Physical Sciences Grade 12 Paper 1 NSC November 2012 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2012 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 12 pages. Hierdie memorandum bestaan uit 12 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/November 2012 NSC/NSS – Memorandum SECTION A QUESTION 1/VRAAG 1 1.1 Frequency/Frekwensie  (1) 1.2 Capacitor/Kapasitor  (1) 1.3 Split ring commutator  Splitringkommutator (1) 1.4 Photons/Fotone  (1) 1.5 Relative velocity/Relatiewe snelheid  (1) [5] QUESTION 2/VRAAG 2 2.1 D  (2) 2.2 C  (2) 2.3 D  (2) 2.4 D  (2) 2.5 A  (2) 2.6 A  (2) 2.7 D  (2) 2.8 C  (2) 2.9 C  (2) 2.10 A  (2) [20] TOTAL SECTION A/TOTAAL AFDELING A: 25 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 2012 NSC/NSS – Memorandum SECTION B/AFDELING B QUESTION 3/VRAAG 3 3.1 Downward/afwaarts  (1) 3.2 3.2.1 Upwards positive/Opwaarts positief: v f = v i + aΔt  = 8 + (-9,8)(4)  = - 31,2 m∙s-1 ∴ v f = 31,2 m∙s-1  Downwards positive/Afwaarts positief: v f = v i + aΔt = - 8 + (9,8)(4)  ∴ v f = 31,2 m∙s-1  (4) 3.2.2 OPTION 1/OPSIE 1 Upwards positive/Opwaarts positief: Δy = v i Δt + ½aΔt2 = (8)(4)  + ½(-9,8)(4)2  = -46,4 m Height of balcony/Hoogte van balkon: 60 – 46,4 = 13,6 m  Downwards positive/Afwaarts positief: Δy = v i Δt + ½aΔt2 = (-8)(4)  + ½(9,8)(4)2  = 46,4 m Height of balcony/Hoogte van balkon: 60 – 46,4 = 13,6 m  OPTION 2/OPSIE 2 Upwards positive/Opwaarts positief: Δy = v i Δt + ½ aΔt2 = (27,13)  (6) + ½ (-9,8)(6)2 = - 13,62 m Height of balcony/Hoogte van balkon: = 13,62 m  Downwards positive/Afwaarts positief: Δy = v i Δt + ½ aΔt2 = (-27,13)  (6) + ½ (9,8)(6)2 = 13,62 m Height of balcony/Hoogte van balkon: = 13,62 m  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 2012 NSC/NSS – Memorandum 3.3 OPTION 1/OPSIE 1 Upwards positive/Opwaarts positief: 27,13 Velocity/snelheid (m∙s-1) 8 0 4 10 Time/tyd(s) -31,2 Marks/ Criteria for graph/Kriteria vir grafiek: Punte Shape has two parallel lines with a gradient.  Vorm het twee ewewydige lyne met gradient. First part of graph starts at v = 8 m∙s-1 at t = 0 s  Eerste deel van grafiek begin by v = 8 m∙s-1 by t = 0 s. Positive marking from QUESTION 3.2.1: Positiewe nasien vanaf VRAAG 3.2.1: First part of the graph extends below the x axis until v = -31,2 m∙s-1  at t = 4 s. Eerste deel van die grafiek verleng onder x-as tot v = -31,2 m∙s-1 by t = 4 s. Graph is discontinuous and object changes direction at 4 s.  Grafiek is nie kontinu nie en voorwerp verander van rigting by 4 s. Second part of graph starts at v = 27,13 m∙s-1 at t = 4 s.  Tweede deel van grafiek begin by v = 27,13 m∙s-1 by t = 4 s. Second part of graph extends below the x axis until t = 10 s.  Tweede deel van grafiek verleng onder x-as tot t =10 s. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 2012 NSC/NSS – Memorandum OPTION 2/OPSIE 2 Upwards negative/Opwaarts negatief: 31,2 Velocity/snelheid (m∙s-1) Time/tyd(s) 0 4 10 -8 -27,13 Marks Criteria for graph/Kriteria vir grafiek: Punte Correct shape as shown (two parallel lines).  Korrekte vorm soos aangetoon (twee ewewydige lyne). First part of graph starts at v = -8 m∙s-1 at t = 0 s  Eerste deel van grafiek begin by v = -8 m∙s-1 by t = 0 s Positive marking from QUESTION 3.2.1. Positiewe nasien vanaf VRAAG 3.2.1. First part of the graph extends above the x axis until v = 31,2 m∙s-1  at t = 4 s. Eerste deel van die grafiek verleng bokant x-as tot v = 31,2 m∙s-1 by t = 4 s. Graph is discontinuous and object changes direction at 4 s.  Grafiek is nie kontinu en voorwerp verander van rigting by 4 s. Second part of graph starts at v = -27,13 m∙s-1 at t = 4 s.  Tweede deel van grafiek begin by v = -27,13 m∙s-1 by t = 4 s. Second part of graph extends above the x axis until t = 10 s.  Tweede deel van grafiek verleng bokant x-as tot t = 10 s. . (6) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 2012 NSC/NSS – Memorandum QUESTION 4/VRAAG 4 4.1 40 m∙s-1 east/oos  (2) 4.2 The total (linear) momentum remains constant/is conserved  in an isolated/a closed system/the absence of external forces/ if the impulse of external forces is zero.  Die totale (liniêre) momentum bly konstant/behoue  in 'n geïsoleerde sisteem/geslote sisteem/ die afwesigheid van eksterne kragte./ indien die impuls van eksterne kragte nul is. (2) 4.3 East positive/Oos positief: Σp i = Σp f  m(20) + 2m(-20)  = (m + 2m)v f  ∴v f = -6,67 m·s-1 ∴v f = 6,67 m·s-1  west /wes  East negative/Oos negatief: Σp i = Σp f  m(-20) + 2m(+20)  = (m + 2m)v f  ∴v f = 6,67 m·s-1  west /wes  (6) 4.4 4.4.1 F Newton’s Third Law of motion/Newton se Derde Bewegingswet  (2) 4.4.2 -½ a / ½ a 1 (Same/Dieselfe F net ), a α  m (2) 4.4.3 Car driver  (Car - driver system) have greater acceleration.  (Car - driver system) have greater change in velocity /greater Δv. Motorbestuurder  (Motor -bestuurder sisteem) het groter versnelling.  (Motor -bestuurder sisteem) het groter verandering in snelheid / groter Δv. (3) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2012 NSC/NSS – Memorandum QUESTION 5/VRAAG 5 5.1 N  F w (3) 5.2 The net (total) work (done on an object) is equal to  the change in kinetic energy (of the object.)  Die netto (totale) arbeid verrig (op 'n voorwerp) is gelyk aan  die verandering in kinetiese energie (van die voorwerp).  (2) 5.3 5.3.1 W net = ΔE k /ΔK  OR/OF F net Δxcosθ = ½ m(v f 2 – v i 2) F net (1,02)cos180° = ½ (1 200)(0 – 202)  F net = 235 294,12 N  (2,35 x 105 N) (4) 5.3.2 OPTION 1 /OPSIE 1 OPTION 2/OPSIE 1 F net Δt = mΔv   v + vf  Δx =  i  Δt  ∴ (-235 294,12)Δt = (1 200)(0 - 20)   2  ∴ Δt = 0,1 s (0,102 s)  20 + 0  1,02 =   Δt   2  Δt = 0,1 s (4) [13] QUESTION 6/VRAAG 6 6.1 Frequency/Frekwensie  (1) 6.2 There is relative motion between the bird and the bird watcher.  Daar is relatiewe beweging tussen die voël en die voëlkyker nie.  (1) 6.3 0,2 m  (1) 6.4 6.4.1 v = fλ  340 = f(0,2)  ∴ f = 1 700 Hz  (3) 6.4.2 v ± vL v fL = f s OR/OF fL = fs  v ± vs v − vs 340 ∴ 1 700  =  (1 650)  340 − v s ∴ v s = 10 m∙s-1  (5) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 2012 NSC/NSS – Memorandum QUESTION 7/VRAAG 7 7.1 Double slit/Dubbelspleet  (1) 7.2 (Alternate) dark and bright/blue bands.  Bright / blue bands of equal broadness (width).  (Afwissellende) donker en helder/blou bande.  Helder / blou bande van gelyke breedte.  (2) 7.3 7.3.1 1 central band 1 sentraleband tan θ = 2 / 2 screen dis tan ce skermafstand 1 (0,22)  ∴ tan θ = 2 1,4  ∴ θ = 4,49°  (3) 7.3.2 OPTION 1/OPSIE 1: OPTION 2/OPSIE 2: mλ mλ sin θ =  sin θ =  a a      (1)( 470 × 10 −9 )  ( −1)( 470 × 10 −9 ) sin4,49 = sin (-4,49°) = a a ∴ a = 6 × 10 −6 m (6 003,67 nm) ∴ a = 6 × 10 −6 m  (6 003,67 nm) (5) 7.4 λ red light > λ blue light  (Degree of) diffraction/sin θ / θ α wavelength ( λ )  λ rooilig > λ bloulig  Diffraksie α golflengte ( λ )  (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 2012 NSC/NSS – Memorandum QUESTION 8/VRAAG 8 8.1 V R=  I 12 1 000 =  I ∴I = 0,01 A  (3) 8.2 12 V  (1) 8.3 Q C=  V Q 120 x 10-6 =  12 ∴Q = 1,44 x 10-3 C  (3) 8.4 8.4.1 Decreases/Verminder  (1) 8.4.2 Increases/Vermeerder  (1) 8.5 8.5.1 Marks/ + Criteria for sketch:/Kriteria vir skets: Punte Parallel lines equally spaced.  Parallelle lyne eweredig gespasieer. Direction from positive plate towards negative plate.(Polarity of plates must be indicated) Rigting vanaf positiewe plaat na negatiewe  plaat.(Polariteit van plate moet aangedui - word) Field curved at the ends of the plates.  Veld gekrom aan einde van die plate. (3) 8.5.2 V E=  d 12 =  12 × 10 − 3 ∴ E = 1 000 V∙m-1  (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 2012 NSC/NSS – Memorandum QUESTION 9/VRAAG 9 9.1 9.1.1 1 1 1 = +  Rp R1 R 2 1 1 = +  60 60 ∴ R p = 30 Ω  (3) 9.1.2 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2: R ext = 30 + 25 = 55 Ω  R tot = (30 + 25) +1,5 = 56,5 Ω Emf/emk = I(R + r)  V = IR ∴12 = I(55 + 1,5)  12 = I(56,5)  ∴I = 0,21 A  ∴I = 0,21 A  (5) 9.1.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 V = IR  V = IR  = (0,21)(30)  = (0,105)(60)  = 6,3 V  = 6,3 V  (3) 9.2 9.2.1 1,5 V  (1) 9.2.2 ∆V gradient/m = ∆I 0,65 - 1,5  = 1,0 - 0  = - 0,85 Ω  (3) 9.2.3 Internal resistance  Interne weerstand (2) 9.2.4 Decreases/Verminder  When I increases/Wanneer I toeneem: “Lost volts”/ Ir increases./“Verlore volts”/Ir neem toe.  V ext = emf – Ir decreases. /V ext = emk – Ir neem af. (3) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 2012 NSC/NSS – Memorandum QUESTION 10/VRAAG 10 10.1 AC /WS  (1) 10.2 V t Marks Criteria for graph/Kriteria vir grafiek: Punte Correct shape as shown; accept more than one cycle.  Korrekte vorm soos aangetoon; aanvaar meer as een siklus. If no/wrong labels: minus 1 mark Indien geen/verkeerde byskifte: minus 1 punt (2) 10.3 OPTION 1/OPSIE 1 OPTION 2 / OPSIE 2 V P ave = V rms Irms /P gem. = V wgk I wgk Vrms / wgk = max/ maks  2 Vmax I rms Vmaks I wgk P ave/gem. = /  30 × 10 3 2 2 =  2 (30 × 103 )Irms / wgk = 2,12 x 104 V 4,45 x 109  =  2 P ave = V rms Irms /P gem. = V wgk I wgk  ∴I rms/wgk = 2,10 x 105 A  9 4 4,45 x 10  = (2,12 x 10 )I rms/wgk ∴ Irms/wgk = 2,10 x 105 A  (5) 10.4 Less loss in (electrical) energy (as heat).  Minder verlies aan (elektriese) energie (as hitte).  (1) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/November 2012 NSC/NSS – Memorandum QUESTION 11/VRAAG 11 11.1 11.1.1 Kinetic energy /Kinetiese energie (E k ) (1) 11.1.2 Frequency /Frekwensie(f) (1) 11.1.3 (Type of) metal (Soort) metaal  (1) 11.2 The minimum frequency needed to emit electrons  from (the surface of) a metal.  Die minimum frekwensie benodig om elektrone vry te stel vanaf (die oppervlak van) 'n metaal. (2) 11.3 9 x 1014 Hz  (1) 11.4 E = W 0 + Ek  Any one /Enige een hf = hf 0 + E k (6,63 x 10-34)(14 x 1014)  = (6,63 x 10-34)(9 x 1014)  + E k ∴ E k =3,32 x 10-19 J  (3,31 x 10-19 J) (4) 11.5 Remains the same/Bly dieselfde  (1) [11] TOTAL SECTION B/TOTAAL AFDELING B: 125 GRAND TOTAL/GROOTTOTAAL: 150 Copyright reserved/Kopiereg voorbehou