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NATIONAL
SENIOR CERTIFICATE/
NASIONALE SENIOR
SERTIFIKAAT
GRADE/GRAAD 12
SEPTEMBER 2021
PHYSICAL SCIENCES P1/
FISIESE WETENSKAPPE V1
MARKING GUIDELINE/NASIENRIGLYN
MARKS/PUNTE: 150
This marking guideline consists of 17 pages./
Hierdie nasienriglyn bestaan uit 17 bladsye.
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Physical Sciences Grade 12 P1 Eastern Cape Mock 2021 Memo
Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape Mock Exam 2021 (English), 17 pages. Read online or download the free PDF.
Physical Sciences · Grade 12 · Eastern Cape Mock Exam · 2021 · English. Memorandum, 17 pages. Read online or download the PDF.
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- 2021
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- Eastern Cape Mock Exam
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- CAPS
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2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021)
GENERAL GUIDELINES/ALGEMENE RIGLYNE
1. CALCULATIONS/BEREKENINGE
1.1 Marks will be awarded for: correct formula, correct substitution, correct
answer with unit.
Punte sal toegeken word vir: korrekte formule, korrekte substitusie, korrekte
antwoord met eenheid.
1.2 No marks will be awarded if an incorrect or inappropriate formula is used,
even though there are many relevant symbols and applicable substitutions.
Geen punte sal toegeken word waar ʼn verkeerde of ontoepaslike formule
gebruik word nie, selfs al is daar relevante simbole en relevante substitusies.
1.3 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 ʼn fout gedurende substitusie in ʼn korrekte formule begaan word,
sal ʼn punt vir die korrekte formule en vir korrekte substitusies toegeken word,
maar geen verdere punte sal toegeken word nie.
1.4 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.
1.5 No penalisation if zero substitutions are omitted in calculations where
correct formula/principle is correctly given.
Geen penalisering indien nulwaardes nie getoon word nie in berekeninge
waar die formule/beginsel korrek gegee is nie.
1.6 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 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 ʼn kandidaat met die korrekte formule
begin en dan die onderwerp van die formule verkeerd verander, sal die punte
vir die formule en korrekte substitusies toegeken word. Die punt vir die
verkeerde numeriese antwoord word verbeur.
1.7 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 slegs vir ʼn formule toegeken indien ʼn poging tot berekening
aangewend is, d.w.s. substitusies is gedoen of ʼn numeriese antwoord is
gegee.
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(EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 3
1.8 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
formules ingestel word en nie vir waardes wat voor ʼn berekening gelys is
nie.
1.9 All calculations, when not specified in the question, must be done to a
minimum of two decimal places.
Alle berekenings, wanneer nie in die vraag gespesifiseer word nie, moet tot
ʼn minimum van twee desimale plekke gedoen word.
1.10 If a final answer to a calculation is correct, full marks will not automatically be
awarded. Markers will always ensure that the correct/appropriate formula is
used and that workings, including substitutions, are correct.
Indien ʼn finale antwoord van ʼn berekening korrek is, sal volpunte nie
outomaties toegeken word nie. Nasieners sal altyd verseker dat die
korrekte/toepaslike formule gebruik word en dat bewerkings, insluitende
substitusies korrek is.
1.11 Questions where a series of calculations have to be made (e.g. a circuit
diagram question) do not necessarily always have to follow the same order.
FULL MARKS will be awarded provided it is a valid solution to the problem.
However, any calculation that will not bring the candidate closer to the
answer than the original data, will not count any marks.
Vrae waar ʼn reeks berekeninge gedoen moet word (bv. ʼn stroombaan-
diagramvraag) hoef nie noodwendig dieselfde volgorde te hê nie.
VOLPUNTE sal toegeken word op voorwaarde dat dit ʼn geldige oplossing vir
die probleem is. Enige berekening wat egter nie die kandidaat nader aan
die antwoord as die oorspronklike data bring nie, sal geen punte tel nie.
2. UNITS/EENHEDE
2.1 Candidates will only be penalised once for the repeated use of an incorrect
unit within a question.
Kandidate sal slegs een keer gepenaliseer word vir die herhaaldelike
gebruik van ʼn verkeerde eenheid in ʼn vraag.
2.2 Units are only required in the final answer to a calculation.
Eenhede word slegs in die finale antwoord op ʼn vraag verlang.
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 sal slegs vir ʼn antwoord en nie vir ʼn eenheid per se toegeken word
nie. Kandidate sal die punt vir die antwoord in die volgende gevalle verbeur:
Korrekte antwoord + verkeerde eenheid
Verkeerde antwoord + korrekte eenheid
Korrekte antwoord + geen eenheid
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4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021)
2.4 SI units must be used except in certain cases, e.g. V.m-1 instead of N.C-1, and
cm•s-1 or km•h-1 instead of m•s-1 where the question warrants this.
SI-eenhede moet gebruik word, behalwe in sekere gevalle, bv. V•m-1 in plaas
van N•C-1, en cm•s-1 of km•h-1 in plaas van m•s-1 waar die vraag dit regverdig.
3. GENERAL/ALGEMEEN
3.1 If one answer or calculation is required, but two are given by the candidate,
only the first one will be marked, irrespective 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 slegs die eerste een nagesien word, ongeag watter een
korrek is. Indien twee antwoorde verlang word, sal slegs 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-1.
For marking purposes, m•s-1 and m/s will also be accepted.
Skei saamgestelde eenhede met ʼn vermenigvuldigingspunt en nie met ʼn punt
nie, byvoorbeeld m•s-1. Vir nasiendoeleindes sal m•s-1 en m/s ook aanvaar
word.
4. POSITIVE MARKING/POSITIEWE NASIEN
Positive marking regarding calculations will be followed in the following cases:
Positiewe nasien met betrekking tot berekeninge sal in die volgende gevalle geld:
4.1 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 of 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 ʼn sekere veranderlike in een subvraag
(bv. 3.1) bereken word en dan in ʼn 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.
4.2 A multistep question of a sub-question: If the candidate has to calculate,
for example, current in die first step and gets it wrong due to a substitution
error, the mark for the substitution and the final answer will be forfeited.
ʼn Vraag met veelvuldige stappe in ʼn subvraag: Indien ʼn kandidaat bv. die
stroom verkeerd bereken in ʼn eerste stap as gevolg van ʼn substitusiefout,
verloor die kandidaat die punt vir die substitusie sowel as die finale antwoord.
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(EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 5
5. NEGATIVE MARKING/NEGATIEWE NASIEN
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 in QUESTION 3.1, and QUESTION 3.1 is incorrect, no marks can be
awarded for QUESTION 3.2. However, if the answer for e.g. QUESTION 3.1 is
based on a calculation, the motivation for the incorrect answer could be considered.
ʼn Verkeerde antwoord, indien dit op ʼn konsepsuele fout gebaseer is, kan
normaalweg nie korrek gemotiveer word nie. Indien ʼn kandidaat gevra word om in
VRAAG 3.2 die antwoord op VRAAG 3.1 te motiveer en VRAAG 3.1 is verkeerd, kan
geen punte vir VRAAG 3.2 toegeken word nie. Indien die antwoord op bv. VRAAG
3.1 egter op ʼn berekening gebaseer is, kan die motivering vir die verkeerde
antwoord in VRAAG 3.2 oorweeg word.
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6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021)
QUESTION/VRAAG 1: MULTIPLE-CHOICE QUESTIONS/
MEERVOUDIGEKEUSE-VRAE
1.1 C (2)
1.2 A (2)
1.3 B (2)
1.4 D (2)
1.5 D (2)
1.6 C (2)
1.7 A (2)
1.8 C (2)
1.9 D (2)
1.10 B (2)
[20]
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(EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 7
QUESTION/VRAAG 2
2.1.1 When a (non-zero) resultant/net force acts on an object, it accelerates in the
direction of the force. The acceleration is directly proportional to the force and
inversely proportional to the mass of the object.
Indien ʼn (nie-nul) resultante/netto krag op ʼn voorwerp inwerk, versnel dit in die
rigting van die krag. Die versnelling is direk eweredig aan die krag en omgekeerd
eweredig aan die massa van die voorwerp. (2)
2.1.2 OPTION/OPSIE 1 OPTION/OPSIE 2
N N
F Fy
cc
f f
T Fx
T
w w
(5)
Accept the following symbols./Aanvaar die volgende simbole:
N FN/Normal/Normaal/Normal force/Normaalkrag
Ff / fk / fr / frictional force/wrywingskrag/kinetic frictional force /
f
kinetiese wrywingskrag
Fg,/mg/weight/FEarth on block/49 N/gravitational force/gewig/Faarde op blok/
w
gravitasiekrag
T Tension / Spanning / FT / Fs
Fapplied/toegepas F / FA / Applied force / Toegepaste krag
Marks awarded for arrow and label/Punte toegeken vir pyltjie en byskrif.
Do not penalise for length of arrows since drawing is not drawn to scale.
Moenie penaliseer vir pyltjie-lengtes nie (diagram is nie volgens skaal nie).
3
Any other additional force(s)/Enige addisionele krag(te)
4
3
If force(s) do not make contact with body. Max./Maks.
4
Indien krag(te) nie kontak maak met voorwerp nie.
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8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021)
2.1.3 OPTION/OPSIE 1 OPTION/OPSIE 2
(To the right is positive) (To the right is negative)
(Na regs positief) (Na regs negatief)
Fnet = ma Fnet = ma Any one
Any one
T – f = ma f – T = -ma
Enige een Enige 1
Fcosθ - T – f = ma T + f – Fcosθ = -ma
T – 10 = 2 (2) 10 – T = 2(-2)
T = 14 N T = 14 N
Fcosθ - T – f = ma T + f – Fcosθ = -ma
°
14 + 15 -Fcos20 = 5(-2)
Fcos20o – 14 – 15 = 5(2)
F = 41,50 N F = 41,50 N (5)
2.2 Gm 1m 2
F=
d2
(6,67 10-11 )( 5,98 1024 )(200)
1 842,50 =
d2
d = 6 579 982,80 m
distance above earth surface/afstand bo die aardoppervlak =
6 579 982,80 – 6,38 x 106
= 199 982,80 m (1,9998280 x 105 m / 2,00 x 105 m) (5)
[17]
QUESTION/VRAAG 3
3.1.1 OPTION/OPSIE 1
UPWARDS POSITIVE UPWARDS NEGATIVE
OPWAARTS POSITIEF OPWAARTS NEGATIEF
vf2 = vi2 + 2aΔy vf2 = vi2 + 2aΔy
(-25)2 = (-20)2 + 2(-9,8) Δy (25)2 = (20)2 + 2(9,8) Δy
Δy = -11,48 Δy = 11,48 m
Δy = 11,48 m
OPTION/OPSIE 2
UPWARDS POSITIVE UPWARDS NEGATIVE
OPWAARTS POSITIEF OPWAARTS NEGATIEF
vf = vi + aΔt vf = vi + aΔt
-25 = -20 + (-9,8)(Δt) 25 = 20 + (9,8)(Δt)
Δt = 0,51 s Δt = 0,51 s
Δy = viΔt + ½gΔt2 Δy = viΔt + ½gΔt2
Δy = (-20)(0,51) + ½(-9,8)(0,51)2 Δy = (20)(0,51) + ½(9,8)(0,51)2
Δy = -11,47 Δy = 11,47 m
Δy = 11,47 m
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(EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 9
OPTION/OPSIE 3
UPWARDS POSITIVE UPWARDS NEGATIVE
OPWAARTS POSITIEF OPWAARTS NEGATIEF
vf = vi + aΔt vf = vi + aΔt
-25 = -20 + (-9,8)(Δt) 25 = 20 + (9,8)(Δt)
Δt = 0,51 s Δt = 0,51 s
v vi v vi
Δy = f Δt Δy = f Δt
2 2
- 25 (-20) 25 20
Δy = x 0,51 Δy = x 0,51
2 2
Δy = -11,48 m Δy = 11,48 m
Δy = 11,48 m
OPTION/OPSIE 4 (ACCEPT/AANVAAR)
UPWARDS POSITIVE UPWARDS NEGATIVE
OPWAARTS POSITIEF OPWAARTS NEGATIEF
vf = vi + aΔt vf = vi + aΔt
-25 = -20 + (-9,8)(Δt) 25 = 20 + (-9,8)(Δt)
Δt = 0,51 s Δt = 0,51 s
Δy = lb + ½bh Δy = lb + ½bh
Δy = 20 x 0,51 + ½(5)0,51 Δy = 20 x 0,51 + ½(5)0,51
Δy = 11,48 m Δy = 11,48 m (3)
3.1.2 UPWARDS POSITIVE UPWARDS NEGATIVE
OPWAARTS POSITIEF OPWAARTS NEGATIEF
vf = vi + aΔt vf = vi + aΔt
-25 = -20 + (-9,8)(Δt) 25 = 20 + (-9,8)(Δt)
Δt = 0,51 s (time to reach to ground) Δt = 0,51 s (time to reach to ground)
(tyd om die grond te bereik) (tyd om die grond te bereik)
vf = vi + aΔt
0 = 12 + (-9,8)(Δt) vf = vi + aΔt
Δt = 1,22 s 0 = -12 + (9,8)(Δt)
(time to reach maximum height) Δt = 1,22 s
(tyd om maks hoogte te bereik) (time to reach maximum height)
t = 1,22 + 0,51 (tyd om maks hoogte te bereik)
t = 1,73 s t = 1,22 + 0,51
t = 1,73 s (5)
3.1.3 OPTION/OPSIE 1
Positive marking from 3.1.1/Positiewe merk vanaf 3.1.1
UPWARDS POSITIVE UPWARDS NEGATIVE
OPWAARTS POSITIEF OPWAARTS NEGATIEF
vf2 = vi2 + 2aΔy vf2 = vi2 + 2a
2 2
0 = 12 + 2(-9,8) Δy 02 = -122 + 2(9,8) Δy
Δy = 7,35 m Δy = -7,35 m
Displacement = - 11,48 + 7,35 Displacement = 11,48 + (- 7,35)
Verplasing = - 4,13 Verplasing = 4,13 m
= 4,13 m (downwards/afwaarts)
(downwards/afwaarts)
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10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021)
OPTION/OPSIE 2
Positive marking from 3.1.1/Positiewe merk vanaf 3.1.1
UPWARDS POSITIVE UPWARDS NEGATIVE
OPWAARTS POSITIEF OPWAARTS NEGATIEF
vf = vi + aΔt vf = vi + aΔt
0 = 12 + (-9,8)(Δt) 0 = -12 + (9,8)(Δt)
Δt = 1,22 s Δt = 1,22 s
2
Δy = viΔt + ½aΔt Δy = viΔt + ½aΔt2
Δy = 12 x1,22 + ½ (-9,8)(1,22)2 Δy = (-12)(1,22) + ½ (9,8)(1,22)2
Δy = 7,35 m Δy = -7,35
Displacement = - 11,48 + 7,35 Displacement = 11,48 + (- 7,35)
Verplasing = - 4,13 Verplasing = 4,13 m
= 4,13 m (downwards/afwaarts)
(downwards/afwaarts)
OPTION/OPSIE 3
Positive marking from 3.1.1 / Positiewe merk vanaf 3.1.1
UPWARDS POSITIVE UPWARDS NEGATIVE
OPWAARTS POSITIEF OPWAARTS NEGATIEF
vf = vi + aΔt vf = vi + aΔt
0 = 12 + (-9,8)(Δt) 0 = -12 + (9,8)(Δt)
Δt = 1,22 s Δt = 1,22 s
v vi v vi
Δy = f Δt Δy = f Δt
2 2
0 (12) 0 (-12)
Δy = x 1,22 Δy = x 1,22
2 2
Δy = 7,32 m Δy = - 7,32 m
Displacement = - 11,48 + 7,32 Displacement = 11,48 + (- 7,32)
Verplasing = - 4,16 Verplasing = 4,16 m
= 4,16 m (downwards/afwaarts)
(downwards/afwaarts)
OPTION/OPSIE 4
Positive marking from 3.1.1 / Positiewe merk vanaf 3.1.1
UPWARDS POSITIVE UPWARDS NEGATIVE
OPWAARTS POSITIEF OPWAARTS NEGATIEF
vf = vi + aΔt vf = vi + aΔt
0 = 12 + (-9,8)Δt 0 = -12 + (9,8)Δt
Δt = 1,22 s Δt = 1,22 s
Area/oppervlakte = ½bh Area/oppervlakte = ½bh
= ½ (1,22)(12) = ½ (1,22)(-12)
= 7,32 m = -7,32 m
Displacement = - 11,48 + 7,32 Displacement = 11,48 + (- 7,32)
Verplasing = - 4,16 Verplasing = 4,16 m
= 4,16 m (downwards/afwaarts) (downwards/afwaarts) (4)
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(EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 11
3.2 Positive marking from 3.1.1 and 3.1.2 / Positiewe merk vanaf 3.1.1 en 3.1.2
Position/Posisie(m)
11,48
m
Time/Tyd(s)
1,73
m
CRITERIA FOR MARKING/MERK KRITERIA
Correct shape/Korrekte vorm
Height indicated/Hoogte aangedui (11,48 m)
Time t indicated/Tyd t aangedui (1,73 s) (3)
[15]
QUESTION/VRAAG 4
4.1 In an isolated system total linear momentum is conserved.
Die totale lineêre momentum bly behoue in ʼn geïsoleerde/geslote sisteem. (2)
4.2.1 ∑pi = ∑pf
Any one/Enige een
mAviA + mBviB = (mA + mB)vf
(2 x viA) + (4 x -5) = (2+4)(-1,67)
viA = 4,99 m.s-1 (East/Ooswaarts )
(4)
4.2.2 POSITIVE MARKING FROM POSITIVE MARKING FROM
4.2.1/Positiewe merk vanaf 4.2.1 4.2.1/Positiewe merk vanaf 4.2.1
OPTION/OPSIE 1 OPTION/OPSIE 2
Fnet.∆t = ∆p Fnet.∆t = ∆p
Any one Any one
Fnet.∆t = m(vf – vi) /Enige een Fnet.∆t = m(vf – vi) /Enige een
Fnet x 0,01 = 2 (-1,67- 4,99) Fnet x 0,01 = 2 [1,67- (-4,99)]
Fnet = - 1 332 N Fnet = - 1 332 N
Fnet = 1 332 N west/left/Wes/links Fnet = 1 332 N west/left/Wes/links
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12 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021)
OPTION/OPSIE 3 OPTION/OPSIE 4
Fnet = ma Any one Fnet = ma
Any one
v f - vi /Enige een v - vi
Fnet = m ( ) Fnet = m ( f ) /Enige een
t t
- 1,67 - 4,99 1,67 - (- 4,99)
Fnet = 2 x ( ) Fnet = 2 x ( )
0,01 0,01
Fnet = -1 332 Fnet = 1 332 N west/left/Wes/links
Fnet = 1 332 N west/left/Wes/links
OPTION 5/OPSIE 5 OPTION 6/OPSIE 5
Fnet.∆t = ∆p Fnet.∆t = ∆p
Any one /Enige Any one /Enige
Fnet.∆t = m(vf – vi) een Fnet.∆t = m(vf – vi) een
Fnet(0,01) = 4(-1,67 – -5) Fnet(0,01) = 4(1,67 – 5)
Fnet = 1 332 Fnet = -1 332
FAB = - FBA FAB = - FBA
Fnet(BA) = 1 332 N west/left Fnet(BA) = 1 332 N west/left
(4)
[10]
QUESTION/VRAAG 5
5.1 Gravitational force / Gravitasiekrag (1)
5.2 12
∆x = = 24 m
sin30o
Wf = f.∆x cos θ
Wf = 35,5 x 24 cos 180o
Wf = - 852 J (4)
5.3 Zero/0 J (1)
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(EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 13
5.4 Positive marking from 5.2/Positiewe merk vanaf 5.2
OPTION/OPSIE 1
Wnet = Ek Any one/Enige een
Wf + W F + W Fg = Ek
f x ∆x cos θ + F xcos θ + mg(h2 – h1) = Ek
-852 + (62,5 x 24 cos180o) + m(9,8)(12-0) = 0
m = 20 kg
OPTION/OPSIE 2
Wnc = Ep + Ek
W f + W F = Ep + Ek Any one/Enige een
f∆xcos θ + F xcos = mg(h2 – h1) + Ek
-852 + (62,5)(24)cos180o = m(9,8)(0 – 12) + 0
m = 20 kg
OPTION 3/OPSIE 3
Wnet = Ek
W f + W F + W w = Ek Any one/Enige een
f∆xcos θ + F xcos θ + mgΔxcos θ = Ek
-852 + (62,5)(24)cos180o + m(9,8)(24)cos60o = 0
m = 20 kg (5)
[11]
QUESTION/VRAAG 6
6.1.1 520 Hz / 520 waves per second (waves.s-1)/golwe per sekondes (golwe.s-1) (1)
6.1.2 The change in frequency (or pitch) observed/detected by a listener because the
listener and the sound source have different velocities relative to the medium of
sound propagation.
Die verandering in frekwensie (of toonhoogte) waargeneem deur ʼn luisteraar
want die luisteraar en die klankbron het verskillende snelhede relatief tot die
medium wat klank voortplant.
OR/OF
The (apparent) changed in observed/detected frequency (pitch) as a result of
relative motion between the sound source and the listener.
Die waarskynlike verandering in die waargenome frekwensie (toonhoogte) as
gevolg van die relatiewe beweging tussen die klankbron en die luisteraar. (2)
6.1.3 TOWARDS Detected frequency is higher than the source frequency
NA. Waargenome frekwensie is hoër as die bron se frekwensie (2)
6.1.4 v ± vL
fL = fs
v ± vs
343
520 = (480)
343 - v s
vs = 26,38 m·s-1 (5)
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14 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021)
6.1.5 Decreases/Neem af
For a constant velocity/speed of sound, if the frequency increases, λ decreases.
Vir ʼn konstante snelheid/spoed van klank, indien die frekwensie toeneem, neem λ
af.
OR/OF
λα at constant velocity/speed/by konstante snelheid/spoed
OR/OF
f α at constant velocity/speed/by konstante snelheid/spoed (2)
6.2 Light from the star is shifted towards longer wavelength (towards the red end of the
spectrum) which indicated that the star is moving away from the earth.
Lig van die ster skuif na ʼn langer golflengte (na die rooi kant van die spektrum)
wat aandui dat die ster wegbeweeg van die aarde af. (2)
6.3 Used to measure the direction and speed of blood flow in arteries and veins.
Word gebruik om die rigting en spoed waarteen bloed in are en slagare vloei, te
meet.
OR/OF
Used to measure the heartbeat of a foetus in the womb.
Word gebruik om die hartklop van ʼn fetus in die baarmoeder te meet. (1)
[15]
QUESTION/VRAAG 7
7.1.1 GAIN/BYGEKRY (1)
7.1.2 Q
n=
qe
5 x 10 -6
n= 19
1,6 x 10 -
n = 3,125 x 1013 (electrons/elektrone) (3)
7.1.3 kQ
E=
r2
9 x 109 x 5 x 10 -5
E=
0,12
E = 4,5 x 106 N.C-1 left/links (5)
7.2.1 Negative/Negatief
Like charges repel each other/Gelyksoortige ladings stoot mekaar af.
OR/OF
The charges repel each other. If sphere A is negative, then sphere B must also be
negative.
Die ladings stoot mekaar af. Indien sfeer A negatief is, dan moet sfeer B ook
negatief wees. (2)
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(EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 15
7.2.2 FE = Tsin30o
FE = 25sin30o
FE = 12,5 N
kQ 1Q 2
FE =
r2
(9 109 )( 5 10 -6 ) Q
12,5 =
0,05 2
Q = 6,94 x 10-7 C (6)
[17]
QUESTION/VRAAG 8
8.1.1 Temperature/temperatuur
Length of the conductors/Lengte van die geleiers (Any two/Enige twee)
Thickness of the conductors/Dikte van die geleiers.
ACCEPT: Type of material
AANVAAR: Tipe materiaal (2)
8.1.2 Gradient is the inverse of the resistance. /
Gradiënt is die inverse van die weerstand
OR/OF
Gradient = (1)
8.1.3 Conductor C./Geleier C.
It has the highest resistance. The higher the resistance of a conductor,
the more heat is produced in the conductor if the current is constant.
Dit het die hoogste weerstand. Hoe hoër die weerstand, hoe meer hitte
geproduseer word in die geleier indien die stroom konstant is. (2)
8.2.1 R =
12
R=
1,5
R=8Ω (3)
8.2.2 OPTION/OPSIE 1 OPTION/OPSIE 2
Rtotal = R + r = I(R+ r)
[8 = (4 + 3) + r ] 12 = 1,5 [(4 + 3) + r]
r=1Ω r=1Ω (4)
8.2.3 W = I2R∆t
W = (1,5)2(3)(180)
W = 1 215 J (3)
8.3.1 Decrease/Neem af. (1)
8.3.2 Increase/Neem toe. (1)
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16 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2021)
8.4 Increase/Neem toe.
Rext decreases. Current through battery increases.
W = I2r∆t / Energy transfer to the battery/work done by battery increases.
Reks neem af. Stroom deur die battery neem toe.
W = I2rΔt / Energie oorgedra aan die battery/arbeid verrig deur die battery neem
toe. (3)
[20]
QUESTION/VRAAG 9
9.1 Mechanical energy to electrical energy.
Meganiese energie na elektriese energie (2)
9.2 AC generator has slip rings and DC generator has a split ring / commutator
WS-generator het sleepringe en GS-generator het ’n splitring / kommutator (1)
9.3
Potential Difference (emf)
(V)
Potensiaalverskil (emk)
CRITERIA FOR MARKING/MERK KRITERIA
Correct shape/Korrekte vorm
Axes labelled correct/Asse korrek
Vmax indicated on graph/
Vmaks aangedui op grafiek (2)
9.4 Pave = VrmsIrms Pgem = VwgkIwgk
2000 = Irms x 230 2000 = Iwgk x 230
Irms = 8,70 A Iwgk = 8,70 A
I I
Irms = max Iwgk = max
2 2
Imax I
8,70 = 8,70 = max
2 2
Imax = 12,30 A Imaks = 12,30 A (4)
9.5 Vmax
Vrms =
2
Vmax
230 =
2
Vmax/Vmaks = 325,27 V (3)
[12]
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(EC/SEPTEMBER 2021) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 17
QUESTION/VRAAG 10
10.1 Work function (of the metal) / Werksfunksie (van die metaal)
hc
Ek(max) = – W0
The intercept on the vertical axis = W 0.
Die afsnit op die vertikale-as = Wo
OR/OF
hc
= W 0 + Ek(max)
The intercept on the vertical axis is equal to the W o
Die afsnit op die vertikale-as is gelyk aan die Wo (3)
10.2 E = W 0 + Ek(max)
hf = W 0 + Ek(max) Any one/Enige een
hf = hf0 + Ek(max)
6,63 x 10-34 x 6,16 x 1014 = 6,63 x 10-34 f0 + 5,6 x 10-20
f0 = 5,32 x 1014 Hz (5)
10.3.1 Remain the same/Bly dieselfde
The gradient is equal to the product of Planck’s constant and the speed of
light in vacuum which are constants.
Die gradient is gelyk aan die produk van Planck se konstante en die spoed
van lig in ʼn vakuum wat konstantes is.
OR/OF
Gradient = hc, which are constants/Gradiënt = hc, wat konstantes is. (2)
10.3.2 Remains the same/Bly dieselfde
Ek(max) (J)
Intensity/Intensiteit
CRITERIA FOR MARKING/MERK KRITERIA
Axes labelled/Asse benoem
Correct shape/Korrekte vorm (3)
[13]
TOTAL/TOTAAL: 150
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