# Physical Sciences Grade 12 Paper 1 EC June 2019 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the EC June 2019, 24 pages. Read online or get the free PDF download, question paper linked when available. - Page: https://www.hlayiso.com/library/phsc-p1-memo-june-ea-latest/8382194b-5329-4102-9a1b-784a1e3ac280 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2019 - Exam period: EC June - Paper: 1 - Curriculum: CAPS - Pages: 24 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 Paper 1 EC June 2019 Memo](https://www.hlayiso.com/library/phsc-p1-memo-june-ea-latest/8382194b-5329-4102-9a1b-784a1e3ac280) ## About this document Physical Sciences · Grade 12 · EC June · 2019. Memorandum, 24 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2019 PHYSICAL SCIENCES P1 MARKING GUIDELINE/ FISIESE WETENSKAPPE V1 NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 20 pages. Hierdie nasienriglyn bestaan uit 20 bladsye. ### Page 2 Downloaded from hlayiso.com 2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) 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 sustitusies. 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 began 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. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V1/FISIESE WETENSKAPPE VI 3 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. 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. Kopiereg verbehou/Kopiereg voorbehou Blaaim om asseblief/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com 4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) 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. 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. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V1/FISIESE WETENSKAPPE VI 5 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 verbehou/Kopiereg voorbehou Blaaim om asseblief/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com 6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) QUESTION/VRAAG 1 1.1 C  (2) 1.2 A  (2) 1.3 B  (2) 1.4 D  (2) 1.5 C  (2) 1.6 A  (2) 1.7 B  (2) 1.8 D  (2) 1.9 A  (2) 1.10 A  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V1/FISIESE WETENSKAPPE VI 7 QUESTION /VRAAG 2 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/resulterende krag op ʼn voorwerp inwerk, veroorsaak dit ʼn versnelling van die voorwerp in die rigting van die netto/resulterende krag. Hierdie versnelling is direk eweredig aan die netto/resultante krag  en omgekeerd eweredig aan die massa van die voorwerp.  (2) 2.2 OPTION/OPSIE 1 OPTION/OPSIE 2 FN  FN  FY  FJane  Ff     FJane Ff   T T   Fg  Fg    Mark awarded for both arrow and label./Punt toegeken vir beide pylpunt en byskrif. Do not penalise for length of forces since drawing is not drawn to scale. Moenie penaliseer vir lengte van kragte want diagram is nie volgens skaal nie. Any other additional force(s)/Enige addisionele krag(te) If force(s) do not make contact with body/Indien krag(te) nie kontak maak met voorwerp nie. Max./Maks. (5) Kopiereg verbehou/Kopiereg voorbehou Blaaim om asseblief/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com 8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) 2.3 Choose East (Right) to be positive/Kies Oos (regs) as positief. 4 kg box/-houer Fnet = ma Fx + fk + T = ma  Any one / Enige een Fx - fk - T = ma FJaneCos30° + fk + T = ma 80x0,866 + (-8,14) + (-T) = 4a 69,28 – 8,14 – T = 4a 61,14 – T = 4a -------- (1) 5 kg box/-houer Fnet = ma T + W = ma T + (-9,8x5) = 5a T – 49 = 5a T - 49 a= ------(2) 5 Subst (2) in (1):/Stel (2) in (1):  T-49  61,14 – T = 4    5  305,7 – 5 T = 4 T – 196 9 T = 501,7 T = 55,74 N  (5) [12] QUESTION/VRAAG 3 3.1 Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.  Elke liggaam in die heelal trek elke ander liggaam aan met ʼn krag wat direk eweredig is aan die produk van hulle massas en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle middelpunte.  (2) 3.2 OPTION/OPSIE 1 OPTION/OPSIE 2 G m 1m 2 GMm F =  gm =  r2 Rm 11 22 6,67 x 10 x 7,35 x 10 m2 6,67 x1011 x 7,35 x1022 126,30=  =  (1,74 x 106 )2 (1,74 x104 )2 m2 = 78,00 kg  = 1,619252874 w = mg 126,30 = m x 1,619252874  m = 78,00 kg  (4) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V1/FISIESE WETENSKAPPE VI 9 QUESTION/VRAAG 4 4.1 Projectile motion is the motion of an object upon which the only force acting is the force of gravity.  Projektielbeweging is die beweging van ʼn voorwerp waarop die enigste krag wat daarop inwerk, gravitasiekrag is.  OR/OF Projectile motion is the motion of an object that experiences only gravitational force.  Projektielbeweging is die beweging van ʼn voorwerp wat slegs gravitasiekrag ondervind.  (2) 4.2 OPTION/OPSIE 1 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf = vi + g∆t  vf = vi + g∆t  0 = (13) + (-9.8 ). ∆t  0 =(-13) + (9.8) .∆t  ∆t = 1,33 s  ∆t = 1,33 s  (3) OPTION/OPSIE 2 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf2 = vi2 + 2a∆y  vf2 = vi2 + 2a∆y  02 = 132 + 2 (-9,8) ∆y 02 = -132 + 2 (9,8) ∆y ∆y = 8,62244898 m ∆y = - 8,62244898 m v f  vi v f  vi ∆y = ∆t ∆y = ∆t 2 2 0  13 0  13 8,62244898 = ∆t  -8,62244898 = ∆t  2 2 ∆t = 1,33 s  ∆t = 1,33 s  (3) OPTION/OPSIE 3 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf2 = vi2 + 2g∆y  vf2 = vi2 + 2g∆y  02 = 132 + 2 (-9,8) ∆y 02 = -132 + 2 (9,8) ∆y ∆y = 8,62244898 m ∆y = 8,62244898 m 1 1 ∆y = vi∆t + g∆t2 ∆y = vi∆t + g∆t2 2 2 1 1 8,62 = 13 ∆t + (- 9,8) ∆t2  -8,62 = -13 ∆t + (9,8) ∆t2  2 2 ∆t = 1,33 s  ∆t = 1,33 s  (3) Kopiereg verbehou/Kopiereg voorbehou Blaaim om asseblief/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com 10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) 4.3 OPTION/OPSIE 1 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf 2 = vi2 +2 g∆y  vf 2 = vi2 +2 g∆y  02  = (13)2 + 2 (-9,8) ∆y  02  = (-13)2 + 2 (9,8) ∆y  ∆y = 8,62 m  ∆y = 8,62 m  OPTION/OPSIE 2 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) 1 1 ∆y = vi∆t + g∆t2  ∆y = vi∆t + g∆t2  2 2 1 1 = 13(1,33)  + (-9,8)(1,33)2  = -13(1,33)  + (9,8)(1,33)2  2 2 = 8,62 m  = -8,62 m = 8,62 m, (upwards/opwaarts)  OPTION/OPSIE 3 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) v f  vi v f  vi ∆y = ∆t ∆y = ∆t  2 2 0  13 0  13 = x 1,33 = x 1,33  2 2 = 8,65 m  = -8,65 m = 8,65 m  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V1/FISIESE WETENSKAPPE VI 11 4.4 4.4.1 OPTION/OPSIE 1 Taking top of building as starting point / Neem toppunt van gebou as beginpunt (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf = vi + a∆t  vf = vi + a∆t  vf = 0 + (-9,8)(3,28 - 1,33)  vf = 0 + (9,8)(3,28 - 1,33)  -1 vf = -19,11 m.s vf = 19,11 m.s-1  -1 vf = 19,11 m.s  OPTION/OPSIE 2 Taking balcony as a starting point Neem balkon as beginpunt (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf = vi + a∆t  vf = vi + a∆t  vf = -13 + (-9,8)(3,28 - 1,66)  vf = 13 + (9,8)(3,28 – 2,66)  -1 vf = -19,08 m.s vf = 19,08 m.s-1  -1 vf = 19,08 m.s  (3) 4.4.2 OPTION/OPSIE 1 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) 1 1 ∆y = h = vi∆t + a∆t2  ∆y = h = vi∆t + a∆t2  2 2 1 1 h = -13(0,62) + (-9,8)(0,62)2 h = 13(0,62)  + (9,8)(0,62)2  2 2 h = - 9,94 m  = 9,94 m  Height of building/Hoogte van Height of building/Hoogte van gebou gebou = h + ∆y = h + ∆y = 8,62 + 9,94  = 8,62 + 9,94  = 18,56 m  = 18,56 m  Kopiereg verbehou/Kopiereg voorbehou Blaaim om asseblief/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com 12 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) OPTION/OPSIE 2 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf2 = vi2 + 2a∆y  vf2 = vi2 + 2a∆y  (-19,11)2 = (-13)2 + 2(-9,8)(∆y1)  (-19,11)2 = (-13)2 + 2(-9,8)(∆y1) h = ∆y = 10,01 m  h = ∆y = 10,01 m  Height of building/Hoogte van gebou Height of building/Hoogte van = ∆y + h gebou = 8,62+ 10,01  = ∆y + h = 18,62 m  = 8,62+ 10,01  = 18,62 m  OPTION/OPSIE 3 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf2 = vi2 + 2a∆y  vf2 = vi2 + 2a∆y  (-19,08)2 = (-13)2 + 2(-9,8)∆y  (-19,08)2 = (-13)2 + 2(-9,8)∆y  h = ∆y = 9,95 m  h = ∆y = 9,95 m  Height of building/Hoogte van gebou Height of building/Hoogte van = ∆y + h gebou = 8,62+ 9,95  = ∆y + h = 18,57 m  = 8,62+ 9,95  = 18,57 m  (6) [18] QUESTION/VRAAG 5 5.1 Downwards / Afwaarts  (1) 5.2 Two (times) OR 2 (times) Twee (keer) OF 2 (keer)  (1) 5.3 Elastic / Elasties  (1) 5.4 4 s  and / en 8 s  (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V1/FISIESE WETENSKAPPE VI 13 5.5 Positive marking from QUESTION 5.4 / Positiewe merk vanaf VRAAG 5.4 Position-time graph / Posisie-tyd grafiek 8 ∆x (m) 0 2 4 6 8 ∆t (s) (Max. height 1) (Max. height 2) (Maks. hoogte 1) (Maks. hoogte 2) Criteria for graph / Kriteria vir die grafiek Marks/Punte The height from which the ball was dropped  Hoogte van waar bal laat val is The times when the ball was at its maximum height – 4 s  and 8 s Die tye wanneer die bal sy maksimum hoogte bereik – 4 s en 8 s Shape of the graph/Vorm van die grafiek  (3) [8] Kopiereg verbehou/Kopiereg voorbehou Blaaim om asseblief/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com 14 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) QUESTION/VRAAG 6 6.1 The total linear momentum of an isolated system is conserved.  Die totale lineêre momentum van ʼn geïsoleerde sisteem bly behoue.  OR/OF In an isolated system the total linear momentum before collision is equal to the total linear momentum after collision.  In ʼn geïsoleerde sisteem is die totale lineêre momentum voor ʼn botsing gelyk aan die totale lineêre momentum na ʼn botsing.  (2) 6.2 pi = pf  mvviv + mcvic = mvvfv +mcvfc 80000 1500 x 30 + 1200 x  = 1500 vfv + 1200 x 25  3600 vfv = 27,78 m.s-1  in the same direction / in dieselfde rigting.  (5) 6.3 OPTION/OPSIE 1 OPTION/OPSIE 2 ∆p = mvf - mvi  ∆p = mvf - mvi  ∆p = 1200 x 25 – 1200 x 80000  ∆p = 1500 x 27,78 – 1500 x 30  3600 ∆p = -3330 kg.m.s-1 ∆p = 3333,33 kg.m.s-1 ∆p = 3330 kg.m.s-1 In the direction of the motion In the direction of the motion In dieselfde rigting as beweging  In dieselfde rigting as beweging  (3) 6.4 Eki = m vic2 + mviv2  80000 2 Eki = x 1200 x ( )+ x 1500 x 302  3600 Eki = 971296,30 J Ekf = m vfc2 + mvfv2 Ekf = x 1200 x (25)2+ x 1500 x 27,782  Ekf = 953796,30 J Eki Ekf  Collision is inelastic / Botsing is onelasties  (5) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V1/FISIESE WETENSKAPPE VI 15 QUESTION/VRAAG 7 7.1 To the right / na regs  For momentum to be conserved:  momentum of trolley B = momentum of trolley A.  momentum of B must be in an opposite direction to the momentum of trolley A  Vir momentum om behoue te bly:  momentum van trollie B = momentum van trollie A.  momentum van trollie B is in die teenoorgestelde rigting as die momentum van trollie A.  (3) 7.2 Let right/East be positive/Laat regs/Oos positief wees Σpi = Σpf mAvAi + mBvBi = mAvAf + mBvBf  Any one / Enige een (mA + mB) vi = mAvAf + mBvBf (0,5 + 0,75) x 0  = 0,5 x (-2,5) + 0,75vBf  vBf = 1,66 m.s-1 (to the right/East)/(na regs/oos)  (4) 7.3.1 21,5 N  According Newton’s Third law of motion,  when the wall exerts a force on the trolley, the trolley simultaneously exerts a force (of 21,5 N) on the wall but in an opposite direction.  Volgens Newton se Derde Bewegingswet,  indien die muur ʼn krag op die trollie uitoefen, oefen die trollie gelyktydig ʼn krag (van 21,5 N) op die muur maar in die teenoorgestelde rigting.  NOTE: If learners just state Newton’s third law without using it to explain, no marks will be awarded LET WEL: As leerders net Newton’s se derde wet sê sonder om teverduidelik, sal geen punte toegeken word. (3) 7.3.2 Fnet ∆t = m ∆v Fnet ∆t = m ∆(vAf - vAi)  Any one / enige -21,5 x 0,1 = 0,5(vAf – 2,5)  -2,15 = 0,5 vAf – 1,25 vAf = -1,8 m.s-1 vAf = 1,8 m.s-1 (to the left/west)(na links/wes)  (5) 7.3.3 Remains the same. / Bly dieselfde.  When the contact time increases, the net force decreases. Indien die kontaktyd toeneem, verminder die netto krag.  (3) [18] Kopiereg verbehou/Kopiereg voorbehou Blaaim om asseblief/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com 16 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) QUESTION/VRAAG 8 8.1 The (total) mechanical energy of an isolated system is conserved. Die (totale) meganiese energie van ʼn geïsoleerde sisteem bly behoue.  (2) 8.2 OPTION/OPSIE 1 OPTION/OPSIE 2 MEB = MEC Wnc = ∆Ek + ∆Ep (EP + EK)B = (EP + EK)C  0 = m vf2 - m vi2 + mgh2 – mgh1  (mgh + m v2)B = (mgh + m v2)B 0 =0 - (0,2)vi2 + 0,2(9,8)(1,2) – 0  0+ (0,2)vi2  = 0,2(9,8)(1,2)+ 0  Vi = 4,85 m.s-1  -1 Vi = 4,85 m.s  (4) 8.3 OPTION/OPSIE 1 OPTION/OPSIE 2 OPTION/OPSIE 3 v v vf = vi + a∆t vf = vi + a∆t ∆x = f i ∆t  2 4,85 = 6 + a x 0,82 4,85 = 6 + a x 0,82 = 6  4,85 x 0,82  a = - 1,40 m.s-2 a = - 1,40 m.s-2 2 1 = 4,45 m  ∆x = vi∆t + a∆t2  vf2 = vi2 + 2a∆x  2 = 6x0,82 + 0,5xax0,822  4,852 = 62 + 2(-1,4) ∆x  = 4,45 m  ∆x = 4,46 m  (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V1/FISIESE WETENSKAPPE VI 17 QUESTION/VRAAG 9 9.1 The net work done on an object is equal to the change in the kinetic energy of the object.  Die netto arbeid verrig op ʼn voorwerp is gelyk aan die verandering in die kinetiese energie van die voorwerp.  OR/OF The amount of work done by a net force  on object is equal to the change in the object’s kinetic energy.  Die hoeveelheid arbeid wat deur ʼn netto krag verrig word op ʼn voorwerp  is gelyk die verandering in die voorwerp se kinetiese energie.  (2) 9.2 OPTION/OPSIE 1 OPTION/OPSIE 2 FN Fapp  FN Fapp  Ff  Ff  Fg   Fg//  Fg┴ (4) Mark awarded for both the arrow and label Punt toegeken vir beide pylpunt en byskrif. Do not penalise for length of forces since drawing is not drawn to scale. Moenie penaliseer vir lengte van kragte want diagram is nie volgens skaal nie. Any other additional force(s)/Enige addisionele krag(te) If force(s) do not make contact with body/Indien krag(te) nie kontak maak met voorwerp nie. Max./Maks. 9.3 OPTION/OPSIE 1 Wnet = ∆EK  Any one / Enige een Wnet = WApp + W// + Wf WApp + W// + Wf = ∆EK 260(8)Cos0°) + 0,16(25)(9,8)(Cos40°)(8)(Cos180°) + 1 25(9,8)(sin40°)(8)(Cos180°)  = (25)vf2 – 0  2 2080 – 240,23 -1259,86 = 12,5 vf² 579,91 = 12,5 vf2 vf = 6,81 m.s-1  OPTION/OPSIE 2 Wnc = ∆EK+ ∆Ep Wapp + Wf = ∆EK+ WFg Any one / Enige een  260(8)(cos0°) + 0,16(25)(9,8)(Cos40°)(8)(cos180°) = 1 25(9,8)(Sin40°)(8)(cos180°) + (25)vf2 – 0  2 vf = 6,81 m.s-1  (6) Kopiereg verbehou/Kopiereg voorbehou Blaaim om asseblief/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com 18 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) 9.4 OPTION/OPSIE 1 OPTION/OPSIE 2 (v i + v f ) v f  vi ∆x = ∆t vav = 2 2 (0 + 6,81) 0  6,81 8= ∆t  vav =  2 2 ∆t = 2,35 s vav = 3,405 W Pav = F.vAv  P=  Pav = 260 x 3,405 ∆t = 885,3 W  P = 2080  2,35 = 885,11 W  (4) [16] QUESTION/VRAAG 10 10.1.1 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 . Die Doppler-effek is die verandering in frekwensie (of toonhoogte) van die klank wat deur die waarnemer waargeneem word  omdat die bron van die klank en die luisteraar verskillende snelhede het relatief tot die medium van die klank.  OR/OF Doppler effect is the apparent change in frequency of a wave when there is relative motion between the source and an observer . Die Doppler-effek is die skynbare verandering in die frekwensie van ʼn golf  wanneer daar relatiewe beweging tussen die bron en luisteraar is.  OR/OF 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 (skynbare) verandering in die waargenome frekwensie (toonhoogte)(golflengte)  as gevolg van die relatiewe beweging tussen ʼn bron en ʼn luisteraar. (2) 10.1.2 750 Hz  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V1/FISIESE WETENSKAPPE VI 19 10.1.3 1) Waves are compressed as Khosi moves towards the source.  2) Wavelength become shorter.  3) Wavelength is inversely proportional to frequency, hence the detected frequency is higher.  1) Golwe word saamgepers soos Khosi na die bron beweeg.  2) Golflengte word korter.  3) Golflengte is omgekeerd eweredig aan frekwensie, dus is die waargeneemde frekwensie hoër.  (3) 10.1.4  When Khosi approaches:/ (Terwyl Khosi nader beweeg): 340  vL 750  = fs  340 750 x 340 fs = …… (1) 340  vL When the Khosi moves away: (Terwyl Khosi wegbeweeg): 340  vL 700  = fs  340 700 x 340 fs = ……… (2) 340  vL (1) = (2) : 750 x 340 700 x 340  340  vL 340  vL 750(340 – vL) = 700(340 + vL) 1450vL = 17000 VL = 11,72 m.s-1 (6) OPTION/OPSIE 1 OPTION/OPSIE 2   750 x 340 700 x 340 fs = 340  11,72  fs = 340  11,72  fs = 725,01 Hz fs = 724,99 Hz (3) Kopiereg verbehou/Kopiereg voorbehou Blaaim om asseblief/Blaai om asseblief ### Page 20 Downloaded from hlayiso.com 20 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE 2019) 10.2.1  Away/Weg van   It shows a red shift. / Dit toon ʼn rooi verskuiwing.   Frequency is decreasing. / Frekwensie neem af.   Wavelength becomes longer. / Golflengte word langer.  (4) 10.2.2 The universe is expanding. / Die heelal is besig om uit te sit.  (1) [20] QUESTION/VRAAG 11 11.1 The force per unit positive charge./ Die krag per eenheid positiewe lading.  (2) 11.2 E= kQ r2  Any one/Enige een Enet = E1 + E2 9 x109 x 6 x106  9 x109 x 4 x10 6 0  (  ) (0,2  x)2  x2 x = 0,09 m  (6) [8] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 21 Downloaded from hlayiso.com ### Page 22 Downloaded from hlayiso.com ### Page 23 Downloaded from hlayiso.com ### Page 24 Downloaded from hlayiso.com ## Related documents - [2019 AFR ENG Gauteng Paper 1 Memo](https://www.hlayiso.com/library/2019-afr-eng-gauteng-paper-1-memo/7e7c9cfc-f2f5-4700-a132-61eae8192020) — Physical Sciences · Memo · 2019 · Eng - [Physical Sciences P1 Nov 2019 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2019-memo-afr-eng/6250ed76-71a4-4f9b-bfe7-95a1f244df41) — Physical Sciences · Memo · 2019 · Eng - [Physical Sciences P1 May June 2019 Afr](https://www.hlayiso.com/library/physical-sciences-p1-may-june-2019-afr/ba254883-c952-4abb-8257-a0f262488497) — Physical Sciences · QP · 2019 · Afr - [Phsc P1 Jun2017 Qp Afr](https://www.hlayiso.com/library/phsc-p1-jun2017-qp-afr/b5663af8-09c3-43af-85ec-615d384497f5) — Physical Sciences · QP · 2017 · Afr - [Phsc P1 Backup Qp Eng June 2016](https://www.hlayiso.com/library/phsc-p1-backup-qp-eng-june-2016/4e8a907f-6a4c-4951-a82f-057b082aad28) — Physical Sciences · QP · 2016 · Eng - [2019Gauteng Paper 2 Memo](https://www.hlayiso.com/library/2019gauteng-paper-2-memo/a25c7c32-611e-418b-a286-f1b20e9709da) — Physical Sciences · Memo · 2019 ## More from Grade 12 Physical Sciences - [PHYSICAL SCIENCE P1 QP JCT 2025 AFR D watermark](https://www.hlayiso.com/library/physical-science-p1-qp-jct-2025-afr-d-watermark/b9429ac5-acbb-4fae-989d-af9e1478a96b) — QP · 2025 · Afr - [PHYSICAL SC P2 QP GR12 SEPTEMBER 2025 AFR1 watermark](https://www.hlayiso.com/library/physical-sc-p2-qp-gr12-september-2025-afr1-watermark/37cff2db-7c1a-4ec8-8ac5-07ec5bcf87a9) — QP · 2025 - [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 KZN](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-kzn/6d225dc8-24c6-4c28-a181-df906d5a2c0e) — QP · 2025 - [Physical Sciences Physics Grade 12 PS01 QP March 2025 Limpopo](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-ps01-qp-march-2025-limpopo/482ee9af-e88a-4818-9120-11ab9e045a1c) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-limpopo/b65e9832-a6be-4312-8f5a-57425d647121) — QP · 2025 ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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