Physical Sciences Grade 12 Paper 1 NSC November 2022 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2022 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 32 pages. Hierdie nasienriglyne bestaan uit 32 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

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

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyn 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. When a resultant/net force acts on an object, the object will accelerate in the direction of the force. The acceleration is directly proportional to the resultant/net force and inversely proportional to the mass of the object.  Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel. Die versnelling is direk eweredig aan die netto krag en omgekeerd eweredig aan die massa van die voorwerp. OR/OF The resultant/net force acting on an object is equal to the rate of change of momentum of the object. (2 or 0) Die resulterende/netto krag wat op 'n voorwerp inwerk is gelyk aan die tempo van verandering van momentum. (2 of 0) (2) 2.2 Accept force diagram/ Aanvaar kragte-diagram: N N f T f T w w Accepted labels/Aanvaarde benoemings Fg /Fw /Fearth on P /weight /mg /12,25 N /gravitational force w Fg /Fw /Faarde op P /gewig /mg /12,25 N /gravitasiekrag T FT /Fstring /Ftou /Ft /tension /spanning /Fs f Ff /fk /(kinetic) friction /(kinetiese) wrywing /1,8 N /Fw N FN /Normal /Fnormal /Fnormaal /Normaal Notes/Aantekeninge  Mark awarded for label and arrow./Punt toegeken vir benoeming en pyltjie.  Do not penalise for length of arrows since drawing is not to scale./Moenie vir die lengte van die pyltjies penaliseer nie aangesien die tekening nie volgens skaal is nie.  Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks 3 4  If everything correct, but no arrows/Indien alles korrek, maar geen pyltjies: Max/Maks 3 4  If force(s) do not make contact with the dot /Indien krag(te) nie met die kolletjie kontak maak nie: Max/Maks 3 4 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 2.3.1 For P/Vir P For P/Vir P RIGHT AS POSITIVE/ LEFT AS POSITIVE/ REGS AS POSITIEF LINKS AS POSITIEF Fnet = ma Any one/ Fnet = ma Any one/ T + f = ma  Enige een T + f = ma  Enige een T – f = ma -T +f = ma T - 1,8 = (1,25)(0,1)  -T + 1,8 = (1,25)(-0,1)  T = 1,93 N  (1,925 N) T = 1,93 N  (1,925 N) (4) 2.3.2 POSITIVE MARKING FROM QUESTION 2.3.1/ POSITIEWE NASIEN VANAF VRAAG 2.3.1. RIGHT AS POSITIVE/REGS AS POSITIEF: For Q/Vir Q Accept/Aanvaar Fnet = ma Sin(90°- θ) Fcosθ - T - f = ma Fcosθ + T + f = ma 7,5cosθ - 1,93 - 2,2 = (2)(0,1)  θ = 54,74° (Range:54,55° - 54,78°) LEFT AS POSITIVE/LINKS AS POSITIEF: For Q/Vir Q Accept/Aanvaar Fnet = ma Sin(90°- θ) -Fcosθ + T + f = ma Fcosθ + T + f = ma -7,5cosθ + 1,93 + 2,2 = (2)(- 0,1)  θ = 54,74°  (Range:54,55° - 54,78°) (3) [13] QUESTION 3/VRAAG 3 3.1 Motion under the influence of gravity/weight/gravitational force only.  Beweging slegs onder die invloed van gravitasie/gewig/swaartekrag. (2 or/of 0) OR/OF Motion in which the only force acting is gravity/weight/gravitational force. Beweging waar die enigste krag wat inwerk, gravitasie/gewig/swaartekrag is. (2 or/of 0) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 3.2.1 Marking criteria/Nasienkriteria  Formula with Δt/Formule met Δt   Correct substitution into formula/Korrekte vervanging in formule   Final answer/Finale antwoord: 1,22 s  (1,22 s to/tot 1,23 s) OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 A-B: B-C: UPWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF OPWAARTS AS POSITIEF vf = vi + aΔt  vf = vi + aΔt  0 = 12 + (-9,8)Δt  -12 = 0 + (-9,8)Δt  Δt = 1,22 s  Δt = 1,22 s  DOWNWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf = vi + aΔt  vf = vi + aΔt  0 = -12 + (9,8)Δt  12 = 0 + (9,8)Δt  Δt = 1,22 s  Δt = 1,22 s  B OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 A-C: A-C: UPWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF OPWAARTS AS POSITIEF vf = vi + aΔt  Δy = viΔt + ½aΔt2  A C -12 = 12 + (-9,8)Δt  0 = (12)Δt + ½(-9,8)Δt2  Δt = 2,45 s Δt = 2,45 s Δtup = 1,23 s  Δtup = 1,23 s  DOWNWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf = vi + aΔt  Δy = viΔt + ½aΔt2  12 = -12 + (9,8)Δt  0 = (-12)Δt + ½(9,8)Δt2  G Δt = 2,45 s Δt = 2,45 s Δtup = 1,23 s  Δtup = 1,23 s  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 B A-B OR/OF B-C: (Emech)Top/Bo = (Emech)25 m (EP + EK)Top/Bo = (EP + EK)25 m (mgh + ½mv2)Top/Bo = (mgh + ½mv2)25 m (9,8)h + 0 = 0 + (½)(12)2 A C h = 7,35 m OPTION 6/OPSIE 6 A-B OR/OF B-C Wnc = K + U Wnc = K + mg(hf - hi) 0 = ½mvf2 - ½mvi2 + mghf - mghi 0 = ½(02 - 122) + (9,8)h G h = 7,35 m OPTION 7/OPSIE 7 A-B OR/OF B-C W net = ΔEk wΔycosθ = ½mvf2 - ½mvi2 (9,8)Δycos180° = 0 - ½(12)2 UPWARDS AS POSITIVE/ Δy = 7,35 m OPWAARTS AS POSITIEF OPTION 8/OPSIE 8 vi + vf A-B: ∆y = ( )∆t  UPWARDS AS POSITIVE/ 2 12 + 0 OPWAARTS AS POSITIEF 7,35 = ( )∆t  v2f = v2i + 2aΔy 2 02 = 122 + 2(-9,8)∆y ∆y = 7,35 m Δt = 1,23 s  DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF DOWNWARDS AS POSITIVE/ v2f = v2i + 2aΔy AFWAARTS AS POSITIEF vi + vf 02 = (-12)2 + 2(9,8)∆y ∆y = ( )∆t  ∆y = -7,35 m 2 -12 + 0 -7,35 = ( )∆t  OPTION 9/OPSIE 9 B-C: 2 UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF Δt = 1,23 s  v2f = v2i + 2aΔy (-12)2 = 02 + 2(-9,8)∆y ∆y = -7,35 m DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF v2f = v2i + 2aΔy (12)2 = 02 + 2(9,8)∆y ∆y = 7,35 m Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 10/OPSIE 10 OPTION 11/OPSIE 11 B A-B: B-C: UPWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF: OPWAARTS AS POSITIEF: vi + vf vi + vf ∆y = ( )∆t  ∆y = ( )∆t  2 2 A C 12 + 0 0 - 12 ∆y = ( )∆t ∆y = ( )∆t 2 2 ∆y = 6∆t ∆y = -6∆t v2f = v2i + 2aΔy v2f = v2i + 2aΔy 0 = (12)2 + 2(-9,8)(6∆t)  -12 = (0)2 + 2(-9,8)(-6∆t)  ∆t = 1,22 s  ∆t = 1,22 s  G DOWNWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF: AFWAARTS AS POSITIEF: vi + vf vi + vf ∆y = ( )∆t  ∆y = ( )∆t  2 2 -12 + 0 12 + 0 ∆y = ( )∆t ∆y = ( )∆t 2 2 ∆y = - 6∆t ∆y = 6∆t v2f = v2i + 2aΔy v2f = v2i + 2aΔy 0 = (-12)2 + 2(9,8)(-6∆t) 122 = 02 + 2(9,8)(6∆t)  ∆t = 1,22 s  ∆t = 1,22 s  OPTION 12/OPSIE 12 A-B: UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF: Fnet∆t = m∆v Any one/ Fnet∆t = m(vf - vi) Enige een -(9,8)∆t = 0 – 12  ∆t = 1,22 s  DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF: Fnet∆t = m∆v Any one/ Fnet∆t = m(vf - vi) Enige een (9,8)∆t = 12 – 0  ∆t = 1,22 s  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 3.2.2 Marking criteria/Nasienkriteria  Formula with vf./Formule met vf.   Correct substitution into formula./Korrekte vervanging in formule.   Correct final answer/Korrekte finale antwoord: 25,18 m·s-1  (25,03 m·s-1 to/tot 25,59 m·s-1)  Correct direction (only if numerical value is given)./Korrekte rigting (slegs indien numeriese waarde gegee is). OPTION 1/OPSIE 1 A-G: A-G: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF: AFWAARTS AS POSITIEF: v2f = v2i + 2aΔy  v2f = v2i + 2aΔy  v2f = (12)2 + 2(-9,8)(-25)  v2f = (-12)2 + 2(9,8)(25)  vf = 25,18 m·s-1 downwards  vf = 25,18 m·s-1 downwards  afwaarts afwaarts OPTION 2/OPSIE 2 B C-G: C-G: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF v2f = v2i + 2aΔy  v2f = v2i + 2aΔy  v2f = (-12)2 + 2(-9,8)(-25)  v2f = (12)2 + 2(9,8)(25)  A C vf = 25,18 m·s-1 downwards  vf = 25,18 m·s-1 downwards  afwaarts afwaarts OPTION 3/OPSIE 3 B-G B-G UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF v2f = v2i + 2aΔy  v2f = v2i + 2aΔy  G v2f = (0)2 + 2(-9,8)(-32,35)  v2f = (0)2 + 2(9,8)(32,35)  vf = 25,18 m·s-1 downwards  vf = 25,18 m·s-1 downwards  afwaarts afwaarts Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 A-G: UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: A-G vi + vf 1 ∆y = ( )∆t  Δy = vi Δt + aΔt2 2 2 12 + vf 1 -25 = ( )3,79 25 = 12Δt + ( 9,8)Δt2 2 B 2 ∆t = 3,79 s vf = - 25,18 m·s-1 vf = 25,18 m·s-1 downwards  afwaarts B-G A C vf = vi + aΔt  vf = 0 + (-9,8)(3,79 - 1,22)  vf = - 25,19 m·s-1 vf = 25,19 m·s-1 downwards  afwaarts C-G G vf = vi + aΔt  vf = -12 + (-9,8)(1,35)  vf = -25,19 m·s-1 vf = 25,19 m·s-1 downwards  afwaarts Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyn A-G: DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: A-G vi + vf 1 ∆y = ( )∆t  Δy = vi Δt + aΔt2 2 2 -12 + vf 1 2 25 = ( )3,79  25 = 12Δt + (9,8)Δt 2 2 ∆t = 3,79 s vf = 25,18 m·s-1 downwards  afwaarts B-G vf = vi + aΔt  = 0 + (9,8)(3,79 - 1,22)  vf = 25,19 m·s-1 downwards  afwaarts C-G vf = vi + aΔt  vf = 12 + (9,8)(3,79 - 2(1,22))  vf = 25,19 m·s-1 downwards  afwaarts OPTION 5/OPSIE 5 C-G: C-G: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ B OPWAARTS AS POSITIEF: AFWAARTS AS POSITIEF: 1 1 Δy = vi Δt + aΔt2 Δy = vi Δt + aΔt2 2 2 1 1 - 25 = -12Δt + (-9,8)Δt2 25 = 12Δt + (9,8)Δt2 2 2 A C Δt = 1,34 s Δt = 1,34 s vi + vf vi + vf ∆y = ( )∆t  ∆y = ( )∆t  2 2 -12 + vf 12 + vf -25 = ( ) 1,34  25 = ( ) 1,34  2 2 -1 G vf = - 25,18 m·s-1 vf = 25,18 m·s  downwards  -1 vf = 25,18 m·s  downwards  afwaarts afwaarts OPTION 6/OPSIE 6 (Emech/meg)i = (Emech/meg)f (EP +EK)i = (EP +EK)f Any one/ 2 2 Enige een (mgh + ½mv )i= (mgh + ½mv )f 2 2 m(9,8)(25) + ½m(12 ) = 0 + ½mvf  vf = 25,18 m·s-1 downwards/afwaarts  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 7/OPSIE 7 nc = ΔE + ΔEp = (Ekf - Eki) + (Epf – Epi) Any one/ 2 2 = (½mvf – ½mvi ) + (mghf -mghi) Enige een 2 2 0 = [½mvf – ½m(12) ] + [0 - m(9,8)(25)]  vf = 25,18 m·s-1 downwards/afwaarts  OPTION 8/OPSIE 8 net = ΔE = (Ekf - Eki) Any one/ 2 = (½mvf – ½mvi ) 2 Enige een 2 2 m(9,8)(25) = ½mvf – ½m(12)  vf = 25,18 m·s-1 downwards/afwaarts  (4) 3.2.3 POSITIVE MARKING FROM QUESTION 3.2.2. POSITIEWE NASIEN VANAF VRAAG 3.2.2. Marking criteria/Nasienkriteria  Substitution into formula to calculate v at the top of the door  Vervanging in formule om v bokant die deur te bereken.  Formula to calculate Δt from top to bottom of door. Formule om Δt te bere en van bo ant tot onder ant van deur.  Substitution to calculate Δt.  Vervanging om Δt te bereken.  Final answer/Finale antwoord: 0,07 to/tot 0,08 s  OPTION 1/OPSIE 1 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: B A-D: D-G: v2f = v2i + 2aΔy vf = vi + aΔt  2 2 vf = (12) + 2(-9,8)(-23,1)  -25,18 = -24,43 + (-9,8)Δt  vf = 24,43 m·s-1 Δt = 0,08 s  A C C-D: D-G: Δy = viΔt + ½aΔt2  v2f = v2i + 2aΔy -1,9 = -24,43Δt + ½(-9,8)Δt2 v2f = (-12)2 + (2)(-9,8)(-23,1)  Δt = 0,08 s  vf = 24,43 m·s-1 D D-G: D-G: vi + vf v2f = v2i + 2aΔy Δy = ( )∆t 2 G (-25,18)2 = (vi)2 + 2(-9,8)(-1,9)  vi = 24,43 m·s-1 -24,43 - 25,18 -1,9 = ( ) Δt  2 B-D v2f = v2i + 2aΔy Δt = 0,08 s  v2f = 0 + 2(-9,8)(-30,447)  = 24,43 m∙s-1 D-G: E(mech top/meg bo) = E(mech bot/meg ond) (Ep + Ek)top/bo = (Ep + Ek)bot/ond (mgh + ½mv2)top/bo = (mgh + ½mv2)bot/ond m(9,8)(1,9) + ½m(vi)2 = 0 + ½m(25,18)2  vi = 24,43 m·s-1 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: B A-D: D-G: v2f = v2i + 2aΔy vf = vi + aΔt  2 2 vf = (-12) + 2(9,8)(23,1)  25,18 = 24,43 + (9,8)Δt  vi = 24,43 m·s-1 Δt = 0,08 s  A C C-D: D-G: Δy = viΔt + ½aΔt2  v2f = v2i + 2aΔy 1,9 = 24,43Δt + ½(9,8)Δt2  v2f = (12)2 + (2)(9,8)(23,1)  Δt = 0,08 s  vf = 24,43 m·s-1 D D-G: D-G: vi + vf v2f = v2i + 2aΔy Δy = ( )∆t 2 G (25,18)2 = (vi)2 + 2(9,8)(1,9)  vi = 24,43 m·s-1 24,43 25,18 1,9 = ( ) Δt  2 B-D Δt = 0,08 s  v2f = v2i + 2aΔy v2f = 0 + 2(9,8)(30,447)  = 24,43 m∙s-1 D-G: E(mech top/meg bo) = E(mech bot/meg ond) (Ep + Ek)top/bo = (Ep + Ek)bot/ond (mgh + ½mv2)top/bo = (mgh + ½mv2)bot/ond m(9,8)(1,9) + ½m(vi)2 = 0 + ½m(25,18)2  vi = 24,43 m·s-1 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne Marking criteria OPTION 2 and 3/Nasienkriteria OPSIE 2 en 3  Either one of the formula to calculate Δt./Enige een van die formules om Δt te bereken.   Substitute into formula to calculate time from A to G or C to G/Vervanging in formule om tyd te bereken tussen A tot G of C tot G   Substitute into formula to calculate time from A to D or C to D/ Vervanging in formule om tyd te bereken tussen A tot D of C tot D   Final answer/Finale antwoord: 0,07 s  (0,07s to/tot 0,08s) B OPTION 2/OPSIE 2 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF: AFWAARTS AS POSITIEF: A-G: A-G: A vf = vi + aΔt vf = vi + aΔt C -25,18 = 12 + (-9,8) Δt 25,18 = -12 + (9,8) Δt Δt = 3,79 s Δt = 3,79 s Any one/ Any one/ Enige een Enige een A-D: A-D: D 1 1 Δy = vi Δt + 2 aΔt2 Δy = vi Δt + 2 aΔt2 -23,1 = (12)Δt + ½(-9,8)Δt2 23,1 = (-12)Δt + ½(9,8)Δt2 G Δt = 3,72 s Δt = 3,72 s Time from top to bottom of door/Tyd Time from top to bottom/Tyd van van bokant tot onderkant van deur: bokant tot onderkant van deur: D-G: D-G: 3,79 – 3,72 = 0,07 s  3,79 – 3,72 = 0,07 s  OPTION 3/OPSIE 3 B UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF: AFWAARTS AS POSITIEF: C-G: C-G: vf = vi + aΔt vf = vi + aΔt -25,18 = -12 + (-9,8) Δt 25,18 = 12 + (9,8) Δt A C Δt = 1,34 s Δt = 1,34 s Any one/ Any one/ C-D: Enige een C-D: Enige een 1 1 y = vi Δt + 2 aΔt2 Δy = vi Δt + 2 aΔt2 D -23,1 = (-12)Δt + ½(-9,8)Δt2 23,1 = (12)Δt + ½(9,8)Δt2 Δt = 1,27 s Δt = 1,27 s G Time from top to bottom of door/Tyd Time from top to bottom of door/Tyd van bokant tot onderkant van deur: van bokant tot onderkant van deur: D-G: D-G: 1,34 – 1,27 = 0,07 s  1,34 – 1,27 = 0,07 s  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF: AFWAARTS AS POSITIEF: G-D: G-D: Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  1,9 = 25,18Δt + ½(-9,8)Δt2  -1,9 = -25,18Δt + ½(9,8)Δt2  Δt = 0,08 s  (0,077 s) Δt = 0,08 s  (0,077 s) (4) 3.3 POSITIVE MARKING FROM QUESTION 3.2.1 AND QUESTION 3.2.2. POSITIEWE NASIEN VANAF VRAAG 3.2.1 EN VRAAG 3.2.2. UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Velocity/Snelheid (m·s-1) 12 1,22 Time/Tyd (s) -25,18 DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: 25,18 Velocity/Snelheid (m·s-1) 1,22 Time/Tyd (s) -12 Criteria for graph/Kriteria vir grafiek Straight line starting at v = 12 m·s-1 with negtive final velocity or straight line starting at v = -12 m·s-1 with positive final velocity.  /Reguitlyn wat begin by v = 12 m·s-1met negatiewe finale snelheid of reguitlyn wat begin by v = -12 m·s-1met positiewe finale snelheid. Straight line cuts time axis at time calculated in Question 3.2.1/  Reguitlyn sny tydas by die tyd bereken in Vraag 3.2.1. Correct final velocity as calculated in Question 3.2.2 is indicated./Korrekte finale snelheid soos uitgewerk in Vraag 3.2.2 is  aangedui. (3) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 2022 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. In an isolated/closed system the total (linear) momentum is conserved/remains constant.  In 'n geïsoleerde/geslote sisteem bly die totale (lineêre) momentum behoue/konstant. (2) 4.2.1 OPTION 1/OPSIE 1 EAST AS POSITIVE/OOS AS POSITIEF ∑ pi = ∑ pf mxvix + myviy = mxvfx + myvfy Any one/Enige een (1,2)(8)  + (0,5)(0) = (1,2)(4) + (0,5)(vfy)  vfy = 9,6 m·s-1  WEST AS POSITIVE/WES AS POSITIEF ∑ pi = ∑ pf  Any one/Enige een mxvix + myviy = mxvfx + myvfy (1,2)(-8)  + (0,5)(0) = (1,2)(-4) + (0,5)(vfy)  vfy = - 9,6 m.s-1 vfy = 9,6 m·s-1 OPTION 2 /OPSIE 2 EAST AS POSITIVE/OOS AS POSITIEF Δpx = - Δpy  Any one/Enige een m(vxf –vxi) = - m(vyf - vyi) 1,2(4 - 8) = - 0,5(vf - 0)  vfy = 9,6 m·s-1  WEST AS POSITIVE/WES AS POSITIEF Δpx = - Δpy   Any one/Enige een m(vxf - vxi) = - m(vyf - vyi) 1,2(- 4 + 8) = - 0,5(vf - 0)  vfy = - 9,6 m.s-1 vfy = 9,6 m·s-1  (4) 4.2.2 OPTION 1/OPSIE 1 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) = 1,2(4 - 8)  Fnet(0,1) = 1,2( - 4 + 8)  Fnet = - 48 N  Fnet = 48 N  Fnet = 48 N  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2 POSITIVE MARKING FROM QUESTION 4.2.1. POSITIEWE NASIEN VANAF VRAAG 4.2.1. EAST AS POSITIVE WEST AS POSITIVE OOS AS POSITIEF WES AS 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) = 0,5(-9,6 – 0)  Fnet (0,1) = 0,5(9,6 - 0)  Fnet = - 48 N Fnet = 48 N  Fnet = 48 N  OPTION 3/OPSIE 3 EAST AS POSITIVE for X WEST AS POSITIVE for X OOS AS POSITIEF vir X WES AS POSITIEF vir X vf = vi + aΔt vf = vi + aΔt -4 = -8 + a(0,1) 4 = 8 + a(0,1) a = - 40 m·s-2 a = 40 m·s-2 Fnet = ma  Fnet = ma  Fnet = (1,2)(-40)  Fnet = (1,2)(40)  Fnet = - 48 N Fnet = 48 N Fnet = 48 N  OPTION 4/OPSIE 4 EAST AS POSITIVE for X WEST AS POSITIVE for X OOS AS POSITIEF vir X WES AS POSITIEF vir X vi + vf vi + vf ∆x = ( )∆t ∆x = ( )∆t 2 2 8+4 -8-4 ∆x = ( )(0,1) ∆x = ( )(0,1) 2 2 Δx = 0,6 m Δx = - 0,6 m FnetΔxcosθ = ½mvf2 – ½mvi2  FnetΔxcosθ = ½mvf2 – ½mvi2  2 2 2 2 Fnet(0,6)cos180º = ½ (1,2)(4) – ½ (1,2)(8)  Fnet(0,6)cos0º = ½(1,2)(-4) – ½(1,2)(-8)  Fnet = 48 N  Fnet = - 48 N  Fnet = 48 N  OPTION 5/OPSIE 5 ∆y Gradient = ∆x ∆v = ∆t 4-8 = 0,1 = -40 m·s-2 Fnet = ma  Fnet = (1,2)(-40)  Fnet = -48 N Fnet = 48 N  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 4.3 POSITIVE MARKING FROM QUESTION 4.2.1/ POSITIEWE NASIEN VANAF VRAAG 4.2.1. OPTION 1/OPSIE 1 Inelastic/onelasties  Ek = ½mv2  ∑ Eki = ½mXvXi2 + ½mYvYi2 = ½(1,2)(8)2 + 0  = 38,4 J ∑ Ekf = ½mXvXf2 + ½mYvYf2 = ½(1,2)(4)2 + ½(0,5)(9,6)2  = 32,64 J ∑ Eki ≠ ∑ Ekf  OPTION 2/OPSIE 2 (Change in Ektotal total /verandering in Ektotaal)) Inelastic/onelasties Ek = ½mv2  ΔEk(X) = ½mv2f - ½mv2i 2 2 = ½(1,2)(4) – ½(1,2)(8)  = - 28,8 J ΔEk(Y) = ½mv2f - ½mv2i 2 2 = ½(0,5)(9,6) – ½(0,5)(0)  = 23,04 J ΔEk(X) ΔEk(Y)  Note/Aantekening: If candidate starts with conservation of kinetic energy/Indien kandidaat begin 4 met behoud van kinetiese energie: max/maks (5) 5 [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/November 2022 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. A force is non-conservative if the work done by the force on an object (which is moving between two points) depends on the path taken.  'n Krag waarvoor die arbeid wat verrig word deur die krag op 'n voorwerp (wat tussen twee punte beweeg,) afhanklik is van die pad wat gevolg word. OR/OF A force is non-conservative if the work it does in moving an object around a closed path is non-zero. 'n Krag is nie-konserwatief wanneer die arbeid wat dit verrig om 'n voorwerp in 'n geslote pad te beweeg, nie nul is nie. (2) Note/Aantekening: 0 -If work done is ommited/ Indien arbeid verrig uitgelaat word: 2 5.2 N F N F OR/OF f f Fg// Fg┴ w Accepted labels/Aanvaarde benoemings w Fw / Fg / mg /117,6 N / gravitational force / gravitasiekrag / weight / gewig F FA /Applied force / T /Toegepaste krag/ FT f Ff / fk / (kinetic) Friction / (kinetiese) wrywing / Fw N FN / Normal / Normaal Notes/Aantekeninge:  Mark awarded for label and arrow./Punt toegeken vir benoeming en pyltjie.  Do not penalise for length of arrows since drawing is not to scale./Moenie vir die lengte van die pyltjies penaliseer nie aangesien die tekening nie volgens skaal is nie.  If w is not shown but F// and Fg┴ are shown, give 1 mark for both./Indien w nie aangetoon is nie maar F// en Fg┴ is getoon, ken 1 punt toe vir beide.  Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks 3 4  If everything is correct, but no arrows/Indien alles korrek is, maar geen pyltjies: Max/Maks 3 • 4  If force(s) do not make contact with the dot /Indien krag(te) nie met die kolletjie kontak maak nie: Max/Maks 3 4 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 5.3 OPTION 1/OPSIE 1 Wnc = ΔEk + ΔEp 1  Any one/Enige een Wnc = 2 m(v2f - v2i ) + mg(hf – hi) = ½(12)(2,252 - 0)  + (12)(9,8)(4,5 - 0)  = 559,58 J  OPTION 2/OPSIE 2 WFg// = Fg//Δxcos WFg = FgΔxcos(90° + θ) = (mgsinθ)Δxcosθ = mgΔx(– sinθ) 4,5 - 4,5 = (12)(9,8)( )Δxcos180° = (12)(9,8)( )Δx ∆x ∆x = - 529,2 J = - 529,2 J Wnet = ΔE W net = W nc+ Wc  = ½mv2f – ½mv2i 30,375 = W nc + (- 529,2)  = ½(12)(2,25)2 – 02  W nc = 559,575 J  (559,58 J) = 30,375 J OPTION 3/OPSIE 3 4,5 Sinθ = ∆x 4,5 Δx = sinθ W net = ΔEk Any one/ WF + W f + W w = ΔEk Enige een W nc + (mgsinθ)∆x(cosβ) = ½mv2f – ½mv2i 4,5 Wnc + (12)(9,8)( )sinθcos180° = ½(12)(2,25)2 – 02  sinθ W nc = 559,58 J  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 20 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 5.4 POSITIVE MARKING FROM QUESTION 5.3. POSITIEWE NASIEN VANAF VRAAG 5.3. Marking criteria for OPTION 1 /Nasienkriteria vir OPSIE 1  Formula for W nc /Formule vir Wnc   Correct substitution of 559,58 J in W nc along inclined plane  Korrekte vervanging van 559,58 J in Wnc langs die skuinsvlak  Correct force equation and substitution of 0 for Fnet OR F = f2 on horizontal plane Korrekte kragvergelyking en vervanging van 0 vir Fnet OF F = f2 op die horisontale vlak.  Relating the two frictional forces (substitution of f1 + 42 for f2). Bring die twee wrywingskragte in verband (vervanging van f1 + 42 vir f2).  Correct answer/Korrekte antwoord: 13,32 m. OPTION 1/OPSIE 1 ALONG THE INCLINE/AB/TEEN DIE SKUINSVLAK W nc = W F + Wf  Any one/Enige een W nc = FΔxcos0° + f1Δxcos180° 559,58 = FΔxcos0° + f1Δxcos180° 559,58 = (F – f1)Δx ....................(1) ALONG THE HORIZONTAL/BC/LANGS DIE HORISONTAAL F - f2 = ma OR/OF F = f2 F - f2 = 0  F - (f1 + 42) = 0 F - f1 = 42....................(2) Substitute/Vervang (2) into/in (1): 559,58 = 42Δx Δx = 13,32 m  Marking criteria for OPTION 2 and 3 /Nasienkriteria vir OPSIE 2 en 3  Correct force equation and substitution of 0 for Fnet OR F = f2 on horizontal plane Korrekte kragvergelyking en vervanging van 0 vir Fnet OF F = f2 op die horisontale vlak.  Relating the two frictional forces (substitution of f1 + 42 for f2). Bring die twee wrywingskragte in verband (vervanging van f1 + 42 vir f2).  Formula for W nc OR W net /Formule vir Wnc OF Wnet  Correct substitution into equation for W nc OR W net on the horizontal plane Korrekte vervanging in Wnc OF Wnet vergelyking langs die skuinsvlak  Correct answer/Korrekte antwoord: 13,32 m. OPTION 2/OPSIE 2 ALONG THE HORIZONTAL/BC/LANGS DIE HORISONTAAL F - f2 = ma F - f2 = 0  OR/OF F = f2 F - (f1 + 42) = 0 F = f1 + 42 ALONG THE INCLINE/AB/TEEN DIE SKUINSVLAK W nc = ΔEK + ΔEP  Any one/ 2 2 Enige een (F – f1)Δxcosθ = [ ½mvf – ½mvi ] + [mghf - mghi] (f1 + 42 – f1)Δxcos0° = [ ½(12)(2,25)2 – 02] + [(12)(9,8)(4,5) – 0]  Δx = 13,32 m  (13,32 m) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 21 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 ALONG THE HORIZONTAL/BC/LANGS DIE HORISONTAAL F - f2 = ma F - f2 = 0  OR/OF F = f2 F - (f1 + 42) = 0 F = f1 + 42 ALONG THE INCLINE/AB/TEEN DIE SKUINSVLAK W net = ΔEK  Any one/ (F – f1 – Fg//)Δxcosθ = [ ½mv2f – ½mv2i ] Enige een 4,5 [(42) – (12)(9,8)( )] Δxcos0o = ½(12)(2,25)2 – 02 ∆x Δx = 13,323214 m  (13,32 m) OPTION 4/OPSIE 4 W nc = ΔEK + ΔEP  Any one/ W nc = [ ½mv2f – ½mv2i ] + [mghf -mghi] Enige een (f1 + 42 – f1)Δxcos0° = 559,575  (42)Δxcos0° = 559,575 Δx = 13,323214 m  (13,32 m) Marking criteria for OPTION 5/Nasienkriteria vir OPSIE 5  Correct force equation and substitution of 0 for Fnet OR F = f2 on horizontal plane/ Korrekte kragvergelyking en vervanging van 0 vir Fnet OF F = f2 op die horisontale vlak.  Relating the two frictional forces (substitution of f1 + 42 for f2). Bring die twee wrywingskragte in verband (vervanging van f1 + 42 vir f2).  Correct substitution to calculate a./Korrekte vervanging om a te bereken. .  Substitution to calculate Fnet./Vervanging om Fnet te bereken.   Correct answer/Korrekte antwoord: 13,32 m. OPTION 5/OPSIE 5 ALONG THE HORIZONTAL/BC/LANGS DIE HORISONTAAL F - f2 = ma F - f2 = 0  OR/OF F = f2 F - (f1 + 42) = 0 F = f1 + 42 ......(1) ALONG THE INCLINE/AB/TEEN DIE SKUINSVLAK v2f = v2i + 2aΔx 2,252 = 0 + 2aΔx  2,53 a= ∆x Fnet = ma F- Fg// - f1 = ma F – mgsinθ – f1 = 12a ....... 2,53 F – (12)(9,8)sinθ – f1 = 12( ) ....(2) ∆x Substitute/Vervang (2) into/in (1): 4,5 42 – (12)(9,8)( ) = 30,38 ∆x ∆x = 13,32 m  (5) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 22 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 Doppler Effect/Doppler-effek  (1) 6.2 Measurement of foetal heartbeat./Meting van die hartklop van 'n fetus. OR/OF Measurement of blood flow./Meting van bloedvloei.  OR/OF Doppler flow meter/Doppler vloeimeter  (1) 6.3 fL fs  OR/OF Directly (proportional)/Direk (eweredig) (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 23 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 6.4 Marking criteria/Nasienkriteria  Doppler formula/Doppler formule   Correct substitution for v and vS./Korrekte vervanging van v en vS.  f  Substitution for = 1,06 OR fL = 1,06 fs OR any set of values for fL and fS so that f f fL= 1,06 fS / Vervanging van = 1,06 OF fL = 1,06 fs OF enige stel waardes vir f fL en fS sodat fL= 1,06 fS   Final answer/Finale antwoord: 20,4 m·s-1  OPTION 1/OPSIE 1 v v v+v fL = fS OR/OF fL = fS v vs v f v v = fs v vs  340 + v  1,06 = 340 vL = 20,4 m·s-1  OPTION 2/OPSIE 2 0-f Gradient = 0-f  0-f 1,06 = 0-f fL = 1,06 fS v v v+v fL = fS  OR/OF fL = fS v vs v 340 + v 1,06 fS = ( 340 ) fS  vL = 20,4 m·s-1  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 24 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 6.5 B  A fL (Hz) 0 fS (Hz) Marking criteria/Nasienkriteria Graph is a straight line starting at the origin./  Grafiek is 'n reguitlyn wat by die oorsprong begin. Gradient of B is greater than gradient of A./  Gradiënt van B is groter as gradiënt van A. (2) [10] QUESTION 7/VRAAG 7 7.1 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The magnitude of the electrostatic force exerted by one point charge on another is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.  Die grootte van die elektrostatiese krag wat een puntlading op 'n ander uitoefen, is direk eweredig aan die produk van die grootte van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. (2) 7.2 Negative/negatief  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 25 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 7.3 FE w Accepted labels/Aanvaarde byskrifte FE Felectrostatic/ F/ FM ON N/ electrostatic force/ FM Felektrostaties/ F/ FM OP N/ elektrostatiese krag/ FM w Fg / w/ mg/ gravitational force / Fw/ weight/ gravity Fg / w/ mg/ gravitasiekrag/ Fw/ gewig/ swaartekrag Notes/Aantekeninge:  Do not penalise for length of arrows./Moenie vir lengte van die pyltjie penaliseer nie.  Any other additional force(s)/Enige addisionele krag(te): Max/Maks 1⁄2  If arrows are omitted but correctly labelled/Indien pyltjies weggelaat is, maar korrek benoem: Max/Maks 1⁄2  If force(s) do not make contact with the dot /Indien krag(te) nie met die kolletjie kontak maak nie: Max/Maks 1⁄2 (2) 7.4 Marking criteria/Nasienkriteria  Correct substitution to calculate weight of M./Korrekte vervanging om gewig van M te bereken.   Coulomb's formula/Coulomb se formule   Substitute/Vervang Fnet = 0/ mg = (equating forces)/ r2 0,02 = (equating forces) r2  Correct substitution into /Korrekte vervanging in  r2 r  Correct final answer (accept negative value)/Korrekte finale antwoord (aanvaar negatiewe waarde): 2,33 x 10-6 C to/tot 2,32 x 10-6 C Fg = mg IF/INDIEN: = (2,04 x 10-3)(9,8)  = 0,02 N F=  r2  9 F= r2  (9 x 10 )( )(8,6 x 10-8 ) -3 2,04 x 10 )(9,8)  =  (0,3)2 Fnet = mg - QM = 2,33 x 10-6 C  r2 9 (9 x 10 )( )(8,6 x 10-8 ) 0 = 0,02 -  (0,3)2 -6 QM = 2,33 x 10 C  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 26 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 7.5.1 Equal/Gelyk  OR/OF Same/Dieselfde (1) 7.5.2 Opposite OR upwards/Teenoorgesteld OF opwaarts (1) 7.6 POSITIVE MARKING FROM QUESTION 7.4. POSITIEWE NASIEN VANAF VRAAG 7.4. Marking criteria/Nasienkriteria  Formula for E./Formule vir E.   Correct substitution for M OR N./Korrekte vervanging vir M OF N.   Subtraction of EM – EN OR EN – EM /Aftrekking van EM – EN OF EN - EM.   Correct final answer/Korrekte finale antwoord: 5,31 x 104 N.C-1 to/tot 5,37 x 104 N·C-1  Correct direction/Korrekte rigting: upwards/opwaarts  UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: E=  r2 (9 x 109 )(2,33 x 10-6 )  (9 x 109 )(8,6 x 10-8 ) Enet = ( 2 ) ( 2 )  (0,4) (0,1) Enet = 131 062,5 - 77 400 = 53 662,5 N·C-1  (5,36 x 104 N.C-1) upwards/towards M opwaarts/na M  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: Enet =  r2 (9 x 109 )(8,6 x 10-8 )  (9 x 109 )(2,33 x 10-6 ) Enet = ( ) ( )  (0,1)2 (0,4)2 Enet = 77 400 – 131 062,5 = - 5,37 x 104 N·C-1 Enet = 53 662,5 N·C-1  (5,36 x 104 N.C-1) upwards/towards M opwaarts/na M  (5) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 27 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature (provided temperature and all other physical conditions are constant).  Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur (mits temperatuur en alle fisiese toestande konstant bly). OR/OF The ratio of potential difference to current is constant at constant temperature. Die verhouding van potensiaalverskil tot stroom is konstant by konstante temperatuur. OR/OF The current in a conductor is directly proportional to the potential difference across the conductor at constant temperature (provided temperature and all other physical conditions are constant). Die stroom in 'n geleier is direk eweredig aan die potensiaalverskil oor 'n geleier by konstante temperatuur (mits temperatuur en alle fisiese toestande konstant bly). (2) NOTE/LET WEL Do not award the mark for addition of 4 if any other value is added to Rp/ Moenie die punt vir bytel van 4 toeken indien enige ander waarde by Rp bygetel word nie. 8.2.1 OPTION 1/OPSIE 1 OPTION2/OPSIE2 1 1 1 1 2 = +  Rp =  p 1 2 1 2 1 1 1 (1)(5) = +  Rp =  p 1 5 1+5 Rp = 0,83 Ω Rp = 0,83 Ω RT = 0,83 + 4  RT = 0,83 + 4 = 4,83 Ω  = 4,83 Ω  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 28 DBE/November 2022 NSC/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 V R =  I 4,83 =  3,5 V = 16,91 V  (16,92 V) OPTION 2/OPSIE 2 2  Rp = 0,83 = 3,5 V2 = 2,91 V 4Ω  Any one/Enige een R4Ω = 4Ω 4= 3,5 V4Ω = 14 V V1 = V2 + V4Ω V1 = 2,91 + 14 = 16,91 V  (16,92 V) (3) 8.2.3 Smaller than/Kleiner as  (1) 8.3.1 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Maximum work done by the battery per unit charge.  Maksimum arbeid verrig deur die battery per eenheidslading. OR/OF Maximum energy supplied by the battery per unit charge. Maksimum energie verskaf deur die battery per eenheidslading. OR/OF The total amount of electric energy supplied by the battery per coulomb/per unit charge. Die totale hoeveelheid elektriese energie verskaf deur die battery per coulomb/per eenheidslading. (2) 8.3.2 No/Nee  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 29 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 8.3.3 The battery has internal resistance. Die battery het interne weerstand. OR/OF Some energy per coulomb of charge/volts is used to overcome internal resistance. 'n Gedeelte van die energie per coulomb lading/volts word gebruik om interne weerstand te oorkom. OR/OF There is a potential drop/lost volts inside the battery. Daar is 'n potensiaalval/verlore volts binne-in die battery. OR/OF ε = Vext + Vint OR/OF ε > Vext (1) 8.4.1 Decreases/Verlaag  (1) 8.4.2 Increases/Verhoog  (1) 8.5 When the voltmeter is connected:  No/very little current through the 1 Ω branch OR Branch with 1 Ω resistor is disabled/bypassed OR A voltmeter has a very high resistance OR The resistance of the parallel branch increases.   (Total) resistance of the circuit increases.   Current in circuit decreases.   Vinternal/ Internal volts/ Vlost decreases.  Therefore, external volts increase for a constant emf. Wanneer die voltmeter geskakel word:  Geen/baie min stroom deur die 1 Ω-tak OF Ta met 1 Ω-weerstand is uitgeskakel OF Voltmeter het baie hoë weerstand OF Die weerstand van die parallelle tak neem toe.  (Totale) weerstand van die stroombaan neem toe.  Stroom in stroombaan neem af.  Vintern/ Interne volts/ Vverlore neem af. Dus neem die eksterne volts toe vir konstante emf. (4) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 30 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1.1 Split ring/commutator/Splitring/kommutator  (1) 9.1.2 Y to/na X OR/OF 0 /no current/geen stroom nie (1) 9.1.3 1 T= f 1 T= 20 T = 0,05 s  (1) 9.1.4 POSITIVE MARKING FROM QUESTION 9.1.3. POSITIEWE NASIEN VANAF VRAAG 9.1.3. Output voltage (V) 90 Uitsetpotensiaalverskil (V) 0 0,025 0,05 Time/Tyd (s) OR/OF Output voltage(V) 0,025 0,05 0 Time/Tyd (s) Uitsetpotensiaalverskil (V) -90 Criteria for graph/Kriteria vir grafiek Correct shape with one full cycle./Korrekte vorm met 1 volledige siklus.  Curve starts at zero to first peak./Kurwe begin by nulpunt tot eerste piek.  Any one of the correct time values at the correct position./Enige een van  die korrekte tyd waardes op die korrekte posisie. Maximum voltage of 90 V OR -90 V/  Maksimum potensiaalverskil van 90 V OF -90 V NOTE/LET WEL: - 1 mark for half cycle/incomplete cycle or more than one cycle -1 punt vir halwe siklus/onvoltooide siklus of meer as een siklus (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 31 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 9.2 Marking criteria/Nasienkriteria  Formula to calculate W ave (do not penalise if subscripts are ommited)./ Formule om Wgem te bereken (moenie penaliseer indien onderskrifte uitgelaat is nie).   Substitution of 220 and 32 in correct equation.  Vervanging van 220 en 32 in die korrekte vergelyking.  Substitution of 120 for Δt/Vervanging van 120 in Δt.   Correct answer in range: 181 500 J to 181 764 J  Korrekte antwoord in gebied: 181 500 J tot 181 764 J  OPTION 1/OPSIE 1 2 rms Δt Wave =  2 220 (120)  = 32  = 181 500 J OPTION 2/OPSIE 2 rms R=  rms 220 32 = rms Irms = 6,88 A (6,875 A) Wave =VrmsIrmsΔt  Wave = Irms2RΔt  = (220)(6,88)(120)  OR/OF = (6,88)2(32)(120)  = 181 632 J  = 181 764,10 J  OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 rms 2 R=  rms Pave = rms 220  2202 32 = Pave =  rms 32 Irms = 6,88 A (6,875 A) Pave = 1 512,5 W (1,51 x103 W) Pave = VrmsIrms Pave = Irms2R = (220)(6,88) = (6,88)2(32) = 1 513,6 W = 1 514,7 W (1 512,5 W) (1,51 x 103 W) (1,51 x 103 W) P=  Δt 1 513,6 = 120  W = 181 632 J  (1,82 x 105 J) (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 32 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1 Light has a particle nature/is quantized  Lig het 'n deeltjie geaardheid/is gekwantiseerd (1) 10.2 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The minimum energy (of incident photons) that can eject electrons from a metal/surface.  Die minimum energie (van invallende fotone) wat elektrone kan vrystel vanuit 'n metaal/oppervlak. (2) 10.3 E = W 0 + Ek(max) hf = hf0 + Ek(max) 1  Any one/ hf = hf0 + mv2max Enige een 2 1 E = W 0 + 2 mv2max (6,63 x 10-34)(5,96 x 1014) = 3,42 x 10-19 + Ek(max)  Ek(max) = 5,30 x 10-20J  (5,32 x 10-20 J) (4) 10.4 q = IΔt = (0,012)(10)  = 0,12 C n= e 0,12  n= 1,6 x 10-19 n = 7,5 x 1017 (electrons/elektrone) number of photons/aantal fotone = n = 7,5 x 1017 (4) 10.5 Increases/Verhoog  More photons strike the surface of the metal per unit time/ at a higher ratehence more (photo) electrons ejected per unit time (resulting in increased current). Meer fotone tref die oppervlak van die metaal per eenheidstyd/ teen ‘n hoër tempo, gevolglik word meer (foto)elektrone per eenheidstyd vrygestel (wat tot 'n verhoogde stroom lei). (3) [14] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou