# Physical Sciences Grade 12 Mpumalanga November 2020 Memo > Grade 12 Physical Sciences memorandum from the Mpumalanga November 2020, 15 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/2020-mpumalanga-physics-memo/e9341184-a243-4f37-9ae2-f5d02abd02db - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2020 - Exam period: Mpumalanga November - Curriculum: CAPS - Pages: 15 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 Mpumalanga November 2020 Memo](https://www.hlayiso.com/library/2020-mpumalanga-physics-memo/e9341184-a243-4f37-9ae2-f5d02abd02db) ## About this document Physical Sciences · Grade 12 · Mpumalanga November · 2020. Memorandum, 15 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) SEPTEMBER 2020 MARKING GUIDELINES / NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consists of 15 pages Hierdie nasienriglyne bestaan uit 15 bladsye Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 2 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 MDE/MDO/September 2020 NSC/NSS QUESTION 1 / VRAAG 1 1.1 B  (2) 1.2 A  (2) 1.3 C  (2) 1.4 D  (2) 1.5 D  (2) 1.6 A  (2) 1.7 C  (2) 1.8 C (2) 1.9 C  (2) 1.10 A  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 3 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 MDE/MDO/September 2020 NSC/NSS QUESTION 2 / VRAAG 2 2.1 When a resultant/net force acts on an object, the object will accelerate in the direction of the force  at an acceleration directly proportional to the force and inversely proportional to the mass of the object.  Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, versnel die voorwerp in die rigting van die krag teen 'n versnelling direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp. Note: If any one of the underlined key words in the correct context is omitted, deduct 1 mark (2) 2.2 F to the right  OR –F (1) 2.3 B experiences no net force (Fnet = 0)  Therefore B will not accelerate  (2) 2.4 N FA  FT  f w (5) Accepted labels / Aanvaarde benoemings w Fg / Fw / force of earth on block / weight / mg / gravitational force FA Fapplied / FT Tension in rope / T N Normal force f friction / fk / kinetic friction / Ff Notes/Aantekeninge:  Any additional forces: deduct 1 mark: max 4⁄5  No labels: deduct 1 mark: max 4⁄5  No arrows: 0⁄5  Force(s) not touching object: deduct 1 mark: max 4⁄5 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 4 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 MDE/MDO/September 2020 NSC/NSS 2.5 OPTION 1 For Block A: For Block B: Fnet = ma Fnet = ma Any one  FB - f = ma FT – 4,7 -11,35 = 4(4,5)  FB – 2,35  = 2(4,5)  FB = 11,35 N FT = 34,05 N  (5) OPTION 2: SYSTEMS APPROACH max 2⁄5 Fnet = ma  FT – f =ma FT – (2,35+4,7) = 6(4,5) FT = 34,05 N  (5) [15] QUESTION 3 / VRAAG 3 3.1 – Yes  Only gravitational force/weight is acting on the ball.  – Ja  Slegs gravitasiekrag/gewig werk op die bal in.  (2) 3.2 OPTION 1 Upwards positive: Downwards positive: vf2 = vi2 + 2a∆y  vf2 = vi2 + 2a∆y  = (8)2 + 2(-9,8)(-21)  = (-8)2 + 2(9,8)(21)  vf = 21,81 m·s-1  vf = 21,81 m·s-1  (3) OPTION 2 Upwards positive: Downwards positive: vf2 = vi2 + 2a∆y vf2 = vi2 + 2a∆y 0 = (8)2 + 2(-9,8) ∆y 0 = (-8)2 + 2(9,8) ∆y ∆y = 3,27 m ∆y = -3,27 m (3,27+21) (3,27+21) vf2 = vi2 + 2a∆y  vf2 = vi2 + 2a∆y  = (0)2 + 2(-9,8)(-24,27)  = (0)2 + 2(9,8)(24,27)  vf = 21,81 m·s-1  vf = 21,81 m·s-1  (3) OPTION 3 (mgh + ½mv2)Roof = (mgh + ½mv2)Ground  m(9,8)(21) + ½(m)(8)2 = 0 + ½(m)v2  v = 21,81 m·s-1  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 5 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 MDE/MDO/September 2020 NSC/NSS OPTION 1 3.3 Upwards Positive Downwards Positive Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  9 = vi (2,24) + ½ (-9,8)(2,24)2  (-9) = vi (2,24) + ½ (9,8)(2,24)2  vi = 14,99 m·s-1  vi = -14,99 vi = 14,99 m·s-1  (4) 3. 4 POSITIVE MARKING FROM QUESTION 3.2 & 3.3 Upwards positive: Downwards positive: 14,99 21,81 Velocity (m·s-1) Velocity (m·s-1) 8    t (s) t (s) -8  -21,81 -14,99 CRITERIA Graph starts at ±8 m·s-1  First line is a straight line through the t-axis.  Line ends at v = ±21,81 m·s-1  Second line correctly drawn starting from v = ±14,99 m·s-1  Two sloping parallel straight lines  (5) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 6 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 MDE/MDO/September 2020 NSC/NSS QUESTION 4 / VRAAG 4 4.1 The total (linear) momentum in a closed system remains constant.  Die totale (liniêre) momentum in 'n geslote sisteem bly behoue. Note: If any one of the underlined key words in the correct context is omitted, deduct 1 mark (2) 4.2.1 OPTION 1 Σpi = Σpf  Any one (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 (0,65)(8) + (0,16)(-20)= (0,65)vf + (0,16)(30)  vf = -4,31 vf = 4,31 m·s-1 West  (Accept: left) OR (0,65)(-8) + (0,16)(20)= (0,65)vf + (0,16)(-30)  vf = 4,31 m·s-1 West/to the left / Wes/na links  (4) 4.2.2 OPTION 1 TO THE RIGHT AS POSITIVE Fnet·∆t = mvf – mvi  Fnet (0,015) = (0,16)(30) – (0,16)(-20)  Fnet = -533,33 Fnet = 533,33 N East/to the right / Oos/na regs  TO THE LEFT AS POSITIVE Fnet·∆t = mvf – mvi  Fnet (0,015) = (0,16)(-30) – (0,16)(20)  Fnet = 533,33 N East/to the right / Oos/na regs  (3) POSITIVE MARKING FROM QUESTION 4.2.1 OPTION 2 For the stick: Fnet·∆t = mvf – mvi  Fnet (0,015) = (0,65)(-4,31) – (0,65)(8)  Fnet = -533,43 ∴(Fnet)Ball = 533,43 N East/to the right / Oos/na regs  (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 7 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 MDE/MDO/September 2020 NSC/NSS QUESTION 5 / VRAAG 5 5.1 ACCEPT N FT  N FT  f f w w┴ w//  Accepted labels / Aanvaarde benoemings w Fg / Fw / force of earth on block / weight / 29,4 N / mg / gravitational force / gewig / gravitasiekrag / krag van aarde op blok N Normal force / FN / Force of incline on block / Normaalkrag / Normaal / krag van skuinsvlak op blok FT Tension / T / force of rope / F / spanning / spankrag / krag van tou op blok f Friction / Kinetic friction / fk / Ff / wrywing / kinetiese wrywingskrag 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 pyle penaliseer nie.  Deduct 1 mark when more than 4 forces drawn. / Trek 1 punt af vir meer as 4 kragte geteken.  Deduct 1 mark if force(s) do not make contact with dot. / Trek 1 punt af indien krag(te) nie met die kolletjie kontak maak nie.  If arrows missing/Indien pylpunte uitgelaat word: Max/Maks 3⁄4 (4) 5.2.1 Weight / Gravitational force / Fg / gewig / Gravitasiekrag (1) 5.2.2 Weight of block R / Gewig van blok R (1) 5.2.3 Tension / Force in rope / Toukrag / Spankrag (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 8 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 MDE/MDO/September 2020 NSC/NSS 5.3 The net/total work done on an object is equal to the change in the object's kinetic energy.  Die netto/totale arbeid verrig op 'n voorwerp is gelyk aan die verandering in kinetiese energie van die voorwerp. ACCEPT: The work done on an object by a resultant/net force is equal to the change in the object's kinetic energy. Die arbeid verrig op die voorwerp deur 'n resulterende/netto krag is gelyk aan die verandering in kinetiese energie van die voorwerp. Note: If any one of the underlined key words in the correct context is omitted, deduct 1 mark (2) 5.4 OPTION 1 Wnet = ΔEk Any one  (W Fg//)P + (W f)P + (W Fg)R = ½mvf2 – ½mvi2 (3)(9,8)sin20°(2)cos180° + (15)(2)cos180°+ (8×9,8)(2)cos0° = ½(11)v2 –0 v = 4,40 m·s-1  (6) OPTION 2 2 h 20° Wnet = ΔEk Any one  h=2sin20° (W Fg)P + (Wf)P (WFg)R = ½mvf2 – ½mvi2 (3)(9,8)(2sin20°)cos180° + (15)(2)cos180°+ (8×9,8)(2)cos0° = ½(11)v2 –0 v = 4,40 m·s-1  (6) OPTION 3 Wnc = ΔEk + ΔEp Any one  (W f)P = [½mvf2 – ½mvi2] + [mghf - mghi]P + [mghf - mghi]R (15)(2)cos180° = [½(11)v2 – 0]  + [(3)(9,8)(2sin20°) - 0] + [0-8(9,8)(2)]  v = 4,40 m·s-1  (6) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 9 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 MDE/MDO/September 2020 NSC/NSS QUESTION 6 / VRAAG 6 6.1 v = fλ  1 500 = 380 λ  λ = 3,95 m  (3) 6.2 Smaller than / Kleiner as  λ∝v  if f stays constant / mits f konstant bly  OR Smaller than / Kleiner as  v = fλ 340 = 380 λ  λ = 0,89 m  (3) 6.3 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 klank wat deur 'n luisteraar waargeneem word omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium waarin die klank voortgeplant word, het. OR The change in frequency (or pitch) of the sound detected by a listener as a result of the relative motion between the source and the observer. Die verandering in frekwensie (of toonhoogte) van die klank waargeneem deur 'n luisteraar as gevolg van die relatiewe beweging tussen die bron en die waarnemer. Note: If any one of the underlined key words in the correct context is omitted, deduct 1 mark (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 10 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 MDE/MDO/September 2020 NSC/NSS 6.4 Marking criteria/Nasienriglyne:  Appropriate formula/Toepaslike formule   Substitution for detector A/Vervanging vir detektor A   Substitution for detector B/Vervanging vir detektor B   Equating difference in frequency/Vergelyking vir verskil in frekwensie   Final answer/Finale antwoord: 7,99 m·s-1  Detector A/Detektor A Detector B/Detektor B v±vL v±vL fL = f  fL = f v±vS S v±vS S 340+vL 1 500+ vL = (380) = (380) 340 1 500 (fL)A - (fL)B = 6,91 340+vL 1 500+ vL (380) - (380) = 6,91 340 1 500 vL = 7,99 m·s-1  (5) [13] QUESTION 7 / VRAAG 7 7.1 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  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.  Note: If any one of the underlined key words in the correct context is omitted, deduct 1 mark  If masses used ( 0⁄2 ) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 11 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 MDE/MDO/September 2020 NSC/NSS 7.2 FT  Accepted labels / Aanvaarde benoemings w Fg / Fw / force of earth on block / weight / 29,4 N / mg / FE  gravitational force / gewig / gravitasiekrag / krag van aarde op blok FT Tension / T / force of rope / F / spanning / spankrag / w krag van tou op blok FE Electrostatic force/Coulomb force/FE Field 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 pyle penaliseer nie.  Deduct 1 mark when more than 3 forces drawn. / Trek 1 punt af vir meer as 3 kragte geteken.  Deduct 1 mark if force(s) do not make contact with dot. / Trek 1 punt af indien krag(te) nie met die kolletjie kontak maak nie.  If arrows missing/Indien pylpunte uitgelaat word: Max/Maks 2⁄3 (3) 7.3 85°  Both spheres experiences the same force  (in magnitude) and have the same mass . / Beide sfere ervaar dieselfde grootte krag en het dieselfde massa. (3) 7.4 kQ1 Q2 FE FE =  r2 85° (9×109 )(q)(2q) = 2  Fg (0,2) 5° F T Fg = FEtan85°  FE = Fgtan5°  9 (9×10 )(q)(2q) (9×109 )(q)(2q) (0,03)(9,8) = 2 tan85°  2  = (0,03)(9,8)tan5°  (0,2) OR (0,2) q = 2,39×10-7 C  q = 2,39×10-7 C  (6) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 12 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 MDE/MDO/September 2020 NSC/NSS QUESTION 8 / VRAAG 8 8.1 Marking criteria/Nasienriglyne: -1 mark for each of the 5 key words omitted in the correct context. -1 punt vir elk van die 5 sleutelwoorde 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 eenheids-lading wat by daardie punt geplaas is, ondervind word. (2) 8.2 Accept/Aanvaar: + - Shape / Vorm  Direction / Rigting  Lines don't cross or touch /Lyne kruis of raak nie.  (3) 8.3 kQ kQ EA =  EB = r2 r2 (9×109 )(0,6×10-6 ) (9×109 )(1,2×10-6 ) =  =  0,092 0,032 = 6,67 × 105 N∙C-1 (to the right) = 1,20 × 107 N∙C-1 (to the right) Enet = EA + EB = 6,67× 105 + 1,20 × 107 [addition of 2] = 1,27 × 107 N∙C-1  (5) 8.4 Smaller than/Kleiner as  (1) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 13 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 MDE/MDO/September 2020 NSC/NSS QUESTION 9 / VRAAG 9 9.1 The rate of flow of charge.  (2 or 0) Die tempo van ladingvloei. (2) 9.2.1 P = I2R  37,5 = (2,5)2 R  R=6Ω (3) 9.2.2 POSITIVE MARKING FROM QUESTION 9.2.1 1 1 1 = + R r1 r2 1 1 = +  2 6 Rp = 1,5 Ω Re = 1,5 + 6  = 7,5 Ω Ɛ = I(R+r) = 2,5(7,5+0,5)  Ɛ = 20 V  (5) 9.3 POSITIVE MARKING FROM QUESTION 9.2.1 and 9.2.2 OPTION 1 180 I= (2,5) = 4,5 A 100 Ɛ = I(R+r) 20 = 4,5(Re + 0,5)  Re = 3,94 Ω 1 1 1 = + R r1 r2 1 1 1 = +  3,94 7,5 R R = 8,32 Ω  [Range: 8,30 Ω - 8,32Ω] (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 14 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 MDE/MDO/September 2020 NSC/NSS QUESTION 10 / VRAAG 10 10.1 Mechanical (energy)/kinetic (energy) TO electrical (energy)  Meganiese (energie)/kinetiese (energie) NA elektriese (energie) (1) 10.2 1,5 OR 1½  (1) 10.3.1 1 f= number of cycles  f=  time T 1,5 1 OR f=  =  3×10−2 2×10-2 = 50 Hz  = 50 Hz  (3) 10.3.2 OPTION 1 Vmax Imax Vrms = Irms = √2 √2 311 23 =  =  √2 √2 = 219,91 V = 16,26 A Pave = VrmsIrms = (219,91)(16,26)  = 3575,74 W  OPTION 2 V I Pave = max2 max  (23)(311) =  2 = 3576,5 W  (5) [10] QUESTION 11 / VRAAG 11 11.1 The minimum energy that an electron in the metal needs to be emitted from the metal surface.  Die minimum energie benodig om 'n elektron uit die oppervlak van 'n metaal vry te stel. Note: If any one of the underlined key words in the correct context is omitted, deduct 1 mark (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asb om ### Page 15 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 MDE/MDO/September 2020 NSC/NSS 11.2 The frequency of red light is less than the threshold frequency of the metal.  Die frekwensie van rooi lig is kleiner as die drumpelfrekwensie van die metaal. OR/OF The energy of a photon of red light is less than the workfunction of the metal.  Die energie van 'n foton van rooi lig is kleiner as die arbeidsfunksie/ werksfunksie van die metaal. (2) 11.3 E = W 0 + Ek(max) hc 1 Any one  =hf0 + mv2max λ 2 (6,63×10-34 )(3×108 ) -9  = (6,63×10-34 )(6,94×1014 )+ ½ (9,11×10-31 )v2max 430×10 vmax = 7,31 × 104 m∙s-1  (4) 11.4.1 W P=  Δt (4,63×10-19 )(5×1020 ) =  60 = 3,86 W  (3) 11.4.2 5×1020 photons eject 5×1020 photo electrons per minute 5×1020 = 8,33×1018 electrons s-1 60 NOTE: Q Any calculation of 8,33×1018  n=  q Enige berekening van 8,33×1018 Q = (8,33×1018) (1,6×10-19) = 1,33 C  (3) 11.4.3 POSITIVE MARKING FROM QUESTION 11.4.2 1,33 A  (1) [15] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou ## Related documents - [PHYS SCIENCES P2 MEMO GR12 SEPT2020 A E](https://www.hlayiso.com/library/phys-sciences-p2-memo-gr12-sept2020-a-e/91c998b5-2ceb-4a4b-b327-c403d1661716) — Physical Sciences · Memo · 2020 - [WC 2020 Gr 12 SEPT MCED Physics P1 COMMON CLUSTER EXAM MEMORANDUM 2](https://www.hlayiso.com/library/wc-2020-gr-12-sept-mced-physics-p1-common-cluster-exam-memorandum-2/b6743005-6dd1-403e-b52a-d4224e11cb5b) — Physical Sciences · Memo · 2020 - [MP 2015 SEPT Physics P 1 Memo](https://www.hlayiso.com/library/mp-2015-sept-physics-p-1-memo/888a8f9f-c0da-4ad1-b3ae-e2ecd133acdd) — Physical Sciences · Memo · 2015 - [mpu physical sciences p2 sept 2014 memo](https://www.hlayiso.com/library/mpu-physical-sciences-p2-sept-2014-memo/b6800da7-3221-43f2-ba55-a68d6aaa7dab) — Physical Sciences · Memo · 2014 - [2020 Free State Chemistry](https://www.hlayiso.com/library/2020-free-state-chemistry/45fa72e9-1b9d-4869-84af-73a2900ddfc8) — Physical Sciences · QP · 2020 - [Physical Sciences P2 Nov 2020 Afr](https://www.hlayiso.com/library/physical-sciences-p2-nov-2020-afr/798fcbd8-dd69-4871-95a6-d42a4af1a841) — Physical Sciences · QP · 2020 · Afr ## More from Grade 12 Physical Sciences - [PHYSICAL SCIENCE P1 QP JCT 2025 AFR D watermark](https://www.hlayiso.com/library/physical-science-p1-qp-jct-2025-afr-d-watermark/b9429ac5-acbb-4fae-989d-af9e1478a96b) — QP · 2025 · Afr - [PHYSICAL SC P2 QP GR12 SEPTEMBER 2025 AFR1 watermark](https://www.hlayiso.com/library/physical-sc-p2-qp-gr12-september-2025-afr1-watermark/37cff2db-7c1a-4ec8-8ac5-07ec5bcf87a9) — QP · 2025 - [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 KZN](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-kzn/6d225dc8-24c6-4c28-a181-df906d5a2c0e) — QP · 2025 - [Physical Sciences Physics Grade 12 PS01 QP March 2025 Limpopo](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-ps01-qp-march-2025-limpopo/482ee9af-e88a-4818-9120-11ab9e045a1c) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-limpopo/b65e9832-a6be-4312-8f5a-57425d647121) — QP · 2025 ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.