# Physical Sciences Grade 12 P1 2015 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Northern Cape November 2015, 22 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/nc-paper-1-memo/1908726c-aca0-4624-97bb-c268a90e10ff
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2015
- Exam period: Northern Cape November
- Paper: 1
- Curriculum: CAPS
- Pages: 22

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 2015 Memo](https://www.hlayiso.com/library/nc-paper-1-memo/1908726c-aca0-4624-97bb-c268a90e10ff)

## About this document

Physical Sciences · Grade 12 · Northern Cape November · 2015. Memorandum, 22 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

NORTHERN CAPE DEPARTMENT OF EDUCATION
NOORDKAAP DEPARTEMENT VAN ONDERWYS

TRIAL EXAMINATION
PROEFEKSAMEN

GRADE 12
GRAAD 12

RR RRAMARASER ARERR AAA LBAEEES,

PHYSICAL SCIENCES: PHYSICS (P1) 4
FISIESE WETENSKAPPE: FISIKA (V1)

LAR ARBUALS ELE IE ABER, FAAAENAAR RAE EAL EERY ,
Se,

SEPTEMBER 2015
MEMORANDUM

PAPAAARRERAOARERERS 2

a

tei

MARKS/ PUNTE: 150

This memorandum consists of 20 pages
Hierdie memorandum bestaan uit 20 bladsye

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

### Page 2

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 2 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO! Sep 2015

GENERAL GUIDELINES / ALGEMENE RIGLYNE

1.

141

1.2

1.3

1.4

1.5

1.6

1.7

1.8

CALCULATIONS/ BEREKENINGE

Marks will be awarded for: correct formula, correct substitution, correct
answer with unit.

Punte sal toegeken word vir: korrekfe formule, korrekte substitusie, korrekte
antwoord met eenheid.

No marks will be awarded if an incorrect or inappropriate formula is used,
even though there may be relevant symbols and applicable substitutions.
Geen punte sal toegeken word waar 'n verkeerde of ontoepaslike formule
gebruik word nie, selfs al is daar relevante simbole en relevante substitusies,

When an error is made during substitution into a correct formula, a mark
will be awarded for the correct formula and for the correct substitutions, but
no further marks will be given.

Wanneer 'n fout gedurende substitusie in 'n korrekte formule begaan word, sal
'n punt vir die korrekte formule en vir korrekte substitusies toegeken word, maar
geen verdere punte sal toegeken word nie

If no formula is given, but all substitutions are correct, a candidate will
forfeit one mark.

Indien geen formule gegee is nie, maar al die substitusies is korrek, verloor
die kandidaat een punt.

No penalisation if zero substitutions are omitted in calculations where
correct formula / principle is given correctly.
Geen penalisering indien nulwaardes nie getoon word nie in berekeninge
waar die formule/beginsel korrek gegee is nie.

Mathematical manipulations and change of subject of appropriate formulae
carry no marks, but if a candidate starts off with the correct formula and then
changes the subject of the formula incorrectly, marks will be awarded for the
formula and the correct substitutions. The mark for the incorrect numerical
answer is forfeited.

Wiskundige manipulasies en verandering van die onderwerp van toepaslike
formules tel geen punte nie, maar indien ‘n kandidaat met die korrekte formule
begin en dan die onderwerp van die formule verkeerd verander, sal punte vir die
formule en korrekte substitusies toegeken word. Die punt vir die verkeerde
numeriese antwoord word verbeur.

Marks are only awarded for a formula if a calculation has been attempted.
i.e. substitutions have been made or a numerical answer given.

Punte word siegs vir ‘n formule toegeken indien 'n poging tot 'n berekening
aangewend is, d.w.s. substitusies is gedoen of 'n numeriese antwoord is gegee.

Marks can only be allocated for substitutions when values are substituted into
formulae and not when listed before a calculation starts.

Punte kan slegs toegeken word vir substitusies wanneer waardes in formule
ingestel is en nie vir waardes wat voor 'n berekening gelys is nie.

Copyright reserved 2 Please turn over

### Page 3

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 3 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
1.9 All calculations, when not specified in the question, must be done to two

decimal places.
Alle berekenings, wanneer nie in die vraag gespesifiseer word nie, moet tot 'n
minimum van twee desimale plekke gedoen word.

2. UNITS/ EENHEDE

2.1 Candidates will only be penalised once for the repeated use of an incorrect
unit within a question or sub-question.
Kandidate sal slegs een keer gepenaliseer word vir die herhaaldelike gebruik van
‘n verkeerde eenheid in ‘n vraag.

2.2 Units are only required in the final answer to a calculation.
Eenhede word siegs in die finale antwoord op 'n vraag veriang.

2.3 Marks are only awarded for an answer and not for a unit per se. Candidates
will therefore forfeit the mark allocated for the answer in each of the following
situations:

- Correct answer + wrong unit

- Wrong answer + correct unit

- Correct answer + no unit

Punte word slegs vir 'n antwoord en nie vir 'n eenheid per se toegeken nie.
Kandidate sal derhalwe die punt vir die antwoord in die volgende gevalle
verbeur:

- Korrekte antwoord + verkeerde eenheid

- Verkeerde antwoord + korrekte eenheid

- Korrekte antwoord + geen eenheid

2.4 SI units must be used except in certain cases, e.g. V-m" instead of N-C”’, and
cm-s" or km-h"' instead of m:s”' where the question warrants this.
Sl-eenhede moet gebruik word, behalwe in sekere gevalle, bv. Vem" in plaas van
N-C’, en em-s of km-h’ in plaas van m-s waar die vraag dit regverdig.

3. GENERAL/ ALGEMEEN

3.1 If one answer or calculation is required, but two given by the candidate, only

the first one will be marked, irmespective of which one is correct. If two
answers are required, only the first two will be marked, etc.
Indien een antwoord of berekening verlang word, maar twee word deur die
kandidaat gegee, sal siegs die eerste een nagesien word, ongeag watter een
korrek is. indien twee antwoorde verlang word, sai siegs die eerste twee
nagesien word, ens.

3.2 For marking purposes, alternative symbols (s,u,t, etc.) will also be accepted.
Vir nasiendoeleindes sal alternatiewe simbole (s, u, t, ens.) ook aanvaar word.

3.3 Separate compound units with a multiplication dot, not a full stop, for
example, m-s". For marking purposes m.s"' and m/s will also be accepted.
Skei saamgestelde eenhede met 'n vermenigvuldigpunt en nie met 'n punt nie,
byvoorbeeld, m-s’. Vir nasiendoeleindes sal m.s en m/s ook aanvaar word.

Copyright reserved 3 Please turn over

### Page 4

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 4 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
4, POSITIVE MARKING/POSITIEWE NASIEN

41

4.2

4.5

Positive marking regarding calculations will be followed in the following cases:
Positiewe nasien met betrekking tot berekenings sal in die volgende gevalle geld:

Sub-question to sub-question: When a certain variable is calculated in one
sub-question (e.g. 3.1) and needs to be substituted in another (3.2 or 3.3),
e.g. if the answer for 3.1 is incorrect and is substituted correctly in 3.2 or 3.3,
full marks are to be awarded for the subsequent sub-questions.

Subvraag na subvraag: Wanneer 'n sekere veranderlike in een subvraag (bv.
3.1) bereken word en dan in 'n ander vervang moet word (3.2 of 3.3), bv. indien
die antwoord vir 3.1 verkeerd is en word korrek in 3.2 of 3.3 vervang, word
volpunte vir die daaropvolgende subvraag toegeken.

A multi-step question in a sub-question: If the candidate has to calculate,
for example, current in the first step and gets it wrong due to a substitution
error, the mark for the substitution and the final answer will be forfeited.

'n Vraag met veelvuldige stappe in ‘'n subvraag: Indien 'n kandidaat
byvoorbeeild, die stroom verkeerd bereken in 'n eerste stap as gevolg van 'n
substitusiefout, verloor die kandidaat die punt vir die substitusie sowel as die
finale antwoord.

Normally an incorrect answer cannot be correctly motivated if based on a
conceptual mistake. If the candidate is therefore required to motivate in
question 3.2 the answer given to question 3.1, and 3.1 is incorrect, no marks
can be awarded for question 3.2. However, if the answer for e.g. 3.1. is
based on a calculation, the motivation for the incorrect answer in 3.2 could
be considered.

'n Verkeerde antwoord, indien dit op ‘n konsepsuele fout gebaseer is, kan
normaalweg nie korrek gemotiveer word nie. Indien 'n kandidaat gevra word om
in VRAAG 3.2 die antwoord op VRAAG 3.1 te motiveer en 3.1 is verkeerd, kan
geen punte vir VRAAG 3.2 toegeken word nie. Indien die antwoord op bv. 3.1
egter op 'n berekening gebaseer is, kan die motivering vir die verkeerde
antwoord in 3.2 oorweeg word.

Copyright reserved 4 Please turn over

### Page 5

Physical Sciences/Gr 12/P1
Fisiese Wetenskappe/ GR 12/ V1

QUESTION 1/ VRAAG 1

1.1
1.2
1.3
1.4
15
1.6
1.7
1.8
1.9

1.10

Copyright reserved

B vv
CvvY
Dvv
A vv
Bvv
A vv
Bvv
Dvv
C vv

Bvv

Downloaded from hlayiso.com

CAPS/KABV

NCDOE/ Sep 2015
NKDVO/ Sep 2015

Please turn over

(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)

(2)
[20]

### Page 6

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 6 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
QUESTION 2/ VRAAG 2

2.1 Normal force is the force or component of a force which a surface exerts on
an object with which it is in contact, “and which is perpendicular to the

surface. “”
Normaalkrag is die krag of komponent van ’n krag wat 'n opperviak op ’n (2)
voorwerp waarmee dit in kontak is, uitoefen en wat loodreg op die opperviak is.
2.2
N/Fw/Fs on of Normal/ normaalv
fF df
Fa ¥
F./W/ Gravity v/Gravitasie
ACCEPT/ AANVAAR
N/Fy/Fs on of Normal’/ Normaal
Fa v
fiFJfn
Fe/WI Gravity “/ Gravitasie (4)

2.3 OPTION 1/ OPSIE 1

Fy -0 =
DFy OR/OF Free = 0 y Any one/Enige een

N+(-mg-Fasin35°=0
N= mg+Fasin35°

N= 215,16 +97,51”
fp =e&N

f, =21291NY

OPTION 2 / OPSIE 2
fe4yNn v

Sf, = umg + F,) ¥ Any one/ Enige een
Sy = ey (mg + Fsin35°)

(4)

Copyright reserved 6 Please turn over

### Page 7

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 7 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015

f, = 0.68 (22 9,8 +170xsin35°)”
=212.91NY

2.4 POSITIVE MARKING FROM 2.3 /POSITIEWE NASIEN VANAF 2.3
OPTION 1/ OPSIE 7

Fm=Fi+f, ORIOF F,=F,-f, ORIOF F,, = Fos35° -f,
F.,, =170c0s35° —212.91¥
Fy = — 73,65 N ORIOF Fy = 73,65 N backwards”/ terugwaarts

The net force is in the opposite direction of motion/ since the net force is in opposite
direction of motion, the mower accelerates backwards.“

Die netto krag is in die teenoorgestelde rigting van die beweging/ aangesien die netto
krag in die teenoorgestelde rigting van beweging is, versnel die grassnyer terugwaarts

OPTION 2/ OPSIE 2

F,=Foos35 ¥

F, =170c0s35 =139.25 NY

Fy<f

Under these circumstances the lawn mower will be accelerating in a backward
direction/It will slow down in a forward direction. ¥

Onder hierdie omstandighede sal die grassnyer in ‘n terugwaartse rigting versnel/ Dit
sal stadiger beweeg in ’n voorwaartse rigting.

QUESTION 3/ VRAAG 3

3.1. Impulse is the product of the force (resultant/net force) acting on an object and
the time the force (resultant/net force) acts on the object.
Impuls is die produk van die krag (resultante/ netto krag) wat op ‘n voorwerp
inwerk en die tyd wat die krag (resultante/ netto krag) op die voorwerp inwerk.

3.2. Impulse (J)=Ap
Impulse (J) = mv tm vi)
Impulse (J) = 0,43(-25 - 20) “
Impulse (J)=—19,35 N-s %
OR/OF
Impulse (J) = 19,35 N’-s ¥ in opposite direction.
3.3 FAt=Ap¥ OR/OF FAt= Ap
(300)At= 19,35 ¥
At= 0,065 s ¥ OR/OF At= 0,07 s

v Anyone/ Enige een

3.4 Inelasticv’/Onelasties
Mass is constant but speed changes /Kinetic energy is not conserved.¥
Massa is konstant maar spoed verander/Kinetiese energie bly nie behoue nie

Copyright reserved q Please turn over

(3)
[13]

(2)

(3)

(3)

(2)
[10]

### Page 8

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 8 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
QUESTION 4/VRAAG 4
4.1 Projectile is an object upon which the only force acting is the force of gravity.
Projektiel is ‘n voorwerp waarop slegs gravitasiekrag inwerk. (2)
4.2 9,8 m-s* “downwards afwaarts (theoretical) (2)
OR/OF

9,6 m:-s* Ydownwardsy (actual)

4.3 OPTION 4/ OPSIE 1
Ay =viA +5 ath)? Y OR/OF  Ay=v,Ar+ 5 eta)?

Ay = 9,8(4,08) ¥ + 5 (-9,8)(4,08)°¥
Ay= - 41,58 mv

OR/ OF
Ay= 41,58 m downwards 7 afwaarts

OPTION 2/OPSIE 2
Ay =viA +55) YORIOF Ay= yas Seay

Ay = -9,8(2.04) ” +5 (-98)02,08)77
Ay= - 41,58 m ¥

OR/ OF

Ay= 41,58 m downwards W/afwaarts

OPTION 3/ OPSIE 3

Aya Wt) 5, v

2
ay - 28+ 30.8) y (408

Ay= - 41,58 m ¥
OR/ OF
Ay= 41,58 m downwards v/afwaarts

OPTION 4/ OPSIE 4

Ay = et¥) oy v

Ay = (-9,8) +9018) Y (2,04)
Ay= - 40,78 m ¥

OR/ OF

Ay= 40,78 m downwards 7 afwaarts

Copyright reserved 8 Please turn over

### Page 9

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 9 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015

OPTION 5/ OPSIE 5
Ay= Area(trapezium)

Ay = area G2),

Ay = £728) + (30,18), CRIM 4.08 - 2,04) ¥
Ay= - 40,78 m ¥

OR OF
Ay= 40,78 m downwards “/afwaarts

OPTION 6/ OPSIE 6
Area= Areag; + Areag2

1 1
Ay = area = om + 3h
Ay = 5 9,8(1,02) ¥ +5 (-30,18)G,06) ¥
Ay=- 41,17mv¥

OR OF
Ay= 41,17 m downwards v/ afwaarts (4)

OPTION 7/ OPSIE 7
1
Ay =ab + 45h
\y apt 3
Ay = -9,8(2,04) ¥ + 5 (2,04)(-20,38) v
Ay=- 40,78 m ¥

OR! OF
Ay=40,78 m downwards ~/afwaarts

Copyright reserved 9 Please turn over

### Page 10

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 10 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
44
A
2
@
7
5
a
Se
SE
e—0 7 >
es 1,02 2,04 4,08 t(s)
7 9
A
40,78 m
2
2
n
9
a
ee
SE
a>
- 40,78
CRITERIA/ KRITERIA MARKS/PUNTE
Graph starting from zero OR 40,78 m/Grafiek begin by nul OF v
40,78m
Correct shape/ korrekte vorm v
Position for maximum height (4,9 m or/ of 5 m) at 1,02 s v
Posisie vir maksimum hoogte (4,9 m or/ of 5 m) by 1,02 s (4)
Final position/ Finale posisie (-40.4 m or 40 m) at/ by 4,08 s v [12]

Copyright reserved 10 Please turn over

### Page 11

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 il NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
QUESTION 5/ VRAAG 5

5.1. The total linear momentum of an isolated/a closed system~ remains constant (is
conserved). ¥

Die totale linére momentum van ‘n geisoleerde/ gesiote sisteem bly konstant
(bly behoue)

OR/ OF
In an isolated systemy the total linear momentum of a system before a
Collision/interaction is equal to the total linear momentum of the system after
the collision. ”
In ’n geisoleerde sisteem is die totale liniére momentum van ’n sisteem voor ’n
botsing/ interaksie gelyk aan die totale liniére momentum van die sisteem na
die botsing

OR/ OF
If the impulse of the external forces acting on a system is zerov the total linear
momentum of the system does not change/remains constant. ¥

As die impuls van die eksterne kragte, wat op ‘n sisteem inwerk, gelyk is aan
nul sal die totale liniére momentum van die sisteem nie verander nie/ konstant
bly.

OR/ OF
If there is no external net force acting on a system of particles», the total linear
momentum of the system is conserved. “

As daar geen eksterne netto krag op ’n sisteem van partikels inwerk nie bly die (2)
totale liniére momentum van die sisteem behoue.

5.2 SS Peeterer voor = S'Paterina
OR/OF

S Prctore/voor = S Panes na

TV betore /voor) + MV afbetore svoory = (Mt My WV syst sist

¥ Anyone/ Enige een

(6x8) + (5x0)¥=(6 + SV va ssict”

Vv =4,36 m:s” to the right ¥/ na regs (4)

5.3. POSITIVE MARKING FROM QUESTION 5.2
POSITIEWE NASIEN VANAF VRAAG 5.2

OPTION 1/OPSIE 1

Wre = AE wy ¥ Any one/ Enige een
Copyright reserved 11 Please turn over

### Page 12

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 12 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
W,, = AE, + AE,

5.4.

Ff, Axcos180° = Sm ~sm} + AE,
Mg NAxcos180° = a ~ mv} + AE, Y¥ Any one/ Enige een

HgmgAxcos|80° = ay — Sm? + AE,

ad 19,82 (-D¥=— 1105)" ~5116436)°¥ +0

H=0.43%
OPTION 2/OPSIE 2
Wra = AE YA f Eni
W., = Eq -E ny one/ Enige een

F,Axcos180° = bn? -L mv?
t ar

My NAxcos180° = ; mv? -+ my? >” Any one/ Enige een

H,,mghxcos180° = sv ~ 5mm?

Mg (11%9,8x2%(-1) v= 2110,5) -51 (4,36) ¥
u.=0.43

1 POSITIVE MARKING FROM QUESTION 5.3
POSITIEWE NASIEN VANAF VRAAG 5.3

SUP ot = ma IFpe=ma

fy +Fg +N=ma ¥ Any one/ Enige een
—f,-F,=ma
-EN-F,=ma

— pg cos30° — mgsin30° = ma
— ugcos30° - gsin30° =a

— (0,43 x11 x 9,8 x cos30°)¥ — (11x 9,8 x sin 30° ¥ = Lla
vi =v? + 2aAxY

0? = (1,5)? +2(-8.55)AxY

Ax=d= 0,26 mv

¥ Any one / Enige een

OR/IOF

Copyright reserved 12 Please turn over

(5)

### Page 13

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 13 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015

SF vt = ma [Fre=ma

F,+Pe+N=ma ¥ Any one / Enige een

-f,-F, = ma

- UN -F, = ma

Oo +1300 oo
pamg.cos 3) mg sin 30 ma } ¥ Any one / Enige een
— vgcos30" ~ gsin30° =a
- (0,43 x 9,8xcos30") “ —(9,8xsin30° VW =a

a= -8,55m-s”

vi =v) + 2adxnY
0? = (15)? + 2(-8.55)Ax ¥
Ax=d= 0,26 mY

5.4.2 Increases “Toeneem
7
Aceleration of the system of blocks decreasesv’. (7
Versnelling van die bloksisteem neem af

(2)
[20]

Copyright reserved 13 Please turn over

### Page 14

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 14 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015

QUESTION 6/ VRAAG 6

6.1 The Doppler effect is the change in frequency (pitch) of sound detected by a
listenerv’ because the sound source and the listener have different velocities
(relative to the medium of sound propagation). “

Die Doppler-effek is die verandering in frekwensie (toonhoogte) van klank soos
waargeneem deur die luisteraar omdat die kKlankbron en die luisteraar
verskillende snelhede besit (relatief tot die medium wat die kiank voortbring)
OR/ OF
The Doppler effect is the change in the observed frequency of a wave” when
the source or the detector moves relative to the transmitting medium. “
Die Doppiler-effek is die verandering in die waargenome frekwensie van “n golf
wanneer die bron of die waarnemer beweeg relatief tot die medium wat die
kKlank dra. (2)

6.2 OPTION 1/OPSIE 1
fr -( |e

VV,

1 000% = 340
340

— |v 930)¥

V.= 23,80 ms!”
OPTION 2/ OPSIE 2

v
f= Fr ~
. fara

340

1 000% = Y (930)
340

-vV,
V.= 23,80 m-s'¥
OPTION 3/ OPSIE 3

f- &£ y oR fe
t2-£ 4-=
v v

930 ¥

qs v
340
V.= 23,80 m-s'¥ (8)

5

v 1000 =

6.3. Increases“/Toeneem (2)
The compressions behind the ambulance are further apart than when it was
approaching”.
Die verdunnings agter die ambulans is verder van mekaar as die verdigtings toe die
ambulans die luisteraar nader.
6.4.1, The stars are moving away “”
Die sterre beweeg weg van die aarde
The spectral lines in the diagrams are shifted towards the red end/red shifted).~”
Die spektrumlyne in die diagramme_ het verskuif na die rooi ent/rooi verskuiwing.
(2)

Copyright reserved 14 Please turn over

### Page 15

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 15 NCDOE/ Sep 2015

Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015

6.4.2. The Universe is expanding.” (1)
Die heeial is besig om uit te sit. [12]

QUESTION 7/ VRAAG 7

7.1 The electric field at a point is the (electrostatic) force experienced per unit
positive charge at this point. “~~”
Die elektriese veld by 'n punt is die (elektrostatiese) krag wat ‘n positiewe (2)
eenheidslading by daardie punt ondervind.
7.2
Criteria for marking
Shape of field lines”
Direction of field lines.
Nasienkriteria
Vorm van die veldlyne
+Q ~Q Rigting van die veldlyne

(2)

7.3 OPTION 1/ OPSIE1

kQ
Ect a

(9x10°)(2x10~)
Que v
(0,5)
E..¢= 7-20x104N-Co‘right /regs

kQ
Ee" a
kQ (9x10°)(2x107°)
| =n ede A acta R
ee (05)?

E_yc=7,2%104 N-C“right/regs

E net = Exc + E 2,0
Positive to the right } ¥ Anyone / Enige een
Et = Ene t Ene

E neto= 7-2X10* +7,2x104¥
=+14,4x10°N-C1Y

OR/ OF
=14,4«10°N-C“ right /regs”

Copyright reserved 15 Please turn over

### Page 16

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 16 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015

OPTION 2/ OPSIE 2
Exe ev
(9x10° }(2x10 6 )
ope
(0,5)
E..c= 7.20x104N- C“right /regs

kQ
Ec

r?
kQ _(9x10°)(2x10*) ,

E,o=
oe (0.5)?
E s,c=14,4x10* N-Cright /regs

E net = Eo + E oye

Positive to the left YAn .
y one/ Enige een
Eee = ~E auc + wtb

E ppto= ~7-2X104 -7,2x10°Y
=-14,4x10'N-ClY

OR/IOF
= 14,4x10'N-C right / regs ¥

OPTION 3/OPSIE 3
Ext ey
(9x10°)(2x10%)
[oi a
(0,5)
Eayg= 120x104 NC“ right /regs
Enc= Eye = 7-20x10*N-C right /regs ¥

E net = Exc + E onc
Positive to the right } v Any one/ Enige een
Ent = Eas + Bac

E veto™ 2X (7.2X10*)¥

= +14,4x10°N-CT

OR/OF

= 14,4104 N-C"'right [regs ¥

Copyright reserved 16 Please turn over

### Page 17

7A

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 17 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
OPTION 4/ OPSIE 4
Bago Rv
(9x10°)(2x10~)
i oe v
(0,5)
Eye 7-20x10*N-C right /regs
Eac= E oye = 7-20X104N- C“right /regs ¥
E vet = Fane + E onc
Positive to the left Y An p
y one / Enige een
Ent = Ea + wtb
E seto= ~7-2X10* — 7,2x10°¥
=-14,4x10°N-ClY
OR/OF
=14,4x104N-C right /regs¥ (6)
OPSION 1/OPSIE 1
Fra = Fis + Fos ¥ ORIOF
+2yuC CL, P +6 pC
pr. KOO, K0,0, | ...... De Ce oo
net 2 2 xX |
Ag hy >
1,0m
= KQ2s _ KQ,0,
re re
8 % ¥ Any one / Enige een
KQ,0, _ KQ,0,
3 os
2.2 ORIOF
2 2 AL=x)? = 6x
hs 2 2
2(x°-2x+1) =6x
nana
2x10% y= 6x10 2x*-4x+2 = 6x"
x” (l-x)v 2x?-2x+1=0
2_ 6
Y den? ee bP? = 4a
2a
4x? +4x-2=0 OR/OF 2+ /2? -4x2x(-1)
V2_ V6 =
—_= 2(2)
x (-x)

x = 0,37 m from charge/ vanaf lading 2 pC ~
OR/OF

Copyright reserved 17

Please turn over

### Page 18

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 18 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/1 CAPS/KABV NKDVO/ Sep 2015

x= 0,63 m from charge/ vanaf lading 6uC”
OPSION 2/OPSIE 2

Fre = Q,Enen2¥

Fra = Qs OQ, _ FQ: +2 uc, rod P +6uC

%3 Les eg a <2 e @--------------------- @-----
9 -KQ2: _ KQ,0, x >|
2 2
ns 13 v An i 1,0m
yone/Enigeeen ©
KQ,Q, = KQ,9,
ns br 3
o_O OR/OF ;
2 2(1-x)’ = 6x’
M3 Nhs 2
262-2x+1) =6x"
a hy
2x10% ve 6x10 2x2-AK+2 = 6x2
x (-x)v 2x”-2x+1=0
2. 6
¥ ax ee Dev b? ~ 400
2a

4x? +4x-2=0 OR/OF 2+ 2? —4x2x(-1)

v2_ V6 =>

ial 2(2)

x (l-x)

x = 0,37 m from charge/ vanaf lading 2 uC “

OR/OF (6)

X= 0,63 m from charge/ vanaf lading 6uCY [16]

QUESTION 8/ VRAAG 8

8.1 When current flows through a voltage source (battery/generator) a resistance to
current flow arises” due to the resistance of the materials (chemicals/conductors)

from which the source is made.~

Wanneer stroom deur ’n volf-kragbron (battery/generator) vioei, ontstaan ’n
weerstand teen stroomvioei as gevolg van die weerstand van die materiaal

(chemikaleé/ geleiers) waarvan die bron gemaak is.

OR/OF

(2)

Internal resistance is the resistance offered to the electron flow” by the

electrolyte/medium of the cell/generator. “

Interne weerstand is die weerstand gebied teen die vioei van elektrone deur die

elektroliet/ medium van die sel/ generator.

8.2 € = Vex Ir
Copyright reserved 18

Please turn over

### Page 19

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1

Fisiese Wetenskappe/ GR 12/ V1

6 = Vout +1(0,10)
Vex = 6- (0,10)1%

83
vv) 4
6Y

8.4

8.4.1 OPTION 1/ OPSI/E 1

W, =PRAt “= 40

(2(4)At = 40v

Wa €(1yRat = 60
At

R=6Qv

8.4.2 POSITIVE MARKING FROM
QUESTION 8.4.1
POSITIEWE NASIEN

VRAAG 8.4.1
OPTION 1/ OPSIE 1

e€={(Rtr) ¥

6 = 1(4+6)+ 0,10)”

1=0,59 AY

8.4.3 POSITIVE MARKING FROM

QUESTION 8.4.1 and 8.4.2
POSITIEWE NASIEN
VRAAG 8.4.1 en 8.4.2

Copyright reserved

19 NCDOE/ Sep 2015
CAPS/KABV NKDVO/ Sep 2015
(2)
0 = 6-(0,10)I
to 60 A
0.1
al
(A)

OPTION 2/ OPSIE 2
W =PRAtY

W, _ PR,At
W, PRA?

40 _ PAA,
60 FRAt
R=60¥

OPTION 2/ OPSIE 2
€=I(Rtr) ¥

6 = I(10)+ 0,10)”
1=0,59 AY

19

(3)

(4)

(3)

(3)

Please turn over

### Page 20

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 2 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
OPTION 1/ OPSIE 1 OPTION 2 /OPSIE 2

€ = Vox +r” V = IRex””

6 = Vex + (0,59)(0,10) “ = (0,59)(10) “

Vext = 5,94 VY =5,9 Vv

8.5 DECREASEY
Total resistance of the circuit decreases “
Current increases”
Vintemal resistance IncreasesY
Vex (voltmeter reading) decreases (Vex = € — Vint)

AFNEEM

Totale weerstand van die stroombaan neem af
Stroom neem toe

Vinterne weerstand Neem toe

Veks(voltmeterlesing) neem af (Voxs = € — Vint)

QUESTION 9/ VRAAG 9

9.1 DC Generator”
Mechanical energy to electrical energy”
GS Generator
Meganiese energie na elektriese energie

9.2 To make the direction of the (induced) current to be the same in every half
cycle/half turn’ “
Om die (geinduseerde) stroom se rigting dieselfde te hou tydens elke
halfsiklus/ halwe rotasie
OR
To keep the (induced) current unidirectional’ “
Om die (geinduseerde) stroom in een rigting te laat vioei

9.3 Graph AY
DC generator becomes an AC generator”
Voltage changes the polarity in every half cycle. ¥
OR

Graph AY

DC generator becomes an AC generator”
The voltage is alternating “

Copyright reserved 2 Please turn over

(4)

[21]

(2)

(2)

(3)

### Page 21

Downloaded from hlayiso.com

Physical Sciences/Gr 12/P1 2 NCDOE/ Sep 2015
Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015
V7 ms /wak
9.4 Pagigem = ————_ ” 2
alg R Priston = Vi ms iwak y
(Vins nats 2 R
P , - V2 (vnw maks \2
svgigem = —— — a
2, ae R
Ce)
6= v2) oy 2»
R 6= v2" v
R=12Q R
R=122
Vims wok
Vem rwagk = R v Vimstwgk = Irmstovwgk tot Rr
V, R
V ims /wok
(—moximaie 2 = Ins Totavwge tat “()
Tnms/wele = _ v2 7 v2 2
R
12
(eY 12 12
V2 —=! (Sv
Imsiwgk =0,71 lims Totaliwgk tot = 142AVv

Iims Totavwgk totaat = 2x(0,71) ¥
=142AV¥

QUESTION 10/ VRAAG 10

10.1. What is the relationship between frequency of the incident radiation” and the
maximum kinetic energy of the ejected electrons? Vv

Wat is die verwantskap tussen frekwensie van die invallende bestraling en die
maksimum kinetiese energie van die vrygestelde elektrone?

10.2
10.2.1 frequency-/ frekwensie

10.2.2 ANY ONEY/ ENIGE EEN
Threshold frequency/ drumpel frekwensie
Work function of the metal/ werkfunksie van die metaal
Potential difference/ potensiaalverskil
Intensity of the incident radiation/ intensiteit van die invallende bestraling

10.3. As the frequency of the incident radiation increases, the kinetic energy also
increases¥ ¥

Soos die frekwensie van die invallende bestraling toeneem, neem die kinetiese
energie ook toe.

Copyright reserved 2 Please turn over

(5)
[12]

(1)

(1)

(2)

### Page 22

Downloaded from hlayiso.com

NCDOE/ Sep 2015

Physical Sciences/Gr 12/P1 3
NKDVO/ Sep 2015

Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV
10.4 OPTION 1/ OPSIE 1

E= W, + Ekman mata
2 Any one /Enige een

34 8
(6.8310 ™)Gx10) yg 49- .ev
200%10

Ex= 1,95 x 10? Jv

OPTION 2/ OPSIE 2
E=W, +Exmawmais *
E _fty

A
E- (6,63 x 10™)(3x10°)

200x 10

E=9,95x1079J
9,95 x10-" =8x10"" +E nay mars

- “19 .
Ex= 1,95 x10" IY (5)

10.5 Remains the same/ No Change. ¥
Bly dieselfde/ Geen verandering (2)

Kinetic energy is independent of intensity of the incident radiation. ~
Kinetiese energie is onafhanklik van die intensiteit van die invallende bestraling

106 DECREASES/ AFNEEM
(1)

[14]
TOTAL /TOTAAL: 150

Copyright reserved 3 Please turn over

## Related documents

- [MP 2015 SEPT Physics P 1 Memo](https://www.hlayiso.com/library/mp-2015-sept-physics-p-1-memo/888a8f9f-c0da-4ad1-b3ae-e2ecd133acdd) — Physical Sciences · Memo · 2015
- [Physical Sciences P1 Nov 2015 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2015-memo-afr-eng/140f25b7-f3c1-406a-a99d-6549fa7ea9e0) — Physical Sciences · Memo · 2015 · Eng
- [Physical Sciences P1 Feb March 2015 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-feb-march-2015-memo-afr-eng/12e143e4-5d6d-42a2-a7d6-a8a4ec637dbd) — Physical Sciences · Memo · 2015 · Eng
- [2015 SEPT Metro South Physics P 1 Memo](https://www.hlayiso.com/library/2015-sept-metro-south-physics-p-1-memo/7ff37983-21da-4189-a413-ec3432102261) — Physical Sciences · Memo · 2015
- [2015 SEPT Winelands Physics P 1 Memo](https://www.hlayiso.com/library/2015-sept-winelands-physics-p-1-memo/69181b1d-95f6-41d8-8658-ef7da76875ba) — Physical Sciences · Memo · 2015
- [2015 SEPT Metro South Physics P 1](https://www.hlayiso.com/library/2015-sept-metro-south-physics-p-1/48a052c6-a1c7-4786-a250-996ebe63e9d7) — Physical Sciences · QP · 2015

## More from Grade 12 Physical Sciences

- [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) — QP · 2025 · Afr
- [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

## 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.