikes the GROUND respectively.

       GIVE AN INDICATION NEXT TO YOUR GRAPH WHICH DIRECTION
       YOU CONSIDER AS POSITIVE.                                                       (2)

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FS/March 2022
Grade 12

The object is again projected vertically upward from the top of the building at
10 m-s- 1 as before. Someone in the building opens a window while the object is
on its way up.

On its way down, the object strikes the top of the window, at a height h, 3,4 s
after it has been projected from the top of the building.

                                               Ground
4.4    Calculate the:

       4.4.1 Magnitude of the velocity of the object when it strikes the top
             of the window.                                                         (3)

       4.4.2 Height, h, above the ground.                                           (5)
                                                                                    [20]

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FS/March 2022
Grade 12

QUESTION 5

The letters A to H in the table below represent eight organic compounds.
                                                 H
                                                 I
                                         HI-C-H
                                    H   H               H           H       H        H
                                    I   I               I           I       I         I
                                H-C-C-C-C                           C       C        C-H
 A    C 4 H 10 0           B        I   I        I                  I                 I
                                    H   H        H      H       H
                                                  H-C-H   H-C-H
                                                                            I
                                                 H-C-H                      H
                                                        I
                                                        H
            H      H   0   H                             H      H       H             H
        I I 11 I                                            I   I       I
      H-C-C-C-C-H                            D       H-C-C-C-C==C
        I I    I                                       I I I I   \
                                                                  H
            H      H       H                           H H H H
 E    3,4-dimethylpentan-2-ol                F       CH3(CH2 )2COOH

 G Ethyl ethanoat@                           H       P@ntan-3-on@

5.1     Define the term saturated compound.                                                (2 )

5.2     Write down the following:

        5.2.1 Letter that represents an UNSATURATED compound.                              (1)

        5.2.2 IUPAC name of B                                                              (3 )

        5.2.3 Letter that represents a FUNCTIONAL ISOMER of compound F.                    (1)

        5.2.4 NAME of the functional group of compound C                                   (1)

        5.2.5 General formula of the homologous series to which compound D
              belongs.                                                                     (1)

5.3     Define the term homologous series.                                                 (2 )

5.4     For compound E:

        5.4.1 To which homologous series does it belong?                                   (1)

        5.4.2 Write down its CONDENSED STRUCTURAL FORMULA.                                 (2)

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FS/March 2022
Grade 12

       5.4.3 Is it a primary, secondary, or tertiary compound?                    (1)

       5.4.4 Explain your answer to question 5.4.3.                               (1)
                                                                                  [16]

QUESTION 6

A learner uses four organic compounds (A, B, C and D) to investigate the effect
of the CHAIN LENGTH on BOILING POINT. The obtained results are shown in
the table below.

                            Condensed structural
              Compound                                Boiling point (° C)
                                 formula
                     A      CH3CH2CH2CH2CH2OH                138
                     B         CH3CH2CH2OH                       96
                     C          CH3CH20H                         77
                     D            CH3OH                          64

6.1    Define the term vapour pressure.                                           (2)

6.2    Write down the INDEPENDENT variable for this investigation.                (1)

6.3    State, with a reason, which ONE (A, B, C or D) of these compounds has
       the HIGHEST vapour pressure.                                               (2)

6.4    Compound A is now compared to pentane.

       6.4.1 Is the boiling point of A HIGHER THAN, LOWER THAN or
             EQUAL TO that of pentane?                                            (1)

       6.4.2 Refer to the TYPES of intermolecular forces to explain the answer
             to question 6.4.1.                                                (4)

6.5    Write down the general conclusion that can be made about the boiling
       points of compounds A, B, C and D.                                         (2)
                                                                                  [12]

                                                                  GRAND TOTAL: 100

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Grade 12                                                                     FS/March 2022

                     DATA FOR PHYSICAL SCIENCES GRADE 12
                            CONTROL TEST - TERM 1
                  GEGEWENS VIR FIS/ESE WETENSKAPPE GRAAD 12
                        KONTROLETOETS - KWARTAAL 1

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

 NAME/NAAM                           SYMBOUS/MBOOL                   VALUE/WAAROE
 Acceleration due to gravity
 Swaartekragversnelling                         g                         9,8 m · s-2

TABLE 2: FORMULAE/ TABEL 2: FORMULES

MOTION/ BEWEGING

 v1 = vi + af1t                           !ix= vif1t + !;,af1t 2
                                                                 or/of 11y = vif1t + !;, af1t 2
                                                       2                                2

 v} = v?- + 2af1x or/of v} = vf + 2af1y
                                                (�+v_i·) 11t or/of 11y = (�
                                          !ix = -                         -
                                                                            +v ·)
                                                                              _i 11t
                                                  2                         2

FORCE/ KRAG

 Fne t = ma                               p =mv

 Fne tf1t = 11p                           11p = mv1 - mvi

 µs -                 µs -
    _ fs(max)     /     _ fs(maks)
         N                   N            µk =-
                                              N

WEIGHT AND ENERGY/ GEWIG EN ENERGIE

 w = mg or/of F9 = mg                     U = mgh or/of Ep = mgh

 K = �mv
     2
         2
           or/of Ek = �mv
                      2
                          2

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��t�        education
                             ���.
                 r;J                  �
                ,�t. ..             ...
                                          r
           � �
            '                                 '   Department of
                                                  Education
                                                  FREE STATE PROVINCE

          CONTROLTEST/KONTROLETOETS

                             GRADE 12/ GRAAD 12

                     PHYSICAL SCIENCES
                    FIS/ESE WETENSKAPPE

                                    MEMORANDUM

                          MARCH 2022/ MAART 2022

                           MARKS: 100 / PUNTE: 100

                           This memorandum consists of eight pages.
                          Hierdie memorandum bestaan uit agt bladsye.

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### Page 14

FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12                                     MEMORANDUM

QUESTION 1 / VRAAG 1

1.1    A✓✓                  1.2     B ✓✓               1.3       C ✓✓            1.4     A✓✓

1.5    D ✓✓                 1.6     D ✓✓               1.7       A✓✓             1.8     B ✓✓

1.9    D ✓✓                 1.10    D ✓✓                                                                 [20]

QUESTION 2 / VRAAG 2

2.1    When a net force is applied to an object, the object accelerates in the
       direction of the net force. Acceleration is directly proportional to
       the net force ✓and inversely proportional to the mass of the object. ✓

       Wanneer 'n resu/terende/netto krag op 'n voorwerp inwerk, versnel
       die voorwerp in die rigting van die netto krag teen 'n versnelling direk
       eweredig aan die netto krag en omgekeerd eweredig aan die massa
       van die voorwerp.                                                                                 (2)
                                                                Accepted labels
2.2
                        N                                  Aanvaarbare b skrifte
                                                 F9 / gravitational force / weight
                                          w                                                       ✓
                                                 F I ravitasiekra I ewi
      fk                      T
                                                 FN / normal force / normal
                                           N                                                      ✓
                                                 FN I normaalkra I normaal
                                                 Fr/tension
                                           T                                                      ✓
                       w                         Frls
                                                 Frriction/f/friction / kinetic frictional
                                                 force
                                           fk                                                     ✓
                                                 F wrywin/f/wrywing/kinetiese
                                                                                                         (4)

2.3    A:
       Fnet = ma                 ✓
       F H + T + fk = ma }                          ---... ✓ both/beide
       80cos 30° -T -7.68 ✓ = Ba
       T = 61,60 -Ba ................................. . ............. ( 1)

       B:
       Fnet = ma
       T + fk = ma
       T-4 9 ✓=Sa
       T = Sa + 4, 9 ..................................................   (2)

       Solve for (1) and (2)
       a = 4,36 m · s- 2 ✓                                                                                (5)
                                                                                                         [11]

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### Page 15

FS/March 2022 / VS/Maart 2022
Grade 12 / Graad 12                        MEMORANDUM

QUESTION 3 I VRAAG 3

3.1    Opposite (direction) ✓                      Dieselfde (rigting)                               (1)

3.2    The total (linear) momentum of an isolated system remains constant
       (is conserved).
       Die totale lineere momentum in 'n geYso/eerde sisteem bly konstant
       (behoue). ✓ ✓                                                                                 (2)

3.3    A system on which the net external force is zero. ✓ ✓
       'n Sisteem waarop die netto, eksterne krag nut is.                                            (2)

3.4    3,5 - 2 = 1,5 s✓                                                                              (1)

3.5    :E p(before/voor) = :E p(after/na) } ✓
       mtVit + mcVic = (mc+mt)Vt
       (2 000)Vit ✓ + 900 x (-4,5) ✓ = (2900)(3 ,2)✓
       Vit =6,665 m·s- 1 (No mar k here. / Geen punt hier.)                                          (4)

3.6    POSITIVE MARKING FROM 3.4. / POSIT/EWE NASIEN VANAF 3.4.

         Fnet�t = L'i p✓                            Fnet�t = Lip ✓
         Fnet( 1,5)✓ = 900(3,2 -(-4,5) ✓            Fnet(1,5) ✓ = 2 000(3,2 -6,665)✓
         Fnet = 4 620 N                             Fnet = -4 620 N
         Fnet = 4 620 N; opposite to car's          Fnet (on car/ op motor)= 4 620 N;
        direction of motion / teenoorqesteld aan    opposite to car's direction of motion I
        motor se beweqinqsriqtinq ✓                 teenoorqesteld aan motor se beweqinqsriqtinq ✓

                                                                                                     (4)

3.7    �:K(before)     = ½mv2 + ½mv2 -------­
                                                                            ✓ Anyone
          (voor)       = ½(2000)(6,665 )2 + ½(900)(-4,5)2 ✓
                                                                            Enigeen
                        = 53 534,725 J

       LK(after)   = ½mv2
         (na)      = ½(2 900)(3,2)2 ✓
                   = 14 848 J
       LK(before/voor) # LK(after/na) ✓ OR/OF
       LK(before/voor) > LK(after/na)

       lnelastic/One/asties ✓                                                                        (5)

3.7    Equal to✓                                   Gelyk aan
       Newton's third law of motion✓               Newton se derde bewegingswet                      (2)
                                                                                                     [21]

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### Page 16

FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12                     MEMORANDUM

QUESTION 4 / VRAAG 4

4.1     An object which has been given an initial velocity and then it moves

        under the influence of the gravitational force only. ✓ ✓
        'n Voorwerp waaraan 'n beginsnelheid gegee is en wat dan slegs
        onder die invloed van die gravitasiekrag beweeg.                                  (2)

4.2.1
         U    ositive / 0      ositief           Down ositive / Af ositief
         Vt 2=Vi 2 + 2a�y ✓                      Vf 2=Vi 2 + 2a�y ✓
         0 ✓= 10 2 + 2(-9,8) �y ✓                0✓ = (-10)2 + 2(9,8) �y ✓
          fly= 5,10m                             �Y= -5,10m

         Max height / Maks hoogte                Max height / Maks hoogte
         = 40 + 5,10                             =40 + 5,10
         =45,10m ✓                               =45,10m ✓
         From maximum height                     From maximum height
         Vanaf maksimum hoogte                   Vanaf maksimum hoogte

         Vt 2=Vi 2 + 2a�y ✓                      Vf 2=Vi 2 + 2afly ✓
         (-10)2✓ =02 + 2(-9,8) �y✓               102 ✓=02 + 2(9,8) �y✓
         �Y=-5,10 m                              �y=5,10m

        Max height / Maks hoogte                Max height / Maks hoogte
        =40 + 5,10                              =40 + 5,10
        =45,10m ✓                               =45,10 m ✓
                                 ✓ both/beide
         Vt = Vi+ a�t                            Vt = Vi+ a�t              bothlbeide
         0 = 10 + (-9,8)�t ✓                     0 = (-10) + (9,8)flt ✓
         �t=1,0204 s                             �t =1,0204 s

         flY=Vi.1.t +�2 a�t2                     �y=Vi�t +�2 a�t2

           =(10)(1,0204) +                         =(-10)(1,0204) +
                    (0,5)(-9,8)(1,0204)2 ✓                    (0,5)(9,8)(1,0204)2 ✓
         �Y=5,10m                                �Y =-5,10m

         Max height / Maks hoogte                Max height / Maks hoogte
         = 40 + 5,10                             = 40 + 5,10

         =45,10m ✓                               =45,10m ✓
                                                                                          (4)

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### Page 17

COM FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12 MEMORANDUM
4.2.2
Up positive / Op positief Down positive / Af positief
w2=vi2+ 2aAy v2 = vi2+ 2ady .
= 10? + 2(-9,8)(-40) ¥ = (-10)? + 2(9,8)(40) v
vi = -29,7321 ms" ve = +29,7321 m:s"
Vi = vie at ¥ Vt = Vi+ aAt ¥
-29,7321 = 10 + (-9,8)At ¥ 29,7321 = -10 + (9,8) AtY
At=4,05s ¥ At=4,05s Vv
FROM MAXIMUM HEIGHT: POSITIVE MARKING FROM 4.2.1 FOR
THE HEIGHT.
VANAF MAKSIMUM HOOGTE: POSITIEWE NASIEN VANAF 4.2.1
VIR DIE HOOGTE.
If At was NOT calculated in 4.2.1: | If At was NOT calculated in 4.2.1:
Vi= vie adt vr= vi+ adt
0 = 10 + (-9,8)At ¥ 0 =-10 + (9,8)At ¥
At = T,0204 s At = T,0204 s
Ay = wit +5 at? ¥ Ay = wit +5 adt? ¥
-45,10 = 0? + (0,5)(-9,8)At? ¥ 45,10 = 0? + (0,5)( 9,8)At? ¥
At = 3,0338 s At = 3,0338 s
At = 3,0338 + 1,02 =4,05s v At = 3,0338 + 1,02 =4,05sv
If At WAS calculated in 4.2.1: If At WAS calculated in 4.2.1:
Ay = vit 6 at? ¥ Ay = vit +3 adt?
-45,10 v= 02 + (0,5)(-9,8)At? ¥ 45,10 v= 02 + (0,5)(9,8)At?2 ¥
At = 3,0338's 7 At = 3,0338's 7
At = 3,0338 + 1,02 =4,05s ¥ At = 3,0338 + 1,02 = 4,05 s ¥
(4)
4.3
Up positive / Op positief | Down positive / Af positief
a ' 1 a H '
1 ' ' 1
' : 9,8
' ' } 1
1 !
0 iti Yt >t 9 Te ¥t2 >t
9,8
H 1 ' \
in L
1 ' u L
' t }
Horizontal line from t = 0 up to ta. v No mark on the value.
Horisontale lyn vanaf t = 0 tot by te. Geen punt op die waarde nie.
Above t-axis if down is positive.
Below t-axis if up is positive. v If indication of direction is not given, the
Bo t-as as af positief is. second mark is forfeited (Max 1/2).
Onder t-as as op positief is. As rigtingaanduiding nié gegee is nie, word

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die tweede punt verbeur (Maks. 1/2).

(2)

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### Page 18

com FS/March 2022/ VS/Maart 2022 Grade 12 / Gr aad 12                                MEMORANDUM

4.4.1

 U       ositive / 0      ositief                  Down ositive / Af ositief
 Vt = Vi + ab.t ✓                                  Vt = Vi + ab.t ✓
     = 10 + (-9,8)(3,4) ✓                             = -10 + (9,8)(3,4) ✓
     = -23,32 m·s-1                                Vt = 23,32 m·s·1

 Vt = 23,32 m-s- 1 downward/afwaarts ✓             Vt = 23,32 m-s· 1 downward/afwaarts✓
                                                                                                   (3)
4.4.2
 U     ositive / 0 ositief                               Down ositive / Af ositief
 b.y = Vib.t +� ab.t2 ✓                                  b.y = Vib.t +� ab.t2 ✓
              2                                                       2
     = (10)(3,4) ✓+ (0,5)( -9,8)(3,4)2 ✓                     = (-10)(3,4) ✓ + (0,5)(9,8)(3,4)2 ✓
     = -22,644 m                                             = 22,644 m
 h = 40 - 22,644 ✓                                       h = 40
                                                             -- 22 644 ✓
   = 17,36 m ✓ (17,356)                                    = 17,36 m ✓ (17,356)
                         POSITIVE MARKING FROM 4.2.2. AND 4.4.1.
                         POSIT/EWE NASIEN VANAF 4.4.2 EN 4.4.1.
 b.y = Vib.t +�2 ab.t2 ✓                    b.y = Vib.t +� ab.t2 ✓
                                                         2
  = (-23,32)(0,65) ✓+ (0,5)(-9,8)(0,65)2 ✓   = (23 ,32)(0,65) ✓+ (0,5)(9,8)(0,65)2 ✓
  = -17,23 m ✓                               = 17,23 m ✓

 h = 17,23 m ✓                                           h = 17,23 m ✓

 Vt 2 = Vi 2 + 2ab.y ✓                                   Vt 2 = Vi 2 + 2ab.y ✓
 (-29,7321)2 ✓= (-23,32)2 + 2(-9,8)b.y ✓                 29,73212 ✓= 23,322 + 2(9,8) b.y ✓

        If ground striking velocity was not used in 4.2.2, that calculation must be shown here to earn
        one of the substitution marks.
        As grondtref-snelheid nie in 4.2.2 gebruik is nie, moet daardie berekening hier gewys word om
        een van die substitusiepunte te kry.

 b.y = 17,36 m ✓ (17,35 m)                               b.y = 17,36 m ✓ (17,35 m)
 h = 17,36 m ✓ 17,35 m                                   h = 17,36 m ✓ 17,35 m

          (Vt+ V·)                                               (Vt+ V·)
 liy =          t lit✓                                   liy =         I lit✓
           2                                                       2
       (-29,7321) + (-23,32)                                   (29,7321 + 23,32)
    =                        ✓(0,65)✓                       =                    ✓(0,65)✓
                 2                                                     2
    = -17,24 m ✓                                            = 17,24 m ✓
 h = 17,24 m ✓                                           h = 17,24 m ✓
                                                                                                   (5)
                                               6t in 4.2.2 minus 3,4 s.                            [20)

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### Page 19

FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12                         MEMORANDUM

QUESTION 5 / VRAAG 5

5.1     Compounds in which there are no multiple bonds between C atoms in their

         hydrocarbon chains. ✓✓
        Verbindings waarin daar geen meervoudige bindings tussen C-atome in hul
        koolwaterstofkettings is nie.                                         (2)

5.2.1 D ✓                                                                                   (1)

5.2.2 4-ethyl-2,5-dimethylheptane         4-etiel-2,5-dimetielheptaan                       (3)
       Marking criteria I Nasienriglyne:
       • Correct stem, i.e. heptane✓
           Korrekte stam, d.i. heptaan
       • Substituents (ethyl & methyl) correctly identified. ✓
           Substituente/sykettings (etiel & metiel) korrek geidentifiseer.
       • IUPAC name completely correct including numbering, sequence,
           hyphens, and commas✓
           IUPAC-naam hee/temal korrek ins/uitende nommers, vo/gorde,
           ko eltekens en kommas.
5.2.3 G ✓                                                                                   (1)

5.2.4 Carbonyl (group) ✓                    Karboniel(groep)                                (1)

5.2.5 CnH2n ✓                                                                               (1)

5.3     A series of organic compounds that can be described by the same general
        formula OR A series of organic compounds in which one member differs
        from the next with a CH2 group.✓✓
        'n Reeks organiese verbindings wat deur diese/fde a/gemene formule
        beskryf kan word OF 'n Reeks organiese verbindings waarin die een lid
        van die volgende verskil met 'n CH2-groep.                              (2)

5.4.1 Alcohols✓                             Alkohole                                        (1)
                          CH3                                               CH3
5.4.2
        CH3-CH-CH-CH-CH3
                           I                         ✓✓
                                                     OR/OF
                                                                            I
                                                                 CH3CHCHCHCH3
                  I
                 OH
                                     I
                                    CH3
                                                                       I        I
                                                                      OH        CH3
                  OR/OF
        CH3CH(OH )CH( CH3)CH (CH3)2                                                         (2)

5.4.3 Secondary ✓                           Sekonder                                        (1)

5.4.4 The carbon atom bonded to the hydroxyl/OH group is bonded to two
      other carbons atoms. ✓
      Die koo/stofatoom wat aan die hidroksie//OH-groep verbind is, is oak aan
      twee ander koolstofatome verbind.                                        (1)
                                                                               [16]

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### Page 20

FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12                   MEMORANDUM

QUESTION 6 / VRAAG 6

6.1       The pressure exerted by a vapour at equilibrium✓with its liquid in
          a closed system✓
          Die druk uitgeoefen deur 'n damp in ewewig met sy vloeistof in
          'n ges/ote sisteem.                                                        (2)

6.2       Chain length/length of carbon chain✓
          Kettinglengte/lengte van koolstofketting                                   (1 )

6.3       D ✓; lowest boiling point.                 D; laagste kookpunt             (2)

6.4.1 Higher than✓                                   Hoer as                         (1)

6.4.2      Marking criteria
           • Compare structures. ✓
           • Compare strength of intermolecular forces. ✓
           • Compare the energy required to overcome intermolecular forces. ✓

           Nasienriglyne
           • Vergelyk strukture. ✓
           • Vergelyk sterkte van intermolekulere kragte.
           • Vergelyk energie benodig om intermolekulere kragte te oorkom. ✓

      •   Structure/Struktuur:
          Between the molecules of A (in addition to London forces) hydrogen forces
          are present. ✓ Between pentane molecules London forces ✓ are present.
          Tussen die molekule van A (bykomend tot Londonkragte) is waterstofbindings.
          Tussen pentaanmolekule is Londonkragte.

      •   Intermolecular forces / lntermolekulere kragte
          Stronger intermolecular forces are present in compound A than in pentane. ✓
          Sterker intermolekulere kragte is teenwoording in verbinding A as in pentaan.
          (Or opposite arguments / Of teenoorgestelde argumente)
      •   Energy/Energie:
          More energy is needed to overcome the intermolecular forces in A. ✓
          Meer energie is nodig om die intermolekulere kragte in A te oorkom.       (4)
          (Or opposite arguments / Of teenoorgestelde argumente)

6.5       Boiling point increases✓with an increase in the chain length / size of the
          molecule.✓
          Kookpunt neem toe met 'n toename in die kettinglengte I grootte van die
          molekuul.                                                                  (2)
                                                                                     [12]
                                               GRAND TOTAL/ GROOTTOTAAL: 100

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