Physical Sciences Grade 12 Paper 1 Mpumalanga Mock 2022 Memo

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Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Sept 2022 September Exam - Mpumalanga Errata Fout by kortvraag 1.10. Moet A wees. En tekenfout by 2.3; moet +f wees. (Antw: F = 56,17N) Error: 1.10 A is correct and sign error in 2.3; should be +f (Anwer: F = 56,17 N)

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) SEPTEMBER 2022 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consists of 15 pages Hierdie nasienriglyne bestaan uit 15 bladsye Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 2 MDE/MDO/September 2022 NSC/ NSS QUESTION 1 / VRAAG 1 1.1 D  (2) 1.2 B  (2) 1.3 B  (2) 1.4 D  (2) 1.5 A  (2) 1.6 B  (2) 1.7 C  (2) 1.8 C  (2) 1.9 A  (2) 1.10 B  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 3 MDE/MDO/September 2022 NSC/ NSS QUESTION 2 / VRAAG 2 2.1 Marking criteria/Nasienkriteria If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. A body will remain in its state of rest or motion at constant velocity unless a non- zero resultant/net/unbalanced force acts on it.  ‘n Liggaam sal in ‘n toestand van rus of beweging teen ‘n konstante snelheid volhard, tensy ‘n nie-nul resulterende/netto/ongebalanseerde krag daarop inwerk. OR/OF A body will remain in its state of rest or uniform motion in a straight line unless a non-zero resultant/net/unbalanced force acts on it.  ‘n Liggaam sal in ‘n toestand van rus of uniforme beweging in ‘n reguit lyn volhard, tensy ‘n nie-nul resulterende/netto/ongebalanseerde krag daarop inwerk. (2) 2. 2 N Accept/Aanvaar F N F T T w┴  w w//  (4) Accepted labels / Aanvaarde benoemings w Fg / Fw / force of earth on block / weight / mg / gravitational force / 29,4 N F Fapplied / FA / Applied force T Tension in rope / FT N Normal force / FN / 25,46 N Notes/Aantekeninge:  Any additional forces: deduct 1 mark: max 3⁄4  No labels: deduct 1 mark: max 3⁄4  No arrows: 0⁄4  Force(s) not touching object: deduct 1 mark: max 3⁄4  Ignore relative sizes of the vectors Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 4 MDE/MDO/September 2022 NSC/ NSS 2.3 F = (Fg//)5kg + (Fg//)3kg - fs = 5(9,8)Sin30° + 3(9,8)Sin30° - 16,97 = 22,23 N (3) 2.4 Marking criteria/Nasienkriteria:  Formula for block P or block Q   Substitution of Fnet for block P   Substitusion of Fnet for block Q   5a OR 3a   Answer: a=5,16 m∙s-2  For Block P: For Block Q: Fnet = ma Fnet = ma Any one  T + (-Fg//) + (-f) = ma F + (-Fg//) + (-T) = ma T – (5×9,8)Sin30° - 4,5  = 5a  85 – (3×9,8)Sin30° - T = 3a T = 5a +29 T = 70,3 – 3a ∴ 5a + 29 = 70,3 – 3a 8a = 41,3 a = 5,16 m∙s-2  (5) 2.5 Marking criteria/Nasienkriteria Q  Both straight lines velocity   For Q: positive gradient and larger than initially P  For P: negative gradient time (3) [17] QUESTION 3 / VRAAG 3 3.1 9,8 m·s-2  downwards / afwaarts  (2) 3.2.1 Note: do not penalize for the omission of zero(s). OPTION 1 Upwards positive: Downwards positive: ∆y = -50-(-1,5) = -48,5 m ∆y = 50-1,5 = 48,5 m vf2 = vi2 + 2a∆y  vf2 = vi2 + 2a∆y  = 0 + 2(-9,8)(-48,5)  = 0 + 2(9,8)(48,5)  vf = 30,83 m·s-1  vf = 30,83 m·s-1  OPTION 2 (mgh+½mv2)top = (mgh+½mv2)P  (mgh+½mv2)top = (mgh+½mv2)P  m(9,8)(50)+0 = m(9,8)(1,5)+½(m)v2 m(9,8)(48,5) + 0 = 0 + ½(m)v2  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 5 MDE/MDO/September 2022 NSC/ NSS v = 30,83 m·s-1  v = 30,83 m·s-1  3.2.2 OPTION 1 Upwards Positive Downwards Positive Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  -48,5 = 0 + ½ (-9,8) Δt2  48,5 = 0 + ½ (9,8) Δt2  Δt = 3,15 s  Δt = 3,15 s  OPTION 2 POSITIVE MARKING FROM Q3.2.1 Upwards Positive Downwards Positive vf = vi +a Δt  vf = vi +a Δt  -30,83 = 0 + (-9,8) Δt  30,83 = 0 + (9,8) Δt  Δt = 3,15 s  Δt = 3,15 s  OPTION 3 Upwards Positive Downwards Positive Fnet∙Δt = mvf – mvi Fnet∙Δt = mvf – mvi m(-9,8) Δt = m(-30,83) – 0 (m×9,8) Δt = m(30,83) – 0 Δt = 3,15 s Δt = 3,15 s (3) 3.3 Marking criteria/Nasienkriteria:  Formula for motion at constant velocity  Substitution for motion at constant velocity  Formula to calculate vi for vertical motion  Substitution to calculate vi  Final answer: 13,28 m∙s-1 Motion at constant velocity: Δx = viΔt + ½aΔt2 OR/ (vf +vi ) OR/ Distance = speed×time 11 = 2,3 Δt  ∆x= ∆t  OF 11 = 2,3 Δt  OF 2 Δt = 4,78 s (2,3+2,3) Δt = 4,78 s 11= ∆t  2 Δt = 4,78 s Vertical motion: Upwards positive Downwards positive Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  -48,5 = vi(4,78) + ½(-9,8)(4,78)2 48,5 = vi(4,78) + ½(9,8)(4,78)2 vi = 13,28 m∙s-1 vi = -13,28 m∙s-1 vi = 13,28 m∙s-1 (5) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 6 MDE/MDO/September 2022 NSC/ NSS QUESTION 4 / VRAAG 4 Marking criteria/Nasienkriteria If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. 4.1 A system on which the resultant/net external force is zero ‘n Sisteem waarin die resulterende/netto eksterne krag nul is. (2) 4.2.1 p = mv  8,4 = m(560)  m = 0,015 kg  (3) 4.2.2 POSITIVE MARKING FROM 4.2.1 Σpi = Σpf  Any one (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 (0,015)(560) + (3)(-2,5)  = (0,015)(80) + (3)(v)  v = -0,1 v = 0,1 m∙s-1  (4) 4.2.3 POSITIVE MARKING FROM QUESTION 4.2.1 OPTION 1 Fnet∙Δt = mvf – mvi Fnet(0,02) = 0,015(80-560)  Fnet = -360 Fnet = 360 N East/Oos Accept: Right/Regs POSITIVE MARKING FROM QUESTION 4.2.2 OPTION 2 Fnet∙Δt = mvf – mvi Fnet(0,02) = 3(0,1 – 2,5)  (Fnet)block = -360 (Fnet)bullet = 360 N East/Oos Accept: Right/Regs (4) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 7 MDE/MDO/September 2022 NSC/ NSS QUESTION 5 / VRAAG 5 5.1 OPTION 1 (Em)A = (Em)B (mgh + ½mv2)A = (mgh +½ mv2)B Any one  m(9,8)(0,7) + ½ m(2)2 = 0 + ½mv2  v = 4,21 m·s-1  OPTION 2 Wnc = ΔEk + ΔEp Any one  0 = [½mvf2 – ½mvi2] + [mghf – mghi] 0 = [½mv2 – ½m(2)2] + [0 – m(9.8)(0,7)]  v = 4,21 m·s-1  (3) 5.2 Marking criteria/Nasien kriteria  Formula for Ek = 11,17  Substitution of mv=5,79  Final answer: v = 3,86 m·s-1 OPTION 1 p = mv = 5,79 ½mv∙v = 11,17 2 Ek = ½mv = 11,17 ½(5,79)v = 11,17 v = 3,86 m·s-1  OPTION 2 p = mv = 5,79 Ek = ½mv2 = 11,17 5,79 1 5,79 2 m= ( ) v =11,17 v 2 v v = 3,86 m·s-1  (3) 5.3 Marking criteria/Nasienkriteria If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The net work done on an object is equal to the change in kinetic energy of the object.  Die netto arbeid op ‘n voorwerp verrig is gelyk aan die verandering in die kinetiese energie van die voorwerp. (2) 5.4 POSITIVE MARKING FROM Q5.2 OPTION 1 OR/OF mv = 5,79 ½mv2 = 11,17 m(3,86)=5,79 ½ m(3,86)2 = 11,17 m = 1,5 kg m = 1,5 kg Wnet = ΔEk Any one  fΔxcosΘ = ½mvf2 – ½mvi2 (3)(X)cos180°= 0 – ½(1,5)(3,86)2 X = 3,72 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 8 MDE/MDO/September 2022 NSC/ NSS OPTION 2 mv = 5,79 ½mv2 = 11,17 m(3,86)=5,79 ½ m(3,86)2 = 11,17 m = 1,5 kg m = 1,5 kg Wnc = ΔEk + ΔEp fΔx cosΘ= [½mvf2 – ½mvi2 ]+ [mghf – mghi] Any one  (3)(X)cos180°= [0 – ½(1,5)(3,86)2] + 0 X = 3,72 m  (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 9 MDE/MDO/September 2022 NSC/ NSS QUESTION 6 / VRAAG 6 6.1 Marking criteria/Nasienkriteria If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The change in frequency/pitch/wavelength of the sound detected by a listener because the sound source and the listerner have different velocities relative to the medium of sound propagation.  Die verandering van frekwensie/toonhoogte/golflengte van die klank waargeneem deur ‘n luisteraar omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium van klankvoorplanting het. OR/OF An (apparent) change in observed frequency/pitch/wavelength as the result of the relative motion between a sounce 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) 6.2 Waves in front of source are more compact/wavelength decreases  More waves per second reaches the detector/listener  Golwe voor die bron kompakteer / golflengte neem af  Meer golwe per sekonde bereik die detector  (2) 6.3.1 Moving towards Moving away v±vL v v±vL v fL = fS  OR fL = fS fL = fS OR fL = f v±vS v-vS v±vS v+vS S 330 330 950 = fs  750 = f  330 –vs 330 +vs s 950(330 − 𝑣𝑠 ) 750(330 + 𝑣𝑠 ) 𝑓𝑠 = 𝑓𝑠 = 330 330 ∴ 950(330-vs) = 750(330+vs) 200(330) = 1700vs vs = 38,82 m∙s-1 (6) 6.3.2 POSITIVE MARKING FROM Q6.3.1 OPTION 1 OPTION 2 OPTION 3 Δt = 12,5-8= 4,5 s Δt = 12,5-8= 4,5 s Δt = 12,5-8= 4,5 s Δx = viΔt + ½aΔt2 (vf +vi ) Distance = speed×time = (38,82)(4,5)  ∆x= ∆t  = (38,82)(4,5)  2 = 174,69 m (38,82+38,82) = 174,69 m = (4,5)  2 = 174,69 m (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 10 MDE/MDO/September 2022 NSC/ NSS QUESTION 7 / VRAAG 7 7.1.1 Marking criteria/Nasienkriteria If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Note: If masses used ( 0⁄2 ) The magnitude of the electrostatic force exerted by one point charge (Q 1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges  and inversely proportional to the square of the distance (r) between them  Die grootte van die elektrostatiese krag wat een puntlading (Q1) op 'n ander puntlading (Q2) uitoefen, is direk eweredig aan die produk van die groottes van die ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle.  (2) 7.1.2 Negative/Negatief (1) 7.1.3 kQ1 Q2 F=  r2 2 (9×109 )Q 57,6 =  (0,1)2 Q = 8×10-6 C  Accept: Q = -8×10-6 C (3) 7.2.1 Marking criteria/Nasienriglyne: -1 mark for each of the 5 key words omitted in the correct context. -1 punt vir elk van die 5 sleutelwoorde weggelaat in die korrekte konteks. The electric field at a point is the (electrostatic) force experienced per unit positive charge placed at that point. Die elektriese veld by 'n punt is die (elektrostatiese) krag wat per positiewe eenheids-lading wat by daardie punt geplaas is, ondervind word. (2) 7.2.2 kQ E=  r2 (9×109 )(8×10-9 ) =  (0,2)2 = 1 800 N∙C-1 to the left (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 11 MDE/MDO/September 2022 NSC/ NSS 7.2.3 Enet = Ex - Ey (9×109 )Q (9×109 )(4×10-9 ) 120 = -  0,22 0,082 Q = 2,55×10-8 C Q n=  e- (2,55×10-8 -8×10-9 ) -2,55×10-8 -(-8×10-9 ) =  OR/OF 1,6×10-19 -1,6×10-19 = 1,1×1011  (5) [17] QUESTION 8 / VRAAG 8 8.1 7,2 (V)  (1) NOTE: Penalize once in Q8.2 for incorrect use of ×10-2 8.2.1 V2 -V1 Gradient= -r = NOTE: Can use any coordinates on the line, eg I2 -I1 (100×10-2 ; 2,4) 7,2-0 = -2  (67,5×10-2 ; 4,0) 0-150×10 (17,5×10-2 ; 6,4) r = 4,8 Ω (2) 8.2.2 POSITIVE MARKING FROM Q8.1 AND Q8.2.1 OPTION 1 OPTION 2 ɛ = I(R+r)  V = IR 7,2 = (117,5×10 )[R+4,8]  -2 1,6 = (117,5×10-2)R R = 1,33 Ω R = 1,36 Ω (3) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 12 MDE/MDO/September 2022 NSC/ NSS QUESTION 9 / VRAAG 9 9.1 The potential difference across a conductor is directly proportional to the current in the conductor  at constant temperature.  Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur. (2) 9.2.1 OPTION 1 OPTION 2 V6Ω = IR R6Ω : R12Ω = 0,5(6)  6 : 12 =3V I6Ω : I12Ω 12 : 6 I12Ω = V⁄R = 3⁄12 = 0,25 A 0,5 : 0,25 I1 = 0,5+0,25=0,75 A I1 = 0,5+0,25 = 0,75 A (3) 9.2.2 POSITIVE MARKING FROM Q9.2.1 V20Ω=IR = 0,75(20) = 15 V P=VI 16 = (15+3)I I2 = 0,89 A (3) 9.2.3 POSITIVE MARKING FROM Q9.2.1 & Q9.2.2 OPTION 1 IT=0,75+0,89 = 1,64 A ɛ = I(R+r)  = (1,64)(10,98+1)  = 19,65 V OPTION 2 IT=0,75+0,89 = 1,64 A ɛ = Ve + Ir  = 18 + (1,64)(1)  = 19,64 V (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 13 MDE/MDO/September 2022 NSC/ NSS QUESTION 10/VRAAG 10 10.1.1 Split ring/Split ring commutator/commutator Splitring/splitring kommutator/kommutator (1) 10.1.2 Anticlockwise/Antikloksgewys (2) 10.1.3 Electrical (energy) TO mechanical/kinetic (energy) Elektriese (energie) NA meganiese/kinetiese (energie) (1) 10.2.1 Marking criteria/Nasienkriteria If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The AC current which dissipates the same amount of energy as an equivalent DC current.  Die WS-stroomsterkte wat dieselfde hoeveelheid energie verbruik as die ekwivalente GS-stroomsterkte. ACCEPT/AANVAAR The DC current which dissipates the same amount of energy as an equivalent AC current. Die GS-stroomsterkte wat dieselfde hoeveelheid energie verbruik as die ekwivalente WS-stroomsterkte. (2) 10.2.2 OPTION 1 OPTION 2 Imax Imax Irms =  Irms =  √2 √2 8 8 =  =  √2 √2 = 5,66 A = 5,66 A Vrms = Irms R Pave = I2rms R = (5,66)(40) = (5,66)2 (40) = 226,4 V = 1280 W  2 Vrms Pavg = VrmsIrms OR/OF Pave   R = (226,4)(5,66) = 1281,42 W (226,4)2 =  40 = 1281,42 W  (5) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 14 MDE/MDO/September 2022 NSC/ NSS QUESTION 11 / VRAAG 11 11.1 Marking criteria/Nasienkriteria If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The minimum energy of light needed to eject electrons from a metal/surface.  Die minimum energie benodig om 'n elektron uit die oppervlak van 'n metaal vry te stel. Note: If referred to frequency: 0⁄2 (2) 11.2 Zinc Photons that have shorter wavelengths will have higher frequencies and thus a higher energy.  OR/OF photons that have longer wavelengths will have lower frequencies, and thus a lower energy  OR/OF 1 𝑊0 ∝ , ℎ&𝑐 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜆0 OR/OF 1 𝑓0 ∝ , 𝑐 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜆0 𝑊0 ∝ 𝑓0 , ℎ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 OR/OF W0 is inversely proportional to λ0  (with h and c staying constant). (3) 11.3 Frequency of photon is less than the threshold frequency of the metal Die frekwensie van die foton is minder as die drumpelfrekwensie van die metaal. OR/OF Energy of the photon is less than the work function of the metal.  Die energie van die foton is minder as die arbeidsfunksie/werksfunksie van die metaal. (2) Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com Physical Sciences P1/Fisiese Wetenskappe V1 15 MDE/MDO/September 2022 NSC/ NSS 11.4 OPTION 1 E = W 0 + Ek(max) 1 2 hf = hf0 + mv 2 max Any one  hc hf = + Ek(max) λ0 (6,63×10-34 )(3×108 ) (6,63×10-34 )(5×1014 ) =  + Ek(max) 10×10-7 Ek(max) = 1,33 × 10-19 J  OPTION 2 c = f0λ0 3×108 = f0(10×10-7) f0 = 3×1014 Hz hf = hf0 + Ek(max) (6,63×10-34 )(5×1014 ) = (6,63×10-34 )(3×1014 ) + Ek(max) Ek(max) = 1,33 × 10-19 J  (4) 11.5.1 Remains the same/Bly dieselfde (1) 11.5.2 Increases/Toeneem (1) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou