# Physical Sciences Grade 12 P1 KwaZulu Natal Mock 2021 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the KwaZulu Natal Mock Exam 2021, 37 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-science-grade-12-trial-2021-p1-and-memo/a5398f16-7086-4494-abb9-af62ed82ab9e
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2021
- Exam period: KwaZulu Natal Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 37

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Physical Sciences · Grade 12 · KwaZulu Natal Mock Exam · 2021. Memorandum, 37 pages. Read online or download the PDF.

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### Page 1

e av +2; RE 7 age Bo Sh ate yeRe ge ee
) RWAZULU-NATAL PROVINCE

é EDUCATION
" REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

PHYSICAL SCIENCES P1 (PHYSICS)

PREPARATORY EXAMINATIONS

SEPTEMBER 2021

150

TIME : 3hours

This question paper consists of 17 pages and 3 data sheets.

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INSTRUCTIONS AND INFORMATION TO CANDIDATES

1. Write your name on the ANSWER BOOK.

2 This question paper consists of TEN questions. Answer ALL the 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

QUESTION 2.1 and QUESTION 2.2.

You may use a non-programmable calculator.

You may use appropriate mathematical instruments.

You are advised to use the attached DATA SHEET.

oo oO

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.

12. Write neatly and legibly.

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Physical Sciences/P1 3 Preparatory Examintion September 2021
NSC

QUESTION 1: MULTIPLE CHOICE

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 (1.1-1.10) in
the ANSWER BOOK, for example 1.11 D.

11 The free-body diagram below shows the relative magnitudes and directions of
all the forces acting on an object.

Normal force

Frictional force Renee 1 E
Gravitational force s
The object is...
A accelerating eastward.
B accelerating westward.
Cc at rest.
D moving with constant velocity. (2)

1.2 Acceleration due to gravity on the surface of the Earth is g. What will be the
acceleration due to gravity on the surface of another planet of the SAME mass
as that of the Earth, but HALF the radius of the Earth?:

A rg

B g

Cc 2g

D 4g (2)

1.3. From the top of a building, ball X is thrown vertically upwards and another ball
Y, is thrown vertically downwards. Both balls are thrown at the same speed.
Ignore the effects of friction.

Which ball(s) will have a greater speed when it hits the ground?
A xX

B Y
Cc X and Y would have the same speed
D

Depends on the mass of the ball. (2)

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1.4. In movies, Superman hovers in mid - air, grabs a villain (bad guy), and throws
him forward. Superman, however, remains stationary. This is not possible in
real life, because it goes against one of the following laws or principles:

A
B
Cc
D

The principle of conservation of energy
Newton’s first law of motion
Newton's Universal law of Gravitation

The principle of conservation of linear momentum (2)

1.5 A force, F, acts in three different directions on a box moving with a velocity, v,
as shown below.

F li
Y, F ‘, |,
A B c

Select the order in which the work done by the force F from minimum
to maximum
A A; B,C
B C;B;A
Cc B,C; A
D CAB (2)

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Physical Sciences/P1 5 Preparatory Examintion September 2021
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1.6 A police car, with its siren on, is moving away at constant speed from a

1.7

stationary observer. The siren emits a sound of constant frequency

Which of the following characteristics, associated with the sound
emitted by the siren is/are CORRECT, as perceived by the observer?

(i) The speed remains the same.

(ii) The frequency increases.

(iii) The wavelength increases.

(iv) The frequency decreases

A (iii) only

B (i), (iii) and (iv)

c (i) and (iii) only

D (i) and (ii) only (2)
A point charge exerts a force of magnitude F on another point charge.

One of the charges is changed to one third and the distance between

the charges is doubled. What is the new electrostatic force in terms of
F?

a
3
BF
12
C8
4
DF
6

(2)

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1.8 The sketch below, show a battery of emf &, connected in a circuit
containing three resistors. Ignore the internal resistance of the battery.

GS)

| LEY)

| A:

WV}

3)

Which ONE of the following combinations best describes the readings of
ammeter Ag3 and voltmeter V2?

Reading on A3 Reading on V2
A ly-lb e-V
B l-l2 E+V,
Cc +h o+Vi
D +h e-V;

1.9 White light is passed through a cold gas, then through a prism as
shown in the sketch.

. Screen
Cold Gas Prism
White light
What type of spectrum is observed on the screen?
A Line absorption spectrum
B Line emission spectrum
Cc Continuous absorption spectrum
D Continuous emission spectrum (2)

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Physical Sciences/P1 7 Preparatory Examintion September 2021
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1.10 Which ONE of the following provides evidence that light behaves as a
particle?
A Light can be diffracted.

B
Cc
D

Light can be refracted.
Light ejects electrons from a metal surface.

The speed of light increases when it travels in a vacuum. (2)
[20]

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QUESTION 2

Two crates of masses 4 kg and 6 kg joined by a light inextensible string, are
placed on a rough inclined plane. A force F pulls on the 6 kg crate such that
the systems accelerates up the plane at 2,5 ms”. The frictional forces exerted
by the surface is 1 N for 4 kg crate and 1,5 N for the 6 kg crate.

2.1

2.2

2.3

2.4

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6k g
4k
30°
State Newton’s Second Law of Motion in words. (2)
Draw a labelled free body diagram for the 6 kg crate. (5)
Calculate the magnitude of the force F, exerted on the 6 kg crate. (5)

The string joining the crates now snaps, whilst the same force F acts
on the 6 kg crate.
How will this change affect the following:

2.4.1 The acceleration of the 6 kg crate. (Choose from INCREASES,

DECREASES or REMAIN THE SAME) (1)
2.4.2 The motion of the of the 4 kg block. (3)
[16]

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Physical Sciences/P1 9 Preparatory Examintion September 2021
NSC

QUESTION 3

A hot — air balloon is rising at a constant speed of 4,9 m’s’. At a height of 80 m
above the ground, a ball is released.
Ignore the effects of friction.

M GROUND

3.1 Write down the acceleration of the ball:
3.1.1 Before it was released (1)
3.1.2 After it is released (2)

3.2 Calculate the:

3.2.1 Position of the ball 0,5 s after it was released relative to the
ground (4)

3.2.2 Time taken by the ball to reach the ground (4)

3.3 Sketch a velocity — time graph for the motion of the ball from the
instant it is released until it reaches the ground.

Indicate the following in your graph:
e Initial velocity of the ball

e The time for the entire motion of the ball (3)
[14]
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QUESTION 4

A car of mass 1200 kg travelling east with a speed of 25 m's”, collides with the
van of mass 1800 kg. The van of mass 1800 kg is stationary on a frictionless
track. The two vehicles move together as a unit after the collision.

N
A
we rE
vy
Sy}
41 State the principle of conservation of momentum in words (2)
4.2 Calculate the speed of the vehicles after the collision. (4)

4.3 Determine by means of a relevant calculation whether the collision was
ELASTIC or INELASTIC. (5)

After the collision the combination of the vehicles move into a real road and
stop, covering a distance of 20 m.

4.4 Calculate the magnitude of the frictional force. (5)
[16]

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Physical Sciences/P1 11 Preparatory Examintion September 2021
NSC

QUESTION 5

A lift arrangement shown below, comprises an electric motor, a cage and its
counterweight. The counterweight moves vertically downwards as the cage
moves upwards. The cage and counterweight move at the same constant

speed.
OC Electric motor

Cage

Counterweight |

The cage, carrying passengers, moves vertically upwards at a constant speed,
covering 60 m in 2 minutes. The counterweight has a mass of 900 kg. The total
mass of the cage and passengers is 1 300 kg. The electric motor provides the
power needed to operate the lift system. Ignore the effects of friction.

5.1 Calculate the work done by the:
5.1.1 Gravitational force on the cage (3)
5.1.2 Counterweight (2)
5.2 Calculate the average power required by the motor to operate the lift

arrangement in 2 minutes as explained above. Assume that there are
no energy losses due to heat and sound. (5)

[10]

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QUESTION 6

A siren is attached to a sea shore that emits sound waves of frequency 755
Hz. A speed boat is travelling with a constant speed away from the shore.

6.1

State Doppler Effect in words .

The speed boat travels 45 min 3s.

6.2

6.3

6.4

Calculate the frequency of the sound waves emitted by the siren as
heard by a person on the boat.

Take the speed of sound in air as 340 ms”.

Write down any two uses of the Doppler Effect.

Spectral lines of star X at an observatory are observed to be red
shifted.

6.4.1. Explain the term red shifted in terms of wavelength.

6.4.2 Will the frequency of the light observed from the star
INCREASE, DECREASE or REMAIN THE SAME?

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(2)

(5)

(1)
[12]

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Physical Sciences/P1 13 Preparatory Examintion September 2021
NSC

QUESTION 7

7.41 Three metal spheres X, Y, and Z carrying charges are placed on the
vertices of a right angled triangle as shown below.

-2,10 x 10°C 1,5x10°C

3x10%m

1,8x 10°C
7.1.1 State Coulomb’s Law in words (2)

7.1.2 Calculate the magnitude of the net electrostatic force on
sphere Z. (8)

7.2 Two identical spheres, A and B, carrying charges +5 uC and -10 uC

respectively, are separated by a distance of 30 mm.
P is a point located at a distance of 15 mm from sphere B as shown

below.
:

30 mm

*p

:

The spheres A and B are allowed to touch each other and then return
to their original positions.

15mm

7.2.1 Define electric field at a point in words (2)
7.2.2 Draw an electric field pattern between spheres A and B, after

touching. (3)

7.2.3 Calculate the magnitude of the net electric field at point P (6)

[21]

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riysical OGiIenvesir | 14 ritpdl awry CxaiNuUl Oepleniver 2uzZ |

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QUESTION 8

A battery with an internal resistance of 0,10 Q is connected to the circuit as
shown in the sketch below. The three light bulbs in the circuit, Ri, Re and Rs,
are all identical, with a power rating of 40 W each.

Ignore the resistance of the wires and that of the ammeter.

8.1 Define emf in words (2)

Switch $ is closed. Now bulb R3 works at optimum condition and the ammeter
reads 2A.

8.2 Calculate the:

8.2.1 TOTAL EXTERNAL RESISTANCE of the circuit (5)
8.2.2 Emf of the battery (3)
8.2.3. The energy dissipated by the bulb labelled R3in 20 minutes (3)

[13]

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Physical Sciences/P1 15 Preparatory Examintion September 2021
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QUESTION 9

9.1 The sketch below represents a simplified version of a generator used
to light up a light bulb. The resistance of the light bulb is 807 Q

Brushes

9.1.1 Name the type of generator in the sketch.

Choose from AC or DC (1)
9.1.2 Name the part labelled P in the sketch. (1)
9.1.3 Write down the principle on which the generator operates. (1)

9.2 The graph of output emf versus time obtained when the above
generator was in use is shown below.

811,13 p-prynnrnnmno :

E ts
mf
(V
9.2.1 Define the term root mean square of an AC voltage. (2)

9.2.2 Calculate the power dissipated through the light bulb connected
to the generator above. (5)

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PiysiGal OUieHces/r 1 10 Preparatory Examintion september 2021
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| Some changes were made to the above generator, a new machine,
shown below, was obtained.

9.3.1 Draw a sketch graph of potential difference versus time

obtained from this machine. (2)
9.3.2 Write down the function of the part labelled Q in the new
machine. (2)
[14]
QUESTION 10

The simplified diagram below illustrates how a cathode emits electrons when
light shines on it.

Incident Light
o>
Cathode

Potential
—__—* Difference * (a)

The incident light has an energy of 2,12 x 10° J and the cathode has a
threshold frequency of 2,21 x 10° Hz.

10.1 Define threshold frequency. (2)

The incident light releases 2,01 x 10° photo-electrons per second from the
cathode.

10.2 Calculate the current flowing through the ammeter. (4)

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Physical Sciences/P1 17 Preparatory Examintion September 2021
NSC

10.3. The brightness of the incident light is now decreased.
What effect will this change have on the current in QUESTION 10.2?
Choose from INCREASES, DECREASES or REMAIN THE SAME
Give a reason for the answer. (2)
10.4 Calculate the maximum kinetic energy of the photo - electrons. (4)

The cathode is replaced by another one with a higher threshold frequency
and the same light was used.

10.5 How does this change affect the kinetic energy of the photoelectrons

released?
Choose from INCREASES, DECREASES or REMAIN THE SAME
Give a reason for your answer (2)
[14]
TOTAL: 150

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Pnysical Sciences/r' 1

NSC

Preparatory Examintion September 2021

DATA FOR PHYSICAL SCIENCES GRADE 12

PAPER 1 (PHYSICS)

GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12

VRAESTEL 1 (FISIKA)

TABLE 1: PHYSICAL CONSTANTS / TABEL 1: FISIESE KONSTANTES

NAME / NAAM

SYMBOL / SIMBOOL

VALUE / WAARDE

Acceleration due to gravity
Swaartekragversnelling

9,8 m-s@

Radius van Aarde

Universal gravitational constant G 6,67 x 10" N-m?-kg*
Universele gravitasiekonstante

Speed of light in a vacuum c 3,0 x 10° m-s™
Spoed van lig in 'n vakuum

Planck's constant h 6,63 x 10 Js
Planck se konstante

Coulomb's constant k 9,0 x 10°N-m?-C*
Coulomb se konstante

Charge on electron e -1,6x 10°C
Lading op electron

Electron mass Me 9,11 x 10°" kg
Elektronmassa

Mass of Earth M 5,98 x 10% kg
Massa van Aarde

Radius of Earth Re 6,38 x 10°m

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Physical Sciences/P1 Preparatory Examintion September 2021
NSC

TABLE 2: FORMULAE / TABEL 2: FORMULES
MOTION / BEWEGING

Vp =Vi taAt Ax =v, At + dadt? or/of Ay = v,At + faAt?

v, +¥ vit
v, =v? +2adxor/ofv,’ =v," +2ady Ax = (“eta or/of Ay = (“eta

FORCE / KRAG

Fie = Ma p=mv

fsqmax) =u N fk = UN

FierAt =Ap w=mg

Ap = mv, -my,

F= Gmym 2 _GM
ar oe

WORK, ENERGY AND POWER / ARBEID, ENERGIE EN DRYWING

W=FAxcos 0 U=mgh or/ofE, =mgh
K= 3 mv? or/of E, = 3 mv? Whet = AK tei Whet = AE
AK = Kr-K, or/of AE, = Exg¢ - Exi
Ww

Whe = AK + AU orlof Whe = AE + AE, ar

Pay = F-Vay / Pgem = F-Vgem

WAVES, SOUND AND LIGHT / GOLWE, KLANK EN LIG

1

v=fr f=

f
viv Cc
f= Lf = =h-
L viv, E=nhf or/of E a

E = Wo + Eximax) Or/of E = Wo + Kimax)  “’*here/waar
3

E=hf and/en Wo=hf. and/en Eximax) = 3 ™Vinax orlof Kimax) = a mVAax

ELECTROSTATICS / ELEKTROSTATIKA

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Physical Sciences/P1 20 Preparatory Examintion September 2021
NSC
k
r r
Ww F
vo E=*
q q
n=2 OR/IOF n=2
e Ge

ELECTRIC CIRCUITS / ELEKTRIESE STROOMBANE

Vv emf (€) =1(R +r)
R=—
I emk (¢) = (R +n)
R, =R,+R2+
14 q=1At
—=—+—+..
R, R, R,
W=Vq
ph
W=VIAt At
P=VI
=/2
W=IRAt P=[’R
v2
V7 At =
We
R R
ALTERNATING CURRENT / WISSELSTROOM
I Pave = Vans Las / Pyemiadeta= Vngk ! wok
Tome Tye =
v2 v2 Pive =TinsR / Premiddeld =TegR
V, V,
Vv. = max i Vv. : — maks v2 v2
™ v2 “ v2 Pave AR i Premiddela a

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                               NATIONAL
                          SENIOR CERTIFICATE

                                  GRADE 12

                  PHYSICAL SCIENCES P1 (PHYSICS)

                   PREPARATORY EXAMINATIONS

                          MARKING GUIDELINE

                            SEPTEMBER 2021

MARKS :    150

TIME: 3 hours

                   This marking guideline consists of 15 pages.

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Physical Sciences/P1           2 Preparatory Examination September 2021
                                NSC

QUESTION 1: MULTIPLE CHOICE

1.1    A                                                         (2)

1.2    D                                                         (2)

1.3    C                                                         (2)

1.4    D                                                         (2)

1.5    D                                                         (2)

1.6    B                                                         (2)

1.7    B                                                         (2)

1.8    A                                                         (2)

1.9    A                                                         (2)

1.10   C                                                         (2)
                                                                  [20]

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Physical Sciences/P1                   3 Preparatory Examination September 2021
                                        NSC

QUESTION 2

2.1   When a resultant/net force acts on an object, the object will accelerate
      in the direction of the force at an acceleration directly proportional to
      the force and inversely proportional to the mass of the object.
                                         OR
      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
      force)(2 or 0).                                                           (2)

2.2
                                                         ACCEPT

                       Accept the following symbols
               F       FA/Fapp/FApplied
               N       FN/Normal/Normal force
               fk      Kinetic friction force/f/Ff/fr
               T       Tension force/FT
               Fg      W/58,8N

       Notes
       • Mark is awarded for label and arrow.
       • Do not penalise for length of arrows.
       • Deduct 1 mark for any additional force.
       • If force(s) do not make contact with body/dot : Max:4/5
       • If arrows missing but labels are there: Max:4/5

                                                                                  (5)

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  Physical Sciences/P1                      4 Preparatory Examination September 2021
                                             NSC

2.3      For the 4 kg crate

         Fnet = ma
         T – f – Fg// = ma
         T – 1 – (9,8)(4)sin30o = 4(2,5)
         T = 30,6 N upwards
 
         For the 6 kg crate
                                                    Any of the two
         Fnet = ma
         F – f – T – Fg// = ma
         F–1,5– 30,6 – (9,8)(6)sin 30o = 6(2,5)
         F = 76,5 N 
                                                                                  (5)

2.4.1    INCREASES                                                               (1)

2.4.2    The 4 kg block will move upwards/forward (for a brief moment), stop
         and then slide down the plane / backward.                             (3)
                                                                               [16]

 QUESTION 3
 3.1.1     Zero                                                                  (1)

 3.1.2     9,8 m.s-2 downwards                                                  (2)

 3.2.1                                        OPTION 1
           UPWARD POSITIVE                     UPWARD NEGATIVE

           Δy= viΔt + ½aΔt2                   Δy= viΔt + ½aΔt2

           Δy= (4,9)( 0,5) + ½(-9,8)(0,5)2    Δy= (-4,9)( 0,5) + ½(9,8)(0,5)2

           Δy = 1,225 m                        Δy = -1,225 m
           Height after 0,5 s = 80 + 1,225    Height after 0,5 s = 80 + 1,225
           = 81,23 m
                                               = 81,23 m
           ∴The ball is 81,23 m above the      ∴The ball is 81,23m above the
           ground                              ground

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Physical Sciences/P1                    5 Preparatory Examination September 2021
                                         NSC

                                          OPTION 2
       UPWARD POSITIVE                            UPWARD NEGATIVE

       vf = vi + aΔt                              vf = vi + aΔt

       vf = 4,9 + (–9,8) (0,5)                    vf = -4,9 + (9,8) (0,5)
       vf = 0 m.s-1                               vf = 0 m.s-1

       vf2 = vi2 + 2aΔy                          vf2 = vi2 + 2aΔy

       0 = 4,92 + 2(–9,8)Δy                      0 = - 4,92 + 2(9,8)Δy

       Δy = 1,225 m                               Δy = -1,225 m
       Height after 0,5 s = 80 + 1,225           Height after 0,5 s = 80 + 1,225
                          = 81,23 m
                                                                        = 81,23 m
       ∴The ball is 81,23 m above the ground      ∴The ball is 81,23 m above the
                                                  ground
                                          OPTION 3
       UPWARD POSITIVE                         UPWARD NEGATIVE

       vf = vi + aΔt                           vf = vi + aΔt

       vf = 4,9 + (–9,8) (0,5)                 vf = 4,9 + (–9,8) (0,5)
       vf = 0 m.s-1                            vf = 0 m.s-1
                v f + vi                                v f + vi
       Δy =              Δt                   Δy =              Δt 
                    2                                       2

                0 + 4,9                                 0 - 4,9
       [ Δy =           0,5 ]                 [ Δy =           0,5 ]
                   2                                       2

       Δy = 1,225 m                            Δy = -1,225 m
       Height after 0,5 s = 80 + 1,225        Height after 0,5 s = 80 + 1,225
                            = 81,23m                               = 81,23 m
       ∴The ball is 81,23 m above the          ∴The ball is 81,23 m above the
       ground                                  ground

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Physical Sciences/P1                 6 Preparatory Examination September 2021
                                      NSC

                                        OPTION 4

      ½mvi2 + mghi = ½mvf2 + mghf 

      ½m(4,9)2 + m(9,8)(80)  = ½m(0)2 + m(9,8)hf 

                     hf = 81,225 m 

                                        OPTION 5

      ½mvi2 + mghi = ½mvf2 + mghf 

      ½m(4,9)2 + m(9,8)(0) = ½m(0)2 + m(9,8)hf 

                     h = 1,225 m

      Height after 0,5 s = 80 + 1,225
                         = 81,23 m
      ∴The ball is 81,23 m above the ground
                                                                         (4)

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Physical Sciences/P1                    7 Preparatory Examination September 2021
                                         NSC

3.2.2                                     OPTION 1
        UPWARDS POSITIVE                      UPWARDS NEGATIVE

        Δy= viΔt + ½aΔt2                     Δy= viΔt + ½aΔt2

        -80 = (4,9)Δt + ½(-9,8)Δt2          80 = (-4,9)Δt + ½(9,8)Δt2

        Δt = 4,57 s                          Δt = 4,57 s

                                          OPTION 2
        UPWARDS POSITIVE                      UPWARDS AS NEGATIVE
        vf2 = vi2 + 2a∆y                      vf2 = vi2 + 2a∆y
        vf2 = (4,9)2 + 2(-9,8)(-80)          vf2 = (-4,9)2 + 2(9,8)(80) 

        vf = -39,9 m.s-1                      vf = 39,9 m.s-1

        vf = vi + a∆t                        vf = vi + a∆t
        -39,9 = 4,9 + (-9.8)∆t               39,9 = -4,9 + (9.8)∆t
        ∆t = 4,57 s                          ∆t = 4,57 s 

        OPTION 3: POSITIVE MARKING FROM QUESTION 3.2.1
        Considering ball from the maximum height
        UPWARDS POSITIVE                             UPWARDS AS NEGATIVE

        Δy= viΔt + ½aΔt2                       Δy= viΔt + ½aΔt2

        -81,23 = (0)Δt + ½(-9,8)Δt2            81,23 = (0)Δt + ½(9,8)Δt2

        Δt = 4,07 s                             Δt = 4,07 s
        Time to reach ground = 0,5 + 4,07      Time to reach ground = 0,5 + 4,07
                            = 4,57 s                                = 4,57 s

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Physical Sciences/P1                                           8 Preparatory Examination September 2021
                                                                NSC

3.3    POSITIVE MARKING FROM Q3.2.2

                                                   UPWARD POSITIVE

                           4,9

                                                                                  4,57
                                              0

        Velocity (m.s-1)
                                                                                          Time (s)

                                                                     OR

                                                    UPWARD NEGATIVE

                           Velocity (m.s-1)
                                              0
                                                                                  4,57
                                                                                             Time (s)

                     -4,9

CRITERIA FOR MARKING OF GRAPH
Correct shape                                                                                        
Indication of initial velocity                                                                       
Indication of the time for the entire motion                                                         
                                                                                                      (3)
                                                                                                     [14]

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Physical Sciences/P1                     9 Preparatory Examination September 2021
                                          NSC

QUESTION 4

4.1   The total linear momentum in an isolated system is conserved
      (accept “closed” instead of “isolated”)
                                             OR
      In an isolated system, the total linear momentum before a collision is
      equal to the total linear momentum after the collision.                       (2)

4.2   ptotal (before) = ptotal (after)
      Σpi =Σpf                           

      m1vi1 + m2vi2 = (m1 + m2)v
      1200 (25) + 1800 (0) = (1200 + 1800)v 
      v = 10 m.s-1.
      ∴ Speed of the cars after is 10 m.s-1
                                                                                      (4)

4.3   ΣKf = ½ (m1 + m2)v2               ΣKi = ½ m1 vi12 + ½m2 vi22
          = ½ (1200 + 1800) 10 2             = ½ (1200 )(25)2 + ½(1800)(0)2
          = 150 000 J                         = 375 000 J
      ΣKf ≠ ΣKi
      ∴Collision is Inelastic 
                                                                                      (5)
      NOTE: If it is assumed that ΣKf = ΣKi at the outset, Max:2/5

4.4   POSITIVE MARKING FROM QUESTION 4.2
      OPTION 1                                     OPTION 2
      vf2 = vi2 + 2a∆y                                     v f + vi
                                                    Δy =             Δt 
                                                                2
      02 = (10)2 + 2a(20)                                0  10 
                                                    20           t 
                                                          2 
      a = - 2,5 m.s-2                              Δt = 4 s
      Fnet = ma                                   vf = vi + aΔt
      Ff = ma                                      0 = 10 + a(4)
      Ff = 3000(-2,5)                             a = - 2,5 m.s-2
      Ff = -7500 N
      ∴ Magnitude of frictional force is 7500 N
                                                                                (5)
                                                                                            [16]

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 Physical Sciences/P1                                    Preparatory Examination September 2021
                                                  NSC

        OPTION 3                                    OPTION 4
                 v f + vi                           Wnet = ΔEk
        Δy =              Δt                                                      
                     2                              f Δx cos θ = ½mvf2 - ½mvi2
              0  10                              f (20) cos 180⁰  = ½(3000)(0)2 - ½(3000)(10)2
        20           t          ∴ Δt = 4 s
              2                                   f = 7500 N 
                        p
        Fnet = f =
                        t
                      m( v f v i )
           f     =                  
                          t
                         ( 0 10 )
                      3000
                  
                           4
                   = -7500 N
        ∴ Magnitude of f is 7500 N
                                                                                                   (4)
                                                                                                 [16]
QUESTION 5

5.1.1   OPTION 1
        W = FΔxcosθ
        Wgravity = mgΔycosθ
        = (1 300)(9,8)(60)cos180o
        = - 764 400 J (-7,64 x105 J)

        OPTION 2
        W = - ΔEp
          = -(1300)(9,8)(60 -0) 
          = - 764 400 J (-7,64 x105 J)
        -1 mark if either negative is omitted
                                                                                                (3)

5.1.2   Wcounterweight = mgΔycosθ
                       = (900)(9,8)(60)cos0o 
                       = 529 200 J (5,29 x 105 J)                                              (2)

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Physical Sciences/P1                            11         Preparatory Examination September 2021
                                                 NSC

5.2    OPTION 1                               OPTION 2
       POSITIVE MARKING FROM 5.1.1            Fmotor = Tright - Tleft
       AND 5.1.2                                     = 1300(9,8) - 900(9,8)
       Wnet = ΔEK                                    = 3920 N
       Wgravity + Wcountweight + Wmotor = 0  Wmotor = Fmotor Δy cos0 
       Wmotor = - (W gravity + W countweight)         = (3920)(60) cos0 
                                                      = 235 200 J
       -764 400 + 529 200 + Wmotor = 0
       ∴Wmotor = 235 200 J

                                                         W
                                            Pave motor =    
                                                         Δt
                                                      235200
                                         Pave motor =         
                                                        120
                                                    = 1960 W                                  (5)
                                                                                                 [10]

QUESTION 6

6.1    The change in frequency (or pitch) of the sound detected by a listener
       because the sound source and the listener have different velocities relative
       to the medium of sound propagation.                                                    (2)

6.2           d            45
       v=             v=       ∴vL = 15 m.s-1
              Δt           3
              v ± vL               v vL
       fL =          fS     fL            fs
              v ± vS                 v
                340 - 15 
       fL = (           )755 
                340 + 0
       fL = 721,69 Hz                                                                          (5)

 6.3   Any two
          Ultrasound waves (to measure the heartbeat of a foetus in the womb).
          Doppler flowmeter (to measure the rate of blood flow)
          Traffic management systems, (especially speed control)
          Radar, (allowing for the tracking of weather systems)
          Astronomy, (where the application of the red-shift and blue-shift of
             light from the stars has revolutionised our understanding of the
             universe)                                                                          (2)

       6.4.1       The spectral lines (light) from the star are shifted towards longer
                   wavelengths.                                                               (2)

       6.4.2       Decrease                                                                    (1)
                                                                                                  [12]

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 Physical Sciences/P1                      12       Preparatory Examination September 2021
                                            NSC

QUESTION 7

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

NB      DO NOT MARK QUESTION 7.1.2. MAX MARK FOR QUESTION 7 WILL BE 13.
        CONVERT THIS MARK TO 21 USING THE CONVERSION TABLE PROVIDED
        AT THE END OF THESE GUIDELINES
7.1.2        q1q
        F=k       2
                2
              r
                                          -6
                      1,5×10 - 6×1,8×10
        FYZ = 9×109 (                          )
                          (3 x10 - 2 )2

        FYZ = 27 N downwards.

                                          -6
                      2,1×10 - 6×1,8×10
        FXZ = 9×109 (                          )
                                      2
                          (6x10 - 2 )
        FAC = 9,45 N at 60o to the vertical
        But FXZ has two perpendicular components i.e. FXZx and FXZy

        FYZx = FYZSin60o                                   Vector diagram
        FYZx = 9,45Sin60o
        FACx = 8,184 N to the left
                                                                    FXZ
        FYZy = FYZCos60o                                                        (FXZ)y
                                                                          60o
        FYZy = 9,45Cos60o
        FACy = 4,725 N upwards
                                                                 (FXZ)x
        Fnet x = 8,814 N to the left

        Fnet y = 27 – 4,725 = 22,275 N downwards                               FYZ

        (Fnet)2 = (Fnet x)2 + (Fnet y)2

        (Fnet)2 = (8,814)2 + (22,275)2

        Fnet = (8,814 2 + 22,275 2 )

        Fnet = 23,73 N                                                                      (8)

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 Physical Sciences/P1                        13         Preparatory Examination September 2021
                                              NSC

7.2.1   The electric field at a point is the (electrostatic) force experienced per
        unit positive charge placed at that point.                                        (2)

7.2.2

                                  A                 B

 CRITERIA FOR MARKING THE ABOVE ELECTRIC FIELD PATTERN
 Correct direction of field lines                                         
 Shape of the electric field lines (At least 4 lines on each sphere)      
 No field lines crossing each other/No field lines inside the spheres     
                                                                                     (3)

7.2.3        Q1+Q2
        Q=
               2

             5 10  6 (-10 10  6 )
        Q=                               
                            2
        = - 2,5 x 10-6 C (-2,5 μC)
           Q
        E=k 2 
           r

                         2,5×10 - 6
        EAP = 9×109 (               )
                          0,045 2
        EQP = 1,11x 107N.C-1 to the left

                     2,5×10-6
                     9
        EBP = 9×10 (          )
                      0,015 2
        EBP = 1,00x 108 N.C-1 to the left
        Enet = EAP + EBP
        Enet = 1,11 x 107 +1,00 x 108
        Enet = 1,11 x 108 N.C-1                                                      (6)
                                                                                     [21]

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 Physical Sciences/P1                                          Preparatory Examination September 2021
                                                   NSC

QUESTION 8

8.1     (Maximum) energy provided (work done) by a battery per coulomb/unit
        charge passing through it                                                          (2)

8.2.1   OPTION 1                                  OPTION 2
        P = I2R                                  Ps = V3I
        40 = 22R3                                40 = V3(2)
                                                  V3= 20 V
        R3 = 10 Ω                                         v3   20
                                                  ∴ R3=      =     = 10 Ω
                                                           I    2
        1    1   1                            1  1                            𝑅1 𝑅2
           =   +                     𝑅𝑃 = (     + )−1                 𝑅𝑃 =
        𝑅𝑃   𝑅1 𝑅2                            𝑅1 𝑅2                          𝑅1 + 𝑅2
        1         1        1                  1       1 −1                   (10)(10)
        𝑅𝑃
             =(
                  10
                       +
                           10
                                )   𝑅𝑃 = (10 + 10)                  𝑅𝑃 =              
                                                                              10+10

        RP = 5 Ω                     RP = 5 Ω                         RP = 5 Ω

        REXT = RP + Rs               REXT = RP + Rs                   REXT = RP + Rs
        REXT = 5 + 10               REXT = 5 + 10                   REXT = 5 + 10
        REXT = 15 Ω                 REXT = 15 Ω                     REXT = 15 Ω           (5)

8.2.2   POSITIVE MARKING FROM QUESTION 8.2.1
        Ɛ = I(R + r) 
        Ɛ = 2(15 + 0,1) 
        Ɛ = 30,2 V 
                                                                                             (3)

        OPTION 1                                      OPTION      2    (Positive   marking
                                                      from Q8.2.1)
8.2.3   W = P x t                                    W = I2Rt
        W = 40 x 20 x 60                             W = (2)2(10)(20 x 60)
        W = 48000 J                                  W = 48000 J                           (3)
                                                                                             [13]

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 Physical Sciences/P1                          15             Preparatory Examination September 2021
                                                NSC

QUESTION 9

9.1.1          AC (generator)                                                            (1)

9.1.2          Slip rings                                                                (1)

9.1.3          (Faraday’s Law of) Electromagnetic induction                              (1)

9.2.1          The rms value of AC is the DC potential difference which dissipates the
                                                                                          (2)
               same amount of energy as AC.  (2 OR 0)

9.2.2          OPTION 1                              OPTION 2
                         V Max                                        VMax
                                                     Imax                   
               VRMS =                                                 R
                             2

                         311,13                                   311,13
               VRMS =                                                  
                             2                                     807

               VRMS = 220,00 V                                = 0,3855 A

                         2
                        VRMS
               Pave =
                        R
                                                                      Imax V max
                                                     Pave                          
                                                                             2
                             2                              311,13  0,3855
               Pave = 220                           Pave =                 
                       807                                         2
               Pave = 59,98 W                       Pave = 59,97 W                      (5)
9.3.1

   Voltage/V
                                                        OR

                                            time/s

                          CRITERIA FOR MARKING THE ABOVE GRAPH
                          Correct shape                  
                          At least one complete cycle    
                                                                                           (2)

9.3.2           (Commutator) allows the induced current to flow in the same direction /
                in one direction in the external circuit                                (2)
                                                                                          [14]

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Physical Sciences/P1                            Preparatory Examination September 2021
                                       NSC

QUESTION 10

10.1 Threshold frequency, fo, is the minimum frequency of light needed to
     emit electrons from a certain metal surface.  (2 or 0)               (2)

10.2 Q = nqe
      Q= 2,01 x 109 X 1,6 x 10-19
      Q = 3,22 x 10-10 C

           Q
      I=      
           Δt

           3,22×10 -10
      I=               
               1

      I = 3,22 x 10-10 A                                                   (4)

10.3 Decreases
      When the intensity of the light is decreased, the number of photons
                                                                            (2)
      per second will decrease

10.4 E = W o + Kmax     
     E = hfo + Kmax
     2,12x10-18  =(6,63x10-34)(2,21x1015) + Kmax
     Kmax = 6,55 x 10-19 J                                                 (4)

10.5 Decreases
  -
      More energy is used to release the electrons. 
              OR
      Work function is greater.                                             (2)
                                                                            [14]

                                                                    TOTAL: 150

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Physical Sciences/P1             17     Preparatory Examination September 2021
                                  NSC

CONVERSION OF MARKS FOR QUESTION 7

MARK             CONVERTED
OBTAINED         MARK OUT
OUT OF 13        OF 21

     0                  0
     1                  2
     2                  3
     3                  5
     4                  6
     5                  8
     6                 10
     7                 11
     8                 13
     9                 15
     10                16
     11                18
     12                19
     13                21

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