# Physical Sciences Grade 12 Mpumalanga March Test 2025 Memo

> Grade 12 Physical Sciences memorandum from the Mpumalanga March Test 2025, 7 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-memo-march-2025-mpumalanga/88039ede-226f-4c60-918e-92c8ef3e11cc
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Mpumalanga March Test
- Curriculum: CAPS
- Pages: 7

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## About this document

Physical Sciences · Grade 12 · Mpumalanga March Test · 2025. Memorandum, 7 pages. Read online or download the PDF.

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## Document text

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### Page 1

education

MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA

FURTHER EDUCATION AND TRAINING

NKANGALA DISTRICT/ DISTRIK

GRADE/ GRAAD 12 |

PHYSICAL SCIENCES
MARCH 2025
CONTROLLED TEST
MEMO

MARKS/ PUNTE:100

TIME/ TYD :2 HOURS/ URE

This marking guideline paper consists 9 of pages
Hierdie nasienriglyn bestaan uit 9 bladsye

Page 1 of 7

### Page 2

QUESTION/ VRAAG 1

11. =A (2)
1.2 vDvv (2)
13 Dvv (2)
14. BYY (2)
15 Avy (2)
16 Cvv (2)

(12]

QUESTION/ VRAAG 2

2A 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 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. Vv ¥

(2)
2.2
Accepted labels
NY F Fa/applied force/35N/
Toegepaste krag
Tv N Normal force/Fy / Normaalkrag
fv F T Tension / Fr / Spanning
we NS f Frictional force / Fi/f, / wrywingskrag
w Fg / Fw /force of earth on block / weight /
mg / gravitational force/ gewig/
gravitasiekrag (5)
2.3.1 Fre=ma
Fret=0 Any one ¥ Marking criteria
Fcos8 —-T-f=0 Formula/ Formule~
Fcoos@ — T — uN =0 Substitution for friction / Vervanging
aH van wrywing¥
Fcos@ — T — (u,mg+35sin20°) = Whole substitution / Hele
35 cos 20 -T-(0.2)(6x9.8+35sin20°) Vv =0V vervangingY
T=18.74 NY Answer with correct units / Antwoord
met korrekte eenheidY (4)
2.3.2. POSITIVE MARKING FROM / POSITIWE MERK VANAF 2.3.1
Fret=ma Any one / Enige een”
Fret=0
T-(mgsin®)-(u,mgcos®) =
18.74 — (mx9.8sin30")-(0.15)(mx9.8cos30")v= 0 ¥
m=3.03 kg ¥ (4)

Page 2 of 7

### Page 3

[15]

QUESTION/ VRAAG 3

3.1. Motion under the influence of the gravitational force/weight ONLY (2)
Vryval is die beweging waartydens die ENIGSTE krag wat op 'n voorwerp inwerk, die
gravitasiekrag is.“ ¥

3.2.

OPTION OPSIE 1 OPTION/ OPSIE 2
UPWARD AS POSITIVE/ OPWAARTS | DOWNWARD AS POSITIVE/
POSITIEF AFWAARTS AS POSITIEF
vz=vi2+2ady ¥
vP=Vvi2+2ady ¥ 292=07+ 2(9,8)AyY
07=292+ 2(-9.8)AyY Ay= 42,91m ¥
Ay= 42,91m ¥
UPWARD AS POSITIVE/ OPWAARTS
DOWNWARD AS POSITIVE/ | POSITIEF
AFWAARTS AS POSITIEF vyr=vie+2ady
v,Z=vi2+2ady ¥ (-29)?=02+ 2(-9,.8)AyY
07=(-29)2+ 2(9,8)AyY Ay= 42,91m ¥
Ay= 42,91m v (3)
OPTION/ OPSIE 1 OPTION/ OPSIE 2
3.3 | UPWARD AS POSITIVE/ OPWAARTS _| UPWARD AS POSITIVE/ OPWAARTS POSITIEF
POSITIEF vey taaty
vevitadty 0 = 29 + (-9.8)At ¥
-29V = 29 + (-9.8)Atv At=2,96s
At = 5,92 s ¥ At = 2x 2,96 V =5,92sv
DOWNWARD AS POSITIVE! DOWNWARD AS POSITIVE/ AFWAARTS AS POSITIEF
AFWAARTS AS POSITIEF ~ e
ViSUPTAREY Vvy=VvitaAt
my 29 v = -29 + (-9.8)At¥
29 v= -29 + (9.8)At ¥ At=5.92sVv
At=5,92sv i
(4)
3.4 POSITIVE MARKING FROM/ POSITIEWE MERK VANAF 3.3

UPWARD AS POSITIVE/ OPWAARTS DOWNWARD AS POSITIVE! AFWAARTS AS
POSITIEF POSITIEF

dt _— si 1
Ay = vidt +5 abt?v Ay = vide +5 ai

v= -1)+ = 12”

“58 ¥= y(5.92-1)+ 4-9,8(5,92-17v | 98” = WE.O2-1 1B.8)16.92-%
vi=12,32_m-s7'V¥ we12,32 ms (4)

Page 3 of 7

### Page 4

3.5

POSITIVE MARKING FROM/ POSITEWE MERK VANAF 3.3 and/ en 3.4

UPWARDS AS POSTIVE
OPWAARTS AS POSITIEF

29

Marking criteria/ Merk riglyne

Graph A start at the initial velocity of 29 m:s~' and
Graph B start at the initial velocity 12,32 m-s-1 and at
t=1s.

Grafiek A begin by 29 m.s EN

Grafiek B begin by 12,32 m.s” by tyd t=1s

Correct shape for graph A and B/ Korrekte vorm vir
| grafieke A en B

The two graph a parallel and graph B below A.
Grafieklyne parallel, B laer as A

12,32 Both A and B end at 5,92 s./Beide eindig by 5.92s
1 5:92 Time/ Tyd (s)
cs} A
oO
£
oO
=
2)
> =
3 n
fee ee ere eee ee
Ss &

DOWNWARD AS POSITIVE/ AFWAARTS AS POSITIEF
A

=

dt
t

5
‘St

Graph A start at the initial velocity of 29 m-s-* and Graph
B start at the initial velocity 12,32 m-s-' and at t=1s.

E i}

= |

oO \

<4 1

oO !

& {

7) '

> '

= |

2 i

@® 1

> it

5.92 __Timet Td

23 Marking criteria/ Merk riglyne.
Grafiek A begin by 29 m.s‘t EN
Grafiek B begin by 12,32 m.s' by tyd t=1s

Correct shape for graph A and B/ / Korrekte vorm vir
grafieke A en B

The two graph a parallel and graph B above A./
Grafieklyne_ parallel, B laer as A

Both A and B end at 5,92 s./ Beide eindig by 5.92s

Page 4 of 7

### Page 5

QUESTION/ VRAAG 4

4.1 In an isolated system the total mechanical energy is conserved/remains constant.

Die totale meganiese energie (som van gravitasie- potensiéle energie en kinetiese
energie) in 'n geslote sisteem bly konstant. Vv

1 2 1 2
Yong + 5 mv i= Dicng + 5mv My

O+ Sx 2xvi2v = 1,2x0,65 + OY

4.2

vi= 3,57 ms" '¥ (4)
4.3 In an isolated system the total linear momentum is conserved/remains constant.

Die totale lineére momentum in 'n geisoleerde sisteem bly konstant (behoue). ¥ ¥ (2)
4.4 POSITIVE MARKING FROM/ POSITIEWEE MERK VANAF 4.2

Y Pi = Y Pr Any one/ Enige een ¥

maviit Mavi2= Mivert Maver

0,4*vb¥ +0=0,4*(-0,36)+1,2% 3,57”

vo=10,35 m:s-'¥

(4)
[12]
QUESTION/ VRAAG 5
5.1 A bond or an atom or a group of atoms that determine(s) the (physical and
chemical) properties of a group of organic compounds
Funksionele groep: 'n Binding of 'n atoom of 'n groep atome wat die fisiese
en chemiese eienskappe van 'n groep organiese verbindings bepaal Vv”
(2)
5.2.1 Av &CY¥ (2)
5.2.2 Ketone / KetoonY Marking ertterta (1)
5.2.3 4-bromo-3,3-dimethlyhexane Correct stem i.e hexane/ Korrekte stamnaam Y¥
4-bromo-3,3-dimetielheksaan ¢ All substituents (bromo and dimethyl) correctly
identified./ Alle substituente (bromo en dimetiel)
korrek geidentifiseerd v
+ Correct IUPAC name / Korrekte IUPAC naam. ¥
(3)
5.2.4 CnHanY (1)
5.2.5 Pent-2-ene/ Pent-2-eenv ¥ (2)

Page 5 of 7

### Page 6

5.3.1 Compound with the same molecular tormula but different functional groups

Organiese molekule met dieselfde molekulére formule, maar_verskillende

funksionele groepe VY (2)
5.3.2 Carbonyl/ KarbonielY (1)
How 2 Marking criteria / Nasienriglyn
H-G-CaG-o vv" . ,
H 4 4 H ¢ Correct functional group/ Korrekte funksionele
groep ¥
v « Whole structure. Hele struktuur ¥
Ete + Correct IUPAC name/ Korrekte IUPAC ¥

(3)
[17]
QUESTION! VRAAG 6

6.1 Temperature at which the vapour pressure equals atmospheric pressure./V ¥
Die temperatuur waarby die dampdruk van die stof gelyk is aan atmosferiese druk. (2)
6.2 Whatis the relationship between chain length/ surface area/ molecular mass and

boiling point of organic compound ?
Wat is die verband tussen kettinglengte/ opperviakte/ molekulére massa en
kookpunt van organiese verbinding? v ¥

Marking criteria/ Merk riglyne

e Must be a question/ Moet ‘n vraag wees (?)
e Dependent variable / Afhanklike veranderlike
e Independent variable / Onafhanklike veranderlike (2)

6.3. e Chain length / surface area/ molecular mass increases from A-C.
Kettinglengte /opperviakte/ molekulére massa neem toe vanaf A-C ¥
e The strength of London forces/ intermolecular forces increases from A-C.
Die sterkte van Londen-kragte/ intermolekulére kragte neem toe vanaf A-C ¥
e More energy needed to overcome intermolecular forces.
Meer energie word benodig om intermolekulére kragte te oorkom Vv (3)
6.5.2 Compound/ Verbinding C ¥ (1)
6.5.3 eBetween molecules of C/ alcohol has hydrogen bonds with one site of bonding
Tussen molekules van C/ alkohol is daar waterstofbindings met een plek van
binding ¥
e Between molecules of propanoic acid/ carboxylic acids has hydrogen bonds with
two sites of bonding Vv (4)

Page 6 of 7

### Page 7

e Intermolecular forces in propanoic acid/carboxylic acids are stronger than in
compoundC /alcohol
Die Intermolekulére kragte in propaansuur/karboksielsure is sterker as in
verbinding C/ alkohol ~

eMore energy is needed to overcome intermolecular forces in carboxylic
acids/propanoic acid than alcohol/ compound C.
Meer energie is nodig om intermolekulére kragte in karboksielsure/ propaansuur te
oorkom as alkohol/ verbinding C W

[12]
QUESTION/ VRAAG 7
7.1.1. The chemical process/reaction in which longer chain hydrocarbon/alkane molecules
are broken down to shorter (more useful) molecules.
Die chemiese proses waarin langer kettingkoolwaterstof-molekule afgebreek word in
korter, meer bruikbare, molekules Vv“ (2)
7.1.2 CeHiav (1)
7.1.3 2C6H14¥+ 1902 — 14H20 + 12CO2” Balancing / BalanseringY (3)
7.2.1 Addition/Addisie (Hydrohalogination/ Hidrohalogenasie)” (1)
7.2.2
Marking criteria/ Merk riglyne
H H H H
| | | | Correct functional group/ Korrekte funksionele
H—C—C—C—C—H groep ¥
| | | | e Whole structure./ Hele struktuur ¥
H Br H H
(2)
7.2.3 Substitution/ Substitusie ( Hydrolysis)” (1)
7.2.4 HBrv (1)
7.2.5 Esterification/ Esterifikasie v (1)
7.2.6 Concentrated Sulphuric acid / Gekonsentreerde Swaelsuur ¥ (1)
7.2.7 Butyl ethanoate / Butieletanoaatv V (2)

[15]
TOTAL/ TOTAAL: 100

Page 7 of 7

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