# Physical Sciences Grade 12 Paper 1 North West June 2025 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the North West June Exam 2025, 15 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-memo-p1-may-june-2025-north-west/e23dfcf7-bc94-442a-8c8d-cc7e968e5fd6
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: North West June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 15

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## About this document

Physical Sciences · Grade 12 · North West June Exam · 2025. Memorandum, 15 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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### Page 1

education

Department:

Education

North West Provincial Government
REPUBLIC OF SOUTH AFRICA

PROVINCIAL ASSESSMENT/
PROVINSIALE ASSESSERING

[ GRADE/GRAAD 12 }

PHYSICAL SCIENCES: PHYSICS (P1)
FISIESE WETENSKAPPE: FISIKA (V1)
JUNE/JUNIE 2025
FINAL MARKING GUIDELINES/NASIENRIGLYNE

SERB RRR TR RR RETEST RRR TTR Re

«
e
.
.
.
.
.
.
.
’

°¢

ETT TTT bd

MARKS/PUNTE: 144

These marking guidelines consist of 15 pages.
Hierdie nasienrigyne bestaan uit 15 bladsye.

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### Page 2

QUESTION 1/VRAAG 1

110 By
12) Dv
13° CW
14 AVY
15° AVY
Te ¢ A
17 CW
18 AVY
19 Dw
110 A VW

Copyright reserved/Kopiereg voorbehou

NW/June/Junie 2025

(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)

(2)
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### Page 3

QUESTION 2/VRAAG 2

2a

2.2

Marking criteria

If any of the underlined key words/phrases in the correct context is
omitted deduct 1 mark. /Indien enige van die onderstreepte
word/frases nie in die korrekte konteks is

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

of the force /and the acceleration is directly proportional to the force and
inversely proportional to the mass of the object. / /Wanneer ‘n netto krag op
‘n voorwerp inwerk, sal die voorwerp versnel in die rigting van die krag en die
versnelling is direk eweredig aan die krag en omgekeerd eweredig aan die

massa van die voorwerp.

Fa

Fu

Fy

NOTE/NOTA
If force — diagram/
Indien ‘n kragtediagram (5/5)

Accept the following symbols: Aanvaar die volgende simbole

| Fu vo | Fx/ N/Normal force/Normaalkrag

TV) ~~ Frit ension/Spanning

Fg Vv w /mg/weight/gravitational force/gravitasiekrag

fv Fk/f/Ff /Force of friction/kinetic frictional force/kinetiese

wrywingskrag

Fav FE/FA/F/Applied force/Toegepaste krag

NOTES/NOTA

Mark awarded for label and arrow. / Punte toegeken vir
benoemimg en pyltjie.

Do not penalise for the length of arrows. / Moet nie
penaliseer vir die lengte van die pyltjies nie.

Any other additional force(s)/ Enige ekstra kragte: Max 4/5

If everything is correct, but no arrows. / Alles korrek maar
geen pyle: Max 4/5

(2)

(5)

### Page 4

2.3

NW/June/Junie 2025

oat Fnet = ma}

(Direction of motio
Car / Kar
-T - fk + Fe = ma}

-T— 1800 + 8000 = 900a v

-T = 900a - 6200

Trailer
T - fk = ma}
T -300 = 150a

Substituting (1) and (2)
T - 300 = 150a
-(900a - 6200) - 300 = 150a v
a=56ms* left v

+/Rigting van beweging as +)

¥ Any one/ enige een

(5)

2.4 Increase Vv ¥

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(2)
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### Page 5

NW/June/Junie 2025

QUESTION 3/VRAAG 3

3.4
Marking criteria
lf any of the underlined key words/phrases in the correct context is

omitted deduct 1 mark. /Indien enige van die onderstreepte word/frases
nie in die korrekte konteks is.

The force that opposes the motion of an object and which acts parallel to the

surface. / / / Die krag wat die geneigdheid vir beweging van ‘n stielstaande

voorwerp reletief / paraliel tot die opperviak teenstaan. (2)
3.2

Fu
tt

Fo

NOTE/NOTA
Freebody — diagram /
vryliggamdiagram (3/3)

Accept the following symbols: Aanvaar die volgende simbole

Fu Vv Fy/ N/Normal force/Normaalkrag

fv F,/f/Ff /Force of friction/kinetic frictional force/kinetiese

wrywingskrag
Fg V w /mg/weight/gravitational force/gravitasiekrag
NOTES/NOTA
e Mark awarded for label and arrow. / Punte toegeken vir benoemimg
en pyltjie.

e Do not penalise for the length of arrows. / Moet nie penaliseer vir
die lengte van die pyltjies nie.

e Any other additional force(s)/ Enige ekstra kragte: Max 2/3

e If everything is correct, but no arrows. / Alles korrek maar geen
pyle. Max 2/3

(3)
3.3
Fai = Fg sin 6 v
=mgsin8 | any onefenige een
= (25) (9,8) sin 30°V
Fgi = 122,5 NV

(3)

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### Page 6

NW/June/Junie 2025

3.4 | OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
tan @ = ps Fgu=f
=0,6 mg sin@ = ps N
@ = tan'(0,6) v Us =mg sin 8/mg cos 8
8 = 30,96°” 6 = tan” (0,6) v
9 = 30,96°v (2)

3.5 Frictional force decreases v “. Frictional force is directly proportional the normal

force/ As the normal force decreases with an increase in the angle of inclination

frictional force will also decrease, “/ Wrywingskrag neem af. Wrywingskrag is direk

eweredig aan normaalkrag/indien die hoek van die helling groter word, sal normaalkrag

en wrywingskrag verminder. (3)
[13]

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NW/June/Junie 2025

QUESTION 4 /VRAAG 4

41

4.2

Marking criteria

lf any of the underlined key words/phrases in the correct context is
omitted deduct 1 mark. /Indien enige van die onderstreepte word/frases
nie in die korrekte konteks is.

tn an isolated/ closed system the total (linear) momentum is conserved/
remains constant. v v/ In ‘n gesiote sisteem bly die totale momentum
behoue/konstant.

4.2.1
Zpi = Ipf Any one V/Enige een
mvic + mvit = mvfc + mvft
(650) (16,67) V_+(1000) (8,33) V = (650) (vfc)+(1000) (10) V
Vic= 14,10 m.s" to the right V / na regs

4.2.2 POSITIVE MARKING FROM QUESTION 4.2.1.
POSITIEWE NASIEN VANAF VRAAG 4.2.1.

If system approched is used (vamvie? +%”m vic = Yamvie? + Yami)
NOTE: Award marks for only the formula and conclusion. (2/6)

Before collision/Voor botsing
Yamvic. + Yarnvie’ ¥

% (650) (16.67) + ¥ (1000) (8,33). ¥
90313,89+34694,45

125008,34 J ¥
After collision/Na botsing

Yernvec + Yarovee

Y%, (650) (14,10)" + % (1000) (10) ¥
64613,25+50 000

114 613,25 Jv

12500,34 J # 114 613,25 J
«. Inelastic v/Onelasies

4.3 Newton's third law. //Newton se derde wet

Copyright reserved/Kopiereg voorbehou

When the car and the truck collide, the force that the car exerts on the truck
is equal in magnitude V but opposite in direction to the force that the car exerts
on the truck. V/Wanneer die kar en die trok bots, sal die krag wat die kar op die
trok uitoefen net so groot wees as die krag wat die trok op die kar uitoefen, maar in

teenoorgestelde rigting.

(6)

(3)
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### Page 8

NW/June/Junie 2025

QUESTION 5 /VRAAG 5

5.1 The product of the resultant/ net force acting on an object and the time the

net force act on an object. / // Die produk van die resultant/netto krag op
die voorwerp en die tyd wat die netto krag op die voorwerp uitgeoefen is. (2)
5.2

Magnitude of net force Frye: (N)

01 02 03 04
At(s)

Marking criteria/ Nasienkriteria
Shape. ¥/ Vorm
Both axis with S.| units. “// Beide asse met SI eenhede. (2)

5.3
5.3.1

OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
Fnet-At = Area under graph Fret-At = 22 x 0.4 W
/Area onder die grafiek = 8,8/2 ¥

=%xbxhv Fnet-At = 4,4 N.s ¥

=%x0,4x22v
Fret-At = 4,4 N.s v

(3)

5.3.2 |bosiTIVE MARKING FROM QUESTION 5.3.1/POSITIEWE NASIEN

FnetAt = Ap }
FnetAt = m (vf - vi) J any one V/enige een

4.4 V = 0,65 (vf-0) V
Vf = 6,77 m.st V

(4)

5.4 Increase. V/ Toeneem (1)

5.5 There must be an increase in change of velocity, to maintain a constant
impulse when mass decreases. V //Daar moet ‘n toename in die verandering van
snelheid wees om 'n konstante impuls te behou as die massa afneem (2)

[14]

QUESTION 6 /VRAAG 6

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### Page 9

6.1

6.2

6.3

6.4

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Negative. V/ Negatief

6.2.1 Gradient= vo- vi ¥
te-t
= 0=(-9,8) ¥
3-2
= 9,8 (m.s7) ¥

6.2.2 (Gravitational) acceleration. “/ (Gravitasie)versnelling.

6.3.1 OPTION 1
Gradient= v2- vi ¥

te-t
9,8= 14.7-vi ¥

2-0
vi=-4,9 ms"

«V1 = 4,9 m.s upwards V /opwaarts

OPTION 2

vi = vi + aAtY
14,7 = vi + (9,8) (2) ¥
vi=-4,9 m.s*
* Vi = 4,9 m.s upwardsY /opwaarts

6.3.2 14,7 m.s‘v downwards. V/ afwaarts
6.3.3 9,8 m.s* V upwards. “/ opwaarts

64.1 Remeved

6.4.2 Remove

6.4.3. Maximum height after the first bounce. ”
Removed

NW/June/Junie 2025

(1)

(3)
(2)

(3)
(2)
(2)

(1)
(1)
(1)
(4)
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### Page 10

QUESTION 7 /VRAAG 7

eA

7.2
7.3

NW/June/Junie 2025

An object which has been given an initial velocity and moves under the
influence of gravitational force only. /V/‘n Voorwerp wat ‘n beginsnelhei

gegee is en dan beweeg slegs onder die invioed van gravitasiekrag

Oms!¥

(2)
(1)

OPTION 1/OPSIE 1
(Upwards as positive)/ Opwaarts
as positief

OPTION 2/OPSIE 2
(Downwards as positive)/ Af
as positief

v2 = vi2 + 2aAy ¥

(0)? = (vi)? + 2(-9,8) (30) ¥
vi = 24,25 m.s'“upwards”/ opwaarts

OPTION 3 /OPSIE 3
(Upwards as positive)/ Opwaarts
as positief

v2 = vi? + 2ahy ¥

ve = (0)? + 2(9,8) (30) V
Vi = 24,25 m.s! Vv downwardsv/
afwaarts

OPTION 4/OPSIE 4
(Downwards as positive)/ Af
as positief

Ay = vidt + “at?
-30 = (0) At+ % (-9,8) At?

-30 = 0-4,9 At?
At=2,47s Both v/Albei

Vr = vi + at

vr = (0) + (-9,8) (2,47) v

vr = 24,21 m.s downwardsv/
Afwaarts

Ay = viAt + “at?

30 =0+4,9 At?
At=2,47s

Both v/ Albei
ve=vit at
(0) = vi + (9,8) (2,47) ¥
Vi = -24,21 m.st
vi= 24,21 m.s-1 upwards v/
opwaarts

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### Page 11

74

Ball/Bal X

Ayx = yt - yi
=yi-0

Ayx= yt

Ball/Bal X
Ayx = viAt+ Yadt? /

= (24,25) (At) + % (-9,8) At Vv
= 24,25 At-4,9 At... nT
Yfx = Yfy

24,25 At-4,9 Av=-4,9 AP + 30/
24,25 At = 30

At=1,24sV

Position m

us)

NW/June/Junie 2025

POSITIVE MARKING FROM QUESTION 7.3./POSITIEWE NASIEN

Ball/Bal Y
Ayy = yt- yi

= yr- 30
Ayy = yr- 30
Ayy + 30 = yyy

Ball/Bal Y
Ayy = viAt + %aAt?

= (0) (At) + % (-9,8) AB V
=-4,9 At. -

(7)

Posisie {m)

Gk hl ts}

Marking criteria/Nasienkriteria

Note: 1 shape labelled (3/3) and No shape labelled (1/3)

Shape for ball X/Vorm van bal X

v

Shape for ball Y/Vorm van bal Y

v

Both axis with S.| units
/Beide asse met S.l eenhede

v

(3)

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### Page 12

NW/June/Junie 2025

QUESTION 8 /VRAAG 8

8.1

8.2

The mechanical energy in an isolated system remains constant. /V /Die totale
meganiese energie bly konstant in 'n geisoleerde sisteem.

Emecha = Emechc
(mghc+Ymvc*)= (mghc+Ymvc2) | Any one // Enige een

(120)(9,8)(8) + % (120)(0)? V= (120)(9.8)(2) — % (120)vc2V

9408 = 2352 + 60 vc?
Ve = 10,84 m.s'/

8.3 POSITIVE MARKING FROM QUESTION 8.2. / POSITIEWE NASIEN

Option 1/Opsie 1
Ware = AEk + AEp
Ws = (amv? - Ymv2) + (mghf - mghi)} Any one V/Enige een
Wr = (A(120)(1,1 2-% (120)(10 84)2) V+((120)(9 8)(5) — (120)(9,8)(2)) V
IW; = - 6977,74 + 3528
IW: = -3449,74 J V
(Accept range/Aanvaar:between/tussen (3449 J — 3456 J)
Option 2/Opsie 2
Emecha = Ep + Ek

=mgh +¥%mv 2 | Any one // Enige een
= (120)(9,8)(5) + %(120)(1,1)2 V
= § 952,6 J
mecha = Mgh

= (120)(9,8)(8)

=9408 JV
Heat generated/Hitte gegenereer = 9408 — 5952,6

= 3455.4 JV
Accept range/Aanvaar: between/tussen (3449 J — 3456 J)
QUESTION 9 /VRAAG 9

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(2)

(4)

(4)
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### Page 13

NW/June/Junie 2025

9.1 | Marking criteria
lf any of the underlined key words/phrases in the correct context is omitted

deduct 1 mark. //ndien enige van die onderstreepte word/frases nie in die
korrekte konteks is nie, trek 1 punt af

A force for which the work done in moving an object between two points
depends on the path taken. ~~/n krag waarvoor die arbeid verrig om 'n
voorwerk tussen twee punte te beweeg, afhanklik is van die roete wat gevolg
word.
(2)
= Option 1/ Opsie 1
Whe = AEk + AEp
Wa + Wr = (Yamve - Yam) + (mghr- mghi) FAny one v / Enige

Wa + (- 5,85 x 108) = (0) + (1750)(9,8)(55) v
Wa= 943250 + 5,84 x 108
Wa = 6 783 250 J

Wa = Fa Ax cos 8 ¥
6 783 250 = Fa (570) cos (0) ¥
Fa= 11 900,44 N v
Accept range: between (11 900,42 N - 11 900,45 N)
Option 2
Wr = f Ax cos 8 ¥
-5,84 x 10° = f (570) cos (180) v
f = 10245,61 J

Faet= ma

Fra =0
-f—Foi+ Fa=0) Anyone “/Enige een
- f-— (1750) (9,8) sin 6+ FA =0

- (10245,61)— (1750) (9,8) (=)+ Fa= 0 v

Fa= 11900, 43Nv

Accept range: between (11 900,42 N - 11 900,45 N) (5)
9.3 Fret = ON v (1)

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### Page 14

9.4

NW/June/Junie 2025

OPTION 1 / OPSIE 1

Whe = AE + AEp
Wa + Wt = (Yamve? - Yamvi7) + (mghr- mghi)F Any one ¥

Wa + (6,44 x 10°) (570) cos (180) v = % (1 100)(25? — 202) v +

(4. 100)(9,8)(0 — 55)v”
Wa= 3 670 800 — 592 900 + 123 750

Wa = 3 201 650 J
OPTION 2/ OPSIE 2

Whet = AE
Wa + Wy + Wegi= (Amve - Zmv2) Any one v
Wa + f Ax cos 6 = mg sin @ Ax cos 9 = (Yemve? - “mvi?)

Wa + (6,44 x 10°) (570) cos (180) v’ + (1100) (9,8) (+) 570 cos (0) v =

Ye (1 100)(25? — 202) v
Wa - 3 660 451,95 + 592 900 = 123 750
Wa = 3 201 650 J v

(5)
[13]

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### Page 15

QUESTION 10 /VRAAG 10

10.1 Doppler effect. //Doppler effek

10.2 fis VEVL fast
vivs
= 340 V (458) v
340-30
fL = 502,32 HzV

10.3 Decrease V/ Afneem

10.4 Equal to V/Gelyk aan

(1)
(4)

(1)

10.5

10.6

The velocity of the fire engine driver relative to the fire engine siren
is zero.’ OR /Die snefheid van de brandweerman relatief tot die
brandweerwa is nul

The fire engine driver has the same velocity as the fire engine. V
OR/ Die brandweerman beweeg teen dieselfde snelheid as die
brandweema.

There is no relative motion between the fire engine driver and the (2)
fire engine. // Daar is geen relatiewe beweging tussen die brandweerwa
en die brandweerman nie.

Use of ultrasound for measuring the rate of blood flow. //Gebruik ultraklank

om bloedvioei te meet

Use of ultrasound to measure the heartbeat of a foetus.V/Gebruik ultrank

om die hartklop van die fetus te meet

To detect weather by analysing doppler shift of radar waves. (3)
v¥/Weervoorspelling deur die annalise van die dopplerskuif van radargolwe.

(Any other correct answers).

When light from a distant star or galaxy is observed to have shifted towards (2)
the red end of the spectrum. V //Wanneer lig van ‘n verafgeleé ster of

sterrestelsel waargeneem word en daar vind rooiverskuiwing van die spektrum

plaas.

[13]

TOTAL/ TOTAAL: 150

Copyright reserved/Kopiereg voorbehou

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