# Physical Sciences Grade 12 KZN April 2021 Memo

> Grade 12 Physical Sciences memorandum from the KZN April 2021, 21 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/phy-april-qp-memo-2021-gr12-kzn/0309e533-00e0-4080-b5a0-0f9965505369
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2021
- Exam period: KZN April
- Curriculum: CAPS
- Pages: 21

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} KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

GRADE 12

PHYSICAL SCIENCES

COMMON TEST

APRIL 2021

MARKS : 100
TIME

: 2 Hours

This question paper consists of 11 pages and 2 data sheets

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Physical Sciences NSC-Grade12 April 2021 Common Test

INSTRUCTIONS AND INFORMATION

A Write your centre number and examination number in the appropriate spaces
on the ANSWER BOOK.

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

. 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

QUESTION 2.1 and QUESTION 2.2.

6. You may use a non-programmable calculator.

ie You may use appropriate mathematical instruments.

8. You are advised to use the attached DATA SHEETS.

9. Show ALL formulae and substitutions in ALL calculations.

10. Round off your final numerical answers to a minimum of TWO decimal places.
11. Give brief motivations, discussions et cetera where required.

2: Write neatly and legibly.

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Physical Sciences NSC-Grade12 April 2021 Common Test

QUESTION 1: MULTIPLE-CHOICE QUESTIONS
Various options are provided as possible answers to the following questions. Choose

the answer and write only the letter (A-D) next to the question number (1.1-1.6) in the
ANSWER BOOK, for example 1.11 D.
1.1. Acar of mass 1000 kg pulls a trailer of mass 500 kg as shown.

- ae
i eee, el

According to Newton’s third law of motion, which ONE of the following statements
is TRUE?

A The car and trailer pull each other with a force that is equal in magnitude
but opposite in direction. Therefore, the net force is zero and the trailer
cannot move.

B The force that the car exerts on the trailer is greater than the force that the
trailer exerts on the car. Therefore, the trailer moves forward.

Cc The action force from the car is quicker than the reaction force from the
trailer, so they move forward.

D The action-reaction forces are equal in magnitude, but the force between
the ground and wheels pushes them forward. (2)

1.2 When the momentum of an object of constant mass is doubled, then its kinetic
energy will be ...

A halved.
B doubled.

~-- 5+ ee C - -~threetimes greater- - - - -  - ----- Hee Sid
D four times greater. (2)

1.3. Two objects are released from the same height at the same time. One object has
TWICE the weight of the other. Neglecting friction, which ONE of the following
statements is CORRECT for the motion?

A The heavier object hits the ground first

B The lighter object hits the ground first.

Cc Both objects hit the ground at the same time

D On hitting the ground, the heavier object has a greater velocity than the

lighter object (2)
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Physical Sciences NSC-Grade12 April 2021 Common Test

1.4 Which ONE of the following compounds is a KETONE?
A ~~ CH3COCH2CH3
B = CHsCOOCH2CH3
C  CH3CH2zCH2CHO
D  CH3sCH(OH)CH2CH3 (2)

1.5 P and Q represent two organic compounds in the reactions below:

REACTION I: ethene 48", p

NaOH(aq)
_—_——
heat

REACTION II: P Q

Which ONE of the following represents compound Q?

A CH2CH2

B  CH3CH3

C  CH3CH20H

D  CH3CH2Br (2)

1.6 CnH2nO2 is the general formula for both...
A Aketone and an aldehyde.

B- Anester and an aldehyde.

C  Aketone and a carboxylic acid.

D  Anester and a carboxylic acid. (2)
[12]

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Physical Sciences NSC-Grade12

QUESTION 2

2.1. A box of mass of 50 kg, placed on a ROUGH horizontal surface,

April 2021 Common Test

is pulled by a

force of 120 N that acts at an angle of 60° to the horizontal. The box moves at a

CONSTANT VELOCITY along the surface.

2.1.1. State Newton's First Law of Motion in words.

moves.

forces acting on the box?

2.1.4 Calculate the normal force acting on the box.

magnitude of the normal force?

EXPLAIN the answer.

2.2. The Sun and the Earth exert a gravitational force on each other.

Sun

2.2.1. State Newton's law of Universal Gravitation in words.

2.2.2. The mass of the Sun is 332 600 times greater than that

(2)
2.1.2. Draw a free-body diagram of ALL the forces acting on the box while it
(4)
2.1.3 The box is travelling at constant velocity. What can be deduced about the
(1)
(3)
2.1.5 The angle of the applied force is now decreased. How will this affect the
Choose from INCREASES, DECREASES or REMAINS THE SAME
(3)
(2)
of the Earth. qhe
distance between the centres of the Sun and the Earth is 1,496 x 10 m.
Calculate the gravitational force that the Sun exerts on the Earth. (4)
[19]

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Physical Sciences NSC-Grade12 April 2021 Common Test

QUESTION 3

A goods train is being assembled in a yard. Car 2, of mass of 92 x 10° kg and moving
with a velocity of 1.3 mss” to the right collides with Car 1 of mass 65 x 10° kg that is
moving to the right with a velocity of 0.80 mes. The two cars are joined together after
the collision. Ignore the effects of friction.

1.3 mes" 0,8 mss" v
- - : ;

Before collision : After collision

3.1. State the principle of conservation of momentum in words (2)
3.2 Calculate the magnitude of the velocity of the cars after the collision. (4)
3.3. Determine, by means of appropriate calculations, whether the collision between the
two cars is elastic or inelastic. (5)
[11]
QUESTION 4

A hot air balloon is rising vertically at a constant velocity of 15 mes’'. When the
balloon is at some unknown height above the ground, a stone is released from the
balloon. The stone is observed to hit the ground with a velocity of 45 mss”.

4.1. Define free fall (2)

4.2 Write down the magnitude and direction of the initial velocity of the stone when it is
released (2)

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Physical Sciences NSC-Grade12 April 2021 Common Test

4.3. Determine how high above the ground the hot air balloon was when the stone was
released from it. (3)

4.4 Determine how long the stone took to hit the ground after being released. (3)

4.5 Sketch the position-time graph for the entire motion of the stone from the moment it
' was released from the hot air balloon to the moment it hits the ground. Use the
ground as a zero reference point.

Indicate the following on your graph:
e Height when the stone is released.

e Time taken to hit the ground. (4)
[14]

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Physical Sciences NSC-Grade12

QUESTION 5 (Start on a new page.)

The letters A to G in the table below represent organic compounds.

* |
H —C—H
roo |
Ho ¢ Cc Cc c¢ ¢
| | | |
H H H H
a
HH ¢ F
|
H
; 1. Pf
ce ace ien ee e
H H H
: Ff fl i.
Ho CG ec Ld
| |
H H
H — C —H
|
n T
D | CyHg0 E | CHsCH(CHs)CH(CHs)CH2CCCHs | F | CHs(CHz)sCH2OH

Use the above table to answer the following questions:
5.1. Write down the:

5.1.1. IUPAC name of compound A.

5.1.2 IUPAC name of compound C.
5.1.3. Name of the homologous series to which compound B belongs.
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April 2021 Common Test

(2)
(2)
(1)

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Physical Sciences NSC-Grade12 April 2021 Common Test

5.2 Compound E is a hydrocarbon.
5.2.1 Define the term hydrocarbon.

5.2.2 Is compound E a saturated or unsaturated compound? Give a reason for
the answer.

5.2.3. Write down the general formula of the homologous series to which
compound E belongs.

5.3. Write down the IUPAC name of a POSITIONAL ISOMER of compound F.
5.4 Dis the molecular formula of TWO functional isomers.
5.4.1. Define the term functional isomer.

5.4.2 Draw the STRUCTURAL FORMULA of ONE of the FUNCTIONAL
isomers of D.

5.4.3. Write down the IUPAC name of the OTHER FUNCTIONAL isomer of D.
(Hint: This is NOT the same compound as mentioned in Question 5.4.2)

(2)

(1)
(2)

(2)

(2)
[18]

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Physical Sciences NSC-Grade12 April 2021 Common Test

QUESTION 6 (Start on a new page.)

The boiling points of TWO organic compounds, A and B, were determined.

FORMULA MOLECULAR MASS (g.mo!")
A CH3(CH2)2COOH 88,1
B CH3(CH2)3CH20H 88,1
6.1. State the definition of boiling point. (2)

The following boiling points were obtained:

137 °C 163 °C
6.2 Write down the boiling point that is most likely to be that of compound A. (2)
6.3 Explain FULLY how you arrived at the answer. (4)
6.4 Write down the IUPAC name of the compound with the LOWER vapour pressure. (2)

6.5 The boiling point of another compound, CH3(CH2)2CH20H was also determined.
How will the boiling point of this compound compare to that of compound B?
Choose from GREATER THAN, LESS THAN or EQUAL TO.
Explain the answer. (3)
[13]

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NSC-Grade12 April 2021 Common Test

QUESTION 7 (Start on a new page.)

The flow diagram shows how COMPOUND X can be used to prepare other organic
compounds. The letters A, B, C and D represent different organic reactions.

Compound X is a HALOALKANE.

CHsCOOCH2CH2CH3
COMPOUND X
Cc A B
D .
CH3CHOHCH3 ALKENE PRIMARY ALCOHOL

(major product)

Use the information in the flow diagram to answer the following questions:

7.1 Write down the type of:
7.1.1 Substitution reaction represented by A. (1)
7.1.2 Elimination reaction represented by C. (1)
7.2 Write down the type of reaction represented by:
724 Be (1)
C2522 “Di (1)
7.3 Consider REACTION GC. Write down
7.3.1 TWO reaction conditions for this reaction. (2)
7.3.2 |\UPAC name of the alkene formed. (1)
7.4 Write down the name or formula of the inorganic reactant for REACTION A. (1)
7.5 Reaction B involves the reaction of an organic compound with a PRIMARY
ALCOHOL.
7.5.1 Define the term primary alcohol. (2)
7.5.2 Write down the name or formula of the catalyst used in REACTION B. (1)
7.5.3 Write down the structural formula of the ORGANIC COMPOUND that
reacts with the primary alcohol in REACTION B. (2)
[13]
TOTAL MARKS: [100]

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ene erm eres cee —

())

1
H

1
3

Li

7

1
; Na
23

TABLE 3: THE PERIODIC TABLE OF ELEMENTS

5 6 7, 8 9 40 1 12) 13  14' °15 #16 «17 | 48
am = (iy) (vil) (vil) (vit)

Atomic number
KEY/SLEUTEL Atoomgetal He

29
Electronegativity ® Cu Symbol

Elekt iwit = Si f
lektronegatiwiteit_y| 63,5 |e imboo

Approximate relative atomic mass
Benaderde relatiewe atoommassa

19

K
39

23/ 24] 25| 26| 27) 28
ij2 V (2 Crj2Mn|2 Fe |2 Co/2 Ni |Z
51| 52; 55| 56| 59| 59

37

Rb
86

“| 42/43] 44! 457) 46
Nb |2 Mo|2 Te |S Ru|a Rh |S Pd
92 | 96 101| 103| 106

55
Cs

73 74 75 76 7 78

Ta} W/ Rej Os| Ir; Pt
181| 184] 186] 190 195

58 | 59 | 60 ] 61 63

Ce | Pr | Nd Eu
144

90 | 91 | 92
Th U

313 NVA 1498V 1 AMFgOle2d Fld -€ TAGVL
SLN3AW3A14 dO 318VL DIGOlMAd 3HL ‘€ ATEVL
ZL @PEIQ-OSN

seouains jealskud

ysa| UOWWOD 1Z0Z Idy

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                                                2
    Physical Sciences                       NSC-Grade12                   April 2021 Common Test

                    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: FISIESE KONSTANTES
           NAME/NAAM          SYMBOL/ SIMBOOL       VALUE / WAARDE
    Acceleration due to gravity
                                                 g                     9,8 m-s -2
    Swaartekragversnelling
    Universal gravitational constant                            6,67 x 10- 11 N·m2·kg-2
                                                 G
    Universe/e gravitasiekonstante
    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
    Coulomb's constant
                                                 k                9,0 X 109 N·m2 -c-2
    Coulomb se konstante
    Charge on electron                               -
                                                 e                   -1 ' 6 X 10-19 C
    Ladinq op electron
    Electron mass
                                                me                  9, 11 X 10-31 kg
    Elektronmassa
    Mass of Earth
                                                 M                  5,98 X 10 24 kg
    Massa van Aarde
    Radius of Earth
                                                RE                   6,38 X 106 m
    Radius van Aarde

    TABLE 2: FORMULAE/ TABEL 2: FORMULES
    MOTION/BEWEGING
    Vf =V j +a�t
                                                        2
                                                          1                     1
                                   /J.x = vi !J.t. + a�t or/of Liy =vi !J.t. + a�t
                                                                                   2

      2        2              2   2
    v 1 =V i +2aill<or/of v 1 =V i +2a�y     tsx =(vi ;v 1 )Mor/of Liy =(vi ; v 1   )M
    FORCE/KRAG

I   Fnet = ma
    fslmax\ = IJ s N
                                             p=mv
                                            fk = IJkN
    Fnet ilt =4'                             W=mg
     4:> = mv 1 -m v;
        Gm 1 m2                               Gm
    F=       2
                                            g=-
           r                                    r2
    WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING
    W=Ffil<COS0                         U= mgh or/ofE p = mgh
                                       Wnet = LiK       or/of Wnet = LiEk
        1
    K=-mv       If
            2 or o E k
                        1
                       =-mv 2
       2                2
                                       Li K = K1- Ki    or/of b.Ek = Ekf - Eki
    Wnc = �K + �U or/of Wnc =�Ek+ �E p  P=W
                                            dt
    Pav = F·Vav / Pgem = F·Vgem

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1 | P a g e pppppppppppPhysical Sciences / P2                             1
April 2021 Common Test
Physical Sciences                                    1                    April 2021 Common Test
                                                NSC-Guideline

                                          NATIONAL
                                     SENIOR CERTIFICATE

                                            GRADE 12

                                    PHYSICAL SCIENCES

                                         COMMON TEST

                                            APRIL 2021

                                    MARKING GUIDELINE

MARKS           : 100

TIME            : 2 hours

                            This Marking Guideline consists of 8 pages.

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2 | P a g e pppppppppppPhysical Sciences / P2
April 2021 Common Test                                                            2

 Physical Sciences                                   2                            April 2021 Common Test
                                                NSC-Guideline

 QUESTION 1

 1.1     D                                                                                            (2)
 1.2     D                                                                                            (2)
 1.3     C                                                                                            (2)
 1.4     A                                                                                            (2)
 1.5     C                                                                                            (2)
 1.6     D                                                                                            (2)
                                                                                                       [12]

 QUESTION 2

 2.1.1    A body will remain in its state of rest or motion at constant velocity unless a non-
          zero resultant/net force acts on it.                                               (2)

 2.1.2

                                N
                                                FA                        FAy
                                                                      N
                      fk                                       fk
                                                                                      FAx

                                  W                                      W
            CRITERIA FOR MARKING                                         Acceptable Labels
 • Mark is awarded for label and arrow.                   N     Normal/FN/FNormal
 • Do not penalise for length of arrows.                  fk    Friction/Ff/fkinetic
 • Deduct 1 mark for any additional force.
                                                          FA    F/120 N/Fapplied
 • If force(s) do not make contact with body/dot
                                                          W     Fg/Fgravity/Gravitational force/490 N
 : Max:3/4
 • If arrows missing but labels are there:
 Max:3/4
                                                                                                     (4)

 2.1.3    Sum of all forces acting on the box is zero/ No net force acting on the box               (1)
 2.1.4    Fnet = ma
          Fnet = 0       
          N+ FAy= W

          N + 120Sin 60o = 50(9,8) 
          N = 386,08 N
                                                                                                     (3)

 2.1.5    INCREASES, If the angle of the applied force decreases, the vertical
          component of the applied force will also decrease. This will affect the normal
          force since the sum of the normal force and vertical component of the applied
          force must be equal to the weight of the object.                                          (3)

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                3 | P a g e pppppppppppPhysical Sciences / P2                   3
                April 2021 Common Test
Physical Sciences                                    3                          April 2021 Common Test
                                              NSC-Guideline

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

2.2.2                  

                                                                
                                                     

         F = 3,55    1026 N                                                                      (4)
                                                                                                 [19]

QUESTION 3

3.1      The total linear momentum of a closed system remains constant (is
         conserved)

                                                    OR

         The total linear momentum before collision is equal to the total linear momentum
         after the collision.                                                                   (2)

3.2      Σpf =Σpi
                                         
         (m1 + m2)v = m1vi1 + m2vi2
         (92 x 103 + 65 x 103)v = 65 x 103 (0,8) + 92 x 103 (1,3)
                                 v = 1,09 m•s-1
         ∴ the velocity of the cars is 1,09 m•s-1
                                                                                                  (4)
         POSITIVE MARKING FROM Q. 3.2
3.3      ΣKf = ½ (m1 + m2)v2                  ΣKi = ½ m1 vi12 + ½m2 vi22
             = ½ (65 x 103 + 92 x 103) 1,09 2     = ½ (65 x 103 )(0,8)2 + ½(92 x 103)(1,3)2
             = 93265,85 J                          = 98540 J
         ΣKf ≠ ΣKi
         ∴Collision is Inelastic                                                                 (5)
                                                                                                 [11]

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                             4 | P a g e pppppppppppPhysical Sciences / P2
                                                                                   4
                             April 2021 Common Test
Physical Sciences                                    4                             April 2021 Common Test
                                                NSC-Guideline
QUESTION 4

4.1      Motion upon which the only force acting is the force of gravity.                          (2)

4.2      15 m•s-1 upwards                                                                          (2)

4.3                                           StoneTrajectory

                                          OPTION 1:From A to D
         UPWARDS POSITIVE                             UPWARDS NEGATIVE
         vf2 = vi2 + 2aΔy                                vf2 = vi2 + 2aΔy
         -452 = 152 + 2(-9,8)Δy                          452 = -152 + 2(9,8)Δy
         Δy = -91,84 m                           Δy = 91,84 m
         ∴ Height above the ground when the ∴ Height above the ground when the
         stone is released is 91,84 m           stone is released is 91,84 m
                                     OPTION 2:From A to D
         UPWARDS POSITIVE                        UPWARDS NEGATIVE
         vf = vi + aΔt                                    vf = vi + aΔt
         -45 = 15 +(-9,8) Δt                              45 = -15 +(9,8) Δt
         Δt = 6,12 s                                      Δt = 6,12 s

         Δy= viΔt + ½aΔt2                                Δy= viΔt + ½aΔt2
         Δy = (15)(6,12) + ½(-9,8)(6,12)2                Δy = (-15)(6,12) + ½(9,8)(6,12)2
         Δy = -91,73 m                           Δy = 91,73 m
         ∴ Height above the ground when the ∴ Height above the ground when the
         stone is released is 91,73 m           stone is released is 91,73 m
                                     OPTION 3:From A to D
         UPWARDS POSITIVE                        UPWARDS NEGATIVE
         vf = vi + aΔt                                    vf = vi + aΔt
         -45 = 15 +(-9,8) Δt                              45 = -15 +(9,8) Δt
         Δt = 6,12 s                                      Δt = 6,12 s

                                                                            

                                                                                      
         Δy = -91,80 m                                    Δy = -91,80 m
         ∴ Height above the ground when the               ∴ Height above the ground when the
         stone is released is 91,80 m                    stone is released is 91,80 m

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                          5 | P a g e pppppppppppPhysical Sciences / P2
                           April 2021 Common Test                             5

Physical Sciences                                  5                          April 2021 Common Test
                                              NSC-Guideline

                                          OPTION 3:From C to D
         UPWARDS POSITIVE                           UPWARDS NEGATIVE
         vf = vi + aΔt                                vf = vi + aΔt
         -45 = -15 +(-9,8) Δt                         45 = 15 +(9,8) Δt
         Δt = 3,06 s                                  Δt = 3,06 s

                                                                      

                                                                            
         Δy = -91,80 m                                Δy = 91,80 m
         ∴ Height above the ground when the           ∴ Height above the ground when the
         stone is released is 91,80 m                stone is released is 91,80 m

         Δy= viΔt + ½aΔt2                            Δy= viΔt + ½aΔt2
         Δy = (-15)(3,06) + ½(-9,8)(3,06)2           Δy = (15)(3,06) + ½(9,8)(3,06)2
         Δy = -91,78 m                      Δy = 91,78 m
         ∴ Height above the ground when the ∴ Height above the ground when the
         stone is released is 91,78 m      stone is released is 91,78 m      (3)

4.4                                       OPTION 1:From A to D
         UPWARDS POSITIVE                      UPWARDS NEGATIVE
         vf = vi + aΔt                         vf = vi + aΔt
         -45 = 15 +(-9,8) Δt                   45 = -15 +(9,8) Δt
         Δt = 6,12 s                           Δt = 6,12 s

                        OPTION 2:From A to B and then C - D
         UPWARDS POSITIVE           UPWARDS NEGATIVE
         vf = vi + aΔt                          vf = vi + aΔt
         0 = 15 +(-9,8) Δt                      0 = -15 +(9,8) Δt
                                      
         Δt = 1,53 s                            Δt = 1,53 s

         vf = vi + aΔt                          vf = vi + aΔt
         -45 = -15 +(-9,8) Δt                   45 = 15 +(9,8) Δt
         Δt = 3,06 s                            Δt = 3,06 s

         Δt = 2(1,53) + 3,06                   Δt = 2(1,53) + 3,06
         Δt = 6,12 s                           Δt = 6,12 s
                      OPTION 3:POSITIVE MARKING FROM 7.2
                                  From A to D
         UPWARDS POSITIVE          UPWARDS NEGATIVE
         Δy= viΔt + ½aΔt2                      Δy= viΔt + ½aΔt2
         -91,73 = (15Δt+ ½(-9,8Δt2             91,73 = (-15Δt+ ½(9,8Δt2
         Δt = 6,12 s                           Δt = 6,12 s                                       (3)
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                                               6 | P a g e pppppppppppPhysical Sciences / P2                6
                                               April 2021 Common Test
Physical Sciences                                                        6                                  April 2021 Common Test
                                                                    NSC-Guideline

4.5      POSITIVE MARKING FROM 7.2 AND 7.3

                                        91,80

                     Position/m

                                                  0
                                                                              6,12       Time/s

                                                  0                      OR

                                  Position/m
                                                                                               Time/s
                                                                         6,12

                             -91,80

NOTE: LEARNERS MAY CHOOSE HEIGHT WHEN THE BALL IS RELEASED AS THE
REFERENCE POINT

                                                              Criteria for marking
Correct shape                                                                                           
Indication of height (9,73 – 9,84 m)                                                                    
Indication of the correct end time                                                                      
Correct choice of the zero point of reference                                                           
                                                                                                                             (4)
                                                                                                                            [14]

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                             7 | P a g e pppppppppppPhysical Sciences / P2
                                                                             7
                             April 2021 Common Test
Physical Sciences                                      7                     April 2021 Common Test
                                                  NSC-Guideline
QUESTION 5

5.1.1 2,4-dimethylhexane                                                                       (2)
5.1.2 1-chloro-2-methylpropane                                                                 (2)
5.1.3 Esters                                                                                    (1)

5.2.1 Compound that consists of carbon and hydrogen only.                                      (2)
5.2.2 Unsaturated. Contains a multiple bond/triple bond between carbon atoms in the
        carbon chain.                                                                          (2)
5.2.3 CnH2n-2                                                                                   (1)
5.3     pentan-2-ol OR pentan-3-ol                                                           (2)
5.4.1 Compounds with same molecular formula but different functional groups.                   (2)
5.4.2
                     H            H           H            O          
                     │            │           │            ║
           H     ―   C       ―    C     ―     C      ―     C      ―   H
                     │            │           │
                     H            H           H

                                                      OR
                                  H           H            O          
                                  │           │            ║
                     H       ―    C     ―     C      ―     C      ―   H
                                              │
                                              H

                     H       ―    C     ―     H
                                  │
                                  H

                                                      OR
                     H            H           O            H
                     │            │           ║            │
           H     ―   C       ―    C     ―     C      ―     C  ―       H
                     │            │                        │
                     H            H                        H

                             Whole structure correct:                2/2
                             Only functional group correct           1/2
                             More than one functional group          0/2

                                                                                                 (2)
5.4.3 POSITIVE MARKING FROM Q5.4.2
        butan-2-one OR butanal OR 2 - Methylpropanal                                         (2)
                                                                                                [18]
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                          8 | P a g e pppppppppppPhysical Sciences / P2   8
                          April 2021 Common Test
Physical Sciences                                    8                    April 2021 Common Test
                                                NSC-Guideline

QUESTION 6

6.1     The temperature at which the vapour pressure of a substance equals the
        atmospheric pressure.                                                               (2)

6.2     163 °C                                                                              (2)
6.3      
         The acid (A) has TWO sites for hydrogen bonding while the alcohol (B) has only
        one site. 
         The carboxylic acid molecules require more energy to overcome the intermolecular
        forces
         Carboxylic acid will have a higher boiling point.                                   (4)
6.4     Butanoic acid.                                                                      (2)

6.5     LOWER.

             _-
        CH3(CH2)2CH2OH or butan-1-ol has a shorter carbon chain/smaller
        surface area than compound B. Strength of the intermolecular forces in
        CH3(CH2)2CH2OH or butan-1-ol is weaker than that in compound B. 
        Therefore lesser energy needed to separate the molecules.  OR
        Compound B has a longer carbon chain/ larger surface area than CH3(CH2)2CH2OH
        or butan-1-ol The intermolecular forces between molecules of compound B
        are therefore strongerand require more energy to separate the molecules.     (3)
                                                                                      [13]

QUESTION 7

7.1.1 Hydrolysis                                                                             (1)
7.1.2 Dehydrohalogenation                                                                    (1)
7.2.1 Esterification                                                                         (1)
7.2.2 Addition/hydration                                                                     (1)
7.3.1 Heat
      Concentrated strong base in ethanol                                                    (2)
7.3.2 Propene/Prop-1-ene                                                                     (1)
7.4   Water/H2O                                                                              (1)
7.5.1 The carbon to which the hydroxyl (OH) group is bonded to, is bonded to ONE other
      carbon atom.                                                                          (2)
7.5.2 sulphuric acid/H2SO4                                                                   (1)
7.5.3
                      H           O
                      │           ║
           H     ―    C     ―     C     ―     O       ―       H
                      │
                      H
                    Whole structure correct:          2/2
                    Only functional group correct    1/2
                    More than one functional group     0/2                                   (2)
                                                                                             [13]

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