# Physical Sciences Grade 12 P1 Free StateNovember 2014 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Free StateNovember 2014, 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/fs-2014-physcie-grade-12-p1-sept-memo/09665baf-624b-4813-a629-3bb1747452eb
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2014
- Exam period: Free StateNovember
- Paper: 1
- Curriculum: CAPS
- Pages: 11

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 Free StateNovember 2014 Memo](https://www.hlayiso.com/library/fs-2014-physcie-grade-12-p1-sept-memo/09665baf-624b-4813-a629-3bb1747452eb)

## About this document

Physical Sciences · Grade 12 · Free StateNovember · 2014. Memorandum, 11 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

education

Department of
Education

FREE STATE PROVINCE

PREPARATORY EXAMINATION
VOORBEREIDENDE EKSAMEN

GRADE 12/GRAAD 12

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

SEPTEMBER 2014

MEMORANDUM

MARKS/PUNTE: 150

This memorandum consists of 11 pages.
Hierdie memorandum bestaan uit 11 bladsye.

Copyright reserved/Kopiereg voorbehou

### Page 2

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 2 FS/VS/September 2014
Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum

SECTION A
QUESTION 1/VRAAG 7

11 Ccyvy (2)
12 Dvwy (2)
1.3 Cvy (2)
14. Cvv (2)
15 Dvv (2)
16 Dvv (2)
17 Byv (2)
18 By (2)
19 Dvv (2)
110 Dvv (2)
[20]

TOTAL SECTION A/TOTAAL AFDELING A: 20

Copyright reserved/Kopiereg voorbehou

### Page 3

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 3

Grade/Graad 12 Prep.Exam./Voorb.Eksam

SECTION B/AFDELING B

QUESTION 2/VRAAG 2

2.1.1

2.1.2

2.2

FS/VS/September 2014
Memorandum

Upwards positive/Opwaarts positief:

Vp =VjtaAty
0 =30v +-9,8At
At = 3,06 sv

Downwards positive/Afwaarts
Ve =Vj taAty

0 =-30V +9,8At

At = 3,06 sv

Upwards positive/Opwaarts positief:

Ay =v At+taat? Vv
= 30(4) v + 4(-9,8)(4)°v
=41,.6m

Height /Hoogte= 41,6 mv”

Downwards positive/Afwaarts
positief:
Ay =v At+taat? v
= - 30(4) v + 14(9,8)(4)2v
=-41,6m
Height/Hoogte = 41,6 mv

OPTION 1/OPSIE 1
Upwards positive/Opwaarts positief:
A

>

Time/tyd(s)
Criteria for graph/Kriteria vir grafiek: pom
Shape of first part of graph up till maximum height. v
Vorm van eerste deel van grafiek tot by maksimum hoogte.
Shape of second part of graph and above zero height at t = 4s. v
Vorm van tweede deel van grafiek en bokant zero hoogte by t= 4s
Ground not zero position (provided everything else is correct): (2)
Grond nie zero posisie (indien die res korrek is): 4%

Copyright reserved/Kopiereg voorbehou

(3)

(4)

### Page 4

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 4 FS/VS/September 2014
Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum
OPTION 2/OPSIE 2
Downwards positive/Afwaarts positief:
4
E
zo a ap
! Time/tyd(s)
v
Criteria for graph/Kriteria vir grafiek: pe
Shape of first part of graph up till maximum height. v
Vorm van eerste deel van grafiek tot by maksimum hoogte.
Shape of second part of graph and above zero height at t = 4 s. v

Vorm van tweede deel van grafiek en bokant zero hoogte byt = 4s

Ground not zero position (provided everything else is correct): 14 (2)
Grond nie zero posisie (indien die res korrek is): 4%

2.3

Positive marking from 2.1.2/
Positiewe merk vanaf 2.1.2.
Upwards positive/Opwaarts positief:
Vv,’ =v,? + 2ady ¥

0_= (24)? + 2(-9,8)Ay~”

Ay = 29,39 m

Total dispalcement = -41,6 +_29,39
Totale verplasing

Positive marking from 2.1.2/
Positiewe merk vanaf 2.1.2
Downwards positive/Afwaarts
positief:

VP =v, + 2ady

O = (-24)? + 2(9.8)Ay~

Ay = -29,39 m

Total dispalcement = 41,6 - Vv 29,39

=-12.21m Totale verplasing
= 12.21 mv =12,21mv
downwards” /afwaarts downwardsv /afwaarts

Copyright reserved/Kopiereg voorbehou

(8)

### Page 5

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 5 FS/VS/September 2014
Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum

2.4

InelasticY
The velocity/speed with which the ball bounces from the ground is less

than/differs from the velocity/speed at which the ball strikes the ground. ¥
The kinetic eneray is not conserved/changes. Y

Onelasties
Die snelheid/spoed waarmee die bal vanaf die grond bons is minder

as/verskillend van die snelheid waarmee dit die grond tref.

Die kinetiese energie word dus nie behou nie/verander. (3)
[17]

QUESTION 3/VRAAG 3

3.1

3.2

3.3

3.4

3.5

Newton’s First Law of motion. Vv A body will remains in its state of rest or
motion at constant velocity” unless a non-zero resultant/net force acts on it.”

Newton se Eerste bewegingswet. 'n Liggaam sal in sy toestand van rus of
konstante snelheid beweging volhard tensy ‘n nie-nul resultante/netto krag
daarop inwerk. (3)

A system in which no net external forces are acting on the spaceship.v”

'n Sisteem waarin geen netto eksteme kragte op ruimtetuig inwerk nie. (2)

F,At =Ap
F,.At=mv, —mv,

(12 000)(30) v= 3500(v;~ vi)
v

Av = 102,86 m-s” (3)

TF fF triction/friction” /Frywing, WryWing

Fg/wiweight/force of gravity” /gewig, gravitasiekrag
(2)

(3500)(9,8) Vv +f= as00(300—600, vv

f = -156 800 N
..f = 156 800 N ¥ (5)
[15]

Copyright reserved/Kopiereg voorbehou

### Page 6

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 6 FS/VS/September 2014
Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum

QUESTION 4/VRAAG 4

41 Newton's Second Law of motion. “ When a resultant/net force acts on an
object, the object will accelarates in the direction of the net force at an
accelaration which is directly proportional to v the force and inversly
proportional to the mass of the object. v

Newton se Tweede bewegingswet. Indien ‘n resulterende/netto krag op 'n

voorwerp inwerk, versnel die voorwerp in die rigting van die netto krag met 'n
versnelling wat direk eweredig is aan die krag en omgekeerd eweredig aan

die massa van die voorwerp. (3)

4.2 fs(max) = UsFnV
= 0,35(500)(9,8)v”

= 1715 NV (3)
4.3
Fappiied/ 10000 NY f/F dF riction/friction”
<= ,_e=_,
Ty
(3)
4.4 Left positive: /links positief: }
Car: Fre = ma ¥
T+f= ma
T v - (0,2)(500)(9,8) “= 500(3,04)v
T= 2500 NY
OR/OF
Truck/Trok: Fret = ma js
Fappiied/Ftoegepas + T+f= ma
10 000 v + T - (0,2)(1500)(9,8) v= 1 500(3,04)v
T = -2500 N
“.T = 2500 NY (5)
[14]

Copyright reserved/Kopiereg voorbehou

### Page 7

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 7 FS/VS/September 2014

Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum

QUESTION 5/VRAAG 5

5.1 Yes, ¥ no other forces other than the gravitational force is acting on the
block.”

Ja, geen ander krag behalwe gravitasiekrag werk op die blok in nie.

5.2 Whet = AK ¥

= %4(3)(6)-0v
= 54JV
5.3 | OPTION 4/ OPSIE 4 OPTION 2/ OPSIE 2
(U + Kya = (U + K)a ¥ Wret = AE/AK
(3)(9.8V(6) + Ov = (3)(9,8)h + (316) || (3)(9,8)Aycos0® v= 14(3)(62) - Ov
h= 4,16 mv
Ay = 1,84m
“h=6-1,84
= 4,16 mv

Positive marking from 5.3/ Positiewe merk vanaf 5.3

5.4 OPTION 1/ OPSIE 4

Wre= AU + AK Y/Wre= AE, + AEx

(f)(5)cos 180° = (3)(9,8)(1) - (3)(9,8)(4,16) VY + % (3)(2)? - % (3)(6)? V
f= 28,18NV

OPTION 2/ OPSIE 2

Whet = AEWAK

W, + Wy = AK

(f\(5)cos180°V + (3)(9,8)(4,16 — 1)cos 0° v= % (3)(2)? - % (3)(6)? ¥
f= 28,18NV

QUESTION 6/VRAAG 6

6.1 Red shift is a shift in the spectra of distant stars ~ towards the longer wave
lengths/red end of the spectrum.”

Rooi verskuiwing is ‘n verskuiwing in die spektra van verafgeleé sterre na die
langer golflengte/rooi kant van die spektrum.

6.2 Moving away from the earth’
The wavelength of the light coming from the distant star has increased/ shift
towards longer wavelengths. ”

Beweeg weg van die aarde af.

Die goifiengte van die lig komende van die ster het toegeneem/verskuif na die
langer golglengtes.

Copyright reserved/Kopiereg voorbehou

(2)

(3)

(4)

(5)

[14]

(2)

(2)

### Page 8

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 8 FS/VS/September 2014
Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum
6.3.1 The difference between the speed of the car and the speed of light is very

large.Y The frequency shift/fincrease is too small to observe.”

Die verskil tussen die spoed van die kar en die van lig is baie groot. Die

frekwensieskuif toename is dus te klein om waar te neem. (2)
6.3.2
32 4, =YEMLt onioF f -——4, v
viv, -V,
340
of, = ————-v (1200) v
t * 340-77,78 © (120)
= 1555,95 Hzv (4)
[10]
QUESTION 7/VRAAG 7
7A Electrostatic force experiencedv per unit positive chargev’ at that point./Die
elektrostatiese krag ervaar per eenheid positiewe lading by daardie punt. (2)
7.2 E, -¢ v
9x10" x12x10° Vv
0,3?
=1,2x 10° N.C1” (3)
7.3
Fry” ;
Note: No deduction of
Fy” marks for labels.
*. Let wel: Geen punte
aftrek vir byskrifte nie. (2)
7.4 kQ,Q
Fy = 72 zy
9x10? x8x 107 x10x10°
7 0,4?
=4,5x10°N
Fret? = Fxy? + Fy?
= (9,6 x 10%)? + (4,5 x 10%)? Fyz
Fret = 1,06 X 10°N S
ety
ny s
4,5x10
Tan@ = ——__v
9,6x10° Fxy
© = 25,11°
©. Fret = 1,06 X 10°N»% 295,11°/ 25,11° North of West/Noord van wes
25,11° upwards from negative x-axis »/opwaarts vanaf negatiewe x - as (6)
[13]

Copyright reserved/Kopiereg voorbehou

### Page 9

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 9 FS/VS/September 2014
Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum

QUESTION 8/VRAAG 8

8.1 21 J of work done /energy transfered” per coulomb/unit charge that is

moved between two points in an electric field./ 21 J arbeid verrig/energie

oorgedra per coulomb/eenheid lading wat tussen twee punte in ‘n elektriese

veld beweeg.
8.2.1 ma v
T Vi= 24-21
Je ee
37 a
1
=3A
V=IR
= (3)(3) ¥
= OV
V2 = 21-9
= 12Vwv
8.2.2 | Positive marking from 8.2.1/ OPTION 2/ OPSIE 2
Positiewe merk vanaf 8.2.1
OPTION 1/ OPSIE 7 E=\(R+9
v 24= 3(R + 1) Vv
R=— R=79
Rp =7-3v
-V =40
=70 eee ee
Rp =7-3¥ R, R, R,
=40 1,.41,1,
Lie aipeniay. 4° 6'R
R, R, R, “R=12Q0
1 11 Ratio of current in parallel
v= —4V branches:2:1/Verhouding van stroom in
4 6R
a parallelle vertakking: 2:1
R=120 Ie=1A
Vp = 21-9 p= IR?
nee = (1) ¥ (12) ¥
pi =12 WY
‘ v
12?
P=
12 ¥
= 12 Wv
8.3 Increasev /vergroot

Total resistance in circuit decrease.v/Totale weerstand in stroombaan verklein.

Total current increasev (and through 3  resistor)./Totale stroom vegroot

Therefore: Power increase according to P= I?R for R (30) staying constant.v/

Daarom: Drywing vergroot ooreenstemmend met P = FR vir R (3 Q) bly

konstant,.

Copyright reserved/Kopiereg voorbehou

(2)

(4)

(8)

(4)

[18]

### Page 10

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 10 FS/VS/September 2014
Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum
QUESTION 9/VRAAG 9
9.1.1 Electromagnetic Inductionv /elektromagnetiese induksie (1)
9.1.2 | Maximumv/ maksimum (1)
9.1.3 Av
The rate of change in the magnetic flux” is a maximumv at position C./ Die
tempo van verandering in die magnetiese vioed is ‘n maksimum by punt C. (3)
9.1.4
v
vw
' Mirror image of graph
/ 4 also correct.
s BC Ts) Spieélbeeld van grafiek
‘= ook korrek.
o

(4)

9.2.1 AC voltages can be stepped up or down by making use of transformers for
different users. OR
Electrical energy can be transmitted over long distances” with low current
and high voltage. (ANY ONE)
WS spanning kan m.b.v transformators verhoog of verlaag word vir
verskillende vebruikers OF
Elektriese energie kan oor lang afstande oorgedra word deur lae stroom en
hoé spanning. (ENIGE EEN) (1)

9.2.2 Pmax = Viaxlmax

300 VY = 80(Imax) ¥
(Imax) = 3,75 A

= 2,65 AV (5)
9.2.3

(3)
[18]

Copyright reserved/Kopiereg voorbehou

### Page 11

Downloaded from hlayiso.com

Physical Sciences P1 (Physics) 1 FS/VS/September 2014
Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum

QUESTION 10/VRAAG 10

10.1.1 Line emission (spectrum) /Lynemissie(spektrum)
(1)

10.1.2  Driffraction gratingy / Diffraksierooster
Prism / Prisma
(Any one/ Enige een) (1)

10.2 The frequency of red vlight is smaller than the threshold frequency’ of the
metal used in photo cell. / Die frekwensie van rooi lig is kleiner as die
drumpelfrekwensie van die metaal gebruik in fotosel.
OR/OF
The energy of the photon of red light is lower than the work function v of the
metal used in photo cell./ Die energie van die foton van rooi lig is minder as die
werksfunksie vir die metaal gebruik in fotosel. (2)

10.3.1 The minimum energy” needed for an electron to be set free from the surface
of a metal.v/ Die minimum energie nodig om ‘n elektron uit die opperviak van

‘n metaal vry te stel. (2)
10.3.2 | OPTION 1/ OPSIE 1 OPTION 1/ OPSIE 1

E=hfv = hiv

= (6,63 x 10°)(1,1 x 10%)v 3x 108V =x 10° (f v

=7.29x1079 J f = 6.67 x 10

Eeh® fo= 1,1 x 10° (given/gegee)
~) No, ¥ fohoton/f foton< fo (cut-off f)
_ 8,63 x10 x3x10°
450x107

=4,42x1079 J
No, ¥ Ephoton/Etoton < Work (5)
Function’ /Werksfunksie

[11]

TOTAL SECTION B/ITOTAAL AFDELING B: 130
GRAND TOTALIGROOTTOTAAL: 150

Copyright reserved/Kopiereg voorbehou

## Related documents

- [PHYS SC P1 MEMO ENGAFR](https://www.hlayiso.com/library/phys-sc-p1-memo-engafr/da9b9ae0-718d-442a-95bc-aa0bfbf308f3) — Physical Sciences · Memo · 2014
- [2014AFR gau physical sciences p1 memo](https://www.hlayiso.com/library/2014afr-gau-physical-sciences-p1-memo/db77042f-0bdc-4e06-8f6b-ce760a7e4148) — Physical Sciences · Memo · 2014
- [2014ENG gau physical sciences p1 memo](https://www.hlayiso.com/library/2014eng-gau-physical-sciences-p1-memo/d8eca56a-f776-4202-84cd-3e2e7e299ae3) — Physical Sciences · Memo · 2014
- [Physical Sciences P1 Nov 2014 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2014-memo-afr-eng/44924678-c33b-42e1-8937-fcfda16ea7b3) — Physical Sciences · Memo · 2014 · Eng
- [wkp kwod physc p1 sept 2014 eng memo](https://www.hlayiso.com/library/wkp-kwod-physc-p1-sept-2014-eng-memo/bb06a52a-4a49-454d-ae6e-5128be28cc24) — Physical Sciences · Memo · 2014 · Eng
- [fs physical sciences p2 gr 12 september 2014](https://www.hlayiso.com/library/fs-physical-sciences-p2-gr-12-september-2014/4dab514d-fa81-4d07-8f7c-8679f92197c1) — Physical Sciences · QP · 2014

## 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.