PROVINCIAL ASSESSMENT
PROVINSIALE ASSESSERING
GRADE/GRAAD 12
PHYSICAL SCIENCES: PHYSICS (P1)
FISIESE WETENSKAPPE: FISIKA (V1)
JUNE/JUNIE 2024
MARKING GUIDELINES/NASIENRIGLYNE
MARKS/PUNTE: 150
These marking guidelines consist of 16 pages including a cognitive grid.
Hierdie nasienriglyne bestaan uit 16 bladsye en ‘n kognitiewe tabel.
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Physical Sciences Grade 12 Paper 1 North West June 2024 Memo
Grade 12 Physical Sciences Paper 1 memorandum from the North West June Exam 2024, 18 pages. Read online or download the free PDF.
Physical Sciences · Grade 12 · North West June Exam · 2024. Memorandum. Read online or download the PDF.
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Physical Sciences P1/Fisiese Wetenskappe V1 2 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
QUESTION 1/VRAAG 1
1.1 B (2)
1.2 D (2)
1.3 C (2)
1.4 C (2)
1.5 A (2)
1.6 A (2)
1.7 B (2)
1.8 C (2)
1.9 D (2)
1.10 B (2)
[20]
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Physical Sciences P1/Fisiese Wetenskappe V1 3 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
QUESTION 2/VRAAG 2
2.1 A body will remain in its state of rest or motion at constant velocity
unless a non-zero resultant/net force acts on it.
‘n Liggaam sal in rus of beweging teen konstante snelheid volhard
tensy ‘n nie-nul resulterende/netto krag daarop inwerk. (2 or 0) (2)
2.2 FN
f T
Fg
Accept the following symbols/aanvaar die volgende simbole
N /Normal/Normal force
FN
Normaal/Normaalkrag
T FT/ Ft /Tension/Spanningskrag
Fg w /mg/weight/gravitational force
Gewig/gravitasiekrag
f
Ff /Force of friction/Wrywingskrag
Notes/Aantekeninge:
• Mark awarded for label and arrow./Punt toegeken vir benoeming en
pyltjie.
• Do not penalise for length of arrows since drawing is not to scale./
Moenie vir die lengte van die pyltjies penaliseer nie aangesien die
tekening nie volgens skaal is nie.
• Any other additional force(s)/Enige ander addisionele krag(te):
Max/Maks 3/4
• If everything is correct, but no arrows/Indien alles korrek is, maar
geen pyltjies: Max/Maks ¾
2.3.1 FY = 0
N + Fg = 0
N – Fg = 0 any one/enige een
N = mg
N = 8 × 9,8
N = 78,4 N (3)
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Physical Sciences P1/Fisiese Wetenskappe V1 4 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
2.3.2 Fnet = ma ✓
(Direction of motion as positive)
(Rigting van beweging as positief.)
Block/Blok A
T + fK = ma
T - fK = ma any one/enige een
T - 23,52 = 8a
T = 8a + 23,52……………(1) ✓
Block/Blok B
F - (T + Fg) = ma
T = F - (ma) - Fg any one/enige een
= 192 - 16a - 16(9,8)…………..(2) ✓
Solving (1) and (2)/Los (1) en (2) op:
8a + 23,5 = 192 - 16a - 156,8
a = 0,49 m.s-2 ✓
Substitute a into (1) or (2)/Vervang a in (1) of (2)
T = 8a +23,52 OR/OF T = 192 - 16a - 156,8
= 8(0,49) + 23,52 = 192 - 16(0,49) -156,8
T = 13,72 N ✓ = 13,72 N (5)
2.4 To the left.
The only force acting of the object is frictional force. (According
Newton’s second law), the body will accelerate in the direction of the
(net) force.
Na links.
Die enigste krag wat op die voorwerp inwerk is wrywingskrag.
(Volgens Newton se tweede wet), die liggaam sal versnel in die (2)
rigting van die (netto) krag.
2.5 Increase
Vermeerder (1)
[17]
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Physical Sciences P1/Fisiese Wetenskappe V1 5 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
QUESTION 3/VRAAG 3
3.1 Marking criteria/Nasienkriteria
If any of the underlined key words/phrases in the correct context is
omitted deduct 1 mark./ Indien enige van die onderstreepte
sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.
The force or the component of a force which a surface exerts on an object
with which it is in contact, and which is perpendicular to the surface.
Die krag of komponent van ‘n krag wat deur die oppervlak uitgeoefen (2)
word op ‘n voorwerp waarmee dit in kontak is, is loodreg op die
oppervlak.
3.2
Accept the following symbols/aanvaar die volgende simbole
FN/Normal/Normal force
N
Normaal/Normaalkrag
T FT/ Ft /Tension
f Ff /Force of friction/Wrywingskrag
w Fg/mg/weight/gravitational force
Fg/mg/gewig/gravitasiekrag
Notes/Aantekeninge:
• Mark awarded for label and arrow./Punt toegeken vir benoeming en
pyltjie.
• Do not penalise for length of arrows since drawing is not to scale./
Moenie vir die lengte van die pyltjies penaliseer nie aangesien die
tekening nie volgens skaal is nie.
• Any other additional force(s)/Enige ander addisionele krag(te):
Max/Maks 3/4
• If everything is correct, but no arrows/Indien alles korrek is, maar
geen pyltjies: Max/Maks 3/4 (4)
3.3 OPTION/OPSIE 1: OPTION/OPSIE 2:
Fnet = ma Fnet = ma
Fnet = 0 any one/enige Fnet = 0 any one/enige een
een T =Fgll + f
T = Fg =(44 x 9,8 Sin 28,66˚) + 87,2
Fg = mg = 294 N
= 30(9,8) (3)
= 294 N
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Physical Sciences P1/Fisiese Wetenskappe V1 6 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
3.4 POSITIVE MARKING FROM QUESTION 3.3
POSITIEWE NASIEN VANAF VRAAG 3.3
To the right as positive/Na regs as positief:
Fnet =ma
Fnet =0 any one/enige een
T - Wsinθ - f = ma
294 - 44x9,8sinθ - 87,2 = 0
θ = 28,66˚ (2)
[11]
QUESTION 4/VRAAG 4
4.1 Marking criteria/Nasienkriteria
If any of the underlined key words/phrases in the correct context is
omitted deduct 1 mark./Indien enige van die onderstreepte
sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt
af.
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. (2)
Die netto/resultante krag wat op ‘n voorwerp uitgeoefen word is
gelyk aan die tempo van momentsverandering in die rigting van die
netto krag.
4.2.1 Ʃpi = Ʃpf
mvi1 + mvi2 = mvf1 + mvf2 any one/enige een
(0,41)(15) + (79)(0) = (0,41)(-12) + (79)vf1
vf1 = 0,14 m.s-1 west/wes (4)
4.2.2 Positive marking from Question 4.2.1/
Positiewe merk vanaf Vraag 4.2.1
FnetΔt = Δp
∆𝑝
Fnet = ∆𝑡 any one/enige een
mvf−mvi
= ∆𝑡
(79)(0,14)−(79)(0)
=
2
= 5,53 N east/oos (3)
4.2.3 Positive marking from Question 4.2.2/
Positiewe merk vanaf Vraag 4.2.2
Fnet = 5,53/4 N west/wes (1)
[10]
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Physical Sciences P1/Fisiese Wetenskappe V1 7 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
QUESTION 5/VRAAG 5
5.1 In an isolated system the total (linear) momentum is
conserved/remains constant.
In 'n geïsoleerde sisteem bly die totale (lineêre) momentum
behoue/konstant. (2 or 0) (2)
5.2.1 FnetΔt = Δp any one/enige een
∆𝑝
Fnet =
∆𝑡
mvf−mvi
= ∆𝑡
111 000−0
7,4 x 105 =
𝑡1 −0
t1 = 0,15 s (4)
5.2.2 Between t = t1 and t = 0,25 s, there is a decrease in
momentum. Therefore, the velocity of the ball decreases
which implies that there is a force opposing the motion of
the ball (i.e friction) (2)
Tussen t = t1 en t = 0,25 s, is daar ‘n afname in momentum.
Daarom sal die snelheid van die bal verminder, dus kan ons
aflei dat daar ‘n krag is wat beweging teenwerk (wrywing).
5.2.3 Ʃpi = Ʃpf
pAi + pBi = pAf + pBf any one/enige een
100 000 + 0 = 88 800 + pBf
pBf = 11 200 kg.m.s-1
but/maar p = mv
11 200 = (15 000)v
v = 0,75 m.s-1 (to the right/na regs) (4)
5.3 According to Newton’s second law, according to momentum
FnetΔt = Δp
Crumple zones causes an increase in collision time which results
in a lesser/smaller force
Volgens Newton se Tweede wet in terme van momentum
FnetΔt = Δp
Frommelsones veroorsaak ‘n verlengde botsingstyd wat ‘n kleiner
krag tot gevolg het. (3)
[15]
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Physical Sciences P1/Fisiese Wetenskappe V1 8 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
QUESTION 6/VRAAG 6
6.1 An object which has been given an initial velocity and then it moves
under the influence of the gravitational force only.
‘n Voorwerp met ‘n aanvanklike snelheid beweeg slegs onder die
invloed van gravitasiekrag. (2 or 0) (2)
6.2.1 OPTION/OPSIE 1 OPTION/OPSIE 2
UPWARDS POSITIVE UPWARDS POSITIVE
OPWAARTS POSITIEF OPWAARTS POSITIEF
vf 2 = vi 2 + 2aΔy vf = vi + aΔt
(- 50)2 =(- 40)2 + 2(-9,8)Δy - 50 = - 40 + (-9,8)Δt
Δy = - 45,92 Δt = 1,02 s
Δy = 45,92 m
downwards/afwaarts Δy = viΔt + ½gΔt2
=(-40)(1,02) + ½(-9,8)(1,02 )2
Δy = 45,90 m (3)
6.2.2 UPWARDS POSITIVE/OPWAARTS POSITIEF
vf = vi + a∆t
-50 = -40 + (-9,8)(Δt)
Δt = 1,02 s (time to reach to ground)/(tyd om die grond te
bereik)
vf = vi + a∆t
0 = 24 + (-9,8)(Δt)
Δt = 2,45 s
(time to reach maximum height)/(tyd om maks hoogte te
bereik)
t = 1,02 + 2,45
t = 3,47 s (4)
6.2.3 Positive marking from 6.1.1/Positiewe merk vanaf 6.1.1
UPWARDS POSITIVE/OPWAARTS POSITIEF
vf 2 = vi 2 + 2aΔy
0 = 242 + 2(-9,8) Δy
Δy = 556,40 m
Displacement/Verplasing
= - 556,40 + 45,90
= - 510,50
= 510,50 m (downwards/afwaarts) (3)
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Physical Sciences P1/Fisiese Wetenskappe V1 9 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
6.3
(3)
[15]
OUESTION 7/VRAAG 7
7.1 free fall/vry val (1)
7.2 Newton's First Law/Newton se Eerste Wet
A body will remain in its state of rest or motion at a constant velocity
unless a non-zero net/resultant force acts on it.
'n Voorwerp sal in sy posisie van rus of beweging bly teen 'n
konstante snelheid tensy 'n nie-nul netto/resulterende krag daarop
inwerk. (3)
7.3 7.3.1 OPTION/OPSIE 1 OPTION/OPSIE 2
(Upwards as (Downwards as
positive/Opwaarts as positive/Afwaarts as
positief) positief)
vf2 = vi2 + 2aΔy vf2 = vi2 + 2aΔy
0 = (16)2 + 2(-9,8)Δy 0 = (-16)2 + 2(9,8)Δy
Δy = 13,06 m Δy = - 13,06 m
Height/Hoogte = 13,06 + 25 Height/Hoogte = 13,06 + 25
= 38,06 m = 38,06 m (4)
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Grade 12 – Marking Guidelines/Nasienriglyne
7.3.2 OPTION/OPSIE 1 OPTION/OPSIE 2
(Whole motion upwards as (Whole motion downwards
positive/Hele beweging as positive/Hele beweging
opwaarts as positief) afwaarts as positief)
Δy = viΔt + ½aΔt2 Δy = viΔt + ½aΔt2
= 16(3,5) + ½(-9,8)(3,5)2 = -16(3,5) + ½(9,8)(3,5)2
= - 4,03 m = 4,03 m
Height/Hoogte = 25 - 4,03 Height/Hoogte = 25 - 4,03
= 20,97 m = 20,97 m
OPTION/OPSIE 3 OPTION/OPSIE 4
(Whole motion upwards as (Whole motion downwards
positive/Hele beweging as positive/Hele beweging
opwaarts as positief) afwaarts as positief)
vf = vi + g∆t vf = vi + g∆t
0 = 16 − 9,8∆t 0 = -16 + 9,8∆t
∆t = 1,63 s ∆t = 1,63 s
Δy = viΔt + ½aΔt2 Δy = viΔt + ½aΔt2
= 0 + ½(-9,8)(3,5 - 1,63)2 = 0 + ½(9,8)(3,5 - 1,63)2
= - 17,13 m = 17,13 m
Height/Hoogte =38,06 - 17,13 Height/Hoogte =38,06 - 17,13
= 20,93 m = 20,93 m
OPTION/OPSIE 5 OPTION/OPSIE 4
(From point of projection (From point of projection
upwards as positive/Vanaf downwards as
punt van projeksie positive/Vanaf punt van
beweging opwaarts as projeksie beweging
positief) afwaarts as positief)
vf = vi + g∆t vf = vi + g∆t
0 = 16 − 9,8∆t 0 = -16 + 9,8∆t
∆t = 1,63 s ∆t = 1,63 s
Δy = viΔt + ½aΔt2 Δy = viΔt + ½aΔt2
= -16(0,24) + ½(-9,8)(3,5 - 3,26)2 = 16(0,24) + ½(9,8)(3,5 - 3,26)2
= - 4,12 m = 4,12 m
Height/Hoogte = 25 - 4,12 Height/Hoogte = 25 - 4,12
= 20,88 m = 20,88 m
RANGE/GEBIED: 20,88 - 20,98 m RANGE/GEBIED: 20,88 - 20,98 m (5)
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Grade 12 – Marking Guidelines/Nasienriglyne
7.3.3 OPTION/OPSIE 1 OPTION/OPSIE 2
(Whole motion upwards as (Whole motion downwards
positive/Hele beweging as positive/Hele beweging
opwaarts as positief) afwaarts as positief)
∆y = vi∆t + ½g∆t2 ∆y = vi∆t + ½g∆t2
-25 = 16Δt + ½(-9,8)Δt2 25 = -16Δt + ½(9,8)Δt2
Δt = 4,42 s Δt = 4,42 s
Note: For options 3 and 4 use any of the equations of
motion containing ∆t to calculate ∆t and substitute into
equation of ∆y = vi∆t + ½g∆t2.
Nota: Vir opsies 3 en 4, gebruik enige van die
bewegingsvergelykings wat ∆t bevat om ∆t te bereken en
vervang in vergelyking van ∆y = vi∆t + ½g∆t2.
OPTION/OPSIE 3 OPTION/OPSIE 4
(From maximum height (From maximum height
upwards as positive/Vanaf downwards as
maksimum hoogte positive/Vanaf maksimum
opwaarts as positief) hoogte afwaarts as
positief)
vf = vi + g∆t
0 = 16 − 9,8∆t vf = vi + g∆t
∆t = 1,63 s 0 = - 16 + 9,8∆t
∆t = 1,63 s
∆y = vi∆t + ½g∆t2
- 38,06= 0Δt + ½(-9,8)Δt2 ∆y = vi∆t + ½g∆t2
Δt = 2,79s 38,06= 0Δt + ½(9,8)Δt2
Δt = 2,79s
Δt = 2,79 + 1,63
= 4,42 s Δt = 2,79 + 1,63
= 4,42 s (4)
7.4 Positive marking from 7.3.3/Positiewe merk vanaf 7.3.3
UPWARDS POSITIVE/OPWAARTS POSITIEF
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Physical Sciences P1/Fisiese Wetenskappe V1 12 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
DOWNWARDS POSITIVE/AFWAARTS POSITIEF
CRITERIA FOR GRAPH/RIGLYNE VIR GRAFIEK
Shape of graph/Vorm van grafiek
Initial velocity/Aanvanklike snelheid
Time to complete motion/Tyd vir volledige beweging (3)
[20]
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Physical Sciences P1/Fisiese Wetenskappe V1 13 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
QUESTION 8/VRAAG 8
8.1 Marking criteria/Nasienkriteria
If any of the underlined key words/phrases in the correct context is
omitted deduct 1 mark./Indien enige van die onderstreepte
sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt
af.
A force for which the work done in moving an object between two points
depends on the path taken.
‘n Krag waar die werk gedoen op ‘n bewegende voorwerp tussen twee
punte, afhanklik is van die pad geneem. (2)
8.2 Frictional force./Gravitasie krag (1)
8.3 fk = µkN
= µk(mgcosθ) any one/enige een
= 0,112 x (70 x 9,8cos30˚)
= 11,85 N (down the slide/glyplank af) (3)
8.4 POSITIVE MARKING FROM QUESTION 8.3
POSITIEWE NASIEN VAN VRAAG 8.3
OPTION/OPSIE 1
Wnet = ΔEk
fΔxcosθ + Fg//Δxsinθ = ½mvf2 - ½mvi2 any one/enige een
(11,85)(2,8)cos180˚+(70)(9,8)(2,8)sin30˚=½(70)vf2 - ½(70)(0,35)2
vf =7,39 m.s-1
OPTION/OPSIE 2:
Wnc = ΔEk + ΔEp
fΔxcosθ = (½mvf2 - ½mvi2) + (mghf + mghi) any one/enige een
(11,85)(2,8)cos180˚= [½(70)vf - ½(70)(0,35) ] + [0 - (70)(9,8)(2,8)]
2 2
vf = 7,39 m.s-1 (5)
8.5 INCREASES
The frictional force is proportional to the normal force , and since the
normal force increases with an increase in the angle of the slide, the
frictional force also increases.
TOENEEM
Die wrywingskrag is direk eweredig aan die normaalkrag, en omdat die
normaalkrag vermeerder as die hoek van die glyplank vermeerder, sal
wrywingskrag ook meer word. (3)
[14]
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Grade 12 – Marking Guidelines/Nasienriglyne
QUESTION 9/VRAAG 9
9.1 The net/total work done (on an object) is equal to the change in the
object's kinetic energy.
Die netto/totale arbeid wat (op ‘n voorwerp) verrig is,is gelyk aan die
verandering in die voorwerp se kinetiese energie.
OR/OF
The work done on an object by a resultant/net force is equal to the
change in the object's kinetic energy.
Die arbeid verrig op ‘n voorwerp deur die resultante/netto krag is gelyk
aan die verandering in die voorwerp se kinetiese energie.
(2 or 0) (2)
9.2 Accept the following symbols/aanvaar die volgende
simbole
air friction f/fair/lugweerstand
F FA/FT/
w Fg/mg/weight/gravitational force
Fg/mg/gewig/gravitasiekrag
Notes/Aantekeninge:
• Mark awarded for label and arrow. /Punt toegeken
vir benoeming en pyltjie.
• Do not penalise for length of arrows since drawing is
not to scale. /
Moenie vir die lengte van die pyltjies penaliseer nie
aangesien die tekening nie volgens skaal is nie.
• Any other additional force(s)/Enige ander
addisionele krag(te):
Max/Maks 2/3
• If everything is correct, but no arrows/Indien alles
korrek is, maar geen pyltjies: Max/Maks 2/3 (3)
9.3 OPTION/OPSIE 1
Wnet = ΔEk
WT + WW + Wf = ΔEk any one/enige een
FTΔycosθ + FgΔycosθ + Wf = 0
(15 000)(50)cos0˚ + (2500)(9,8)(50)cos180˚ + Wf = 0
Wf = 475 000 J (5)
OPTION/OPSIE 2:
UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF
Fnet = ma
F + f + w = ma
15 000 - f - (2500)(9,8) = 0
f = 9 500 N Both equations/Beide vergelykings
Wf = fΔycosθ
= 9500(50)cos180˚
= 475 000 J (5)
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Grade 12 – Marking Guidelines/Nasienriglyne
9.4 W
P=
∆t
475 000
=
(30 ×60)
= 263,89 W
(3)
[13]
QUESTION 10/VRAAG 10
10.1 Marking criteria/Nasienkriteria
If any of the underlined key words/phrases in the correct context is
omitted deduct 1 mark./ Indien enige van die onderstreepte
sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt
af.
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 frekwensie (toonhoogte) van die waargenome
klank deur die luisteraar agv die klankbron en die luisteraar wat
verskillende snelhede relatief tot mekaar het.
OR/OF
An (apparent) change in (observed/detected) frequency (pitch), as
a result of the relative motion between a source and an observer
(listener).
'n (Skynbare) verandering in(waargenome) frekwensie (toonhoogte),
as gevolg van die relatiewe beweging tussen die bron en 'n
waarnemer/luisteraar. (2)
10.2 EQUAL TO /GELYK AAN
The driver of the taxi is at rest relative to the source of sound.
Die bestuurder van die taxi beweeg nie relatief tot van die bron van
klank nie.
OR/OF
No relative motion between the taxi driver and taxi.
Geen relatiewe beweging tussen bestuurder en taxi nie. (2)
10.3 OPTION/OPSIE 1: OPTION/OPSIE 2:
340 340
383 =340+ v 400 418,5 =340 - v 400
s s
vs = 15,09 m.s-1 vs = 15,03 m.s-1 (5)
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Grade 12 – Marking Guidelines/Nasienriglyne
10.4 ANY TWO
• It is used (in flow meters) in medical science to measure the
speed and direction (velocity) of blood flow.
• movement of the heart of a foetus.
• To find the rate of blood flow (Doppler scanning)
• To see the unborn child (Ultra sound scanning)
• To hear the heart of a foetus (Ultra sound scanning)
• It is used in medical sonography to generate images (and
sounds) of flowing blood.
• To detect blood clotting (Doppler ultrasound test)
ENIGE TWEE
• Dit word (in vloeimeters) gebruik in mediese wetenskap om die
volgende te meet:
• Die spoed en rigting (snelheid) van bloedvloei.
• Beweging van 'n fetus se hart
• Om die tempo van bloedvloei te meet (Doppler skandering)
• Om 'n ongebore baba te sien (Ultraklankskandering)
• Om die hart van 'n fetus te hoor (Ultraklankskandering)
• Dit word gebruik in mediese sonografie om beelde (en klanke) te
vorm van vloeiende bloed.
• Om vorming van bloedklonte op te spoor. (Doppler-
ultraklamktoets) (2)
10.5 Doppler effect/Doppler effek (1)
10.6 DIAGRAM 2
Star Y’s frequency is lower than star X/Ster Y se frekwensie is laer
as Ster X (3)
Star Y is redshifted/Ster Y ervaar rooiverskuiwing
[15]
TOTAL/TOTAAL: 150
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Physical Sciences P1/Fisiese Wetenskappe V1 NW/June/Junie 2024
Grade 12 – Marking Guidelines/Nasienriglyne
SUBJECT: Physical Sciences P1 ASSESSMENT TASK: GRADE 12 - June exams
QUESTION ANALYSIS GRID 2024
Cognitive Levels Topic Difficulty Levels
Momentum&Impulse
Vertical Projectile Work, Energy and
Newton's Laws Doppler effect
TOTAL
Moderate Difficult
QUESTION Mark
1 2 3 4 Easy
motion Power
1.1 2 2 2 2 2
1.2 2 2 2 2 2
1.3 2 2 2 2 2
1.4 2 2 2 2 2
1.5 2 2 2 2 2
1.6 2 2 2 2 2
1.7 2 2 2 2 2
1.8 2 2 2 2 2
1.9 2 2 2 2 2
1.10 2 2 2 2 2
Ques 1 20 2 10 8 0 4 4 2 4 6 20 10 6 4
2.1 2 2 2 2 2
2.2 4 4 4 4 4
2.3.1 3 3 3 3 3
2.3.2 5 5 5 5 5
2.4 2 2 2 2 2
2.5 1 1 1 1 1
Ques 2 17 2 7 3 5 17 0 0 0 0 17 5 5 7
3.1 2 2 2 2 2
3.2 4 4 4 4 4
3.3 3 3 3 3 3
3.4 2 2 2 2 2
Ques 3 11 2 4 5 0 11 0 0 0 0 11 2 6 3
4.1 2 2 2 2 2
4.2.1 4 4 4 4 4
4.2.2 3 3 3 3 3
4.2.3 1 1 1 1 1
Ques 4 10 2 1 7 0 0 10 0 0 0 10 3 7 0
5.1 2 2 2 2 2
5.2.1 4 4 4 4 4
5.2.2 2 2 2 2 2
5.2.3 4 4 4 4 4
5.3 3 3 3 3 3
Ques 5 15 2 5 8 0 0 15 0 0 0 15 6 9 0
6.1 2 2 2 2 2
6.2.1 3 3 3 3 3
6.2.2 4 4 4 4 4
6.2.3 3 3 3 3 3
6.3 3 3 3 3 3
Ques 6 15 2 3 10 0 0 0 15 0 0 15 2 6 7
7.1 1 1 1 1 1
7.2 3 3 3 3 3
7.3.1 4 4 4 4 4
7.3.2 5 5 5 5 5
7.3.3 4 4 4 4 4
7.4 3 3 3 3 3
Ques 7 20 1 3 16 0 0 0 20 0 0 20 4 3 13
8.1 2 2 2 2 2
8.2 1 1 1 1 1
8.3 3 3 3 3 3
8.4 5 5 5 5 5
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8.5 3 3 3 3 3
Ques 8 14 2 4 3 5 0 0 0 14 0 14 8 6 0
9.1 2 2 2 2 2
Ques 6 15 2 3 10 0 0 0 15 0 0 15 2 6 7
7.1 1 1 1 1 1
7.2 3 3 3 3 3
7.3.1 4 4 4 4 4
Physical
7.3.2 Sciences
5 P1/Fisiese Wetenskappe
5 V1 5 NW/June/Junie
5 2024
5
7.3.3 4 Grade 412 – Cognitive grid/Kognitiewe
4 tabel 4 4
7.4 3 3 3 3 3
Ques 7 20 1 3 16 0 0 0 20 0 0 20 4 3 13
8.1 2 2 2 2 2
8.2 1 1 1 1 1
8.3 3 3 3 3 3
8.4 5 5 5 5 5
8.5 3 3 3 3 3
Ques 8 14 2 4 3 5 0 0 0 14 0 14 8 6 0
9.1 2 2 2 2 2
9.2 3 3 3 3 3
9.3 5 5 5 5 5
9.4 3 3 3 3 3
Ques 9 13 2 6 0 5 0 0 0 13 0 13 2 6 5
10.1 2 2 2 2 2
10.2 2 2 2 2 2
10.3 5 5 5 5 5
10.4 2 2 2 2 2
10.5 1 1 1 1 1
10.4 3 3 3 3 3
Ques 10 15 3 7 5 0 0 0 0 0 15 15 5 5 5
SUMMARY
QUES 1 20 2 10 8 0 4 4 2 4 6 20 10 6 4
QUES 2 17 2 7 3 5 17 0 0 0 0 17 5 5 7
QUES 3 11 2 4 5 0 11 0 0 0 0 11 2 6 3
QUES 4 10 2 1 7 0 0 10 0 0 0 10 3 7 0
QUES 5 15 2 5 8 0 0 15 0 0 0 15 6 9 0
QUES 6 15 2 3 10 0 0 0 15 0 0 15 2 6 7
QUES 7 20 1 3 16 0 0 0 20 0 0 20 4 3 13
QUES 8 14 2 4 3 5 0 0 0 14 0 14 8 6 0
QUES 9 13 2 6 0 5 0 0 0 13 0 13 2 6 5
QUES 10 15 3 7 5 0 0 0 0 0 15 15 5 5 5
Total marks 150 20 50 65 15 32 29 37 31 21 150 47 59 44
Norm marks 150 22,5 45 67,5 15 32 30 32 32 24 150 45 60 45
Total % 14,6% 30,6% 44,6% 10,0%
Norm % 100 15 30 45 10 100 30 40 30
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