# Physical Sciences Grade 12 P1 NSC June 2012 Memo (Afrikaans)

> Grade 12 Physical Sciences Paper 1 memorandum from the NSC June 2012 (Afrikaans), 9 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-science-p1-sg-english-and-afrikaans-june-2012-memo/0977e089-b0f0-4d9a-84ce-7a9d8adacf92
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Language: Afrikaans
- Document type: Memorandum
- Year: 2012
- Exam period: NSC June
- Paper: 1
- Curriculum: CAPS
- Pages: 9

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 P1 NSC June 2012 Memo (Afrikaans)](https://www.hlayiso.com/library/physical-science-p1-sg-english-and-afrikaans-june-2012-memo/0977e089-b0f0-4d9a-84ce-7a9d8adacf92)

## About this document

Physical Sciences · Grade 12 · NSC June · 2012 · Afrikaans. Memorandum, 9 pages. Read online or download the PDF.

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

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

Downloaded from hlayiso.com

NTT

basic education
Department:
Basic Education
REPUBLIC OF SOUTH AFRICA

SENIOR CERTIFICATE EXAMINATION
SENIORSERTIFIKAAT-EKSAMEN

PHYSICAL SCIENCE P1/NATUUR- EN SKEIKUNDE V1
oe __PHYSICSIFISIKA
STANDARD GRADE/STANDAARDGRAAD

pee

EDUCATION
MARKS: 150
pote -05- 25 PUNTE: 150
TE BAG X (10
PRETORIA 0001 _

PUBLIC EXAMINATION

This memorandum consists of 9 pages.
Hierdie memorandum bestaan uit 9 bladsye.

Vs Ga ya —
aS a MALLS (9-9 S-VAL

Please turn over/Blaai om asseblief

0
Copyright reserved/Kopiereg voorbehou Aetl

|
|
|

### Page 2

Downloaded from hlayiso.com

Physical Science/Natuur- en Skeikunde/SG/P1/V1 2 DBE/2012
SCE/SSE ~ Memorandum

QUESTION 1/VRAAG 1

11 +A 12 B 13° 4A 14 A ik ©

is) (3) eh 18 C 19 A 10 B

1.11 B 1.12 C 1.13 D 1.14 A 1.15 C

[15x3 = 45]

[General remarks > Accept all equations / formulae as currently used in NCS / CAPS
~> Where applicable use g = 9,8 m.s® also acceptable
> Rounding off to whole numbers or 1 decimal also accepted
> Equivalent wording in all cases to be accepted

QUESTION 2/VRAAG 2

2.1. * the resultant force on the block is zero/resulterende krag op blok is nul OR/OF

* all the forces acting on the block are balanced/al die kragte wat op die blok
inwerk, is gebalanseerd. ORIOF
e Itis at rest/dit is in rus.

e All forces will form a closed triangle / a/ die kragte vorm ‘n geslote driehoek
vv (2)

43.cm

40°
10 cm 10cm (8)

v Measure 10 cm (50 N) — must be ‘horizontal’
Meet die 10 cm (50 N) horisontaal
¥ Measure 140° (or 40° from horizontal)
Meet die 140° (of die 40° vanaf die horisontaal)
v Measure 13 cm (force in XY)/Meet 13 cm (krag in XY)
v¥ Complete parallelogram or triangle/Voltooi parallelogram of driehoek

v¥ Measure R/Meet R (8.1 cm to 8.5 cm)
¥ Convert R to N/Skakel R om na N_: Range: 40,5 N to
Y Direction/Rigting O0°%lrigting 0°/due north/reg noort

PARTMENT OF BASIC
42 g2.BH EDUCATION

= 41,5 Nin a direction of 0°/rigting 0°/due north/reg ngord 2012 -05- 25

PRIVATE BAG X 110

F=mg v F=mg Y iH PRETORIA 0001
Y v ¥ PUBLIC EXAMINATIONS
41,5=mx10 at 5= mx9,8 a
m = 4,15 kg¥ m = 4,23 kg¥ (4)
[14]
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

### Page 3

Downloaded from hlayiso.com

Physical Science/Natuur- en Skeikunde/SG/P1/V1 3 DBE/2012
SCE/SSE — Memorandum
QUESTION 3/VRAAG 3 :
34 =u + atv v=utatyv VY Av _6Y v=at
a=——=—
v _ 0 marks
6 = a(4) =15ms
a=15ms? ¥

a=1,5ms*

(4)

3.2 | For1 secondat constant velocity/

112
= = v
s ut + pat

: a
= (6)(1) + gO

=6mY
OR/OF
s = vt= 6)(1)=6 mY
OR/OF

v +
s= (Ite CS =6m¥

Vir 1 sekond teen konstante snelheid

If the displacement to come to
restis 12 m/As die verplasing om tot
rus te kom, 12 m is
y2 =u +2as %
v
(0% es 2(-1)s
s=18mv
Cyclist need more than 18 mto
stop/Fietsryer benodig meer as

vv

18m omte stop OR/OF
Total displacement/

Totale verplasing

=6+18

=24mv _

For1 second at constant velocity/
Vir 1 sekond teen konstante snelheid

: 1
=ut+—at? Vv
s=ut 3%

= 61) + ZO)?
=6m/’
ORIOF
s=vt=(6)(1)=6m”%
OR / OF

vv
utv 6+6
$= (1 = me =6m

Copyright reserved/Kopiereg voorbBO8 0S- 25
PRIVATE BAG * 10

PRETORIA 0001 me

Cyclist need more than 18 mto
stop/Fietsryer benodig meer as
18m om te stop

If the displacement to come to
restis12 m/As die verplasing om tot
rus te kom, 12 m is

y2=u2 +2as %
v
Cores 2(a)12
=-15 ms? ¥

Cyclist has a constant deceleration of -1 ms"?

he must have a deceleration of - 1,5 m.s"2to
stop in time /He does not stopintime vv
Fietsryer het 'n konstante negatiewe versnelling

van-1 m.s”?, hy moet 'n negatiewe versnelling

-2 :
| .van-1,5m.s"“ he om betyds te stop : Hy stop dus

+ Gie betyds nie.

Please turn over/Blaai om asseblief

ane ELAMINATONS

PUBLIC EX

— K

### Page 4

Downloaded from hlayiso.com

Physical Science/Natuur- en Skeikunde/SG/P1/V1

4 DBE/2012

SCE/SSE - Memorandum

For 1 second at constant velocity/
Vir 1 sekond teen konstante snelheid

Le
=ut+ cat" v
s=ut+ sal

- 2
= (6)(1) + 30)

=6m Vv

OR/OF

s= v= (6)(1)=6 mY
OR / OF

wa
utv 6+6
s= (t= (SM 6m %

3.3 v

If the displacement to come to
restis 12 m/As die verplasing om tot
rus te kom, 12m is
a
s=utt+ 2at
v v v
12 =6t+ see
t=255s ¥
v=utat
0=6+(-1)t
t=6s¥%
Cyclist must complete final 12 min6 s but has
only 2,55 s v
Fietsryer moet laaste 12 min6 s voltooi, het net 2,55 s
Cyclist will not stop in time / stop nie betyds nie

(8)

v v

Cyclist continues to move forward at same speed/velocity due to inertia/Newton4

Fietsryer beweeg steeds vorentoe teen selfde spoed/snelheid a.g.v. traagheid

v

v

An object will stay at rest or carry on moving at a_ constant velocity in a straight line

unless acted on by an external resultant / net force (Newton1) ”
n Voorwerp sal in rus bly of teen n konstante spoed in n reqguit lyn volhard tensy
daar in eksterne resulterende / netto krag daarop inwerk (Newton1) (3)

Copyright reserved/Kopiereg voorbehou

[15]

DEPARTMENT OF BASIC

EDUCATION

202 -05- 25

PHaVATE BAG x 110
PRETORIA 0001

PUBLIC EXAMINATIONS
err L

Please turn over/Blaai om asseblief

Go? I

### Page 5

Downloaded from hlayiso.com

Physical Science/Natuur- en Skeikunde/SG/P /V1 5 DBE/2012
SCE/SSE — Memorandum
QUESTION 4/VRAAG 4
If 15 m.s” used without being calculated — full marks
as 15 m.s' gebruik word sonder ‘n berekening — volpunte ifg=9,8m. g2 is used
" = As g=9,8m.s"
1 v=u* gt gebruik:
s=ut+ zat'v = 0 + (10)(1,5) Answer / antwoord:
v 2 =15m st ae
= 0+ 3001.5) — $= 11,03m
2 ¥5+0”% s=11,03m
= 11,25 mv Beuk + 298” aes ns \
4 @ <0) (toys) |} Y= 14,28 m. (4)
s =17,25m v
4.2
on if g = 9,8 ms? is used
_ g = 9,8 m.s~ is use
= 0 + (10)(18) ae 7 As g=9,8m.s?
=15ms7¥ 2 gebruik:
OR/OF VA 25 = “5 Ye Answer / antwoord:
v? =u? 4 298 ee v=147 ms"
= (0) + 2(10)(11,25) DEPARTMENT OFLBASIC
v =15m.st% EDUCATION
2012 -05- 25 tt
PRIVATE BAG X 110
QUESTION S/VRAAG 5 PRETORIA 0004
PUBLIC EXAMINATIONS

5.1 Aresultant force atting on a mass proauces an Acceleration in the direction of the
y nice. The acceleration is directly propoftional to the resultant force and is

¥ inversely propotional to the mass.

nResulterende krag wat op n massa inwkrk. sal die massa laat versnel in die
rigting van die resulterende krag. Die versnelling is direk eweredig aan die krag en

v omgekeerd eweredig aan die massa.

OR / OF
v v

The rate of change in momentum of an object is directly proportional to the
resultant / net force acting on the object and the rate of change of momentum is in
the direction of the resultant / net force.”
Die tempo van momentumverandering van ‘n voorwerp is direk eweredig aan die
resulterende/netto krag en die momentumverandering is in die rigting van die
resulterende / netto krag.

F (3)
aa— 1mark only
m

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

### Page 6

Downloaded from hlayiso.com

Physical Science/Natuur- en Skeikunde/SG/P1/V1 6 DBE/2012
SCE/SSE — Memorandum

=ma” v

5.2 Fres = ane ov Fres = Ma = Fappliea — Friction
7 4 v v
= 110N 55x 2= 115 — Friction

Fres = Friction + Fapptied

wee aaa,
110 = Friction + 115 ¥ Friction = 115-110

Friction = 5N  ¥

on = v > :
Ee Feton N a ne Friction = 5 N in the opposite a
friction = 9 Nin ; direction of being pulled v I
opposite direction of being
pulled
QUESTION 6/VRAAG 6
v v ¥
6.1 In an isolated system the total (linear) momentum remains a constant in
magnitude and direction, or
v v v
In an isolated system the total momentum before a collision is equal to the total
momentum after the collision. (3)
v v v
Inn geisoleerde stelsel bly die totale (liniére) momentum konstant in grootte en
tigting v v
of

Inn geisoleerde stelsel is die totale momentum voor fn botsing gelyk aan die totale
momentum na die botsing.

6.2 | (total momentuM)pefore = (total MomentuM)arer
(mp + mMs)U = Mp_V + MeV

v v v
(50 410)2 = 50v + 10(-3)
v = 3 ms" Vto the right vina regs

(6)

Pbefore = Pafter

“DEPARTMENT OF BASIC
EDUCATION

wl

momentuMpbefore = MOoMEntuUMatter

12 -05 25 19)

| PRIVATE BAG A 109
j PRETORIA 0001

PUBLIC EXAMINATION

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

### Page 7

Downloaded from hlayiso.com

Physical Science/Natuur- en Skeikunde/SG/P 1/V1 if DBE/2012
SCE/SSE — Memorandum

QUESTION 7/VRAAG 7

7.1. The sum “of the gravitational potential and kinetic energy of the object. v (2)

Die som v van die gravitasie potensiéle energie en die kinetiese energie van die
voorwerp ¥

Law of conservation of mechanical energy will also be accepted
Wet van behoud van meganiese enerige sal ook aanvaar word.

a2 j
total mechanical Energy soomm = total Mechanical ENerY vest polht of motion

_ v
(E, + Ey )atsoomm = (E, + Ey atiowest point of motion /laagste punt van beweging

(mgh +5 coon = (mgh +3mv?)

at the lowest point of motion laagste punt van beweging
YoY A
(0,15)(10)(0,1) + 0 = (0,15)(40)(0) + 3(015)v")

=1,41 mst”

Viowest point of moiion/iaagste punt van beweging

If 9,8 m.s” used then the answer is : 14ms7
(6)

Note: E,(top) = E,(bottom) will lose only 1 mark can still get 5
E,(bo) = E,(onder) verloor 1 put maar kan nog 5 punte verdien

v=./2gh zero

o
6

7.3. Equations of motion only apply to motion in a straight line. ~v” OR The bob does
not move at uniform acceleration. OR To apply equations of motion the bob must
move under uniform acceleration. (2)
Bewegingsvergelykings is slegs van toepassing op liniére beweging. VY OF Die
massastuk beweeg nie teen uniforme versnelling nie OF Om van toepassing te
wees, moet die massastuk uniform versnel.

[10]

DEPARTMENT OF BASIC
EDUCATION

2012 -05- 25

PRIVATE BAG & 110
PRETORIA 0001

L_PUBLIC EXAMINATIONS |

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Bo

### Page 8

Downloaded from hlayiso.com

Physical Science/Natuur- en Skeikunde/SG/P 1/V1 8 DBE/2012
SCE/SSE — Memorandum

QUESTION 8/VRAAG 8
v v v v
8.1 Phas an excess of electrons /P het in oormaat elektrone/

(2)

v
8.2 The force (electrostatic / force of attraction or repulsion) which one charge exert

v
on the other is directly proportional to the product of the charges and is inversely
v
proportional to the square of the distance between them.

v
Die elektrostatiese (aantrekkende of afstotende) krag wat een lading op n ander

v v
uitoefen, is direk eweredig aan die produk van die ladings en omgekeerd eweredig
aan die kwadraat_van die afstand tussen hulle.

(3)

Curved electric field
line impt. on outside

Geboé elektriese veld
en lyne belangrik op
buitekant

¥ field lines between charges/veldiyne tussen die ladings

Y field lines to left and right of the charges/veldlyne links en regs (buite die) van ladings

Y direction: from Q to P/rigting: van Q na P toe

Y field lines not going into spheres but touching surface and without touching each other
/veldlyne nie binne sfere nie, maar raak opperviak en sonder om aan mekaar te raak

If dots are used for spheres award full marks for last bullet

(4)

8.4
kqyqo y DEPARTMENT OF BASIC
fa EDUCATION
Jo v
_ (9x10°)(6x1079)(3x1079)
ma, 2012 -05- 25 F
7 PRIVATE BAG X 110
=18x10-5N % PRETORIA 0001
PUBLIC EXAMINATIONS |
8.5 . 9% 49-9 gy (3)
new charge = 2% 107 +310” =-15x1079 C [18]
2v
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief f

### Page 9

Downloaded from hlayiso.com

Physical Science/Natuur- en Skeikunde/SG/P1/V1 9
SCE/SSE — Memorandum

DBE/2012

QUESTION 9/VRAAG 9
9.1
f Sv
ae —— Rtotal series = 14 2=30
R Rea Rio Rat tt
iy R Rea Reon
a rey a 60 series
[ 3 1)
“2 “ Ihe =20
= v
R effective) parallel = = 2 Q
voivy v
92.1 emf=V=IR=6x2=12V
OR /O)
amis Ry
= (20) = 13
9.2.2 | = Vert ThA rs Sled
10+20 R +R ov
~~ Gof Vacross"2 2
aay,
“4A x42 8V
=x12=
V, Flea * Roo ¥
=4x2”
=8VVY

9.3.1 Remains the same/bly dieselfde

9.3.2 Decreases/neem af vv

9.3.3 Decreases/neem af,

Copyright reserved/Kopiereg voorbehou

A
Rtotal series = 14 2=30
Rproduct v
Reffec = R
sum
_ (6x3) ¥
“6 +3) Vv
R =2Q v
(8)
(4)

=
PARTMENT OF BASI
DEP EDUCATION -

ani? -05- 25°)

PRIVATE BAG X 110
PRETORIA 0001

PUBLIC EXAMINATION:
ee

(2)
(2)

(2)

[21]
Question 1: 45
Question 2-9: 105
Grand Total: 150

Ke

## Related documents

- [Physical Science P2 (SG) (English and Afrikaans) June 2012 Memo](https://www.hlayiso.com/library/physical-science-p2-sg-english-and-afrikaans-june-2012-memo/230c1a41-a0da-4842-92a1-abb8df5e31ac) — Physical Sciences · Memo · 2012 · Afr
- [Gr 12 Physical Sciences P1 (English and Afrikaans) June 2024 Possible Answers](https://www.hlayiso.com/library/gr-12-physical-sciences-p1-english-and-afrikaans-june-2024-possible-answers/f78ca708-f7f0-4bc5-8ff9-36594b1b1052) — Physical Sciences · Memo · 2024 · Afr
- [PHYS SC P1 MEMO GR12 JUNE2021 Afrik English](https://www.hlayiso.com/library/phys-sc-p1-memo-gr12-june2021-afrik-english/51075955-8380-4def-a361-3f9b02ab715c) — Physical Sciences · Memo · 2021 · Afr
- [Physical Science Bill P1 (HG) 2013 Memo](https://www.hlayiso.com/library/physical-science-bill-p1-hg-2013-memo/8307a6e5-2dc5-4bdb-8d5d-f16160258425) — Physical Sciences · Memo · 2013
- [PHYSICAL SCIENCE P1 QP JCT 2025 AFR D watermark](https://www.hlayiso.com/library/physical-science-p1-qp-jct-2025-afr-d-watermark/b9429ac5-acbb-4fae-989d-af9e1478a96b) — Physical Sciences · QP · 2025 · Afr
- [PHYSICAL SCIENCES P1 GR12 QP JUNE2022 Afrikaans](https://www.hlayiso.com/library/physical-sciences-p1-gr12-qp-june2022-afrikaans/7da2a467-33ba-4d42-9de8-a501165ff1e0) — Physical Sciences · QP · 2022 · Afr

## More from Grade 12 Physical Sciences

- [PHYSICAL SC P2 QP GR12 SEPTEMBER 2025 AFR1 watermark](https://www.hlayiso.com/library/physical-sc-p2-qp-gr12-september-2025-afr1-watermark/37cff2db-7c1a-4ec8-8ac5-07ec5bcf87a9) — QP · 2025
- [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — QP · 2025
- [Physics Grade 12 NSC P1 QP September 2025 KZN](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-kzn/6d225dc8-24c6-4c28-a181-df906d5a2c0e) — QP · 2025
- [Physical Sciences Physics Grade 12 PS01 QP March 2025 Limpopo](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-ps01-qp-march-2025-limpopo/482ee9af-e88a-4818-9120-11ab9e045a1c) — QP · 2025
- [Physics Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-limpopo/b65e9832-a6be-4312-8f5a-57425d647121) — QP · 2025
- [Physical Sciences Physics Grade 12 NSC QP P1 May June 2025 Mpumalanga](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-qp-p1-may-june-2025-mpumalanga/d9c5f1c3-bd6c-468e-a50a-f6b40782c695) — QP · 2025

## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.