PHY SC P1 MEMO GR 12 SEPT 2025 E+A

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2025 PHYSICAL SCIENCES P1 MARKING GUIDELINE/ FISIESE WETENSKAPPE V1 NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 15 pages./ Hierdie nasienriglyn bestaan uit 15 bladsye.

2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) QUESTION/VRAAG 1 1.1 D  (2) 1.2 B  (2) 1.3 C  (2) 1.4 D  (2) 1.5 A  (2) 1.6 D  (2) 1.7 A  (2) 1.8 D  (2) 1.9 C  (2) 1.10 B  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 3 QUESTION/VRAAG 2 2.1 A body will remain at rest or motion at constant velocity unless a non-zero net/resultant force acts on it.  ʼn Liggaam sal in rus bly of aanhou beweeg teen konstante snelheid tensy ʼn netto/resultante krag op die liggaam inwerk. (2) 2.2 Accepted labels/Aanvaarde benoeming w Fg/Fw/weight/gravitational force/  gewig/gravitasiekrag T FT/Tension/Force in string/  Spanning/ Krag in die tou (2) 2.3 Gravitational force > Tension force  Gravitasiekrag > Spanningskrag (1) 2.4.1 Fnet = ma Fnet = T – f Any one/Enige een ✓ Fnet = Fg – T 3 kg block / 3 kg blok 3 x 9,8 – T ✓= 3 x 2 ∴ T = 23,4 N N 5 kg block / 5 kg blok Any one/Enige een (3 x 2 or/of 5 x 2)✓ T – fk = 5 x 2 fk = 23,4 – 10 ✓ (5) ∴ fk = 13,4 ✓ 2.4.2 fk = μkN✓ = μkFg = μk mg 13,4 = μk (5)(9,8)✓ (3) ∴ μk = 0,27✓ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) 2.5 The force that the rope exerts on the box and the force that the box exerts on the rope.  Die krag wat die tou op die houer uitoefen en die krag wat die houer op die tou uitoefen. OR/OF The force that the Earth exerts on the box and the force that the box exerts on the Earth.  Die krag wat die Aarde op die houer uitoefen en die krag wat die houer op die Aarde uitoefen. (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 5 QUESTION/VRAAG 3 3.1 The motion of an object where the only force acting on the object is the gravitational force.  Die beweging van ʼn voorwerp waar die enigste krag wat op die voorwerp inwerk die gravitasiekrag is. (2) 3.2.1 Upwards as positive/Opwaarts as positief Ball/Bal A: Δy = viΔt + ½aΔt2  –Δy = (–2,5) Δt + ½(–9,8) Δt2  Ball/Bal B: – (Δy – 5,2) = 0 + ½(–9,8) Δt2  –Δy = –5,2 + ½(–9,8) Δt2 (–2,5) Δt + ½(–9,8) Δt2 = – 5,2 + ½(–9,8) Δt2  (equating/gelykstelling ΔyA to/na ΔyB) Δt = 2,08 s  Upwards as negative/Opwaarts as negatief Ball/Bal A: Δy = viΔt + ½aΔt2  Δy = (2,5) Δt + ½(9,8) Δt2  Ball/Bal B: (Δy– 5,2) = 0 + ½(9,8) Δt2  Δy = 5.2 + ½(9,8) Δt2 (2,5) Δt + ½(9,8) Δt2 = 5,2 + ½(9,8) Δt2  (equating/gelykstelling ΔyA to ΔyB) Δt = 2,08 s  (5) 3.2.2 POSITIVE MARKING FROM 3.2.1/POSITIEWE NASIEN VANAF VRAAG 3.2.1 Upwards as positive/Opwaarts as Upwards as negative/Opwaarts positief as negatief v f = vi + a t  v f = vi + a t  Vf = –2,5 + (–9,8)(2,08)  vf = 2,5 + (9,8)(2,08)  vf = –22,884 vf = 22,884 m·s–1, vf = 22,884 m·s–1, downwards/afwaarts  downwards/afwaarts  (3) 3.2.3 POSITIVE MARKING FROM 3.2.2/ POSITIEWE NASIEN VANAF VRAAG 3.2.2 OPTION/OPSIE 1 Upwards as positive/Opwaarts as Upwards as negative/Opwaarts positief as negatief Ball/Bal A: Ball/Bal A: v 2f = v i2 + 2ay  v 2f = v i2 + 2ay  –22,8842 = –2,52 + 2(–9,8)( y)  22,8842 = 2,52 + 2(9,8)( y)  y = –26,40 m y = 26,40 m Height/Hoogte Y = –26,40 + 5,2 Height/Hoogte Y = 26,40 – 5,2 Height/Hoogte Y = 21,20 m  Height/Hoogte Y = 21,20 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) POSITIVE MARKING FROM 3.2.1/ POSITIEWE NASIEN VANAF VRAAG 3.2.1 OPTION/OPSIE 2 Upwards is positive/Opwaarts as positief Upwards is negative/Opwaarts as negatief 2 Δy = viΔt + ½aΔt  Δy = viΔt + ½aΔt2  2 Δy = (–2,5) x 2,08 + ½(–9,8) 2,08  Δy = (2,5) 2,08 + ½(9,8) 0,2082  Δy = –26,40 Δy = 26,40 Height/Hoogte Y = –26,40 + 5,2 Height/Hoogte Y = 26,40 – 5,2 Height/Hoogte Y = 21,20 m  Height/Hoogte Y = 21,20 m  OPTION/OPSIE 3 Ball/Bal B POSITIVE MARKING FROM 3.2.1/ POSITIEWE NASIEN VANAF VRAAG 3.2.1 Upwards is positive / Opwaarts as Upwards is negative/ Opwaarts positief as negatief Δy = viΔt + ½aΔt2 Δy = viΔt + ½aΔt2  Δy = 0 x 2,08 + ½(–9,8) 2,082 Δy = 0 2,08 + ½(9,8) 0,2082  Height/Hoogte Y = 21,20 m Height/Hoogte Y = 21,20 m  (3) 3.3 Marks/ Criteria for graph/Kriteria vir grafiek Punte Shape parallel lines/Beide lyne parallel.  B starts on (0,0) and A from 2,5/ B begin by (0,0) en vanaf 2,5  2,08 s and /en 22,884 m·s–1 indicated / aangedui (2,08, 22,884)  (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 7 QUESTION/VRAAG 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. (2) 4.2 Right as positive/Regs as positief Σpi = Σpf (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 Any one/Enige een ✓ (mvi) 1 + (mvi)2 = (m1 + m2) vf (4 000 x 32,17) + (2 000 x 25)  = (6 000) vf  vf = 29,78 m.s–1 in the original direction/in die oorspronklike rigting  (4) 4.3 Inelastic/Onelasties  (1) 4.4.1 equal to F/gelyk aan F  (1) 4.4.2 Newton’s Third Law of motion/Newton se derde Bewegingswet.  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 Bewegingswet Wanneer voorwerp A ʼn krag op voorwerp B uitoefen, oefen voorwerp B gelyktydig ʼn krag van gelyke grootte en in die teenoorgestelde rigting op voorwerp A uit. (3) [11] QUESTION/VRAAG 5 5.1 Accepted labels/Aanvaarde benoeming w Fg/Fw/weight/gravitational force  gewig/gravitasiekrag F FA/Applied Force/  Toegepastekrag N FN/N/Normal force/  Normaalkrag (3) 5.2 No , it is not an isolated system  Nee, dit is nie op geїsoleerde stelsel/sisteem nie OR/OF No , there is an external force (applied force) acting on the object  Nee, daar is ʼn eksterne krag (toegepastekrag) wat op die voorwerp inwerk. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) 5.3 Gravitational force/weight/Gravitasiekrag/gewig ✓ (1) 5.4 The force is perpendicular to the displacement. ✓ Die krag is loodreg aan die verplasing. OR/OF Therefore, cos θ = 90°/daarom is cos θ = 90° ✓ (1) 5.5 From B to A ✓ (1) Van B tot A 5.6 OPTION/OPSIE 1 Wnet = ∆K Wnet = ∆Ek Any one/Enige een ✓ W// + WF = ∆K Fg//·Δx·Cosθ + F·Δx·Cosθ = ½mvf2 – ½mvi2 / (20 x 9,8 x Sin30º)(4,2)Cos180º + F(4,2)Cos0º = ½(20)(12,2)2 – ½(20)(13,5)2  411,6)(–1) + F(4,2)(1) = 1488,4 – 1822,5 F = 18,45 N  OPTION/OPSIE 2 Wnc = ΔEk + ΔEp Any one/Enige een ✓ F·Δx·Cosθ = ½mvf2 – ½mvi2 + mghf – mghi F(4,2)Cos0º= [½(20)(12,2)2 – ½(20)(13,5)2]  + [(20)(9,8)(4,2 Sin30º)  – 0] F(4,2)(1) = 1488,4 – 1822,5 + (196)(4,2)(0,5) F = 18,45 N  (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 9 QUESTION/VRAAG 6 6.1 v =f✓ 330 = (440)  ✓  = 0,75 m ✓ (3) 6.2 440 Hz ✓ (1) 6.3 The (apparent) change in frequency/pitch  of the sound detected by a listener because the listener and the sound source have different velocities relative to the medium of sound propagation.  Die (skynbare) verandering in frekwensie/toonhoogte van die klank wat deur ʼn luisteraar waargeneem word want die luisteraar en die klankbron het verskillende snelhede relatief aan die medium van klank voortplanting. (2) 6.4 330 m∙s–1 ✓ (1) 6.5 v  vL fL = fs  v  vs 330 ✓ = 440  330−8 ✓ fL = 450,93 Hz  (5) [12] QUESTION/VRAAG 7 7.1 It is the electrostatic force experienced per unit positive magnitude of the charge placed at that point.  Dit is die elektrostatiese krag wat deur ʼn eenheidpositiewe-lading wat by daardie punt geplaas is, ondervind word. (2) 7.2 Marking Criteria/Nasienkriteria Correct Shape/Korrektevorm  Arrows pointing towards P and Q/Pyltjies wys na P en Q  Field lines don’t touch/cross and perpendicular on charge/Veldlyne raak nie/kruis en loodreg op lading  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) 7.3 kQ E= 2  r1 (9 10 9 )(5 10 - 9 )  5,22 10 5 = r2 r = 9,285 x 10–3 ∴d = 9,285 x 10–3 – 8 x 10–3 m ✓ = 1,29 x 10–3 m ✓(or 1,29 mm) (4) 7.4 The magnitude of the electrostatic force exerted by one point charge on another point charge is directly proportional to the product of the charges  and inversely proportional to the square of the distance between them.  Die grootte van die elektrostatiese krag wat deur een puntlading op ʼn ander puntlading uitgeoefen word, is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. (2) 7.5 𝐅 = 𝒌𝑸𝟏𝑸𝟐  𝟐 𝒓 (𝟗×𝟏𝟎𝟗 )(𝟓×𝟏𝟎−𝟗 )(𝟓×𝟏𝟎−𝟗 ) 𝐅𝟏 =  (𝟎,𝟎𝟎𝟖)𝟐  𝐅𝟏 = 3,52×10–3 N (west/wes OR/OF left/links) (𝟗×𝟏𝟎𝟗 )(𝟓×𝟏𝟎−𝟗 )(𝟔×𝟏𝟎−𝟗 ) 𝐅𝟐 =  (𝟎,𝟎𝟏𝟐)𝟐 𝐅𝟐 = 1,88 × 10–3 N (south/suid OR/OF down/af) Fnet 2 = (3,52×10–3)2 + (1,88 × 10–3)2  ∴Fnet = 3,98 X 10–3 N  Range (3,98×10–3 N – 3,99×10–3 N) (6) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 11 QUESTION/VRAAG 8 8.1 emf ( ε ) = IRext + Ir ✓/ emf ( ε ) = V + Vint When the current increases, Ir (lost volts) increases ✓ IVext (terminal voltage/pd)( voltage of the load) decreases ✓ Since emf ( ε ) is the same/constant ✓ Wanneer die stroom toeneem, neem Ir (verlore volts) toe IRekst (terminaal spanning/ spanning van die las) neem af emk ( ε ) is dieselfde/konstant (4) 8.2 Group/Groep 2 ✓ [Accept/Aanvaar r2] (1) 8.3 Δy gradient = ✓ Use any coordinate values Δx Gebruik enige koördinaatwaardes 𝟒−𝟖 = 𝟒−𝟐 ✓ gradient = –2 Ω r=2Ω✓ (3) 8.4 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 OR/OF The ratio of potential difference across a conductor to current through the conductor is constant at constant temperature. Die verhouding van potensiaalverskil oor ʼn geleier tot stroom deur die geleier is konstant by konstante temperatuur. (2) 8.5 OPTION/OPSIE 1 𝑅 𝑅 1 1 1 Rext= Rx+ Rparallel OR/OF 𝑅𝑒𝑥𝑡 = 𝑅𝑥 + (𝑅 1+𝑅2 ) OR/OF =𝑅 +𝑅 ✓ 2 1 𝑅𝑝 1 2 Any one/Enige een 1 1 1 = + ✓ ∴ R p = 2,67Ω R // 4 8 V R ext = ✓ I 10 ✓ Rext = 1,5 = 6,67Ω 6,67 = Rx+ 2,67 ✓ Rx = 6,67 – 2,67 =4Ω✓ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

12 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) OPTION/OPSIE 2  R 1R 2  1 1 1 Rext= Rx+ Rparallel OR/OF R ext = R x +   R + R  OR/OF 𝑅𝑝 =𝑅 +𝑅 ✓  2 1  1 2 Any one/Enige een 1 1 1 = + ✓ ∴ R= 2,67Ω R // 4 8 Vp = IRp ✓ = (1,5)(2,67) ✓ = 4,00 V OR/OF 4,01 V VRx = 10 – 4 OR/OF 10 – 4,01 =6V OR/OF 5,99 V VRx =IRx 6 = (1,5)RX ✓ OR/OF 5,99 = (1,5)RX ✓ ∴RX = 4 Ω ✓ OR/OF 3,99 Ω ✓ (6) 8.6 OPTION/OPSIE 1 POSITIVE MARKING FROM 8.5/ POSITIEWE NASIEN VANAF VRAAG 8.5 Emf/emk = I(R + r) ✓ 12 = (1,5)(2,67+ 4 + r) ✓ OR/OF (1,5)(2,67 + 3,99 + r) ∴r = 1,33 Ω (1,33 Ω – 1,34 Ω) ✓ OPTION/OPSIE 2 Vlost/ verlore = Ir ✓ 2= (1,5) r ✓ ∴r = 1,33 Ω ✓ (3) OR/OF 8.7 W =VI Δt  For the same potential different and time, internal I4Ω is greater than I8Ω  W𝛼I Vir dieselfde potensiaal verskil en tyd, is I4Ω groter as I8Ω V2 Energy: W = Δt  For the same potential difference and time  W∝ 1 is R R greater for the smaller resistance than for the larger resistance.  V2 Energie: W = Δt Vir dieselfde potensiaalverskil en tyd W∝ 1 is groter vir die R R kleiner weerstand as vir die groter weerstand (3) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 13 QUESTION/VRAAG 9 9.1 The rms voltage of AC is the AC potential difference which dissipates produces the same amount of energy as an equivalent DC potential difference  Die wgk spanning van WS is die WS-potensiaalverskil wat dieselfde hoeveelheid energie verbruik/oordra as ʼn ekwivalente GS-potensiaalverskil. (2) 9.2 Mechanical energy to electrical energy.  Meganiese energie na elektriese energie (1) 9.3 It can be stepped up or stepped down/is easier to transmit  with less energy lost AC can be converted to DC but DC cannot be converted to AC Dit kan verhoog of verlaag word/is makliker om oor te dra met minder energie wat verlore raak WS kan na GS omgeskakel word maar kan nie na WS herlei word nie (1) 9.4 OPTION/OPSIE 1 Vmax I Vrms =  Any one/Enige een Irms = max 2 2 311 21 Vrms =  I rms =  2 2 = 219,91 V = 14,85 A Pave = VrmsIrms ✓ = (219,91)(14,85)  = 3265,66 W ✓ OPTION/OPSIE 2 Pave = Vmax I max ✓✓ 2 Pave = 311 x 21 ✓✓ 2 Pave = 3265,5 W ✓ OPTION/OPSIE 3 I V Irms = max ✓ R= 2 I 21 = √2 ✓ 311 R = 21 ✓ = 14,85 A Pave = I2rmsR ✓ R = 14,81Ω = (14,85)2 (14,81) ✓ Pave = 3265,94✓ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

14 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) OPTION/OPSIE 4 V R= I Pave = V2rms ✓ 311 R R = 21 ✓ = 219,912 ✓ R = 14,81Ω 14,81 Vmax Pave = 3 265,39 W ✓ Vrms = ✓ 2 311 ✓ Vrms = 2 Vrms = 219,91 V Range [ 3 265,39 V – 3 265,94] (6) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 15 QUESTION/VRAAG 10 10.1 The energy of the photons of red light is greater than the work function of the metal in the photocell.  Photo electrons are ejected (from the metal surface)  Die energie van die fotone van rooi lig is groter as die arbeidsfunksie van die metaal in die fotosel. Foto-elektrone word vrygestel (vanaf die metaal oppervlak) OR/OF The frequency of the red light is higher than the threshold/cut-off frequency of the metal in the photocell.  Photo electrons are ejected (from the metal surface)  Die frekwensie van die rooi lig is hoër as die drumpel/afsnyfrekwensie van die metaal in die fotosel. Foto-elektrone word vrygestel (vanaf die metaal oppervlak) (2) 10.2 Increase/Toeneem  (1) 10.3 Stays the same  The change in colour/frequency only affected the kinetic energy of the photo electrons.  Only the intensity of the light affected the number of photo electrons emitted per time unit.  The intensity of the light stays the same and therefore the number of photo electrons emitted per unit time /current stays the same.  Bly dieselfde Die verandering in kleur/frekwensie het slegs ʼn invloed op die kinetiese energie van die foto-elektrone. Slegs die intensiteit van die lig het ʼn invloed op die aantal foto-elektrone wat per tydseenheid vrygestel word Die intensiteit van die lig bly dieselfde en daarom bly die aantal foto- elektrone wat per eenheid tyd/stroom vrygestel word dieselfde (4) 10.5 E = Wo + Ek (max) c h = hfo + ½mvmax2 Any one/Enige een λ 6,63 x 10-34  3  108  6,63 x 10-34(f )  + ½(9,11 x 10-31)(4,78 x 105)2 = o 4,5  10−7   fo = 5,10 x 1014 Hz  (6) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief