Physical Sciences Grade 12 P1 Gauteng Prelim 2024 Memo (Afr)

Downloaded from hlayiso.com PREPARATORY EXAMINATION 2024 MARKING GUIDELINE PHYSICAL SCIENCES: PHYSICS (PAPER 1) (10841) 16 pages

Downloaded from hlayiso.com 10841/24 QUESTION 1/VRAAG 1 1.1 B  (2) 1.2 C  (2) 1.3 B  (2) 1.4 B  (2) 1.5 A  (2) 1.6 C  (2) 1.7 C (2) 1.8 D  (2) 1.9 C  (2) 1.10 B  (2) [20] 2

Downloaded from hlayiso.com 10841/24 QUESTION 2 / VRAAG 2 2.1 The net force acting on a body is the vector sum  of two or more forces  acting on that body. Die netto krag wat op ʼn liggaam inwerk is die vektor som  van twee of meer kragte  wat op die liggaam inwerk. OR/OR A single vector having the same effect as two or more vectors together. Die enkel vektor wat dieselfde effek het as twee of meer vektore saam. (2) Marking criteria/ Nasienkriteria: Mark for scientifically correct explanation of the net force. Exact wording is not required. Merk vir wetenskaplik korrekte verduideliking van netto krag. Bewoording hoef nie identies soos bo te wees nie. 2.2 Accepted labels/ Aanvaarde benoemings w Fg/Fw/weight/gravitational force/gewig/gravitasiekrag N N/FN/Normal Force/Normaalkrag f fs/fk/friction /wrywing T2 T/T/FA/Force applied/tension at an angle/or its components/ Krag toegepas/spanning teen ʼn hoek T1 T/T/FA/Force applied/tension/ Krag toegepas/spanning Notes: • Mark awarded for label and arrow. • Do not penalise for length of arrows, drawing is not to scale. • Any other additional force(s) Max 4�5 • If force(s) do not make contact with body Max 4�5 • If no arrows indicated and all forces correctly drawn deduct 1 mark 4�5 • Diagram must preferably not be in pencil Aantekeninge: • Punt toegeken vir benoeming en pyltjie. • Moenie penaliseer vir lengte van pyle nie; skets is nie volgens skaal nie. • Enige ander addisionele krag(te) Maks 4�5 • Indien krag(te) nie kontak maak met liggaam nie Maks 4�5 • Indien geen pyle nie, maar alle kragte is reg geteken, trek 1 punt af Maks 4�5 • Diagram moet verkieslik nie in potlood wees nie. (5) 3

Downloaded from hlayiso.com 10841/24 2.3 OPTION 1/ OPSIE 1: OPTION 2/ OPSIE 2: CAR/kar: FNet =m x a  Car/kar: Fnet = ma = +T1 – f – Fg//  T1 - Fg∥ - Ff =1a 1a = +T1 – 1,5 – (1)(9,8)(sin 20 °) T1 – 3,35 -1,5 =1a 1a = +T1 – 5,5318 ……..(1)  T1 = 4,85 +a  TRAIN/Trein: FNET =ma Train /trein: Fnet = ma = +T2x – f – Fg// – T1 FHORIZ -T1 - Fg∥ - Ff =2 a 2a = +(20 x cos 10 °) – 2 – (2)(9,8)(sin 20 °) – T1 19,696-T1 -6,704-2=2a 2a = 11,49 – T1 ……..(2)  1+2  T1 =10,992-2a  1a = +T1 – 5,5318 ……..(1) CAR T1 =TRAIN T1  / 2a = 11,49 – T1 ……..(2) Kar T1= Trein T1 3a = 6,141 4,85+1a =10,992 - 2a a = 2,0468 m.s-2 a=2,047 m.s-2 (5) OPTION 3/ OPSIE 3: Fnet = ma = +T1 – f – Fg//  Marking criteria/ 1a = +T1 – 1,5 – (1)(9,8)(sin 20 °) Nasienkriteria:  Formula/formula T1 = +1a + 4,85 ……..(1)   Substitution for 1 kg/ Fnet = ma = +T2x – f – Fg// – T1 Vervanging vir 1 kg T1 = – 2a +(20)(cos 10°) – 2 – (2)( 9,8)(sin 20 °)  Substitution for 2 kg/ T1 = – 2a + 10,99..…..(2)  Vervanging vir 2 kg T1 = T1   Equating 1 and 2/ 1a + 4,85 = – 2a + 10,99 Gelykstelling van 1 en 2 3a = 6,141  Answer a = 2,05 m.s-2 / a = 2,05 m.s-2  (5) Antwoord a = 2,05 m.s-2 2.4 Remains the same  • The normal force stays the same.  • The frictional force is directly proportional to the normal force ∴ remains the same.  OR • Since the friction force is dependent on the normal force, • and the NORMAL FORCE DOES NOT CHANGE. OR (3) 4

Downloaded from hlayiso.com 10841/24 • The coefficient of the surface remains constant, the incline stays constant, thus the normal stays constant. • The frictional force is directly proportional to the coefficient, thus remains constant. Bly dieselfde  • Die normaalkrag bly dieselfde.  • Die wrywingskrag is direk eweredig aan die normaalkrag ∴ bly dieselfde.  OF • Aangesien die wrywingskrag afhanklik is van die normaalkrag, • en die NORMAALKRAG VERANDER NIE. OF • Die koëffisiënt van die oppervlak bly konstant, die helling bly konstant, dus bly die normaalkrag konstant. • Die wrywingskrag is direk eweredig aan die koëffisiënt, dus bly dit konstant [15] QUESTION 3/VRAAG 3 3.1 Free fall is the motion of a body when the only force acting on it is its gravitational force.  (2 or zero) Vryval is beweging waartydens die enigste krag wat op ʼn voorwerp inwerk, die gravitasiekrag is (2 of nul) (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. 3.2 OPTION 1/OPSIE 1: (down positive/af OPTION 2/OPSIE 2: (up positive/op positief) positief) 1 2 1 ∆y= vi ∆t+at  ∆𝐲𝐲 = vi ∆t + a∆t 2 2 2 1 1 2 44 =0 + (9,8)(6t)2  -44=0+ (-9,8)(6t) 2 2 44=176,4 t2 –44 = – 176,4 t2 44 t = 0,499 s/0,5 s t=� =0,499 s  176,4 (4) 5

Downloaded from hlayiso.com 10841/24 OPTION 3/OPSIE 3: (up positive/op Marking criteria/ Nasienkriteria: positief) 1  Formula / Formule x= vi t+ at2 2  Substitution left /Vervanging links 1  (x and g same signs!)/ Vervanging −44=0+ (-9,8)t2 2 (x en g dieselfde tekens!) –44 = – 4,9 t2  Substitution right / Vervanging regs  Answer (accept 0,5 s or 0,499 s)/ t =3s÷6 Antwoord (aanvaar 0,5 s of 0,499 s) = 0,499 s = 0,5 s 3.3 Positive marking from QUESTION 3.2 Positiewe merk van VRAAG 3.2 OPTION 1/OPSIE 1: (down positive/af OPTION 2/OPSIE 2: (up positive/op positief) positief) 1 2 1 ∆𝑦𝑦 = 𝑣𝑣𝑖𝑖 ∆𝑡𝑡 + 𝑎𝑎𝑡𝑡  ∆y = vi ∆t + a∆t 2  2 2 1 1 = 0 + 2 (9,8)(3 𝑥𝑥 0,499)2  = 0 + 2 (−9,8)(3 x 0,5)2  = 10,98 𝑚𝑚 = - 11,03 m ∴ ℎ = 44 − 10,98  = 33,02 𝑚𝑚  = 11,03 m down ∴ hB =44-11,03 =32,98 m  OPTION 3/OPSIE 3: (down positive/af Marking criteria/ Nasienkriteria: positief)  Formula / Formule 𝒗𝒗𝒇𝒇 = 𝒗𝒗𝒊𝒊 + 𝒂𝒂∆𝒕𝒕  Substitution left / Vervanging = 0 + (9,8 x 1,5) links = 14,7 𝑚𝑚. 𝑠𝑠 −1 Mark for both  (x/y and g same signs!) / equations 𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2𝑎𝑎∆𝑦𝑦  Vervanging 14,72 = 0 + 2(9,8)∆𝒚𝒚  (x/y en g dieselfde tekens!) ∆𝒚𝒚 = 11,025 𝑚𝑚  h – x answer / h – x antwoord ∴ 𝒉𝒉𝑩𝑩 = 𝟒𝟒𝟒𝟒 − 𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎  = 𝟑𝟑𝟑𝟑, 𝟗𝟗𝟗𝟗 𝒎𝒎  (4) 6

Downloaded from hlayiso.com 10841/24 OPTION 4/OPSIE 4: (down positive/af positief) 𝒗𝒗𝒇𝒇 = 𝒗𝒗𝒊𝒊 + 𝒂𝒂∆𝒕𝒕 = 0 + (9,8)(1,5) = 14,7 𝑚𝑚. 𝑠𝑠 −1  Mark for both equations 1 𝑥𝑥 = 𝑣𝑣𝑖𝑖 𝑡𝑡 + 2 𝑎𝑎∆𝑡𝑡 2  1 = (14,7 𝑥𝑥 1,5) + 2 (9,8)1,52  = 33,075 m (33,08 m)  OPTION 5/OPSIE 5: (down positive/af positief) 𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2𝑎𝑎∆𝑥𝑥 = 0 + 2(9,8)(44) Mark for both equations 𝑣𝑣𝑓𝑓 = 29,37 m.s-1 Punt vir albei vergelykings 𝑣𝑣𝑓𝑓 = 𝑣𝑣𝑖𝑖 + 𝑎𝑎𝑎𝑎  29,37 = 0 + 9,8 t  𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 3 s = 6t t = 0,5 s  3.4 OPTION 1/OPSIE 1: up positive Marking criteria/ Nasienkriteria: Positive marking from QUESTIONS 3.2 and 3.3 for values: Positiewe merk van VRAAG 3.2 en 3.3 position (m)  44 m initial height/ aanvanklike hoogte  Height after bounce/ hoogte na bons  Time for bounce (6t or 3 s as calculated) Tyd vir bons (6t of 3 s soos bereken)  Time at C before bounce (3t or 1,5 s as calculated) Tyd by C voor bons (3t of 1,5 s soos bereken) 3t 6t  Correct shape/ Regte vorm time(s) Accept contact time – did not specify hard or soft ball /Aanvaar kontak tyd – vraag spesifiseer nie harde of sagte bal nie 7

Downloaded from hlayiso.com 10841/24 OPTION 2/OPSIE 3: down positive/ af positief 3t 6t time (s) position (m) (5) [15] QUESTION 4 4.1 The total linear momentum of an isolated system remains constant (is conserved). Die totale lineêre momentum van ʼn geïsoleerde sisteem bly konstant (behoue). Marking criteria/ Nasienkriteria: If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. If closed system is used – no marks Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Indien geslote sisteem gebruik word – geen punte. (2) 4.2 ∑𝐩𝐩𝐛𝐛𝐛𝐛𝐛𝐛𝐛𝐛𝐛𝐛𝐛𝐛/𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗 = ∑𝐩𝐩𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚/𝒏𝒏𝒏𝒏 Marking criteria: 𝒎𝒎𝑩𝑩 𝒗𝒗𝒊𝒊𝒊𝒊 + 𝒎𝒎𝒃𝒃 𝒗𝒗𝒊𝒊𝒊𝒊 = 𝑴𝑴𝒗𝒗𝒇𝒇   Formula/formula (0,15)(0) + (0,05)(200)  = (0,2)vf  Substitution right/ 𝒗𝒗𝒇𝒇 = 𝟓𝟓𝟓𝟓 𝒎𝒎. 𝒔𝒔−𝟏𝟏  Vervanging regs  Substitution left / Vervanging links  Answer/ Antwoord (4) 8

Downloaded from hlayiso.com 10841/24 4.3 INELASTIC  / ONELASTIES Marking criteria/ Nasienkriteria: 1 1  INELASTIC/ ONELASTIES ∑Ek before = mB v2 + mb v2  2 2 = ½ (0,15)(0)2 + ½ (0,05)(200)2  Formula / Formule = 1000 J  Substitution BEFORE / 1 Vervanging VOOR Ekafter = m(B+b) v2  Substitution AFTER/ 2 1 200 Vervanging NA = � � x (50)2  2 1000  Answer/ Antwoord = 250 J Since total Ek before collision ≠ total Ek after collision, the collision is INELASTIC. Aangesien Ek voor botsing ≠ Ek na botsing is die botsing, ONELASTIES (4) 4.4 Positive marking from QUESTION 4.2 and Marking criteria/ Nasienkriteria: 4.3 / Positiewe merk van VRAAG 4.2 en 4.3  Formula/ Formule ∆p FNET =   Substitution/ Vervanging ∆t  Answer/ Antwoord (0,05)(195) - (0,05)(200) -5=  ∆t ∆t=0,05 s (3) [13] QUESTION 5 5.1 The (total) net work done is equal to the change in kinetic energy.  OR The work done by a net force is equal to the change in kinetic energy. Die netto arbeid verrig op ʼn voorwerp is gelyk aan die verandering in kinetiese energie van die voorwerp. OF Die netto arbeid verrig deur ‘n netto krag is gelyk aan die verandering in kinetiese energie (2) Marking criteria/ Nasienkriteria: If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. If work is omitted, then zero marks Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Indien werk uitgelaat word, geen punte. 9

Downloaded from hlayiso.com 10841/24 5.2 Accepted labels w Fg/Fw/weight/gravitational force N N/FN/Normal Force f fs/Ff/friction (3) 5.3 A non-conservative force is the force for which the work done is dependent on the path taken.  (2) Marking criteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark 5.4 OPTION 1/OPSIE 1: Wnc = ∆ Ep + ∆ Ek  FfΔxcos180o = (53)(9,8)(1,5 - 0) + ½ (53)(02 – 62) 20Δxcos180o = 779,1 – 954 Marking criteria/Nasienriglyn: -20x = -174,9  Formula/Formule Δx = 8,745 m  Substitution (Wnc) / Vervanging van Wnc 1,5  Substitution Ep/ Vervanging van Ep Sinθ =  8,745  Substitution Ek/ Vervanging van Ek θ = 9,88o   Angle substitution/Hoek vervanging  answer OPTION 2/OPSIE 2: Wnet = ∆Ek  Fnet = f + Fg// (friction and Fg// must have the same sign) = 20 + (53 x 9,8 x sin θ) 1 1 Wnet = Fnet x cos 180° = 2 𝑚𝑚 𝑣𝑣𝑓𝑓2 − 2 𝑚𝑚 𝑣𝑣𝑖𝑖2 1,5 (20 + (53 x 9,8 x 𝑥𝑥 )) x cos 180° = 0 – (½ x 53 x 62)  – 20 x = – 174,9 x = 8,745 m  (6) 10

Downloaded from hlayiso.com 10841/24 1,5 sinθ =  8,745 = 9,88 °  (accept range 9,87 – 9,88) [13] QUESTION 6/VRAAG 6 6.1 The change in frequency (or pitch) 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 (of toonhoogte) van die klank waargeneem deur ʼn luisteraar omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium waarin die klank voortgeplant word, het (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. 6.2 Graph B.  The frequency recorded is lower than the emitted frequency.  OR Lower frequency gives a longer wavelength, thus moving away.  OR Frequency decreases compared to emitted frequency Grafiek B.  Die waargenome frekwensie is laer as die bron se frekwensie. OF Laer frekwensie gee ʼn langer golflengte, dus beweeg weg. OF Frekwensie is laer as die bron frekwensie (2) 11

Downloaded from hlayiso.com 10841/24 6.3 OPTION 1/OPSIE 1: 𝒗𝒗 ∓ 𝒗𝒗𝑳𝑳 𝒗𝒗 ∓ 𝒗𝒗𝑳𝑳 𝒇𝒇𝑳𝑳 = 𝒙𝒙 𝒇𝒇𝑺𝑺 𝒇𝒇𝑳𝑳 = 𝒙𝒙 𝒇𝒇𝑺𝑺  𝒗𝒗 ∓ 𝒗𝒗𝑺𝑺 𝒗𝒗 ∓ 𝒗𝒗𝑺𝑺 𝟑𝟑𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑𝟑𝟑 9𝟎𝟎𝟎𝟎 = 𝒙𝒙 𝒇𝒇𝑺𝑺  𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = 𝒙𝒙 𝒇𝒇𝑺𝑺  𝟑𝟑𝟑𝟑𝟑𝟑 + 𝒗𝒗𝑺𝑺 𝟑𝟑𝟑𝟑𝟑𝟑 − 𝒗𝒗𝑺𝑺 𝟗𝟗𝟗𝟗𝟗𝟗 𝒙𝒙 𝟑𝟑𝟑𝟑𝟑𝟑 + 𝒗𝒗𝑺𝑺 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝒙𝒙 𝟑𝟑𝟑𝟑𝟑𝟑 − 𝒗𝒗𝑺𝑺 𝒇𝒇𝑺𝑺 = 𝒇𝒇𝑺𝑺 = 𝟑𝟑𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑𝟑𝟑 𝑓𝑓𝑠𝑠 = 𝑓𝑓𝑠𝑠 Marking criteria: (1100) 340− 𝑣𝑣𝑠𝑠 (900) 340+ 𝑣𝑣𝑠𝑠 =  340 340  Complete Doppler formula 𝑣𝑣𝑠𝑠 = 34 𝑚𝑚. 𝑠𝑠 −1   Substitution (towards) / Vervanging (na)  Substitution (away)  Equate fs (5) 6.4 6.4.1 B  Red shifted / Rooi verskuiwing (1) 6.4.2 High frequency  Short wavelength  Hoë frekwensie  Kort golflengte  (2) [12] QUESTION 7 7.1 The electric field at a point is the electrostatic force experienced per unit positive charge placed at that point.  Die elektriese veld by ʼn punt is die elektrostatiese krag wat per eenheidspositiewe- lading wat by daardie punt geplaas is, ondervind word.  (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. 7.2 Take direction of EA as positive Enet = EA – EB  for subtraction Marking criteria/ Nasienkriteria: kQA kQB = – formula r2 r2  Formula /Formule (9x109 )(6x10-6 ) (9x109 )(3x10-6 )  Substitution QA/ Vervanging QA =  –   Substitution QB/ Vervanging QA 42 12  Answer with direction / Antwoord met = – 23 625 N·C –1 rigting. Enet is 23 625 N·C –1 to the left/ Direction of EB can also be taken as towards P  positive. Signs will then be different. 12

Downloaded from hlayiso.com 10841/24 OPTION 2: 𝑘𝑘𝑄𝑄𝐴𝐴 𝐸𝐸 =  𝑟𝑟 2 = 9 x 109 x 6 x 10-6  (4)2 = 3 375 N.C-1 to the right 𝑘𝑘𝑄𝑄𝐵𝐵 𝐸𝐸 = 𝑟𝑟 2 = 9 x 109 x 3 x 10-6  (1)2 = 27 000 N.C-1 to the left Enet = EA – EB (take right as positve)  = 3 375 – 27 000 = –23 625 = 23 625 N·C-1 to the left  (5) 7.3 Positive marking from QUESTION 7.2 /Positiewe merk van VRAAG 7.2 F E=  Marking criteria: q F  Formula to calculate F 23 625 = -19  1,6x10  Substitution / Vervanging  Substitution mass of proton F is 3,78 x 10-15 N  Answer for acceleration Fnet = ma 3,78 x 10-15 = 1,67 x 10-27a  a = 2,26 x 1012 m∙s-2  (4) 1 7.4 Larger acceleration  as mass of electron is smaller. (a α ) 𝑚𝑚 Opposite direction  1 Groter versnelling omdat die massa van die elektron kleiner is. (a α 𝑚𝑚) Teenoorgestelde rigting  (2) [13] 13

Downloaded from hlayiso.com 10841/24 QUESTION 8/VRAAG 8 8.1 The maximum energy provided by a battery per unit charge passing through it. Die maksimum energie wat ʼn battery lewer per eenheidslading wat daardeur vloei (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 8.2 8.2.1 3V (1) 8.2.2 0,4 A  (1) 8.3 Positive marking from QUESTION 8.2 Marking criteria/ Nasienkriteria: Positiewe nasien van VRAAG 8.2 V  Formula/Formule R= I   Substitution/ Vervanging 6+3  Answer/Antwoord =  0,2 = 45 Ω  (3) 8.4 Positive marking from QUESTION 8.2 Positiewe nasien van VRAAG 8.2 OPTION 1/OPSIE 1: Marking criteria/ Nasienkriteria: 1 1 1  Formula RP / Formule RP = +  R p R1 + R 2 R 3  Substitution / Vervanging 1 1  Substitution RT = +   Formule EMF/ 15+7,5 45  Substitution Rp = 15 Ω  Answer RT = Rs +Rp = 4 + 15  = 19 Ω emf = I(R+r)  = (0,6)(19 + 1)  = 12 V  (6) 14

Downloaded from hlayiso.com 10841/24 OPTION 2: Vp = 6+3 = 9V  Ip = 0,2 + 0,4 = 0,6 A  𝑉𝑉𝑝𝑝 𝑅𝑅𝑃𝑃 = 𝐼𝐼𝑝𝑝 9 = 0,6 = 15 Ω  𝑅𝑅𝑇𝑇 = 𝑅𝑅𝑠𝑠 + 𝑅𝑅𝑃𝑃 = 4 + 15 =19 Ω emf = I(R+r)  = (0,6)(19 + 1)  = 12 V  8.5 Gradient = 𝛥𝛥𝛥𝛥=V2  𝛥𝛥𝛥𝛥 Marking criteria: 600−0 V2 = 6−0   Substitution top/ Vervanging bo V = 10 V   Substitution bottom/ Vervanging onder  V2 answer / V2 antwoord OR Δy V2 Gradient = =V2  from P = Δx R 800-200 V2 =  8-2 V = 10 V  (4) [17] 15

Downloaded from hlayiso.com 10841/24 QUESTION 9 9.1 AC  as there are sliprings  WS omdat daar sleepringe is (2) 9.2 Mechanical energy to electrical energy  Meganiese energie na elektriese energie (1) 9.3 a to b  a na b (1) 9.4 OPTION 1: Imax Irms = Marking criteria: √2 Imax = (4,55)√2   Formula  First substitution = 6,434 A  Second substitution V  Answer R= max  Imax 150 =  6,434 = 23,31 Ω  (4) OPTION 2: Calculate Vrms and use rms values to calculate resistance. 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 𝑉𝑉rms = √2 150 = √2  = 106,07 V 𝑉𝑉 𝑅𝑅 = 𝐼𝐼rms  (subscripts optional) rms = 106,07  4,55 = 23,31 Ω  9.5 W = PΔt  = (400)(3 x 60)  = 72 000 J  (3) 16

Downloaded from hlayiso.com 10841/24 9.6 9.6.1  double the emf/amplitude/300  half the frequency Do not penalise if more than one wave is shown. dubbel die emk/amplitude die helfte van die frekwensie Moenie penaliseer as meer as een golf geteken is nie (2) 9.6.2 With a factor 4W  ε α1/Δt  if N and flux stay constant. If the rotation doubles, the induced emf will double  W α V2 if R and Δt stay constant  and energy emitted will quadruple (4x) Note: Correct calculations can be accepted as an explanation. (4) [17] QUESTION 10 /Vraag 10 10.1 The minimum energy that an electron in the metal needs to be emitted from the metal surface.  (exact words not needed as definition was not asked) Die minimum energie benodig om ʼn elektron uit die oppervlak van ʼn metaal vry te stel.  (presiese woorde is nie nodig nie aangesien definisie nie gevra word nie) (2) 17

Downloaded from hlayiso.com 10841/24 10.2 E = Wo + Ek max  Marking criteria / hf = hfo + Ek max Nasienkriteria: (6,63 x 10-34)(6,3 x 1014)  = Wo + 1,2 x 10-19   Formula/Formule Wo = 2,98 x 10-19 J ’  Substitution left / Vervanging links  Substitution right/ Vervanging links (4)  Answer /Antwoord 10.3 Positive marking from QUESTION 10.2 OPTION 1/OPSIE 1 YES  /Ja Wo = hfo  2,98 x 10-19 = (6,63 x 10-34)fo  fo = 4,49 x 1014 Hz  Electrons will be emitted as the threshold frequency is lower than the frequency of the radiation.  OPTION 2/OPSIE 2 𝐸𝐸 = ℎ𝑓𝑓  = (6,63 x 10-34) (3 x 1016 )  = 1,99 x 10-17 J  𝐸𝐸 > 𝑊𝑊0  1,99 x 10-17 J > 2,98 x 10-19 J Yes  electrons will be emitted from this metal.  OPTION 3/OPSIE 3: 𝐸𝐸 = 𝑊𝑊0 + 𝐸𝐸𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘  (6,63 x 10-34) (3 x 1016 )  = 2,98 x 10−19 + 𝐸𝐸𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘  𝐸𝐸𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 = 1,96 x 10-17 J  𝐸𝐸 > 𝑊𝑊0  1,96 x 10-17 J > 2,98 x 10-19 J Yes  electrons will be emitted from this metal.  (6) 10.4 When UV light shines on metal, energy is transferred to electrons on surface of metal . If plate is positive, electrons will be attracted to the plate  as opposite charges attract one another  OR Plate is positive, few electrons are available  to be emitted by the UV light.  No change in charge of the plate will be observed.  (3) 18

Downloaded from hlayiso.com 10841/24 Wanneer UV lig op die metaal skyn, word energie oorgedra na die elektrone in die oppervlakte van die metaal. As die plaat positief gelaai is, sal die elektrone aangetrek word na die plaat toe aangesien teenoorgestelde ladings mekaar aantrek OF Plaat is positief, dus min elektrone beskikbaar om deur die UV lig vrygestel te word nie. Geen verandering in die lading van die plaat sal waargeneem word nie. [15] TOTAL: 150 19