Physical Sciences Grade 12 Paper 1 NSC November 2014 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2014 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 20 pages. Hierdie memorandum bestaan uit 20 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/November 2014 NSC/NSS – Memorandum QUESTION 1/VRAAG 1 1.1 A  (2) 1.2 A  (2) 1.3 D  (2) 1.4 C  (2) 1.5 B  (2) 1.6 C  (Accept/ Aanvaar R) (2) 1.7 A  (2) 1.8 D  (2) 1.9 A  (2) 1.10 C  (2) [20] QUESTION 2/VRAAG 2 2.1 When a resultant (net) force acts on an object, the object will accelerate in the direction of the force. This acceleration is directly proportional to the force and inversely proportional to the mass of the object. Wanneer 'n resulterende (netto) krag op 'n voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel. Hierdie versnelling is direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp. OR/OF The net force acting on an object is equal to the rate of change of momentum  of the object (in the direction of the force). (2 or 0) Die netto krag wat op 'n voorwerp inwerk is gelyk aan die tempo van verandering in momentum van die voorwerp (in die rigting van die krag).(2 of 0) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 2014 NSC/NSS – Memorandum 2.2 T2  T1  w (3) 2.3 OPTION 1/OPSIE 1 F net = ma For 5 kg block/Vir 5 kg-blok T 2 + (-mg) + (-T 1 )= ma 250 – (5)(9,8) – T 1 = 5 a  201 – T 1 = 5 a T 1 = 201 - 5a........(1) For 20 kg block/Vir 20 kg-blok T 1 + (-mg) = ma.......(2) T 1 + [-20(9,8)]  = 20a 5 = 25 a a = 0,2 m∙s-2 upwards/opwaarts ∴ T 1 = 201 - 5(0,2)  = 200 N  OR/OF T 1 = 20(9,8) + 20(0,2)  = 200 N  (6) OPTION 2 /OPSIE 2 F net = ma For 5 kg block/Vir 5 kg-blok T 2 + (-mg) + (-T 1 )= ma 250 - (5)(9,8) - T 1 = 5 a  201 - T 1 = 5a T 1 = 201 - 5a........(1) For 20 kg block/Vir 20 kg-blok , T 1 + (-mg) = ma.......(2) T 1 + [-20(9,8)]  = 20a (1) x 4 : 4T 1 = 804 – 20a ∴T 1 - 196 = 804 – 4T 1  ∴5T 1 = 1 000 ∴T 1 = 200 N  (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 2014 NSC/NSS – Memorandum OPTION 3/OPSIE 3 F net = ma For 5 kg block/Vir 5 kg-blok T 2 + (-mg) + (-T 1 )= ma 250 - (5)(9,8) - T 1 = 5 a  201 - T 1 = 5 a T 1 = 201 - 5a........(1) 201 − T1 ∴a= 5 For 20 kg block/Vir 20 kg-blok , T 1 + (-mg) = ma.......(2) T 1 + [-(20)(9,8)]  = 20a 201 − T1 ∴T 1 - 196 = 20( ) 5 ∴T 1 = 200 N (6) 2.4 Q (1) [12] QUESTION 3/VRAAG 3 3.1 An object moving / Motion under the influence of gravity / weight / gravitational force only (and there are no other forces such as friction). (2 or/of 0) ('n Voorwerp wat / Beweging slegs onder die invloed van swaartekrag / gewig / gravitasiekrag (en daar is geen ander kragte soos wrywing nie). (2) 3.2 OPTION 1/OPSIE 1 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: ∆y = v i ∆t + ½ a∆t2 ∆y = v i ∆t + ½ a∆t2 0 = 15 ∆t + ½ (-9,8) ∆t2 0 = -15 ∆t + ½ (9,8) ∆t2 ∆t = 3,06 s ∆t = 3,06 s It takes/Dit neem 3,06 s It takes/Dit neem 3,06 s (4) OPTION 2/OPSIE 2 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: v f = v i + a∆t v f = v i + a∆t 0 = 15 +(-9,8)∆t 0 = -15 +(9,8)∆t ∆t = 1,53 s ∆t = 1,53 s It takes (2)(1,53) = 3,06 s It takes/Dit neem 3,06 s (4) OPTION 3 / OPSIE 3 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: v f = v i + a∆t v f = v i + a∆t -15 = 15 + (-9,8)∆t 15 = -15 + (9,8)∆t ∆t = 3,06 s ∆t = 3,06 s (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 2014 NSC/NSS – Memorandum OPTION 4/OPSIE 4 Upwards positive/Opwaarts positief: Downwards positive /Afwaarts positief: F net ∆t = ∆p F net ∆t = ∆p mg ∆t = m (v f -v i ) mg ∆t = m (v f -v i ) (0 - 15) 0 − ( −15)  ∆t = ∆t = -9,8  9,8  ∆t = 1,53 s ∆t = 1,53 s It takes/Dit neem (2)(1,53s) = 3,06 s It takes/Dit neem (2)(1,53s) = 3,06 s (4) OPTION 5/OPSIE 5 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: F net ∆t = ∆p F net ∆t = ∆p mg ∆t = m (v f -v i ) mg ∆t = m (v f -v i ) − 15 − (15)  15 − ( −15) ∆t = ∆t = − 9,8  9,8  = 3,06 s ∆t = 3,06 s (4) OPTION 5/OPSIE 6 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: v f = v i + 2aΔy  2 2 v f 2 = v i2 + 2aΔy  For ball A/Vir bal A For ball A/Vir bal A 0 = (15)2 + 2 (-9,8)∆y 0 = (-15)2 + 2 (9,8)∆y ∆y A = 11,48 m ∆y A = -11,48 m  v + vi   v + vi  Δy =  f ∆t Δy =  f ∆t  2   2   15 + 0   − 15 + 0  11,48 =  ∆t  -11,48 =  ∆t   2   2  Δt = 1,53 s Δt = 1,53 s It takes/Dit neem (2)(1,53s) = 3,06 s It takes/Dit neem (2)(1,53s) = 3,06 s Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 2014 NSC/NSS – Memorandum 3.3 OPTION 1/OPSIE 1 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: v f = v i + 2aΔy  2 2 v f 2 = v i2 + 2aΔy  For ball A/Vir bal A For ball A/Vir bal A 0 = (15)2 + 2 (-9,8)∆y 0 = (-15)2 + 2 (9,8)∆y ∆y A = 11,48 m ∆y A = -11,48 m When A is at highest point When A is at highest point Wanneer A op hoogste punt is Wanneer A op hoogste punt is ∆y B = v i ∆t + ½ a∆t2 ∆y B = v i ∆t + ½ a∆t2 = 0 + ½ (-9,8) (1,53)2 = 0 + ½ (9,8) (1,53)2 ∆y B = -11,47 m ∆y B = 11,47 m ∆y B = 11,47 m downward/afwaarts ∆y B = 11,47 m downward/afwaarts Distance/Afstand = y A + y B Distance/Afstand = y A + y B = 11,47 + 11,48  = 11,48 + 11,47  = 22,95 m = 22,95 m OPTION 2/OPSIE 2 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: At maximum height v f = 0: At maximum height v f = 0: By maksimum hoogte v f = 0: By maksimum hoogte v f = 0: Ball/Bal A Ball/Bal A ∆y A = v i ∆t + ½ a∆t2 ∆y A = v i ∆t + ½ a∆t2 = 15 (1,53)  + ½ (-9,8) (1,53)2 = (-15) (1,53) + ½ (9,8) (1,53)2 = 11,48 m = -11,48 m When A is at highest/point When A is at highest point Wanneer A op hoogste punt is Wanneer A by hoogste punt is ∆y B = v i ∆t + ½ a∆t2 ∆y B = v i ∆t + ½ a∆t2 = 0 + ½ (-9,8) (1,53)2 = 0 + ½ (-9,8) (1,53)2 ∆y B = -11,47 m ∆y B = -11,47 m ∆y B = 11,47 m downward/afwaarts ∆y B = 11,47 m downward/afwaarts Distance/Afstand = y A + y B Distance/Afstand = (y A + y B ) = 11,48 + 11,47 = 11,48 + 11,47  = 22,95 m = 22,95 m (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2014 NSC/NSS – Memorandum OPTION 3/OPSIE 3 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: Ball A/Bal A Ball A/Bal A ∆y A = v i ∆t + ½ a∆t2 y A = v i ∆t + ½ a∆t2 ∆y A = 15(1,53) + ½ (-9,8) (1,53)2 ∆y A = -15 (1,53)  + ½ (9,8) (1,53)2 = 11,48 m = -11,48 (m) = 11,48 m upward/opwaarts For ball B/Vir bal B v f = v i + a∆t For ball B/Vir bal B v f = 0 + (-9,8)(1,53) v f = v i + a∆t v f = 14,99 m∙s-1 v f = 0 + (9,8)(1,53) v f = 14,99 m∙s-1 v f = v i + 2a∆x 2 2 v f = v i + 2a∆x 2 2 14,992 = 0 + 2(-9,8) ∆y B  ∆y B = -11,47 (m) 14,992 = 0 + 2(9,8) ∆y B  = 11,47 m downward/afwaarts ∆y B = 11,47 (m) Distance/Afstand = (y A + y B ) Distance/Afstand = (y A + y B ) = 11,48 + 11,47 = 11,48 + 11,47 = 22,95 m = 22,95 m (7) OPTION 4/OPSIE 4 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: Ball A/Bal A Ball A/Bal A   vi + vf (15 + 0) vi + vf (-15 + 0) ∆y A = Δt  = (1,53)  ∆y A = Δt  = (1,53)  2 2 2 2 = 11,48 m = -11,48 (m) = 11,48 m upwards/opwaarts For ball B/Vir bal B v f = v i + a∆t v f = v i ∆t + a∆t = 0 +( -9,8) (1,53) = 0 +( 9,8) (1,53) = -15 m∙s-1 = 15 m∙s-1   vi + vf (0 - 15) × 1,53 vi + vf (0 + 15) × 1,53 ∆y = Δt =  ∆y = Δt =  2 2 2 2 = -11,47 m = 11,47m = 11,47 m downward/afwaarts Distance/Afstand = (y A + y B ) Distance/Afstand = y A + y B = 11,48 + 11,47 = 11,48 + 11,47 = 22,95 m = 22,95 m (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 2014 NSC/NSS – Memorandum OPTION 5/OPSIE 5 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: Ball A/Bal A Ball A/Bal A W net = ∆K W net = ∆K OR/OF OR/OF ½ m( v2 f – v2 i ) = mg(h f – h i )cosθ ½ m( v2 f – v2 i ) = mg(h f – h i )cosθ 2 o ½ m(0 -15 ) = m(9,8)h f cos180  ½ m(0 -152) = m(9,8)h f cos180o  h = 11,48 m h = 11,48 m OR/OF OR/OF For Ball B when A is at highest point./ For Ball B when A is at highest point./ Vir Bal B wanneer A by sy hoogste Vir Bal B wanneer A by sy hoogste punt punt is. is. v f = v i + a∆t v f = v i + a∆t = 0 +( -9,8) (1,53) = -15 m∙s-1 = 0 +(9,8) (1,53) = 15 m∙s-1   vi + vf (0 - 15) × 1,53 vi + vf (0 + 15)(1,53) ∆y = Δt =  ∆y = Δt =  2 2 2 2 =-11,48 m = 11,48 m downward/afwaarts = 11,48 m downward/afwaarts Distance/Afstand = y A + y B Distance/Afstand = y A + y B = 11,48 + 11,48 = 11,48 + 11,48 = 22,96 m = 22,96 m (7) OPTION 7/OPSIE 7 Upwards positive/Opwaarts positief: Downwards positive/Afwaarts positief: Ball A Ball A ½ m v2 i + mgh i = ½ m v2 f + mgh f  ½ m v2 i + mgh i = ½ m v2 f + mgh f  ½ m(152)  +0 = ½ m(0) + m(9,8)h  ½ m(152)  +0 = ½ m(0) + m(9,8)h  h = 11,48 m h = 11,48 m OR/OF OR/OF For Ball B when A is at highest point. For Ball B when A is at highest point. Vir Bal B wanneer A by sy hoogste Vir Bal B wanneer A by sy hoogste punt punt is. is. v f = v i + a∆t v f = v i + a∆t = 0 +( -9,8) (1,53) = 0 +(9,8) (1,53) = -15 m∙s-1 = 15 m∙s-1 vi + vf vi + vf ∆y = Δt ∆y = Δt 2 2   (0 - 15)(1,53) (0 + 15)(1,53) =  =  2 2 =-11,48 m = 11,48 m downward/afwaarts = 11,48 m downward/afwaarts Distance/Afstand = y A + y B Distance/Afstand = y A + y B = 11,48 + 11,48 = 11,48 + 11,48 = 22,96 m = 22,96 m (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 2014 NSC/NSS – Memorandum 3.4 15 v (m·s-1) 0 1,53 3,06 t (s) -15 (4) [17] CONSIDER MOTION DOWNWARD AS POSITIVE/BESKOU BEWEGING AFWAARTS AS POSITIEF 15 v (m·s-1) 0 1,53 3,06 -15 Criteria/Kriteria Marks/Punte Graph starts at correct Initial velocity shown./Grafiek  begin by korrekte beginsnelheid aangetoon. Time for maximum height shown (1,53 s)./Tyd vir  maksimum hoogte aangetoon.(1,53 s) Time for return shown (3,06 s) /Tyd om terug te keer  (3,06) aangetoon. Shape/Vorm: Straight line extending beyond 3,06 s/  Reguitlyn wat verby 3,06 s strek. (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 2014 NSC/NSS – Memorandum QUESTION 4/VRAAG 4 4.1 p = mv = 50(5) = 250 kg∙m∙s-1 (downward/afwaarts) OR/OF p = mv = 50(-5) = - 250 kg∙m∙s-1 = 250 kg∙m∙s-1 (downward/afwaarts) (3) 4.2 The product of the (net) force and the time interval (during which the force acts) (2 or 0) Die produk van die (netto) krag en die tydinterval (waartydens die krag inwerk) (2 of 0). (2) 4.3 OPTION 1/OPSIE 1 ∆p = F net ∆t ∆p = F net ∆t ∆p = F net ∆t 0 - 250 = F net (0,2) 250 - 0 = F net (0,2) 50(0 – (-5))= F net (0,2) F net = -1 250 N F net = 1 250 N  F net = 1 250 N  = 1 250 N  (3) OPTION 2/OPSIE 2 m(v f – v i ) = F net ∆t m(v f – v i ) = F net ∆t 50 (0 - 5) = F net (0,2) 50 (5 - 0) = F net (0,2) F net = -1 250 N F net = 1 250 N  = 1 250 N  (3) OPTION 3 /OPSIE 3 v f = v i + a∆t v f = v i + a∆t 0 = 5 + a(0,2)  5 = 0 + a(0,2)  a = -25 m∙s-2 a = 25 m∙s-2 F net = ma F net = ma = 50 (-25) = 50 (25) = - 1 250 N = 1 250 N = 1 250 N (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 2014 NSC/NSS – Memorandum 4.4 Greater than/Groter as (1) 4.5 For the same momentum change,  the stopping time (contact time) will be smaller (less)  ∴ the (upward) force exerted (on her) is greater. Vir dieselfde verandering in momentum, sal die stilhoutyd (kontaktyd) kleiner wees ∴ die (opwaartse)krag wat (op haar) uitgeoefen word, sal groter wees. (3) [12] QUESTION 5/VRAAG 5 5.1.1 In an isolated/closed system,  the total mechanical energy is conserved / remains constant  In 'n geïsoleerde/geslote sisteem bly die totale meganiese energie behoue / bly konstant. OR/OF The total mechanical energy of a system is conserved/ remains constant  in the absence of friction. Die totale meganiese energie van 'n sisteem bly behoue/bly konstant in die afwesigheid van wrywing. OR/OF The total mechanical energy of a system remains constant provided the net work done by external non conservative forces is zero. Die totale meganiese energie van 'n sisteem bly konstant, mits die arbeid verrig deur eksterne nie-konserwatiewe kragte, nul is. OR/OF In the absence of a non-conservative force, the total mechanical energy is conserved/remains constant In die afwesigheid van ‘n nie-konserwatiewe krag, bly die totale meganiese energie behoue / konstant OR/OF In an isolated/closed system,  the sum of kinetic and gravitational potential energy is conserved / remains constant  In 'n geïsoleerde/geslote sisteem bly som van kinetiese en gravitasionele potensiële energie behoue / bly konstant. 5.1.2 No/Nee (1) 5.1.3 OPTION 1/OPSIE 1 Along AB/Langs AB Along AB/Langs AB E mechanical at A = E mechanical at B W net = ΔE k  (E p + E k ) A = (E p + E k ) B  F g Δhcosθ = ½ m(v f 2 – v i 2) (mgh + ½ mv2) A = (mgh + ½ mv2) B (10)(9,8)(4)cos0o = ½ (10)(v f 2 – 0)  (10)(9,8)(4) + 0 = 0 + ½ (10) v f 2 v f = 8,85 m∙s-1 v f = 8,85 m∙s-1 (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/November 2014 NSC/NSS – Memorandum Along AB/Langs AB W nc = ΔK + ΔU  0 = ½ (10)(v f 2 – 0) + 10(9,8)(4 – 0)  v f = 8,85 m∙s-1 Substitute 8,85 m∙s-1 in one of the following options Vervang 8,85 m∙s-1 in een van die volgende opsies Along BC/Langs BC Along BC/Langs BC W net = ∆K W nc = ∆K + ∆U  f∆x cosθ = ∆K f ∆xcosθ = ∆K + ∆U f(8)cos 180o = ½ (10)(0 – 8,852)  f(8)cos 180 = ½ (10)(0 - 8,852) + 0 f = 48,95 N f = 48,95 N (Accept/ Aanvaar 49 N) OPTION 2/OPSIE 2 Along AC/Langs AC W nc = ∆K + ∆U f∆xcosθ = ∆K + ∆U (f)(8)(cos180o) = (0 - 0)  + 10 (9,8)(0 - 4)  f = 49 N  (6) 5.2.1 f k = µ k N = µ k mgcosθ = (0,19)(300)(9,8) cos 25o = 506,26 N (3) 5.2.2 OPTION 1/OPSIE 1 F FN app f θ Fg F net = 0  F app + (-F g sinθ) + (-f) = 0 F app - (300)(9,8)sin 25o - 506,26 = 0 F app = 1 748,76 N P ave = Fv ave  = 1748,76 x 0,5 = 874,38 W  (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 2014 NSC/NSS – Memorandum OPTION 2/OPSIE 2 W f + W app + W N + W g = 0  F∆xcosθ + F app ∆x cosθ + 0 + F g ∆x cosθ = 0 (506,26∆x cos180o) + (F app ∆x cos0) + 300(9,8)Δxcos115o= 0 F app = 1748,76 N P ave = Fv ave  = (1748,76 ) (0,5) = 874,38 W (6) OPTION 3/OPSIE 3 W f + W app + W N + W g = 0 F∆xcosθ + F app ∆x cosθ +0 + F g sinθ∆x cosθ = 0 (506,26∆x cos 0)  + (F ap ∆x cos0) + 300 (9,8)sin 25o Δxcos180) = 0 F app = 1 748,76 N P ave = Fv ave  = (1 748,76 )(0,5) (6) = 874,38 W [18] QUESTION 6/VRAAG 6 6.1.1 An (apparent) change in observed/detected frequency (pitch), (wavelength) as a result of the relative motion between a source and an observer (listener). 'n Skynbare verandering in waargenome frekwensie (toonhoogte),(golflengte) as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer/luisteraar. (2) 6.1.2 Towards/Na Observed/detected frequency is greater than the actual frequency. Waargenome frekwensie is groter as die werklike frekwensie. (2) 6.1.3 v ± vL v fL = f s OR/OF fL = fs  v ± vs v − vs 343  (1200)= 1130  343 - v s v s = 20,01 m∙s-1 Accept/Aanvaar: (19,42 – 20,01 m∙s-1) (5) 6.2 The star is approaching the earth.  Die ster nader die aarde. OR/OF The earth and the star are approaching (moving towards) each other. Die aarde en die ster nader mekaar. The spectral lines in diagram 2 are shifted towards the blue end/blue shifted. (2) Die spektrumlyne in diagram 2 het verskuif na die blou ent/blou verskuiwing [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 2014 NSC/NSS – Memorandum QUESTION 7/VRAAG 7 7.1 To ensure that charge does not leak to the ground/insulated.  Om te verseker dat die lading nie na die grond toe lek nie/isoleer. (1) Notes/Aantekeninge Accept/Aanvaar In order retain original charge/To insulate the charges./ Om oorspronklike lading te behou/ Om lading te isoleer. 7.2 QR + Q S + 8 + (-4) Net charge/Netto lading = = = 2 µC 2 2 (2) 7.3 Marks/ Criteria for sketch:/Kriteria vir skets: Punte Correct direction of field lines  Korrekte rigting van veldlyne Shape of the electric field  Vorm van elektrieseveld No field line crossing each other / No field lines inside the spheres/  Geen veldlyne wat maekaar kruis nie / Geen veldlyne binne sfeer nie (3)  T  7.4 FS on T ● F R on T (2) 7.5 OPTION 1/OPSIE 1 QQ F = k 12 2  r (1× 10 -6 )(2 × 10 -6 )  F ST = (9 × 10 9 ) = 0,45 N / 4,5 x 10-1 N left/links (0,2)2  OR/OF 1 1 F TS = F RT = (1,8) = 0,45 N 4 4 (2 × 10 -6 )(1× 10 -6 ) F RT = 9 × 10 9 ×  = 1,8 N right/regs (0,1) 2 OR/OF F RT = 4F ST = 4(0,45) = 1,8 N right /regs F net = F ST + F RT = 1,8 + (-0,45)  = 1,35 N or towards sphere S / na sfeer or/of right/regs S (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 2014 NSC/NSS – Memorandum OPTION 2/OPSIE 2 kQ (9 × 10 9 )(2 × 10 −6 ) ER = 2 = = 1,8 x 106 N.C-1 right/regs r (0,1) 2 kQ (9 × 10 9 )(2 × 10 −6 ) Es = = = 4,5 x 105 N.C-1 left/links r2 (0,2) 2 E net = 1,8 x 106 - 4,5 x 105 = 1,35 x 106 N.C-1right/regs F = EQ = (1,35 x 106)(1 x 10-6)  = 1,35 N towards sphere S / na sfeer S right/regs  (6) 7.6 Force experienced per unit positive charge placed at that point. Krag ondervind per eenheid positiewe lading by daardie punt. (2) 7.7 OPTION 1/OPSIE 1 F 1,35  6 E= = - 6 = 1,35 x 10 N∙C-1 q 1× 10 (3) OPTION 2/OPSIE 2 kQ (9 × 10 9 )(2 × 10 −6 ) ER = 2  = 2 = 1,8 x 106 N∙C-1 right/regs r (0,1) kQ (9 × 10 9 )(2 × 10 −6 ) Es = 2 = 2 = 4,5 x 105 N∙C-1 left/links r (0,2) E net = 1,8 x 10 - 4,5 x 105 = 1,35 x 106 N∙C-1 6 (3) OPTION 3/OPSIE 3 F 1,8  E= = = 1,8 x 106 N∙C-1 q 1× 10 -6 F 0,45 E= = = 4,5 x 105 N∙C-1 q 1× 10 -6 E net = 1,8 x 106 - 4,5 x 105 = 1,35 x 106 N∙C-1 (3) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 2014 NSC/NSS – Memorandum QUESTION 8/VRAAG 8 8.1.1 Keep the temperature (of battery) constant. Hou die temperatuur (van battery) konstant (1) 8.1.2 Grafiek van potensiaalverskil teenoor stroom Marks/ Criteria for drawing line of best fit:/Kriteria vir teken van lyn van beste pas: Punte ALL points correctly plotted (at least 4 points)  ALLE punte korrek gestip (ten minste 4 punte) Correct line of best fit if all plotted points are used ( at least 3 point)  Korrekte lyn van beste pas indien alle punte gebruik word (ten minste 3 punte) (3) 8.1.3 7,2 V (Accept any readings between 7,0 V and 7,4 V or the value of the y-intercept /Aanvaar enige lesing tussen 7,0 V en 7,4 V of die waarde van die y-afsnit (1) 8.1.4 ΔV Slope/Helling = ΔI 0 - 7,2  = =-9 0,8 - 0  r = 9 Ω (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/November 2014 NSC/NSS – Memorandum 8.2.1 OPTION 1/OPSIE 1 P = VI  100 = 20(I) I=5A (3) OPTION 2/OPSIE 2 V2 P=  R (20) 2 100 = R R=4Ω V = IR 20 = I(4)  I=5A (3) OPTION 3/OPSIE 3 V2 P=  R (20) 2 100 = R R=4Ω P=I2R 100 = I2(4)  I=5A 8.2.2 OPTION 1/OPSIE 1 V2 P=  R (20) 2 R=  150 = 2,67 Ω  (3) OPTION 2/OPSIE 2 P = VI 150 = (20)I I = 7,5 A V = IR 20 = (7,5)R  R = 2,67 Ω  OR/OF P = I2R 150 = (7,5)2R R = 2,67 Ω  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/November 2014 NSC/NSS – Memorandum OPTION 3/OPSIE 3 IX : IY 5 : 7,5 1 : 1,5 RX : RY 1,5 : 1 4 : 2,67 Ω (3) 8.2.3 OPTION 1/OPSIE 1 P = VI OR/OF P = I2R 150 150 I 150W =  = 7,5 A I150W =  = 7,5 A 20 2,67 I tot = (5 + 7,5)  ε = I(R + r)  24 = 12,5(R + r) 24 = V ext + V ir 24 = 20 + 12,5(r )  r = 0,32 Ω  (5) OPTION 2/OPSIE 2 V = Ir I tot = (5 + 7,5)  (24 - 20) = 12,5 r 4 ∴r = 12,5 r = 0,32 Ω (5) OPTION 3/OPSIE 3 1 1 1 = + R R1 R 2 1 1 1 ( 4)(2,67) = + OR/OF R // = R // 4 2,67 4 + 2,67 ∴R || =1,6 Ω 20 I tot = = 12,5 A 1,6 ε = I(R + r)  24 = 12,5(R + r) 24 = V ext + V ir 24 = 20 + 12,5(r )  r = 0,32 Ω  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/November 2014 NSC/NSS – Memorandum OPTION 4/OPSIE 4 P = VI 250 = (20)I I = 12,5 A V = Ir 4 = (12,5)r r = 0,32 Ω  (5) 8.2.4 Device Z is a voltmeter . Toestel Z is 'n voltmeter (1) 8.2.5 Device Z should be a voltmeter (or a device with very high resistance) because it has a very high resistance and will draw very little current.  The current through X and Y will remain the same hence the device can operate as rated. Toestel Z moet 'n voltmeter wees (of 'n toestel met 'n baie hoë weerstand) omdat dit 'n baie hoë weerstand het en baie min sal stroom trek Die stroom deur X en Y sal dieselfde bly, gevolglik kan die toestel werk soos ontwerp. (2) [22] QUESTION 9/VRAAG 9 9.1 Electromagnetic induction / Elektromagnetiese induksie (1) 9.2 Rotate the coil faster/Increase the number of coils/ Increase the strength of the magnetic field. Roteer die spoel vinniger/Verhoog die aantal spoele / Verhoog die sterkte van die magneetveld. (1) 9.3 Slip rings/Sleepringe (1) 9.4.1 It is the value of the voltage in a DC circuit that will have the same heating effect as an AC circuit.  Dit is die waarde van die potensiaalverskil in 'n GS-stroombaan wat dieselfde verhittingseffek het as 'n WS-stroombaan (2) 9.4.2 Vmax V rms =  2 339,45 =  2 (3) V rms = 240,03 V  [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com QUESTION 10/VRAAG 10 10.1 The minimum frequency (of a photon/light) needed to emit electrons  from (the surface of) a metal. (substance)  Die minimum frekwensie (van 'n foton/lig) benodig om elektrone vanaf die (oppervlakte van)'n metaal (stof) vry te stel. (2) 10.2 OPTION 1/OPSIE 1 E = Wo + E k(max) 1 2 E = Wo + mv max Any one / Enige een 2 c 1 2 h = hf0 + mv max λ 2 (6,63 × 10-34 )(3 × 108 ) 1 = (6,63 × 10-34 )(5,548 × 1014 )  + (9,11 × 10-31 )(5,33 × 105 )2  λ 2 λ = 4 x 10 m -7 (5) OPTION 2/OPSIE 2 E = Wo + E k(max) 1 2 E = Wo + mv max Any one / Enige een 2 1 hf = hf0 + mv max 2 2 1 (6,63 x 10-34)f = (6,63 × 10-34 )(5,548 × 1014 )  + (9,11 × 10-31 )(5,33 × 105 )2  2 f = 7,5 x 1014 Hz c = fλ 3 x 108 = (7,5 x 1014)λ λ = 4 x 10-7 m  (5) 10.3 Smaller (less) than  Kleiner (minder) as (1) 10.4 The wavelength/frequency/energy of the incident light (photon/hf) is constant. Die golflengte/frekwensie/energie van die invallende lig (foton/hf) is konstant Since the speed is larger, the kinetic energy is larger  the work function/W 0 /threshold frequency smaller.  Aangesien die spoed vergroot, is die kinetiese energie groter, is die arbeidsfunksie / W 0 / drumpel frekwensie kleiner (3) [11] GRAND TOTAL/GROOTTOTAAL: 150 Copyright reserved/Kopiereg voorbehou