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

> Grade 12 Physical Sciences Paper 1 memorandum from the NSC June 2012 (Afrikaans), 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-science-p1-hg-english-and-afrikaans-june-2012-memo/e4632c7d-9db7-4367-afc4-33c738ae3a8f
- 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: 10

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## About this document

Physical Sciences · Grade 12 · NSC June · 2012 · Afrikaans. Memorandum, 10 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

basic educatio

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n

Department:
Basic Education

SENIOR CERTIFICATE EXAMINATION
SENIORSERTIFIKAAT-EKSAMEN

REPUBLIC OF SOUTH AFRICA

PHYSICAL SCIENCE P1__
__ NATUUR- EN SKEIKUNDE V1

_HIGHER GRADE/HOER GRAAD

BOR es

MEMORANDUM

DEPARTMENT OF BASIC
EDUCATION

2012 -05- 25

PRIVATE BAG X 110
PRETORIA 0001

PUBLIC EXAMINATIONS

This memorandum consists of 10 pages.
Hierdie memorandum bestaan uit 10 bladsye.

WR STR eG
hp

WALL

Copyright reserved/Kopiereg voorbehou

hastens ceeencecess iis piiiaaisssains Sesissss

MARKS: 200
PUNTE: 200

Vio s| 10} 1.
Please turn over/Blaai om asseblief

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Physical Science/Natuur- en Skeikunde HG/P1/V1 2 DBE/2012
SCE/SSE —- Memorandum

QUESTION 1/VRAAG 1
B

14 iw A 13°C 14. 8B 15° 4A
16 8B 17 A 18 D 19 A 110 C
111A 112 B 113 D 114 A 115 D

[15x 4=60]

[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
>Equivilant wording in all cases to be accepted

QUESTION 2/VRAAG 2

v
force BC/krag BC/Tac

2.4 force AB/krag AB/ Tas 60° Y Or any other correct angle/ enige

a ander korrekte hoek

Man’s weight/Gravitational force of earth Fe/Fearn/ W (4)
Man se gewig/Aantrekkingskrag van die aarde/Fo/F ade! W

OR/ OF
v
force BC/krag BC/Tac Man’s weight/Gravitational force of earth Fe/Featn/ W ¥
Man se gewig/Aantrekkingskrag van die aardelF G/F aarde! W
force AB/krag AB/Tas ¥

OR / OF.

y-component of force BC/krag BC/Tac
¥ for both components / beide komponente
x-component of force BC/krag BC/Tac

4
force AB/krag AB/Tas

v

Man’s weight/Gravitational force of earth Fe/Feann/ W
‘Man se gewig/Aantrekkingskrag van die aarde/Fg/Faardel W

Note: *No arrows no marks / geen pylpunte geen punte
+ Vectors/ arrows not starting at same point / vektore begin nie in een punt (- 1)
«If any extra forces OR BC and components ( — 1} / as enige ander kragte OF BC
en komponente (- 1)

2.2. Horizontal / x component/horisontaal / x komponent / Fac cos 60°V ¥ (2)

ee
DEPARTMENT OF BASIC
: EDUCATION
Copyright nine voorbehou
2012-08-25
' PRIVATE BAGX 10
PRETORIA 0004
PUBLIC EXAMINATIONS

Please turn over/Blaai om asseblief

h

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Physical Science/Natuur- en Skeikunde HG/P1/V1 3 DBE/2012
SCE/SSE— Memorandum

v
2.3. When three forces acting at the same point are in equilibrium they can be
represented in magnitude and direction v by the sides of a triangle taken “

in order./
Wanneer drie kragte op dieselfde punt inwerk wat in ewewig is, kan
hulle voorgestel word in grootte en rigting deur die sye van fn driehoek (3)
2.4 Facey = 400 = Faccos60° vor 400
Fes | Fecy | Fee= 800 ¥ 80 Fac
Fac y = 800c0s30° v : a
= 692,82 Nv Fac = 800 N
5 400? + Facyy)* = 8007
ee) Feoyy) = 69282N ¥
v
cos60°= a = y
oa o_ BCIY) (4)
Fac = 800N tan60°=— oy y
singo° = 200) . Fecty) “ Facty) = 400 x tan 60° = 692,82N
Fac B00 v |
Facty) = 800 x sin60°= 692,82N

2.5 Fg = Faciy yeuilibriunt/ ewewig Fe = Faciy) equilibriun’/ ewewig
F=mx10= 692,82N Fg =mx98 = 692,82N 3)
m = 69,28 kg y m= 70,70 kg v [16]
DEPARTMENT OF BASIC
EDUCATION
QUESTION 3/VRAAG 3
a1 AB vv 2012 ~05- 25 (2)
PRIVATE BAG X 110
32 va p-200" va PRETORIA 0001
v(velocity) = gradient = -00= 2007 e 0,67m.8" ‘PUBLIC EXAMINATIONS)
v=0,67m.s~" opposite to direction of original motion he ee of motion
| AS
v(velocity) = gFadient= 200-9 4 o6Tms 1 rahe ahaa aa
v=0,67ms 1 opposite to direction of original motion 2, zero

Because the graph shifted a value of 205 could have been used — answer: 0,68 m.s"
Omdat die grafiek geskuif het kan ‘n waarde van 205 gebruik word — antwoord:
0,68 ms

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### Page 4

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

3.3

eoceoe

3.4
3.5

3.6

3.7

3.8

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DEVY

v

EF vv

4 DBE/2012
SCE/SSE ~ Memorandum
Because the graph shifted the following answers will be acceptable:
Omdat die grafiek geskuif het sal die volgende antwoorde aanvaar word:
Only / slegs BC VVVY (4) or
BC vv and DEVY (4) or
(2) or
BC and any other set of points / BC en enige ander stel puntev ¥ (2)
v v
200 + 200 = 400m OR/OF 205 +205 =410m (3)
(2)
1.2 vtu
$= —at«v =
ut + 3 a s=( 5 yt
v v. v
200 = (0x 200) + “ (ay(2002) 200 = rs = 2 500
2 2 (5)
a=001ms-2v v=2ms"
= v4 “
a A og v2 Fu2 +28”
2 =0+ (a)(200) 2 _
, 2* =0 + 2(a)(200)
= 2
ee a=001 ms? v

v=urat v? =u? + 2asV
re v i v
= 0 + (0,01)(200) = 0 + 2(0,01)(200)
v=2ms' v v=2ms'v¥
ay
2 |
v
- 4
(m.s") vv
0,67
0 v

“DEPARTMENT OF BASIC
EDUCATION

| 2012 -05- 25

| PrReesTE BAG A 110
PRETORIA 0001

“PUBLIC EXAMINATIONS

100 200 300 400 500

> 6
te)” (6)

Note! If 205 is used the answers in 3.6 and 3.7 will still be the same

Copyright reserved/Kopiereg voorbehou

Ple:

As 205 gebruik word sal die antwoorde van 3.6 en 3.7 dieselfde bly

[31]

ase turn over/Blaai om asseblief

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Physical Science/Natuur- en Skeikunde HG/P1/V14 5 DBE/2012
SCE/SSE -— Memorandum
QUESTION 4/VRAAG 4 y
4.1 Aresultant/net force acting on a mass produces an acceleration in the direction
of the force. The acceleration is directly proportional to the (resultant/net) force

and is inversely proportional to the mass./'n Resulterende/netto krag wat inwerk
op 'n massa veroorsaak 'n versnelling in die rigting van die krag. Die versnelling is
direk eweredig aan die (resulterende/netto) krag en is 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 v
Die tempo van momentumverandering van ‘n voorwerp is direk eweredig aan die
resulterende/netto krag en die tempo van momentumverandering is in die rigting
van die resulterende / netto krag.

(3)
4.2 v v v
T T
v P — Q oo 7
v
én 3N < 6)
Note: ¢ Any additional horizontal forces(— 1)
/enige ekstra horisontale kragte (- 1)
¢ If NOT separate diagrams - linked system (— 1)
! as NIE aparte diagramme nie ~ gekoppelde stelse! (- 1)
¢ If no arrows then (-1) / As daar nie pylpunte is nie (-1)
43 v v P and Qa system: (4)
: Fres = T -2 = m.a Fes = 55-2-3 =ma=20.1+mi1iv
T-2 =201v” m= 30 kg
T=22NV yv A
For Q: Fyes = 55-T-3 = 30.1
T=22N v
Ad 7, / 7, PandQa system: (4)
Fres = 55 ~T — 3y= ma Fres = 55-2-3 = Meal”
55-22 -3 =m1v Mp+ay = 50 kg ¥
m=30kg~ | | mass of Q: 50 — 20 = 30 kg “
[16]
DEPARTMENT OF BASIC
EDUCATION
|
Copyright reserved|Kopiereg voorbehou Please turn over/Blaai om asseblief
| 2012 -05- 25

Fiylyat Ei BAG X +
RRETORIA 0001 Ge =
PUBLIC EXAMINATIONS |

### Page 6

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

QUESTION 5/VRAAG 5

5.1
Fem) v ae ow)
_ 80(0-20) v _ 0-20
_— 0,2
= -8000N “ o = -100ms? Vv
F = 8000N/ in the opposite direction F on = (80)(-100) = -8000N
of the motion of the car / backwards F=8000N in the opposite direction

of the motion of the car / backwards

(5)

v a v
5.2 The time taken to bring the car to a stop increases and this decreases the
“force on the driver because the force is inversely proportional to the time.

m Av Ap
ee ee
t orlof 2 At orfof

Fe F At= mAv

Die tyd wat dit neem om die motor tot stilstand te bring neem toe en dit
verminder die krag op die bestuurder want die krag is omgekeerd eweredig
aan die tyd.

mAv _ Ap

Bate ooh aN orlct

F At=mAv

(4)
19]

QUESTION 6/VRAAG 6

61

Any two particles in the universe exert a force of attraction on each other which is
vdirectly proportional to the product of the masses of the particles and inversely”

proportional to the square of the distance between the two particles.

(Objects / bodies and mass / masses will also be acceptable)

Enige twee deeitjies in die heelal oefen ‘n aantrekkingskrag op mekaar uit en die

krag is direk eweredig aan die produk van die massas van die deeltjies en

omgekeerd eweredig aan die kwadraat van die afstand tussen die deeltjies.

(3)

DEPARTMENT OF BASIC 1
EDUCATION

| 2012 -05- 25

; : i, PRIVATE BAG X 110
Copyright reserved/Kopiereg voorbehou PRETORIA 0001 Please turn over/Blaai om asseblief k

“PUBLIC EXAMINATIONS By

i

### Page 7

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Physical Science/Natuur- en Skeikunde HG/P1/V1 7 DBE/2012
SCE/SSE- Memorandum
Mm
6.2 37 and Fy = w =mg
r
GMmv_ ev
g =o, 4)
r2
6.3 GM
=e Vv G8OMpianet
Qplanet 2 Qearth = — U
y yt
= Gao 16S od _ BAMpIanet _ 80 ae
(1 a 80 ye * 5 Searttiaarde Taey 4 5Qptanefplanee
val
(3)
[10]
QUESTION 7/VRAAG 7
a (FE, +E av =(E, + Ew If equations of Motion are used
1 1 zero marks are obtained,
(mgh +5? ay = (mgh +5 MV" Dany E,(top) = Ex(bottom)
64)(10)(42) +0 4
Vw = 49 mS" (to the right) 4)
DEPARTMENT OF BASIC
Whet = AEx : EDUCATION
(2 i
F,.Ax.cos@ = =m(v4* - Vv; ii
9 i i
P 2 i 2012 -05- 25
(0,4 *9,8)(1,2)cos0° = 500 anu? -0) | PRIVATE BAG X 110
. PRETORIA 0001
Vv; = 4,85 m.s"to the right v PUBLIC EXAMINATIONS
7.2

total Ppefore collision/ voorbot sing = totalPaftercollision /nabotsing

(mv + MV)before/voor = (M+ M)Vatter/na
(mv y ae ee
(0,449) +0 = (0,4+05)v

v=22mst y

(my +M2 Watter/na

v

If before and after not in equation lose 1 mark for formula
As voor en na nie in vergelyking nie - verloor 1 punt

Copyright reserved/Kopiereg voorbehou

6)

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### Page 8

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Physical Science/Natuur- en Skeikunde HG/P1/V1 8 DBE/2012
SCE/SSE — Memorandum

7A 7.2

7.3 (6)

Wv
tv ip a
E, (before) = yim? = 3 (0.4)(497) 0s v

ied A 3 v
E, (after) = mv? = 5(0,4+05)(22°) = 2.2J
AE, =48-2,2=26) v 115)

QUESTION 8/VRAAG 8

v

The force (electrostatic / force of attraction or repulsion) which one charge
exerts on the other is
v

8.1

directly proportional to the product of the charges and is inversely
v

proportional to the square of the distance between them. (3)
Die krag (elektrostatiese / aantrekkende of afstotende krag) wat een lading
op ‘n ander uitoefen is direk eweredig aan die produk van die ladings en
omgekeerd eweredig aan die kwadraat van die afstand tussen hulle.

8.2 negative, “v field lines directed towards the charge. positive test charge will be
attracted, the direction in which a positive test charge will experience a force in a
field, electric field lines point towards negative charge (any one)”
Negatief, veldiyne is gerig na die lading toe, positiewe toetsiading sal
aangetrek word, rigting waarin ‘n positiewe toetslading ‘n krag sal ondervind in ‘n
veld, elektriese veldiyne wys in rigting van ‘n negatiewe lading (enige een van)

(3)

8.3

a Fg
+4x107> + (-2x1075)

v6
=+1x107°C
2v

new charge/nuwelading =

(3)

84 | 8.4.1 Sphere A, YY Ais positively charged / electrons moving from negative to
positive / A is now less positively chargedv
Sfeer A, A is positief gelaai / elektrone beweeg van negatief na positief/ A is
nou minder positief gelaai (3)

8.4.2 3x 10°C of charge was transferred/3 x 10° C van die lading is oorgedra:
Ignore the signs / ignoreer die tekens

v
x10°°
number of electrons / aantal elektrone =>,
1,6x10° a
=1,875 10" electrons a
[15]
DEPARTMENT OF BASIC : :
Copyright reserved/Kopiereg VoopbeHgE ATION Please turn over/Blaai om asseblief
2012 -05- 25 fo
PRIVATE BAG X 110
PRETORIA 0001
C BRIO OY ARAIMATIORNC

### Page 9

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Physical Science/Natuur- en Skeikunde HG/P1/V1 9

QUESTION 9/VRAAG 9

v

SCE/SSE ~ Memorandum

v

9.1 The current in a resistor is directly proportional to the potential difference
across the resistor if the temperature of the resistor remains constant.”

Die stroom deur ‘n resistor is direk eweredig aan die potensiaalverskil

oor die resistor as die temperatuur van die resistor konstant bly.

9.2 The resistance / opposition / difficulty experienced by charge when moving

through the cell or battery vv

Die weerstand / opposisie / ondervind deur lading as dit deur die se! of battery

beweeg.
9.3
1 i . Jy R : R(product) /
Roarallel Ry Ro parallel “R(sum)
tv, tv _ (10x15)
40° 15 (10+15)v
Rparetlel =8Q v Roarellel =6QY

DBE/2012

(3)

(2)

(4)

576 DEPARTMENT OF BASIC
60. EDUCATION
60
=9,6 Watt
ead 2012 -05- 25
R PRIVATE BAG X 110
vv v2 PRETORIA 0001 (5)
86=—E, PUBLIC EXAMINATIONS |
V=12V¥
p-W Wert ¥
t =2
_ 876 576 =! 15.60 (4)
“ony v "se “to . v
=9,6 Watt hoa = 5 = 12K
=/2p =|2 Yv
P=I?R=1*.15 i, =08+12=2A
lysq = 08 AY
12 v
=—=I12A
loa = 79 = 12

v
Iq =08+12=2A

Copyright reserved/Kopiereg voorbehou

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Physical Science/Natuur- en Skeikunde HG/P1/V1 10
SCE/SSE — Memorandum

layiso.com

DBE/2012

fs

96 | emPaIRtinYy Vostornatokstem = Emf — (2 x 1,5) ¥
24 V=2(R+6)+(2x15) -%4_av
24 =2R+12+3 = 21V (6)
2R =9 14 06¥
R=45Q0v Rexternal/ekstern = “3” ee
R = Rosternaviekstem ~ 6 = 10,5-6 = 4,50 V
9.7 R increase, | decreases, less “lostvolts’, (4)
V, increases v
[28]
QUESTION/VRAAG 1: 60
QUESTION/VRAAG 2-9: 140
GRAND TOTAL/GROOTTOTAAL: 200

| eerie Ba aNATO
EXAMINATIONS

aot -05° 25

BAG X10

RIA 0001 ee

PRIVATE

PUBLIC |

| eenanornennooet

Copyright reserved/Kopiereg voorbehou

ff

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