Physical Sciences Grade 12 Paper 1 NSC November 2021 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2021 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 This marking guidelines consists of 26 pages. Hierdie nasienriglyne bestaan uit 26 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 A  (2) 1.2 B  (2) 1.3 D  (2) 1.4 B  (2) 1.5 C  (2) 1.6 D  (2) 1.7 B or/of F (2) 1.8 A or/of V1 (2) 1.9 D  (2) 1.10 D  (2) [20] Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 2021 NSC/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 is omitted deduct 1 mark. /Indien enige van die onderstreepte sleutel woorde/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 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 /unbalanced 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/ongebalanseerde krag daarop inwerk. (2) 2.2 F Accept force diagram/ N Aanvaar kragtediagram fk • N F w fk Accept/Aanvaar N F • w fk wll w Ty labels/Aanvaarde benoemings Accepted equa Fg /pe tion/196 N / gravitational force Fw / weight / mg w eq here. Fg / Fw / gewig / mg / 196 N / gravitasiekrag FA /uat Applied force F FT /ion Toegepaste krag her Friction /(kineties)wrywing / Ff / f /18 N / Fw / fw fk (kinetic) N FN /e.Normal / Normaal / 169,74 N Notes/Aantekeninge:  Mark awarded for label and arrow, but penalise only once if arrows are omitted/Punt toegeken vir benoeming en pyltjie, maar penaliseer slegs een keer indien pyle uitgelaat is.  Do not penalise for length of arrows, drawing is not to scale. /Moenie vir die lengte van die pyltjies penaliseer nie, die tekening is nie volgens skaal nie.  Any other additional force(s) deduct 1 mark. / Enige ander addisionele krag(te) trek 1 punt af.  If force(s) do not make contact with body deduct 1 mark. /Indien krag(te) nie met die voorwerp kontak maak nie, trek 1 punt af. (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 2.3 OPTION 1/OPSIE 1 Positive up the incline/Positief opwaarts teen skuinsvlak Fnet = ma F+ fk + wll = ma NOTE/LET WEL Any one/Enige een Fnet = 0  F+ (-fk) + (-wll) = ma F - (fk + wll) = ma F = fk + w||  F - [18 + (20)(9,8)(sin30º)]  = 0  F = 116 N  OPTION 2/OPSIE 2 Positive up the incline/Positief opwaarts teen skuinsvlak Wnet = ΔEk  NOTE/LET WEL FΔxcos0º + fΔxcos180º + wΔxcos120º = 0  Wnet = 0  FΔx = 18Δx + (20)(9,8)Δx(0,5) FΔx = fΔx + wΔx(0,5)  F = 116 N  (4) 2.4 POSITIVE MARKING FROM QUESTION 2.3 / POSITIEWE NASIEN VANAF VRAAG 2.3 116 N / f + w||  Down the incline/opposite to direction of motion /Teen die helling af / in teenoorgestelde rigting van beweging ACCEPT/AANVAAR: Downwards/down/Afwaarts/af (2) 2.5 POSITIVE MARKING FROM QUESTION 2.4 / POSITIEWE NASIEN VANAF VRAAG 2.4 OPTION 1/OPSIE 1 Up the incline positive/Teen skuinsvlak op positief Fnet = ma -116 = 20a  a = -5,80 m∙s-2 vf2 = vi2 + 2aΔx  vf = vi +Δt vf = vi +Δt 0 =(2)2 + (2)(-5,8)Δx 0 = 2 + (-5,8)Δt 0 = 2 + (-5,8)Δt Δx = 0,34 m  Δt = 0,34 s Δt = 0,34 s OR/OF OR/OF FnetΔt = m(vf – vi) FnetΔt = m(vf – vi) (-116)Δt = (20)(0 – 2) (-116)Δt = (20)(0–2) Δt = 0,34 s Δt = 0,34 s Δx = viΔt + ½ aΔt2   v  vf  =(2)(0,34) + ½(-5,8) (0,34)2 Δx   i  Δt  = 0,34 m   2  2  0  0,34   2  = 0,34 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 1/OPSIE 1 Down the incline positive /Teen skuinsvlak af positief Fnet = ma 116 = 20a  a = 5,80 m∙s-2 vf2 = vi2 + 2aΔx  vf = vi + aΔt vf = vi + aΔt 0 =(-2)2 + (2)(5,8)Δx 0 = -2 + (5,8)Δt 0 = -2 + (5,8)Δt Δx = -0,34 m Δt = 0,34 s Δt = 0,34 s Distance = 0,34 m  OR/OF OR/OF Afstand FnetΔt = m(vf – vi) FnetΔt = m(vf – vi) (116)Δt = (20)(0 – (-2)) (116)Δt =(20)(0-(-2)) Δt = 0,34 s Δt = 0,34 s  v  vf  Δx = viΔt + ½ aΔt2  Δx   i  Δt  = (-2)(0,34) + ½(5,8)(0,34)2  2  = -0,34 m  2  0  0,34  Distance/Afstand = 0,34 m   2  = -0,34 m  Distance/Afstand = 0,34 m  OPTION 2/OPSIE 2 Wnet = ∆EK Fnet∆xcosθ = ½m(vf2 - vi2) Any one/Enige een (116)∆xcos180º = ½(20)(02 – 22)  ∆x = 0,34 m  OPTION 3/OPSIE 3 Wnet = ∆EK Wf + W wll = ½mvf2 - ½mvi2 Any one/Enige een f∆xcosθ + (mgsin30º)∆xcosθ = ½m(vf2 - vi2) (18)∆xcos180º + (20)(9,8)sin30º∆xcos180º= ½(20)(02 – 22)  ∆x = 0,34 m  OPTION 4/OPSIE 4 Wnet = ∆EK Wf + W w = ½mvf2 - ½mvi2 Any one/Enige een f∆xcosθ + mg∆xcos120° = ½m(vf2 - vi2) (18)∆xcos180º + (20)(9,8)∆xcos120º= ½(20)(02 – 22)  ∆x = 0,34 m  OPTION 5/OPSIE 5 Wnc = ΔEp + ΔEk Any one/Enige een f∆xcosθ = mg(hf - hi) + ½m(vf2 – vi2) 18∆xcos180º = 20(9,8)∆x + ½(20)(0 – 22)  2 -18∆x = 196∆xsin30º - 40 ∆x = 0,34 m  (4) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 No/Nee  ANY ONE/ENIGE EEN:  Gravitational force is not the only force acting on the balloon. /There are other forces acting on the balloon.  Gravitasiekrag is nie die enigste krag wat op die ballon inwerk nie./Daar is ander kragte wat op die ballon inwerk.  Its acceleration is not 9,8 m∙s-2/is zero. Sy versnelling is nie 9,8 m∙s-2/is nul.  It has constant velocity/no acceleration. Dit het 'n konstante snelheid/geen versnelling nie. (2) 3.2.1 OPTION 1/OPSIE 1 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf2 = vi2 + 2aΔy  vf2 = vi2 + 2aΔy  (-62,68)2 = vi2 + 2(-9,8)(-200)  (62,68)2 = vi2 + 2(9,8)(200)  vi = 2,96 m∙s-1  vi = -2,96 m∙s-1 = 2,96 m∙s-1  OPTION 2/OPSIE 2 (Emech/meg)200 m = (Emech/meg)bottom/onder (EP +EK)200 m = (EP +EK)bottom/onder Any one/Enige een (mgh + ½ mv2)200 m = (mgh + ½ mv2)bottom/onder m(9,8)(200) + ½m(v2) = 0 + ½m(62,68) 2  vi = 2,96 m∙s-1  NOTE/LET WEL Mass may be omitted during substitution. Massa mag uitgelaat word tydens vervanging. OPTION 3/OPSIE 3 Wnc = ΔEp + ΔEk Any one/Enige een 0 = mg(hf - hi)+ ½m(vf2 – vi2) 0 = m(9,8)(0 - 200) + ½m(62,68 – vi2)  2 vi = 2,96 m∙s-1  NOTE/LET WEL Mass may be omitted during substitution. Massa mag uitgelaat word tydens vervanging. OPTION 4/OPSIE 4 Wnet = ΔEk FnetΔxcos  = ½m(vf2 – vi2) Any one/Enige een mgΔxcos  = ½m(vf2 – vi2) m(9,8)(200) = + ½m(62,682 – vi2)  vi = 2,96 m∙s-1  NOTE/LET WEL Mass may be omitted during substitution. Massa mag uitgelaat word tydens vervanging. (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2021 NSC/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 Δt of stone A  Formule om Δt van klip A te bereken  Substitution to calculate Δt of stone A  Vervanging om Δt van klip A te bereken  Final answer/Finale antwoord: 6,70 s  Accept/Aanvaar: (6,69 to/tot 6,7) NOTE: The calculation of Δt for A might be split up into two parts. LET WEL: Die berekening van Δt vir A kan in twee dele opgedeel word. OPTION 1/OPSIE 1 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf = vi + aΔt  vf = vi + aΔt  -62,68 = 2,96 + (-9,8)Δt  62,68 = -2,96 + 9,8Δt  Δt = 6,70 s  (6,698) Δt = 6,70 s  (6,698) OPTION 2/OPSIE 2 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF Δy = viΔt + ½ aΔt2  Δy = viΔt + ½ aΔt2  -200 = (2,96)Δt + ½ (-9,8)Δt2  200 = (-2,96) Δt + ½ (9,8)Δt2  Δt = 6,70 s  (6,697) Δt = 6,70 s  (6,697) OPTION 3/OPSIE 3 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF  v  vf   v  vf  Δy   i  Δt  Δy   i  Δt   2   2    2,96  ( 62,68)    2,96  62,68  -200    Δt  200    Δt   2   2  Δt = 6,70 s  (6,698) Δt = 6,70 s  (6,698) OPTION 4/OPSIE 4 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF From 200 m upwards: From 200 m upwards: Vanaf 200 m opwaarts: Vanaf 200 m opwaarts: vf = vi + aΔt  v = vi + aΔt  0 = 2,96 + (-9,8)Δt  0 = -2,96 + (9,8)Δt  Δt = 0,3 s (0,302) Δt = 0,3 s (0,302) From max h downwards: From max h downwards: Vanaf maks h afwaarts: Vanaf maks h afwaarts: vf = vi + aΔt vf = vi + aΔt -62,68 = 0 + (-9,8)Δt 62,68 = 0 + (9,8)Δt Δt = 6,40 s (6,369) Δt = 6,40 s (6,369) tA = 0,3 + 6,40 = 6,7 s  tA = 0,3 + 6,40 = 6,7 s  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF From 200 m upwards: From 200 m upwards: Vanaf 200 m opwaarts: Vanaf 200 m opwaarts: vf = vi + aΔt  vf = vi + aΔt  0 = 2,96 + (-9,8)Δt  0 = -2,96 + (9,8)Δt  Δt = 0,3 s (0,302) Δt = 0,3 s (0,302) From 200 m downwards: From 200 m downwards: Vanaf 200 m afwaarts: Vanaf 200 m afwaarts: vf = vi + aΔt vf = vi + aΔt -62,68 = -2,96 + (-9,8)Δt 62,68 = 2,96 + (9,8)Δt Δt = 6,09 s (6,094) Δt = 6,09 s (6,094) tA = 2(0,3) + 6,09 = 6,69 s  tA = 2(0,3) + 6,09 = 6,69 s  OPTION 6/OPSIE 6 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF FnetΔt = m(vf – vi)  FnetΔt = m(vf – vi)  mgΔt = m(vf – vi) mgΔt = m(vf – vi) gΔt = vf - vi gΔt = vf - vi (-9,8)Δt = (-62,68) – (2,96)  (9,8)Δt = 62,68 – (-2,96)  Δt = 6,69 s  Δt = 6,69 s  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 3.2.3 POSITIVE MARKING FROM QUESTION 3.2.1 and QUESTION 3.2.2/ POSITIEWE NASIEN VANAF VRAAG 3.2.1 en VRAAG 3.2.2 Marking criteria/Nasienkriteria  Formula to calculate Δy of stone B  Formule om Δy van klip B te bereken  Substitution of t = 1,7 s  (tA – 5) Vervanging van t = 1,7 s (tA – 5)  Substitution to calculate Δy of stone B  Vervanging om Δy van klip B te bereken  Substitution to calculate Δy of balloon  Vervanging om Δy van ballon te bereken  Calculating distance between balloon and stone B  Berekening van afstand tussen ballon en klip B  Final answer/Finale antwoord: 14,16 m  (14,11 to/tot 14,16) OPTION 1/OPSIE 1 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF Stone B/Klip B: Stone B/Klip B: Δy = viΔt + ½ aΔt2   Δy = viΔt + ½ aΔt2  = 2,96(6,7 - 5)+ ½(-9,8)(6,7- 5)2  = -2,96(6,7 - 5) + ½(9,8)(6,7 - 5)2  = -9,13 m (-9,09 to/tot -9,13) = 9,13 m (9,09 to/tot 9,13) Distance travelled by stone B: 9,13 m Distance travelled by stone B: 9,13 m Afstand afgelê deur klip B: 9,13 m Afstand afgelê deur klip B: 9,13 m Hot-air balloon/Lugballon Hot-air balloon/Lugballon Δy = viΔt + ½ aΔt2 Δy = viΔt + ½ aΔt2 = 2,96(6,7 - 5)  + 0 = -2,96(6,7 - 5)  + 0 = 5,03 m = -5,03 m Distance travelled by hot-air balloon/ Distance travelled by hot-air balloon/ Afstand afgelê deur lugballon: 5,03 m Afstand afgelê deur lugballon: 5,03 m Distance between hot-air balloon and Distance between hot-air balloon and stone B/Afstand tussen lugballon en stone B/Afstand tussen lugballon en klip B = 9,13 + 5,03  klip B = 9,13 + 5,03  = 14,16 m  (14,11 - 14,16) = 14,16 m  (14,11 - 14,16) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF Stone B/Klip B: Stone B/Klip B: vf = vi + aΔt vf = vi + aΔt   = 2,96 + (-9,8)(6,70 – 5) = -2,96 + (9,8)(6,70 – 5) = - 13,7 m∙s-1 = 13,7 m∙s-1 vf = vi2 + 2aΔy  2 vf = vi2 + 2aΔy  2 (-13,7)2 = (2,96)2 + 2(-9,8)Δy  (13,7)2 = (-2,96)2 + 2(9,8)Δy  Δy = -9,13 m Δy = 9,13 m Distance travelled by stone B: 9,13 m Distance travelled by stone B: 9,13 m Afstand afgelê deur klip B: 9,13 m Afstand afgelê deur klip B: 9,13 m Hot-air balloon/Lugballon Hot-air balloon/Lugballon Δy = viΔt + ½ aΔt2 Δy = viΔt + ½ aΔt2 = -2,96(6,70 - 5) + 0  = -2,96(6,70 - 5) + 0  = -5,03 m = -5,03 m Distance travelled by hot-air balloon/ Distance travelled by hot-air balloon/ Afstand afgelê deur lugballon: 5,03 m Afstand afgelê deur lugballon: 5,03 m Distance between hot-air balloon and Distance between hot-air balloon and stone B/Afstand afgelê deur lugballon stone B/Afstand afgelê deur lugballon en klip B = 9,13+ 5,03  en klip B = 9,13+ 5,03  = 14,16 m  (14,11 - 14,16) = 14,16 m  (14,11 - 14,16) OPTION 3/OPSIE 3 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF Stone B/Klip B: Stone B/Klip B: vf = vi + aΔt vf = vi + aΔt = 2,96 + (-9,8)(6,70  – 5)  5) = -2,96 + (9,8)(6,70 – = - 13,7 m∙s-1 = 13,7 m∙s-1  v  vf   v  vf  Δy   i  Δt  Δy   i  Δt   2   2    2,96  ( 13,7)    2,96  (13,7)    (6,70 – 5)    (6,70 – 5)   2   2  = -9,13 m = 9,13 m Distance travelled by stone B: 9,13 m Distance travelled by stone B: 9,13 m Afstand afgelê deur klip B: 9,13 m Afstand afgelê deur klip B: 9,13 m Hot-air balloon/Lugballon Hot-air balloon/Lugballon Δy  v i Δt  21 at 2 Δy  v i Δt  21 at 2 = 2,96(6,70 - 5) + 0  = -2,96(6,70 - 5) + 0  = 5,03 m = -5,03 m Distance travelled by hot-air balloon/ Distance travelled by hot-air balloon/ Afstand afgelê deur lugballon: 5,03 m Afstand afgelê deur lugballon: 5,03 m Distance between hot-air balloon and Distance between hot-air balloon and stone B/Afstand afgelê deur lugballon stone B/Afstand afgelê deur lugballon en klip B = 9,13 + 5,03  en klip B = 9,13+ 5,03  = 14,16 m  (14,11 - 14,16) = 14,16 m  (14,11 - 14,16) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 UPWARDS POSITIVE/ DOWNWARDS POSITIVE/ OPWAARTS POSITIEF: AFWAARTS POSITIEF: Stone B/Klip B: Stone B/Klip B: vf = vi + aΔt vf = vi + aΔt   = 2,96 + (-9,8)(6,70 – 5) = -2,96 + (9,8)(6,70 – 5) = - 13,7 m∙s-1 = 13,7 m∙s-13 Balloon's height after 5 s: 214,8 m Balloon's height after 5 s: 214,8 m Ballon se hoogte na 5 s: 214,8 m Ballon se hoogte na 5 s: 214,8 m Emech/meg)214,8 m = (Emech/meg)1,7 s (Emech/meg)214,8 m = (Emech/meg)1,7 s (EP +EK)214,8 m = (EP +EK)1,7 s  (EP +EK)214,8 m = (EP +EK)1,7 s  (mgh+½ mv2) = (mgh+½ mv2)1,7s (mgh +½ mv2) = (mgh + ½ mv2)1,7s (9,8)(214,9)+½(2,96)2 = (9,8)(214,8)+½(2,96)2 = (9,8)h+½(13,7)2  (9,8)h+½(13,7)2   h = 205,67 m  h = 205,67 m Distance travelled by stone B/ Distance travelled by stone B/ Afstand afgelê deur klip B: Afstand afgelê deur klip B: 214,8 – 205,67 = 9,13 m 214,8 – 205,67 = 9,13 m Hot-air balloon/Lugballon Hot-air balloon/Lugballon Δy  v i Δt  21 at 2 Δy  v i Δt  21 at 2 = 2,96(6,70 - 5)  + 0 = -2,96(6,70 - 5)  + 0 = 5,03 m = -5,03 m Distance travelled by hot-air balloon/ Distance travelled by hot-air balloon/ Afstand afgelê deur lugballon: 5,03 m Afstand afgelê deur lugballon: 5,03 m Distance between hot-air balloon and Distance between hot-air balloon and stone B/Afstand tussen lugballon en stone B/Afstand tussen lugballon en klip B: 9,13 + 5,03  = 14,16 m  klip B: 9,13 + 5,03  = 14,16 m  (14,11 to/tot 14,16) (14,11 to/tot 14,16) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF Stone B/Klip B: Stone B/Klip B: vf = vi + aΔt vf = vi + aΔt   = 2,96 + (-9,8)(6,70 – 5) = -2,96 + (9,8)(6,70 – 5) = - 13,7 m∙s-1 = 13,7 m∙s-1 Wnet = ∆EK  Wnet = ∆EK  Fnet∆xcosθ = ½mvf2 - ½mvi2 Fnet∆xcosθ = ½mvf2 - ½mvi2 = ½m(vf2- vi2) = ½m(vf2- vi2) (9,8)∆hcos 0° = ½(13,72 – 2,962)  (9,8)∆hcos0° = ½(13,72 – 2,962)  ∆h = 9,13 m ∆h = 9,13 m Distance travelled by stone B/ Distance travelled by stone B/ Afstand afgelê deur klip B: 9,13 m Afstand afgelê deur klip B: 9,13 m Hot-air balloon/Lugballon Hot-air balloon/Lugballon Δy  v i Δt  21 at 2 Δy  v i Δt  21 at 2 = 2,96(6,70 - 5)  + 0 = -2,96(6,70 - 5)  + 0 = 5,03 m = -5,03 m Distance travelled by hot-air balloon/ Distance travelled by hot-air balloon/ Afstand afgelê deur lugballon: 5,03 m Afstand afgelê deur lugballon: 5,03 m Distance between hot-air balloon and Distance between hot-air balloon and stone B/Afstand tussen lugballon en stone B/Afstand tussen lugballon en klip B: 9,13 + 5,03  = 14,16 m  klip B: 9,13+ 5,03  = 14,16 m  (14,11 to/tot 14,16) (14,11 to/tot 14,16) OPTION 6/OPSIE 6 Using relative velocities/Deur relatiewe snelhede te gebruik: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF Δy  v i Δt  2 at  1 2 Δy  v i Δt  21 at 2    1    1  = (2,96 - 2,96)(1,7) + (-9,8)(1,7)2 = (2,96 - 2,96)(1,7) + (9,8)(1,7)2 2 2 = -14,16 m = 14,16 m  Distance between hot-air balloon and stone B/Afstand tussen lugballon en klip B: 14,16 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 7/OPSIE 7 UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF Δy = viΔt + ½ aΔt2   = (2,96)(1,7) + ½ (-9,8)(1,7)2  = -9,13 m Distance travelled by stone B: 9,13 m Afstand afgelê deur klip B: 9,13 m DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF Δy = viΔt + ½ aΔt2  = (-2,96)(1,7) + ½(9,8)(1,7)2  = 9,13 m Height of stone B from the ground = 200 + 14,8 – 9,13 = 205,63 m Hoogte van klip B vanaf die grond: Height of balloon from the ground = 200 + (6,7)(2,96) = 219,83 m Hoogte van ballon vanaf die grond: Distance between B and the balloon = 219,83 – 205,63  = 14,16 m Afstand tussen B en die ballon: (6) 3.3 UPWARDS POSITIVE DOWNWARDS POSITIVE OPWAARTS POSITIEF AFWAARTS POSITIEF Balloon/ Position/Posisie (m) ballon 0 Position/Posisie (m) Time/Tyd (s) A A Balloon/ Ballon 0 Time/tyd (s) Criteria for graph/Kriteria vir grafiek Correct shape for stone A not starting from 0 m./Korrekte vorm vir klip A  wat nie by 0 m begin nie. Correct shape and initial position for hot-air balloon. /Korrekte vorm en  aanvanklike posisie vir lugballon. Gradient for hot-air balloon is higher than that of stone A until stone A reaches the maximum height./Gradiënt vir lugballon is groter as dié vir klip  A totdat klip A sy maksimum hoogte bereik. Both graphs starting at the same position and ending at the same time. /  Beide grafieke begin by dieselfde posisie en eindig by dieselfde tyd. (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark. /Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. NOTE/LET WEL If “total” is omitted: minus 1 mark / Indien “totaal” uitgelaat is: minus 1 punt A collision in which both the total momentum and total kinetic energy are conserved. 'n Botsing waar die totale momentum en die totale kinetiese energie behoue bly. (2) 4.2 OPTION 1/OPSIE 1 ∑EKi = ∑EKf ½m1v i21 + ½m2v i22 = ½m1v 2f 1 + ½m2v 2f 2 Any one/Enige een 2 2 2 2 ½mxv + ½myv ix iy = ½mxv fx + ½myv fy 2 2 ½(10)(2) + ½(2)v  = 0 + 36  iy vy = ± 4 m∙s-1 vy = 4 m∙s-1 west/wes ACCEPT/AANVAAR: left/links OPTION 2/OPSIE 2 EKi = ½ mYvf2 36 = ½ (2) vf2 vf = 6 m∙s-1 ∑pi = ∑pf m1v1i + m2v2i = m1v1f + m2v2f Any one/Enige een mxvxi + myvyi = mXvxf + myvyf (10)(2) + (2)vy  = 0 + (2)(6)  vy = -4 m∙s-1 vy = 4 m∙s-1 west/wesACCEPT/AANVAAR: left/links OPTION 3/OPSIE 3 EKi = ½ mYvf2 36 = ½ (2) vf2 vf = 6 m∙s-1 ΔpX = -ΔpY Any one/Enige een mx(vXf - vXi) = - mY(vYf – vYi) (10)(0 – 2) = -(2)(6 – vY)  vYf = -4 m∙s-1 vy = 4 m∙s-1 west/wes ACCEPT/AANVAAR: left/links (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 4.3 POSITIVE MARKING FROM QUESTION 4.2 FOR Y; OPTIONS 1, 3 and 6 POSITIEWE NASIEN VANAF VRAAG 4.2 VIR Y; OPSIES 1, 3 en 6 OPTION 1/OPSIE 1 EAST POSITIVE/OOS POSITIEF: WEST POSITIVE/WES POSITIEF: For Y/Vir Y: For Y/Vir Y: FnetΔt = Δp Any one/ FnetΔt = Δp Any one/ FnetΔt = m(vf – vi) Enige een FnetΔt = m(vf – vi) Enige een Fnet (0,1) = 2{6 - (-4)}  Fnet (0,1) = 2(-6 - 4)  Fnet = 200 N  Fnet = -200 N Fnet = 200 N  OPTION 2/OPSIE 2 EAST POSITIVE/OOS POSITIEF: WEST POSITIVE/WES POSITIEF For X/Vir X: For X/Vir X: FnetΔt = Δp Any one/ FnetΔt = Δp Any one/ FnetΔt = m(vf – vi) Enige een FnetΔt = m(vf – vi) Enige een Fnet (0,1) = 10(0 - 2)  Fnet (0,1) = 10{0 - (-2)}  Fnet = -200 N Fnet = 200 N  Fnet = 200 N  OPTION 3/OPSIE 3 EAST POSITIVE/OOS POSITIEF: WEST POSITIVE/WES POSITIEF For Y/Vir Y: For Y/Vir Y: vf = vi + aΔt vf = vi + aΔt 6 = -4 + a(0,1) -6 = 4 + a(0,1) a = 100 m∙s-2 a = -100 m∙s-2 Fnet = ma  Fnet = ma  = 2(100)  = 2(-100)  = 200 N  = -200 N Fnet = 200 N  OPTION 4/OPSIE 4 EAST POSITIVE/OOS POSITIEF: WEST POSITIVE/WES POSITIEF For X/Vir X: For X/Vir X: vf = vi + aΔt vf = vi + aΔt 0 = 2 + a(0,1) 0 = -2 + a(0,1) -2 a = -20 m∙s a = 20 m∙s-2 Fnet = ma  Fnet = ma  = 10(-20)  = 10(20)  = -200 N Fnet = 200 N  Fnet = 200 N  OPTION 5/OPSIE 5 EAST POSITIVE/OOS POSITIEF: For X/Vir X:  v  vi  vf = vi + aΔt vf2 = vi2 + 2aΔx Δx =  f  Δt  2  0 2 0 = 2 + a(0,1) 0 = (2)2 + 2(-20)Δx =  (0,1)  2  a = -20 m∙s-2 Δx = 0,10 m = 0,10 m Wnet = ΔEk  FnetΔxcosθ = ½ m(vf2 – vi2) Fnet(0,1)cos180º = ½ (10)(02 – 22)  Fnet = 200 N  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 WEST POSITIVE/WES POSITIEF: For X/Vir X:  v  vi  vf = vi + aΔt vf2 = vi2 + 2aΔx Δx =  f  Δt  2   0  ( 2)  0 = -2 + a(0,1) 0 = (-2)2 + 2(20)Δx =   (0,1)  2  a = 20 m∙s-2 Δx = -0,10 m = - 0,10 m Wnet = ΔEk  FnetΔxcosθ = ½ m(vf2 – vi2) Fnet(0,1)cos180º = ½ (10)(02 – 22)  Fnet = 200 N  OPTION 6/OPSIE 6 EAST POSITIVE/OOS POSITIEF: For Y/Vir Y:  v  vi  vf = vi + aΔt vf2 = vi2 + 2aΔx Δx =  f  Δt  2  64 6 = -4 + a(0,1) (6)2 = (-4)2 + 2(100)Δx =  (0,1)  2  a = 100 m∙s-2 Δx = 0,10 m = 0,10 m Wnet = ΔEk  FnetΔxcosθ = ½ m(vf2 – vi2) Fnet(0,1)cos0° = ½ (2)(62 – (-4)2)  Fnet = 200 N  OPTION 6/OPSIE 6 WEST POSITIVE/WES POSITIEF: For Y/Vir Y:  v  vi  vf = vi + aΔt vf2 = vi2 + 2aΔx Δx =  f  Δt  2  64 -6 = 4 + a(0,1) (-6)2 = (4)2 + 2(-100)Δx =  (0,1)  2  a = -100 m∙s-2 Δx = -0,10 m = -0,10 m Wnet = ΔEk  FnetΔxcosθ = ½ m(vf2 – vi2) Fnet(0,1)cos0° = ½ (2)((-6)2 – (4)2)  Fnet = 200 N  (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/November 2021 NSC/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 is omitted deduct 1 mark. /Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. ACCEPT/AANVAAR For isolated system:  Closed system/Geslote sisteem.  Only conservative forces act on the system/Slegs konserwatiewe kragte werk in op die sisteem.  No external forces act on system/Geen eksterne kragte werk in op die sisteem. The total mechanical energy in an isolated system remains constant / the same.  Die totale meganiese energie in 'n geïsoleerde sisteem bly konstant / dieselfde. OR/OF The sum of the kinetic and gravitational potential energies in an isolated system remains constant/the same. Die som van die kinetiese en gravitasie potensiële energie in 'n geïsoleerde/geslote sisteem bly konstant/dieselfde. (2) 5.2 NOTE/LET WEL  Mass may be omitted during substitution. /Massa mag uitgelaat word tydens vervanging.  If equations of motion are used. Max 1/3 for correct answer. / Indien bewegingsvergelykings gebruik word. Maks 1/3 vir korrekte antwoord. OPTION 1/OPSIE1 E P/mech top/meg bo = E Q/mech ground /meg grond (Ep + Ek)P/top/bo = (Ep + Ek)Q/bottom/onder Any one/Enige een 2 2 (mgh + ½mv )P/top/bo = (mgh + ½mv )Q/bottom/onder (2)(9,8)(5) + 0 = 0 + ½(2)v 2f  vf = 9,90 m∙s-1  (9,899) OPTION 2/OPSIE2 ∆Ep + ∆EK = 0 Any one/Enige een (mghf - mghi) + ½m(vf2 – vi2) = 0 0 - (2)(9,8)(5) + ½(2)(vf2 – 0)  = 0 vf = 9,90 m∙s-1  (9,899) (3) 5.3 POSITIVE MARKING FROM QUESTION 5.2. POSITIEWE NASIEN VANAF VRAAG 5.2. OPTION 1/OPSIE 1 Wnet = ∆EK Wf = ½mvf2 - ½mvi2 Any one/Enige een WN + Wf + W w= ½mvf2 - ½mvi2 f∆xcosθ = ½m(vf2- vi2) f(10)cos180° = ½(2)(42 – 9,902)  f = 8,2 N  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2 Wnc = ∆EK + ∆Ep Wf = ½mvf2 - ½mvi2 Any one/Enige een 2 2 WN + Wf = ½mvf - ½mvi f∆xcosθ = ½m(vf2 – vi2) + mg(hf – hi) f(10)cos180° = ½(2)(42 – 9,902) + 0  f = 8,2 N  (4) 5.4 LEFT NEGATIVE/LINKS NEGATIEF Fnet∆t = ∆p ACCEPT/AANVAAR Fnet∆t = mvf - mvi Any one/Enige een Impulse/Impuls = mΔv Fnet∆t = m(vf - vi) -14 = 2(vf – 4)  vf = -3 m∙s-1 ∆EK = ½mvf2 - ½mvi2  = ½(2)[(-3)2 - 42]  Do not penalise if +3 is substituted. = -7 J  Moenie penaliseer indien +3 vervang is. ACCEPT/AANVAAR ∆EK = ½mvf2 - ½mvi2  = ½(2)[(0)2 – (-3)2]  Do not penalise if +3 is substituted. = -9 J  Moenie penaliseer indien +3 vervang is. RIGHT NEGATIVE/REGS NEGATIEF Fnet∆t = ∆p Fnet∆t = mvf - mvi Any one/Enige een ACCEPT/AANVAAR Fnet∆t = m(vf - vi) Impulse/Impuls = mΔv 14 = 2(vf – (-4))  vf = 3 m∙s-1 ∆EK = ½mvf2 - ½mvi2  Do not penalise if +4 is substituted. = ½(2)[(3)2 – (-4)2]  Moenie penaliseer indien +4 vervang is. = -7 J  ACCEPT/AANVAAR ∆EK = ½mvf2 - ½mvi2  Do not penalise if +3 is substituted. = ½(2)[(0)2 – (-3)2]  Moenie penaliseer indien +3 vervang is. = -9 J  (5) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 v = fλ  340 = 680λ  λ = 0,5 m  (3) 6.2 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutel woorde/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 listener have different velocities relative to the medium of sound propagation.  Die verandering in frekwensie/toonhoogte/golflengte van die klank waargeneem deur 'n luisteraar omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium van klank voortplanting het. OR/OF 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) 6.3.1 Decreased/Afgeneem  (1) 6.3.2 Increased/Toegeneem  (1) 6.4 POSITIVE MARKING FROM QUESTION 6.1 / POSITIEWE NASIEN VANAF VRAAG 6.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 v  vL v v  vL v fL  fs OR fL  fs  fL  fs OR fL  fs  v  vs v  vs v  vs v  vs v v v +0 fL  = ( ) f λL λL v - vs s  340 =  0,5  0,05   340 340  0  340 =   680  = 0,5  0,05  340  v s  0,45 = 755,56 Hz 340  340  0  =   680  340  0   0,45  340  v s  755,56    680   340  v s  vs = 34 m∙s-1  (33,67 - 34,04) vs = 34 m∙s-1  (33,67 - 34,04) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 20 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 v  vL v v  vL v fL  fs OR fL  fs  fL  fs OR fL  fs  v  vs v  vs v  vs v  vs v  v 0  v v1 = v2   fsλ1 = fLλ2  L  v - v s   s (600)(0,5) = fL(0,45)  1  v 0  1 fL = 755,56 Hz     L  v - v s   s     340  0  1  340  0  1  755,56    680  =    340  v s  0,5  0,05  340  v s  0,5 1  340  0  1 vs = 34 m∙s-1  (33,67 - 34,04) =   0,45  340  v s  0,5 vs = 34 m∙s-1  (33,67 - 34,04) (5) [12] QUESTION 7/VRAAG 7 7.1.1 Added/Toegevoeg  (1) 7.1.2 NOTE/LET WEL Ignore signs of the charges./ Ignoreer tekens van die ladings. Q n=  qe  1,95 x 10 6 =   1,6 x 10 19 = 1,22 x 1013  (1,21875 x 1013) (3) 7.1.3 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The (electrostatic) force experienced per unit positive charge placed at that point. Die (elektrostatiese) krag per eenheid positiewe lading wat by die punt geplaas is. NOTE/LET WEL (1 mark for:/1 punt vir:) An electric field is a region of space in which an electric charge experiences a force. 'n Gebied in die ruimte waarin 'n elektriese lading 'n krag ondervind. (2) 7.1.4 kQ E  r2 (9  109 )(1,95  10 6 ) =  (0,5)2 = 7,02 x 104 N·C-1  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 21 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 7.2 OPTION 1/OPSIE 1 Marking criteria/Nasienkriteria:  Coulomb's Law formula/Coulomb se formule  kQ1Q 2  Correct substitution for Fq1 OR Fq2 into  r2 kQ1Q 2 Korrekte substitusie van Fq1OF Fq2 in r2  Correct substitution of 1,38 N for F(net) /Korrekte substitusie van 1,38 N vir F(net)  Subtracting (vector addition) electrostatic forces /Aftrek (vektoraddisie) van elektrostatiese kragte   Final answer/Finale antwoord: 1,11 x 10-7 C  (1,106 x 10-7 C) FE(net) = Fq2 + Fq1  kQ1Q2   kQ1Q2  1,38 =   +   r2   r2   (9  109 )(1,95  106 )q2    (9  109 )(1,95  10 6 )q2  1,38 =   2  +   2   ( 0,03)   ( 0,05)  q2 = 1,11 x 10-7 C  (1,106 x 10-7 C) OPTION 2/OPSIE 2 Marking criteria/Nasienkriteria: kQ  E  r2  Correct substitution of 7,08 x 105 N·C-1/Korrekte substitusie van 7,08 x105 N·C-1 kQ 2  Correct substitution for Eq1 OR Eq2 into  r2 kQ 2 Korrekte substitusie van Eq1 OF Eq2 in 2 r  Subtracting electric fields/Aftrek van elektriese velde   Final answer/Finale antwoord: 1,11 x 10-7 C  (1,106 x 10-7 C) F 1,38 E = q 1,95  10 6 = 7,08 x 105 N∙C-1 (707692,30) Enet = Eq2 + Eq1   kQ   kQ  7,08 x 105 =   2 2  +   2 1    r   r   (9  109 )q2   (9  109 )q1  =   2  +   2    ( 0,03)   ( 0,05)  q2 = 1,11 x 10-7 C  (1,106 x 10-7 C)  (5) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 22 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1.1 12 V  (1) 8.1.2 0 (V)  (1) 8.2 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The rate at which work is done or energy is expended/transferred. Die tempo waarteen arbeid verrig word of energie oorgedra word. (2) 8.3 OPTION 1/OPSIE 1 P = I2R  5,76 = (1,22)R  R=4Ω OPTION 2/OPSIE 2 P = VI 5,76 = V(1,2) V = 4,8 V V2 V = IR  P=  4,8 = (1,2)R  R R=4Ω ( 4,8) 2 5,76 =  R R=4Ω (3) 8.4 POSITIVE MARKING FROM QUESTION 8.3 POSITIEWE NASIEN VANAF VRAAG 8.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 1 1 1 R 1R 2   Rp  Rp R1 R2 R1  R 2 1 1 1 (6)(8,4)    Rp   Rp 6 8,4 6  8,4 Rp = 3,5 Ω Rp = 3,5 Ω RT = 3,5 + 4  RT = 3,5 + 4 = 7,5 Ω  = 7,5 Ω  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 23 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 8.5 POSITIVE MARKING FROM QUESTION 8.3 POSITIEWE NASIEN VANAF VRAAG 8.3 CALCULATE VP/BEREKEN VP CALCULATE V2/BEREKEN V2 Marking criteria/Nasienkriteria Marking criteria/Nasienkriteria  Formula/Formule:  Substitution to calculate Ibranch or ratio V = IR  of Rbranch/Vervanging om Itak of  Substitution to calculate Vp / verhouding van Rtak te bereken.  Vervanging om Vp te bereken.  Substitution to calculate V2 / Vervanging om V2 te bereken.  Final Answer/Finale antwoord: 3 V OPTION 1/OPSIE 1 V I= Vp = IR R = (1,2)(3,5)  4,2 = 4,2 V =  8,4 OPTION 2/OPSIE 2 = 0,5 A Px = VI 5,76 = V(1,2) V2 = IR  Vx = 4,8 V = (0,5)(6)  8,4 I6 Ω = 1,2 =3V 14,4 = 0,7 A V6 Ω = IR OR/OF = (0,7)(6)  = 4,2 V OPTION 3/OPSIE 3 R2,4 : R6 = 2,4 : 6  ε = I(R + r) = 2:5 12 = 1,2(7,5 + r) V 2,4 : V 6 = 1,2 : 3  r = 2,5 Ω V2 = 3 V  Vp = 12 – 1,2(2,5 + 4) = 4,2 V CALCULATION OF I8,4Ω AND V2/BEREKENING VAN I8,4Ω EN V2 OPTION 4/OPSIE 4  6   3,5  I8,4Ω =   (1,2) OR/OF   (1,2)  14,4   8,4  = 0,5 A  V2 = IR  = (0,5)(6)  =3V OPTION 5/OPSIE 5 Vx = IR Vext = IRext = (1,2)(4) = (1,2)(7,5) = 4,8 V =9V Vp = 9 – 4,8 = 4,2 V V8,4Ω = IR 4,2 = I(8,4) I = 0,5 A V2 = IR  = (0,5)(6)  =3V (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 24 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 8.6 Decreases/Neem af  Total resistance decreases. / Totale weerstand neem af.  Current increases. /Stroom neem toe.  Vinternal /Internal voltage (“lost volts”) increases. /Interne potensiaalverskil neem toe.  Vexternal /external voltage decreases. /Eksterne potensiaalverskil neem af. NOTE/LET WEL Do not penalise if “total” is omitted. / Moenie penaliseer indien “totaal” uitgelaat is nie. (4) [19] QUESTION 9/VRAAG 9 9. 1 Slip rings/Sleepringe  ACCEPT/AANVAAR Split ring/slip ring commutator /splitring/sleepring kommutator (1) 9. 2 Y to/na X  (2) 9.3 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The AC potential difference which dissipates the same amount of energy as an equivalent DC potential difference. Die WS-potensiaalverskil wat dieselfde hoeveelheid energie verbruik as die ekwivalente/soortgelyke GS-potensiaalverskil. ACCEPT/AANVAAR The DC potential difference which dissipates the same amount of energy as an equivalent AC potential difference. Die GS-potensiaalverskil wat dieselfde hoeveelheid energie verbruik as die ekwivalente/soortgelyke WS-potensiaalverskil. (2) 9.4 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 V V Imax/maks = max/maks V2 Vrms/wgk = max /maks Pave = rms 2 R R 100 100 100 2 =  =  2 25 = 2 = 200 W = 70,71 V =4A 25 Vrms /w gk I Pave = VrmsIrms  Irms/wgk =  Irms/wgk = max /maks  R  100  2 200 =   Irms  70,71 4  2 =  =  25 Irms = 2,83 A  2 = 2,83 A  = 2,83 A  ACCEPT/AANVAAR If subscipts omitted in V = IR Indien onderskrifte uitgelaat is in V = IR (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 25 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 9.5 POSITIVE MARKING FROM QUESTION 9.4 / POSITIEWE NASIEN VANAF VRAAG 9.4 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 2 Vrms/w gk Pave  VrmsIrms  Pave/ gem  Irms/ 2 w gkR  Pave/ gem   R = (70,71)(2,83) = (2,83)2(25)  70,712  = 200,22 W    = 200,11 W  25 = 200,00 W  (200 W) OPTION 4/OPSIE 4 I Irms/wgk = max /maks 2 Imax 2,83 = 2 Imax/maks = 4 A V I Pave/gem = max/maks max/maks  2 (100)( 4) =  2 = 200 W  (3) 9.6 200 V (V) 0 0,025 0,05 0,075 0,1 t (s) -200 Marking criteria/Nasienkriteria  2 waves  2 golwe  Period of wave is 0,05 s  Periode van golf is 0,05 s  Amplitude = 200 V  (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 26 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The minimum frequency of light needed to eject electrons from a metal / surface.  Minimum frekwensie van lig benodig om elektrone vanaf 'n metaal / oppervlak vry te stel. (2) 10.2 Greater than/Groter as  (2) 10.3 OPTION 1/OPSIE 1 E = W o + Ek(max)  1 fx = ( ) (23,01x10-19)  + 10,40 x 1014  6,63 x 10 -34 = 4,51 x 1015 (Hz) (45,1 x 1014 Hz ) OPTION 2/OPSIE 2 1 m=  h f x  10,4  1014  1 19 =  23,01 10  0  6,63  10 34 fx = 4,51 x 1015 (Hz) (45,1 x 1014 Hz) OPTION 3/OPSIE 3 E = W o + E k(max)  hf = hf0 + E k(max) 6,63 x 10-34 fx = (6,63 x 10-34)(10,40 x 1014)+ 23,01 x 10-19  fx = 4,51 x 1015 (Hz) (45,1 x 1014 Hz ) (5) 10.4 10.4.1 No effect/Geen effek nie  (1) 10.4.2 Increases/Verhoog  (1) 10.4.3 No effect/Geen effek nie  (1) [12] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou