Physical Sciences Grade 12 P1 Gauteng June 2025 Memo (Afr)

JUNE EXAMINATION JUNIE EKSAMEN GRADE/GRAAD 12 2025 MARKING GUIDELINES/ NASIENRIGLYNE PHYSICAL SCIENCES/ FISIESE WETENSKAPPE (PAPER/VRAESTEL 1) 15 pages/bladsye

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 QUESTION 1/VRAAG 1 1.1 C  (2) 1.2 D  (2) 1.3 B  (2) 1.4 A  (2) 1.5 C (2) 1.6 A  (2) 1.7 B  (2) 1.8 D  (2) 1.9 C  (2) 1.10 C  (2) [20] 2

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 QUESTION 2/VRAAG 2 2.1 The force that opposes the motion of a moving object relative to a surface.  (2 or zero) Die krag wat die beweging van 'n bewegende voorwerp relatief tot 'n oppervlak teenwerk. (2 of nul) (2) 2.2 No components are accepted Accepted labels/Aanvaarde byskrifte Marks/Punte 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 Normal (force)/Fnormal FN  Normaal (krag)/Fnormaal Fk/Ff/Friction f  Fk/Ff/Wrywing 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 1 punt af vir enige addisionele kragte. Punt word toegeken vir pyltjie en byskrif. Indien alles korrek, maar geen pyltjies, trek 1 punt af (4) 3

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 2.3 Marking criteria/Nasienkriteria • Any one of the correct formulae/enige een van die korrekte formules  • Fx substitution/Fx invervanging  • Fy substitution/Fy invervanging  • Substitution for 0 or equating equation 1 and 2/Invervanging van 0 of gelykstel van vergelykings 1 en 2  • Final answer /Finale antwoord (Range/Gebied 37,59 N – 37,77 N) OPTION 1/OPSIE 1: Fnet = ma Fnet = 0  Any one/Enige een Tcos θ – fk = ma Tcos 40o – µkN = ma Tcos 40o – (0,31)(Fg – Tsin40o)  = 0  T cos 40o – (0,31)((12 x 9,8) – Tsin40o) =0 T = 37,77 N  OPTION 2/OPSIE 2: fk= µkN  Tcos 40o = 0,31FN (0,31FN ) T=  ....... equation 1/vergelyking1 cos 40o Fy = Fg - FN Tsin 40o = (12x9,8) - FN 117,6−FN T=  …… equation 2/vergelyking 2 sin 40𝑜𝑜 (0,31FN ) 117,6−FN = (equating the two equations/gelykstelling van die cos 40𝑜𝑜 sin 40𝑜𝑜 twee vergelykings) 117,6 - FN = 0,839 (0,31FN) FN = 93,32 N (0,31)(93,32) T= cos 40o T = 37,77 N  (5) 4

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 2.4 2.4.1 OPTION 1/OPSIE 1:  v + vf  Δx =  i  Δt   2  (10)+(10) 24 = ( ∆ t)  2 ∆ t = 2,4 s  OPTION 2/OPSIE 2: Δx = v i Δt + 21 a∆t 2  Accept/Aanvaar Δx = vΔt 1 24 = (10) ∆ t + 2 (0) ∆ t2  ∆ t = 2,4 s  (3) 2.4.2 REMAINS THE SAME/BLY DIESELFDE  (1) 2.5.1 DECREASE/VERLAAG  (1) 2.5.2 As θ increases, the Fy will increase/Soos θ toeneem, sal die Fy toeneem  FN = Fg – Fy / N α f / normal force decreases / normaalkrag verminder and μk is constant/en μk is konstant  (3) [19] QUESTION 3/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 ’n krag direk eweredig aan die produk van hul massas en omgekeerd eweredig aan die kwadraat van die afstand tussen hul middelpunte. (2) 5

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 3.2 3.2.1 OPTION 1/ OPSIE 1 OPTION 2/ OPSIE 2 Fg = mg m m2 w = ma F = G 12 r 5,38 x 106 = m(9,8)  �6,67𝑥𝑥10−11 ��5,98𝑥𝑥1024 �(m) 5,38 x 106 =  m = 548 979,59 kg (5,49 x 105 kg) (6,38𝑥𝑥106 )2 m = 548 979,59 kg (5,49 x 105 kg) (2) 3.2.2 m 1m 2 F=G  r2 �6,67𝑥𝑥10−11 ��5,98𝑥𝑥1024 �(m)  6 5,04 x 10  = (9 000+6,38𝑥𝑥106 )2  m = 515 785,7 kg  (5,16 x 105 kg) (5) 3.2.3 POSITIVE MARKING FROM 3.2.1 AND 3.2.2/POSITIEWE NASIEN VANAF 3.2.1 en 3.2.2 Fuel mass = Total mass – mass of rocket (at 9000m) Brandstofmassa = Totale massa – massa van vuurpyl (op 9000m) = 548 979,59 – 515 785,70  = 33 193,89 kg  (3,32 x 104 kg) range/gebied 33 000 – 33 193,89 (2) 3.3 INCREASE/VERHOOG  (2) [13] QUESTION 4/VRAAG 4 4.1 The block B/die blok B Only gravitational force is acting on the block./Slegs gravitasiekrag werk op die blok in.  (2) 4.2.1 Take down as negative/Neem af as Take down as positive/Neem af as negatief: positief: vf = vi + aΔt  vf = vi + aΔt  0 = 5 + (−9,8)Δt  0 = −5 + (9,8)Δt  Δt = 0,51 s  Δt = 0,51 s  (3) 4.2.2 Take down as negative/Neem af as Take down as positive/Neem af as negatief: positief: vf2 = vi2 + 2aΔx  vf2 = vi2 + 2aΔx  = 52 + 2(−9,8)(−50)  = (−5)2 + 2(9,8)(50)  vf = 31,7 m.s-1 (downwards)  vf = 31,7 m.s-1 (downwards)  (3) 6

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 4.2.3 POSITIVE MARKING FROM 4.2.1 AND 4.2.2/ POSITIEWE NASIEN VANAF 4.2.1 EN 4.2.2 OPTION 1/OPSIE 1: OPTION 1/OPSIE 1: Take down as negative/Neem af as Take down as positive/Neem af as negatief: positief: 1 1 Δy = vi Δt + aΔt 2  Δy = vi Δt + aΔt 2  2 2 1 2 1 −50 = 5Δt + 2 (−9,8)Δt  50 = −5Δt + 2 (9,8)Δt 2  Δt = 3,745 s  Δt = 3,745 s  OPTION 2/OPSIE 2: OPTION 2/OPSIE 2: Take down as negative/Neem af as Take down as positive/Neem af as negatief: positief: 1 1 Δy = vi Δt + aΔt 2  Δy = vi Δt + aΔt 2  2 2 1 1 −50 = 0 + 2 (−9,8)Δt 2  50 = 0 + 2 (9,8)Δt 2  Δt = 3,19 s Δt = 3,19 s t = 3,19 + 0,51 T = 3,19 + 0,51 = 3,7 s  = 3,7 s  OPTION 3/OPSIE 3: OPTION 3/OPSIE 3: Take down as negative/Neem af as Take down as positive/Neem af as negatief: positief: vf = vi + aΔt  vf = vi + aΔt  −31,70 = 5 + (−9,8)Δt  31,70 = 5 + (9,8)Δt  Δt = 3,745 s  Δt = 3,745 s  (3) 4.3 POSITIVE MARKING FROM 4.2/POSITIEWE NASIEN VANAF 4.2 Marking criteria/Nasienriglyne:  initial height/aanvanklike hoogte (50 m)  time at maximum height/tyd by maksimum hoogte (4.2.1)  time reaching the ground/tyd om grond te bereik (4.2.3)  ground as zero reference point/grond as nulverwysingspunt  correct shape of graph/korrekte vorm van grafiek Take down as negative/Neem af as Take down as positive/Neem af as negatief: positief: - (5) [16] 7

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 QUESTION 5 /VRAAG 5 5.1 5.1.1 The TOTAL linear momentum of an isolated system remains constant (is conserved).  do not accept closed system (2 or zero) Die TOTALE lineêre momentum van ‘n geïsoleerde sisteem bly konstant (behoue). Moet nie geslote sisteem aanvaar nie. (2 of nul) (2) 5.2.1 p = mv 30 000 = (2 000)v  (3) v = 15 m∙s-1  right/regs  5.2.2 POSITIVE MARKING FROM Q5.2.1 / POSTIEWE NASIEN VANAF V5.2.1 OPTION 1/ OPSIE 1 Ʃ pi = Ʃ pf pi(car A) + pi( car B) = pf(car A) + pf(car B)  Any one/Enige een mvi(car A) + mvi( car B) = mvf(car A) + mvf(car B) (2 000)(15) + m(0) = (2 000)(5) + m(20) (30 000) + 0 = (10 000) + m(20)  m = 1 000 kg OPTION 2 / OPSIE 2 ΔpA = - ΔpB  Any one/Enige een mvi(car A) - mvf(car A) = - ( mvf(car B) - mvi( car B) ) (2 000)(5) - (2 000)(15)  = -(m(20) - m(0))  m = 1 000 kg If the negative in the formula is omitted (0/4) Indien die negatief in formule uitgelaat word (0/4) (4) 5.3 ∆ p= 20 000 kg∙m∙s-1  Arrow to the left/pyl na links  (2) (Do not penalise length of arrow/Moet nie lengte van pyltjie penaliseer nie) 8

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 5.4 OPTION 1/ OPSIE 1 OPTION 2/ OPSIE 2 FnetΔt = mΔv  Any one/ FnetΔt = mΔv  Any one/ FnetΔt = Δp Enige een FnetΔt = Δp Enige een Fnet(0,5) = 2000(5 – 15)  Fnet(0,5) = 1000(20 – 0)  OR / OF Fnet(0,5) = (10 000 – 30 000) Fnet(0,5) = 20 000 Fnet = - 40 000 Fnet = 40 000 N  Fnet = 40 000 N  (3) 5.5 For the same change in momentum/Vir dieselfde verandering in momentum During impact, the airbag increases the contact time/Tydens impak verhoog 1 die lugsak die kontaktyd.  (Fnet α ) t The net force will decrease./Die netto krag sal afneem.  (3) [17] QUESTION 6/VRAAG 6 6.1 A force for which the work done in moving an object between two points depends on the path taken.  (2 or zero) ‘n Krag waarvoor die arbeid verrig om ’n voorwerp tussen twee punte te laat beweeg, afhanklik is van die roete wat gevolg word. (2 of nul) (2) 6.2 (Kinetic) frictional force  and pulling force / applied force  (Kinetiese) wrywingskrag en trekkrag / toegepaste krag. (2) 6.3 OPTION 1/OPSIE 1 Wnc = ΔEk + ΔEp  Any one/Enige een WFA + Wfk = Ekf – Eki + 0 1 1 FΔxcosθ + fkΔxcosθ = (20)vf2 - (20)(0)2 2 2 (100)(4)cos0o + (65)(4)cos180o = 10vf2 vf = 3,74 m∙s-1  9

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 OPTION 2/OPSIE 2 Wnet = ΔEk  Any one/Enige een WFA + Wfk = Ekf – Eki 1 1 FΔxcosθ + fkΔxcosθ = (20)vf2 - (20)(0)2 2 2 (100)(4)cos0o + (65)(4)cos180o = 10vf2 vf = 3,74 m∙s-1 (5) 6.4 POSITIVE MARKING FROM 6.3/POSITIEWE NASIEN VANAF 6.3 OPTION 1/ OPSIE 1 OPTION 2/OPSIE 2 Pave = Fvave 𝑣𝑣 +𝑣𝑣 Δx= 𝑓𝑓2 𝑖𝑖 𝛥𝛥𝛥𝛥 3,74+0 0+3,74 =(100)( ) 4= 2 𝛥𝛥𝛥𝛥 2 = 187 W  Δt = 2,139s W P = Δt  Any one/ (Range/Gebied 187 – 187, 08 W) enige een FΔx cosθ = Δt (100)(4)cos0 =  2,139 = 187 W  (3) 6.5 6.5.1  Accept w / mg Aanvaar w / mg (2) 6.5.2 No , The Fnet will be less (decrease) / the w will do negative work and remove energy from the system. Nee, die Fnet sal minder wees (afneem) / die gewig sal negatiewe arbeid verrig en die energie uit die sisteem verwyder. (2) [16] 10

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 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 sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The (apparent) change in frequency (or pitch) (of the sound) detected by a listener because the source and the listener have different velocities relative to the medium of propagation.  Die (skynbare) verandering in die frekwensie (of toonhoogte) (van die klank) waargeneem deur 'n luisteraar omdat die bron en die luisteraar verskillende snelhede relatief tot die voortplantingsmedium 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. (2) 7.2 Away/Weg van  (1) 7.3 fL = v ± vL fS  v ± vS do not accept any other formula/moet 88  = 340  90 geen ander formule aanvaar nie 340+ vS vS = 7,73 m.s-1  (5) [8] 11

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 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 sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them.  If masses in place of charge – zero marks Die grootte van die elektrostatiese krag wat een puntlading (Q1) op ‘n ander puntlading (Q2) uitoefen, is direk eweredig aan die produk van die groottes van die ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle. Indien massa in plek van ladings – geen punte. (2) 8.2 Marking criteria/Nasienriglyne • correct formula for Coulomb’s Law/korrekte formule vir Coulomb se wet  • substitution of 0,866 in Fnet/vervanging van 0,866 in Fnet  • Substitution for C on A/vervanging vir C op A  • Substitution for B on A/vervanging vir B op A  • Correct final answer/Korrekte finale Antwoord  2 2 2 Fnet = FConA + FBonA 2 kQC QA 2 kQB QA 2 Fnet =� � +� �  r2 r2 2 2 9𝑥𝑥109 (7x10−6 )(5x10−6 ) 9𝑥𝑥109 (5x10−6 )(2x10−6 ) 0,866  = � 2 �  + � �  d2 0,52 d = 0,63 m  (5) 8.3 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 sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. 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 eenheid positiewe-lading wat by daardie punt geplaas is, ondervind word. (2) 12

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 8.3.2  EA EB OR/OF EB Enet EA OR/OF Enet (2) 8.3.3 Marking criteria/Nasienriglyne • correct formula /korrekte formule  • substitution for both QA and QB/ vervanging van beide QA en QB  • Subtraction/aftrek  • Correct final answer/Korrekte finale Antwoord  kQA kQB Enet = 2 −  r r2 9×109 (2×10−6 ) 9×109 (5×10−6 ) = −  0,22 0,72 = 358 163,27 N.C-1  OR 3,58 x 105 N.C-1 (4) 8.3.4 The statement is true  The net electric field will always be the sum of EB and EA. The two electric fields will always be in the same direction.  Die stelling is waar. Die netto elektriese veld sal altyd die som van EB en EA wees, die twee elektriese velde sal altyd in die selfde rigting wees. (2) 8.4.1 Directly proportional/direk eweredig.  (1) 8.4.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 F ΔF E=  gradient/gradiënt = Δq  q 0,005 0,005 − 0 =  =  1 x 10−6 1,0 x 10−6 − 0 = 5 000 N · C−1  = 5 000 N · C −1  (3) [21] 13

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 QUESTION 9/VRAAG 9 9.1 The rate at which work is done  (2 or zero) Die tempo waarteen arbeid verrig word. (2 of 0) (2) 9.2.1 1,2 A  (2) 9.2.2 OPTION 1/OPSIE 1: OPTION 3/OPSIE 3: P = VI  P = I2 R  P = 3,6(2,4)  = 2,42 (1,5)  = 8,64 W  = 8,64 W  OPTION 2/OPSIE 2: OPTION 4/OPSIE 4: 1 1 1 1 1 𝑃𝑃𝑅𝑅2 = 𝐼𝐼 2 𝑅𝑅  = + = + R R1 R2 3 3 R = 1,5 Ω = (1,2)2(3)  = 4,32 W V2 P=  Ptotal = PR2 + PR3 R 3,62 =  = 4,32 + 4,32 1,5 = 8,64 W  = 8,64 W  (3) 9.3.1 OPTION 1/OPSIE 1: OPTION 2/OPSIE 2: W = I2 Rt  V2 t W=  = 1,262 (3)(600)  R 3,782 (600) =  = 2857,68 J  3 = 2857,68 J  OPTION 3/OPSIE 3: W = VIt  = 3,78(1,26)(600)  = 2857,68 J  (3) 9.3.2 DECREASE  No current will go through the circuit. Voltmeter has a very high resistance and will not let current pass through.  VERLAAG Geen stroom sal deur die stroombaan beweeg nie. Die voltmeter het ’n baie hoë weerstand en sal nie die stroom deurlaat nie (2) 14

PHYSICAL SCIENCES: PHYSICS MARKING GUIDELINES FISIESE WETENSKAPPE: FISIKA NASIENRIGLYNE (PAPER/VRAESTEL 1) GR12 0625 9.4 The maximum energy provided/work done by a battery per unit charge passing through it.  (2 or zero) Die maksimum energie wat ’n battery lewer/arbeid verrig per eenheidslading wat daardeur vloei. (2 of nul) (2) 9.5 CIRCUIT A/STROOMBAAN A: CIRCUIT B/STROOMBAAN B: ε = I(R + r)  ε = I(R + r) ε = 2,4(3 + 1,5 + r)  ……(1) ε = 1,26(3 + 3 + 3 + r)  ……(2) (1) = (2)  2,4(4,5 + r) = 1,26(9 + r) r = 0,47 Ω  ε = 11,94 V  (6) [20] TOTAL/TOTAAL: 150 15