Physical Sciences Grade 12 Paper 1 NSC November 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2019 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 24 pages. / Hierdie nasienriglyne bestaan uit 24 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 2 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 C  (2) 1.2 A  (2) 1.3 A  (2) 1.4 D  (2) 1.5 B  (2) 1.6 D  (2) 1.7 A  (2) 1.8 B  (2) 1.9 D  (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/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 NOTE: -1 mark for each key word/phrase omitted in the correct context LET WEL: -1 punt vir elke sleutel woord/frase in die korrekte konteks weggelaat (2) 2.1 When a resultant/net force acts on an object, the object will accelerate in the direction of the force with an acceleration that is directly proportional to the force and inversely proportional to the mass of the object.   Wanneer 'n resultante/netto krag op 'n voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel teen 'n versnelling wat direk eweredig is aan die 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 (in the direction of the resultant/net force.)  Die netto krag wat op 'n voorwerp inwerk is gelyk aan die tempo van verandering van momentum van die voorwerp (in die rigting van die resulterende/netto krag.) (2) 2.2 N ACCEPT/AANVAAR T N N T T f f OR/OF f w w┴ w// w Accept the following symbols/Aanvaar die volgende simbole. N F N /Normal/Normaal/Normal force/Normaalkrag /16,97 N F f / f k / f r / frictional force/wrywingskag/kinetic frictional force/kinetiese f wrywingskrag F g,/ mg/weight/F Earth on block /19,6 N/gravitational force/gewig/F aarde op blok / w gravitasiekrag T Tension/Spanning/F T / F A / F / F s 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 force(s) do not make contact with body/dot /Indien krag(te) nie met die voorwerp / kolletjie kontak maak nie: Max./Maks: 3 4 • If arrows missing/Indien pyltjies uitgelaat is: 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 2019 NSC/NSS –Marking Guidelines/Nasienriglyne 2.3 For the 2 kg (P) block/Vir die 2 kg (P) blok: Marking criteria/Nasienriglyne F net = ma • Formula/Formule  T + (- w ll ) + (- f k ) = ma Any one/Enige een • Substitution of /vervanging van T – (w ll + f k ) = ma w ll for/vir 2 kg: (2)(9,8)sin30°  T – (2)(9,8)sin30° – 2,5 = 2a  • Substitution of -2,5 N /Vervanging T – 9,8 – 2,5 = 2a van -2,5 N  T – 12,3 = 2a ……….(1) • 2a OR/OF 3a  • Calculate/Bereken F x : For the 3 kg (Q) block/Vir die 3 kg (Q) blok: 40 cos 25°  (40 Sin 65°) F x + (-T) + (- w ll ) = ma • Substitution of/vervanging van w ll F x – (T + w ll ) = ma for/vir for 3 kg: (3)(9,8)sin30°  [40 cos 25° – T – (3)(9,8)sin30°]  = 3a • Left hand side substitution for 36,25 – T – 14,7 = 3a 3 kg/Linkerkant vervanging vir 3 21,55 – T = 3a ………….(2) kg  • Final answer/Finale antwoord: 9,25 = 5 a 1,85 m∙s-2 -2 a = 1,85 m∙s  (8) Systems Approach (Massless String Approximation / Sisteembenadering (Massalose Tou Benadering) (Max 5/8 marks / Maks 5/8 punte) F net = ma F x + (- w ll ) + (- f k ) = ma Any one/Enige een F x – (w ll + f k ) = ma 40cos25°  – (5)(9,8)sin30° – 2,5 = 5a a = 1,85 m∙s-2  2.4 Greater than/groter as  - F net increases. /F net neem toe.  ACCEPT/AANVAAR There is no friction. /Daar is geen wrywing nie. OR/OF The surface is smooth / Die oppervlak is glad (2) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 5 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 (Motion during which) the only force acting is the force of gravity.  (Beweging waartydens) die enigste krag wat inwerk gravitasiekrag is. (2 or/of 0) (2) 3.2 Marking criteria/Nasienriglyne: • Any appropriate formula for Δy/Enige toepaslike formule vir Δy • Whole substitution to calculate 5,1 m /Hele vervanging om 5,1 m te bereken  • 40 + answer from calculation/antwoord van berekening  • Final answer/Finale antwoord: 45,10 m (Accept/aanvaar 45,1 m) OPTION 1/OPSIE 1 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: v f 2 = v i 2 + 2aΔy  0 = (10)2 + (2)(- 9,8)Δy  Δy = 5,10 m (5,102 m) Height = 40 + 5,10  = 45,10 m  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: v f 2 = v i 2 + 2aΔy  0 = (-10)2 + (2)( 9,8)Δy  Δy = - 5,10 m (5,102) Height = 40 + 5,10  = 45,10 m  OPTION 2/OPSIE 2 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: v f = v i + aΔt 0 = (10) + (- 9,8)Δt Accept swopping of vi and vf Δt = 1,02 s Aanvaar die omruiling van vi en vf Δy = v i Δt + ½ aΔt2 OR/OF = (10)(1,02) + ½ (-9,8)(1,02)2  v + vf  = 5,10 m Δy =  i  Δt   2   10 + 0  = (1,02)   2  = 5,10 m Height = 40 + 5,10  = 45,10 m  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: v f = v i + aΔt 0 = (-10) + ( 9,8)Δt Accept swopping of vi and vf Δt = 1,02 s Aanvaar die omruiling van vi en vf Δy = v i Δt + ½ aΔt2  = (-10)(1,02) + ½ (9,8)(1,02)2  = 5,10 m Height = 40 + 5,10  = 45,10 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 6 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Δy = v i Δt + ½ aΔt2 0 = (10) Δt + ½(-9,8)Δt2 Δt = 2,04 s ½ Δt Δy = v i Δt + ½ aΔt2 = (10)(1,02) +½ (-9,8)(1,02)2 = 5,10 m Height = 40 + 5,10 = 45,10 m  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: Δy = v i Δt + ½ aΔt2 0 = (-10) Δt + ½(9,8)Δt2 Δt = 2,04 s ½ Δt Δy = v i Δt + ½ aΔt2 = (-10)(1,02) + ½ (9,8)(1,02)2  = - 5,10 m Height = 40 + 5,10  = 45,10 m  OPTION 4/OPSIE 4 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Δy = v i Δt + ½ aΔt2 0 = (10)Δt + ½(-9,8)Δt2 Δt = 2,04 s  v + vf  ½ Δt Accept swopping of vi and vf Δy =  i  Δt  Aanvaar die omruiling van vi en vf  2   10 + 0  = (1,02)   2  = 5,10 m Height = 40 + 5,10  = 45,10 m  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: Δy = v i Δt + ½ aΔt2 0 = (-10) Δt + ½(9,8)Δt2 Δt = 2,04 s  v + vf  ½ Δt Accept swopping of vi and vf Δy =  i  Δt   2  Aanvaar die omruiling van vi en vf  - 10 + 0  = (1,02)   2  = - 5,10 m Height = 40 + 5,10  = 45,10 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 7 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 E (mech/meg)roof/dak = E (mech/meg)top/bo (E p + E k ) roof/dak = (E p + E k ) top/bo Any one/Enige een (mgh + ½ mv2) roof/dak = (mgh + ½ mv2) top/bo [m(9,8)(0) + ½m (10)2 = m(9,8) (h)+ 0)]  h = 5,10 m Height = 40 + 5,10  = 45,10 m  OPTION 6/OPSIE 6 W net = ΔE k  wΔxcosθ = ½ mv f 2 – ½ mv i 2 (m)(9,8)Δxcos180o = 0 – ½ m(10)2 Δx = 5,10 m Height = 40 + 5,10  = 45,10 m  OPTION 7/OPSIE 7 W nc = ΔE p + ΔE k  0 = m(9,8)(h f – 0) + ½ m( 0 – 102)  h f = 5,10 m Height = 40 + 5,10  = 45,10 m  OPTION 8/OPSIE 8 Marking criteria/Nasienriglyne: • Appropriate formula/Toegepaste formule  • Substitution left/Vervanging links  • Substitution right/Vervanging regs • Final answer/Finale antwoord: 45,10 m  E (mech/meg)roof/dak = E (mech/meg)top/bo (E p + E k ) roof/dak = (E p + E k ) top/bo Any one/Enige een (mgh + ½ mv2) roof/dak = (mgh + ½ mv2) top/bo m(9,8)(40) + ½m (10)2 = m(9,8) (h)+ 0)  h = 45,10 m  (4) 3.3 9,8 m·s-2  downwards/afwaarts  (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 8 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne 3.4 Marking criteria/Nasienriglyne • Calculation/use of 10,26 m./Berekening/gebruik van 10,26 m.  • Appropriate formula to calculate Δt/Toepaslike formule om Δt te bereken • Substitution for stone A/Vervanging vir klip A  • Substitution for stone B/Vervanging vir klip B  • Calculating time difference between two stones. /Berekening van tydverskil tussen klippe.  • Final answer/Finale antwoord: 1,34 (s)  OPTION 1/OPSIE 1 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Displacement from roof to meeting point /Verplasing vanaf dak tot ontmoetingspunt = -40 + 29,74 = - 10,26 m Stone/Klip A Δy A = v i Δt + ½aΔt2  -10,26 = 10Δt + ½(-9,8)Δt2  Δt = 2,79 s Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy -10,26 = 0 + ½(-9,8)Δt2  = 02 + 2(-9,8)(-10,26) Δt = 1,45 s (1,447 s) v f = -14,18 m·s-1 x = 2,79 – 1,45  = 1,34 (s)  v f = v i + aΔt -14,18 = 0 + (-9,8)Δt OR/OF Δt = 1,45 s [-10,26 = 0(2,79 – x) + ½ (-9,8)(2,79 – x)2] x = 1,34 (s) x = 2,79 – 1,45  = 1,34 (s)  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: Displacement from roof to meeting point /Verplasing vanaf dak tot ontmoetingspunt = 40 - 29,74 = 10,26 m Stone/Klip A Δy A = v i Δt + ½aΔt2  10,26 = -10Δt + ½(9,8)Δt2  Δt = 2,79 s Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy 10,26 = 0 + ½(9,8)Δt2  = 02 + 2(9,8)(10,26) Δt = 1,45 s (1,447 s) v f = 14,18 m·s-1 x = 2,79 – 1,45  = 1,34 (s)  v f = v i + aΔt 14,18 = 0 + (9,8)Δt OR/OF Δt = 1,45 s [-10,26 = 0(2,79 – x) + ½ (-9,8)(2,79 – x)2] x = 1,34 (s)  x = 2,79 – 1,45  = 1,34 (s)  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 9 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Displacement from roof to meeting point /Verplasing vanaf dak tot ontmoetingspunt= - 40 + 29,74 = - 10,26 m Displacement of stone A from max height to meeting point/ Verplasing van klip A vanaf maksimum hoogte tot ontmoetingspunt = -15,36 m Stone/Klip A v f = v i + aΔt 0 = 10 +(-9,8)Δt Δt = 1,02 s Δy A = v i Δt + ½aΔt2  -15,36 = 0 + ½(-9,8)Δt2  Δt = 1,77 s Δt tot = 1,77 + 1,02 = 2,79s Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy -10,26 = 0 + ½(-9,8)Δt2  = 02 + 2(-9,8)(-10,26) Δt = 1,45 s (1,447 s) v f = -14,18 m·s-1 x = 2,79 – 1,45  = 1,34 (s)  v f = v i + aΔt -14,18 = 0 + (-9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: Displacement from roof to meeting point /Verplasing vanaf dak tot ontmoetingspunt= 40 - 29,74 = 10,26 m  Displacement of ball A from max height to meeting point/ Verplasing van bal A vanaf maksimum hoogte tot ontmoetingspunt = 15,36 m Stone/Klip A v f = v i + aΔt 0 = -10 + (9,8)Δt Δt = 1,02 s Δy A = v i Δt + ½aΔt2  15,36 = 0 + ½(9,8)Δt2  Δt = 1,77 s Δt tot = 1,77 + 1,02 = 2,79s Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy 10,26 = 0 + ½(9,8)Δt2  = 02 + 2(9,8)(10,26) Δt = 1,45 s (1,447 s) v f = 14,18 m·s-1 x = 2,79 – 1,45  = 1,34 (s)  v f = v i + aΔt 14,18 = 0 + (9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 10 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Displacement of stones A and B from roof to meeting point/Verplasing van klippe A en B vanaf dak tot by ontmoetingspunt = - 40 + 29,74 = - 10,26 m Stone/Klip A v f = v i + aΔt 0 = 10 + (-9,8)Δt Δt = 1,02 s Δy A = v i Δt + ½aΔt2 - 10,26 = -10 + ½(-9,8) Δt2  Δt = 0,75 s Δt tot = 1,02 + 1,02 + 0,75 = 2,79 s Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy -10,26 = 0 + ½(-9,8)Δt2  = 02 + 2(-9,8)(-10,26) Δt = 1,45 s (1,447 s) v f = -14,18 m·s-1 x = 2,79 – 1,45 = 1,34 (s)  v f = v i + aΔt -14,18 = 0 + (-9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: Displacement of stones A and B from roof to meeting point/Verplasing van klippe A en B vanaf dak tot by ontmoetingspunt = 40 - 29,74 = 10,26 m Stone/Klip A v f = v i + aΔt 0 = -10 + (9,8)Δt Δt = 1,02 s Δy A = v i Δt + ½aΔt2  10,26 = 10 + ½ (9,8)Δt2  Δt = 0,75s Δt tot = 1,02 + 1,02 + 0,75 = 2,79 s Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy 10,26 = 0 + ½(9,8)Δt2  = 02 + 2(9,8)(10,26) Δt = 1,45 s (1,447 s) v f = 14,18 m·s-1 x = 2,79 – 1,45 = 1,34 (s)  v f = v i + aΔt 14,18 = 0 + (9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 11 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Displacement from roof to meeting point /Verplasing vanaf dak tot ontmoetingspunt = - 40 + 29,74 = - 10,26 m Stone/Klip A Δy A = v i Δt + ½aΔt2 - 5,10 = 0 + ½(-9,8)Δt2 Δt = 1,02 s Δy A = v i Δt + ½aΔt2  - 10,26 = -10 + ½(-9,8)Δt2  Δt = 0,75s Δt tot = 1,02 + 1,02 + 0,75 = 2,79 s Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy -10,26 = 0 + ½(-9,8)Δt2  = 02 + 2(-9,8)(-10,26) Δt = 1,45 s (1,447 s) v f = -14,18 m·s-1 x = 2,79 – 1,45 = 1,34 (s)  v f = v i + aΔt -14,18 = 0 + (-9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: Displacement from roof to meeting point /verplasing vanaf dak tot by ontmoetingspunt = 40 - 29,74 = 10,26 m Stone/Klip A Δy A = v i Δt + ½aΔt2 5,10 = 0 + ½(9,8)Δt2 Δt = 1,02 s Δy A = v i Δt + ½aΔt2  10,26 = 10 + ½(9,8)Δt2  Δt = 0,75s Δt tot = 1,02 + 1,02 + 0,75 = 2,79 s Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy 10,26 = 0 + ½(9,8)Δt2  = 02 + 2(9,8)(10,26) Δt = 1,45 s (1,447 s) v f = 14,18 m·s-1 x = 2,79 – 1,45 = 1,34 (s)  v f = v i + aΔt 14,18 = 0 + (9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 12 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Displacement from roof to meeting point /Verplasing vanaf dak tot ontmoetingspunt = - 40 + 29,74 = - 10,26 m Displacement of stone A from max height to meeting point/ Verplasing van klip A vanaf maksimum hoogte tot ontmoetingspunt = -15,36 m Stone/Klip A v f 2 = v i 2 + 2aΔy v f 2 = (0)2 + (2)(- 9,8)(-15,36) v f = - 17,35 m·s-1 v f = v i + aΔt -17,35 = 0 + (-9,8)Δt  Δt = 1,77 s  Δt tot = 1,02 + 1,77 = 2,79 (s) Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy -10,26 = 0 + ½(-9,8)Δt2  = 02 + 2(-9,8)(-10,26) Δt = 1,45 s (1,447 s) v f = -14,18 m·s-1 v f = v i + aΔt x = 2,79 – 1,45  = 1,34 (s)  -14,18 = 0 + (-9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: Displacement from roof to meeting point /Verplasing vanaf dak tot ontmoetingspunt= 40 - 29,74 = 10,26 m Displacement of stone A from max height to meeting point/ Verplasing van klip A vanaf maksimum hoogte tot ontmoetingspunt = 15,36 m Stone/Klip A v f 2 = v i 2 + 2aΔy v f 2 = (0)2 + (2)(9,8)(15,36) v f = - 17,35 m·s-1 v f = v i + aΔt 17,35 = 0 + (9,8)Δt  Δt = 1,77 s  Δt tot = 1,02 + 1,77 = 2,79 (s) Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy 10,26 = 0 + ½(9,8)Δt2  = 02 + 2(9,8)(10,26) Δt = 1,45 s (1,447 s) v f = 14,18 m·s-1 x = 2,79 – 1,45  = 1,34 (s)  v f = v i + aΔt 14,18 = 0 + (9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 13 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne OPTION 6/OPSIE 6 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Displacement from roof to meeting point/Verplasing vanaf dak tot by ontmoetingspunt = - 40 + 29,74 = - 10,26 m Stone/Klip A v f 2 = v i 2 + 2aΔy v f 2 = (-10)2 + (2)(- 9,8)(-10,26) v f = - 17,35 m·s-1 v f = v i + aΔt -17,35 = -10 + (-9,8)Δt   Δt = 0,75 s Ball A: Δt = 1,02 + 1,02 + 0,75 = 2,79 (s) Stone/Klip B Δy B = v i Δt + ½ aΔt2 v f 2 = v i 2 + 2aΔy -10,26 = 0 + ½ (-9,8)Δt2  = 02 + 2(-9,8)(-10,26) Δt = 1,45 s (1,447 s) v f = -14,18 m·s-1 x = 2,79 – 1,45 = 1,34 (s)  v f = v i + aΔt -14,18 = 0 + (-9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: Displacement from roof to meeting point/Verplasing vanaf dak tot by ontmoetingspunt = 40 - 29,74 = 10,26 m Stone/Klip A v f 2 = v i 2 + 2aΔy v f 2 = [(10)2 + (2)(9,8)(10,26) v f = 17,35 m·s-1 v f = v i + aΔt 17,35 = 10 + (9,8)Δt   Δt = 0,75 s Ball A: Δt = 1,02 + 1,02 + 0,75 = 2,79 (s) Stone/Klip B Δy B = v i Δt + ½aΔt2 v f 2 = v i 2 + 2aΔy 10,26 = 0 + ½(9,8)Δt2 = 02 + 2(9,8)(10,26) Δt = 1,45 s (1,447 s) v f = 14,18 m·s-1 x = 2,79 – 1,45  = 1,34 (s)  v f = v i + aΔt 14,18 = 0 + (9,8)Δt Δt = 1,45 s x = 2,79 – 1,45  = 1,34 (s)  (6) 3.5.1 d  Accept / Aanvaar ( 0 – e; 0 – d; d – e) (1) 3.5.2 a (1) 3.5.3 f (1) 3.5.4 c (1) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 14 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 NOTE: -1 mark for each key word/phrase omitted in the correct context. LET WEL: -1 punt vir elke sleutelwoord/frase weggelaat in die korrekte konteks. Isolated system is a system on which the resultant/net external force is zero. Geïsoleerde sisteem is 'n sisteem waarop die resultante / netto eksterne krag nul is. OR/OF Isolated system is one that has no net / external force acting on it. ‘n Geïsoleerde stelsel is een wat geen netto eksterne krag het wat daarop inwerk nie. (2) 4.2.1 p = mv  Note: p and v must have the same sign 24 = m (480)  Let wel: p en v moet dieselfde tekens hê m = 0,05 kg  (3) 4.2.2 Marking criteria/Nasienriglyne • Appropriate formula including F net or W net. /Toepaslike formule wat F net of W net insluit.  • Substitutions/Vervangings  • Final answer/Finale antwoord: 2 000 N  • Correct direction/Korrekte rigting: west or left/Wes of links  POSITIVE MARKING FROM QUESTION 4.2.1 POSITIEWE NASIEN VANAF VRAAG 4.2.1 OPTION 1/OPSIE 1 F net ∆t = ∆p F net ∆t = (p bullet ) f - (p bullet ) i Any one/Enige een F net ∆t = (mv bullet ) f - (mv bullet ) i F net (0,01)  = (0,05)(80) – 24  or/of (0,05)(80) – (0,05)(480) F net = -2 000 N F net = 2 000 N  west/wes  OPTION 2/OPSIE 2 v f = v i + aΔt 80 = 480 + a(0,01)  a = - 40 000 m∙s-2 F net = ma  = (0,05)(-40 000)  = - 2 000 N F net = 2 000 N  west/wes  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 15 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3  v +vf  Δx =  i  Δt  2  480 + 80 = (0,01) 2 = 2,80 m v f = v i 2 + 2aΔx 2 W net = ∆K  (80)2 = (480)2 + 2a(2,80)  F net Δxcosθ = ½mv f 2 - ½mv i 2 a = - 40 000 m∙s-2 F net (2,80)cos0°  = ½(0,05)(802 - 4802)  F net = -2 000 N F net = ma  F net = 2 000 N  west/wes  = (0,05)(-40 000)  = - 2 000 N OR/OF 2 2 F net = 2 000 N  west/wes  F net (2,80)cos180°  = ½(0,05)(80 - 480 )  F net = 2 000 N  west/wes  (5) [10] QUESTION 5 5.1 Note: -1 mark for each key word/phrase omitted in the correct context. Let Wel: -1 punt vir elke sleutelwoord/frase weggelaat in die korrekte konteks. IF: The word "work" is omitted - 0 marks. INDIEN: Die woord "arbeid" uitgelaat is - 0 punte. A conservative force is a force for which the work done (in moving an object between two points) is independent of the path taken.  'n Konserwatiewe krag is 'n krag waarvoor die arbeid wat verrig is (om 'n voorwerp tussen twee punte te beweeg) onafhanklik is van die pad wat gevat word. OR/OF A conservative force is a force for which the work done in moving an object in a closed path is zero. ‘n Konserwatiewe krag is ‘n krag waarvoor die arbeid verrig om ‘n voorwerp in ‘n geslote pad te beweeg, nul is. (2) 5.2 Gravitational (force)/Gravitasiekrag  ACCEPT/AANVAAR: Gravitation /Gravity /Gravitasie / Weight /Gewig (1) 5.3 No/Nee  There is friction/non-conservative force (doing work)/It is not isolated system. Daar is wrywing/nie konserwatiewe krag (wat arbeid verrig)./Dit is nie ’n geïsoleerde sisteem nie. OR/OF The net work done by the non-conservative forces is not zero/Die netto arbeid deur die nie-konserwatiewe kragte is nie nul nie.  (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 16 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne 5.4 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 E P = mgh  W w = -ΔE p  = (1,8)(9,8) (1,5)  (1,8)(9,8)(h – 0)cos180o = -(E pA – E p(ground) ) = 26,46 J  (1,8)(9,8)(1,5)(-1) = -E pA  E p = 26,46 J OR/OF W = FΔxcosθ = mgΔhcosθ Any one/Enige een = (1,8)(9,8) (1,5)cos0o  = 26,46 J (3) 5.5 POSITIVE MARKING FROM QUESTION 5.4 / POSITIEWE NASIEN VANAF VRAAG 5.4 OPTION 1/OPSIE 1 W nc = ∆K + ∆U Any one/Enige een W f = ½m(v f 2 – v i 2) + mg(h f – h i ) = ½(1,8)(42 – 0,952) + (0 – 26,46)  = -12,87 J  OPTION 2/OPSIE 2 W net = ∆K W f + W g = ½mv f 2 - ½mv i 2 Any one/Enige een W f + mgh = ½m(v f 2- v i 2) W f + mgh = ½mv f 2 - ½mv i 2 W f + 26,46 = ½(1,8)[(4)2 - (0,95)2]  W f = -12,87 J (- 12,872 J)  OPTION 3/OPSIE 3 E (mech/meg)A = E (mech)B - W f (E p + E k ) A = (E p + E k ) B - W f Any one/Enige een 2 2 (mgh + ½ mv ) A = (mgh + ½ mv ) B - W f 26,46 + ½(1,8)(0,952)  = 0 + ½(1,8)(42) - W f  W f = -12,87 J  (4) 5.6 W net = 0 (J ) / zero (1) [13] QUESTION 6/VRAAG 6 6.1 Doppler effect/Doppler-effek  (1) 6.2 (Q): (records sounds with) longer period/ longer time per wave / lower frequency. (Q): (teken klank aan met) langer periode / langer tyd per golf / laer frekwensie. OR/OF P: (records sounds with) shorter period/ shorter time per wave / higher frequency.  P: (teken klank aan met) korter periode/ korter tyd per golf / hoër frekwensie. ACCEPT/AANVAAR (Q): longer wavelength. /P: shorter wavelength. (Q): langer golflengte./P: korter golflengte het. (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 17 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne 6.3 OPTION 1/OPSIE 1 1 1 f= = -4  = 5,88 x 102 = 588,24 Hz  T 17 x 10 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 dis tan ce λ speed = v= time T dis tan ce λ 340 = −4 340 = 25,5 × 10 17 × 10 −4 Distance = 0,867 m ∴λ = 0,578 m Distance = 1 ½ λ ∴λ = 0,578 m v = fλ  v = fλ  340 = f(0,578)  340 = f(0,578)  f = 588,24 Hz  f = 588,24 Hz  (3) 6.4 POSITIVE MARKING FROM QUESTIONS 6.2 AND 6.3. POSITIEWE NASIEN VANAF VRAE 6.2 EN 6.3 Do not penalise if 10-4 is again omitted. /Moenie penaliseer indien 10-4 weer uitgelaat is nie. OPTION 1/OPSIE 1 1 f=  = 5,56 x 10 2 = 555,56 Hz 18 x 10 - 4 v ± vL v fL = fs  OR/OF fL = fs v ± vs v + vs  340 555,56 =  588,24  340 + v v = 20 m∙s-1  Range/Gebied 19,57 – 20,09 m·s-1 OPTION 2/OPSIE 2 1 fL =  18 x 10 - 4 v ± vL v fL = f s  OR/OF fL = fs v ± vs  v + v s 1   340  1 −4 =   −4  18 × 10  340 + v  17 × 10 v = 20 m∙s-1  Range/Gebied 19,57 – 20,09 m·s-1 (6) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 18 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1.1 Positive/Positief  (1) 7.1.2 Marking criteria/Nasienriglyne: • Appropriate formula /Toepaslike formule • Whole substitution Hele vervanging/ • Final answer/finale antwoord: 2,26 x 10-6 C  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 kQ 1Q 2 kQ F=  E= r2 r2 (9 x 10 9 )(6 x 10 −6 ) Q (9 x 10 9 )(6 x 10 −6 ) 3,05 =  = 0,2 2 0,2 2 Q = 2,26 x 10-6 C  = 1,35 x 106 N·C-1 (2,259 x 10-6 C) F = Eq 3,05 = (1,35 x 106)q q = 2,26 x 10-6 N (3) 7.1.3 T Accepted labels/Aanvaarde benoemings F g / F w / weight / mg / gravitational force w F g / F w / gewig / mg / gravitasiekrag T F T / tension / spanning FE Electrostatic force/ Coulomb force/ F E Field /F x on Y / F E /F 3,05 N w Elektrostatiese krag/ Coulombkrag 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 force(s) do not make contact with dot /Indien krag(te) nie met die kolletjie kontak maak nie: Max/Maks 2 3 • If arrows missing/Indien pyltjies uitgelaat word: Max/Maks 2 3 (3) 7.1.4 F net = 0 F E = Tsin10° F E = Tcos80° [3,05 = Tsin10°] [IF /INDIEN Tcos10o = 3,05 ( 1 )] 3 OR/OF [3,05 = Tcos80°] [IF/INDIEN Tsin80o = 3,05 ( 1 )] 3 T = 17,56 N  (17,564 N) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 19 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne 7.2.1 Marking criteria/Nasienriglyne -1 mark for each key word/phrase omitted in the correct context. -1 punt vir elke sleutelwoord/frase weggelaat in die korrekte konteks. The electric field at a point is the (electrostatic) force experienced per unit positive charge placed at that point.  Die elektriese veld by 'n punt is die (elektrostatiese) krag wat per positiewe eenheidslading wat by die punt geplaas is, ondervind word. [IF the word “unit” or phrase “positive charge” is omitted in this definition: -1 for each INDIEN die woord “eenheid” of frase “positiewe lading” uitgelaat is in hierdie definisie: -1 vir elk] OR/OF The electric field at a point is the (electrostatic) force experienced by a UNIT positive charge placed at that point.  Die elektriese veld by 'n punt is die (elektrostatiese) krag wat deur ‘n positiewe EENHEIDSlading wat by die punt geplaas is, ondervind word. [If “UNIT” is omitted in this definition, then 0 marks. Indien “EENHEIDS” uitgelaat word in hierdie definisie, dan 0 punte.] (2) 7.2.2 OPTION 1/OPSIE 1 Electric field at M due to A (+2 x10-5 C): kQ EA = 2  r (2 × 10 -5 ) = 9 × 10 9  (0,2) 2 = 4,5 x106 N∙C-1 Electric field at M due to B (-4 x10-5 C): kQ EB = 2 OR/OF q B = 2q A r ( 4 × 10 -5 ) = 9 × 10 9  E B = 2E A  (0,2) 2 = 9 x106 N∙C-1 = 9 x 106 N∙C-1 E net at M = E A + E B = (4,5 x 106 + 9 x 106)  = 1,35 x 107 N∙C-1  to the right/na regs/towards B/na B /away from A / weg vanaf A OPTION 2/OPSIE 2 Net electrostatic force at M / Netto elektrostatiese krag by M kQ1Q 2 kQ1Q 2 kQ F net = 2 + 2 If/Indien F = 2 Max/Maks 2 r r r 6 9 −5 9 −5 (9 × 10 )(2 × 10 )q (9 × 10 )( 4 × 10 )q = 2 +  (any one/ enige een) (0,2) (0,2)2 = 4,5 x 106q + 9 x 106 q = 1,35 x 107q N F net = E net q  1,35 x 107q = E net q E net = 1,35 x 107 N.C-1 to the right/na regs /towards B / na B (6) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 20 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 (Maximum) energy provided (work done)  by a battery per coulomb / unit charge passing through it.  (Maksimum) energie verskaf (arbeid verrig) deur 'n battery per coulomb/eenheidlading wat daardeur beweeg. ACCEPT/AANVAAR: The reading on a voltmeter connected across a battery when there is no current/ in an open circuit. Lesing op 'n voltmeter oor 'n battery as daar geen stroom is nie (2) 8.2 13 V  (1) 8.3.1 V Marking criteria/Nasienriglyne: R =  • Appropriate formula/Toepaslike formule  I • Whole substitution/Hele vervanging  10,5 5,6 =  • Final answer/Finale antwoord: 1,88 A  I I = 1,88 A  (1,875 A) (3) 8.3.2 POSITIVE MARKING FROM QUESTION 8.3.1. POSITIEWE NASIEN VANAF VRAAG 8.3.1 OPTION 1 OPTION 2 P = VI  P = I2R  = (10,5)(1,88)  = (1,88)2(5,6)  = 19,74 W  (19,688 W) = 19,79 W  (19,688 W) OPTION 3 V2 P=  R 10,5 2 =  5,6 = 19,69 W  (19,688 W) (3) 8.3.3 POSITIVE MARKING FROM QUESTIONS 8.2 AND 8.3.1. POSITIEWE NASIEN VANAF VRAE 8.2 EN 8.3.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Ɛ = I(R + r)  V 13 = 1,88 (5,6 + r)  r = int ernal  I r = 1,31 Ω  (1,31 – 1,33 Ω) 2,5 =  1, 88 = 1,33 Ω  ( 1,31 – 1,33 Ω) OPTION 3/OPSIE 3 Ɛ = V ext + V int 13 = 10,5 + V int V int = 2,5 V V int = Ir  2,5 = (1,88)r r = 1,31 Ω  (1,31 – 1,33 Ω) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 21 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne 8.4.1 Decreases/Neem af  - V internal resistance /Internal volts increase  V interne weerstand /Interne volts neem toe (2) 8.4.2 Marking criteria/Nasienriglyne • Formula/Formule Ɛ = I(R + r)  • Correct substitution into/ Korrekte vervanging in Ɛ = I(R + r)  • Substitution of values into R p formula/Vervanging van waarde van R p in formule • Halving value of R 2X /Halvering van waarde van R 2X  • Final answer/Finale antwoord: 1,49 Ω  Range/Gebied: 1,46 Ω – 1,49 Ω POSITIVE MARKING FROM QUESTIONS 8.2 AND 8.3.3 POSITIEWE NASIEN VANAF VRAE 8.2 EN 8.3.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Ɛ = I(R + r)  Ɛ = I(R + r)  13 = 4 (R ext + 1,31)  13 = 4(R ext + 1,31)  R ext = 1,94 Ω (1,92 Ω) R ext = 1,94 Ω (1,92 Ω) 1 1 1 R1 R 2 = + Rp = RP R1 R 2 R1 + R 2 5,6 R 2 1 1 1 1,94 =  = +  5,6 + R 2 1,94 5,6 R 2 R 2 = 2 , 97 Ω (2,92 Ω) R 2 = 2,97 Ω (2,92 Ω) 1 1 X = (2 , 97)  X = (2 , 97)  2 2 = 1, 49 Ω  (1,46 – 1,49 Ω) = 1, 49 Ω  (1,46 – 1,49 Ω) OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 Ɛ = I(R + r)  Ɛ = I(R + r)  13 = 4(R ext + 1,31)  13 = 4(R ext + 1,31)  R ext = 1,94 Ω (1,92 Ω) R ext = 1,94 Ω (1,92 Ω) 1 1 1 R1 R 2 = + Rp = RP R1 R 2 R1 + R 2  1 1 1   (5,6)(2 X)  1,94 = 5,6 + 2 X  1,94 = 5,6 + 2 X      X = 1, 49 Ω  (1,46 – 1,49 Ω) X = 1, 49 Ω  OPTION 5/OPSIE 5 OPTION 6/OPSIE 6 Ɛ = I(R + r)  Ɛ = I(R + r)  V ext = 13 – (4)(1,31)  V ext = 13 – (4)(1,31)  = 7,76 V V ext = 7,76 V V p = IR 5,6 7,76 7,76 = I(5,6) I 5,6Ω = = 1,39 A 5,6 I 5,6Ω = 1,37 A I 2x = 4 – 1,39 = 2,61 A I T = I 2X + I 5,6 4 = I 2x + 1,37 V 2x = I 2X R 2X V X = I XR X I 2X = 2,63 A [7,76 = (2,61)2X] 3,88=(2,61)R X  V = IR 2X 2X = 2,97 Ω R X = 1,49 Ω [7,76 = (2,63)2X] X = 1,49 Ω X = 1,46 Ω Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 22 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne OPTION 7/OPSIE 7 Ɛ = I(R + r)  V ext = 13 – (4)(1,31)  = 7,76 V V ext = IR ext −1  1 1  7,76 = (4)  +   2 X 5,6  X = 1,48 Ω  (5) [19] QUESTION 9/VRAAG 9 9.1 9.1.1 DC/GS  (1) 9.1.2 NOTE: -1 mark for each key word/phrase omitted in correct context. LET WEL: -1 punt vir elke sleutel woord/frase weggelaat in die korrekte konteks. Emf is induced as a result of change of magnetic flux (linked) with the coil.  Emk word geïnduseer as gevolg van verandering van die magnetiese vloedkoppeling. (2) 9.1.3 POSITIVE MARKING FROM QUESTION 9.1.1 POSITIEWE NASIEN VANAF VRAAG 9.1.1 V (V) V (V) t (s) t (s) OR/OF Marking criteria for graph: Correct shape Korrekte vorm  One complete cycle/Een volledige siklus  V (V) t (s) V(V) t (s) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 23 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne 9.2.1 The AC potential difference/voltage which dissipates the same amount of energy  as DC potential difference.  Die WS potensiaalverskil/spanning wat dieselfde hoeveelheid energie verbruik as GS potensiaalverskil/spanning OR/OF (The rms value of AC is) the DC potential difference/voltage which dissipates the same amount of energy  as AC potential difference/voltage.  Dit is die GS potensiaalverskil/spanning wat dieselfde hoeveelheid energie verbruik as WS potensiaalverskil/spanning. (2) 9.2.2 OPTION 1/ OPTION 2 / OPTION 3 / OPTION 4 / OPSIE 1 OPSIE 2 OPSIE 3 OPSIE 4 V2 W = I2R ∆ t 2 Pave = Irms R V W= Δt  500 = I2 (200)(10) R = rms  R 500 Irms I = I rms = 0,5 A = I2 rms V2 10 V 500 = (10)  Pave = VrmsIrms  200 = rms  200 (200) 0,5 500 I rms = 0,5 A = V rms = 100 V V = V rms = 100 V 10 Pave = VrmsIrms  V rms (0,5) Vmax 500 Vmax = V rms (0,5) V rms =  V rms =  V rms = 100 V 10 2 2 V  Vmax V rms = max  V rms = 100 V 100 =  Vmax 2 2 100 =  2 Vmax V V max = 141,42 V V rms = max  V max = 141,42 V 100 = 2  2 V max = 141,42 V 100 = Vmax  2 V max = 141,42 V OPTION 5 / OPTION 6 / OPTION 7 / OPTION 8 / OPSIE 5 OPSIE 6 OPSIE 7 OPSIE 8 V 2 P W 500 Pave = Irms2 R P ave = rms  P ave = max  P= = R 2 ∆t 10 500 500 2 = 50 W = V 500 P 10 = rms  = max  10 200 10 2 V2 I2 rms (200) P ave = rms  V rms = 100 V P max = 100 W R I rms = 0,5 A Vmax 2 Vmax 2 Vrms I V rms =  P max =  50 =  I rms = max 2 R 200 2 V V2 V rms = 100 V Imax 100 = max  100 = max  0,5 = 2 200 V 2 V rms = max  V max = 141,42 V V max = 141,42 V 2 I max = 0,71 A Vmax 100 =  V max = I max R  2 V max = 141,42 V = (0,71)(200) = 142 V (5) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 24 DBE/November 2019 NSC/NSS –Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1 Note: -1 mark for each key word/phrase omitted in correct context. Let Wel: -1 punt vir elke sleutel woord/frase weggelaat in die korrekte konteks. The process whereby electrons are ejected from a metal / surface when light (of suitable frequency) is incident on that surface.  Die proses waarby elektrone vanaf 'n (metaal)oppervlak vrygestel word wanneer lig (van geskikte frekwensie) daarop skyn/inval. (2) 10.2 7,48 x 10-19 (J)  - E = W o + E k(max) (= W o + ½mv2 max)  When/Wanneer E k(max) = 0 / v = 0 / v2 = 0 / E = Wo / W o is the y-intercept / is die y-afsnit  (3) 10.3 Mass (of photo-electron)/Massa (van foto-elektron)/m  ACCEPT/AANVAAR: ½m (1) 10.4 OPTION 1/OPSIE 1 Gradient = ½m 11,98x 10 -19- 7,48 x 10 -19  -31 = ½(9,11 x 10 )  X - 0 X = 0,9879  (0,99 or 0,988) ACCEPT/AANVAAR X = 0,9879 x 1012 (m2∙s-2) POSITIVE MARKING FROM 10.2/POSITIEWE NASIEN VANAF 10.2 OPTION 2/ OPSIE 2 E = W o + Ek(max) Any one / Enige een E = W o + ½mv2 (max) 11,98 x 10-19  = 7,48 x 10-19 + ½(9,11 x 10-31) v2 [or/of ½(9,11 x 10-31)X] 2 4,5 x 10-19 = 4,56 x 10-31v v2 = 0,9868 x 1012 X/v2 = 0,9868  (0,99) Range/gebied (0,9868 – 0,9879 / 9,87 x 1011 – 9,88 x 1011) ACCEPT/AANVAAR: X = 0,9868 x 1012 (m2∙s-2) / 9,868 x 1011 (m2∙s-2) (5) 10.5.1 Remains the same /Bly dieselfde  (1) 10.5.2 Increases / Neem toe  (1) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou