# Physical Sciences Grade 12 P1 2025 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Mpumalanga March Test 2025, 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-p1-p2-memo-march-2025-mpumalanga/de54708c-c366-462b-adec-e5ae89b50760
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Mpumalanga March Test
- Paper: 1
- Curriculum: CAPS
- Pages: 12

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## About this document

Physical Sciences · Grade 12 · Mpumalanga March Test · 2025. Memorandum, 12 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

EHLANZENI DISTRICT

PHYSICAL SCIENCES

GRADE 12

PHYSICS & CHEMISTRY (P1 & P2)

MARCH CONTROL TEST (MARKING GUIDELINES)

2025

MARKS: 100

This marking guidelines consists of 10 pages

Copyright reserved Please turn over

### Page 2

Physical Sciences 2 March/Ehlanzeni 2025

Grade 12
QUESTION 1
11 Byv (2)
12 Avv (2)
13 Dvv (2)
14 Cvv (2)
[08]
QUESTION 2
2.1 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 beweeg
tensy 'n nie-zero resultante/netto krag op dit inwerk. vv (2)
2.2.1 Zero/0 ¥ (1)
2.2.2 EASTY (1)
2.3
Fr/N T
v
eS
fm v
v
Fo/w
(4)
2.4

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### Page 3

Physical Sciences 3 March/Ehlanzeni 2025
Grade 12

Marking criteria / Merkriglyne

e Note: If Fy/w does not start from the centre deduct 1 mark.
Nota: Indien_Fo/w nie vanaf die middelpunt begin nie, trek 1 punt af.

2.5.1 Upwards as positive Upwards as negative
Opwaarts as positief Opwaarts as negatief
=mav
Foot = ma Fret = ma ¥

T+(Fo) + (F) = ma -T+F)+F=ma

TASiG.8 121) 50 T +(5)(9,8) + (21) = 5(0) ¥
T=70NY
T=70NY
(3)
2.5.2 Fret = ma
T +(-Fg) + (-F) =ma
70cos60°+ (-0.25(9.8m — 70sin60°)) Y= 0 ¥
mM = 20,472 kg ¥ (3)
2.6 INCREASESY
e The net force acting on the object will increase v
TOENEEM ¥
e Die netto krag wat op die voorwerp inwerk sal verhoog.¥
(2)
[19]

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### Page 4

Physical Sciences 4 March/Ehlanzeni 2025

Grade 12
QUESTION 3
3.1 Motion during which the only force acting on an object is the gravitational force VY
Beweging waartydens die enigste krag wat op die voorwerp inwerk, gravitasie krag is.
vv

(2)
3.2.1 80mv (1)
3.2.2 9,8 m-s? ¥ downwards / afwaarts ¥ (2)
3.3.1
Option 1
Upwards as positive Upwards as negative
Opwaarts as positief Opwaarts as negatief
Vi = Vi + adt ¥ Vi= Vi + adt ¥
0= vi+ (-9,8)(0,6)” O0= vi + (9,8)(0,6)¥
vi = 5,88 m-s' upwards / opwaarts ¥ vi = -5,88 m-s?
..Vi = 5,88 m-s upwards / opwaarts
v
Option 2
Upwards as positive Upwards as negative
Opwaarts as positief Opwaarts as negatief
1 13
Ay= viat+ saat v Ay=viAt+aaty
1 2 1 2
O=vi(1,2)+5(-9.8)(1,2)'Y O=wi(1,2)+5(9.8)(1,2)'v
vi = 5,88 m-s"' upwards / opwaarts V vi = -5,88 m-s™
..Vi = 5,88 m-s' upwards/opwaarts
Y
(3)
3.3.2
Option 1 Option 2
h= 80-55 v h= -55+80v
h=25mv h=25mv (2)
3.3.3
Upwards as positive Upwards as negative
Opwaarts as positief Opwaarts as negatief
V? = ve+2aAy¥ V? = v?+2adyv
(0) = v? + 2(-9,8)(60)v (0)? =v? + 2(9,8)(-60)v
vi= 34.293 m-s'upwards /opwaarts VY | vi = 34.293 m-supwards/opwaarts ¥ (3)

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### Page 5

Physical Sciences 5

3.3.4

Grade 12

March/Ehlanzeni 2025

Upwards as positive
Opwaarts as positief

Upwards as negative
Opwaarts as negatief

Ve = v?+2aAy
v? = (5,88)? + 2(-9,8)(-80)”
vi = 40,032 m-s! downwards
Vi=Vvi+ adt ¥
-40,032 = 5,88 + (-9,8)At ¥
At = 4,685 s

Vi =Vi+ adt
0 = 34,293 + (-9,8)At ¥
At = 3,499 s

wt = 4,685 + 3,499 = 8,184 (s) ¥

vy =v +2aay
v? = (-5,88)* + 2(9,8)(80)¥
vi = 40,032 m-s! downwards
Vi = Vi + aAt ¥
40,032 = -5,88 + (9,8)At ¥
At = 4,685 s

vi = vit adt
= -34,293 + (9,8) At ¥
At = 3,499 s

“t= 4,685 + 3,499 = 8,184 (s) ¥

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(5)

### Page 6

Physical Sciences 6 March/Ehlanzeni 2025
Grade 12

3.4

Marking criteria / Merkriglyne
e Correct shape and ends at time, t. ¥ / Korrekte vorm en eindig by tyd t Vv

e Initial velocity ¥ / Aanvanklike snelheid ¥

e Velocity at which it leaves the ground after the first bounce.“ / Snelheid
waarteen dit die grond verlaat na die eerst bons ¥

e Time taken to reach maximum height for the first time.~/ Tyd wat dit neem om
die maksimum hoogte vir die eerste keer te bereik. ¥

Upwards as positive / Opwaarts as positief

Time / Tvd (s)

Velocity / Snelheid (m-s"')

Upwards as negative

Velocity (m-s")

Time(s)

-5,88

4
-34,29)

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### Page 7

Physical Sciences 7
Grade 12

QUESTION 4

41

4.2.1

4.2.2

4.2.3

_1 2 1 2
) Ex, =5mv* + 5mv
38,5 =" yay + 1 3)2¥
"2 2

m=5kgv¥

March/Ehlanzeni 2025

Is a system on which the net external force is zero VY (2 or 0)
Is ‘n sisteem waarvan die netto eksterne kragte zero is VV 2 of 0)

Note: if direction is specified in the final answer, deduct 1 mark.
Nota: /ndien rigting gespesifiseer word in die finale antwoord, trek 1 punt af

East as positive / Oos as positief

East as negative / Oos as negatief

Ap = m(v;— vi) ¥
“12,5 = 2(vi—4) ¥
vi = - 2,25 m-s!

2. Wi = 2,25 m-st ¥

Ap = m(v;— vi) ¥
12,5 = 2(vi—(-4)) ¥

vi = 2,25 m-s'¥

Option 1

/ Opsie 1

East as positive / Oos as positief

East as negative / Oos as negatief

(2)(4) + (5)(-3) = (2)(-2,25) + 5v2 W
V2 = -0,5 m-s™

“. V2 = 0,5 m-s' west / wes ¥

(2)(-4) + (5)(3) = (2)(2,25) + 5v2 ¥

V2 = 0,5 m-s! west / wes ¥

Option 2

/ Opsie 2

East as positive / Oos as positief

East as negative / Oos as negatief

Ap = m(v;— vi) ¥
12,5 = 5(vi— (-3)) ¥

vi =- 0,5 m-s"

. Vj = 0,5 m-s" west ¥

Ap = m(v;— vi) ¥
12,5 = 5(v— 3) ¥

vi = 0,5 m-s" west ¥

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[22]

### Page 8

Physical Sciences 8 March/Ehlanzeni 2025
Grade 12

4.3 EQUAL TOV
The trolleys exert forces of equal magnitude on each other and the time of contact during
the collision is the same v (Accept: the trolleys have equal magnitude of change in

momentum).

GELYK AAN ¥

Die trollies oefen kragte uit op mekaar met gelyke grootte en die kontaktyd gedurende
die botsing is dieselfde. v (Aanvaar: die trollies het gelyke grootte van verandering in
momentum).

(2)
[12]

QUESTION 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 VY
‘n Binding of ‘n atom of ‘n groep van atome wat die fisiese en chemiese eienskappe van
‘n groep organiese verbindings bepaal Vv

(2)
5.2.1 CHsCH2CH(CHs)CHO VY

Marking criteria / Merk riglyne
e Ending with CHO v / Eindig met CHOV
e Whole structure correct and CHs in brackets v / Hele struktuur korrek
en CHs in hakies ¥ (2)

5.2.2 3-methyl”butanal ¥

Marking criteria / Merk riglyne
e 3-methyl/ 3-metiel ¥

e _butanal / butanaal Vv (2)

5.3.1 SECONDARY (HALO-ALKANE) ¥

The carbon atom with the halogen atom or the functional group is bonded to two other
carbon atoms v

SEKONDERE (HALO-ALKAAN) ¥

Die koolstofatoom met die halogen atoom of die funksionele groep is gebind aan twee

ander koolstofatome v (2)

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### Page 9

Physical Sciences 9 March/Ehlanzeni 2025
Grade 12

5.3.2 4-bromo-2,2-dimethylpentane / 4-bromo-2,2-dimetielpentaany V ¥

Marking criteria

¢ Correct stem i.e. pentane. / Correct stam, pentaan V

e All substituents (bromo and dimethyl) correctly identified.“ / Alle
substituente (bromo en dimetiel) is korrek geidentifiseer. ¥

e IUPAC name completely correct including numbering, sequence,
hyphens and commas. ¥/ /UPAC naam volledig korrek, insluitden
nommering, volgende, koppelpunte en kommas ¥

5.4.1

(1)
5.4.2 2C4Hi0 + 1302 > 8CO2 + 10H20

Marking criteriab/ Merkriglyne

e Correct reactants (CaHio and Oz) “ / Korrekte reaktante (CaHi0 and Oz) ¥
e Correct products (COz and H2O) ¥ / Korrekte produkte (COz and H2O) ¥
e Correct balancing without multiples and fractions Y / Korrekte balansering

sonder veelvoude en breuke ¥

e Marking rule 6.3.10 is applied / Merkreé! 6.3.10 is toegepas

5.5.1 Carboxyl (group) ¥ / Karboksiel (groep) ¥ (1)
5.5.2 H H o
wt oy
Id

Marking criteria / Merkriglyne
e Functional group correct v / Funksionele groep is korrek ¥

e Whole structure correct Y / Hele struktuur is korrek. V

(2)
[18]

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### Page 10

Physical Sciences 10 March/Ehlanzeni 2025
Grade 12
QUESTION 6

6.1

Marking criteria / Merkriglyne

If any one of the underlined key words/phrases in the correct context (temperature) is
omitted, deduct 1 mark.

Indien enige van die onderstreepte sleutelwoorde/ frases in die korrekte konteks
(temperatuur) uitgelaat word, trek 1 punt af.

The temperature at which the vapour pressure of a substance equals atmospheric
pressure Vv ¥

Die temperatuur waarby die dampdruk van ‘n stof gelyk is aan atmosferiese druk. VV

(2)

6.2.1 London forces / induced dipole forces / dispersion forces

London kragte / geinduseerde dipoolkragte / dispersie kragte ¥ (1)

6.2.2 Sv
Has a higher boiling point than R v (or R has a lower boiling point then S)

Sv
Het ‘n hoér kookpunt as R v (of R het ‘n laer Kookpunt as S)

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### Page 11

Physical Sciences 11 March/Ehlanzeni 2025
Grade 12
6.3

rking criteria / Merkriglyn

¢ @Q(2-bromobutane) Y / Q (2-bromobutaan) ¥

e State London forces in P ¥ / Noem Londonkragte in PV

¢ State dipole-dipole forces in Q V / Noem dipool-dipool kragte in QV

¢ Compare strengths of IMFs/compare energies required v / Vergelyk sterkte
van IMK / vergelyk energie benodig ¥

¢ Q(2-bromobutane) Y / Q (2-bromobutaan) ¥
¢ Compound P has London forces” and compound Q has dipole-dipole forcesY
¢ Compound Q has stronger intermolecular forces than P/Compound P has weaker
intermolecular forces than Q ¥
OR
e More energy is required to overcome the intermolecular forces in compound

Ql/Less energy is required to overcome the intermolecular forces in compound
Py

¢ @Q(2-bromobutaan) ~
e Verbinding P het Londonkragte “ en verbinding Q het dipool-dipool kragte ~
¢ Verbinding @ het sterker intermolekulére kragte as P/Verbinding P het swakker

intermolekulére kragte as Q ~

OF

Meer energie word benodig om die intermolekulére kragte te oorkom in
verbinding Q/Minder energie word benodig om die intermolekulére kragte te
oorkom in verbinding PY

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(4)
[09]

### Page 12

Physical Sciences 12 March/Ehlanzeni 2025

Grade 12
QUESTION 7
7.1.1. Addition Y (accept: hydrogenation) / Addisie Y (aanvaar: hidrogenasie) (1)
7.1.2 Use Pt/Ni/Pd as a catalyst ¥ / Gebruik Pt/Ni/Pd as katalis ¥ (1)
7.1.3
Marking criteria / Merkriglyne
¢ Correct structural formula of ethene ¥ / Korrekte struktuurformule van
eteen ¥
© H—-Hor/of He v/
¢ Correct structural formula of ethane V / Korrekte struktuurformule van
etaan ¥
H oH
HH y - | |v”
n—tat “4 + H—H/H,; —— ana
H H
(3)
7.2.1 Substitution’ (Accept: Halogenation) / Vervanging V ( Aanvaar: halogenering) (1)
7.2.2. Halo-alkanes/Alkyl halides Y / Halo-alkane / alkielhaliedes V (1)
7.2.3. Hydrogen chloride Y (do not accept: HC?) / Waterstofchloried Vv | Moenie HCt
aanvaar nie) (1)
7.3.1 Hydrohalogenation / Hidrohalogenering ¥ (1)
7.3.2 1-chloroethane / 1-chloroetaan VY (Accept: chloroethane / aanvaar: chloroetaan) (2)
7.4 Elimination / Eliminasie ¥ (1)
[12]

GRAND TOTAL: 100

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