# Physical Sciences Grade 11 P1 2025 Memo > Grade 11 Physical Sciences Paper 1 memorandum from the Eastern Cape November Exam 2025, 12 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sc-p1-memo-nov-2025-e-a-watermark/ab3b23c2-3309-44f6-8a96-b01d7a06aa02 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 11 - Document type: Memorandum - Year: 2025 - Exam period: Eastern Cape November Exam - Paper: 1 - Curriculum: CAPS - Pages: 12 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Physical Sciences](https://www.hlayiso.com/library/grade-11/physical-sciences) › [Physical Sciences Grade 11 P1 2025 Memo](https://www.hlayiso.com/library/physical-sc-p1-memo-nov-2025-e-a-watermark/ab3b23c2-3309-44f6-8a96-b01d7a06aa02) ## About this document Physical Sciences · Grade 11 · Eastern Cape November Exam · 2025. Memorandum, 12 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 11 NOVEMBER 2025 PHYSICAL SCIENCES P1/ FISIESE WETENSKAPPE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 12 pages./ Hierdie nasienriglyn bestaan uit 12 bladsye. ### Page 2 2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2025) QUESTION/VRAAG 1 1.1 A  (2) 1.2 C  (2) 1.3 B  (2) 1.4 C  (2) 1.5 D  (2) 1.6 D  (2) 1.7 B  (2) 1.8 C  (2) 1.9 C  (2) 1.10 D  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 (EC/NOVEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 3 QUESTION/VRAAG 2 2.1 A physical quantity having both magnitude and direction.  ʼn Fisiese hoeveelheid met beide grootte en rigting. (2) 2.2 2.2.1 For the x-components/ For the y-components / Vir die x-komponente Vir die y-komponente F1 F1 Fx = 700 cos60o  Fy = 700 sin60o  Fx = 700 sin30o Fy = 700 cos30o = 350 N right / regs = 606,22 N upward/opwaarts F2 F2 Fx = 800 cos40o  Fy = 800 sin40o  Fx = 800 sin50o Fy = 800 cos50o = 612,84 N right / regs = 514,23 N downward/afwaarts F3 = Fx = 650 N left / links F3 = Fy = 0 N Fx = 350 + 612,84 – 650 = 312,84 N right Fy = 606,22 – 514,23 = 91,99 N /regs  upward/ opwaarts  Fnet2 = Fy2 + Fx2 Fnet2 = (312,84)2+ (91,99)2 Fnet = 326,08 N  91,99 N 91,99 tan θ =  312,84 θ = 16,39° θ Direction / rigting = 73,7o  312,84 N (9) 2.2.2 W = mg W = 25 x 9,8  W = 245 N  (2) 2.3 The resultant force is not equal to zero.  Die resulterende krag is nie gelyk aan nul nie. (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2025) QUESTION/VRAAG 3 3.1 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.  Wanneer ʼn resulterende/netto krag op ʼn voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel teen ʼn versnelling direk eweredig aan die krag en omgekeerd eweredig aan die massa (van die voorwerp). OR/OF This acceleration is directly proportional to the net force  and inversely proportional to the mass.  Die versnelling is direk eweredig aan die netto krag en omgekeerd eweredig aan die massa. (2) 3.2 3.2.1 Truck / Vragmotor: N = mg = (4 500)(9,8) = 44 100 N fk = µkN  8 820 = µ (44 100)  µk = 0,2  (3) 3.2.2 POSITIVE MARKING FROM QUESTION 3.3.1 / POSITIEWE NASIEN VANAF VRAAG 3.3.1 Bakkie: N = mg = (1 250)(9,8) = 12 250 N  fk = µkN fk = (0,2)(12 250)  fk = 2 450 N  (3) 3.2.3 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 Fnet = F - fk – T  Fnet = T - fk  0 = 11 270 + (- 8 820) - T  0 = T – 2 450  T = 2 450 N  T = 2 450 N  (3) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 (EC/NOVEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 5 QUESTION/VRAAG 4 4.1 The force that opposes the motion of a moving object relative to a surface.  Die krag wat die beweging van ʼn bewegende voorwerp relatief tot ʼn oppervlak teenwerk. (2) 4.2 fk Ff, frictional force / Wrywingskrag T FT, tension, tension force / Spanning / spanningskrag N FN, normal force / normaalkrag F FA, Fapplied /toegepas, FApp W Fg, gravitational force / gravitasiekrag NOTE / LET WEL: If arrows do not touch the dot deduct 1 mark./ As pyle nie die kolletjie raak nie, trek 1 punt af. Any additional force(s), deduct 1 mark./ Enige addisionele krag(te), trek 1 punt af. (5) 4.3 4.3.1 N = Fg – Fsin𝜃𝜃  N = 10 x 9,8 – 50 sin30°  N = 73 N  (3) 4.3.2 POSITIVE MARKING FROM QUESTION 4.3.1 / POSITIEWE NASIEN VANAF VRAAG 4.3.1 Fnet = ma Fnet = T - fk Fnet = Fcosθ - T - fk Any one / Enige een  Fnet = Fx - T - fk Any one / Enige een (10a or/of 5a)  T – 9  = 5a ……….. 1 [50cos30o - T – (0,18 x 73)  ]  = 10a ……….. 2 a = 1,41 m.s–2  (7) 4.4 Increases.  The magnitude of the vertical component of the applied force will decrease  (as the weight will remain constant) and hence an increase in the magnitude of the normal force.  Toeneem. Die grootte van die vertikale komponent van die toegepaste krag sal afneem (soos die gewig konstant bly) en dus ʼn toename in die grootte van die normaalkrag. (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2025) 4.5 (According to Newton First law, the car will be moving at a constant velocity). When breaks are applied the seat of the child will continue in its state of constant velocity  (in the direction of the windscreen which can be fatal), when the child is secured the belt will exert a net force in opposite direction  that will accelerate the child back to the safety position in the child’s seat.  (Volgens Newton se eerste wet sal die motor teen ʼn konstante snelheid beweeg). Wanneer remme toegepas word, sal die kind se sitplek in sy toestand van konstante snelheid bly (in die rigting van die voorruit wat noodlottig kan wees), wanneer die kind vasgemaak is, sal die gordel ʼn netto krag in die teenoorgestelde rigting uitoefen wat die kind terug na die veiligheidsposisie in die kinderstoel sal versnel. (3) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 (EC/NOVEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 7 QUESTION/VRAAG 5 5.1 A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force acts on it.  ʼn Liggaam sal in sy toestand van rus of beweging teen konstante snelheid behou, tensy ʼn nie-nul resulterende/netto krag daarop inwerk. (2) 5.2 T w (2) 5.3 5.3.1 fk = μkN fk = μkFgcosθ Any one / Enige een  fk = μkmgcosθ fk = (0,25)(8)(9,8)cos20o  = 18,42 N  (3) 5.3.2 Fnet = ma Fnet = Fg – T Fnet = T – F – fk – Fg// Any one / Enige een  0 = T – 6 x 9,8  T = 58,8 N 0 = [58,8 – F – 18,42 – 8 x 9,8sin20o ]  F = 13,57 N  (5) 5.4 vf2 = vi2 + 2aΔy  = 02 + 2(3,6)(2)  = 3,79 m∙s-1  (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2025) QUESTION/VRAAG 6 6.1 Newton’s 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, oefen voorwerp B GELYKTIGDIG ʼn teenoorgestelde krag van gelyke grootte op voorwerp A uit. (3) 6.2 F = Gm1m2  r2 250  = (6,67 x 10-11)(5,98 x 1024)(500)  (d)2 d2 = 7,98 x 1014 d = 2,82 x 107 m  distance above earth surface / afstand bo die aarde se oppervlakte = 2,82 x 107 – 6,38 x 106 = 2,182 x 107 m  (5) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 (EC/NOVEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 9 QUESTION/VRAAG 7 7.1. The electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges  and inversely proportional to the square of the distance between them.  Die elektrostatiese aantrekkings- of afstotingskrag tussen twee ladings is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. (2) 7.2 F = kQ1Q2  F2 = (9 x 109)(100 x 10-6)2  d2 (0,03)2 F2 = 100 000 N right / regs F1 = (9 x 109)(100 x 10-6)2  (0,06)2 Fnet = FA + FB F1 = 25 000 N right / regs Fnet = 25 000 + 100 000  Fnet = 125 000 N  right / regs  (6) 7.3 7.3.1 Electric field at a point is the electrostatic force experienced per unit positive charge placed at that point.  Die elektrieseveld by ʼn punt is die elektrostatiese krag wat per eenheid positiewe lading wat by daardie punt geplaas word, ervaar word. (2) 7.3.2 CRITERIA FOR MARKING/ NASIENKRITERIA Correct field direction / Korrekte veldrigting  Lines starting from sphere and not crossing each  other / Lyne begin vanaf die sfeer en kruis nie mekaar nie (2) 7.3.3 kQ E=  r2 (9 × 109 )(2 × 10−6 ) E=  (0,1) 2 E = 1,8 x 106 N·C-1 left / links  (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2025) QUESTION/VRAAG 8 8.1 North pole / Noordpool  (2) 8.2 Towards Q / Na Q  (2) 8.3 8.3.1 The magnitude of the induced emf across the end of a conductor is directly proportional to the rate of change in the magnetic flux linkage with the conductor.  Die grootte van die geïnduseerde emk oor die punt van ʼn geleier is direk eweredig aan die tempo van verandering in die magnetiesevloed-koppeling in die geleier. (2) 8.3.2 Pull the solenoid out of the magnetic field faster. Increase the number of turns on the coil. Increase the area of the coil. Use a stronger magnetic field. (Any TWO)  Trek die solenoïde vinniger uit die magneetveld. Verhoog die aantal windings op die spoel. Vergroot die oppervlakte van die spoel. Gebruik ʼn sterker magnetiese veld. (Enige TWEE) (2) 8.3.3 Φ = BA cosθ  = 0,6 (0,08) cos 0° = 0,048 Wb  (3) 8.3.4 Ɛ = -N∆Φ  ∆t Ɛ = - 350 (0 - 0,048 )  0,25 Ɛ = 67,2 V  (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 (EC/NOVEMBER 2025) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 11 QUESTION/VRAAG 9 9.1. Ohmic conductor.  The graph is a straight line from the origin.  The current is directly proportional to the potential difference across the conductor.  Ohmiese geleier. Die grafiek is ʼn reguit lyn vanaf die oorsprong. Die stroom is direk eweredig aan die potensiaalverskil oor die geleier. (3) 9.2 9.2.1 Current / Stroom  (1) 9.2.2 Temperature / Temperatuur  Length of conductor / Lengte van die geleier  any TWO/ Thickness of conductor / Dikte van geleier enige TWEE Accept Resistance / Aanvaar Weerstand (2) 9.3 9.3.1 Gradient / gradiënt = △𝐼𝐼 △𝑉𝑉  0,6−0,2 Gradient / gradiënt = 2,5−0,85  Gradient / gradiënt = 0,24  (3) 9.3.2 POSITIVE MARKING FROM QUESTION 9.3.1 / POSITIEWE NASIEN VANAF VRAAG 9.3.1 1 R = 0,24  R = 4,2 Ω P = I2R  150 = 62R  R = 4,2 Ω  The conductor will be suitable.  Resistance of conductor is equal to the resistance required.  Die geleier sal geskik wees. Die weerstand van die geleier is gelyk aan die weerstand wat benodig word. (6) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 12 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2025) QUESTION/VRAAG 10 10.1 Emf / Emk  OR/OF Electromotive force/Elektromotoriese krag (1) 10.2 1 1 1 = +  R p R1 R 2 𝟏𝟏 𝟏𝟏 𝟏𝟏 = +  𝐑𝐑𝐑𝐑 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏+𝟏𝟏𝟏𝟏 Rp = 7,37 Ω Rext = 7,37 + 5  = 12,37 Ω emf = I(R + r)  12 = I(12,37 + 0,6)  I = 0,93 A  (6) 10.3 OPTION/OPSIE 1 OPTION/OPSIE 2 V = IR  V = IR  V = (0,93)(12,37)  V = (0,93)(0,6)  V = 11,50 V  Vint = 0,558 V V = 12 – 0,582 Range / Gebied (11,44 – 11,50) V = 11,44 V  (3) 10.4 Increase / Verhoog  (1) 10.5 Total resistance increases,  the total current decreases  so the lost volts decreases. Vext = Ɛ ‒ Vint  Die totale weerstand neem toe, die totale stroom neem af dus sal die verlore volt / spanning afneem. 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