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

Downloaded from hlayiso.com JUNE EXAMINATION JUNIE EKSAMEN GRADE/GRAAD 12 2024 MARKING GUIDELINES/ NASIENRIGLYNE PHYSICAL SCIENCES: PHYSICS/ FISIESE WETENSKAPPE: FISIKA (PAPER/VRAESTEL 1) 13 pages/bladsye

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 QUESTION/VRAAG 1 1.1 B  (2) 1.2 B  (2) 1.3 B  (2) 1.4 D  (2) 1.5 D  (2) 1.6 A  (2) 1.7 B  (2) 1.8 A  (2) 1.9 C  (2) 1.10 B  (2) [20] QUESTION/VRAAG 2 2.1 The force or the component of a force which a surface exerts on an object with which it is in contact, and which is perpendicular to the surface. ✓✓ Die krag of komponent van ʼn krag wat ʼn oppervlak op ʼn voorwerp waarmee dit in kontak is, uitoefen en wat loodreg op die oppervlak is. (2) Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. 2.2   Accepted labels/Aanvaarde byskrifte Fg/Fw/weight/mg/gravitational force do not accept gravity w Fg/Fw/gewig/gravitasiekrag moet nie gravitasie aanvaar nie. FT/tension T FT/spanning/spankrag Deduct 1 mark for any additional force. Mark is given for both arrow and label. If everything is correct, but no arrows, deduct a mark (2) 2

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 Trek een punt af vir enige addisionele kragte. Punte word gegee vir beide pylpunt en byskrif. Indien alles korrek is, maar geen pyle nie, trek ʼn punt af. 2.3 In x-direction for mass M:/In x-rigting vir massa M: Fnet = ma  Any one/Enige een T + f – wII = 0 2(9,8)  + f – 10(9,8)sin30°  = 0  OR 2(9,8)  - f – 10(9,8)sin30°  = 0  f = –2(9,8) + 10(9,8)sin30° = –19,6 + 49 = 29,4 N (up the slope/teen die helling op)  (5) 2.4 2.4.1 Accepted labels  Aanvaarde byskrifte  Fg/Fw/weight/mg/gravitational force w Fg/Fw/gewig/mg/gravitasiekrag Ffriction/Ff/friction/fk f Fwrywing/Ff/wrywing/fk N FN/Normal force/Normaalkrag Deduct 1 mark for any additional force. Mark is given for both arrow and label DO NOT ACCEPT COMPONENTS IN FREE BODY DIAGRAM  Trek 1 punt af vir addisionele kragte. Punte word gegee vir beide pyl en byskrif. MOET NIE KOMPONENTE AANVAAR IN DIE VRYLIGGAAMDIAGRAM NIE (3) 2.4.2 Fnet = ma  wII – f = ma 10(9,8)sin30° – 25  = 10 a  a = 2,4 m·s –2  (4) 2.5 2.5.1 vf = vi + a∆t  = 0 + (9,8)(0,5)  A learner can use the option of -9,8 vf = 4,9 m·s –1 (down / afwaarts)  The same answer will be obtained. OR Fnet Δt = Δp ✓ 2(9,8)(0,5) = 2vf – 2vi ✓ vf = 4,9 m.s-1 (down / afwaarts) ✓ (3) 3

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 2.5.2 REMAIN THE SAME/BLY DIESELFDE ✓ The object is moving under the force of gravity only ✓and through the same height.✓ That will have no effect on the final velocity. OR When using equations of motion and falling at gravitational acceleration only, the mass has no influence on the final velocity. The height will though. Die voorwerp beweeg slegs onder die grativitasiekrag en deur dieselfde hoogte. Dit sal geen effek op die finale eindsnelheid hê nie. OF Wanneer bewegingsvergelykings gebruik word en slegs teen gravitasie- versnellings val, het die massa geen invloed op die finale eindsnelheid nie. Die hoogte sal wel. (3) [22] QUESTION/VRAAG 3 3.1 Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses ✓and inversely proportional to the square of the distance between their centres. ✓ Elke liggaam in die heelal trek elke ander liggaam aan met ʼn krag direk eweredig aan die produk van hul massas en omgekeerd eweredig aan die kwadraat van die afstand tussen hul middelpunte. (2) Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. 3.2 Weight/Gewig w = ma = 899 (9,8) ✓ = 8 810,2 N ✓ (2) 3.3 GM g = ✓ r2 6,67  10-11(6,42  1023 ) ✓ = (0,5325  6,38  106 )2 ✓ = 3,71 m·s-2 ✓ The gravitational acceleration on Earth is 9,8 m·s-2 while on Mars it is 3,71 m·s-2. (range of 3,70 – 3,72) Die grativitasieversnelling op Aarde is 9,8 m·s-2 terwyl dit op Mars 3,71 m·s-2 is. (gebied van 3,70 – 3,72) (4) 4

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 3.4 No/Nee ✓ Mass represents the matter an object is made up of and will not differ, no matter where in the universe you are. ✓ OR The mass will be the same on both planets. Massa verteenwoordig die materie waaruit ʼn voorwerp bestaan en wat nie sal verskil nie, ongeag waar in die heelal jy is. OF Die massa sal dieselfde wees op beide planete. (2) [10] 4.1 Motion of an object is free fall. ✓ Beweging van ʼn voorwerp. Object given an initial velocity is projectile. ✓ ʼn Voorwerp wat begin met ʼn beginsnelheid. OR Accept if definitions are given. / Aanvaar indien definisies gegee word OR Free fall and projectile motion is the same. Vryval en projektielbeweging is dieselfde. (2) 4.2 4.2.1 OPTION/OPSIE 1 (down as OPTION/OPSIE 2 (up as positive/op as positive/af as positief) positief) vf = vi + a∆t  vf = vi + a∆t  0 = (–11) + (9,8)∆t  0 = (11) + (–9,8)∆t  Δt = 1,12 s  Δt = 1,12 s  (3) 4.2.2 Positive marking from QUESTION OPTION 2 down as negative/OPSIE 4.2.1 2 af as negatief OPTION 1 down as positive/ Positiewe nasien vanaf VRAAG 4.2.1 OPSIE 1 af as positief vf = vi + a∆t  vf = vi + a∆t  0 = vi +(9,8)(1,12 – 0,72)  0 = vi + (–9,8)(1,12 – 0,72)  0 = vi + (9,8)(0,40) vi = 3,92 m·s-1 ✓ upwards/opwaarts ✓ vi = 3,92 m·s-1 ✓ upwards/opwaarts  (4) 5

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 4.2.3 Positive marking from Positive marking from 4.2.1 and QUESTION 4.2.1 and 4.2.2/Positiewe nasien vanaf 4.2.2./Positiewe nasien vanaf VRAAG 4.2.1 en 4.2.2. VRAAG 4.2.1 en 4.2.2. OPTION 2 (upwards positive) OPTION 1 (down as positive) Ball A height above the ground/ Ball A height above the ground: OPSIE 2 (op as positief) OPSIE 1 (af as positief) Bal A hoogte bo die grond Bal A hoogte bo die grond 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑦 ✓ 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑦 ✓ 0 = (11)2 + 2(−9,8)∆𝑦 ✓ 0 = (–11)2 +2(9,8)∆y ✓ ∆y = 6,17 m ∆y = 6,17 m ✓ Ball B height above the building/ Ball B height above the building/ Bal B hoogte bo die gebou: Bal B hoogte bo die gebou: 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑦 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑦 0 = (3,92)2 +2(–9,8)∆y ✓ 0 = (–3,92)2 +2(9,8)∆y ✓ ∆y = 0,784 m ∆y = 0,784 m Height of building/Hoogte van die Height of building/Hoogte van die gebou: gebou: ✓ ✓ Height/Hoogte = 6,17 – 0,784 Height/Hoogte = 6,17 – 0,784 = 5,39 m (5,388 m) ✓ = 5,39 m ✓ OPTION/OPSIE 3 OPTION/OPSIE 4 Ball A above the ground/ Ball A above the ground/ Bal A bo die grond: Bal A bo die grond: v f + vi 1 ∆x = ( 2 ) t ✓ ∆𝑥 = 𝑣𝑖 ∆𝑡 + 2 𝑎∆𝑡 2 ✓ 0 + 11 ∆x A = ( 2 ) x (1,12) ✓ ∆xA = 11 × 1,12 1 + 2(–9,8) (1,12)2 ✓ ∆x = 6,16 m ∆xA = 6,17 m Ball B above the building/ Ball B above the building/ Bal B bo die gebou: Bal B bo die gebou: v f + vi 1 ∆x = ( 2 ) t ✓ ∆𝑥 = 𝑣𝑖 ∆𝑡 + 2 𝑎∆𝑡 2 ✓ 0 + 3, 92 ∆xB = ( 2 ) (0,4) ✓ ∆xB = 3,92 × 0,4 + ✓ 1 (–9,8) (0,4)2 ∆x = 0,784 m 2 ∆xB = 0,784 m Height of building/Hoogte van die Height of building/Hoogte van die gebou: gebou: ✓ ✓ Height/Hoogte = 6,16 – 0,784 Height/Hoogte = 6,17 – 0,784 = 5,38 m (5,376 m) ✓ = 5,39 m (5,386 m) ✓ (6) 6

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 Positive marking from QUESTION 4.2 for answers of QUESTION 4.3/ Positiewe nasien van VRAAG 4.2 vir antwoorde van VRAAG 4.3 4.3 4.3.1 6,16 m ✓ (1) 4.3.2 5,38 m ✓ (1) 4.3.3 0,72 s ✓ (1) 4.3.4 1,12 s ✓ (1) [19] QUESTION/VRAAG 5 5.1 5.1.1 The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. ✓✓ Die totale meganiese energie (som van gravitasie-potensiële energie en kinetiese energie) in ʼn geïsolleerde sisteem bly konstant. (2) Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct /If momentum is mentioned 0/2 1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Indien momentum genoem word 0/2 5.1.2 OPTION/OPSIE 1: Emech i = Emech f 1 1 Any one/Enige een m X v 2Xi + m X gh X = m X v 2Xf + m Xghh 2 2 1 0 + 950(9,8)(3,5) = (950)(v 2 ) + 950(9,8)(1) 2 v = 7 m  s−1 OPTION/OPSIE 2: Emech i = Emech f Any one/Enige een 1 2 1 2 m X v Xi + m X gh X = m X v Xf + m Xghh 2 2 1 0 + 950(9,8)(3,5-1) = (950)(v 2 ) + 950(9,8)(0) 2 v = 7 m  s−1 7

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 OPTION 3/OPSIE 3 Wnc = Ek + Ep 1 1 Any one/Enige een 2 2 0 = ( m X v Xf - m X v Xi ) + (m Xghh - m Xgh X ) 2 2 1 0 = ( (950)(v 2 ) − 0) + (950(9,8)(1) - 950(9,8)(3,5)) 2 v = 7 m  s −1 (4) 5.1.3 Positive marking from QUESTION 5.1.2/Positiewe nasien vanaf VRAAG 5.1.2 Fnet∆t = m∆v ✓ Fnet(0,1) = (950)(0 – 7) ✓ Fnet = – 66 500 N Fnet = 66 500 N ✓ OR vf = vi + aΔt Fnet = ma ✓ 0 = 7 + a(0,1) = 950(-70) ✓ a = - 70 m.s-2 = - 66 500 N = 66 500 N ✓ (3) 5.1.4 Positive marking from 5.1.3/positiewe nasien vanaf 5.1.3 66 500 N to the left/west/na links/wes ✓ OR Fball on wall Newtons third law. ✓ When object A exerts a force on object B, object B SIMULTANEOUSLY exerts an oppositely directed force of equal magnitude on object A. ✓ Newton se derde wet. Wanneer voorwerp A ʼn krag op voorwerp B uitoefen, sal voorwerp B GELYKTYDIG ʼn krag van gelyke grootte en in die teenoorgestelde rigting op voorwerp A uitoefen. (3) 8

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 5.2 1 Ek after = mv 2 2 1 146 = (75 + 2 + 0,5)v 2 2 v = 1,94 m  s−1 pbefore = pafter Any one/Enige een mlearner v leaerner i + mball v ball i = (mlearner + mball )v f (75+2)(2) + 0,5(v) = 77,5(1,94) OR m1v1 + m2v2 = (m1+m2)vf v = -7,3 v ball i = 7,3 m  s−1 Range 7,13 – 7,3 (5) [17] QUESTION/VRAAG 6 6.1   Accepted labels/Aanvaarde byskrifte Fg/Fw/weight/mg/gravitational force w Fg/Fw/gewig/mg/gravitasiekrag FT/tension T FT/spanning/spankrag Deduct 1 mark for any additional force. Mark is given for both arrow and label. If everything is correct, but no arrows, deduct a mark Trek een punt af vir enige addisionele kragte. Punte word gegee vir beide pylpunt en byskrif. Indien alles korrek is, maar geen pyle nie, trek ʼn punt af. (2) 6.2 Tension/Spanning ✓ (1) 9

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 6.3 OPTION/OPSIE 1: WF = Fxcos Any one/Enige een = mgxcos = (85)(9,8)(11)cos180o = - 9 163 J OPTION/OPSIE 2: Ww = - EP Any one/Enige een = - (mgh - 0) = - (85)(9,8)(11) = - 9 163 J (3) 6.4 The net work done on an object is equal to the change in the object's kinetic energy. ✓✓ (2 or 0) OR The work done on an object by a net force is equal to the change in the object's kinetic energy. Die netto arbeid verrig op ʼn voorwerp is gelyk aan die verandering in kinetiese energie van die voorwerp. OF Die arbeid verrig op die voorwerp deur ʼn netto krag is gelyk aan die verandering in kinetiese energie van die voorwerp. (2) 6.5 Wnet = K 1 1 Any one/Enige een Fnet xcos =mv 2f - mv i2 2 2 1 maxcos = (85)v 2f - 0 2 o (85)(0,75)(11) cos0 = 42,5v 2f vf = 4,06 m  s−1 (5) 10

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 6.6 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 vf = vi + a∆t Fnet + w = Fmotor 4,06 = 0 + (0,75)∆t ✓ 85(0,75) + 85(9,8) ✓ = Fmotor Fmotor = 896,75 N t = 5,41 s Pave = Fvave ✓ Positive marking from QUESTION 6.3 for 0+4,06 Work done by gravitational force./Positiewe = (896,75) ✓ 2 ✓ nasien vanaf VRAAG 6.3 vir arbeid verrig = 1 820,4 W deur gravitasiekrag. 1 820,4 % efficiency = 𝑥 100 ✓ 30 000 Wnet = ΔEk = 6,07% ✓ -9163 + Wmotor = ½ (85)( 4,062 – 02) ✓ Wmotor = 9 863,55 J 𝑊 P= 𝑡 ✓ 9 863,55 = ✓ 5,41 = 1 823,21 W 1 823,21 % efficiency = 30 000 𝑥 100 ✓ = 6,08% ✓ (6) [19] QUESTION/VRAAG 7 7.1 7.1.1 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. OR/OF An (apparent) change in (observed/detected) frequency/pitch as a result of the relative motion between a source and an observer (listener). 'n (Skynbare) verandering in (waargenome) frekwensie/toonhoogte as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer/luisteraar. Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. (2) 11

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 7.1.2 Towards/Na ✓ • The apparent frequency was higher/bigger ✓ • The wave fronts were closer together ✓ and • The wavelength was shorter/smaller. ✓ • Die skynbare frekwensie was hoër, • Die golffronte was nader aan mekaar en • Die golflengte was korter. (4) 7.1.3 v ± vL fL = fs v ± vs 340 405 = 360 340 - v s v s = 37,78 m  s−1 (4) 7.2 7.2.1 360 (Hz) ✓ (1) 7.2.2 Less than/Minder as 360 (Hz) ✓ OR < 360 (Hz) Any value less than 360 given is accepted. / Enige waarde minder as 360 word aanvaar. (1) 7.2.3 340 (m·s-1 ) ✓ (1) [13] QUESTION/VRAAG 8 8.1 Positive/Positief ✓ (1) 8.2 Q Q NOTE: do not penalize if –1,6 is n= ✓ or n = substituted, but answer must be e qe positive. Q ✓ NOTA: moenie penaliseer indien 1238 = 1,6 x 10−19 –1,6 vervang, maar antwoord moet positief wees. Q = 1,981 x 10−16 C ✓ (3) 8.3 An electric field is a region of space in which an electric charge experiences a force. ✓✓ (2 or 0) ʼn Elektriese veld is ʼn area of gebied waarin ʼn elektriese lading ʼn krag sal ondervind. (2 of 0) (2) 12

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 8.4 Criteria for sketch/Kriteria vir skets Marks/ Punte Correct direction of electric field – away from positive and ✓ towards negative. Do not penalize if + and – are not indicated. /Korrekte rigting van elektriese veld –weg van positief en na negatief. Moenie penaliseer indien + en – nie aangedui word nie. Correct shape of the electric field lines between, above and on ✓ the outside of the charges. Lines curve in the correct direction./ Korrekte vorm van die elektriese veldlyne, tussen, bo en aan die buitekant van die ladings. Lyne buig in die regte rigting. No field lines crossing each other. Field lines touch the charge, ✓ start at an angle perpendicular to the charge, do not go inside the charge./Geen veldlyne wat mekaar kruis nie. Veldlyne raak aan die ladings, begin by ʼn hoek loodreg op die lading, gaan nie binne die lading nie. 8.5 A✓ The field is strongest closest ✓to the charged sphere. Point A is closer to sphere X than point B. The electric field is inversely proportional to the 1 distance squared. OR E  ✓/ r2 Die veld is die sterkste, naaste aan die gelaaide sfeer. Punt A is nader aan sfeer X as punt B. Die elektriese veld is omgekeerd eweredig aan die afstand 1 kwadraat. OF E  r2 (3) 8.6 kQ X E= r2 9x109 (2x10−9 ) = (0,03)2 = 20 000 N  C−1 left / links (3) 13

Downloaded from hlayiso.com PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0624 8.7 8.7.1 FYX FZX Marking criteria/Nasienkriteria: Marks/Punte Line and arrow with label//Lyn en pyl met byskrif ✓ Line and arrow with label/Lyn en pyl met byskrif ✓ (2) 8.7.2 Marking criteria/Nasienriglyne • Correct formula for Coulombs law/Korrekte formule vir Coulomb wet • Correct substitution for FYX/Korrekte vervanging vir FYX • Correct substitution for Fzx/Korrekte vervanging vir FZX • Addition of forces/Optel van kragte • Correct final answer/Korrekte finale antwoord kQY Q X kQZQ X Fnet = 2 + r r2 9x109 (2,8x10 −9 )(2x10 −9 ) (9x109 )(3,2x10 −9 )(2x10 −9 ) = + (0,03)2 (0,01)2 = 6,32 x 10−4 N (5) [22] QUESTION/VRAAG 9 9.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. ✓✓ Die potensiaalverskil oor ʼn geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur. (2) 9.2 Lamp ✓ or bulb/gloeilamp of lamp (1) 9.3 12 V ✓ (1) 9.4 12 V ✓ (1) 9.5 0V✓ (1) 9.6 Stays the same/Bly dieselfde ✓ (1) 9.7 Stays the same/Bly dieselfde ✓ (1) [8] TOTAL/TOTAAL: 150 14