Physical Sciences P1 May June 2022 MG Afr Eng_hlayiso.com_.pdf

Downloaded from hlayiso.com SENIOR CERTIFICATE EXAMINATIONS/ NATIONAL SENIOR CERTIFICATE EXAMINATIONS SENIORSERTIFIKAAT-EKSAMEN/ NASIONALE SENIORSERTIFIKAAT-EKSAMEN PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) 2022 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 28 pages./ Hierdie nasienriglyne bestaan uit 28 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 B  (2) 1.2 B  (2) 1.3 B  (2) 1.4 A  (2) 1.5 D  (2) 1.6 C  (2) 1.7 D  (2) 1.8 D  (2) 1.9 A  (2) 1.10 C  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 Marking criteria/Nasienkriteria: If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per word/frase. A body will remain in its state of rest or motion at constant velocity unless a (non-zero) resultant/net force/unbalanced force acts on it.  'n Liggaam sal in sy toestand van rus of beweging teen konstante snelheid volhard, tensy 'n (nie-nul) resulterende/netto krag/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 force acts on it.  'n Liggaam sal in sy toestand rus of uniforme beweging in 'n reguit lyn volhard, tensy 'n (nie-nul) resulterende/netto krag daarop inwerk. (2) 2.2 ACCEPT/AANVAAR T N T N f • f w w N Ty • f Tx w Accepted symbols/Aanvaarde simbole N FN /Normal/Normal force/Normaal/Normaalkrag/Fbuoyant f Ff /fk/frictional force/wrywingskrag/kinetic frictional force/kinetiese wrywingskrag/300 N w Fg/mg/Weight/Gewig/FEarth on man/ FAarde op man/Fw/Gravitational force/ Gravitasiekrag/ 686 N T Tension/Spanning/FTension/FSpanning /FT/FS / ACCEPT/ AANVAAR F/Fapplied/Ftoegepas Notes/Aantekeninge  Mark is awarded for label and arrow./Punt word toegeken vir byskrif en pyltjie.  Do not penalise for length of arrows./Moenie vir die lengte van die pyltjies penaliseer nie.  Deduct 1 mark for any additional force /Trek 1 punt af vir enige addisionele krag  If T is not shown but TY and TX are shown give 1 mark for both. Indien T nie aangetoon is nie maar TY en TX is getoon, ken 1 punt toe vir beide. (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 2.3 OPTION 1/OPSIE 1 Fnet = ma OR/OF Any one/Enige een Fnet = 0  Tcos50° – Ff = ma Tcos50° - 300 = 0  OR/OF Tcos50° = 300  Tcos50° = Ff  T = 466,72 N (468,75 N) OPTION 2/OPSIE 2 Wnet = ΔEk  TΔxcos0º + fΔxcos180º = 0  OR/OF Tcos50° – 300  = 0 Wnet = 0  T = 466,72 N (468,75 N) TΔxcos50º = - fΔxcos180º  NOTE/AANTEKENING Can use sin40º instead of cos50º. Kan ook sin40º i.p.v. cos50º gebruik. (4) 2.4 Increases/Neem toe  Fnet increases / Fnet is not zero / Tx > f / Tcos50º > f  Fnet neem toe / Fnet is nie nul nie / Tx > f / Tcos50º > f (2) 2.5 Marking criteria Options 1 & 2/Nasienkriteria Opsies 1 & 2  Substitution to calculate a/Vervanging om a te bereken   Formula to calculate Fup/water/Formule om Fop/water te bereken   Substitution to calculate Fup/water/Vervanging om Fop/water te bereken   Final answer/Finale antwoord: 679,20 N  OPTION 1/OPSIE 1 DOWNWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ AFWAARTS AS POSITIEF: OPWAARTS AS POSITIEF v f  v i  2ay v f  v i  2ay 2 2 2 2 0 = (16)2 + 2a (0,8)  0 = (-16)2 + 2a(-0,8)  a = -160 m∙s-2 a = 160 m∙s-2 Fnet = ma Fnet = ma Any one/Enige een Any one/Enige een Fg – Fup/op = ma -Fg + Fup/op = ma (4)(9,8) - Fup/op = (4)(-160)  -(4)(9,8) + Fup/op = (4)(160)  Fup/op = -679,20 N Fup/op = 679,20 N  Fup/op = 679,20 N  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2 DOWNWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ AFWAARTS AS POSITIEF OPWAARTS AS POSITIEF  v  vf   v  vf  Δx   i  Δt Δx   i  Δt  2   2   16  0   - 16  0  0,8   t -0,8 =   Δt  2   2  t  0,1 s ∆t = 0,1 s vf = vi + aΔt vf = vi + aΔt 0 = 16 + a(0,1)  0 = -16 + a(0,1)  a = -160 m∙s-2 a = 160 m∙s-2 OR/OF OR/OF ΔyB = viΔt + ½aΔt2 ΔyB = viΔt + ½aΔt2 -0,8 = (-16)(0,1) + ½(a)(0,1)2  0,8 = (16)(0,1) + ½(a)(0,1)2  a = 160 m∙s-2 a = -160 m∙s-2 Fnet = ma Fnet = ma Any one/Enige een Any one/Enige een -Fg + Fup/op = ma Fg – Fup/op = ma (4)(9,8) - Fup/op = (4)(-160)  -(4)(9,8) + Fup/op = (4)(160)  Fup/op = -679,20 N Fup/op = 679,20 N  Fup/op = 679,20 N  USING ENERGY PRINCIPLES/GEBRUIK VAN ENERGIE BEGINSELS Marking criteria OPTIONS 3 to 5/Nasienkriteria OPSIES 3 to 5  Formula / Formule   Substitution / Vervanging   Final answer/Finale antwoord: 679,20 N  OPTION 3/OPSIE 3 Wnet = K Any one/Enige een FnetΔxcosθ = ½mvf2 - ½mvi2 (4)(9,8)(0,8)cos0° + Fup/op(0,8)cos180°  = ½(4)(0 - 162)  Fup/op = 679,20 N  OPTION 4/OPSIE 4 Wnc = K + U Any one/Enige een Fup/opΔxcosθ = ½m(vf2 – vi2) + mg(hf – hi) Fup/op(0,8)cos180°  = ½(4)(0 - 162) + (4)(9,8)(0 - 0,8)  Fup/op = 679,20 N  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5  v  vf  Δx   i  Δt  2   16  0  0,8   t  2  t  0,1 s Fnet∆t = ∆p Fnet∆t = (ptube/band)f - (ptube/band)i Any one/Enige een (Fg - Fup/op)∆t = m(vtube/band(f) - vtube/band(i)) [(4)(9,8)  - Fup/op](0,1)  = (4)(0 – 16)  Fup/op = 679,20 N  (5) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per word/frase. An object which has been given an initial velocity and then it moves under the influence of the gravitational force only/ is in free fall. 'n Voorwerp waaraan 'n beginsnelheid gegee is en wat dan slegs onder die invloed van die gravitasiekrag beweeg/in vryval is. (2) 3.2.1 OPTION 1/ OPSIE 1 OPTION 2/ OPSIE 2 UPWARDS AS POSITIVE/ Motion from top/Beweging van bo OPWAARTS AS POSITIEF UPWARDS AS POSITIVE/ vf = vi + aΔt  OPWAARTS AS POSITIEF 0 = 15 + (-9,8)Δt  vf = vi + aΔt  Δt = 1,53 s  -15 = 0 + (-9,8)Δt  Δt = 1,53 s  DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF DOWNWARDS AS POSITIVE/ vf = vi + aΔt  AFWAARTS AS POSITIEF 0 = -15 + (9,8)Δt  vf = vi + aΔt  Δt = 1,53 s  15 = 0 + (9,8)Δt  Δt = 1,53 s  OPTION 3/ OPSIE 3 OPTION 4/ OPSIE 4 UPWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF OPWAARTS AS POSITIEF vf = vi + aΔt  Δy = viΔt + ½aΔt2  -15 = 15 + (-9,8)Δt  0 = (15)Δt + ½(-9,8)Δt2  Δt = 3,06 s Δt = 3,06 s Δt up = 1,53 s  Δt up = 1,53 s  DOWNWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF AFWAARTS AS POSITIEF Δy = viΔt + ½aΔt2  vf = vi + aΔt  0 = (-15)Δt + ½(9,8)Δt2  15 = -15 + (9,8)Δt  Δt = 3,06 s Δt = 3,06 s Δtup = 1,53 s  Δt up = 1,53 s  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Marking criteria OPTIONS 5 to 7/Nasienkriteria OPSIES 5 tot 7  Any formula relating Δy and Δt/ OR/OF Fnet∆t = m∆v  Substitution to calculate Δt/Vervanging om Δt te bereken   Final answer/finale antwoord: 1,53 s  OPTION 5/OPSIE 5 OPTION A/OPSIE A (Emech)Top/Bo = (Emech)30 m UPWARDS AS POSITIVE/ (EP +EK)Top/Bo = (EP +EK) 30 m OPWAARTS AS POSITIEF (mgh + ½mv2)Top/Bo = (mgh + ½mv2)30 m Δy = viΔt + ½aΔt2 (9,8)h + 0 = 0 + (½)(15)2 11,48 = (15)Δt + ½(-9,8)Δt2  h = 11,48 m Δt = 1,53 s  OPTION 6/OPSIE 6 DOWNWARDS AS POSITIVE/ Wnc = K + U AFWAARTS AS POSITIEF Wnc = K + mg(hf - hi) Δy = viΔt + ½aΔt2  0 = ½mvf2 - ½mvi2 + mghf - mghi 11,48 = (-15)Δt + ½(9,8)Δt2  0 = ½(0 – 152) + (9,8)h Δt = 1,53 s  h = 11,48 m OPTION B/OPSIE B  v  vf  OPTION 7/OPSIE 7 Δy   i  Δt  Wnet = ΔEk  2  wΔycosθ = ½mvf2 – ½mvi2  15  0  (9,8)Δycos180o = 0 – ½(15)2 11,48   t   2  Δy = 11,48 m Δt = 1,53 s  OPTION 8/OPSIE 8 OPTION 9/OPSIE 9 UPWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF OPWAARTS AS POSITIEF  v  vf  Fnet∆t = m∆v Any one/ Δy   i  Δt  Fnet∆t = m(vf - vi) Enige een  2  -(9,8)∆t = 0 – 15   15  0  ∆t = 1,53 s  ∆y   t  2  ∆y = 7,5 ∆t DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF Fnet∆t = m∆v v f  v i  2ay Any one/ 2 2 Fnet∆t = m(vf - vi) Enige een 0 = (15)2 + 2(-9,8)(7,5∆t)  (9,8)∆t = 15 – 0  ∆t = 1,53 s  ∆t = 1,53 s  DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF  v  vf  Δy   i  Δt  2    15  0  ∆y   t  2  ∆y = -7,5∆t v f  v i  2ay  2 2 0 = (-15)2 + 2(9,8)(-7,5∆t)  ∆t = 1,53 s  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 3.2.2 POSITIVE MARKING FROM QUESTION 3.2.1/ POSITIEWE NASIEN VANAF VRAAG 3.2.1 Marking criteria/Nasienkriteria  Formula to calculate Δy/Formule om Δy te bereken   Substitution to calculate Δy/Vervanging om Δy te bereken   Substitution of/Vervanging van 30 m   Final answer/Finale antwoord: 41,48 m  NOTE/AANTEKENING vf and vi can be swopped vf en vi kan omgeruil word OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 UPWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF OPWAARTS AS POSITIEF Δy = viΔt + ½aΔt2  vf2 = vi2 + 2aΔy  = (15)(1,53) + ½(-9,8)(1,53)2  0 = (15)2 + (2)(-9,8)Δy  = 11,48 m Δy = 11,48 m Height/hoogte = 11,48 + 30  Height/hoogte = 11,48 + 30  = 41,48 m  = 41,48 m  DOWNWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF AFWAARTS AS POSITIEF Δy = viΔt + ½aΔt2  vf2 = vi2 + 2aΔy  = (-15)(1,53) + ½(9,8)(1,53)2  0 = (-15)2 + (2)( 9,8)Δy  = -11,48 m Δy = -11,48 m Height/Hoogte = 11,48 + 30  Height/Hoogte = 11,48 + 30  = 41,48 m  = 41,48 m  OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 UPWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF OPWAARTS AS POSITIEF  v  vf  vf2 = vi2 + 2aΔy  Δy   i  Δt  vf2 = (-15)2 + (2)(-9,8)(-30)   2  vf = -28,51 m·s-1  15  0  y   (1,53)   2  vf2 = vi2 + 2aΔy = 11,48 m (-28,51)2 = (0)2 + (2)(-9,8)Δy  Δy = 41,48 m Height/Hoogte = 11,48 + 30  Height/Hoogte = 41,48 m  = 41,48 m  DOWNWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf2 = vi2 + 2aΔy   v  vf  vf2 = (15)2 + (2)(9,8)(30)  Δy   i  Δt   2  vf = 28,51 m·s-1  - 15  0  y   (1,53)  vf2 = vi2 + 2aΔy  2  (28,51)2 = (0)2 + (2)(9,8)Δy  = -11,48 m Δy = 41,48 m Height/Hoogte = 11,48 + 30  Height/Hoogte = 41,48 m  = 41,48 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 5/ OPSIE 5 (Emech)Top/Bo = (Emech)30 m (EP +EK)Top/Bo = (EP +EK) 30 m Any one/Enige een 2 2 (mgh + ½ mv )Top/Bo = (mgh + ½ mv )30 m (9,8)h + 0 = (9,8)(30) + (½ )(15)2  h = 41,48 m  OPTION 6/OPSIE 6 Wnet = ΔEk Any one/Enige een wΔxcos180° = ½m( v 2f  v i2 ) (9,8)(Δx)cos180° = ½(0 - 152)  Δx = 11,47 m Height above the ground/hoogte bokant grond= 30 + 11,47 = 41,48 m  OPTION 7/OPSIE 7 Wnc = ΔEp + ΔEk Any one/Enige een 0 = mg(hf - hi)+ ½m(vf2 – vi2) 0 = (9,8)(hf - 0) + ½(0 – 152)  h = 11,47 m Height above the ground/hoogte bokant grond= 30 + 11,47 = 41,48 m  (4) 3.3 POSITIVE MARKING FROM QUESTION 3.2.2/ POSITIEWE NASIEN VANAF VRAAG 3.2.2 Marking criteria/Nasienkriteria  Formula/Formule   Substitute to calculate ΔyB /Vervang om ΔyB te bereken   Substitute to calculate ΔyC/Vervang om Δyc te bereken   Substitute/Vervang Δt + 0,5 or/of Δt – 0,5   Equating yB and yC   Final answer/Finale antwoord: 1,71 s  OPTION 1/OPSIE 1 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF Take yC as height of disc above ground at meeting point / neem hoogte yC as die hoogte van teiken bokant grond by ontmoetingspunt: ΔyC = viΔt + ½aΔt2  yC - 30 = 15Δt + ½(-9,8)Δt2  yC = 15Δt - 4,9Δt2 + 30….. (1) Take yB as height of ball above ground at meeting point/Neem hoogte yB as die hoogte van bal bokant grond by ontmoetingspunt: ΔyB = viΔt + ½aΔt2 yB - 0 = 40(Δt - 0,5) + ½(-9,8)(Δt - 0,5)2  yB = 44,9t – 21,225 – 4,9t2…..(2) At meeting point/By ontmoetingspunt: yC = yB 15Δt - 4,9Δt2 + 30 = 44,9t – 21,225 – 4,9t2  Δt = 1,71 s  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF Take yC as height of disc above ground at meeting point/neem hoogte yC as die hoogte van teiken bokant grond by ontmoetingspunt: ΔyC = viΔt + ½aΔt2  yC - 30 = -15Δt + ½(9,8)Δt2  yC = -15Δt + 4,9Δt2 + 30….. (1) Take yB as height of ball above ground at meeting point/neem hoogte yB as die hoogte van bal bokant grond by ontmoetingspunt: ΔyB = viΔt + ½aΔt2 yB - 0 = -40(Δt - 0,5) + ½(9,8)(Δt - 0,5)2  yB = -44,9t + 21,225 + 4,9t2…..(2) At meeting point/By ontmoetingspunt: yC = yB ∴ -15Δt + 4,9Δt2 + 30 = -44,9t + 21,225 + 4,9t2  Δt = 1,71 s  OPTION 2/OPSIE 2 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: ΔyC = viΔt + ½aΔt2  = 15Δt + ½(-9,8)Δt2  = 15Δt - 4,9Δt2….. (1) ΔyB = viΔt + ½aΔt2 = 40(Δt - 0,5) + ½(-9,8) (Δt - 0,5)2  ΔyC in terms of/in terme van ΔyB: 30 + ΔyC = 40(Δt - 0,5) + ½(-9,8)(Δt - 0,5)2 ΔyC = - 4,9Δt2 + 44,9Δt - 51,225 …..(2) Equate (1) and (2)/Stel (1) en (2) gelyk: 15Δt = 44,9Δt - 51,225  Δt = 1,71 s  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: ΔyC = viΔt + ½aΔt2  = -15Δt + ½(9,8)Δt2  = 4,9Δt2 - 15Δt …..(1) ΔyB = viΔt + ½aΔt2 = -40(Δt - 0,5) + ½(9,8)(Δt - 0,5)2  ΔyC in terms of/in terme van ΔyB: 30 + ΔyC = (-40)(t – 0,5) + ½(9,8)(Δt – 0,5)2 ΔyC = 4,9Δt2 - 44,9Δt + 51,225…..(2) Equate (1) and (2)/Stel (1) en (2) gelyk: -15Δt = -44,9Δt + 51,225  Δt = 1,71 s  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF ΔyC = viΔt + ½aΔt2  = 15(Δt + 0,5) + ½(-9,8)(Δt + 0,5)2  = -4,9t2 + 10,1Δt + 6,275…..(1) ΔyB = viΔt + ½aΔt2 = 40Δt + ½(-9,8)Δt2  ΔyC in terms of/in terme van ΔyB: 30 + Δy = 40Δt + ½(-9,8)t2 Δy = - 4,9Δt2 + 40t - 30 …..(2) Equate (1) and (2)/Stel (1) en (2) gelyk: 10,1Δt + 6,275 = 40Δt – 30  Δt = 1,21 s ΔtTOT = 1,21 + 0,5 = 1,71 s  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF ΔyC = viΔt + ½aΔt2  = -15(Δt + 0,5) + ½(9,8)(Δt + 0,5)2  = 4,9Δt2 - 10,1Δt + 6,275…..(1) ΔyB = viΔt + ½aΔt2 = -40Δt + ½(9,8)Δt2  ΔyC in terms of/in terme van ΔyB: 30 + Δy = -40Δt + ½(9,8)t2 Δy = -4,9Δt2 - 40Δt - 30…..(2) Equate (1) and (2)/Stel (1) en (2) gelyk: -10,1Δt + 6,275 = - 40Δt – 30  Δt = 1,21 s ΔtTOT = 1,21 + 0,5 = 1,71 s  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 Marking criteria: Nasienkriteria:  Formula: ΔyC = viΔt + ½aΔt2   Formule: ΔyC = viΔt + ½aΔt2   Substitute to calculate ΔyB   Vervang om ΔyB te bereken   Substitute to calculate ΔyC   Vervang om Δyc te bereken   Substitute Δt = 0,5   Vervang Δt = 0,5   Adding ΔyB and ΔyC   Som van ΔyB en ΔyC   Final answer: 1,71 s   Finale antwoord: 1,71 s  UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF Displacement of C after 0,5 s/Verplasing van C na 0,5 s ΔyC = viΔt + ½aΔt2  = 15(0,5) + ½(-9,8)(0,5)2 ΔyC = 6,28 m Velocity of C after 0,5 s/ Snelheid van C na 0,5 s vf = vi + aΔt = 15 + (-9,8)(0,5) = 10,10 m·s-1 Displacement of C at meeting point/Verplasing van C by ontmoetingspunt ΔyC = viΔt + ½aΔt2 = 10,1Δt + ½(-9,8)Δt2  = 10,1Δt – 4,9Δt2 Displacement of B at meeting point/Verplasing van B by ontmoetingspunt ΔyB = viΔt + ½aΔt2 = 40Δt + ½(-9,8)Δt2  = 40t – 4,9t2 At meeting point/By ontmoetingspunt: ΔyC + ΔyB = -[10,1Δt – 4,9Δt2] + 40Δt – 4,9Δt2 36,28 = -10,10Δt + 40Δt Δt = 1,21 s Δttot = 1,21 + 0,5 = 1,71 s  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF Displacement of C after 0,5 s/Verplasing van C na 0,5 s ΔyC = viΔt + ½aΔt2  = -15(0,5) + ½(9,8)(0,5)2 ΔyC = -6,28 m Velocity of C after 0,5 s/ Snelheid van C na 0,5 s vf = vi + aΔt = -15 + (9,8)(0,5) = -10,10 m·s-1 Displacement of C at meeting point/Verplasing van C by ontmoetingspunt ΔyC = viΔt + ½aΔt2 = -10,1Δt + ½(9,8)Δt2  = -10,1Δt + 4,9Δt2 Displacement of B at meeting point/Verplasing van B by ontmoetingspunt ΔyB = viΔt + ½aΔt2 = -40Δt + ½(9,8)Δt2  = -40t + 4,9t2 At meeting point/By ontmoetingspunt: ΔyC + ΔyB = -[-10,1Δt + 4,9Δt2] - 40Δt + 4,9Δt2 -36,28 = 10,10Δt - 40Δt Δt = 1,21 s Δttot = 1,21 + 0,5 = 1,71 s  (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 3.4 POSITIVE MARKING FROM QUESTIONS 3.2.1 AND 3.3 / POSITIEWE NASIEN VANAF VRAE 3.2.1 EN 3.3. UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF Marking criteria:  Initial velocities 40 and 15 and straight lines  Velocity/Snelheid (m·s-1)  B starting at 0,5 s  40  Parallel lines with negative gradient  B (ball/bal)  Time at which disc reaches maximum height (answer from 3.2.1) 1,53 s   Time at which B hits C (answer from 3.3) 1,71 s  15 Nasienkriteria: C (disc/skyf)  Aanvanklike snelhede 40 en 15 en reguitlyne  0,5 1,53 1,71 Time/  B begin by 0,5 s Tyd (s)  Parallelle lyne met negatiewe gradiënt   Tyd wanneer skyf maks hoogte bereik (antwoord van 3.2.1) 1,53 s   Tyd wanneer B vir C tref (antwoord van 3.3) 1,71s  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF Marking criteria:  Initial velocities -40 and -15 Velocity/Snelheid (m·s ) -1 and straight lines  0,5 1,53  B starting at 0,5 s  1,71 Time/  Parallel lines with positive gradient  C (disc/skyf) Tyd (s)  Time at which disc reaches maximum height -15 (answer from 3.2.1) 1,53 s   Time at which B hits C (answer from 3.3) B (ball/bal) 1,71 s  Nasienkriteria: -40  Aanvanklike snelhede -40 en -15 en reguitlyne   B begin by 0,5 s  Parallelle lyne met positiewe gradiënt   Tyd wanneer skyf maks hoogte bereik (antwoord van 3.2.1) 1,53 s   Tyd wanneer B vir C tref (antwoord van 3.3) 1,71s   (5) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 4 /VRAAG 4 4.1 A system on which the resultant/net external force is zero./'n Sisteem waarop die resultante/netto eksterne krag nul is.  (2 or/of 0) (2) 4.2 1. According to Newton 3rd Law  the rocket exerts a force on the toy cart to the left/opposite to direction of motion.  OR 2. The toy cart exerts a force on the rocket to the right and the rocket exerts a force on the toy cart to the left/opposite to direction of motion.  OR 3. The rocket experiences a change in momentum to the right , the toy cart experiences a change in momentum to the left.  OR 4. ∆ptoy cart = -∆procket  OR 5. Total momentum is conserved / remains constant.  The momentum of the rocket increases. Therefore, the momentum of the toy cart must decrease.  OR 6. The rocket experiences an impulse to the right  therefore, the toy cart experiences an impulse to the left.  OR 7. Impulserocket = -Impulsetoy cart  1. Volgens Newton se derde wet  oefen die vuurpyl 'n krag op die speelgoedwaentjie na links uit/ teen die bewegingsrigting.  OF 2. Die speelgoedwaentjie oefen 'n krag op die vuurpyl na regs  en die vuurpyl oefen ‘n krag op die speelgoedwaentjie na links/teen die bewegingsrigting . OF 3. Die vuurpyl ondervind 'n verandering in momentum na regs , die speelgoedwaentjie ondervind 'n verandering in momentum na links.  OF 4. ∆pspeelgoedwaentjie = -∆pvuurpyl OF 5. Totale momentum bly behoue.  Die momentum van die vuurpyl neem toe. Dus moet die momentum van die speelgoedwaentjie af neem.  OF 6. Die vuurpyl ondervind 'n impuls na regs  dus ondervind die speelgoedwaentjie 'n impuls na links.  OF 7. Impulsvuurpyl = -Impulsspeelgoedwaentjie  (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 4.3 OPTION 1/OPSIE 1 RIGHT AS POSITIVE/REGS AS POSITIEF ∑pi = ∑pf (m1+ m2)vi = m1v1f + m2v2f Any one/Enige een mvi = m1v1f + m2v2f (20 + m2)2,5 = 20(0,6) + m2(30)  m2 = 1,38 kg  LEFT AS POSITIVE/LINKS AS POSITIEF ∑pi = ∑pf (m1+ m2)vi = m1v1f + m2v2f Any one/Enige een mvi = m1v1f + m2v2f (20 + m2)(-2,5) = 20(-0,6) + m2(-30)  m2 = 1,38 kg  OPTION 2/OPSIE 2 RIGHT AS POSITIVE/REGS AS POSITIEF ∆ptoy cart/speelgoedwaentjie = -∆procket/vuurpyl Any one/Enige een m1(v1(f) – v1(i)) = - m2(v2(f) - v2(i)) (20) (0,6 – 2,5)  = - (m)(30 - 2,5)  m2 = 1,38 kg  LEFT AS POSITIVE/LINKS AS POSITIEF ∆ptoy cart/speelgoedwaentjie = -∆procket/vuurpyl Any one/Enige een m1(v1(f) – v1(i)) = - m2(v2(f) - v2(i)) (20) [-0,6 – (-2,5)]  = - (m)[(-30 – (-2,5)]  m2 = 1,38 kg  (5) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: -1 punt per word/frase. The net/total work done on an object is equal to the change in the object's kinetic energy.  Die netto/totale arbeid op 'n voorwerp is gelyk aan die verandering in die voorwerp se kinetiese energie. OR/OF The work done on an object by a net force is equal to the change in the object's kinetic energy. Die arbeid verrig op 'n voorwerp deur 'n netto krag is gelyk aan die verandering in die voorwerp se kinetiese energie. (2) 5.2 Fnet opposite to direction of displacement x.  / Both frictional force and gravitational force are in opposite direction of displacement x. Fnet teenoorgesteld tot rigting van verplasing x. / Beide wrywingskrag en gravitasie krag is teenoorgesteld tot die verplasing x. OR/OF K is negative. / The final K is zero. / Ek decreases. K is negatief. / Die finale K is nul. / Ek neem af. OR/OF Wnet = Fnet xcosθ and/en θ = 180°/cosθ = -1 (1) 5.3 OPTION 1/OPSIE 1 W net = K W w + W f = ½mvf2 - ½mvi2 Any one/Enige een 2 2 mgsinθ∆xcosθ + W f = ½mvf - ½mvi (30 000)(9,8)sin28°xcos180° + (31 000)xcos180° = ½(30 000)(02 – 332)  x = 96,64 m  OPTION 2/OPSIE 2 W nc = K + U W f = K + mg(hf - hi) Any one/Enige een fxcosθ = ½mvf2 - ½mvi2 + mghf - mghi 31 000x cos180° = ½(30 000)(02 – 332)  + 30 000(9,8)(x sin28° - 0)  x = 96,64 m  OPTION 3/OPSIE 3 W net = K W w + W f = ½mvf2 - ½mvi2 Any one/Enige -Ep + W f = ½mvf2 - ½mvi2 -mg(hf - hi) + W f = ½mvf2 - ½mvi2 -(30 000)(9,8)(xsin28° - 0) + (31 000)xcos180°= ½(30 000)(02 – 332)  x = 96,64 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 W net = K W w + Wf = ½mvf2 - ½mvi2 Any one/Enige 2 2 mg∆xcosθ + W f = ½mvf - ½mvi (30 000)(9,8)xcos118° + (31 000)xcos180°= ½(30 000)(02 – 332)  x = 96,64 m  OPTION 5/OPSIE 5 Fnet = ma Fnet = Fw// + f = (30 000)(9,8)sin 28° + 31 000 = 169 024,64 N Wnet = Ek Any one/Enige een Fnet∆xcosθ = ½mvf2 - ½mvi2 169 024,64∆x cos 180° = ½ (30 000) (02 – 332)  ∆x = 96 64 m  (5) 5.4 Ascending/Opgaande  1. Ascending/Opgaande: Fnet(A) = Fw// + f  Descending/Afgaande: Fnet(D) = Fw// - f Fnet(A) > Fnet(D)  OR/OF 2. Ascending: Fw// and f are both acting in the opposite to direction of displacement.  Descending: only f is acting in the opposite direction of displacement. Net force for ascending greater than net force for descending.  Opgaande: Fw// en f werk beide teen die rigting van verplasing. Afgaande: slegs f werk teen die rigting van verplasing. Die netto krag opgaande is groter as die netto krag afgaande. OR/OF 3. Ascending: Fnet acts opposite to the direction of motion.  Descending: Fnet acts downwards in the direction of motion.  Opgaande: Fnet werk teen die bewegingsrigting. Afgaande: Fnet werk afwaarts in die bewegingsrigting. OR/OF 4. Ascending: Fw(//) acts opposite to the direction of motion  Descending: Fw(//) acts downwards in the direction of motion Net force for ascending greater than net force for descending.  Opgaande: Fw(//) werk teen die bewegingsrigting. Afgaande: Fw(//) werk afwaarts in die bewegingsrigting. Die netto krag opgaande is groter as die netto krag afgaande. (3) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 20 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per word/frase. The (apparent) change in frequency (or pitch) (of the sound) detected by a listener because the source and the listener have different velocities relative to the medium of propagation.  Die (skynbare) verandering in die frekwensie (of toonhoogte) (van die klank) waargeneem deur 'n luisteraar omdat die bron en die luisteraar verskillende snelhede relatief tot die voortplantingsmedium het. OR/OF An (apparent) change in observed/detected frequency/pitch as a result of the relative motion between a source and an observer/listener. 'n (Skynbare) verandering in waargenome frekwensie/toonhoogte as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer/ luisteraar. (2) 6.2 v=f  340 = (880) λ  λ = 0,39 m (0,386)  (3) 6.3 OPTION 1/OPSIE 1 v v v+v fL = fS  OR/OF fL = fS v v v 340 + 10  fL = 880  340 fL = 905,88 Hz POSITIVE MARKING FROM QUESTION 6.2/ POSITIEWE NASIEN VANAF VRAAG 6.2 OPTION 2/OPSIE 2 v=f  340 + 10  = fL(0,39)  fL = 897,44 Hz  (4) 6.4 B Frequency (Hz) Marking criteria/Nasienkriteria B parallel with A and above A.  A B parallel aan A en bokant A. 2 or/of 0 (2) Time (s) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 21 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1.1 Marking criteria/Nasienkriteria: If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per word/frase. 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 ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle. (2) 7.1.2 Criteria for graph/Kriteria vir grafiek: P Correct shape  Korrekte vorm Correct direction from P to T.  Korrekte rigting van P na T. Lines must not cross and must touch spheres.  Lyne mag nie kruis nie en moet die sfere raak. NOTE/AANTEKENING: If the net electric field pattern is drawn T for two like charges: 0⁄ Indien die netto elektriese veldpatroon 3 vir twee gelyksoortige ladings geteken is: (3) 7.1.3 positive/ positief  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 22 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 7.1.4 Marking criteria/Nasienkriteria  Equation for Coulomb's law./Vergelyking vir Coulomb se wet.   Correct substitution into Coulomb's equation for for FTP.  Korrekte vervanging in Coulomb se vergelyking vir FTP.  Correct substitution into Coulomb's equation for FTS.  Korrekte vervanging in Coulomb se vergelyking vir FTS.  Correct substitution into resultant force equation (Pythagoras equation).  Korrekte vervanging in resultante krag vergelyking (Pythagoras vergelyking).  Substitute into Q = ne. /Vervang in Q = ne.   Final answer/Finale antwoord: 3,05 x1013  2 2 2 Fnet = FTP + FTS kQ 1Q 2 2 kQ 1Q 2 2 = ( 2 ) + ( ) r r2  2  2 9 9 (9 10 )(3 10-6 )(3 10-6 ) (9 10 )(3 10-6 ) 2  102 = ( ) + ( ) 0,12 0,152 QS = 4,887 x 10-6 C QS = ne 4,887 x 10-6 = n(1,6 x10-19)  n = 3,05 x1013  electrons/elektrone (6) 1 1 7.2.1 E is directly proportional to ./E is direk eweredig aan 2 .  r 2 r OR/OF 1 E 2 r (1) 7.2.2 ΔE ACCEPT/AANVAAR Gradient =  kQ 1 E 2  Δ 2 r r E - (0) 680  680  = EA = 1 0,042  - (0)  0,042 = 4,25 x 105 N∙C-1 EA = 4,25 x 105 N∙C-1 OR/OF y = mx + c / y = mx 1 EA = 680( ) 0,042 = 4,25 x 105 N∙C-1 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 23 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 7.2.3 Greater than / Groter as  The gradient is equal to kQ./The gradient is proportional to Q.  Graph of sphere B has a steeper gradient than graph of sphere A.  Die gradiënt is gelyk aan kQ./Die gradiënt is proporsioneel aan Q. Grafiek vir sfeer B het 'n steiler gradiënt as die grafiek vir sfeer A. OR/OF 1 For the same , E is greater for sphere B.  r2 1 Vir dieselfde , is E groter vir sfeer B. r2 (3) [20] QUESTION 8/VRAAG 8 8.1 A conductor (resistor) which obeys Ohm's law./'n Geleier wat Ohm se wet gehoorsaam.  (2 or/of 0) OR/OF V always directly proportional to I at constant temperature.  (2 or/of 0) V is altyd direk eweredig aan I by konstante temperatuur. OR/OF V = constant / k / constant at constant temperature.  (2 or/of 0) I V = konstant / k / konstant bly by ‘n konstante temperatuur. I OR/OF A conductor for which the resistance remains constant at constant temperature when voltage or current change.  (2 or/of 0) ‘n Geleier waar die weerstand konstant bly by ‘n konstante temperatuur wanneer die potensiaalverskil of die stroom verander. (2) 8.2.1 R=  3,2 4 =  I = 0,8 A  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 24 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 8.2.2 POSITIVE MARKING FROM QUESTION 8.2.1. / POSITIEWE NASIEN VANAF VRAAG 8.2.1. OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 ε = (R+r)   V8 = IR  = (0,8)(8) = 0,8 (4 + 8 + 0,5) = 6,4 V = 10 V  Vext = 3,2 + 6,4 = 9,6 V OR/OF ε = (R+r)  Vint = Ir  = (0,8)(0,5)  = 0,8 (4 + 8) + 0,8 x 0,5  = 0,4 V = 10 V  ε = I(R + r) Any one/Enige een = Vext + Vint = 9,6 + 0,4  = 10 V  (4) 8.3.1 POSITIVE MARKING FROM QUESTION 8.2.2./ POSITIEWE NASIEN VANAF VRAAG 8.2.2. OPTION 1/OPSIE 1 OPTION 2 /OPSIE 2 Vint = Ir Vint = 10 - 8,8 1,2 = I(0,5)  = 1,2 V I = 2,4 A V Vint = Ir Rext = 1,2 = I(0,5)  I 8,8 I = 2,4 A = V 2,4 Iseries branch = R = 3,67 Ω (3,667) 8,8 = 12R 8+4 RP  = 0,73 A (0,733) 12  R 12R 3,67 =  IR = 2,4 - 0,73  12  R = 1,67 A (1,667) V R = 5,29 Ω (5,28)  R= IR OR/OF 8,8 =  1,67 1 1 1 = 5,27 Ω (5,28)    Rp R1 R2 1 1 1    3,67 R 12 R = 5,29 Ω (5,28)  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 25 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 8.3.2 There is a short circuit /Daar is ‘n kortsluiting.  The resistance of the connected wire is very low. / The total resistance decreases.  Die weerstand van die verbindingsdrade is baie klein. /Die totale weerstand neem af. 1  I  , current delivered by the battery is very high.  R 1 I  , stroom gelewer deur die battery is baie groot. R  Higher current produces more heat.  Hoër stroom produseer meer hitte. OR/OF Any one of the following equations can be used to explain the effect of current on heat/Enigeen van die volgende vergelykings kan gebruik word om die effek van stroom op hitte te verdudidelik: 2 2 2 2 W =I RΔt / W = Δt / W = Δt / P = R / P = / P = VI (3) R R [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 26 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1.1 Electrical to mechanical/kinetic/rotational  Elektries na meganies/kineties/rotasie (1) 9.1.2 DC/GS  (1) 9.1.3 Ensures continuous rotation of the coil.  Verseker aanhoudende rotasie van spoel. OR/OF Ensures change in direction of the current in the coil.  Verseker verandering van rigting van stroom in spoel. (1) 9.2 QUESTIONS 9.2.1 AND 9.2.2/VRAE 9.2.1 EN 9.2.2 Only penalise once if subscripts are omitted. Penaliseer slegs een keer indien onderskrifte uitgelaat is. 9.2.1 Marking criteria/Nasienkriteria:  Correct formula to calculate resistance.  Korrekte formule om weerstand te bereken.  Substitute into formula to calculate resistance.  Vervang in formule of weerstand te bereken.  Final answer/Finale antwoord: 484 to/tot 493,83 Ω  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 V2 Pave  VrmsIrms Pave  VrmsIrms Pave  rms  R 100 = 220Irms 100 = 220Irms 2202 Irms = 0,45 A (0,455) Irms = 0,45 A (0,455) 100   R R = 484 Ω  Vrms Pave  Irms 2 R Irms   R 100 = (0,45)2R  220 R = 493,83 Ω  0,45   R R = 488,89 Ω  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 27 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 9.2.2 POSITIVE MARKING FROM QUESTION 9.2.1/ POSITIEWE NASIEN VANAF VRAAG 9.2.1. Marking criteria: Nasienriglyne:  Uses power of Y in circuit (80 W) to  Gebruik drywing van Y in calculate I rms of the circuit.  stroombaan (80 W) om Iwgk te  Determines V rms across RZ in the circuit.  bereken.   Uses I rms and Vrms across RZ in the circuit to  Bepaal Vwgk oor RZ in die calculate resistance RZ.  stroombaan.   Use of any one relevant power equation.   Gebruik Iwgk en Vwgk oor RZ in  Uses RZ and 220 V to calculate X.  die stroombaan om weerstand  Final answer for X.  RZ te bereken.   Accept range:  Gebruik van enige drywing- 846,07 W to 856,03 W formule.  Gebruik RZ en 220 V om X te bereken.   Finale antwoord vir X.  Aanvaar gebied: 846,07 W tot 856,03 W For resistor Y/Vir resistor Y X for Z/X vir Z: Pave  Irms 2 R V2 X = Pave  rms  80 = 2rms (484)  R Irms = 0,407 A 220 2 =  57,08 OR/OF = 847,93 W  V2 Pave  rms R OR/OF V2 V 80  rms Irms  rms 484 R Vrms = 196,77 V 220 = 57,08 Vrms = 3,85 A Irms  R 196,77 X = Pave  Irms 2 R   484 = (3,85)2(57,08)  = 0,407 A = 846,07 W  For/Vir Z OR/OF Vrms = 220 – 196,77  V Irms  rms = 23,23 V R 220 Vrms = Irms  57,08 R = 3,85 A 23,23 [ 0,407  ] R X = Pave  VrmsIrms  R = 57,08 Ω = (220)(3,85)  = 847 W  Range/Gebied: 56,66 Ω to/tot 57,13 Ω (6) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 28 DBE/2022 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1.1 Marking criteria/Nasienkriteria: If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per word/frase. The process whereby electrons are ejected from a (metal) surface when light of suitable frequency is incident on that surface.  Die proses waartydens elektrone vrygestel word vanaf 'n (metaal) oppervlak wanneer lig van geskikte frekwensie invallend is op die oppervlak. (2) 10.1.2 For one photon/Vir een foton: NOTE/LET WEL E = hf  = (6,63 x 10-34)(1,2 x 1015)  Wo = hfo 0 2 = 7,96 x 10-19J No of e = No of photons/Hoeveelheid fotone Total energy of photons Totale energie van fotone = / Energy of one photon Energie van een foton 9 1,75 x 10 =  7,96 x 1019 = 2,2 x 109  (2,198 x 109) (4) 10.1.3 POSITIVE MARKING FOR E ONLY FROM QUESTION 10.1.2/ POSITIEWE NASIEN VIR SLEGS E VANAF VRAAG 10.1.2 E = W0 + K max 1 Any one/Enige een hf = hfo + 2 mv2max 1 7,96 x 10-19  = (6,63 x 10-34)(9,09 x 1014)  + (9,11 x 10-31)v2max  2 vmax = 6,51 x 105 m∙s-1  (5) 10.2 An atom (electron) in higher (excited) energy state/level returns to a lower energy state/level.  Energy is released as light (photons/frequencies of light are released).  'n Atoom (elektron) in ‘n hoër (opgewekte) energie toestand/vlak keer terug na ‘n laer energievlak (grondvlak). Energie word vrygestel as lig (fotone/frekwensies van lig word vrygestel). (2) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou