# Physical Sciences Grade 12 P1 Western Cape Mock 2025 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Western Cape Mock Exam 2025, 15 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physics-grade-12-nsc-p1-memo-september-2025-western-cape/fe333498-f7d9-46f2-9550-3d347624c06a
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Western Cape Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 15

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## About this document

Physical Sciences · Grade 12 · Western Cape Mock Exam · 2025. Memorandum, 15 pages. Read online or download the PDF.

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## Document text

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### Page 1

Western Cape
Government

FOR YOU

Education

| CAPE WINELANDS EDUCATION DISTRICT |

PHYSICAL SCIENCES PAPER 1
MARKING GUIDELINE
GRADE 12

fi N

COMMON TRIAL EXAMINATION
SEPTEMBER 2025

MARKS: 150
TIME: 3 hours

This exam paper consists of 14 pages.

### Page 2

QUESTION 1/ VRAAG 1

4.1

1.2

1.3

14

1.5

1.6

17

1.8

1.9

4.10

D

D

vv

vv

vv

vv

Vv

vv

Vv

### Page 3

QUESTION 2/ VRAAG 2

24

2.2

2.3

When a net force acts on an object, the object will accelerate in the direction of the

force and the acceleration is directl roportional_to_the force and_inversel
proportional to the mass of the object. vv

(Deduct one mark if any of the underlined phrases are omitted.)

Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, versnel die voorwerp in
die rigting van die krag teen 'n versnelling direk eweredig aan die krag en omgekeerd

eweredig aan die massa van die voorwerp.

(Trek een punt af indien enige van die onderstreepte frases weggelaat word.) (2)

Note: 1 mark for each arrow WITH its label, deduct 1 mark for every extra force
included.

Let wel: 1 punt vir elke pyl MET sy byskrif, trek 1 punt af vir elke ekstra krag wat
ingesluit is.

AAA (5)

FOS pW

= [,mgcos8, Vv mark for either

(0, 2)(2)(9,8)(cos40°) v

= 3,0Ndownthe slope v (3)

### Page 4

2.4 POSITIVE MARKING FROM QUESTION 2.3/POSITIEWE NASIEN VANAF VRAAG 2.3:

2a:
F =ma v

net

TF, — Fg, = ma

T — F, — mgsin® = ma

5kg:

F =ma
net

F,—F,—Tcos® = ma

(70 — 17,15 — Tcos20°)¥ = 5a

(T — 3 — (2x9, 8xsin40’))¥ = 2a¥ (forboth2a& | 52,85 — Tcos20° = 5a ...eq2
5a)
T= 2@ + 15,599 ..eq 1
Substitute equation 1 into equation 2:
52,85 — (2a + 15,599)cos20° = Sav
38,19 = (5 + 2cos20°)a
a=5,55ms Vv (6)
[16]

QUESTION 3/ VRAAG 3

3.1 An object which has been given an initial velocity and then it moves under the

influence of the gravitational force only. WY (2 or 0)

'n Voorwerp waaraan 'n beginsnelheid gegee is en wat dan slegs onder die

invioed van die gravitasiekrag beweeg. (2 of 0)
3.2 Downwards as positive Upwards as positive
Ay = v At + tadt® Vv Ay = vdt + tant’ Vv
2 1 2
1,8 =v,0,5)+>Q9,8)0,5)¥ - 18 =v(0,5)+>(- 9,8)(0,5)v
=1 ~
val 15 ms v v =~ 1,15 mrs '
sspeed = 1,15 ms! Vv
3.3 ve = ve + 2aAy Vv vy = ve + 2ady Vv
0 =v + 209,8)(- 0,9) v O=v + 2— 9,8)(0,9)v

v.= 17,64 v.= 17,64

i i

### Page 5

v= 4,2 ms ‘upwards Vv v= 4,2 m-s ' upwards ¥ (3)

3.4 v=, t adt Vv v =v, + adt Vv
0 =— 4,2 + 9,84 v +MARKING 0=424¢ 9,8)At Vv +MARKING
At = 0,435 At = 0,43s
t=0,5+ 200,43) v t=0,5 + 200,43) v
t=1,36s v (4)
t= 1,36s v
3.5 r" 4
4,2
a ‘ : =
pits : / ty) 228
™ op/ 1,36 :
-4,2|--------
4 v

Criteria for marking/ Nasienkriteria:
e Lines are correctly drawn (parallel)/ Lyne is korrek getrek (parallel) W
e Initial velocity correctly indicated/ Aanvangssnelheid korrek aangedui “
e Velocity of the ball as it leaves the ground./ Snelheid van die bal soos dit
die grond verlaat. ¥
e Time when the ball strikes the ground after the first bounce (0,5 s)/

Tyd wanneer die bal die grond tref na die eerste bons (0,5 s).” (4)
[16]

### Page 6

QUESTION 4/ VRAAG 4

4.1

4.2

The total linear momentum of an isolated systemW_remains constant / is conserved

Vv

Die totale lineére momentum in 'n geisoleerde sisteem “bly konstant / behoue. ¥

OR

Accept: In an isolated system, ¥ the total linear momentum before collision is

equal to the total linear momentum after collision Vv

Aanvaar: In 'n geisoleerde stelsel, Y is die totale lineére momentum voor botsing
gelyk aan die totale lineére momentum na botsing. ¥

(2)

Right as +:

Z Pinitiasy = Z Prinay ~ OR

MyVin + M2Vi2 = (M, + My)v_

(5)(4) + (2)-1)v =(6 +2) (vw
“V,= 2,57 ms" Vv

Left as +:

E Piinitiay = Z Pinar v

MiVin + MoVin = (M, + Ma)Vy

(5)(-4) + (2)(1)v =(+2)(y)¥
OV =-2,57ms'¥

If no subscripts:mv + mv = mv + mv

max. 3/4

4.3 POSITIVE MARKING FROM 4.2/ POSITIEF NASIEN VANAF 4.2

OPTION 1:

F At =mAv vW or F At = Ap
net net

F (0, 3)¥ = (7) — 2,57)¥

F = —59,97N

net

magnitude of Fe 59,97 NW

OPTION 2:
0, =, + adt

= 2,57 + a(0,3)v

a 59,99 N
magnitude of Eni = 59,99NY
Range/gebied: 59,97 N to/tot 60,20 N

(4)

### Page 7

QUESTION 5/ VRAAG 5

(4)
[10]

5.1.1. The total mechanical energy in an isolated systemv is conserved/remains constant. ¥

Die totale meganiese energie in 'n geisoleerde stelsel¥ word behoue/bly konstant. Y (2)

5.1.2

OPTION 1:

E wcinitiat =E al

M(final)

mv,” + m(9,8)(1,3) ¥ = m(9,8)(5)¥ +0

vy = 852ms! upwards/opwaarts Vv

i

OPTION 2:

net

m(9, 8)(3, 7)cos180° ¥ = 0 — mv’ v

v = 852m. s- upwards/opwaarts ¥

QPTION:3:
Wre = AE, + AB,
= mv? - mv) + (mgh;, - mgh))

=$m(0) - S mv,” v+ m(9,8)(5) - M(9,8)(1,3)¥

B= 8,52 ms upwards/opwaarts W

i

5.1.3 DECREASE / AFNEEM v

Friction will remove energy from the system.v /

Wrywing sal energie uit die stelsel verwyder.
OR/OF

Some of the object's kinetic energy would be converted into heat and sound due to

friction, leaving less energy available to convert into gravitational potential energy./

Van die voorwerp se kinetiese energie sal as gevolg van wrywing in hitte en klank

omgeskakel word, wat minder energie beskikbaar laat om in gravitasiepotensiéle

energie om te skakel.
OR/OF

### Page 8

Friction opposes motion, causing the object to lose energy more quickly as it moves

upward. /
Wrywing staan beweging teé, wat veroorsaak dat die voorwerp vinniger energie

verloor soos dit opwaarts beweeg.

OR/OF

With less energy available to overcome gravity, the object would not rise as high,
resulting in a lower maximum height. /
Met minder energie beskikbaar om gravitasiekrag te oorkom, sal die voorwerp nie so
hoog styg nie, wat 'n laer maksimum hoogte tot gevolg sal hé.

OR/OF
Increased F.. downwards/increased acceleration downwards/loses velocity faster,
resulting in a lower maximum height./

Fret afwaarts neem toe/versnelling afwaarts neem toe/verloor snelheid vinniger, wat lei tot 'n
laer maksimum hoogte. (2)

5.2.1 Non-conservative/ Nie-konserwatiewe krag ¥
It is a force for which the work done in moving an object between two points depends

on the path taken. ¥
Dit is 'n krag waarvoor die arbeid verrig om 'n voorwerp tussen twee punte te beweeg,

afhanklik is van die roete wat gevolg word. (2)

5.2.2 W = AE, + AE LV
ne K P

(4, 493)" — 4, 4)(0,6)) ¥ + (1,4)(9,8)0 - 1,5) ¥

friction
friction = — 14,532]
friction = F -rictionA*C0S®
— 14,532 = F__. (1, 8)¥cos180° v
friction
_ = 807NY
friction

w= AE.V

net

### Page 9

adhe _ dopa
We +W,, =ymv, ym,
F Axcos180 + mgAxsinBbcos0 = zm, = mv;

= B07NV

friction
>

uv, = v + 2aax
(3)W = (0,6) + 2a(1,8) ¥
2

a=2,4ms_

F =mav
net
F = A924av
F = 3,36N
net
Ess = H, parallel ~ F,
F =F sind —F
net g f
3,36 = (1,49, 8)GSv - F,
3,36 = 11,43 - F,
F,=8,07N ¥ (6)
[16]

QUESTION 6/ VRAAG 6

6.1

The apparent change in frequency (or pitch)“ detected by a listener because the source and

the listener have different velocities relative to the medium of propagation. ¥

Die skynbare verandering in frekwensie (of toonhoogte) wat deur 'n luisteraar waargeneem
word omdat die bron en die luisteraar verskillende snelhede het relatief tot die

voortplantingsmedium.
OR/OF

An apparent change in detected frequencyY as a result of the relative motion between
asource and a listener. ¥

'n Skynbare verandering in waargenome frekwensie as gevolg van die relatiewe beweging

tussen 'n bron en 'n luisteraar. (2)

62 v=fa Vv

340 = f(0.013)  v
f = 26153,85Hz Vv (3)

### Page 10

viv,

63 f==f v

vtv 7s

340
f= (261 53,85)¥ positive marking from Q6.2

f, = (340 + v_, )(76, 92308824)

vtv,
f, = viv, f
340
31500¥ = [ai |(s40 + v,,.)(76, 92308824) ¥

115600+340v
car
340—v
coi

i}

409, 4999398

5

139229,9795 — 409, 4999398v = 115600 + 340v_
23629, 97953 = 749, 49993980

v= 31,53ms VY OR/OF v= 113,51km.h \_ (Note: 100km.h | = 27,78m.s ’)
.. The car is speeding./Die kar oorskry die spoedgrens Varese)
(6)
[11]
QUESTION 7/ VRAAG 7

7.1.1. The magnitude of the electrostatic force exerted by one point charge on another point
charge is directly proportional to the product of the magnitude of the charges and
Die grootte van die elektrostatiese_kraq wat deur een puntlading op 'n ander

puntlading uitgeoefen word, is direk eweredig aan die produk van die grootte van die
ladings en omgekeerd eweredi n die kwadraat van die afstand tussen hull

Deduct one mark if any of the underlined phrases are omitted./Trek een punt af indien

enige van die onderstreepte frases weggelaat word. (2)
71.2 F=-G Vv If Q, is written as a negative
" 9. ~9 charge, no mark for final answer.
-7 — (9x10°)(3,8 x10 )Q ; ; ‘
8,54x 10 © = ————_—+ Vv Indien Q, as 'n negatiewe lading
(0,4) geskryf is, geen punt vir die finale
antwoord nie

Q, =4x10° Cv (3)

10

### Page 11

7.2.1 The electric field at a point is the electrostatic force experienced per unit positive charge

placed at that point. vw

Die elektriese veld by 'n punt is die elektrostatiese krag wat per eenheidspositiewe-

lading wat by daardie punt geplaas is, ondervind word. (2 of 0) (2)
Te
Right as positive Left as positive
kQ,
kQ, kQ, E k 2
= +—+v
E et = 2 Ba 2 ov net Be e
9. —6, 9. -6.
_ (oxdo’yeext0™ Hit (9x10°}(3x10 7°) =- (fe 16x10 yy fee XSx10 J
(ay ay ™ "
Vv v
= 3375 — 27000 = — 3375 + 27000
—1 a
= 23 625 N.C to the left/ na links v = 23625N.C to the left/ na links¥

7.2.3 POSITIVE MARKING FROM Q7.2.2/POSITIEF NASIEN VANAF V7.2.2

OPTION 1/OPSIE 1
E= = Vv OR/OF F = Eq
F = (1,6 x 10 )(23 625)¥
= 3,78 x 10 °N
F
a= —
— 3.78 x10"
1,67x10

a=2,26x10'ms Vv

(4)

OPTION 2/OPSIE 2

= I | #2,
net ? r
F = (9x10°)(3x10 (6x10) _(9x10°)(6x10*)(,6x10-")
net ay’ ay
v
F  =3,78x10 N
net
F =ma

net
3,78 x10 = (1,67 x 10 Wa
a = 2,26x10'ms Vv

Marking Criteria:
e Correct formula
Calculate force
Substitute mass in F = ma
Correct answer

Nasienkriteria:
e Korrekte formule
e Bereken krag
e Vervang massa in F = ma
e Korrekte antwoord

7.2.4 Greater than/Groter as v

(4)
(1)

[16]

A:

### Page 12

QUESTION 8/ VRAAG 8

8.1.1

8.2.1

8.2.2

The potential difference across a conductor is directly proportional to the current Y

in the conductor at constant temperature. Y
Deduct one mark if any of the underlined phrases are omitted.

Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by
konstante temperatuur.

Trek een punt af indien enige van die onderstreepte frases weggelaat word. (2)
Diagram 1: Diagram 2:
R=ty Vi =emf —V,
yo
é =6~4
wid =2Vv
= 3.0
v,
i= 7
= 4 Vv
= 0,67A
v
Ry = a
4
=a 4
(6)
12,6V¥ (1)
POSITIVE MARKING FROM Q8.2.1/POSITIEF NASIEN VANAF V8. 2.1

OPTION 1/OPSIE 1

e=1(R+ nV

12,6¥ =1(5 + 0,08)” 1 = 2,484
= IR

ext
v= (2,48)(5)¥
V=12,4Vv
ext

12

### Page 13

V

R R
=~ xV iv OR/OF V, = z

x
1 Ri+R, total Ri+R, total
Sv
Se OS:
Ve aa lbn 5+0,0B¥ 12, 6W
= 12,4Vv (5)

headlamps

8.2.3 Decrease (brightness) /Verminder (helderheid)Y¥

Rexternar decrease/ Roexsteme VerminderY

|, increase/ |, vermeerderY

Vinternal resistance (Of Vig.) increase OR V,,, decrease (emf remain constant)//
Vinterne weerstand (OF Vyertora) Vermeerder OF V,.. verminder (as emk konstant bly)
P (Power) decrease (according to P = V7/R )¥ /

P (Drywing) verminder (volgens P = V7/R) (5)
[19]

13

### Page 14

QUESTION 9/ VRAAG 9

9.1.1 AC/ Alternating current/WS/Wisselstroom YW (1)

9.1.2 DC/ Direct currentY / GS / Gelykstroom

Due to the presence of the split ring/commutator. ¥

A.g.v. die teenwoordigheid van die splitring/kommutator. (2)

9.2.1 The AC generator has slip rings Y whilst a DC generator has a split ring/commutator. ¥
Die WS generator het sleepringe, terwyl die GS generator splitring/kommutator het. (2)

9.2.2 Increase in voltage and decrease in current. Y
Verhoging in potensiaalverskil en verlaging in stroom.

Loss in power/energy decreases. (P = I?R) ¥

Verlies aan drywing/energie verminder (P = (?R) (2)
93.1 01s ov (1)
_ Vmax _ 84,8 4 _
9.3.2 Vises 7a = 59,96V ¥ (2)
w
933 Pp =m oy
ave
40 = 38 cov Positive marking from Q9.3.2
R = 89,880 ¥ (3)
[13]

14

### Page 15

QUESTION 10/ VRAAG 10

c

8
3x10

= aa Vv
284 x 10

(2,06 x 10 H2)w @)

10.2 It is the minimum energy that an electron in the metal needs to be emitted from the

metal surfaceY ¥

Dit is die minimum energie benodig om 'n elektron uit die opperviak van 'n metaal
vry te stel.
Deduct one mark if any of the underlined phrases are omitted./Trek een punt af indien

enige van die onderstreepte frases weggelaat word. (2)

10.3 POSITIVE MARKING FROM Q10.1/ POSITIEF NASIEN VANAF V10.1

E=W)+ E aie OR/OF hf = Wit 4mv’ Vv (Any one of the formulae)

(6,63x10 )(1,06x10%)v = WW, + £(9,11x10"")(1,48x109 ¥
W. = 6,93 x 10° J v(Range/Gebied: 6,90 x 10°? J to/tot 6,93 x 10"

J)
:. The metal is zinc/ Die metaal is sink ¥ (5)
10.4.1 No effect /Geeneffek v (1)
10.4.2 No effect / Geeneffek v (1)
10.4.3 Increases / Toeneem Vv (1)

[13]

TOTAL/TOTAAL: 150 MARKS/PUNTE

15

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