# Physical Sciences Grade 12 Free State March Test 2025 Memo

> Grade 12 Physical Sciences memorandum from the Free State March Test 2025, 15 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-memo-march-2025-free-state/177e6d75-e3f1-4726-aede-d61aeeed85a8
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Free State March Test
- Curriculum: CAPS
- Pages: 15

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## About this document

Physical Sciences · Grade 12 · Free State March Test · 2025. Memorandum, 15 pages. Read online or download the PDF.

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## Document text

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### Page 1

education

Department of
Education

FREE STATE PROVINCE

CONTROL TEST/
KONTROLETOETS

GRADE/GRAAD 12
PHYSICAL SCIENCES
MARCH 2025

CONTROLLED TEST

MARKING GUIDELINE/
NASIENRIGLYN.

This MARKING GUIDELINE consists of 15 pages/ Hierdie NASIENRIGLYN bestaan uit
15 bladsye.

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### Page 2

Marking Guideline/Nasienriglyn
Grade/Graad 12

QUESTION 1 / VRAAG 1

1.1
12
1.3
1.4
1.5
1.6
1.7
1.8
1.8

1.10

Copyright reserved/Kopiereg voorbehou

Byv
Cvv
Dvv
Dvv
Bvv
Avy
Byv
Cvv
Cv

Avy

FS/March/Maart 2025

[20]

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### Page 3

FS/March/Maart 2025

Marking Guideline/Nasienriglyn
Grade/Graad 12

QUESTION 2 / VRAAG 2

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

direction of the force, at an acceleration directly proportional to the force

and inversely proportional to the mass of the object.v Vv

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.

¥

N
22
Tv
fv
w

v

ACCEPTED LABELS/AANVAARBARE BYSKRIFTE

w | F, /weight/gravitational force/ 39,2 N/

F iriction!/F¢/friction/fx/ wrywingskrag

f
N | Fy/Frormai/Normal force/Normaalkrag/Normaal
T

F;/Tension/Spanning

NOTE:
- 1mark awarded for label and arrow.
- Deduct 1 mark ONCE for any additional forces.
- Donot consider the length of the forces.

LET WEL:
- 1 punt toegeken vir etiket en pyl.
- Trek 1 punt EEN keer af vir enige bykomende kragte.
- Moenie die lengte van die kragte in ag neem nie.

(2)

(4)

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### Page 4

FS/March/Maart 2025
Marking Guideline/Nasienriglyn
Grade/Graad 12

2.3
Toolbox B/ Gereedskapkis B

Fnet = ma —— —...
T+(-f,) =ma

T- KN = ma

T- 0,2x4x9.8v = 4x5,852V

T = 31,248 N |

——————— Anyone/ Enige eenv

Toolbox A/ Gereedskapkis A
Fnet = ma ————_——
F + (-T)+ (- fk) =ma
F-T—pxN =ma (6)
F - 31,248 — 0,3x6x9.8Y = 6x5,852V
F=84NV
2.4 Force of toolbox A on earth. v v
Krag van gereedskapkis A op aarde (2)
[14]

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### Page 5

FS/March/Maart 2025

Marking Guideline/Nasienriglyn
Grade/Graad 12

QUESTION 3 / VRAAG 3

3.1

3.2.2

3.2.3

3.3

Momentum is the product of an object's mass and its velocity. vv

Momentum is die produk van die massa en snelheid van 'n voorwerp. (2)
(2 or/of 0)

West / Westwards Vv ¥

Wes / Weswaarts (2)
0 (N)/Zero ¥ (1)

Velocity/momentum is constant between 0 s and 10 s¥
Change in momentum/Ap = 0 ¥

«According to Fret = 2 Yi Fiat =ON

Snelheid/momentum is konstant tussen 0s en 10s
Verandering in momentum/Ap = 0

-.Volgens Fnet =e Fnet=0N

ru

OR/OF

Velocity is constant between 0 s and 10 s¥
Change in velocity/Av = 0 OR Acceleration/a = 0 ¥
«According Fret = ma ¥ ,Fnet = ON

Snelheid is konstant tussen 0s en 10s
Verandering in snelheid/Av = 0 OF Versnelling/a = 0
-.Volgens Fnet = ma ,Fnet = 0 N

OR/OF
Gradient of graph between 0 s and 10s =0 ¥
Gradient of graph = 2 v

:. Accoding to Fret = 2 Y , Fret= 0

Gradiént van grafiek tussen 0 s en 10s =0
Gradiént van grafiek = 2
-Volgens Fnet =2, Fnet = 0 N (3)

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### Page 6

FS/March/Maart 2025

Marking Guideline/Nasienriglyn
Grade/Graad 12

3.4.1

3.4.2

3.4.3

Fret = Ap
Fret = pr - pi
Fret (5)¥_ = 40000 - (- 60000) v
Fret = 20000 N

v Any one/Enige een

Fret = ma

20000 = m(4)v
m = 5000 kg¥ (5)

POSITIVE MARKING FROM QUESTION 3.4.1
POSITIEWE NASIEN VANAF VRAAG 3.4.1

p=mvY

(- 60000) = (5000)v~

v=-12ms"

v=12 ms! West/Westward/Wes/Weswaartv (3)

POSITIVE MARKING FROM QUESTION 3.4.1 and 3.4.2.
POSITIEWE NASIEN VANAF VRAAG 3.4.1 and 3.4.2.
Ex = smv? v

¥ Ex = SMarucks Verve? od FM can Vicar)?

© Ex = 3(5000)(12)? + =(2000)(20)?v

> Exi = 360 000 + 400 000
> Exi = 760 000 J

> Ee #D Exw

« Inelastic / onefastiesY (4)
[20]

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### Page 7

QUESTION 4 / VRAAG 4

4.1. NOv NEE

FS/March/Maart 2025

Gravitational force is NOT the only force acting on the hot-air ballon/
Acceleration of the ballon is equal to zero/ Net force acting on the hot-air

ballon is equal to zero.”

(2)

Gravitasiekrag is NIE die enigste krag wat op die warmlugballon inwerk
nie/Versnelling van die ballon is gelyk aan nul/ Netto krag wat op die
warmlugballon inwerk is gelyk aan nul.

4.2 12m-s‘VupwardsY
12 m-s" opwaarts

(2)

4.3 POSITIVE MARKING FROM QUESTION 4.2
POSITIEWE NASIEN VANAF VRAAG 4.2.

OPTION/OPSIE 1
Up/Op +ve
Vy = VitaAt Y
0 =12+(-9,8)AtY
At=1,22s Vv

OPTION/OPSIE 2

Up/Op -ve

Vy = VitaAt Y

0 =-12+(9,8)AtY
At=1,22 s ¥ (3)

4.4 POSITIVE MARKING FROM QUESTION 4.2 and 4.3.
POSITIEWE NASIEN VANAF VRAAG 4.2 and 4.3.

OPTION/OPSIE 1

Up/Op +ve

Max. height (camera)/Maks. hoogte
(kamera)

Ay = viAtt Laat? v
= 12(1,22)+2(-9.8)(1,227) v
=7,35m
Height (hot-air balloon) after 1,22 s
Hoogte (warmlugballon) na 1,22 s

Ay = vat + Saat?
= 12(1,22) +3.(0)(1,227) v
= 14,64 m

:. Distance/Afstand = 14,64 —7,35 v
=7,29mv

Copyright reserved/Kopiereg voorbehou

OPTION/OPSIE 2

Up/Op -ve

Max. height (camera)/Maks. hoogte
(kamera)

Ay = viAtt Laat? v
= -12(1,22)+4 (9.8)(1,227) v
=-7,35m
Height (hot-air balloon) after 1,22 s
Hoogte (warmlugballon) na 1,22 s

Ay = viAt + Saat?
= 4 2
= -12(1,22) +5(0)(1,22°) ¥

= -14,64 m
-. Distance/Afsiand = -14,64 — (-7,35) v
=-7,29m
=7,29mv

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### Page 8

March/Maart 2025

OPTION/OPSIE 3
Up/Op +ve
Max. height (camera)/Maks. hoogte
(kamera)
v? = ve+2aAy ¥
0? = 127+2(-9,8)Ay ¥
Ay =7,35m
Height (hot-air balloon) after 1,22 s
Hoogte (warmlugballon) na 1,22
s Ay=vjAt+ saat?

= 12(1,22) +3(0)(1,22) ¥

OPTION/OPSIE 4
Up/Op -ve
Max. height (camera)/Maks. hoogte
(kamera)
v? = ve+2ady ¥
0* = -127+2(9,8)Ay V
Ay =7,35m
Height (hot-air balloon) after 1,22 s
Hoogte (warmlugballon) na 1,22 s
Ay = viAt + Saat?

= -12(1,22) +3.(0)(1,22") v

Height (hot-air balloon) after 1,22 s
Hoogte (warmlugballon) na 1,22s

Ay = vat + Saat?
= 12(1,22) +4.(0)(1,227) v

=14,64m =-14,64m

-. Distance/Afstand = 14,64 —7,35 v «. Distance/Afstand = -14,64 — (-7,35) ¥
=7,29m¥ =-7,32m
=7,29mv
OPTION/OPSIE 5 OPTION/OPSIE 6
Up/Op +ve Up/Op -ve
Max. height (camera)/Maks. hoogte || Max. height (camera)/Maks. hoogte
(kamera) (kamera)
; a
Ay = (“*)xat v Ay = ()xat v
-12+0
Ay = (= )x1,22v Ay = (= )x1,22v
Ay =7,32m Ay =-7,32m

Height (hot-air balloon) after 1,22 s
Hoogte (warmlugballon) na 1,22 s

Ay = v,At + saat?
“a 1 2
= -12(1,22) +5(0)(1,22 yy

= 14,64m = -14,64m
«. Distance/Afstand = 14,64 — 7,32 Vv :. Distance/Afstand = -14,64 — (-7,32) Vv
=7a2m = -7,32m
=7,32mv¥
Note: RANGE FINAL ANSWER (7,29 m to 7,35 m)
Let wel: REEKS FINALE ANTWOORD (7,29 m tot 7,35 m) (5)

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4.5

March/Maart 2025

OPTION/OPSIE 1

Up/Op +ve

Time taken by the camera to reach the ground/ Tyd wat die kamera neem
om die grond te bereik

Ay = viAt + saat v
-100¥ = 12At+ 5(-9,8)At? v
M=5,91s

Time taken by the jogger to cover 15 m/ Tyd wat die joggie neem om 15m
af te lé

Ay=vAt+ Saat?
15¥=3At+ 5 (O)Ar ¥
At=5s

Yes, the jogger will catch the camera Vv
Ja, die joggie sal die kamera vang

OPTION/OPSIE 2

Up/Op -ve

Time taken by the camera to reach the ground/ Tyd wat die kamera neem
om die grond te bereik

Ay = viAt+ Saat? v
100¥ =-12At + 5(9,8)At? v
At =5,91s

Time taken by the jogger to cover 15 m/ Tyd wat die joggie neem om 15m
af te lé

Ay=vjAtt+ saat?
15v= 3at + 2 (Oat? v
At=5s

Yes, the jogger will catch the camera ¥
Ja, die joggie sal die kamera vang

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### Page 10

March/Maart 2025

OPTION/OPSIE 3:

POSITIVE MARKING FROM 4.3 AND 4.4

POSITIEWE NASIEN VANAF VRAAG 4.3 AND 4.4.

Up/Op +ve

Time taken by the camera to reach the ground from Max. height/

Tyd wat die kamera neem om die grond vanaf die Maksimum hoogte te bereik:

1 2
Ay = vjAt + past

-(100+7,35)¥ = OAt+ 5(-9,8)At? v
At =4,681s
At (total)= 4,681+ 1,22
=5,90s

Time taken by the jogger to cover 15 m/ Tyd wat die joggie neem om 15m
af te lé

Ay = vjAt + sary
15v= Sat + 2(O)at’ ¥
At=5s

Yes, the jogger will catch the camera ¥
Ja, die joggie sal die kamera vang

OPTION/OPSIE 4:

POSITIVE MARKING FROM 4.3 AND 4.4

POSITIEWE NASIEN VANAF VRAAG 4.3 AND 4.4.

Up/Op -ve

Time taken by the camera to reach the ground from the Max. height/

Tyd wat die kamera neem om die grond vanaf die Maksimum hoogte te bereik:

1 2
Ay = vjAt + pant

(100+7,35)¥ =OAt+ +(9,8)at? v
At = 4,681 s
At (total)= 4,681+ 1,22
=5,90s

Time taken by the jogger to cover 15 m/ Tyd wat die joggie neem om 15 m
af te lé

Ay = viAt+ saary

15¥= Bat + 1 (O)AP v
At=5s

Yes, the jogger will catch the camera ¥
Ja, die joggie sal die kamera vang (6)

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### Page 11

FS/March/Maart 2025

Marking Guideline/Nasienriglyn
Grade/Graad 12

4.6 POSITIVE MARKING FROM QUESTION 4.3
POSITIEWE NASIEN VANAF VRAAG 4.3

OPTION/OPSIE 1
Up/Op +ve

a

v(m:s")

1,22 t s)

OPTION/OPSIE 2
Up/Op -ve
7

v (ms)

4,22 t(s)

¥

Criteria for graph/ Kriteria vir grafiek Marks
Graph starts at/ Grafiek begin by v = 12 m-s" att=0s. v
Maximum height reached/Maksimum hoogte bereik (v = 0 m-s‘') at//by t | ¥
=1,22s.

v

Correct shape as shown: straight line with gradient and section of the
graph after max height longer than before max height./ Korrekte vorm
soos getoon: reguit lyn met gradiént en snit van die grafiek na
maksimum hoogte langer as voor maksimum hoogte.

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(3)
[21]

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### Page 12

QUESTION 5 / VRAAG 5

5.1.1 Wv
5.1.2) 27
5.1.3 V¥

5.2 Hydroxyl (group) ~
Hidroksiel (groep)

5.3.1 Propan-1-ol/ 1-propanol. ¥ ~ (2 or 0)
Propaan-1-of

5.3.2 6-bromo-3-ethyl-2-methylheptane/ 6-bromo-3-etiel-2-metielheptaan

Marking criteria:
e Correct stemi.e. heptane. ¥
e Substituents (bromo, ethyl and methyl) correctly
identified. v
e IUPAC name completely correct including
numbering, sequence, hyphens and commas. ¥
Nasienkriteria:
e Korrekte stam d.i. heptaan. ~
e Substituente (bromo, etiel en metiel) korrek
geidentifiseer. “
e /UPAC-naam heeltemal korrek insluitende

nommering, volgorde, koppeltekens en kommas. “

H H H
| Cc Cc C oTH
H | H
H—C—H
ae
H

Marking criteria:
e Correct functional group (triple bond) drawn. “
e Whole structure correct.”
Nasienkriteria:
e Korrekte funksionele groep (drievoudige binding)
getrek
e Hele struktuur korrek

(1)
(1)
(1)
(1)

(2)

(3)

(2)

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### Page 13

FS/March/Maart 2025
Marking Guideline/Nasienriglyn
Grade/Graad 12

5.5.1 CaHanO2 ¥ (1)
5.5.2 CHsCH2CH2CH2CH2COOH OR CH3(CH2)aCOOH
OR
CHs(CH2)s— FO OR  CH3CH2CH2CH2CH2 —C——-OH
j j
Marking criteria: Nasienkriteria:
e Functional group for e Funksionele groep van
carboxylic acids is correct karboksielsuur is korrek
(carboxyl group) “” (karboksiel groep) ¥
e Whole structure is correct e Hele struktuur is korrek
(carbon with 4 bonds, all H (koelstof met 4 bindings, al H-
atoms shown) “ atome word getoon) “
(2)
[14]

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### Page 14

FS/March/Maart 2025

Marking Guideline/Nasienriglyn
Grade/Graad 12

QUESTION 6 / VRAAG 6

6.1

6.2

6.3

6.4

6.5

The temperature at which the solid and liquid phases of a substance

are at equilibrium. v ¥

Die temperatuur waarby die vaste- en vioeistoffases van 'n stof in ewewig

is. (2)

(Type of) Functional group/Type of Intermolecular forces/

Homologous series.” (1)
(Tipe van) Funksionele groep/Tipe intermolekulére kragte/

Homoloé reeks.

SECONDARY ’Y, carbon bonded to hydroxyl group (OH’) is bonded to two
other carbon atoms. “” (2)
SEKONDER, koolstof gebind aan hidroksielgroep (OH-) is gebind aan

twee ander kooistofatome

Liquid. (1)
Vioeistof

Ahas london forces only”, D has london forces and dipole-dipole forces”,
B has london forces, dipole-dipole forces and hydrogen bonds.”

FROM Ato D to B:

- Strength of intermolecular forces increases.v
- Energy required to overcome the strength on intermolecular forces
increases.v

Therefore the melting point of D is greater than that of A but less
than that of B.

A het slegs Londense magte, D het Londense kragte en dipool-dipoolkragte,
B het Londense kragte, dipool-dipoolkragte en waterstofbindings.

VAN A tot Dna B:
- Die sterkte van intermolekulére kragte neem toe.

- Energie wat nodig is om die sterkte op intermolekulére kragte te oorkom,
neem toe.

Daarom is die smeltpunt van D groter as dié van A maar minder
as dié van B.

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### Page 15

FS/March/Maart 2025
Marking Guideline/Nasienriglyn
Grade/Graad 12

OR/OF

A has london forces only”, D has london forces and dipole-dipole forces”,
B has london forces, dipole-dipole forces and hydrogen bonds.”

FROM B to DtoA:

- Strength of intermolecular forces decreases. ¥
- Energy required to overcome the strength on intermolecular forces
decreases.v

Therefore the melting point of D is greater than that of A but less
than that of B.

A het slegs Londense magte, D het Londense kragte en dipool-dipoolkragte,
B het Londense kragte, dipool-dipoolkragte en waterstofbindings.

VAN B tot Dna A:

- Die sterkte van intermolekulére kragte neem af.
- Energie wat benodig word om die krag op intermolekulére kragte te
oorkom, neem af.

Daarom is die smeltpunt van D groter as dié van A maar minder
as dié vanB (5)
[11]
Grand Total: 100 marks
Groottotaal: 100 punte

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