# Physical Sciences Grade 12 P1 NSC November 2025 Memo (Afr) > Grade 12 Physical Sciences Paper 1 memorandum from the NSC November Exam 2025 (Afrikaans), 20 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-12-physical-sciences-2025-nsc-november-exam-memorandum-paper-1-afrikaans/b6617982-c662-4e74-a254-d9c7f8e89d85 - Updated: 2026-09-28 - Subject: Physical Sciences - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2025 - Exam period: NSC November Exam - Paper: 1 - Curriculum: CAPS - Pages: 20 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 P1 NSC November 2025 Memo (Afr)](https://www.hlayiso.com/library/grade-12-physical-sciences-2025-nsc-november-exam-memorandum-paper-1-afrikaans/b6617982-c662-4e74-a254-d9c7f8e89d85) ## About this document Physical Sciences · Grade 12 · NSC November Exam · 2025 · Afrikaans. Memorandum, 20 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 NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2025 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 20 pages. Hierdie nasienriglyne bestaan uit 20 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 B  (2) 1.2 B  (2) 1.3 C  (2) 1.4 C  (2) 1.5 A  (2) 1.6 D  (2) 1.7 D  (2) 1.8 B  (2) 1.9 C  (2) 1.10 C  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. When the net/resultant force acts on an object, the object will accelerate in the direction of the force and the acceleration is 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 OR/OF The net/resultant force acting on an object is equal to the rate of change of momentum of the object (in the direction of the net force). (2 or 0) Die netto/resulterende krag wat op 'n voorwerp inwerk, is gelyk aan die tempo van verandering van momentum van die voorwerp (in die rigting van die netto krag). (2) 2.2 Accept force diagram/ Aanvaar kragtediagram: N F N F fsmax s fsmax w w Accepted labels/Aanvaarde byskrifte Fg /Fw /Fearth on P /weight /mg /49 N /gravitational force w Fg /Fw /Faarde op P /gewig /mg /49 N /gravitasiekrag F FA /18 N / Applied force / Toegepaste krag fsmax Ff / fs /f /Friction (force) / (static) friction / (statiese) wrywingskrag N FN /Normal (force) /Fnormal /Fnormaal /Normaal(krag) Notes/Aantekeninge • Mark awarded for label and arrow./Punt toegeken vir byskrif en pyltjie. • fk not accepted as label/fk word nie aanvaar nie • Do not penalise for length of arrows since drawing is not to scale./Moenie vir die lengte van die pyltjies penaliseer nie aangesien die tekening nie volgens skaal is nie. • Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks: 3⁄ 4 • If everything correct, but no arrows/Indien alles korrek, maar geen pyltjies: Max/Maks: 3⁄ 4 • If force(s) do not make contact with the dot /Indien krag(te) nie met die kolletjie kontak maak nie: Max/Maks: 3⁄4 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 2.3.1 Fx = Fcos θ OR/OF OR/OF 15 = 18 cos θ  9,95 = 15 tan θ  9,95 = 18 sin θ  θ = 33,56º  θ = 33,56º  θ = 33,56º  (2) NOTE/NOTA: If answer only/Indien slegs antwoord 1/2 2.3.2 POSITIVE MARKING FROM QUESTION 2.3.1/POSITIEWE NASIEN VANAF VRAAG 2.3.1 Marking criteria/Nasienkriteria • Formula / Formule fsmax = µsN  • Correct substitution of fsmax / Korrekte vervanging van fsmax  • Correct calculation of Fy / Korrekte berekening van Fy  • Subtraction for the calculation of N from Fg thus: (5)(9,8) –(18sin 33,56°) Aftrek vir die berekening van N van Fg • Correct final answer/ Korrekte finale antwoord: 0,38  OPTION 1/OPSIE 1 fsmax = µsN fsmax = µs(mg – Fsin θ)  Anyone/Enigeen 15 = µs [(5)(9,8) − 18sin33,56°] µs = 0,38 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 Fnet = 0 FN + Fy − Fg = 0 Fy = √(18)2 − (15)2  FN + Fy = Fg  = 9,95 N FN = (5)(9,8) –(18sin 33,56°) Fnet = 0 FN = 39,05 N FN + Fy − Fg = 0 FN = Fg − Fy fsmax = µsN  FN = (5)(9,8) –(9,95) 15 = µs(39,05) FN = 39,05 N µs = 0,38 fsmax = µsN  15 = µs(39,05) µs = 0,38 NOTE: If N calculation NOT shown (3⁄5) NOTA: Indien berekening van N nie getoon (3⁄5) (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 2.4 Increases ✓ Neem toe • Vertical component of F decreases ✓ • Vertikale component van F • Normal force increases ✓ neem af • • Normaalkrag neem toe µs remains constant, fsmax is directly proportional to N✓ • µs bly konstant, fsmax direk eweredig aan N OR OF Decreases  Neem af • Horizontal component of F increases  • Horisontale komponent van F • The block will move (Fnet will neem toe increase) • Die blok sal beweeg (Fnet sal • fk < fsmax  toeneem) • fk < fsmaks OR OF Remains the same  Bly dieselfde  • Explain using relevant calculation • Verduidelik deur van ‘n gepaste berekening gebruik te maak NB: In the calculation, candidates must NB: In die berekening moet die choose a µk < µs value and θ that kandidaat ‘n waarde µk < µs en θ together with the µk value will give an wat saam met µk waarde ‘n answer of 15 N for fk antwoord van 15 N gee vir fk (4) [17] QUESTION 3/VRAAG 3 3.1 Motion under the influence of weight/gravitational force only.  Beweging slegs onder die invloed van gewig/gravitasiekrag. (2 or/of 0) NOTE: If projectile defined 0/2 NOTA: Indien projektiel gedefinieer 0/2 (2) 3.2 Marking criteria/Nasienkriteria • Formula with Δt/Formule met Δt  • Correct substitution into formula/Korrekte vervanging in formule  • Correct final answer / Korrekte finale antwoord: 1,53 s  NOTE for 3.2 and 3.3: If energy principles used (1⁄3) for the final answer NOTA vir 3.2 en 3.2: Indien energiebeginsels gebuik (1⁄3) vir finale antwoord OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 B A-B: B-C: UPWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF OPWAARTS AS POSITIEF vf = vi + aΔt  vf = vi + aΔt  A C 0 = 15 + (−9,8)Δt  −15 = 0 + (−9,8)Δt  Δt = 1,53 (s)  Δt = 1,53 (s)  DOWNWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf = vi + aΔt  vf = vi + aΔt  G 0 = −15 + (9,8)Δt  15 = 0 + (9,8)Δt  Δt = 1,53 (s)  Δt = 1,53 (s)  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 A-C: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf = vi + aΔt  vf = vi + aΔt  −15 = 15 + (−9,8)Δt  15 = −15 + (9,8)Δt  Δt = 3,06 Δt = 3,06 p = 3,06 ÷ 2 p = 3,06 ÷ 2 p = 1,53 (s)  p = 1,53 (s)  (3) 3.3 POSITIVE MARKING FROM QUESTION 3.2/POSITIEWE NASIEN VANAF VRAAG 3.2 Marking criteria/Nasienkriteria • Formula with vf./Formule met vf.  • Correct substitution into formula./Korrekte vervanging in formule.  • Correct final answer/Korrekte finale antwoord: 17,97 m·s-1  • [RANGE: 17,93 – 18,33] OPTION 1/OPSIE 1 A-G: OPTION 2/OPSIE 2 A-G: UPWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF: OPWAARTS AS POSITIEF: 2 2 vf = vi + aΔt  vf = vi + 2aΔy 2 vf = (15) + 2(−9,8)(−5)  2 = (15) + (−9,8)(3,36)  vf = −17,97 = −17,93 vf = 17,97 (m·s )  -1 = 17,93 (m·s-1)  DOWNWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF: AFWAARTS AS POSITIEF: B vf = vi + aΔt  v2f = v2i + 2aΔy  = (−15) + (9,8)(3,36)  v2f = (−15)2 + 2(9,8)(5)  = 17,93 (m·s-1)  vf = 17,97 (m·s-1)  A C OPTION 3/OPSIE 3 C-G: C-G: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF v2f = v2i + 2aΔy  v2f = v2i + 2aΔy  v2f = (−15)2 + 2(−9,8)(−5)  v2f = (15)2 + 2(9,8)(5)  vf = −17,97 vf = 17,97 (m·s-1)  G vf = 17,97 (m·s-1)  OPTION 4/OPSIE 4 B-G: B-G UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF v2f = v2i + 2aΔy v2f = v2i + 2aΔy 02 = (−15)2 + 2(−9,8)Δy 02 = (15)2 + 2(9,8)Δy Δy = −11,48 m Δy = 11,48 m From B to C = −11,48 + (−5) From B to C = 11,48 + 5 = −16,48 m = 16,48 m v2f = v2i + 2aΔy  v2f = v2i + 2aΔy  v2f = (0)2 + 2(−9,8)(−16,48)  v2f = (0)2 + 2(9,8)(16,48)  vf = −17,97 vf = 17,97 (m·s-1)  vf = 17,97 (m·s-1)  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 C-G: C-G: B UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF: AFWAARTS AS POSITIEF: Δy = vi Δt + ½aΔt2 Δy = vi Δt + ½aΔt2 A C −5 = −15∆t + ½(−9,8)∆t 2 5 = 15Δt + ½(9,8)Δt2 Δt = 0,3 s Δt = 0,3 s vi + vf vi + vf ∆y = ( )∆t  ∆y = ( )∆t  2 2 −15 + vf 15 + vf G −5 = ( ) (0,3)  5=( ) (0,3)  2 2 vf = − 18,33 vf = 18,33 (m·s-1)  vf = 18,33 (m·s-1)  OPTION 6/OPSIE 6 UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE OPWAARTS AS POSITIEF: AFWAARTS AS POSITIEF: A-G A-G vi + vf vi + vf ∆y = ( )∆t  ∆y = ( )∆t  2 2 15 + vf −15 + vf −5 = ( )(3,36)  5=( ) (3,36)  2 2 vf = −17,98 vf = 17,98 (m·s-1)  vf = 17,98 (m·s-1)  OR/OF OR/OF B-G B-G vf = vi + aΔt  vf = vi + aΔt  vf = 0 + (−9,8)(3,36 − 1,53)  = 0 + (9,8)(3,36 − 1,53)  vf = −17,93 vf = 17,93 (m·s-1)  vf = 17,93 (m·s-1)  OR/OF OR/OF C-G C-G vf = vi + aΔt  vf = vi + aΔt  vf = −15 + (−9,8)(0,3)  vf = 15 + (9,8)(3,36 − 2(1,53))  vf = −17,94 vf = 17,94 (m·s-1)  vf = 17,94 (m·s-1)  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 3.4 Inelastic.  The height decreased/smaller.  The velocity with which the ball strikes the ground is not the same as the velocity with which it left the ground. / Kinetic energy is converted to heat and sound energy / Kinetic energy is NOT conserved. / The kinetic energy is not the same before and after the ball strikes the ground  ACCEPT: Kinetic energy is lost. NOTE: DO NOT ACCEPT ANY CALCULATION AS AN EXPLANATION. Onelasties. Die hoogte neem af/kleiner. Die snelheid waarmee die bal die grond tref is nie dieselfde as die snelheid waarmee dit die grond verlaat nie. / Kinetiese energie is omgeskakel na hitte en klank / Kinetiese energie is nie behoue nie./ Die kinetiese energie is nie dieselfde voor en na die bal die grond tref nie. AANVAAR: Kinetiese energie is verloor LET WEL: GEEN BEREKENINGE WORD AANVAAR AS VERDUIDELIKING NIE. (3) 3.5 POSITIVE MARKING FROM QUESTION 3.2 AND 3.3 POSITIEWE NASIEN VANAF VRAAG 3.2 EN 3.3 Criteria for graph/Kriteria vir grafiek • Straight line starting at v = 15 m·s-1 with negative final velocity or straight line starting at v = –15 m·s-1 with positive final velocity.  Reguitlyn wat begin by v = 15 m·s-1met negatiewe finale snelheid of reguitlyn wat begin v = –15 m·s-1 met positiewe finale snelheid. • Straight line cuts time axis at time calculated in Question 3.2/  Reguitlyn sny tydas by die tyd bereken in Vraag 3.2. • Correct velocity as calculated in Q 3.3 and the ball is on the ground/  Korrekte snelheid bereken in Vr 3.3 en die bal is op die grond • Velocity of the second line is less than calculated in question Q 3.3, lines are drawn parallel and ends at v = 0 /  Snelheid van tweede lyn is minder as die waarde bereken in Vr 3.3, lyne is parallel geteken en eindig by v = 0 NOTE: Accept if no contact time is shown on the graph. NOTA: Aanvaar indien geen kontaktyd op die grafiek getoon word (4) UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF: Velocity/Snelheid (m·s-1) 15 1,53 Time/Tyd (s) -17,97 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF: 17,97 Velocity/Snelheid (m·s-1) 1,53 Time/Tyd (s) -15 [15] QUESTION 4/VRAAG 4 4.1 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The product of the resultant/net force acting on an object and the time the (net) force acts on the object. Die produk van die resulterende/netto krag wat op 'n voorwerp inwerk en die tyd wat die (netto) krag op die voorwerp inwerk. OR/OF The product of the resultant/net force and the contact time. Die produk van die resulterende/netto krag en die kontak tyd (2) 4.2.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Δpf FnetΔt = Δp Gradient/Fnet = Δ(Δt) Fnet(0,01) = 0 – pi  ........... eq (1)  4,5 – 0 = Fnet(0,03) = 4,5 – pi  ........... eq (2) 0,03 – 0,01  = 225 N Fnet = 225 N to the right / 4,5 – Fnet(0,03) = 0 – (0,01)Fnet opposite to the original direction of Fnet = 225 N to the right/ the ball  opposite to the original direction of the Regs / teenoorgestelde rigting as die ball  oorspronklike rigting van die bal Regs / teenoorgestelde rigting as die oorspronklike rigting van die bal Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 FnetΔt = Δp Fnet(0,01) = 0 – 0,15vi  ........... eq (1) Fnet(0,03) = 4,5 – 0,15vi  ........... eq (2) 4,5 – Fnet(0,03) = 0 – (0,01)Fnet Fnet = 225 N to the right/ opposite to the original direction of the ball  Regs / teenoorgestelde rigting as die oorspronklike rigting van die bal (3) 4.2.2 POSITIVE MARKING FROM 4.2.1 POSITIEWE NASIEN VANAF VRAAG 4.2.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 FnetΔt = Δp FnetΔt = Δp  Any one /  Any one / FnetΔt = pf − pi FnetΔt = pf − pi enige een enige een FnetΔt = mvf − mvi FnetΔt = mvf − mvi (225)(0,03) = 4,5 − (0,15)vi  (225)(0,01) = 0 − (0,15)vi  vi = −15 m·s -1 vi = −15 m·s-1 vi = 15 m·s -1 vi = 15 m·s-1 OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 pf = FnetΔt + pi Fnet = ma 4,5 = 225(0,03) + pi  225 = 0,15a  pi = -2,25 a = 1500 m∙s-2 pi = mvi  pf = mvf −2,25 = 0,15vi  4,5 = 0,15vf vi = −15 m·s-1 vf = 30 m∙s-1 vi = 15 m·s-1  ∆v a=  ∆t 30 − vi 1500 =  0,03 − 0 vi = 15 m∙s-1  (4) 4.3 A B pf (kg·m·s−1)  (2 or 0) Δt (s) NOTE: If labels A and B not indicated (0⁄2) Award full marks for solid lines with labels. NOTA: Indien lyne nie benoem A en B (0⁄2) Volpunte toegeken vir soliede benoemde lyne (2) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The net/total work done (on an object) is equal to the change in the object's kinetic energy.  Die netto/totale arbeid wat (op 'n voorwerp) verrig is, is gelyk aan die verandering in die voorwerp se kinetiese energie. OR/OF 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 ‘n voorwerp) deur die resultante/netto krag is gelyk aan die verandering in die voorwerp se kinetiese energie. (2) 05.2 Wnet = ΔEk WFA = ΔEk 1 1 FA∆xcosθ = 2mvf2 − 2mvi2  Any one / Enige een Wnc = ΔEp + ΔEk (12)(7)cos 0° = Ekf – 0  Ekf = 84 J  (3) 5.3 ACCEPT/AANVAAR: f f N N w w Accepted labels/Aanvaarde benoemings w Fw / Fg / mg /gravitational force / 29,4 N/ gravitasiekrag / weight / gewig f Ff / fk / (kinetic) Friction / 21 N (kinetiese) wrywing / Fw N FN / Normal (force) / Normaal (krag) Notes/Aantekeninge: • Mark awarded for label and arrow./Punt toegeken vir benoeming en pyltjie. • Do not penalise for length of arrows since drawing is not to scale./Moenie vir die lengte van die pyltjies penaliseer nie aangesien die tekening nie volgens skaal is nie. • Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks 2⁄3 • If everything is correct, but no arrows/Indien alles korrek is, maar geen pyltjies: Max/Maks 2⁄3 • If force(s) do not make contact with the dot /Indien krag(te) nie met die kolletjie kontak maak nie: Max/Maks 2⁄3 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 55.4 POSITIVE MARKING FROM 5.2/POSITIEWE NASIEN VANAF VRAAG 5.2 MARKING CRITERIA/NASIENKRITERIA: • Correct substitution to calculate W Fg/Korrekte vervanging om WFg te bereken  • Correct substitution to calculate W f/Korrekte vervanging om Wf te bereken  • Correct substitution to calculate ΔEk/Korrekte vervanging om ∆Ek te bereken  • Correct final answer/Korrekte finale antwoord  • Conclusion/Gevolgtrekking  NOTE: ACCEPT for all options: cos70° of sin20° NOTA: AANVAAR vir alle opsies: cos70° of sin20° OPTION 1/OPSIE 1 Wnet = ΔEk WFg + W fk = Ekf − Eki Fg//∆xcosθ + fk ∆xcosθ = Ekf − Eki mgsinθ∆xcosθ + fk∆xcosθ = Ekf − Eki (3)(9,8)sin20°∆xcos0°+ 21∆xcos180°= 0 − 84 ∆x = 7,68 m  7,68 m > 6,8 m The crate will pass point C.  OPTION 2/OPSIE 2 W net = ΔEk W Fg + W fk = ΔEk Fg//∆xcosθ + fk ∆xcosθ = ΔEk mgsinθ∆xcosθ + fk∆xcosθ = Ekf − Eki (3)(9,8)sin 20°(6,8)cos 0°+ (21)(6,8)cos180°= Ekf − 84 Ekf = 9,58 J  OR vf = 2,52 m·s-1 9,58 J > 0 J OR 2,52 m·s-1 > 0 m·s-1 The crate will pass point C.  OPTION 3/OPSIE 3 h sin 20° = 6,8 h = 2,326 m W nc = ΔEk + ΔEp fk ∆xcosθ = + ΔEk + ΔEp (21)(6,8)cos180°= (Ekf – 84)  – (3)(9,8)(2,326)  Ekf = 9,58 J OR vf = 2,52 m·s-1 9,58 J > 0 J OR 2,52 m·s-1 > 0 m·s-1 The crate will pass point C.  OPTION 4/OPSIE 4 Relative to C: Ep at B = mgh = (3)(9,8)(6,8)sin 20° = 68,38 J W nc = ΔEk + ΔEp (21)(6,8)cos180° = (Ekf – 84)  + (0 – 68,38)  Ekf = 9,58 J  The crate will pass point C.  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 Wnet = ΔEk Fnet∆xcosθ = Ekf − Eki [(3)(9,8)sin20° − 21]✓∆xcos0° = 0 − 84 ∆x = 7,68 m  7,68 m > 6,8 m The crate will pass point C.  OR/OF [3(9,8)sin20° - 21] ✓(6,8)cos0° ✓ = Ekf − 84 ✓ Ekf = 9,58 J  The crate will pass point C.  (5) [13] QUESTION 6/VRAAG 6 6.1 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. 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 'n luisteraar omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium waarin die klank voortgeplant word, het. OR/OF Is an (apparent) change in (observed/detected) frequency (pitch) as a result of the relative motion between a sound source and an observer. Die (skynbare) verandering in (waargenome) frekwensie (toonhoogte) as gevolg van relatiewe beweging tussen die klankbron en ‘n luisteraar (2) 6.2.1 Velocity/speed/vs of the ambulance/source  Snelheid van die ambulans/klankbron (1) 6.2.2 Frequency/wavelength of the sound produced by the ambulance siren / fs / speed of sound/ density/temperature of air/detector/listener is stationary  Frekwensie/golflengte van die klank voortgebring deur die sirene van die ambulans / fs / spoed van klank / digtheid/temperatuur van lug / detektor/luisteraar staan stil (1) 6.3 As the velocity of the ambulance increases, the detected frequency decreases.  Soos die snelheid van die ambulans toeneem, sal die waargenome frekwensie afneem. NOTE: If inversely proportional indicated NOTA: Indien omgekeerd eweredig (0⁄2) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 6.4 Ambulance approaching/Ambulans beweeg nader: V ± VL V fL = fS OR/OF fL = fS V ± VS V − VS  v 1298 = fS v − 25 vfS = 1298(v – 25) .......... eq. (1)  Either/ Enige Ambulance moving away/Ambulans beweeg weg: V ± VL V fL = fS OR/OF fL = fS V ± VS V + VS  Either/ Enige  v 1115 = fS v + 25 vfS = 1115(v + 25) ........ eq. (2) (1) = (2) OR/OF (1) ÷ (2) 1298(v – 25) = 1115(v + 25)  OR/OF 1,164(v – 25) = v + 25 v = 329,64 m∙s-1  (6) [12] QUESTION 7/VRAAG 7 7.1 A region of space in which an electric charge experiences a force.  'n Gebied in die ruimte waarin 'n elektriese lading 'n krag ondervind (2 or 0) (2) 7.2 Marking criteria/Nasienkriteria: Direction / Rigting  Field lines radially outward/Veldlyne radiaal uitwaarts  + Two or more field lines inside sphere / not touching sphere / cross each other: deduct 1 mark Twee of meer veldlyne binne sfeer / raak nie aan sfeer / kruis: trek 1 punt af If more than one charge drawn/Indien meer as een lading geteken: 0⁄2 (2) 7.3 kQ E=  r2 (9×109 )(3×10–7 ) 1,08×106  =  r2 r = 0,05 m (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 7.4 NOTE: Ignore if negative charges are substituted. NOTA: Ignoreer indien negatiewe ladings invervang is OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 kQXQY kQ kQ FXY =  Enet = + r2 r2 r2 9 -7 9 -7 (9 × 109 )(3 × 10-7 )(5 × 10-7 ) (9 × 10 )(5 × 10 ) (9 ×10 )(4 ×10 ) = 2  = 2 − 2  (0,05) (0,05) (0,04) = 0,54 N Enet = 450 000 N·C-1 kQXQZ Fnet = qE FXZ = r2 Fnet = (3 x 10-7)(450 000)  (9 × 109 )(3 × 10-7 )(4 × 10-7 ) Fnet = 0,14 N = 2  (0,04) = 0,68 N No/Nee  Fnet(E) >/≠ Fnet  Fnet(E) = FXZ + FXY = 0,68 − 0,54  = 0,14 N No/Nee  Fnet(E) >/≠ Fnet  OPTION 3/OPSIE 3 kQXQY kQXQZ Fnet(E) = r2 − r2  (9 x 10 )( 3 x10-7 )(5 x 10-7 ) 9 (9 x 109 )(3 x 10-7 )( 4 x 10-7 ) = 2 − 2  (0,05) (0,04) = 0,14 N 0,0427 = 0,14 – f  f = 0,0923 N No/Nee  (6) 7.5 Decrease/Neem af  (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 Maximum/ Total energy provided / work done by a battery/ cell per unit/ coulomb of charge passing through it.  Die maksimum/ totale energie gelewer/ arbeid verrig deur battery per eenheids/ coulomb-lading wat daardeur vloei. (2 or 0) (2) 8.2 V R=  I V 0,1 =  120 V = 12 V  (3) 8.3.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 P = I2R  V2 V2 15 = I 2(12)  P= P= R R I = 1,12 A  V2 V2 15 = 15 = 12 12 V = 13,42 V V = 13,42 V P = VI ✓ V = IR ✓ 15 = (13,42)I ✓ 13,42 = I(12) ✓ I = 1,12 A ✓ I = 1,12 A ✓ (3) 8.3.2 POSITIVE MARKING FROM QUESTION 8.3.1 POSITIEWE NASIEN VANAF VRAAG 8.3.1 A1 = (1,12)(2) = 2,24 A  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 8.4 POSITIVE MARKING FROM QUESTION 8.2 and 8.3.2 POSITIEWE NASIEN VANAF VRAAG 8.2 en 8.3.2 [RANGE: 13,42 V – 13,47 V] OPTION 1/OPSIE 1 1 1 1 R1 R2 = + Rp = Rp R1 R2 R1 + R2 1 1 1 (12)(12) = + ✓ Rp = ✓ Rp 12 12 12 + 12 Rp = 6 Ω Rp = 6 Ω When S1 closed and S2 open/Wanneer S1 gesluit en S2 oop is: Ԑ1 = I(R + r)✓ = 2,24(6 + r) ✓ When S1 open and S2 closed/Wanneer S1 oop en S2 gesluit is: Ԑ2 = I(R + r) = 120(0,1 + r) ✓ Ԑ1 = Ԑ2 (equating) 2,24(6 + r) = 120(0,1 + r) ✓ r = 0,012 Ω Ԑ = I(R + r) OR/OF Ԑ = IR + Ir = 2,24(6 + 0,012) = 120(0,1+ 0,012) = 13,46 V ✓ = 13,47 V ✓ OPTION 2/OPSIE 2 Ԑ1 = IR + Ir ✓ = 12 + 120r ✓ V2 OR/OF Vp = IR OR/OF VP = IR P= = 2,24(6) ✓ = 1,12(12) ✓ R V2 = 13,44 V = 13,44 V 15 = ✓ 12 V = 13,42 V Ԑ2 = Vext + Ir = 13,42 + 2,24r ✓ Ԑ1 = Ԑ2 (equating) 12 + 120r = 13,42 + 2,24r ✓ r = 0,012 Ω Ԑ1 = IR + Ir Ԑ2 = IR + Ir = 12 + (120)(0,012) = 13,42 + (2,24)(0,012) = 13,47 V ✓ = 13,45 V ✓ (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 8.5 Decrease • Total/External resistance decreases  • Total current increases.  • Vinternal increases • Voltage across the light bulbs decreases / Vexternal decreases.  NOTE: Calculation given as explanation 1⁄5 Neem af • Totale/Eksterne weerstand neem af • Totale stroom neem toe • Vintern neem toe • Potensiaalverskil oor die gloeilamp neem af / Vekstern neem af NOTA: Berekening gegee as verduideliking 1⁄5 ACCEPT/AANVAAR: NOTE/NOTA: max: 3⁄5 Mark for INCREASES will ONLY be awarded if explanation is correct Punt vir TOENEEM sal SLEGS toegeken word indien verduideliking korrek is Refer to the explanation first to check if the candidate is referring to Q8.2 Verwys na verduideliking eerste om te bepaal of die kandidaat verwys Vr8.2 Increases ✓ • Initial reading of A1 = 0 A when switch S1 is open ✓ • Now both switch closed the current is flowing through the circuit. ✓ Toeneem • Aanvanklike lesing op A1 = 0 A wanneer skakelaar S1 oop is • Nou is beide skakelaars gesluit en stroom vloei deur die stroombaan (5) [20] QUESTION 9/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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The AC potential difference /voltage that produces/dissipates the same amount of energy as an equivalent DC potential difference/voltage.  Die WS-potensiaalverskil wat dieselfde hoeveelheid energie verbruik/oordra as 'n ekwivalente GS-potensiaalverskil. NOTE: If DC written first/ energy or heating effect is omitted/ alternating current is defined: 0⁄2 LET WEL: Indien GS eerste geskryf/ energie of verhittingseffek uitgelaat is/ wisselstroom gedefinieer : 0⁄2 (2) 9.2 N/North/Noord  (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 9.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Vmax Vmax Vrms = Vrms = √2 √2 311,11 311,11 = ✓ = ✓ √2 √2 = 219,99 V = 219,99 V V2rms V2rms Pave =  Pave =  R R 219,99 2 219,99 2 =  =  60 60 = 806,59 W = 806,59 W = 0,80659 kW = 0,80659 kW E = PΔt cost = PΔt × tariff = 0,80659(1,5) = 0,80659(1,5)(3,33)  = 1,21 kWh ✓ for both/beide = R4,03  cost = E × tariff = 1,21(3,33) = R4,03  (5) OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 Pave = VrmsIrms  Vmax 311,11 220 Vrms = √2 =( )  ( 60 )  311,11 √2 = 806,67 W =  √2 = 0,8067 kW = 219,99 V cost = PΔt × tariff V2 = 0,8067(1,5)(3,33)  W = R ∆t  Cost = E x tariff = R4,03  (219,99 )2 = (1,5)(10-3 )  = (1,21)(3,33)  60 = 1,21 kWh = R4,03  9.4 Induced emf/ B A 0 Geïnduseerde emk (V) 4 Time/Tyd (s) Marking criteria/Nasienkriteria: Amplitude of graph B bigger than that of graph A  Amplitude van grafiek B groter as dié van A Period of B is half of A/Periode van B helfte van A  Correct shape and one cycle(Cosine graph)/  Vorm en een siklus(cos-grafiek) (3) 9.5 By replacing slip rings with split ring commutator / split ring / commutator. Vervang sleepringe met splitring kommutator / splitring / kommutator (1) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 20 Physical Sciences P1/Fisiese Wetenskappe V1 20 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1.1 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 sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Minimum energy needed for the electrons 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 threshold frequency defined NOTA: Indien drumpelfrekwensie gedefinieer: 0⁄2 (2) 10.1.2 1,045 × 1015 Hz  (1) 10.1.3 Greater than  • The maximum kinetic energy of electrons from sodium is smaller. • E is constant / same frequency used / same light used (EK(max) is smaller; W0 is larger). Groter as • Die maksimum kinetiese energie van die elektrone van natrium sal kleiner wees. • E is konstant/ dieselfde frekwensie gebruik/dieselfde lig gebruik is (sal Ek(maks) kleiner wees; W0 groter) (3) 10.1.4 No  • The intensity of light does not have any effect on the ejection of electrons for metal M OR • The frequency of light has not changed / Energy of a photon has not changed. Nee • Die intensiteit van die lig het geen invloed op die vrystelling van elektrone vir metaal M nie OF • Die frekwensie van lig het nie verander nie/ Energie van fotone het nie verander (2) 10.2.1 OPTION 1/OPSIE 1 Accept E3 – E1 OR E1 – E3 in all options E = hf  Aanvaar: E3 – E1 OF E1 – E3 in alle 5,7 x 10-19 – 1,3 x 10-19 = (6,63 × 10–34) f  opsies f = 6,64 × 1014 Hz  OPTION 2/OPSIE 2 E = hf ✓ E = E1 – E3 4,4 x 10-19 = (6,63 x 10-34)f ✓ = 1,3 x 10-19 – 5,7 x 10-19 ✓ f = 6,64 x 1014 Hz ✓ = – 4,4 x 10-19 f must be positive/ f moet positief wees (4) 10.2.2 No  The energy does not correspond to any transition in this atom  Nee Die energie stem nie ooreen met enige oorgang in hierdie atoom nie (2) [14] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou ## Related documents - [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) — Physical Sciences · QP · 2025 · Afr - [Gr 12 Physical Sciences P1 (English and Afrikaans) June 2024 Possible Answers](https://www.hlayiso.com/library/gr-12-physical-sciences-p1-english-and-afrikaans-june-2024-possible-answers/f78ca708-f7f0-4bc5-8ff9-36594b1b1052) — Physical Sciences · Memo · 2024 · Afr - [PHYS SC P1 MEMO GR12 JUNE2021 Afrik English](https://www.hlayiso.com/library/phys-sc-p1-memo-gr12-june2021-afrik-english/51075955-8380-4def-a361-3f9b02ab715c) — Physical Sciences · Memo · 2021 · Afr - [Physical Sciences P1 Nov 2019 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2019-memo-afr-eng/6250ed76-71a4-4f9b-bfe7-95a1f244df41) — Physical Sciences · Memo · 2019 · Eng - [Physical Sciences P1 Nov 2015 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2015-memo-afr-eng/140f25b7-f3c1-406a-a99d-6549fa7ea9e0) — Physical Sciences · Memo · 2015 · Eng - [Physical Sciences P1 Nov 2014 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2014-memo-afr-eng/44924678-c33b-42e1-8937-fcfda16ea7b3) — Physical Sciences · Memo · 2014 · Eng ## More from Grade 12 Physical Sciences - [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 - [Physical Sciences Physics Grade 12 NSC QP P1 May June 2025 Mpumalanga](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-qp-p1-may-june-2025-mpumalanga/d9c5f1c3-bd6c-468e-a50a-f6b40782c695) — 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. 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