Physical Sciences Grade 12 Paper 1 EC June 2017 Memo

Downloaded from hlayiso.com NATIONAL/NASIONALE SENIOR CERTIFICATE/SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2017 PHYSICAL SCIENCES P1/ FISIESE WETENSKAPPE V1 MEMORANDUM MARKS/PUNT: 150 This memorandum consists of 16 pages./ Hierdie memorandum bestaan uit 16 bladsye.

Downloaded from hlayiso.com 2 PHYSICAL SCIENCES P1 (EC/JUNE 2017) 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. Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2017) 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 no 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 Please turn over

Downloaded from hlayiso.com 4 PHYSICAL SCIENCES P1 (EC/JUNE 2017) 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. Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2017) FISIESE WETENSKAPPE V1 5 QUESTION/VRAAG 1 1.1 D ✓✓ (2) 1.2 B ✓✓ (2) 1.3 A ✓✓ (2) 1.4 B ✓✓ (2) 1.5 A ✓✓ (2) 1.6 A ✓✓ (2) 1.7 B ✓✓ (2) 1.8 C ✓✓ (2) 1.9 A ✓✓ (2) 1.10 A ✓✓ (2) [20] QUESTION/VRAAG 2 2.1 2.1.1 It is called a projectile/Dit word ʼn projektiel genoem ✓ (1) 2.1.2 OPTION/OPSIE 1 OPTION/OPSIE 2 Data: vi = 15 m·s , -1 [for upward motion] g = - 9,8 m·s-2 [vir opwaartse beweging] hmax = ? vf = vi + g Δt vf = 0 m.s-1 at max height 0 = 15 + (– 9,8) Δt ✓ (by maks hoogte) 15 (Let downward be negative) Δt = = 1,53s 9,8 (Laat afwaarts negatief wees) 1 vf2 = vi2 + 2gΔy ✓ Δy = viΔt + gΔt2 ✓ 2 1 02 ✓ = (15)2 + 2(-9,8)( Δy) ✓ =(15)(1,53)+ 2 (− 9,8)(1,53)2 ✓ 19,6Δy = 225 ∴ Δy = hmax = 11,48 m ✓ ∴ Δy = hmax = 11,48 m ✓ OPTION/OPSIE 3 vf = vi + g Δt 0 = 15 + (– 9,8) Δt ✓ 15 Δt = 9,8 = 1.53s vf + vi Δy = Δt ✓ 2 0+ 15 = × 1,53 ✓ 2 ∴ Δy = hmax = 11,48 m ✓ (4) Copyright reserved Please turn over

Downloaded from hlayiso.com 6 PHYSICAL SCIENCES P1 (EC/JUNE 2017) 2.1.3 OPTION/OPSIE 1 OPTION/OPSIE 2 Δt = ? vf = vi + g Δt ✓ [for upward motion] [vir opwaartse beweging] 1 Δy = viΔt + gΔt2 ✓ 0 = 15 + (– 9,8) Δt ✓ 2 1 15 0 = 15Δt + (-9,8)Δt2 ✓ Δt = = 1,53s 2 9,8 -9,8 Δt2 + 30Δt = 0 -Δt(9,8Δt – 30) = 0 ∴ total time / totale tyd = 2(1,53) 30 Δt = = 3,06s✓ 9,8 ∴ Δt = 3,06s ✓ (3) 2.1.4 Δy = 8 m Δt = ? 1 Δy = vi Δt + gΔt2 ✓ 2 1 8✓ = 15 Δt + (-9,8)Δt2 ✓ 2 4,9 Δt2 -15 Δt + 8 = 0 Using a quadratic formula to find the roots: Gebruik ʼn kwadratiese formule of die wortels te bereken: −𝑏±√𝑏2 −4𝑎𝑐 Δt = 2𝑎 −(−15)±√(−15)2 −4(4,9)(8) = 2(4,9) 15 ±√225−156,8 = 9,8 15 ±8,26 = 9,8 ∴ Δt = 2,37s ✓or Δt = 0,69s ✓ Both values of Δt are acceptable/Albei waardes vir Δt aanvaarbaar (5) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2017) FISIESE WETENSKAPPE V1 7 2.1.5 Position vs time graph / Posisie vs tyd grafiek t (s) Both axes labelled /Albei asse met byskrifte ✓ All points plotted as directed/ Alle punte geplot soos gevra ✓✓ Correct shape ✓ NOTE: Take away a mark if not all points are plotted : For the maximum height accept y value = 11.48 m or 11.5 m on the graph AANDAG: Neem 1 punt weg indien nie alle punte geplot nie. : Vir die maksimum hoogte, aanvaar y = 11,48 of 11,5 m op die grafiek. (4) 2.2 OPTION/OPSIE 1 OPTION/OPSIE 2 (Take downwards as positive) (Take downwards as positive) (Afwaarts positief) (Afwaarts positief) Stone/Klip 1: Stone/Klip 1: 1 1 Δy1 = vi Δt + gΔt2 For both✓✓ Δy1 = vi Δt + gΔt2 ✓ 2 2 1 1 = 0 × Δt + (9,8)Δt2 ✓ = 0 × Δt + (9,8)Δt2 ✓ 2 2 Δy1 = 4,9 Δt2 {For Δt = t} Δy1 = 4,9 Δt2 {For Δt = t + 2} ∴ Δy1 = 4,9 t2 ∴ Δy1 = 4,9 (t + 2)2 Stone/Klip 2: Stone/Klip 2: 1 1 Δy2 = vi Δt + gΔt2 Δy2 = vi Δt + gΔt2 For both ✓ 2 2 /Vir beide 1 1 = 30 Δt + (9,8)Δt2 ✓ {Δt =t – 2} = 30 Δt + (9,8)Δt ✓ {Δt = t} 2 2 2 ∴ Δy2 = 30(t – 2) + 4,9(t – 2)2 ∴ Δy2 = 30(t) + 4,9(t)2 But / maar Δy1 = Δy2, ∴ But / maar Δy1 = Δy2, ∴ 4,9 t2 = 30(t – 2) + 4,9(t – 2)2 ✓ 4,9 (t + 2)2 = 30(t) + 4,9(t)2 ✓ 10,4t = 40,4 10,4t = 19,6 ∴ t = 3,89s ∴ t = 1,89s ✓✓ For second stone / Vir tweede klip For second stone / Vir tweede klip t – 2 = 3,89 – 2 ✓ 1,89s = 1,89s✓ (6) [23] Copyright reserved Please turn over

Downloaded from hlayiso.com 8 PHYSICAL SCIENCES P1 (EC/JUNE 2017) QUESTION/VRAAG 3 3.1 3.1.1 Impulse is the product of the net force acting ✓on an object and the time the net force acts on the object ✓. OR It is a measure of how hard ✓and for how long does a net force act✓ on an object Impuls is die produk van die netto krag ✓ wat op ʼn voorwerp inwerk en die tyd wat die netto krag op die voorwerp inwerk. ✓ OF Dit is die maatstaf vir hoe hard ✓ en vir hoe lank ʼn netto krag op ʼn voorwerp inwerk. ✓ (2) 3.1.2 OPTION/OPSIE 1 OPTION/OPSIE 2 Fnet Δt = m Δv ✓ Impulse = Change in momentum = 0,045(45 – 0) ✓ Impuls = verandering in momentum = 2,03 N·s✓ Δp = mΔv ✓ = 0,045(45 – 0) ✓ = 2,03 kgm·s-1 Impulse / Impuls = 2,03 N·s ✓ (3) 3.1.3 m Δv Fnet = Δt 2,03 = ✓ 3,5 × 10−3 = 580 N ✓ (2) 3.2 3.2.1 Take direction towards the wall as positive/ Neem rigting na die muur as positief m = 60 g = 0,060 kg vi = 12 m·s-1 vf = -10 m·s-1 Δp = ? Δp = m(vf – vi) ✓ = 0,060✓ (-10 – 12) ✓ = -1,32 kg·m·s-1 Δp = 1,32 kg·m·s-1✓ away from the wall/weg vanaf die muur ✓ (5) 3.2.2 1,32 N·s or/of 1,32 kg·m·s-1 ✓ (1) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2017) FISIESE WETENSKAPPE V1 9 3.3 3.3.1 The total linear momentum of an isolated ✓system remains constant / is conserved. ✓ OR/OF The total linear momentum of an isolated system before collision ✓is equal to the total linear momentum after collision. ✓ Die totale lineêre momentum van ʼn geslote sisteem ✓ bly konstant / bly behoue ✓ OF Die totale lineêre momentum van ʼn geslote sisteem voor ʼn botsing ✓ is gelyk aan die totale liniêre momentum na die botsing ✓ (2) 3.3.2 There is a need to calculate the velocity of block m just before collision/Bereken die snelheid van blok m voor die botsing: Em top/bo = Em bottom/onder (EP + EK) top/bo = (EP + EK) bottom/onder ✓ 1 mgh + 0 = 0 + mv2 2 2gh = v2 v = √2gℎ = √2(9,8)(3,6) ✓ = 8,40 m·s-1 ✓ m1v1i + m2v2i = m1v1f + m2v2f ✓ 2,2 × 8,4 + 7 × 0 = 2,2 × 0 + 7 v2f ✓ v2f = 2,64 m·s-1 ✓ (6) [21] Copyright reserved Please turn over

Downloaded from hlayiso.com 10 PHYSICAL SCIENCES P1 (EC/JUNE 2017) QUESTION/VRAAG 4 4.1 4.1.1 OPTION/OPSIE 1 OPTION/OPSIE 2 Ff FN Ff FN Fg F𝐹G┴ Fg// g┴ Fg Both components (Fg// and 𝐹𝑔┴ )  Albei komponente (Fg// en 𝐹𝑔┴ )  FN  FN  Ff  Ff  (3) 4.1.2 Fg// = mg Sin θ Fg┴ = - mg Cos θ (opposite direction of FN)/ (teenoorgestelde rigting as FN) Applying Newton’s Second law for the motion parallel to the slope: Toepassing van Newton se Tweede Wet van die beweging parallel aan die skuinsvlak:  for one of the two/ Fg// + Ff = ma vir een van die twee mg Sin θ - µkFN = ma ……….. (1) Applying Newton’s Second law for the motion perpendicular to the slope/Toepassing van Newton se Tweede Wet van die beweging loodreg aan die skuinsvlak:: FN + FG┴ = ma FN – mg Cos θ = ma …………. (2) (Since there is no motion perpendicular to the slope a = 0 m.s-2) (Geen beweging loodreg aan skuinsvlak dus a = 0 m.s-2) ∴ FN = mg Cos θ  (Substitute/Vervang FN in (1)) mg Sin θ - µk(mg Cos θ) = ma (dividing by m) g Sin θ - µk(g Cos θ) = a  (Substitute/Vervang µk = 0,10, g = 9,8 m·s-2 and/en θ = 30°) 9,8 Sin30° – 0,10 × 9,8 × Cos30° = a ∴ a = 4,00 m·s-2  (5) 4.1.3 vf = vi + aΔt  =0+4×4 = 16 m·s-1  (3) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2017) FISIESE WETENSKAPPE V1 11 4.2 4.2.1 When a net force/resultant force acts on an object, it produces the acceleration of the object in the direction of the net force/resultant force. This acceleration is directly proportional to the net/resultant force  and inversely proportional to the mass of the object.  Indien ʼn netto krag op ʼn voorwerp inwerk, versnel die voorwerp in die rigting van die netto krag. Die versnelling is direk eweredig aan die netto krag  en omgekeerd eweredig aan die massa van die voorwerp.  (2) 4.2.2 Take down wards as negative/Afwaarts as negatief For the Elevator/Vir die hysbak Fnet = FT – mEg = mEaE = - mEa (aE = - a) FT – mEg = - mEa  ----------- (1) For the counterweight/Vir die teengewig Fnet = FT – mCg = mCaC = mCa (aE = a) FT – mCg = mCa  ………… (2) (2) – (1) : mEg - mCg = mEa + mCa g(mE - mC) = a(mE + mC)  g(mE − mC ) a = 9,8(1150 –1000) = a(1150 +1000) (mE + mC ) 9,8(1150− 1000) 1470 = 2 150a = OR/OF a = 0,68 m·s-2 (1150+ 1000) = 0,68 m·s-2  (6) 4.2.3 OPTION/OPSIE 1 OPTION/OPSIE 2 For the counterweight For the Elevator FT – mCg = mCa  FT – mEg = -mEa  FT – 1000(9,8) = 1000(0,68)  FT – 1150(9,8) = -1150(0,68)  FT = 10 480 N  FT = 10 488 N (3) [22] Copyright reserved Please turn over

Downloaded from hlayiso.com 12 PHYSICAL SCIENCES P1 (EC/JUNE 2017) QUESTION/VRAAG 5 5.1 W = FΔxCosθ  = 600 × 30 × Cos0° = 600 × 30 × 1 = 18 000 J  (3) 5.2 5.2.1 Wf = Ff ΔxCosθ  = 50 × 6 × Cos180° = 50 × 6 × -1 = - 300 J  (3) 5.2.2 OPTION/OPSIE 1 OPTION/OPSIE 2 {Positive marking from/ Merk {Positive marking from/ Merk positief vanaf 5.2.1} positief vanaf 5.2.1} Wnet = W f + W HC  Fnet = Ff + FappCos 60° = Ff ΔxCosθ + FappΔxCosФ = - 50 + 300 × 0,5  = - 300  + 300 × 6 × Cos 60°  = 100 N = - 300 + 900 Wnet = Fnet Δ × Cosθ  = 600 J  = 100 × 6 × 1  = 600 J  (4) 5.3 5.3.1 The net work done on an object is equal to the change in the kinetic energy of the object.  OR The amount of work done by a net force  on object is equal to the change in the object’s kinetic energy.  Die netto arbeid op ʼn voorwerp verrig is gelyk  aan die verandering in kinetiese energie van die voorwerp.  OF Die hoeveelheid arbeid verrig deur ʼn netto krag  op ʼn voorwerp is gelyk aan die verandering in die voorwerp se kinetiese energie  (2) 5.3.2 W net = ΔEK  Wf = EKf - EKf 1 1  for any of the 2/ Ff ΔxCosθ = mvf2 - mvi2 2 2 vir enige van die 2 1 1 6 000(Δx)Cos 180°  = (800)(0) - (800)(20,5)2  2 2 - 6000 Δx = - 168 100 The braking distance/ Remafstand = Δx = 28,02 m  (5) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2017) FISIESE WETENSKAPPE V1 13 5.4 5.4.1 Wnet = ΔEK  1 = m(vf2 - vi2) 2 1 = (80)(252 - 02)  2 = 40(625) = 25 000 J  = 25 kJ (3) 5.4.2 fK = µKFN  = 0,34(mg Cosθ) 1.09 = 0,34 × 80 × 9,8  ×  1,20 = 242,13 N (4) 5.4.3 FN FN F app/toeg  Fapp/toeg  Fg// Fg┴  Fg  Type equation here.  (4) 5.4.4 OPTION/OPSIE800 1 Wnet = W App + W7// + W f = FappΔxCosθ + FG//ΔxCosФ + FfΔxCosФ = FappΔxCosθ + mgSinδΔxCosФ + FfΔxCosФ 0.5 = 450×1,2×Cos0°+80×9,8× ×1,2Cos180°+242,13×1,2×Cos180° 1.2 0,5 = 450 × 1,2 × 1 + 80 × 9,8 × 1,2 × 1,2 × -1 + 242,13 × 1,2 × -1 = 540 – 392 – 290,56 Wnet = - 142,56 J OPTION/OPSIE 2 Fnet = Fapp + Fg// + Ff  for any of the two = Fapp + mgSinδ+ Ff vir enige van die twee 0,5 = 450 – (80 × 9,8 × 1,2 + 242,13)  = - 118,80 N Wnet = FnetΔ × Cosθ = 118,80 × 1,2 × -1 = -142,56 J (4) 5.4.5 EP = mgh s = 80 × 9,8 × 0,5  = 392 J (3) [35] Copyright reserved Please turn over

Downloaded from hlayiso.com 14 PHYSICAL SCIENCES P1 (EC/JUNE 2017) QUESTION/VRAAG 6 6.1 6.1.1 Doppler effect is 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. OR Doppler effect is the apparent change in frequency of a wavewhen there is relative motion between the source and an observer. OR Doppler Effect is an (apparent) change in observed/detected frequency (pitch), (wavelength)  as a result of the relative motion between a source and an observer (listener) . Die Doppler effek is die verandering in frekwensie (of toonhoogte) van die klank waargeneem deur ʼn luisteraar want dit klankbron en luisteraar het verskillende snelhede relatief tot die medium van die voortplanting van die klank.  OF Die Doppler effek is die skynbare verandering in die frekwensie van ʼn golf  as daar relatiewe beweging is tussen die bron en die waarnemer.  OF Die Doppler effek is ʼn waarskynlike verandering in die waargenome frekwensie (toonhoogte)(golflengte)  as gevolg van die relatiewe beweging tussen die bron en die luisteraar  (2) 6.1.2 Towards the Listener. (Na die luisteraar) The frequency of the sound waves heard by the listener is greater than the frequency of the sound waves emitted by the ambulance.  The compressions in front of the source are closer together because the source is moving towards the previously emitted wavefront when the next wavefront is sent resulting in a decrease in wavelength and a sound of higher pitch is heard. Die frekwensie van die klankgolwe gehoor deur die luisteraar is hoër as die frekwensie van die klankgolwe uitgestraal deur die ambulans.  Die samepersing aan die voorkant van die bron is nader aanmekaar want die bron beweeg na die uitgestraalde golffront wanneer die vorige golffront gestuur word en veroorsaak ʼn afname in die golflengte en die hoër toonhoogte word gehoor.  (4) 6.1.3 v  vL fL  fs  v  vs 340  400  350  340  vs 340 340  vs  350 400 v s = 42,5 m·s-1 (4) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2017) FISIESE WETENSKAPPE V1 15 6.2 6.2.1 v  vL fL  fs  v  vs When the car approaches/ 343 Soos die motor naderbeweeg:  450  fs  343  vs 450(343  vs ) fs  343 When the car moves away/ 343 Soos die motor weg beweeg:  390  fs  343  v s 390(343  v s ) fs  343 450(343  vs ) 390(343  vs ) =  for either of 343 343 the two/ vir enige van die twee 154350 - 450 v s = 133770 + 390 v s v s = 24,5 m·s-1 (7) 6.2.2 OPTION/OPSIE 1 OPTION/OPSIE 2 450(343  vs ) 390(343  v s ) fs   fs   343 343 450(343  24,5) 390(343  24,5) fs   fs   343 343 f s = 417,86 Hz  f s = 417,86 Hz  (3) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com 16 PHYSICAL SCIENCES P1 (EC/JUNE 2017) QUESTION/VRAAG 7 7.1 7.1.1 The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is directly proportional to the product of the charges  and inversely proportional to the square of the distance (r) between them.  OR The magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges  and inversely proportional to the square of the distance between them.  Die grootte van die elektrostatiese krag wat deur een puntlading (Q1) op ʼn ander puntlading (Q2) uitoefen, is direk eweredig aan die grootte van die produk van die lading  en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle.  OF Die grootte van die elektrostatiese krag tussen twee puntladings is direk eweredig aan die produk van die massas van die ladings  en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle.  (2) 7.1.2 Electrostatic force exerted by Q1 on Q2/ Elektrostatiese krag uitgeoefen deur Q1 op Q2: Q1 Q2 F=k 2  r (9×109 )(4×10−9 )(2×10−9 ) =  0,04 2  = 4,5 × 10-5 N, to the East/na die Ooste Electrostatic force exerted on Q2 by Q3/Elektrostatiese krag uitgeoefen deur Q2 op Q3: Q2 Q3 F=k 2 r (9×109 )(2×10−9 )(6×10−9 ) =  0,062  = 3,0 × 10-5 N, to the East Both forces are towards the same direction/ Albei kragte is in dieselfde rigting: The net electrostatic force/ Die netto elektrostatiese krag Fnet = 4,5 × 10-5 N + 3,0 × 10-5 N  = 7,5 × 10-5 N  (To the East) (7) [9] TOTAL/TOTAAL: 150 Kopiereg voorbehou Blaai om asseblief