NATIONAL
SENIOR CERTIFICATE/
NASIONALE
SENIOR SERTIFIKAAT
GRADE/GRAAD 12
JUNE/JUNIE 2021
PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1
MARKING GUIDELINE / NASIENRIGLYN
(EXEMPLAR/EKSEMPLAAR)
MARKS/PUNTE: 150
This marking guideline consists of 14 pages./
Hierdie nasienriglyn bestaan uit 14 bladsye.
Skip to contentAll 2021 papers 
%20June%202024%20Possible%20Answers_hlayiso.com_--f78ca708-f7f0-4bc5-8ff9-36594b1b1052/v1-a152cbabd59bdb3c6c42/card.webp)



%20June%202024%20Possible%20Answers_hlayiso.com_--7e563c7f-5d30-46c2-a50f-2942505df8e3/v1-c8add7667fd054150e1b/card.webp)
View all





Physical Sciences Grade 12 P1 Eastern Cape June 2021 Memo
Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape June 2021 (Afrikaans), 14 pages. Read online or download the free PDF.
Physical Sciences · Grade 12 · Eastern Cape June · 2021 · Afrikaans. Memorandum, 14 pages. Read online or download the PDF.
- Subject
- Physical Sciences
- Grade
- Grade 12
- Language
- Afrikaans
- Document type
- Memorandum
- Year
- 2021
- Exam period
- Eastern Cape June
- Paper
- 1
- Curriculum
- CAPS
- Pages
- 14
- File size
- 1.2 MB
Free to read online, no sign-up needed
Loading document…
Loading document…
1 of 14
This exam
Eastern Cape June 2021
Questions in this paper
Document textSearch extracted text and jump to a page.
2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 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 verkeerde 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 ʼn berekening
aangewend is, d.w.s. substitusies is gedoen of ʼn numeriese antwoord is
gegee.
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 formule
ingestel word en nie vir waardes wat voor ʼn berekening gelys is nie.
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
(EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 3
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
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.
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
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, no 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 Subquestion to subquestion: When a certain variable is calculated in one
subquestion (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 subquestions.
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 in a subquestion: 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.
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 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 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 3.1 is verkeerd, kan geen punte vir VRAAG
3.2 toegeken word nie. Indien die antwoord op bv. 3.1 egter op ʼn berekening
gebaseer is, kan die motivering vir die verkeerde antwoord in 3.2 oorweeg word.
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
(EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 5
QUESTION/VRAAG 1
1.1 B (2)
1.2 C (2)
1.3 C (2)
1.4 D (2)
1.5 A (2)
1.6 C (2)
1.7 B (2)
1.8 A (2)
1.9 D (2)
1.10 B (2)
[20]
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
QUESTION/VRAAG 2
2.1 When object A exerts a force on object B, object B simultaneously exerts an
oppositely directed force of the same magnitude on object A.
Wanneer voorwerp A ʼn krag op voorwerp B uitoefen, oefen voorwerp B ʼn krag
gelyk in grootte en in teenoorgestelde rigting op voorwerp A uit. (2)
2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
FA on/op B FN FA on/op B
FN
Ff
Ff
Fg
Fg for both
Fg components/vir
beide komponente
(4)
2.3 Data:
MB = 25 kg
fB = 15 N
a = 1,5 m.s-2
FA on B = ?
FB on A = ?
Fnet = ma
FA on B + fB + Fg = ma Any one/Enige een
FA on B + fB + mgSin35°= ma
FA on B + (- 15) + (- 25x9,8x0,574) = 25x1,5
FA on B – 155,63 = 37,50
FA on B = 193,13 N
FB on A = - FA on B
FB on A = - 193,13 N
FB on A = 193,13 N down the slope/ teen die helling af (6)
2.4 Fnet = ma
FNceba + FB on A + fA + Fg = ma Any one/Enige een
FNceba + FB on A + fA + mgSin35°= ma
FNceba + (- 193,13) + (- 4,5) + (- 10x9,8x0,574) = 10x1,5
FNceba – 253,88 = 15
FNceba = 268,88 N (4)
[16]
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
(EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 7
QUESTION/VRAAG 3
3.1 When a net/resultant force acts on an object of mass m, the object
accelerates in the direction of the net force. The accelereation is directly
proportional to the net force and inversely proportional to the mass.
Wanneer ’n netto krag op ’n voorwerp inwerk, versnel die voorwerp in die
rigting van die netto krag. Die versnelling is direk eweredig aan die netto
krag en omgekeerd aan die massa van die voorwerp. (2)
3.2 Fnet = ma
Fg + FN + FY = ma Any one/Enige een
Fg + FN + Fa Sin40° = ma
Fg + (-623,3) + (-500Sin40°) = mx0
Fg - 623,3 – 321,39 = 0
Fg = 944,69 N
mg = 944,69
mx9,8 = 944,69
m = 96,40 kg (5)
3.3 For 96,4 kg box / vir 96,4 kg houer
Fnet = ma
FX + fk + T = ma Any one/Enige een
Fa Cos40° + f + T = ma
500 Cos40° + (-115) +(- T) = 96,4a
383,02 – 115 – T = 96,4a
268,02 - T = 96,4a
T = 268,02 – 96,4a .............. (1)
For 15 kg box / vir 15 kg houer
Fnet = ma
Any one/Enige een
Fg + T = ma
mg + T = ma
-15x9,8 + T = 15a
-147 + T = 15a (Substitute/Stel in vir T)
-147 + (268,02 – 96,4a) = 15a
121,02 = 111,40a
a = 1,09 m.s-2 (7)
[14]
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
QUESTION/VRAAG 4
4.1 It (projectile) is an object upon which the only force acting is the force of
gravity.
Dit (projektiel) is ’n voorwerp waarop die enigste krag wat daarop inwerk, die
gravitasiekrag is.
OR/OF
Projectile is an object that experiences only gravitational force.
Projektiel is ’n voorwerp wat slegs die gravitasiekrag ondervind. (2)
4.2 OPTION 1 /OPSIE 1
(upwards positive) (downwards positive)
(opwaarts positief) (afwaarts positief)
vf2 = vi2 + 2gΔy vf2 = vi2 + 2gΔy
02 = 52 + 2 (-9,8) Δy 02 = (-5)2 + 2 (9,8) Δy
Δy = 1,28 m Δy = -1,28 m
Δy = 1,28 m, (upward)
OPTION 2/OPSIE 2
(upwards positive) (downwards positive)
(opwaarts positief) (afwaarts positief)
vf = vi + g∆t vf = vi + g∆t
0 = (5) + (-9,8 ). ∆t 0 = (5) + (-9,8 ). ∆t
∴∆t = 0,51 s ∴∆t = 0,51 s
Δy = viΔt + 1 gΔt2 Δy = viΔt + 1 gΔt2
2 2
= (5)(0,51) + 1 (- 9,8) (0,51)2 = (-5)(0,51) + 1 (9,8)(0,51)2
2 2
∴Δy = 1,28 m Δy = - 1,28 m
∴Δy = 1,28 m, (upwards) (3)
4.3 The maximum height reached by ball A (Δymax)
Die maksimum hoogte bereik deur bal A ( Δymax) = 1,28 + 5 = 6,28 m
Maximum height reached by ball B is given by:
Makisum hoogte bereik deur bal B is:
(upwards positive) (downwards positive)
(opwaarts positief) (afwaarts positief)
vf2 = vi2 + 2gΔy vf2 = vi2 + 2gΔy
02 = 82 + 2 (-9,8) Δy 02 = (-8)2 + 2 (9,8) Δy
ΔyB = 3,27 m Δy = -3,27 m
ΔyB = 3,27 m, (upward)
Since ΔyB < Δymax
OR 3,27 m < 6,28 m Since ΔyB < Δymax
∴ Ball B did not reach the maximum OR 3,27 m < 6,28 m
height. ∴ Ball B did not reach the maximum
∴ Bal B het nie die maksimum height.
hoogte bereik nie. ∴ Bal B het nie die maksimum hoogte
bereik nie. (5)
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
(EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 9
4.4 (upwards positive) (downwards positive)
(opwaarts positief) (afwaarts positief)
vf = vi + g∆t vf = vi + g∆t
0 = (5) + (-9,8 ). ∆t 0 = (5) + (-9,8 ). ∆t
∴∆t = 0,51 s ∴∆t = 0,51 s
vf = vi + g∆t vf = vi + g∆t
= (8) + (-9,8 ). 0,51 = (-8) + (9,8 ). 0,51
∴ vf = 3,00 m.s-1 vf = -3,00 m.s-1
∴ vf = 3,00 m.s-1 (4)
4.5 (Upwards positive)
For ball A/ Vir bal A
vf = vi + g∆t
vf = (5) + (-9,8 ). ∆t ...... (1)
For ball B/ Vir bal B
vf = vi + g∆t
vf = (8) + (-9,8 ). ∆t ...... (2)
Equating (1) and (2)/ Stel (1) gelyk aan (2):
Since ball A will be moving downwards then/Aangesien bal A afwaarts
beweeg dat:
-(5 – 9,8.∆t) = 8 – 9,8.∆t
19,6.∆t = 13
∆t = 0,66 s
∆t = 0,011min (5)
[19]
QUESTION/VRAAG 5
5.1 A system on which the net external forces acting is zero.
Dit is ’n sisteem waarop die netto eksterne krag wat op die sisteem inwerk,
nul is. (2)
5.2 Let east be positive/Laat oos positief wees
p = mv
= 1 140 x 30
= 34 200 kg.m.s-1, (east/oos) (3)
5.3 Σpi = Σpf
mvviv + mcvic = mvvfv +mcvfc
1 650 viv+ 1 140 x 30 = 1 650 x - 12 + 1 140x 10
1 650 viv = 11 400 – 34 200 – 19 800
viv = - 25,82
viv = 25,82 m.s-1 , west (5)
5.4 Principle of conservation of linear momentum
The total linear momentum of an isolated system remains constant
(preserved).
Beginsel van behoud van lineêre momentum.
Die totale lineêre momentum in ’n geïsoleerde sisteem bly konstant
(behoue). (3)
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
5.5 ∆p = m(vf - vi)
∆p = 1 650 (-12 – (-25,82))
∆p = 22 803 kg.m.s-1 (3)
[16]
QUESTION/VRAAG 6
6.1 The net/resultant force acting on an object is equal to the rate of change in
momentum of the object.
Die netto/resulterende krag wat op ’n voorwerp inwerk, is gelyk aan die tempo
van verandering van momentum van die voorwerp. (2)
6.2 Total Ek Before = Ek tree + Ek car
1 1 Any one/Enige een
= m v 2tree + m 2
v car
2 2
1 1
= (m)(02) + (1 125)(25)2
2 2
= 351 562,5 J (4)
6.3 Impulse of the tree/Impuls van die boom:
Let original direction be positive/Laat die oorspronklike rigting positief
wees
Fnet∆t (car) = m(vf - vi)
= 1 125(-6 – 25)
= - 34 875 N
Fnet∆t (car) = 34 875 N, backwards/terugwaarts
Fnet∆t (tree) = 34 875 N , original direction of the car/ oorspronklie rigting van
die motor. (4)
6.4 m(v f - v i )
Fnet (car) =
Δt
1 125(-6 - 25)
Fnet (car) =
3,5
Fnet (car) = - 9 964 N
Fnet (car) = 9 964 N, backwards/terugwaarts (3)
6.5 When the airbags deploy during collision, the contact time of the driver with the
airbag increases. According to the equation Fnet∆t = ∆p, when contact time
increases the net force experienced by the driver decreases. Then, the extent
of injury is reduced.
Wanneer die lugsak gedurende die botsing ontplooi, sal die kontaktyd van die
bestuurder met die lugsak verhoog. Volgens die vergelyking Fnet∆t = ∆p ,
wanneer die kontaktyd verhoog sal die netto krag wat deur die bestuurder
ervaar word verlaag. Dus sal die omvang van die beserings verminder word. (2)
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
(EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 11
6.6 Decrease
The final velocity will now be zero. Then, according to the equation:
∆p = m(vf – vi), it follows that ∆p will decrease. If ∆p decreases then Fnet
decreases as they are directly proportional to each other.
Verminder
Die finale snelheid sal nou nul wees. Volgens die vergelyking: ∆p = m(vf – vi)
dit volg dat ∆p sal verlaag. As Δp afneem dan sal Fnet verlaag aangesien
hulle direk eweredig is aanmekaar. (4)
[19]
QUESTION/VRAAG 7
7.1 Gm1m2
F=
r2
(6,67 x10-11 )( 3,4 x103 )( 5,98 x1024 )
F=
(2,32 x106 )2
= 251 959,05
= 2,52 x 105 N (4)
7.2 Newton’s Law of Universal Gravitation
Any two objects or particles in the universe attract each other with a force
which is directly proportional to the product of their masses and inversely
proportional to the square of the distance between their centres.
Newton se Universele Gravitasiewet
Enige twee liggaam of deeltjie in die heelal trek mekaar aan met ’n krag wat
direk eweredig aan die produk van hul massas en omgekeerd eweredig
aan die kwadraat van die afstand tussen hul middelpunte aan. (3)
7.3 Gm1m2
F1 = =F
r2
m
Gm1 2
F2 = 2
2
r
2
Gm1m2
F2 = 2
r2
4
Gm1m2
F2 =2 =2F
r2
F2 = 2(2,52 x 105 N)
F2 = 5,04 x 105 N (3)
[10]
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
12 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
QUESTION/VRAAG 8
8.1 8.1.1 The net work done on an object is equal to the change in the kinetic
energy of the object.
Die netto arbeid deur ‘n voorwerp verrig is gelyk aan die verandering in
kinetiese energie van die voorwerp.
OR /OF
The amount of work done by a net force on an object is equal to the
change in the object’s kinetic energy.
Die arbeid verrig deur ’n netto krag is gelyk aan die verandering in
kinetiese energie van die voorwerp. (2)
8.1.2 OPTION 1/OPSIE 1
Wnet = W F1 + W F2 + W// + W f
= F1ΔxCos180° + F2ΔxCos0° + mgSin30°ΔxCos180° + fΔxCos180°
= (20)(20)(-1) + (100)(20)(1) + (8)(9,8)(0,5)(20)(-1) + (8,5)(20)(-1)
= -400 + 2000 – 784 – 170
= 646 J
OPTION 2/OPSIE 2
Fnet = F2 + F1 + Fg + f
= F2 + F1 + mgSin30°+ f
= 100 + (-20) + {-(8)(9,8)(0,5)} + (-8,5)
= 32,3 N
Wnet = FnetΔxCos θ
= Fnet ΔxCos0°
= (32,3) (20)(1)
= 646 J (4)
8.1.3 Wnet = ∆Ek
Wnet = Ekf – Eki Any one/Enige een
1 1
Wnet = mv 2f - mv i2
2 2
1 1
646 = (8)v 2f - (8)(0)
2 2
(2)(646)
v 2f =
8
= 161,5
vf = 12,71 m.s-1 (4)
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
(EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 13
8.2 1 1
∆Ek = mv 2f - mv i2
2 2
1 1
= (1 500)(26)2 - (1 500)(0)2
2 2
= 507 000 J
ΔEk
P=
Δt
507000
=
4
= 126 750 W
= 126,75 kW (5)
[15]
QUESTION/VRAAG 9
9.1 9.1.1 It is the apparent change in frequency (or pitch/wavelength) of the
sound detected by a listener because the sound source and the listener
have different velocities relative to the medium of sound
propagation.
Dit is die verandering in frekwensie (toonhoogte/golflengte) van die
klank waargeneem deur ’n luisteraar omdat die klankbron en die
luisteraar verskillende snelhede relatief tot die medium waarin die
klank voortgeplant word, het.
OR/OF
An apparent change in observed/detected frequency (pitch/
wavelength) as a result of relative motion between a source and an
observer.
Dit is oënskynlike verandering in frekwensie (toonhoogte/golflengte)
wat waargeneem word as gevolg van die relatiewe beweging tussen die
bron en waarnemer. (2)
9.1.2 v ± vL
fL = fS
v ± vS
Any one/Enige een
v vL
fL = fS
v
340 vL
0,85x = (x)
340
289x = (340 – vL)(x)
289 = 340 – vL
vL = 51 m.s-1 (4)
9.1.3 When the listener moves away from the source the sound waves that
reach him have longer wavelength and a sound with a lower pitch is
registered because of a lower frequency.
As die luisteraar van die bron af wegbeweeg, het die klankgolwe wat
hom bereik langer golflengte en ’n klank met ’n laer toonhoogte
geregistreer weens 'n laer frekwensie. (2)
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
14 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
9.1.4 It would be the same
Dit sal dieselfde bly (2)
9.1.5 FL = 0,85fS
= 0,85 x 190
= 161,5 Hz (2)
9.2 To measure the velocity of blood flowing through blood vessels.
To scan a fetus/foetus.
Om die snelheid van bloed wat deur are vloei, te meet.
Om ’n fetus te skandeer. (2)
[14]
QUESTION/VRAAG 10
10.1 The magnitude of the electrostatic force exerted by one point charge (Q) on
another point charge (Q) is directly proportional to the product of the
magnitudes of the charges and inversely proportional to the square of the
distance (r) between them.
Die grootte van die elektrostatiese krag wat een puntlading (Q 1) op ’n ander
puntlading (Q2) uitoefen, is direk eweredig aan die produk van ladings en
omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle. (2)
10.2 East is positive/Oos is positief
kQ A QN
FAN =
r2
(9 x109 )(4 x10 9 )(2 x10 9 )
=
(0,014)2
= 0,0005 N, East/Oos
kQBQN
FBN =
r2
(9 x109 )(8 x10 9 )(2 x10 9 )
=
(0,049)2
= 0,00006 N, West/Wes,
Fnet = FAN + FBN
= 0,0005 – 0,00006
= 0,00044 N, East/Right / Oos/regs (5)
[7]
TOTAL/TOTAAL: 150
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
Study this subject
Recommended for this subject

Question paper
Physical Sciences Grade 12 P1 Eastern Cape June 2022 QP
%20June%202024%20Possible%20Answers_hlayiso.com_--f78ca708-f7f0-4bc5-8ff9-36594b1b1052/v1-a152cbabd59bdb3c6c42/card.webp)
Memorandum
Physical Sciences Grade 12 P1 Gauteng June 2024 Memo (Afr)

Memorandum
Physical Sciences Grade 12 P1 North West November 2021 Memo

Question paper
Physical Sciences Grade 12 P1 Eastern Cape June 2025 QP

Question paper
Physical Sciences Grade 12 P1 NSC June 2021 QP (Eng)
%20June%202024%20Possible%20Answers_hlayiso.com_--7e563c7f-5d30-46c2-a50f-2942505df8e3/v1-c8add7667fd054150e1b/card.webp)
Memorandum
Physical Sciences Grade 12 P2 Gauteng June 2024 Memo (Afr)
More from Grade 12 Physical Sciences

Question paper
Physical Sciences Grade 12 P2 Eastern Cape Mock 2025 QP

Question paper
Physical Sciences Grade 12 P2 Gauteng Mock 2025 QP

Question paper
Physical Sciences Grade 12 P1 KZN Mock 2025 Question Paper

Question paper
Physical Sciences Grade 12 Limpopo March Test 2025 QP

Question paper
Physical Sciences Grade 12 P1 Limpopo Mock 2025 QP

Question paper