# Physical Sciences Grade 12 P1 KwaZulu Natal Mock 2022 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the KwaZulu Natal Mock Exam 2022, 33 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-science-grade-12-sept-2022-p1-and-memo/efaa3a48-ccac-4aac-a5b8-92b76870fcab
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: KwaZulu Natal Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 33

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### Page 1

LI

4 ? ss Baas Sangh
KYVAQUILUSRIATAL PROVENCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE
GRADE 12

m PHYSICAL SCIENCES P1 (PHYSICS) *
a PREPARATORY EXAMINATION :
i SEPTEMBER 2022 :
VOCS CPO Ce errr reer rer rrrr rr? tt ri
MARKS: 150

TIME: 3 hours

This question paper consists of 15 pages and 3 data sheets.

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Physical Sciences P1

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INSTRUCTIONS AND INFORMATION TO CANDIDATES

1,

10.

1:43

12:

Write your name on the ANSWER BOOK.

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

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

Number the answers correctly according to the numbering system used in this

question paper.
x

Leave ONE line between two sub questions, for example between G
QUESTION 2.1 and QUESTION 2.2. .

y oY
‘ou may use a non-programmable calculator. ss
You may use appropriate mathematical instruments. Ro
You are advised to use the attached DATA woe
Show ALL formulae and substitutions in AL bleulatons
Round off your final numerical soos minimum of TWO decimal places.
Give brief motivations, aoc et cetera where required.
Write neatly and ACP o
w
ay

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Physical Sciences P1

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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.

At

12

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Which ONE of the following physical quantities is a measure of the inertia of a
body?

A Mass.

B Velocity.

Cc Energy.

D Acceleration. RS

The distance between the centres of two planets, P and Q is rTae mass of P is

TWICE that of Q. S
xs

ie

Consider the following tat ents regarding the gravitational force exerted by
the planets. KS)

(i) The fore Wild always be attractive.

(ii) The fo OO ered by P on Q is twice that exerted by Q on P.

(ii) TI ete that the bodies exert on each other is independent of their
mat Ss
h

(iv) TI force exerted by P on Q is equal in magnitude but apposite in
direction to the force exerted by Q on P.

Which statement(s) is/are TRUE?

(i) only

(i) and (ii)
(i) and (iv)
(ii) and (iii)

CoOU>

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Physical Sciences P14

1.3

1.4

1.5

1.6

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A ball is thrown vertically upwards and it returns to the starting point.

Which ONE of the following statements regarding the rate of change of velocity
of the ball is TRUE?

A Its direction for the upward and downward motionsiare different.

B It is always directed downwards during the motion:

Cc It is zero at the maximum height reached.

D It is in the same direction as the velocity of the ball. (2)

A vehicle is moving horizontally at a constant velocity on a frictionless path. The
kinetic energy of the vehicle is K and the magnitude of the momentum is p. The
speed of the vehicle can be given as:

K
A ‘et
Pp
Pp
. K
ec
2p
2K
D par
p (2)
The engine.of &,Car does work, W, to increase its velocity from 0 to v.
The work done by the engine to increase the velocity from v to 2y, is
A W.
B 2w
Cc 3w
D 4w (2)
A taxi sounds it hooter whilst it is moving at a constant velocity away from a
stationary listener. Compared to the sound waves detected by the taxi driver, the
sound waves detected by the observer has a greater ...
A wavelength.
B frequency.
c energy.
D speed. (2)

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1.7 Sketches I, II and Ill, not drawn to scale, are different ways in which three point
charges are arranged. The charges are identical in MAGNITUDE and the
distance ‘d’ between the charges is the same.

®
i iy! .
+q d -q d “q
il ¢ oe 8
-q d +q d -q
q
Ui d
a
+q d -q

In which of these sketches will the magnitude of the NET electrostatic force
experienced by the + q charge be LARGEST and SMALLEST respectively?

LARGEST NET FORCE on +q_ | SMALLEST NET FORCE on +q
| ll
Ul Ut
Hi |

Ul Ul

DO a>

1.8 Three identical resistors X, Y and Z are connected as shown in the circuit
below.

LZ

If the power in X is P, then the power in Y is ...

A 2P
BOY

cP

dp % (2)

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1.9 The graph below shows the relationship between maximum kinetic energy
(Kmax) of the photoelectrons and frequency of incident photons during an
experiment on the photoelectric effect.

A
=
S
5 * >
G - Frequency (Hz)
ZG

v

The point labelled G on the graph represents the ...

threshold wavelength.

maximum frequency.

threshold frequency.

work function. (2)

lekens De 2

1.10 Which ONE of the following graphs best represents the relationship between
the maximum kinetic energy of the emitted photoelectrons and the intensity
of the incident light when light is shown on a metal surface?

A B

4
>
a o
2 oO
© c
5 io)
3 2
3 3
2 — Ui.
= a4 i ‘
me intensity intensity
B D
c £ 3
f
2 o
g 2
= intensity = intensity
(2)
[20]

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QUESTION 2

A 10 kg block is connected to another block of unknown mass ‘m' by a light
inextensible string. The blocks are placed on a rough horizontal surface. When a
force of 45 N is applied to the 10 kg at an angle of 30° to the horizontal, as shown
in the sketch below, the blocks accelerate at 2,5 m's? to the LEFT.

10 kg string m

The block of mass m kg experiences a constant frictional force of magnitude 1.N
whilst moving.

2.1 Draw a labelled free body diagram indicating ALL the forces acting on the

10 kg block while it is accelerating. (5)
2.2 Define the term normal force. (2)
2.3 Calculate the magnitude of the normal force acting on the 10 kg block. (4)

The co-efficient of kinetic friction (yx), between the 10 kg block and the surface
is 0,125.

2.4 Calculate the unknown mass ‘m’. (6)

[17]

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QUESTION 3

A small stone is projected vertically upwards from the ground with a speed of
20 ms“. It passes the roof of an 8 m tall building on its way up and reaches its
maximum height at A. On its way down, the stone passes the roof at point B as
shown in the diagram below. Ignore the effects of air friction.

caer:
"
1
i}
}
SA el = eee TRE
1
I
O
a 8m
'
1

3.1. Write down the acceleration of the stone at point A. (2)

3.2 Using equations of motion, calculate the:

3.2.1 Time it takes for the stone to reach point A (3)
3.2.2 Magnitude of the velocity of the stone at the instant it reaches pointB = (3)

3.2.3. Total time it takes for the stone to move from the ground until it reaches
point B on its way down (4)

3.3. Sketch a velocity — time graph for the motion of the stone from the instant it is
projected-until it reaches point B on its way down.
Indicate the following in your graph:
e Initial velocity
e Time taken to reach maximum height

e The final velocity at point B (4)
[16]

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QUESTION 4

Two friends on wheel chairs are playing on a surface that has NEGLIGIBLE friction.
The combined mass of girl A and her wheelchair is 60 kg whilst that of
girl B and her wheelchair is 85 kg.

Girl A moves to the right at a velocity of 0,60 m-s"! and collides with girl B, who is
stationary. After the collision, girl A and her wheelchair move with a velocity of 0,20
ms" to the left in the same straight line.

BEFORE AFTER

0,20 mst

0,60 ms"
py

4.1 State the principle of conservation of momentum in words (2)

4.2 Calculate the velocity of the friend on wheelchair B immediately after
the collision? (5)

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

AE

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QUESTION 5

A block of mass 2 kg was dropped from point X and reaches point Y. Neglect friction.

2kg
| xX
5m

ie

5.1. State the principle of conservation of mechanical energy in words. (2)

5.2 Use the principle of conservation of mechanical energy to calculate the
speed of the block when it reaches point Y. (4)

The same block was then released from X again, but travelled along a path XZ
which is frictionless, aS shown below

2 kg _X
if
5m
5.3 State the work — energy theorem in words (2)

5.4 Use the work — energy theorem to calculate the speed of block when it
reaches point Z. (4)

5.5 How should the answers of QUESTIONS 5.2 and 5.4 compare according to
physics principles?
Choose from: EQUAL or UNEQUAL
Explain the answer. (3)

[15]

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QUESTION 6

A policeman with a speed detector is standing on a straight section of a road where

the speed limit is 80 km-h’. The hooter of a car moving on this section of the road
with a constant UNKNOWN speed emits sound waves of frequency 400 Hz.

The wavelength of the sound waves detected by the police man is 0,80 m.
The speed of sound in air is 340 ms.

6.1. State the Doppler Effect in words. (2)
6.2 Calculate the frequency of the sound wave detected by the policeman. (3)

6.3. Determine, with the aid of a suitable calculation, whether the car is
EXCEEDING the speed limit or not. (5)

6.4 How would your answer to QUESTION 6.2 be affected if the car travels
at a LOWER constant speed? Choose from GREATER THAN, LESS THAN

or THE SAME AS. (1)
6.5 State two uses of the Doppler effect in medicine (2)
[13]

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QUESTION 7

In FIGURE 1 below, two IDENTICAL, negative point charges are placed a distance X
metres apart. The charges exert an electrostatic force of magnitude 1984 N on each
other.

xX
+ : >
@ oes tebete hese veecceus -e
-Q -Q
Figure 1

In FIGURE 2, the charges are separated a further 0,8 m. The electrostatic force the
charges now exert on each other is of magnitude 124 N.

. (X + 0,8) m
+¢ >
Sih soe 8 tie: peca tno eit Rn pose aNd wee ees as ee aS ae e
-Q -Q
Figure 2
7.1. State Coulomb’s Law in words. (2)
7.2 Calculate the distance X. (4)
7.3. Calculate the magnitude of the charges. (3)

7.4 The charges are returned to their original positions as in FIGURE 1
(X metres apart). A proton is now placed at point P which is 0,4 m to the
left of -Q as shown below.

l+0,4 me aan) >|

625 ee eee seat nas -e

Pp -Q -Q
7.4.1 Define electric field at a point in words. (2)
Calculate the magnitude of the:
7.4.2 NET electric field at point P (5)
74.3 NET force experienced by the proton at P (3)

[19]

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Physical Sciences P4

QUESTION 8

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8.1 Abattery is connected to two IDENTICAL resistors, R, in two separate experiments

as shown below.

EXPERIMENT 1

O
R
R

LY

EXPERIMENT 2

©)

The internal resistance and emf of the battery are UNKNOWN. The ammeter

and voltmeter readings in both experiments are shown in the table below.

AMMETER READING (A)

VOLTMETER READING (V)

EXPERIMENT 1

6,13

18,40

EXPERIMENT 2

1,78

20,58

8.1.1 Define emf in words.

8.1.2 Calculate the emf of the battery

8.1.3 Calculate the resistance of R

8.1.4 One of the resistors in EXPERIMENT 1 is now disconnected.
How will this affect the power dissipated by the battery?

Choose from: INCREASE, DECREASE or REMAIN THE SAME.

Explain the answer by making reference to a suitable formula for

power.

(4)

8.2 Accurrent of 4,4 A passes through an oven of resistance 50 Q. Calculate the
cost of electricity consumed by the oven for 6 hours if the price of electricity

is R1,69 per kWh.

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QUESTION 9
9.1 The sketch below represents a simplified version of a DC electric motor.

coil

9.1.1. Name the component that ensures that the coil (armature) rotates

continuously in the SAME direction. (1)
9.1.2 In which direction does the coil rotate? Choose from: CLOCKWISE

or ANTICLOCKWISE. (2)
9.1.3 Write down the energy conversion that occurs in a motor. (2)

9.2 The above motor is CONVERTED to operate as an ac generator and the
output of the generator is shown in the sketch graph below.

na et a

7 ie t(s)

9.2.1 Write down TWO changes that need to be made for the above

conversion. (2)
9.2.2 Calculate the rms voltage. (3)
[10]

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QUESTION 10

The work functions of two metals are listed in the table below.

Work function
Metal
(J)
Sodium 2,46 x 10°18
Platinum 6,35 x 10°18
10.1. Define work function. (2)

The sodium surface is illuminated with light having a wavelength of 400 x 10°m.
10.2 Calculate the energy of a photon of the incident light. (3)
10.3 Calculate the maximum speed of the ejected electrons. (4)

10.4 The same light is now incident on a platinum metal.

Will electrons be ejected from the platinum metal?
Choose from: YES or NO

Write down a reason for your answer (2)

[11]
TOTAL: 150

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September 2022 Preparatory Examination

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 9,8 m-s?
Swaartekragversnelling
Universal gravitational constant G 6,67 x 10°11 N-m2-kg?
Universele gravitasiekonstante
Speed of light in a vacuum c 3,0 x 108 m-s*
Spoed van lig in 'n vakuum
Planck's constant h 6,63 x 10°34 J's
Planck se konstante
Coulomb’s constant k 9,0 x 109N-m?-C2
Coulomb se konstante
Charge on electron e -1,6 x 10°79C
Lading op electron
Electron mass Me 9,11 x 10°31 kg
Elektronmassa
Mass of Earth M 5,98 x 10% kg
Massa van Aarde
Radius of Earth Re 6,38 x 10°m
Radius van Aarde

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TABLE 2: FORMULAE / TABEL 2: FORMULES

MOTION / BEWEGING

Vp=vjitadt

Ax =v At+ 4adt? or/of Ay = v,At+ taat?

v, =v; +2aAxorlofv,” =v,’ + 2aAy

ax-(% = atorotay -(“ ~ Jas

FORCE / KRAG
Fret =ma p=mv
fs(max) = UsN fk = UkN
F_,At=Ap w=mg
Ap=mv, —mv,
Gm,mz GM
re 2 ae

WORK, ENERGY AND POWER / ARBEID, ENERGIE EN DRYWING

W=FAxcos 0 U=mgh or/ofE, =mgh
= A i
K= i mv? eiore. = pm? Whret = AK or/of Wret = AE
AK = Kr—- Ki or/of AEk = Ext - Eki
Whe = AK + AU or/of Whe = AEk + AEp p=

Pay = F-Vav 1 Pgem = F-Vgem

WAVES, SOUND AND LIGHT / GOLWE, KLANK EN LIG

v=fr

Te!
f

vty,

viv,

f.

s

f=

E=hf orlof E=h=

E = Wo + Eximay or/of E = Wo + Kimax)

where/waar

E=hf and/en Wo=hfo and/en Ekimax = i mVivex orlof  Kimax) = 5 MVitax

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ELECTROSTATICS / ELEKTROSTATIKA

kQ4Q2 kQ

Fa sci2 bat

r2 r2

vow pak

q q
n=2 OR/IOF n=2
e€ qe

ELECTRIC CIRCUITS / ELEKTRIESE STROOMBANE

i Vv emf (e) =I (R +15)
rl emk (e) =(R +n)
R, =R, +R, +
4 4 4 q=IAt
eS tot...
R, R, R,
W=V
i _w
W=VIAt At
P=VI
= |2
W=PRAt P=PR
2
_ Vat pov
We = R

ALTERNATING CURRENT / WISSELSTROOM

Pave =Vimslams / Premiddeta = V wok Tek

I I

Lao / Two = maks.
rms wa
V2 v2 Paglia! f° -Pactiniad= le R
AVA Vv,
Vins =F 1 Vg = SEE Vi Vk
V2 m V2 Pies =R 1 Pyemiddeld = =

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

GRADE 12

MARKS: 150

This marking guideline consists of 15 pages.

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Physica se iloa ey” from S tanmere. phySies.£O. ryperember 2022 Preparatory Examination

QUESTION 1
1. Avy (2)
12 Cvv (2)
13. BYv (2)
14. Dvv (2)
(2)
(2)
L7 Dvv (2)
18 BYY (2)
19 Dvv (2)
110 BY (2)
[20]

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QUESTION 2
21
Fapp.v
Fapplied Formal we Frormal¥
v v
F OR Fapp.h uy
Feravityl~ Feravityl~
nocert the fokowing symbols If components of Fapp and
Normal : $
Fapplied | 45,0N/F/Fa/Fapp Fapp oe eran
T Frension/Tensional force LAU
Fx f</f/friction/Fr
Ww. Fg/mg/98N
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

6)

2.2 Normal is the force or the component of a force which a surface exerts on
an object with which it is in contact, and which is perpendicular to the
surface. Vv (2)

NOTE:

If any one of the underlined key words in the correct context is omitted deduct 1 mark.

2.3 Fnet=mav
Fv +N—Fyg=0

45sin30° + N — 10(9,8) v = OY
N = 75,50 NY (4)

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POSITIVE MARKING FROM QUESTION 2.3
OPTION 1

CONSIDER 10 kg BLOCK
TAKING LEFT TO BE POSITIVE
v
Fret = ma
Fu -T-—Fr =ma
45cos 30°Y — T— 0,125(75,50) “ = 10(2,5)”

T=4,53 N
CONSIDER BLOCK m

Fret = ma

T - Fr = m(2,5)
4,53 —1¥ = m(2,5)
m= 1,41 kg¥

OPTION 2

CONSIDER 10 kg BLOCK
TAKING LEFT TO BE NEGATIVE

Fnet = ma a Me

-Fi+ T + Fa =ma
-45cos 30°V + T + 0,125(75,50) Y = -10(2,5)”
T=-4,53 N

CONSIDER BLOCK m

Fret = ma

T - Fix = m(2,5)

- 4,53+1v = m(-2,5)
m = 1,41 kgv¥

OPTION 3
If masses added then max 3/6
e¢ Fret =ma¥

e 45cos 30° v
° 141kgv

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m

(6)
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(2)

(3)

QUESTION 3

3.1 9,8 ms? ¥ downwards ¥

3.2.1 | (downward positive) (upwards positive)
v= vit alt ¥ v= vit adt ¥
O = (-20) + 9,8At v 0 = 20 + (-9,8)Atv
At = 2,04s Vv At = 2,04 sv

3.2.2 OPTION1

(downward positive)

(upwards positive)

vit= vit+ 2aAy ¥

vee = (-20)? +2 x (9,8)(-8) Y
vi = 15,60 m-s! ¥

vit= vie+ 2aAy ¥

ve? = (20)? +2 x (-9,8)(8) “
vi= 15,60 m-st ¥

OPTION 2

Considering motion from maximum height

(downward positive)

(upwards positive)

vie= viet Zadys
0 =20* + 2(9,8) Ay
Ay = 20,44) m

vit= vit+ 2aAy
0 = 20? + 2(-9,8) Ay
Ay = 20,41 m

On its way down Ay = 20,41 — 8 = 12,41

m

vit= vie+ 2ady ¥
ve= 0+ 2(9,8)( 12,41) ¥
Vi = 15,60 m-s? ¥

OPTION 3

vit= vit+ 2aAy ¥
ve= O+ 2(-9,8)(-12,41) ¥
vi = 15,60 mst ¥

Considering motion from ground to the roof

(downward positive)

(upwards positive)

Ay = vi At + % aAt®
8 = 20 At + % (-9,8)At2
At= 0,45s

v= vit alt ¥

Vi_=_(-20) + 9,8(0,45) ¥
vi = 15,60 m-st ¥

Ay = vi At + % aAt?
-8 =- 20 At+ % (9,8)At2
At= 0,45s

vr=vit alt ¥

v= (-20) + 9,8(0,45) W
vr = 15,60 m-st ¥

(3)

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3.2.3

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OPTION 1
(downward positive) (upwards positive)
Ay = vi At + % alt? ¥ Ay = vi At + % alt? ¥
-8 v =-20 At + % (9,8)(At)? ¥ 8 v= 20 At + % (-9,8) (At)?v
At = 3,63s or At = 0,45s At = 3,63s or At = 0,45s
At = 3,63s ¥ At = 3,63s ¥
OPTION 2:

POSITIVE MARKING FROM QUESTION 3.2.2

(downward positive)

(upwards positive)

Time taken to reach maximum height
from the ground.
v= vit aAt ¥

0 = (-20) + 9,8At ¥
At = 2,04 s

From the maximum height to the top of
the building

v= vit aAt

15,60 = 0 + 9,8Atv

At=1,59s

The total time from the ground to the top
of the building:
Attota = 2,04 +1,59 = 3,63 s ¥

Time taken to reach maximum
height from the ground.
v= vit alt ¥

O = 20 + (-9,8)AtY
At =2,04s

From the maximum height to the top
of the building

v= vit alt

-15,60 = 0 - 9,8Atv”

At=1,59s

The total time from the ground to the
top of the building:
Attotal = 2,04 +1,59 = 3,63 s ¥

OPTION 3:

POSITIVE MARKING FROM QUESTION 3.2.2

(downward positive)

(upwards positive)

w= vit alt ¥
15,.60V =-20+9,8At v
At=3,63sv¥

v= vit alt YW
-15,60Y = 20 +(-9,8)At ¥
At =3,63s v

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3.3. POSITIVE MARKING FROM QUESTION 3.2.1 AND 3.2.2

(upwards positive) (downward positive)
4
15,60
20 Nis | EEA AAR AARC AAAS
é E
= = : —
3 0 . 8 (e) ey 3,63 t(s)
g t(s) $
-15,60 ae
v
v
Criteria for marking the graph Marks
Correct shape (straight line) having a positive and negative velocities with line | “
longer before 2,04 s than after 2,04 s
Graph starts at v = 20 m:s/-20 mst andt=0s v
Graph cuts t-axis at 2,04 s atv = 0 ms-1 v
Graph shows the v = -15,60m-s7/15,60 m:stat t = 3,63 s v
(4)
[16]

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September 2022 Preparatory Examination

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QUESTION 4

4.1 The total linear momentum of an isolated system remains constant (is

conserved). VY = (2)
4.2  OPTION1 OPTION 2
Taking right to be positive Taking left to be positive

Zpi =Zpr iW Zpi =Zpr by
Mavia + MBViB = Mavra + MBVsB. Mavia + MBViB = MavtA + MBVsB
60 (0,60)+85 (0) v= 60(-0,20)+85 60 (-0,60) + 85 (0) Y = 60 (0,20) + 85

vey vie
vig = 0,57 ms V to the right¥ vig = - 0,57 mst
Vig = 0,57 ms* v to the right¥ (5)

4.3 POSITIVE MARKING FROM QUESTION 4.2

EKIE A maAvA? + ama vee EKf = Ye ma via? + Yom vB?
= ¥% (60)(0,6)2 +0V = ¥% (60)(0,2)2 + 4(85)(0,57)2”
= 10,80 J = 15,01 J

XKr# ZKi¥ OR Total Ek before # Total Ek after
-. Collision is Inelastic “

NOTE:

If learners starts with 2Kr = ZKi, then minus one mark.

(6)
[12]

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QUESTION 5

5.1 The total mechanical energy (sum of gravitational potential energy and kinetic
energy) in an isolated system remains constant. Vv“

~
5.2  Ema= Ene

(K +U)x = (K+U)y

(% mv? +mgh)x = “amv? +mgh)y
0,5)(2)(0) + (2)(9,8)(5) W = 0,5(2)v? +0”
v=9,90 mstv

5.3 The net work done on an object is equal to the change in the object's kinetic
energy VV

OR

The work done on an object by a net force is equal to the change in the object's

kinetic energy. vv

NOTE:

If any one of the underlined key words in the correct context is omitted deduct
1 mark.

5.4 OPTION1 OPTION 3
Whet = AKY Whe = AEp + AEk ¥
mgsin30°(—>—) v= 1A(2v2_- OV OY = 2x 9,8 (0-5) + #(2)\v2 -OY
v=9,90 ms!v v=9,90 mslv
OPTION 2
Whet = AKY
mgcos300°( = ) v= ~(2)v?_- OY

sin30°
v=9,90 ms!v

5.5 Equal ¥
Work done by the conservative force does not depend on the path taken. VW”
OR: gravitational force is the only force doing work on the block. It is a
conservative forcev Vv

September 2022 Preparatory Examination

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(4)

(2)

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[15]

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QUESTION 6

6.1 Doppler effect as the (apparent) change in frequency (or pitch) of the sound
detected by a listener, because the sound source and the listener have different

velocities relative to the medium of sound propagation. vv (2)
62 v=fAv

340 = £(0,80) Vv

f= 425 Hzv (3)

6.3. POSITIVE MARKING FROM QUESTION 6.2

viv
i=

t. fv
vtv,

340 v
(340 - v,) “
Vs = 20 mst
2 Vs = 72 kmh
The car was not exceeding the speed limit. W

NOTE:
Statement must be marked in relation to answer. (5)
6.4 Less than’ (1)

6.5 Blood flow meter”
Detecting foetal heartbeatv (2)

[13]

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QUESTION 7

“A

7.2

7.3

September 2022 Preparatory Examination

m

The magnitude of the electrostatic force exerted by one point charge (Q1) on
another point charge (Qz) is directly proportional to the product of the

magnitudes of the charges and inversely proportional to the square of the

distance (r) between them vv

(2)

NOTE: -_ =

If any one of the underlined key words in the correct context is omitted deduct

1 mark.

p — KQi1Q2 y
r2

POSITIVE MARKING FROM QUESTION 7.2

OPTION 1

p= kQiQ2 v
r2

9x 10° Q*
984 = 0277

Q = 1,27 x104 CY

OPTION 2

i kQiQ2 v
r2

9 2
a4 = 9x 10° Q’ :
(0,27 +08

Q=1,26 x 104 Cv

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

NOTE:

If any one of the underlined key words in the correct context is omitted deduct
1 mark.

POSITIVE MARKING FROM QUESTION 7.3

TAZ eo kOy
r2

v
X10°)(1,26x10*
2 am 7 7087500 N-C* to the right

9x10°)(1,26x10%) v
Eop = a = 2526175,095 N:C* to the right

1P.

Enet = E1p + Ep
= 7087500 + ¥2526175,095 NOTE: The mark is for the addition
= 9,61x10 N-C* to the right “ (5)

7.4.3, OPTION 1:

POSITIVE MARKING FROM QUESTION 7.4.2

E-Fv
q
9616375,10 = ——. v
nm 1,6x1079

F = 1,54 x 10°? N to the rightY

OPTION 2:

POSITIVE MARKING FROM QUESTION 7.3

F — KQ1Q2
2

_ k(1,26 x10%)(1,6 x10") yf
ead (0,4)
= 1,134 x 107? N to the right
k(1,26 x10%)(1,6 x10?)
Fas appease
(0,67)
= 4,04 x 10 ?° N to the right
Fret = Fai + Fq2 = 1,134 x 10°! + 4,04 x 10 “8
= 1,54 x 10°1? N to the rightY (3)
[19]

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Physica se iloa ey” from S tanmere. phiysies.com 2022 Preparatory Examination

QUESTION 8

8.1.1 Emf is the maximum energy provided by a battery per unit charge passing

through it. vv (2)
NOTE:
If any one of the underlined key words in the correct context is omitted deduct
1 mark.
8.1.2 | EXPERIMENT 2 EXPERIMENT 1
E=1(R+)NY E=1(R+r)
€ = 20,58 + 1,78r v------------ (1) € = 18,40 + 6,13rv’ --------- (2)
r=0,500
€ = 20,58 + 1,78(0,5) OR € = 18,40 + 6,13(0,5) ¥
€ = 21,47 VW (5)

8.1.3 POSITIVE MARKING FROM QUESTION 8.1.2

OPTION 1 OPTION 2
EXPERIMENT 2 EXPERIMENT 1
car ery E=I1(R+n)nv
21,47 = 1,78(R+0,5) ¥ 21,47 = 6,13(R+0,5) ¥
R = 17;56°0 R=30
R, =R, +R; +... erg iy
R, 1 R,
11,56 = Ri+R2v 1 de,
— = + Vv
2,99 RR
R=5,78 OY R=6QY
OPTION 3 OPTION 4
V = IRext¥ V = IRext¥
20,58 = 1,78.2RY 18,40 = 6,13 R/2vV
R=5,78 OY R=6 Ov (3)

8.1.4 DecreaseY
If Riis removed in circuit X, the total resistance will increase V. For a constant
emfv, from P= V7/Rv,, power will decrease. (4)

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8.2 OPTION1

kWh = PAt
1000
_ | Rat
1000
_ (44°)(50)(6) |
1000
= 5,808 units

Cost of electricity = 1,69 x 5,808 “
= R9,82v

OPTION 2
V=IR

=4,4x 50
=220 V

1000

1000
220°
Cea?

= —90_y
1000

= 5,808 units

Cost of electricity = 1,69 x 5,808 v
= R9,82v

OPTION 3

kwh = Pat
1000

- ViAt

~ 1000

_ (220)(4,4)(6) |
1000

=5,808 units

Cost of electricity = 1,69 x 5,808 “”
= R9,82Y

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September 2022 Preparatory Examination

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Physica se iloa ey” from S tanmere. phiysies.com 2022 Preparatory Examination

QUESTION 9

9.1.1 Split rings (commutator) (1)
9.1.2 Anticlockwisev ¥ (2)
9.1.3. Electrical to mechanicalv ¥ (2)

9.2.1 ANY TWO
The DC power supply can replaced by a load / resistor. “
The split rings (commutator) is replaced by the slip rings. ”

The coil should be rotated. (2)
9.2.2 V,
Vins = pe
v2
31113
Vio= v
rms J2
Vrms = 220 VY (3)
[10]
QUESTION 10
10.1 Work function issthe»minimum energy that an electron in the metal needs to be
emitted from the meta! surface. vv (2)
ee ‘
10.2 he “ as
ES OR c=Af E=hf
- 34) 8
= es ie Le, 3x 10°=400x10°V =66x10%.7,5x 10"
400x10
E=4,97 x10 Jv f=7,5 x 10%Hz = 4,97 x 10° Jv (3)
10.3. POSITIVE MARKING FROM QUESTION 10.2
1 2
E=Wo + =Mviawx %
2
al
4,97 x 1079 v= 2,46 x 1079 v+ 5 (QAIK10 Vr
Vimax) = 742322,57 ms lv (4)
10.4 Nov
The work function of platinum is higher than the energy of the photon” (2)
[11]
TOTAL: 150

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