# Physical Sciences Grade 12 Paper 1 Mpumalanga June 2025 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Mpumalanga June Exam 2025, 17 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-mpumalanga/100394f3-7dc4-4283-b4ab-5a243840e8d2
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Mpumalanga June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 17

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## About this document

Physical Sciences · Grade 12 · Mpumalanga June Exam · 2025. Memorandum, 17 pages. Read online or download the PDF.

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### Page 1

education

MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE
NASIONALE
SENIOR SERTIFIKAAT

GRADE/GRAAD 12

PHYSICAL SCIENCES: PHYSICS P41
FISIESE WETENSKAPPE: FISIKA V1

JUNIE/JUNE 2025

MARKING GUIDELINES/NAS/ENRIGLYNE

MARKS/PUNTE: 150

These marking guidelines consists of 17 pages
Hierdie nasienriglyne bestaan uit 17 bladsye

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### Page 2

QUESTION 1 / VRAAG 1

4.4
4:2
1.3
1.4
1.5
1.6
flak
1.8
1.9
1.10

Copyright reserved/Kopiereg voorbehou

D vv
G vv
DvyY
B vv
B vv
Avy
Dvv
Bvv
ByY

Dvv

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(2)
(2)
(2)
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(2)
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QUESTION 2/ VRAAG 2

2.1. Marking criteria/Nasienkriteria
\f any one of the underlined key words/phrases in the correct context is omitted,
deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die
korrekte konteks uitgelaat is, trek 1 punt af.
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 resultant/net force
and inversely proportional to the mass of the object. vv

Wanneer ‘n resulterende/netto krag op ‘n voorwerp inwerk, versnel die voorwerp
in die rigting van die krag ieen 'n versnelling direk eweredig aan die
resultante/netto_krag en omgekeerd eweredig aan die massa van die voorwerp.
OR/OF

The resultant/net force acting on an object is equal to the rate of change of
momentum of the object. (2 or 0)

Die resulterende/netto krag wat op ‘n voorwerp inwerk is gelyk aan die tempo

van verandering van momentum. (2 of 0) (2)
2:2
Nw Tv
fv
wv
(4)
Accepted labels / Aanvaarde benoemings
Ww Fy / Fw/ force of earth on block / weight / mg / gravitational force /
11760N
T Tension / Fr
f Frriction / Fr / frictional force /wrywing
N Normal force / Fn

Notes/Aantekeninge:
e Mark awarded for label and arrow.
e Any additional forces: deduct 1 mark: max 3/4

e No labels: deduct 1 mark: max 3/4

e If everything correct, but no arrows: deduct 1 mark: 3/4
e Force(s) not touching object: deduct 1 mark: max 3/4
e Ignore relative sizes of the vectors

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### Page 4

2.3

2.4

2.5

2.6

27

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f=uNv

= (0.2)(1200)(9,8)cos20° v
=2210,16N¥ (3)

YESHA ¥ (1)

For trailer:
Fret = ma P

T + (4) + ((Fot) =ma > Any one/Enige een ¥
T—2210,16 Y —(1200)(9,8)sin20° v= (1 200)(2,5) v
T — 6232,317 = 3000

T =9 232,22NY
(5)
Marking criteria/Nasienkriteria
If any one of the underlined key words/phrases in the correct context is
omitted, deduct 1 mark. Andien enige van die onderstreepte
sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.
A body will remain in its state of rest or motion at constant velocity unless a
non-zero resultant/net force acts on it.vv
‘n Liggaam sal in sy toestand van rus of beweging teen konstante snelheid
volhard, tensy 'n nie-nul resulterende/netto krag daarop inwerk. (2)
For trailer:
Fret = ma
T + (-f) + ((-Fgi) = ma
T= 2210,16 - (1200)(9,8)sin20° v= 0 v
T —6232,317 =0
T =6232,32N
For bakkie:
Fret = ma
F + (-f) + ((-Fol) + (-T) = ma
£ — (0,3)2700)(9,8) cos20° v —(2700)(9,8)sin20° v - 6232,32 = 0
F — 7459,28 — 9049,85 - 6232,32 =0
F = 22 741,45 NV
(5)
[22]

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QUESTION 3/ VRAAG 3

3.1

3.2

3.3.1

3.3.2

3.4

9,8 ms? V downwards/afwaarts Vv

Free fall is the motion during which the only force acting on an object is the

gravitational force.. v Y (2 or 0)

Vryval is die beweging waartydens die enigste krag wat op 'n voorwerp inwerk,

die gravitasiekrag is. (2 or 0)

OPTION 1/OPSIE 1
Upwards positive/Opwaarts positief

ve = vi2+ Zady ¥

vi? = (0)? + 2(-9,8)(-92,4) ¥

w=42.56m:s' v

Downwards positive/Afwaarts
positief

ve = vit+ 2aAy

ve? = (0)? + 2(9,8)(92,4)

vi = 42.56 m:s*

POSITIVE MARKING FROM Q3.2
POSITIEWE MERK VANAF Q3.2

OPTION 1/OPSIE 1

Upwards positive/Opwaarts positief
vf =vi+ at v

-42,56 = 0 + (-9,8)At v

At=4,34s v

Downwards positive/Afwaarts positief
vf = vi + aAt

42,56 = 0 + (9,8)At

At= 4,345

OPTION 2/OPSIE 2

Upwards positive/Opwaarts positief
Ax=viAt + %aAt? ¥

-92,4 = OAt + % (-9,8)Ar v
At=4,34s v

Downwards positive/Afwaarts positief
Ax = viAt + % aAt?

92,4 = OAt + % (9,8)At?

At =4,34s

POSITIVE MARKING FROM Q3.2 & 3.3
POSITIEWE MERK VANAF Q3.2 & 3.3

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Downwards positive/Afwaarts positief

;
42,56
vim-s")
0 —>
4,34 t(s)
¥
Marking criteria for graph Mark
Nasienriglyne vir grafiek Punt
Correct shape of graph v
| Korrekte vorm van grafiek
Final velocity v
Eindsnelheid
Time at which it reaches the ground v
Tyd wat dit neem om die grond te tref

Upwards positive/Opwaarts positief

4
v(m-s"t)
4,34
-42,56 t(s)
(3)
3.5 Same/Dieselfde v

Mass has no influence/Gravitational acceleration is the same/it is in a free fall v

Massa het geen invloed nie/Gravitasie versnelling is dieselfde/dit is in vryval (2)
[15]

QUESTION 4/ VRAAG 4

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41 Impulse is the product of the resultant/net force acting on an object and the time
the net force acts on the object. vv (2 or 0)
Impuls is die produk van die resulterende/netto krag wat op 'n voorwerp inwerk
en die tyd wat die netto krag op die voorwerp inwerk.. (2 of 0)
(2)
4.2 Frer At = Ap
Fret-At = mvr — mvi } v Any one/enige een
Impulse = mvs — mvi
= (0,15)(3,65) — (0,15)(-6,5) ~
= 1,52 kg: m-s/N:s VW (3)
4.3. | Ewtop = Emsotrom
Extop + Eptor = Exsottom + Epgotrom } v Any one/enige een
Yamv? + mgh = Y%mv? + mgh
0 + (0,15)(9,8)h » = 4(0,15)(6.5)? “+ 0
h=2,16m
Initial height dropped from is 2,16 m.
YBh=%x 2,16 ¥ = 0,72m
The ball does not meet minimum requirementsY because it only reached
0,65 m.
(5)
[10]

QUESTION 5/ VRAAG 5

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### Page 8

5.1

5.2

5.3

5.4

MDE/June Junie2025
NSC/NSS-Marking Guideline/Nasienriglyn

Marking criteria/Nasienkriteria

\f any one of the underlined key words/phrases in the correct context is omitted,
deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die
korrekte konteks uitgelaat is, trek 1 punt af.

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

Die totale meganiese energie (som van gravitasie- potensiéle energie en

kinetiese energie) in 'n geslote sisteem bly konstant.

Nv

FY

wv

Accepted labels / Aanvaarde benoemings

w Fg / Fw/ force of earth on block / weight / mg / gravitational force /2450 N

F Applied force / Force of trolley A on trolley B

N Normal force / Fn

Notes/Aantekeninge:
e Mark awarded for label and arrow.

e Any additional forces: deduct 1 mark: max 2/5

e No labels: deduct 1 mark: max 2/5

e If everything correct, but no arrows: deduct 1 mark: 2/5
e Force(s) not touching object: deduct 1 mark: max 2 fe
e Ignore relative sizes of the vectors

Emtop = Emsottom

Extop + Eprop = Exsotrom + Epsottom } v Any one/enige een
Yamv? + mgh = Yamv2 + mgh

0 + (300)(9,8)(30) v= %4(300)v? V+ 0

v= 24,25 ms! Vv

POSITIVE MARKING FROM Q5.3
POSITIEWE MERK VANAF Q5.3

TAKE TO THE RIGHT AS POSITIVE/NEEM REGS AS POSITIEF

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(3)

(4)

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Zpi = Xpr .
(mya + (mvia= (mvi1 + (mvij2_ Y Any one/enige een
(300)(24,25) v +0 = (300 + 250)vi v
vr = 13,23 m-s"' ¥ to the right/na regs ¥

5.5 | POSITIVE MARKING FROM Q5.3 & Q5.4

POSITIEWE MERK VANAF Q5.3 & Q5.4

Whet = AEk t Any one

fAx cos@ + fgiAx cos© = Yamver? — Yamvi2

(10) Axcos180°v + (550)(9,8)sin 30° Axcos180° v= 0 = [4(550)(13,23)?] v

Ax=17,79mv (5)
[19]

QUESTION 6/ VRAAG 6

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6.1.1 Doppler effect as 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.v v (2 or 0)

Doppler-effek neer as die verandering in frekwensie (of toonhoogte) van die
klank waargeneem deur 'n luisteraar omdat die klankbron en die luisteraar
verskillende snelhede relatief tot die medium waarin die klank voortgeplant

word, het. (2 of 0) (2)
6.1.2. Higher than v
Hoér as (1)
6.1.3 —
‘t
els Y OR feels
335 +0
v
ae -) Sap
f, = 903,97 Hz v
(5)
614) yasy Accept : ve%
t At
d
20=-v
3
d=60m¥ (3)
6.2.1 Yvv (2)
6.2.2 Away from the Earth/Weg van die Aarde v Vv (2)
[15]
QUESTION 7 / VRAAG 7

7.1 | Marking criteria/Nasienkriteria
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Note: If masses used ( %, )

If any one of the underlined key words/phrases in the correct context is
omitted, deduct 1 mark. /Indien enige van die onderstreepte
sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.

The magnitude of the electrostatic force exerted by one point charge (Q1) on
another point charge (Qz2) is directly proportional to the product of the

magnitudes of the charges Y and inversely proportional to the square of the
distance (r) between them ¥

Die grootte van die elektrostatiese krag wat een puntlading (Q:) op 'n ander

puntlading (Q2) uitoefen, is direk eweredig aan die produk van die groottes van
die ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen

hulle.
G2
Fev
Fg v
Accepted labels / Aanvaarde benoemings
Fe Electrostatic force between charges X and Y
Elektrostatiese krag tussen ladings X en Y
Fy Gravitational force on X
Gravitasiekrag op X
Notes/Aantekeninge:
e Any additional forces: deduct 1 mark: max i,
e No labels: deduct 1 mark: max "Yo
e No arrows: 0/2
e Force(s) not touching object: deduct 1 mark: max 1,
e Ignore relative sizes of the vectors
7.3 kQ1Q
F)=mg p= ee

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(2)

(2)

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= (40x 103) (9,8) ¥
= 0,392 N

Fy =Fe

(9x10°)q?
(0,3)

0,392 =

q =2x10°C v

Charge on Y original = 4 x 10° C v

(9x10°) q?

v
(0,3)°

QUESTION 8 / VRAAG 8

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[10]

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8.1 The electric field at a point is the electrostatic force experienced per unit
positive charge placed at that point.“ v (2 or 0)

Die efektriese veld by 'n punt is die elektrostatiese krag wat per
eenheidspositiewe-lading wat by daardie punt gepiaas is, ondervind word.
(20r0) (2)

Criteria for sketch/Kriteria vir skets Marks/Punte
Correct shape as shown. v
Korrekte vorm soos getoon.
Direction away from positive to negative. v4
Rigting weg van positief na negatief.
Field lines start on spheres and do not cross for correct v
diagram.
Veldlyne begin op elke sfeer en kruis nie vir korrekte diagram. (3)

8.2

8.3 Added / Bygevoeg V
(1)

84.1[ 7 @y
e
—4 x 107°

n= —16 x10-%

= 2,5 x 10" electrons/elektrone V (3)

8.4.2 | Marking criteria/Merk kriteria
e Mark for correct formula to calculate E
e Mark for substituting into formula for charge A
e Mark for substituting into formula for charge C

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| _® Mark for correct direction

MDE/June Junie2025
NSC/NSS-Marking Guideline/Nasienriglyn

« Mark for correct use of positive and negative signs
e Mark for the final answer

kQ

72
r.

B=

Enet=Ea tEgtEc

_ x 10°)(1 x 1078) e (9 x 10°)(2,5 x 107°) - (9 x 10°)(4 x 107°)
~ (0,016)? v (0,014)? ~ (0,01)2

Enet = 2, 098 x 10° N.C* ¥ left ¥

QUESTION 9/ VRAAG 9

9.1

Copyright reserved/Kopiereg voorbehou

The maximum energy provided by a battery per unit charge passing through
it. vv

(2 or 0)

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(6)
[15]

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Die maksimum energie wat ‘n battery lewer per eenheidslading wat daardeur
vloei. (2 of 0)

92 fi a,i1,1

R= 2,860 v (3)

9.3 | OPTION 1/OPSIE 1
Diagram B
e The total resistance of the circuit increasesY
e Total current decreasesY
e The Vint decreasesY
e therefore Vext increases

OPTION 2/OPSIE 2

Diagram A

The total resistance of the circuit decreasesY
Total current increasesY

The Vint increasesY

therefore Vext decreases (3)

9.4.1 | POSITIVE MARKING FROM Q9.2
POSITIEWE MERK VANAF Q9.2

1=3,50A v

(3)

9.4.2 | POSITIVE MARKING FROM Q9.2 & 9.4.1
POSITIEWE MERK VANAF Q9.2 & 9.4.1

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Diagram A
E=1(R+r) v

© =3,5 (2,86 +r) Vv....... 1
Diagram B

12
#55 72,18 A

€=1(R+n
© =2,18 (5,5 4r) W......2

Equation 1 = Equation2 ¥
3,5 (2,86 +r) = 2,18(5,5 + r)

r=15Q Vv

9.4.3 | POSITIVE MARKING FROM Q9.2 & 9.4.1 & 9.4.2
| POSITIEWE MERK VANAF Q9.2 & 9.4.1 & 9.4.2
€=l(R+09
= 3,5 (2,86 +1,5) V
= 15,26 V v
OR
E=1(R+5)
= 2,18 (5,5 +1,5) ¥
=15,26Vv

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9.4.4

9.5

MDE/une Junie2025
NSC/NSS-Marking Guideline/Nasienriglyn

POSITIVE MARKING FROM Q9.2 & 9.4.1 & 9.4.2 & 9.4.3
POSITIEWE MERK VANAF Q9.2 & 9.4.1 & 9.4.2 & 9.4.3

OPTION 1/OPSIE 1
P=PRV
= (2,18)? (5,5)¥
= 26,14 WY

OPTION 2/OPSIE 2
P
P= a2"?

5,5

P = 26,18WvY

ll
|
x

OPTION 3/OPSIE 3
P=Viv
= (12) (2,18)”

= 26,16 Wv

The voltmeter reading INCREASES. v
Total resistance increases and total current decreases, Vv
Vint decreases “

* Die voltmeterlesing NEEM TOE

* Totale weerstand neem toe en totale stroom neem af
+ Vint neem af

TOTAL/TOTAAL

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