# Physical Sciences Grade 12 Paper 1 Gauteng Mock 2019 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Gauteng Mock Exam 2019, 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2019gauteng-vraestel-1-memo/50fef74b-e45d-4f07-8ac1-68087039e4ea
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2019
- Exam period: Gauteng Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 12

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## About this document

Physical Sciences · Grade 12 · Gauteng Mock Exam · 2019. Memorandum, 12 pages. Read online or download the PDF.

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## Document text

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### Page 1

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BEES, GAUTENG PROVINCE

3 *y' EDUCATION
‘wr REPUBLIC OF SOUTH AFRICA

PREPARATORY EXAMINATION
VOORBEREIDENDE EKSAMEN

2019

MARKING GUIDELINES
NASIENRIGLYNE

10841
PHYSICAL SCIENCES: PHYSICS (PAPER 1) /
FISIESE WETENSKAPPE: FISIKA (VRAESTEL 1)

12 pages / bladsye

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GAUTENG DEPARTMENT OF EDUCATION /
GAUTENGSE DEPARTEMENT VAN ONDERWYS
PREPARATORY EXAMINATION /
VOORBEREIDENDE EKSAMEN

PHYSICAL SCIENCE: PHYSICS /
FISIESE WETENSKAPPE: FISIKA
(Paper 1 / Vraestel 1)

MARKING GUIDELINES /

NASIENRIGLYNE

QUESTION / VRAAG 1

11 Dyvv 2
1.2 Dvv 2
1.3 Dvy A CORRECT 2
1.4 BYY = ANDC 2
1.5 Avv ANDB 2
1.6 Dvv 2
1.7 AVY 2
1.8 BYY 2
1.9 AVY 2
1.10 Dv¥v¥ 2

[20]

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QUESTION / VRAAG 2

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2.1.1. When a net force acts on an object, the object will accelerate in the direction of the
force” and the acceleration is directly proportional to the force and inversely

T
¥ Tension/
Spanning

2.3

proportional to the mass~ of the object.
OR

The net (or resultant) force acting on an object is equal to the rate of change of
momentum of the object in the direction of the net force. v¥

Wanneer n_netio krag op ‘n voorwerp inwerk sal die voorwerp versnel in die rigting van
die netto krag. Die versneiling is direk eweredig aan die krag en omgekeerd eweredig

aan die massa van die voorwerp.
OF

Die netto krag wat inwerk ap in voorwerp is gelyk aan die tempo van verandering van
momentum van die voorwerp in die rigting van die netto krag.

N/ Fx / Normal / Normaal ~

Ni Fu / Normal ”

» ZONTFa

12N

o Fag onaey
Fo/w v¥

3 kg block / 3 kg blok
Fret = mav

F,-T -f.= 3a

20 -T-1.2 = 3av
18,8 —T = 3a
T=188-3a.... (1)
2 kg block | 2 kg blok
Fret = ma

18,8 -3a = 2a
a= 3,76 m:s*
T = 2(3.76) v= 7,52 NY

Set the two equations equal “

3
10841 Physical Science P1 Memo Bilcorrected (2)

Stel die twee vergelykings gelyk
aan mekaar

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2.2.1 Each body in the universe attracts every other body with a force that is directly
proportional to the product of their masses” and inversely proportional to the square of
the distance between their centres. v
Elke liggaam in die heelal trek elke ander liggaam aan met n krag wat direk eweredig
is aan die produk van hul massas en omgekeerd eweredig aan die kwadraat van die
afstand tussen hulle middelpunte. 2

2.2.2 To the right as positive. Neem na regs as positief

Force attracting mto earth,Fe = -G2%E (Ignore if the negative is missing.)
Krag wat m aantrek na aarde for both correct expressions
9 ; mMy vir albei korrekte stellings
Force attracting mto moon, Fu = Goat
Krag wat m aantrek na die maan
Net force on m, Fe + Fu = -GUe 4 ort = ov
Netto krag op m
- GUE + G mM yy 0
2. Ont 0; <a =0v 4
[20]
QUESTION / VRAAG 3
3.1 The motion of an object in the force of gravity only. “”
Die beweging van n voorwerp in gravitasiekrag alleen. 2
3.2 2/twiceY twee maal 1
3.3.1
OPTION / OPSIE 1 OPTION / OPSIE 2 OPTION / OPSIE 3
ve=vP + 2ahy ¥ V=vy + aAty vy=zyeo+ aAtv
(107 v =O? + (2)-9,8)Ay ¥ 10 =0 + (-9,8)At ¥ ~10_= 0+ (-9,B)At ¥
“Ay = -5,1m ~At= 1,025 “At= 1,02 5
ieh=51m v oN, ~~
0 10 Ay =viAt + % adt
Ay = y= ( ) x1,02 ¥| ay= 0+ (A)-9,8)(4,02)~
—_s =-5,1m
--5tm he5,1m¥
oh=51mv¥
4
OPTION / OPSIE 4 OPTION / OPSIE 5
Woet = AK any one (U+K) =(U+K) }
Woray = AK pow (mgh + % mv) =(mgh + % mv}, 2
mghcos® = % mv? -% mv? 2
(0,29,8)h(1) ¥ = (%4)\(0,2)(10"-0 ¥ ((0,2)(9,8)h +0) ¥ = (0 +(%X{0,2)(10)) ¥
uh=5,1mv ohe5,.mv
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3.3.2

3.4

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OPTION / OPSIE 1 ; OPTION / OPSIE 2
ve =v? + 2aAyv VYyavyieo+ aAt
(OY = 7.5? + (2\(-9,8)Ay ¥ 0_=7.5 + (-9,8)At V

v Ay =2.87m 4% - At = 0.775

ie. h=2.87m
Ay=vAt + %adt?
ay

= 7.5 (0.77) _+ {(%\-9.8)0.77) ¥
= 2.87 mY
« h=2.87m

OPTION / OPSIE 3

Woe: = AK

Woray = AK Y
mghcos@ = % mv? -% mv?

(0,2)(9,8)h(-1) = (4)(0,2)(0)" — (4)(0,27.5)? ¥
oh=2.87mv

5,1 mv (POSITIVE MARKING FROM 3.3.1)

QUESTION / VRAAG 4

4.1

4.2.2

4.2.2

4.3

¥
Isolated (system) or closed system
Geisoleerde sisteem of gesiote sisteem

Fret = Ap ¥

Fret (0,1) ¥ = 14 000 — 30 000
Fret =-160 000 N
Fnet = 160 000 NY

Xp; = Lp; } Any one/ Enige een | ¥

My¥4i F MV = MyVagt MoV2¢

30 000 + (9000)(-1.5)“ = 14000 + (9000) ve) ¥
Ve =0.28m's" ¥

Foes Poet

20 20.1 20,2 203° t(s) on 20 adi 202 20,3 t{s)

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QUESTION / VRAAG 5

5.1 5.1.1 A force for which the work done in moving an object between two points
depends on the path taken. vv
in Krag waarvoor die arbeid verrig om n voorwerp tussen twee punte te
beweeg, is afhanklik van die roete wat gevolg word. 2

5.1.2 Wre= AEp + AEKY

fAx cos 6 = AE, + AE,

18 x Ax cos 180° v= 5x9.8x(3-0) ¥ + % x 5x(0-9.907)

Ax = 5.4458 (m) ¥

se 3

e= sin ; Gaal”

6 = 33.43°v 7
Include solution Wnet = Ack SEE LAST PAGE

5.2 5.2.1 The work done on an object by a net force” is equal to the change in the

object's kinetic energy.“ OR
The net work done on an object by a force” is equal to the change in the

object's kinetic eneragy.~ OR
The net work” done on an object is equal to the change in the object's kinetic

energy. ~

Die netto arbeid verrig op ’n voorwerp deur 'n netto krag is gelyk aan die

verandering in die voorwerp se kinetiese energie.

Of

Die arbeid verrig deur in netto krag is gelyk aan die verandering in die

voorwerp se kinetiese energie.

Of

Die netto arbeid verrig op n voorwerp is gelyk aan die verandering in die

voorwerp se kinetiese energie. 2

5.2.2 Wnet = AEkY
Weg + Wea = FoAxcos® +Wapp = 2 Mp — 2 MVi2
(987)(9.8)(46) cos180° + Wapp = (9870.9) — OY
Wea = 445339.34 JY 4
[45]

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1084119

QUESTION / VRAAG 6

6.1 The change in frequency (or pitch) of the sound detected by a listener,“

because the sound source and the listener have different velocities relative to

the medium of sound propagation.”

Die verandering in die waarneembare frekwensie van die klankgolwe as gevolg
van die relatiewe beweging tussen die snelhede van die luisteraar en die klank
Se voortplanting. 2

62, _ (v+v,)

fs Vv

*  (vitv,)°
gov = (940+9)  sagv
(340—v,)
Vs = 41,099 m-s! v 5
63 63.1 910Hz~ 1

6.3.2 Positive marking from 6.2
(v+v,)

= LS lo! £,
(v+v,)
340-0 v
= ——— 800V
f 340+ 41000"
= 713,73 Hz 4
6.4  800Hz 1

[13]

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QUESTION / VRAAG 7

7.1

7.2

7.3

7.4

The magnitude of the electrostatic force exerted by one point charge (Q,) on
another point charge (Q2) is directly proportional to the product of the

magnitudes of the charges and inversely proportional to the square of the
distance (r) between them”

if mass is used in definition 9/,

Die grootte van die elektrostatiese krag uitgeoefen deur een puntiading op n
ander puntlading is direk eweredig aan die produk van die grooites van die
ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle.

Indien massas gebruik is in definsie o%, 2
v e Arrow and
F correct labelv
a8 © Pylen korrekte
byskrif
v
Fa 2
kQ,Q 9x10°x4x10°x5x10° .
F, = As v= ¥ = 4,5x104N
'Q2 due to O3 r? (20x10°)"
kQ,Q 9x10°x2x10°x4x10°
F ert = ¥ =3,2x104N
Q2 due to 1 r2 a 5x1 0° y *
Fret= + (4,5x10~4)? + (3,2x10-4)? V = 5,522 x 107 N ¥ 5
Take the right is positive / Regs is positief
go -3
E-anc = KQy = x10" x2x10"" = 28 800 (N--C" to the left / na links)
r (25x10)
kQ 9x10°x4x10° 4 .
Eanc= = = —— VY = 360 000(N-C"' to the right / na regs:
4nc 2 (10x10) ( ig gs)
Enet = Eonc - Eanc Y This mark is for subtracting / Punt vir aftrek.
= 360 000 - 28 000
= 331200 N-C" to the right / na regs v 5

[14]

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QUESTION / VRAAG 8
8.1 Resistance within / offered by the cell” : 1
Weerstand gebied deur die sel self.
8.2 Temperature” 1
Temperatuur
8.3 Volts (V) V Potential difference / Potensiaal verskil 1
8.4 Internal resistance / interne weerstand VOR / OF r 1
85 7,2 1
8.6 slope / helling = AV/ Alv 3
=0-7.2“/08-0v
=-9
r=9 Ov
8.7 POSITIVE MARKING FROM 8.5 4
7,2V
[10]
QUESTION / VRAAG 9
91 10VY (1)
9.2 OV OR/ OF zero (1)
93 141 + a RRs 10x5
Rp i 1 R,= RotRs = 10+5v=3,3330
= To + z v
= Y= av
Ry = 3,33 Rex = 3,333 +2 5,333 0
Rext = 3,33 +2 =5,330Y (3)

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9.4

9.5

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POSITIVE MARKING FROM 9.3

= Vext
Rext = 1 v

Vext _ _8 =

Rext 5.333 =15(4) v
emf = IRext + Ir y
10=8 + 1l5rv

r=1330 Vv

l=

Option / Opsie 2

Vy = loRy “= 1,5 “x 3,333 = 4,9995 V
Ve=1,5x2=3V

Vext = VirtVs = 4,9995 +3 =7,9995 V

Vint = & - Vex V= 10-7,9995V = 2,00 V

Vint 2
Tint = =—= 13340 Vv
in I 15 (5)

increases ~ (Negative marking)

vermeerder (negatiewe nasien)

Circuit resistance increases v therefore (circuit) current decreasesv lost volts
decreases / Vin decreases.”

Stroombaan se weerstand vermeerder ¥ daarom sal die stroom afneem ¥ en die

verlore volts sal verminder of die Vintern sal verminder. “ (4)
[14]

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QUESTION / VRAAG 10

10.1. 10.1.1 Slip rings ¥
Sleepringe (1)

10.1.2  coil/ solenoid ¥
spoe! / solenoid (1)

10.2 Connects the generator to the external circuit.
OR Generated current exits the generator through C. “
OR Allows movement and conductivity between coil and external circuit.

Verbind die generator aan die eksterne stroombaan
OF Gegenereerde siroom veriaat generator deur D
OF Laat beweging toe en geleiding tussen spoel en eksterne stroombaan (1)

10.3 parallel” (1)
10.4 = Imax/maks =25mA

nx
= v2 ¥

25x10>

v2"
17,678 x10 Av (3)

lms

10.5 Positive marking from 10.4
Pay/gem = Inns J wok” xRv¥
= (17,678 x 40°)? x100
= 3,125x107 W v (3)
[10]

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QUESTION / VRAAG 11

11.1

11.2

11.3

11.4

11.5

Is the minimum energy that an electron in the metal needs to

be emitted from the metal surface. “ ”

Dit is die minimum energie wat n elekiron in die metaal benodig om vrygestel te
word uit die metaal se opperviakte.

Frequency of green light is below threshold frequency (f.). ““
OR

Energy of green light lower than work function.
OR
Too low frequency.

Frekwensie van groen lig is onder die drumpelfrekwensie
OF
Energie van groen lig is laer as die werksfunksie.

Die frekwensie is te laag.

Negative marking
INCREASE ~“
Vermeerder

Negative marking

more photons sirike the metal per unit time’ hence more electrons ejected per
unit timev’ thus rate of flow of charge increases hence increased current
reading.

Meer fotone tref die metaal per eenheid tyd daarom word meer elektrone
vrygestel per eenheid tyd dus vermeerder die vioei van die lading dus word die
stroom lesing verhoog.

hf
he Y any one / enige een

3.0x10°
= p4yy, She vy
6,63 x 10 x 500x107
=3,98x 10? Uv

E

"

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