Physical Sciences Grade 12 P1 2015 Memo

Downloaded from hlayiso.com NORTHERN CAPE DEPARTMENT OF EDUCATION NOORDKAAP DEPARTEMENT VAN ONDERWYS TRIAL EXAMINATION PROEFEKSAMEN GRADE 12 GRAAD 12 RR RRAMARASER ARERR AAA LBAEEES, PHYSICAL SCIENCES: PHYSICS (P1) 4 FISIESE WETENSKAPPE: FISIKA (V1) LAR ARBUALS ELE IE ABER, FAAAENAAR RAE EAL EERY , Se, SEPTEMBER 2015 MEMORANDUM PAPAAARRERAOARERERS 2 a tei MARKS/ PUNTE: 150 This memorandum consists of 20 pages Hierdie memorandum bestaan uit 20 bladsye Copyright reserved/Kopiereg voorbehou Please turn over/Biaai om asseblief

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 2 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO! Sep 2015 GENERAL GUIDELINES / ALGEMENE RIGLYNE 1. 141 1.2 1.3 1.4 1.5 1.6 1.7 1.8 CALCULATIONS/ BEREKENINGE Marks will be awarded for: correct formula, correct substitution, correct answer with unit. Punte sal toegeken word vir: korrekfe formule, korrekte substitusie, korrekte antwoord met eenheid. No marks will be awarded if an incorrect or inappropriate formula is used, even though there may be relevant symbols and applicable substitutions. Geen punte sal toegeken word waar 'n verkeerde of ontoepaslike formule gebruik word nie, selfs al is daar relevante simbole en relevante substitusies, 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 'n fout gedurende substitusie in 'n korrekte formule begaan word, sal 'n punt vir die korrekte formule en vir korrekte substitusies toegeken word, maar geen verdere punte sal toegeken word nie 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. No penalisation if zero substitutions are omitted in calculations where correct formula / principle is given correctly. Geen penalisering indien nulwaardes nie getoon word nie in berekeninge waar die formule/beginsel korrek gegee is nie. 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 the 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 ‘n kandidaat met die korrekte formule begin en dan die onderwerp van die formule verkeerd verander, sal punte vir die formule en korrekte substitusies toegeken word. Die punt vir die verkeerde numeriese antwoord word verbeur. 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 siegs vir ‘n formule toegeken indien 'n poging tot 'n berekening aangewend is, d.w.s. substitusies is gedoen of 'n numeriese antwoord is gegee. 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 is en nie vir waardes wat voor 'n berekening gelys is nie. Copyright reserved 2 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 3 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 1.9 All calculations, when not specified in the question, must be done to two decimal places. Alle berekenings, wanneer nie in die vraag gespesifiseer word nie, moet tot 'n minimum van twee desimale plekke gedoen word. 2. UNITS/ EENHEDE 2.1 Candidates will only be penalised once for the repeated use of an incorrect unit within a question or sub-question. Kandidate sal slegs een keer gepenaliseer word vir die herhaaldelike gebruik van ‘n verkeerde eenheid in ‘n vraag. 2.2 Units are only required in the final answer to a calculation. Eenhede word siegs in die finale antwoord op 'n vraag veriang. 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 word slegs vir 'n antwoord en nie vir 'n eenheid per se toegeken nie. Kandidate sal derhalwe 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" instead of N-C”’, and cm-s" or km-h"' instead of m:s”' where the question warrants this. Sl-eenhede moet gebruik word, behalwe in sekere gevalle, bv. Vem" in plaas van N-C’, en em-s of km-h’ in plaas van m-s waar die vraag dit regverdig. 3. GENERAL/ ALGEMEEN 3.1 If one answer or calculation is required, but two given by the candidate, only the first one will be marked, irmespective 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 siegs die eerste een nagesien word, ongeag watter een korrek is. indien twee antwoorde verlang word, sai siegs 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". For marking purposes m.s"' and m/s will also be accepted. Skei saamgestelde eenhede met 'n vermenigvuldigpunt en nie met 'n punt nie, byvoorbeeld, m-s’. Vir nasiendoeleindes sal m.s en m/s ook aanvaar word. Copyright reserved 3 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 4 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 4, POSITIVE MARKING/POSITIEWE NASIEN 41 4.2 4.5 Positive marking regarding calculations will be followed in the following cases: Positiewe nasien met betrekking tot berekenings sal in die volgende gevalle geld: 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 or 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 'n sekere veranderlike in een subvraag (bv. 3.1) bereken word en dan in 'n 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. A multi-step question in a sub-question: If the candidate has to calculate, for example, current in the first step and gets it wrong due to a substitution error, the mark for the substitution and the final answer will be forfeited. 'n Vraag met veelvuldige stappe in ‘'n subvraag: Indien 'n kandidaat byvoorbeeild, die stroom verkeerd bereken in 'n eerste stap as gevolg van 'n substitusiefout, verloor die kandidaat die punt vir die substitusie sowel as die finale antwoord. 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 to 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 in 3.2 could be considered. 'n Verkeerde antwoord, indien dit op ‘n konsepsuele fout gebaseer is, kan normaalweg nie korrek gemotiveer word nie. Indien 'n 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 'n berekening gebaseer is, kan die motivering vir die verkeerde antwoord in 3.2 oorweeg word. Copyright reserved 4 Please turn over

Physical Sciences/Gr 12/P1 Fisiese Wetenskappe/ GR 12/ V1 QUESTION 1/ VRAAG 1 1.1 1.2 1.3 1.4 15 1.6 1.7 1.8 1.9 1.10 Copyright reserved B vv CvvY Dvv A vv Bvv A vv Bvv Dvv C vv Bvv Downloaded from hlayiso.com CAPS/KABV NCDOE/ Sep 2015 NKDVO/ Sep 2015 Please turn over (2) (2) (2) (2) (2) (2) (2) (2) (2) (2) [20]

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 6 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 QUESTION 2/ VRAAG 2 2.1 Normal force is the force or component of a force which a surface exerts on an object with which it is in contact, “and which is perpendicular to the surface. “” Normaalkrag is die krag of komponent van ’n krag wat 'n opperviak op ’n (2) voorwerp waarmee dit in kontak is, uitoefen en wat loodreg op die opperviak is. 2.2 N/Fw/Fs on of Normal/ normaalv fF df Fa ¥ F./W/ Gravity v/Gravitasie ACCEPT/ AANVAAR N/Fy/Fs on of Normal’/ Normaal Fa v fiFJfn Fe/WI Gravity “/ Gravitasie (4) 2.3 OPTION 1/ OPSIE 1 Fy -0 = DFy OR/OF Free = 0 y Any one/Enige een N+(-mg-Fasin35°=0 N= mg+Fasin35° N= 215,16 +97,51” fp =e&N f, =21291NY OPTION 2 / OPSIE 2 fe4yNn v Sf, = umg + F,) ¥ Any one/ Enige een Sy = ey (mg + Fsin35°) (4) Copyright reserved 6 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 7 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 f, = 0.68 (22 9,8 +170xsin35°)” =212.91NY 2.4 POSITIVE MARKING FROM 2.3 /POSITIEWE NASIEN VANAF 2.3 OPTION 1/ OPSIE 7 Fm=Fi+f, ORIOF F,=F,-f, ORIOF F,, = Fos35° -f, F.,, =170c0s35° —212.91¥ Fy = — 73,65 N ORIOF Fy = 73,65 N backwards”/ terugwaarts The net force is in the opposite direction of motion/ since the net force is in opposite direction of motion, the mower accelerates backwards.“ Die netto krag is in die teenoorgestelde rigting van die beweging/ aangesien die netto krag in die teenoorgestelde rigting van beweging is, versnel die grassnyer terugwaarts OPTION 2/ OPSIE 2 F,=Foos35 ¥ F, =170c0s35 =139.25 NY Fy<f Under these circumstances the lawn mower will be accelerating in a backward direction/It will slow down in a forward direction. ¥ Onder hierdie omstandighede sal die grassnyer in ‘n terugwaartse rigting versnel/ Dit sal stadiger beweeg in ’n voorwaartse rigting. QUESTION 3/ VRAAG 3 3.1. Impulse is the product of the force (resultant/net force) acting on an object and the time the force (resultant/net force) acts on the object. Impuls is die produk van die krag (resultante/ netto krag) wat op ‘n voorwerp inwerk en die tyd wat die krag (resultante/ netto krag) op die voorwerp inwerk. 3.2. Impulse (J)=Ap Impulse (J) = mv tm vi) Impulse (J) = 0,43(-25 - 20) “ Impulse (J)=—19,35 N-s % OR/OF Impulse (J) = 19,35 N’-s ¥ in opposite direction. 3.3 FAt=Ap¥ OR/OF FAt= Ap (300)At= 19,35 ¥ At= 0,065 s ¥ OR/OF At= 0,07 s v Anyone/ Enige een 3.4 Inelasticv’/Onelasties Mass is constant but speed changes /Kinetic energy is not conserved.¥ Massa is konstant maar spoed verander/Kinetiese energie bly nie behoue nie Copyright reserved q Please turn over (3) [13] (2) (3) (3) (2) [10]

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 8 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 QUESTION 4/VRAAG 4 4.1 Projectile is an object upon which the only force acting is the force of gravity. Projektiel is ‘n voorwerp waarop slegs gravitasiekrag inwerk. (2) 4.2 9,8 m-s* “downwards afwaarts (theoretical) (2) OR/OF 9,6 m:-s* Ydownwardsy (actual) 4.3 OPTION 4/ OPSIE 1 Ay =viA +5 ath)? Y OR/OF Ay=v,Ar+ 5 eta)? Ay = 9,8(4,08) ¥ + 5 (-9,8)(4,08)°¥ Ay= - 41,58 mv OR/ OF Ay= 41,58 m downwards 7 afwaarts OPTION 2/OPSIE 2 Ay =viA +55) YORIOF Ay= yas Seay Ay = -9,8(2.04) ” +5 (-98)02,08)77 Ay= - 41,58 m ¥ OR/ OF Ay= 41,58 m downwards W/afwaarts OPTION 3/ OPSIE 3 Aya Wt) 5, v 2 ay - 28+ 30.8) y (408 Ay= - 41,58 m ¥ OR/ OF Ay= 41,58 m downwards v/afwaarts OPTION 4/ OPSIE 4 Ay = et¥) oy v Ay = (-9,8) +9018) Y (2,04) Ay= - 40,78 m ¥ OR/ OF Ay= 40,78 m downwards 7 afwaarts Copyright reserved 8 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 9 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 OPTION 5/ OPSIE 5 Ay= Area(trapezium) Ay = area G2), Ay = £728) + (30,18), CRIM 4.08 - 2,04) ¥ Ay= - 40,78 m ¥ OR OF Ay= 40,78 m downwards “/afwaarts OPTION 6/ OPSIE 6 Area= Areag; + Areag2 1 1 Ay = area = om + 3h Ay = 5 9,8(1,02) ¥ +5 (-30,18)G,06) ¥ Ay=- 41,17mv¥ OR OF Ay= 41,17 m downwards v/ afwaarts (4) OPTION 7/ OPSIE 7 1 Ay =ab + 45h \y apt 3 Ay = -9,8(2,04) ¥ + 5 (2,04)(-20,38) v Ay=- 40,78 m ¥ OR! OF Ay=40,78 m downwards ~/afwaarts Copyright reserved 9 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 10 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 44 A 2 @ 7 5 a Se SE e—0 7 > es 1,02 2,04 4,08 t(s) 7 9 A 40,78 m 2 2 n 9 a ee SE a> - 40,78 CRITERIA/ KRITERIA MARKS/PUNTE Graph starting from zero OR 40,78 m/Grafiek begin by nul OF v 40,78m Correct shape/ korrekte vorm v Position for maximum height (4,9 m or/ of 5 m) at 1,02 s v Posisie vir maksimum hoogte (4,9 m or/ of 5 m) by 1,02 s (4) Final position/ Finale posisie (-40.4 m or 40 m) at/ by 4,08 s v [12] Copyright reserved 10 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 il NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 QUESTION 5/ VRAAG 5 5.1. The total linear momentum of an isolated/a closed system~ remains constant (is conserved). ¥ Die totale linére momentum van ‘n geisoleerde/ gesiote sisteem bly konstant (bly behoue) OR/ OF In an isolated systemy the total linear momentum of a system before a Collision/interaction is equal to the total linear momentum of the system after the collision. ” In ’n geisoleerde sisteem is die totale liniére momentum van ’n sisteem voor ’n botsing/ interaksie gelyk aan die totale liniére momentum van die sisteem na die botsing OR/ OF If the impulse of the external forces acting on a system is zerov the total linear momentum of the system does not change/remains constant. ¥ As die impuls van die eksterne kragte, wat op ‘n sisteem inwerk, gelyk is aan nul sal die totale liniére momentum van die sisteem nie verander nie/ konstant bly. OR/ OF If there is no external net force acting on a system of particles», the total linear momentum of the system is conserved. “ As daar geen eksterne netto krag op ’n sisteem van partikels inwerk nie bly die (2) totale liniére momentum van die sisteem behoue. 5.2 SS Peeterer voor = S'Paterina OR/OF S Prctore/voor = S Panes na TV betore /voor) + MV afbetore svoory = (Mt My WV syst sist ¥ Anyone/ Enige een (6x8) + (5x0)¥=(6 + SV va ssict” Vv =4,36 m:s” to the right ¥/ na regs (4) 5.3. POSITIVE MARKING FROM QUESTION 5.2 POSITIEWE NASIEN VANAF VRAAG 5.2 OPTION 1/OPSIE 1 Wre = AE wy ¥ Any one/ Enige een Copyright reserved 11 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 12 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 W,, = AE, + AE, 5.4. Ff, Axcos180° = Sm ~sm} + AE, Mg NAxcos180° = a ~ mv} + AE, Y¥ Any one/ Enige een HgmgAxcos|80° = ay — Sm? + AE, ad 19,82 (-D¥=— 1105)" ~5116436)°¥ +0 H=0.43% OPTION 2/OPSIE 2 Wra = AE YA f Eni W., = Eq -E ny one/ Enige een F,Axcos180° = bn? -L mv? t ar My NAxcos180° = ; mv? -+ my? >” Any one/ Enige een H,,mghxcos180° = sv ~ 5mm? Mg (11%9,8x2%(-1) v= 2110,5) -51 (4,36) ¥ u.=0.43 1 POSITIVE MARKING FROM QUESTION 5.3 POSITIEWE NASIEN VANAF VRAAG 5.3 SUP ot = ma IFpe=ma fy +Fg +N=ma ¥ Any one/ Enige een —f,-F,=ma -EN-F,=ma — pg cos30° — mgsin30° = ma — ugcos30° - gsin30° =a — (0,43 x11 x 9,8 x cos30°)¥ — (11x 9,8 x sin 30° ¥ = Lla vi =v? + 2aAxY 0? = (1,5)? +2(-8.55)AxY Ax=d= 0,26 mv ¥ Any one / Enige een OR/IOF Copyright reserved 12 Please turn over (5)

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 13 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 SF vt = ma [Fre=ma F,+Pe+N=ma ¥ Any one / Enige een -f,-F, = ma - UN -F, = ma Oo +1300 oo pamg.cos 3) mg sin 30 ma } ¥ Any one / Enige een — vgcos30" ~ gsin30° =a - (0,43 x 9,8xcos30") “ —(9,8xsin30° VW =a a= -8,55m-s” vi =v) + 2adxnY 0? = (15)? + 2(-8.55)Ax ¥ Ax=d= 0,26 mY 5.4.2 Increases “Toeneem 7 Aceleration of the system of blocks decreasesv’. (7 Versnelling van die bloksisteem neem af (2) [20] Copyright reserved 13 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 14 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 QUESTION 6/ VRAAG 6 6.1 The Doppler effect is the change in frequency (pitch) of sound detected by a listenerv’ 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 (toonhoogte) van klank soos waargeneem deur die luisteraar omdat die kKlankbron en die luisteraar verskillende snelhede besit (relatief tot die medium wat die kiank voortbring) OR/ OF The Doppler effect is the change in the observed frequency of a wave” when the source or the detector moves relative to the transmitting medium. “ Die Doppiler-effek is die verandering in die waargenome frekwensie van “n golf wanneer die bron of die waarnemer beweeg relatief tot die medium wat die kKlank dra. (2) 6.2 OPTION 1/OPSIE 1 fr -( |e VV, 1 000% = 340 340 — |v 930)¥ V.= 23,80 ms!” OPTION 2/ OPSIE 2 v f= Fr ~ . fara 340 1 000% = Y (930) 340 -vV, V.= 23,80 m-s'¥ OPTION 3/ OPSIE 3 f- &£ y oR fe t2-£ 4-= v v 930 ¥ qs v 340 V.= 23,80 m-s'¥ (8) 5 v 1000 = 6.3. Increases“/Toeneem (2) The compressions behind the ambulance are further apart than when it was approaching”. Die verdunnings agter die ambulans is verder van mekaar as die verdigtings toe die ambulans die luisteraar nader. 6.4.1, The stars are moving away “” Die sterre beweeg weg van die aarde The spectral lines in the diagrams are shifted towards the red end/red shifted).~” Die spektrumlyne in die diagramme_ het verskuif na die rooi ent/rooi verskuiwing. (2) Copyright reserved 14 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 15 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 6.4.2. The Universe is expanding.” (1) Die heeial is besig om uit te sit. [12] QUESTION 7/ VRAAG 7 7.1 The electric field at a point is the (electrostatic) force experienced per unit positive charge at this point. “~~” Die elektriese veld by 'n punt is die (elektrostatiese) krag wat ‘n positiewe (2) eenheidslading by daardie punt ondervind. 7.2 Criteria for marking Shape of field lines” Direction of field lines. Nasienkriteria Vorm van die veldlyne +Q ~Q Rigting van die veldlyne (2) 7.3 OPTION 1/ OPSIE1 kQ Ect a (9x10°)(2x10~) Que v (0,5) E..¢= 7-20x104N-Co‘right /regs kQ Ee" a kQ (9x10°)(2x107°) | =n ede A acta R ee (05)? E_yc=7,2%104 N-C“right/regs E net = Exc + E 2,0 Positive to the right } ¥ Anyone / Enige een Et = Ene t Ene E neto= 7-2X10* +7,2x104¥ =+14,4x10°N-C1Y OR/ OF =14,4«10°N-C“ right /regs” Copyright reserved 15 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 16 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 OPTION 2/ OPSIE 2 Exe ev (9x10° }(2x10 6 ) ope (0,5) E..c= 7.20x104N- C“right /regs kQ Ec r? kQ _(9x10°)(2x10*) , E,o= oe (0.5)? E s,c=14,4x10* N-Cright /regs E net = Eo + E oye Positive to the left YAn . y one/ Enige een Eee = ~E auc + wtb E ppto= ~7-2X104 -7,2x10°Y =-14,4x10'N-ClY OR/IOF = 14,4x10'N-C right / regs ¥ OPTION 3/OPSIE 3 Ext ey (9x10°)(2x10%) [oi a (0,5) Eayg= 120x104 NC“ right /regs Enc= Eye = 7-20x10*N-C right /regs ¥ E net = Exc + E onc Positive to the right } v Any one/ Enige een Ent = Eas + Bac E veto™ 2X (7.2X10*)¥ = +14,4x10°N-CT OR/OF = 14,4104 N-C"'right [regs ¥ Copyright reserved 16 Please turn over

7A Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 17 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 OPTION 4/ OPSIE 4 Bago Rv (9x10°)(2x10~) i oe v (0,5) Eye 7-20x10*N-C right /regs Eac= E oye = 7-20X104N- C“right /regs ¥ E vet = Fane + E onc Positive to the left Y An p y one / Enige een Ent = Ea + wtb E seto= ~7-2X10* — 7,2x10°¥ =-14,4x10°N-ClY OR/OF =14,4x104N-C right /regs¥ (6) OPSION 1/OPSIE 1 Fra = Fis + Fos ¥ ORIOF +2yuC CL, P +6 pC pr. KOO, K0,0, | ...... De Ce oo net 2 2 xX | Ag hy > 1,0m = KQ2s _ KQ,0, re re 8 % ¥ Any one / Enige een KQ,0, _ KQ,0, 3 os 2.2 ORIOF 2 2 AL=x)? = 6x hs 2 2 2(x°-2x+1) =6x nana 2x10% y= 6x10 2x*-4x+2 = 6x" x” (l-x)v 2x?-2x+1=0 2_ 6 Y den? ee bP? = 4a 2a 4x? +4x-2=0 OR/OF 2+ /2? -4x2x(-1) V2_ V6 = —_= 2(2) x (-x) x = 0,37 m from charge/ vanaf lading 2 pC ~ OR/OF Copyright reserved 17 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 18 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/1 CAPS/KABV NKDVO/ Sep 2015 x= 0,63 m from charge/ vanaf lading 6uC” OPSION 2/OPSIE 2 Fre = Q,Enen2¥ Fra = Qs OQ, _ FQ: +2 uc, rod P +6uC %3 Les eg a <2 e @--------------------- @----- 9 -KQ2: _ KQ,0, x >| 2 2 ns 13 v An i 1,0m yone/Enigeeen © KQ,Q, = KQ,9, ns br 3 o_O OR/OF ; 2 2(1-x)’ = 6x’ M3 Nhs 2 262-2x+1) =6x" a hy 2x10% ve 6x10 2x2-AK+2 = 6x2 x (-x)v 2x”-2x+1=0 2. 6 ¥ ax ee Dev b? ~ 400 2a 4x? +4x-2=0 OR/OF 2+ 2? —4x2x(-1) v2_ V6 => ial 2(2) x (l-x) x = 0,37 m from charge/ vanaf lading 2 uC “ OR/OF (6) X= 0,63 m from charge/ vanaf lading 6uCY [16] QUESTION 8/ VRAAG 8 8.1 When current flows through a voltage source (battery/generator) a resistance to current flow arises” due to the resistance of the materials (chemicals/conductors) from which the source is made.~ Wanneer stroom deur ’n volf-kragbron (battery/generator) vioei, ontstaan ’n weerstand teen stroomvioei as gevolg van die weerstand van die materiaal (chemikaleé/ geleiers) waarvan die bron gemaak is. OR/OF (2) Internal resistance is the resistance offered to the electron flow” by the electrolyte/medium of the cell/generator. “ Interne weerstand is die weerstand gebied teen die vioei van elektrone deur die elektroliet/ medium van die sel/ generator. 8.2 € = Vex Ir Copyright reserved 18 Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 Fisiese Wetenskappe/ GR 12/ V1 6 = Vout +1(0,10) Vex = 6- (0,10)1% 83 vv) 4 6Y 8.4 8.4.1 OPTION 1/ OPSI/E 1 W, =PRAt “= 40 (2(4)At = 40v Wa €(1yRat = 60 At R=6Qv 8.4.2 POSITIVE MARKING FROM QUESTION 8.4.1 POSITIEWE NASIEN VRAAG 8.4.1 OPTION 1/ OPSIE 1 e€={(Rtr) ¥ 6 = 1(4+6)+ 0,10)” 1=0,59 AY 8.4.3 POSITIVE MARKING FROM QUESTION 8.4.1 and 8.4.2 POSITIEWE NASIEN VRAAG 8.4.1 en 8.4.2 Copyright reserved 19 NCDOE/ Sep 2015 CAPS/KABV NKDVO/ Sep 2015 (2) 0 = 6-(0,10)I to 60 A 0.1 al (A) OPTION 2/ OPSIE 2 W =PRAtY W, _ PR,At W, PRA? 40 _ PAA, 60 FRAt R=60¥ OPTION 2/ OPSIE 2 €=I(Rtr) ¥ 6 = I(10)+ 0,10)” 1=0,59 AY 19 (3) (4) (3) (3) Please turn over

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 2 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 OPTION 1/ OPSIE 1 OPTION 2 /OPSIE 2 € = Vox +r” V = IRex”” 6 = Vex + (0,59)(0,10) “ = (0,59)(10) “ Vext = 5,94 VY =5,9 Vv 8.5 DECREASEY Total resistance of the circuit decreases “ Current increases” Vintemal resistance IncreasesY Vex (voltmeter reading) decreases (Vex = € — Vint) AFNEEM Totale weerstand van die stroombaan neem af Stroom neem toe Vinterne weerstand Neem toe Veks(voltmeterlesing) neem af (Voxs = € — Vint) QUESTION 9/ VRAAG 9 9.1 DC Generator” Mechanical energy to electrical energy” GS Generator Meganiese energie na elektriese energie 9.2 To make the direction of the (induced) current to be the same in every half cycle/half turn’ “ Om die (geinduseerde) stroom se rigting dieselfde te hou tydens elke halfsiklus/ halwe rotasie OR To keep the (induced) current unidirectional’ “ Om die (geinduseerde) stroom in een rigting te laat vioei 9.3 Graph AY DC generator becomes an AC generator” Voltage changes the polarity in every half cycle. ¥ OR Graph AY DC generator becomes an AC generator” The voltage is alternating “ Copyright reserved 2 Please turn over (4) [21] (2) (2) (3)

Downloaded from hlayiso.com Physical Sciences/Gr 12/P1 2 NCDOE/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV NKDVO/ Sep 2015 V7 ms /wak 9.4 Pagigem = ————_ ” 2 alg R Priston = Vi ms iwak y (Vins nats 2 R P , - V2 (vnw maks \2 svgigem = —— — a 2, ae R Ce) 6= v2) oy 2» R 6= v2" v R=12Q R R=122 Vims wok Vem rwagk = R v Vimstwgk = Irmstovwgk tot Rr V, R V ims /wok (—moximaie 2 = Ins Totavwge tat “() Tnms/wele = _ v2 7 v2 2 R 12 (eY 12 12 V2 —=! (Sv Imsiwgk =0,71 lims Totaliwgk tot = 142AVv Iims Totavwgk totaat = 2x(0,71) ¥ =142AV¥ QUESTION 10/ VRAAG 10 10.1. What is the relationship between frequency of the incident radiation” and the maximum kinetic energy of the ejected electrons? Vv Wat is die verwantskap tussen frekwensie van die invallende bestraling en die maksimum kinetiese energie van die vrygestelde elektrone? 10.2 10.2.1 frequency-/ frekwensie 10.2.2 ANY ONEY/ ENIGE EEN Threshold frequency/ drumpel frekwensie Work function of the metal/ werkfunksie van die metaal Potential difference/ potensiaalverskil Intensity of the incident radiation/ intensiteit van die invallende bestraling 10.3. As the frequency of the incident radiation increases, the kinetic energy also increases¥ ¥ Soos die frekwensie van die invallende bestraling toeneem, neem die kinetiese energie ook toe. Copyright reserved 2 Please turn over (5) [12] (1) (1) (2)

Downloaded from hlayiso.com NCDOE/ Sep 2015 Physical Sciences/Gr 12/P1 3 NKDVO/ Sep 2015 Fisiese Wetenskappe/ GR 12/ V1 CAPS/KABV 10.4 OPTION 1/ OPSIE 1 E= W, + Ekman mata 2 Any one /Enige een 34 8 (6.8310 ™)Gx10) yg 49- .ev 200%10 Ex= 1,95 x 10? Jv OPTION 2/ OPSIE 2 E=W, +Exmawmais * E _fty A E- (6,63 x 10™)(3x10°) 200x 10 E=9,95x1079J 9,95 x10-" =8x10"" +E nay mars - “19 . Ex= 1,95 x10" IY (5) 10.5 Remains the same/ No Change. ¥ Bly dieselfde/ Geen verandering (2) Kinetic energy is independent of intensity of the incident radiation. ~ Kinetiese energie is onafhanklik van die intensiteit van die invallende bestraling 106 DECREASES/ AFNEEM (1) [14] TOTAL /TOTAAL: 150 Copyright reserved 3 Please turn over