# Physical Sciences Grade 12 Free State Prelim 2024 QP (Afr) > Grade 12 Physical Sciences question paper from the Free State Prelim Exam 2024 (Afrikaans), 19 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/fsphysicalsciencesp1prepsept2024mgeng-afr/b6949bed-03b5-40bf-b617-1395d1469896 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Question paper - Year: 2024 - Exam period: Free State Prelim Exam - Curriculum: CAPS - Pages: 19 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 Free State Prelim 2024 QP (Afr)](https://www.hlayiso.com/library/fsphysicalsciencesp1prepsept2024mgeng-afr/b6949bed-03b5-40bf-b617-1395d1469896) ## About this document Physical Sciences · Grade 12 · Free State Prelim Exam · 2024. Question paper, 19 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 PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN GRADE/GRAAD 12 PHYSICAL SCIENCES P1: (PHYSICS) FISIESE WETENSKAPPE V1: (FISIKA) SEPTEMBER 2024 MARKS/PUNTE: 150 MARKING GUIDELINE NASIENRIGLYNE This marking guideline consists of 19 pages. Hierdie nasienriglyne bestaan uit 19 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne QUESTION/VRAAG 1 1.1 B  1.2 D  1.3 A  1.4 C  1.5 A  1.6 C  1.7 B  1.8 C  1.9 A  1.10 D  [10 x 2 = 20] QUESTION/VRAAG 2 2.1.1 Marking criteria/Nasienkriteria: If any of the underlined keywords/phrases in the correct context is omitted, deduct 1 mark. Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks weggelaat word, trek 1 punt af. A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force acts on it.  'n Liggaam sal in sy toestand van rus of beweging teen konstante snelheid bly, tensy 'n nie-nul resulterende/nettokrag daarop inwerk. (2) 2.1.2 TA = 12,2 sin 40  = 7,84 N  (2) 2.1.3 POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 2.1.2 Fg = TA = 7,84 N TA Fg = mg  Fg 7,84 = m (9,8)  m = 0,8 kg  TB (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne 2.2.1 Marking criteria/Nasienkriteria: If any of the underlined keywords/phrases in the correct context is omitted, deduct 1 mark. Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks weggelaat word, trek 1 punt af. The force acting parallel to a surface and opposes the tendency of motion of a stationary object relative to the surface. Die krag wat parallel met 'n oppervlak inwerk en die neiging van beweging van 'n stilstaande voorwerp relatief tot die oppervlak. (2) 2.2.2 N fs Fg Acceptable labels/Aanvaarde byskrifte N FN / Fnormal/normaal / Normal force/Normaalkrag Fg Weight/gewig / w / mg / Fgravitation/gravitasie fs static friction/statiese wrywing Notes/Nota: • Mark is awarded for label and arrow. • Do not penalise for the lengths of vectors. • If arrows do not touch the dot: Max (2⁄3) • Any additional force(s): Max (2⁄3) • No arrows drawn with correct labels: Max (2⁄3) • Punt word toegeken vir byskrift en pyl. • Moenie penaliseer vir die lengtes van vektore nie. • As pyle nie aan die punt raak nie: Maks (2⁄3) • Enige bykomende krag(te): Maks (2⁄3) • Geen pyle met korrekte byskrifte nie: Maks (2⁄3) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne 2.2.3 Marking criteria/Nasienkriteria: • Any correct formula for/Enige korrekte formule vir Fnet  • Correct substitution for/Korrekte substitusie vir Fg//  • Correct substitution for/Korrekte substitusie vir fs  • Correct final answer./Korrekte finale antwoord.  OPTION/OPSIE 1 + DOWNWARDS/OPWAARTS Fnet = ma Fg// − fs = ma any one/enige een (mgsinθ) − (μS ∙ mgcosθ) = 0 [m(9,8)sin30°] − [(μS )(m)(9,8)(cos30°)] = 0 4,9 m − 8,49 m(μS ) = 0 μS = 0,58 OPTION/OPSIE 2 + UPWARDS/OPWAARTS Fnet = ma −Fg// + fs = ma any one/enige een −(mgsinθ) + (μS ∙ mgcosθ) = 0 [−m(9,8)sin30°] + [(μS )(m)(9,8)(cos30°)] = 0 −4,9 m + 8,49 m(μS ) = 0 μS = 0,58 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne 2.2.4 Marking criteria/Nasienkriteria: • Correct formula for calculating a/Korrekte formule vir berekening van a • Correct substitution for the whole substitution  • Correct formula for calculating/Korrekte formule vir berekening van Fnet  • Correct substitution for/Korrekte substitusie vir Fg// and/en fk  • Correct final answer./Korrekte finale antwoord.  OPTION/OPSIE 1 + DOWNWARDS/AFWAARTS ∆y = vi ∆t + ½a∆t 2  2,5 = (0)(4) + ½(a)(42 )  a = 0,3125 m ∙ s −2 Fnet = ma Fg// − fk = ma any one/ (mgsinθ) − (μk ∙ mgcosθ) = ma enige een [m(9,8)sin30°] − [(μk) (m)(9,8)(cos30°)] = m(0,3125)  4,9 m − 8,49 m(μ𝑘 ) = 0,3125m μ𝑘 = 0,54 OPTION/OPSIE 2 + UPWARDS/OPWAARTS ∆y = vi ∆t + ½a∆t 2  −2,5 = (0)(4) + ½(−a)(42 )  a = 0,3125 m ∙ s −2 Fnet = ma −Fg// + f𝑘 = ma  any one/ ( ) − mgsinθ + (μ𝑘 ∙ mgcosθ) = ma enige een [−m(9,8)sin30°] + [(μ𝑘 )(m)(9,8)(cos30°)] = m(0,3125)  −4,9 m + 8,49 m(μ𝑘 ) = 0,3125m μ𝑘 = 0,54 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne 2.2.5 fsmax f (N) fs fk FA (N) Marking criteria/Nasienkriteria Showing fs, fs(max) and fk/Toon fs, fs(maks) en fk  Correct labels of f and FA/Korrekte byskrifte f and FA  • Correct shape./Korrekte vorm. • fs straight line from origin./ fs reguit lyn vanaf oorspring.  • fk lower and parallel to x-axis./fk laer en parallel aan x-as. (3) [24] QUESTION/VRAAG 3 3.1 Marking criteria/Nasienkriteria: If any of the underlined keywords/phrases in the correct context is omitted, deduct 1 mark. Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks weggelaat word, trek 1 punt af. An object which has been given an initial velocity and then it moves under the influence of the gravitational force only.  'n Voorwerp wat 'n beginsnelheid gegee is en dan beweeg dit slegs onder die invloed van die gravitasiekrag. (2) 3.2 Marking criteria/Nasienkriteria: • Any correct answer leading to the distance at which the ball strikes the glass roof.  Enige korrekte antwoord wat lei tot die afstand waarteen die bal die glasdak tref. • Correct substitution of/Korrekte substitusie vir vf.  • Correct substitution into/ Korrekte substitusie in vi + 2a∆y.  • Correct final answer./Korrekte finale antwoord.  OPTION/OPSIE 1 OPTION/OPSIE 2 + DOWNWARDS/AFWAARTS + UPWARDS/OPWAARTS vf2 = vi2 + 2a∆y  vf2 = vi2 + 2a∆y (7,84𝟐 ) = 02 + 2(9,8)∆y (7,84𝟐 ) = 02 + 2(−9,8)∆y ∆y = 3,14 m ∆y = −3,14 m ∆y = 3,14 m Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne OPTION/OPSIE 3 OPTION/OPSIE 4 + DOWNWARDS/AFWAARTS + UPWARDS/OPWAARTS ∆y = vi ∆t + ½a∆t 2  ∆y = vi ∆t + ½a∆t 2  ∆y = (0)(0,8) + ½(9,8)(0,8)2  ∆y = (0)(0,8) + ½(−9,8)(0,8)2  ∆y = 3,14 m ∆y = −3,14 m ∆y = 3,14 m OPTION/OPSIE 5 OPTION/OPSIE 6 v +vf Distance = area under the graph ∆y = ( i ) ∆t 2 Afstand =oppervlakte onder die 0 + 7,84 grafiek ∆y = ( ) 0,8 2 = ½bh ∆y = 3,14 m = (½)(0,8) (7,84)  = 3,14 m (4) 3.3 0,2 s  (1) 3.4 Marking criteria/Nasienkriteria: • Any correct answer leading to the F net at which the ball strikes the glass roof.  Enige korrekte antwoord wat lei na die F net waarteen die bal die glasdak tref. • Correct substitution of/Korrekte substitusie van vf + vi.  • Correct substitution into/Korrekte substitusie in ∆t.  • Correct final answer./Korrekte finale antwoord.  OPTION/OPSIE 1 OPTION/ 2 Fnet ∙ ∆t = ∆𝑝 v𝑓= v𝑖 + 𝑎∆t vf − vi  Any one/Enige een 1,53 = 7,84 + 𝑎(0,2) Fnet = m ( ) ∆t 𝑎 = −31,55 𝑚 ∙ 𝑠 −2 1,53 − 7,84 Fnet = 0,175 ( ) 0,2 Fnet = 𝑚𝑎 Fnet = −5,52 N = (0,175)(−31,55) Fnet = 5,52 N upwards/𝑜𝑝𝑤𝑎𝑎𝑟𝑡𝑠 Fnet = −5,52 N Fnet = 5,52 N upwards/𝑜𝑝𝑤𝑎𝑎𝑟𝑡𝑠  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne OPTION/OPSIE 3 OPTION/OPSIE 4 v𝑓= v𝑖 + 𝑎∆t 1,53 = 7,84 + 𝑎(0,2) Fnet ∙ ∆t = ∆𝑝  Any one/ 𝑎 = −31,55 𝑚 ∙ 𝑠 −2 Fnet ∙ ∆t = m(vf − vi ) Enige een Fnet (0,2) = 0,175(1,53 − 7,84) ∆x = vi ∆t + ½a∆t 2 Fnet = −5,52 N ∆x = (7,84)(0,2) + ½(−31,55)(0,22 ) Fnet = 5,52 N upwards/𝑜𝑝𝑤𝑎𝑎𝑟𝑡𝑠 ∆x = −0,937 m 𝑊𝑛𝑒𝑡 = ∆𝐸𝐾  𝐹𝑛𝑒𝑡 ∆𝑥𝑐𝑜𝑠 𝜃 = ½m(𝑣𝑓2 − 𝑣𝑖2 ) 𝐹𝑛𝑒𝑡 (−0,937)𝑐𝑜𝑠 0° = ½(0,175)(1,532 − 7,842 ) Fnet = 5,52 N upwards/𝑜𝑝𝑤𝑎𝑎𝑟𝑡𝑠  (4) 3.5 Marking criteria/Nasienkriteria: • Any correct answer leading to the speed at which the ball strikes the ground. /Enige korrekte antwoord wat lei tot die spoed waarteen die bal die grond tref. • Correct substitution of/Korrekte substitusie van vi.  • Correct substitution to calculate./Korrekte substitusie om vf te bereken.  • Correct final answer./Korrekte finale antwoord.  OPTION/OPSIE 1 OPTION/OPSIE 2 + DOWNWARDS/AFWAARTS + UPWARDS/OPWAARTS vf = vi + a∆t vf = vi + a∆t vf = (1,53) + (9,8)(0,7) vf = (−1,53) + (−9,8)(0,7) vf = 8,39 m ∙ s−1  vf = −8,39 m ∙ s−1 vf = 8,39 m ∙ s−1  OPTION/OPSIE 3 OPTION/OPSIE 4 + DOWNWARDS/AFWAARTS + UPWARDS/OPWAARTS ∆y = vi ∆t + ½a∆t 2 ∆y = vi ∆t + ½a∆t 2 ∆y = (1,53)(0,7) + ½(9,8)(0,72 )  ∆y = (−1,53)(0,7) + ½(−9,8)(0,72 )  ∆y = 3,472 m ∆y = −3,472 m vf2 = vi2 + 2a∆y vf2 = vi2 + 2a∆y vf2 = (1,53)2 + 2(9,8)(3,472)  vf2 = (−1,53)2 + 2(−9,8)(−3,472)  vf = 8,39 m ∙ s−1  vf = −8,39 m ∙ s−1 vf = 8,39 m ∙ s−1  (3) 3.6 Equal/Gelyk  (1) 3.7 Newton's third law of motion/Newton se derde bewegingswet  (1) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne QUESTION/VRAAG 4 4.1 The total linear momentum in an isolated system is conserved.  Die totale lineêre momentum in 'n geïsoleerde sisteem bly behoue. (2 or/of 0) (2) 4.2 Marking criteria/Nasienkriteria: • Any correct formula for conservation of momentum./Enige korrekte formule vir behoud van momentum. • Correct substitutions./Korrekte substitusie.  • Correct final answer./Korrekte finale antwoord.  POSITIVE TO THE RIGHT/POSITIEF NA REGS ∑pi = ∑pf any one/enige een mtvit + msvis = (mt + ms)vf (45)(1,5) + (5)(-3) = (45 + 5) vf  vf = 1,05 m∙s-1 right/regs  POSITIVE TO THE LEFT/POSITIEF NA LINKS ∑pi = ∑pf mtvit + msvis = (mt + ms)vf Any one/Enige een (45)(-1,5) + (5)(3) = (45 + 5) vf  vf = -1,05 m∙s-1 vf = 1,05 m∙s-1 right/regs  (4) [6] QUESTION/VRAAG 5 5.1 The total mechanical energy in an isolated system is conserved.  Die totale meganiese energie in 'n geïsoleerde stelsel bly behoue. (2 or/of 0) (2) 5.2 Yes/Ja  (1) 5.3 No friction/Resultant force is zero and thus no loss in energy.  Geen wrywing/Resulterende krag is nul en dus geen verlies aan energie nie. OR/OF Only conservative forces are present./Slegs konserwatiewe kragte is teenwoordig. OR/OF Mechanical energy is conserved./Meganiese energie bly behoue. (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne 5.4 Marking criteria/Nasienkriteria: • Any correct formula for energy./Enige korrekte formule vir energie.  • Correct substitutions./Korrekte substitusies.  • Correct final answer./Korrekte finale antwoord.  OPTION/OPSIE 1 ∑EM(A) = ∑EM(D) (EK + EP )A = (EK + EP )B  any one/enige een (½mv 2 + mgh)A = (½mv 2 + mgh)B 0 + (0,4 x 9,8 x 0,8) = (½ x 0,4 x v 2 ) + (0,4 x 9,8 x 0,3) vf = 3,13 m ∙ s −1  OPTION/OPSIE 2 Wnc = ∆K + ∆U  any one/enige een 0 = ½m(vf2 − vi2 ) + mg(hf − hi ) 0 = ½(0,4)(vf2 − 0) + (0,4)(9,8)[0,3 − 0,8] vf = 3,13 m ∙ s −1  OPTION/OPSIE 3 𝑊𝑛𝑒𝑡 = ∆𝐸𝐾 𝑊𝑛𝑒𝑡 = ½m(vf2 − vi2 )  any one/ enige een 2 2 mg∆ycosθ=½m(vf − vi ) (0,4)(9,8)(0,5)cos0° = ½(0,4)(vf2 − 0) vf = 3,13 m ∙ s −1  (4) 5.5 POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 5.4 Marking criteria/Nasienkriteria: • Correct formula for work./Korrekte formule vir arbeid.  • Correct substitutions./Korrekte substitusies.  • Correct final answer./Korrekte finale antwoord.  OPTION/OPSIE 1 Wnet = ∆EK  any one/enige een Wnet = 1⁄2 mvf2 − 1⁄2 mvi2 f(3) cos 180°  = (0) − 1⁄2 (0,4)(3,13)2  f = 0,65 N OPTION/OPSIE 2 Wnc = ∆EK + ∆E𝑝  any one/enige een Wnet = ½m(vf2 − vi2 ) + mg(hf − hi ) f(3) cos 180°  = ½(0,4)(0 − 3,132 ) + (0,4)(9,8)(0,3 − 0,3) f = 0,65 N (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne QUESTION 6 6.1 Frequency/Frekwensie  (1) 6.2 OPTION/OPSIE 1 Approaches/Benaderings: V±VL fL = f V±VS S 𝐕  any one/enige een 𝐟𝐋 = 𝐟 𝐕−𝐕𝐒 𝐒 343 890 = f  343−VS S 890(343 − vS ) = 343fS ………………………... (1) Moves Away/Beweeg weg: V±VL fL = f V±VS S any one/enige een 𝐕 𝐟𝐋 = 𝐟 𝐕+𝐕𝐒 𝐒 343 760 = f  343+V S S 760(343 + vS ) = 343fS ………………………... (2) (1) - (2): (305270 − 890vS ) − (260680 + 760vS ) = 0 vS = 27,02 m ∙ s −1  (1) and/en (2) both/beide = 343 fs 890(343 − vS ) = 760(343 + vS ) vS = 27,02 m ∙ s −1  (6) 6.3 POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 6.2 OPTION/OPSIE 1 OPTION/OPSIE 2 305270 − 890(27,02) = 343fS  260680 + 760(27,02) = 343fS  fS = 819,89 Hz fS = 819,89 Hz (2) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne QUESTION/VRAAG 7 7.1.1 Marking Criteria/Nasienkriteria: If any of the underlined keywords/phrases in the correct context is omitted, deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks weggelaat word, trek 1 punt af. The magnitude of the electrostatic force exerted by one point charge on another point charge is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.  Die grootte van die elektrostatiese krag wat uitgeoefen word deur een puntlading op 'n ander puntlading is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. (2) 7.1.2 Q1 on Q2 is left./Q1 op Q2 is links.  Q3 on Q2 is upwards./Q3 op Q2 opwaarts.  (2) 7.1.3 Fnet F3on2 F2on1 Marking criteria/Nasienkriteria: • Correct calculation of/Korrekte berekening van F12 /F32 • Correct formula for Coulomb's law./Korrekte formule van Coulom se wet . • Correct substitution of/Korrekte substitusie van F12 / F32 • Correct calculation for/Korrekte berekening van F12 / F32 • Correct calculation for/ Korrekte berekening van Q1 • Correct calculation for/ Korrekte berekening van Q3 OPTION/OPSIE 1 F2on1 = Fnet cos 56° = 6,02 × 10−4 cos 59,3° = 3,07 × 10−4 N kQ1 Q 2 F2on1 =  r2 (9×109 )Q1 (4×10−9 ) 3,07 × 10−4  =  (0,015)2 Q1 = 1,92 × 10−9 𝐶 ∴ Q 3 = 3 (1,92 × 10−9 ) = 5,76 × 10−9 C Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne OPTION/OPSIE 2 F3on2 = Fnet sin 56° = 6,02 × 10−4 sin 59,3° = 5,18 × 10−4 N kQ3Q2 F3on2 =  r2 (9×109 )Q3 (4×10−9) 5,18 × 10−4  = 2  (0,02) Q 3 = 5,76 × 10−9 C ∴ Q1 = 1⁄3 (5,76 × 10−9 ) = 1,92 × 10−9 C OPTION/OPSIE 3 Marking criteria/Nasienkriteria: • Correct formula for/Korrekte formule vir Fnet  • Correct substitution of/Korrekte substitusie van Fnet  • Correct substitution of/Korrekte substitusie van F12  • Correct substitution of/Korrekte substitusie van F32 • Correct calculation for/Korrekte berekening van Q1 • Correct calculation for/Korrekte berekening van Q3 𝐹𝑛𝑒𝑡 2 = (𝐹12 )2 + (𝐹32 )2 2 2 kQ1 Q2 2 kQ 3 Q 2  any one/enige een 𝐹𝑛𝑒𝑡 = ( 2 ) + ( 2 ) r r 2 2 −4 2 (9 × 109 )Q1 (4 × 10−9 ) (9 × 109 )3Q3 (4 × 10−9 ) (6,02 × 10 )  = ( )  + ( ) (0,015)2 (0,02)2 Q1 = 1,92 × 10−9 𝐶 ∴ Q 3 = 3 (1,92 × 10−9 ) = 5,76 × 10−9 C (6) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne QUESTION/VRAAG 8 8.1.1 Marking criteria/Nasienkriteria: If any of the underlined keywords/phrases in the correct context is omitted, deduct 1 mark. Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks weggelaat word, trek 1 punt af. The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature.  Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur. OR/OF The ratio of potential difference to current is constant provided the temperature remains the same. Die verhouding van potensiaalverskil tot stroom is konstant mits die temperatuur dieselfde bly. (2) 8.1.2 OPTION/OPSIE 1 1 OPTION 2/OPSIE 2 1 1 1 R1 × R 2 = +  RP =  R P R1 R 2 R1 + R 2 1 1 13×10 = +  =  7+6 4+6 13+10 R P = 5,65 Ω R P = 5,65 Ω (4) 8.1.3 POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 8.2 OPTION/OPSIE 1 1 OPTION 2/OPSIE 2 ℰ = I(R ext + rint )  VT RT =  15 = I(5,65 + 0,45)  I I = 2,46 A 15 (5,65 + 0,45) =  I I = 2,46 A (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne 8.1.4 POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 8.3 Marking criteria/Nasienkriteria: • Correct formula for calculating Vint /Vext /Korrekte formule vir berekening Vint /𝑉𝑒𝑥𝑡 • Correct substitution/Korrekte substutisie  • Correct calculation for Vext/Korrekte berekening vir Vext  • Correct substitution of I7Ω & 6Ω /Korrekte substutisie van I7Ω & 6Ω • Correct substitution of I4Ω & 6/ Korrekte substutisie van I4Ω & 6 • Correct calculation for VA/Korrekte berekening vir VA  • Correct calculation for VB/Korrekte berekening vir VB  • Correct calculation for V2/Korrekte berekening vir V2  OPTION/OPSIE 1 Vint Vext r=  R=  I I Vint Vint 0,45 =  5,65 =  2,46 2,46 Vint = 1,11 V Vext = 13,89 V VT = Vext + Vint 15 = Vext + 1,11 Vext = 13,89 V Current through/Stroom deur Current through/ Stroom deur 7𝛀 & 𝟔𝛀 4𝛀 & 𝟔𝛀 V V R= R= I I 13,89 13,89 13 =  10 =  I I I = 1,07 A I = 1,39 A Potential difference across/ Potential difference across/ Potensiaalverskil oor 7𝛀 Potensiaalverskil oor 4𝛀 V V 𝑅7𝛺 = R= I I 𝑉 𝑉 7=  4 =  1,07 1,39 V = 7,49 V V = 5,56 V VA = 13,89 − 7,49 VB = 13,89 − 5,56 = 6,40 V = 8,33 V V2 = V𝐵 − V𝐴 V2 = 8,33 − 6,40 V2 = 1,93 𝑉 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne OPTION/OPSIE 2 Vint Vext r=  R=  I I Vint Vint 0,45 =  5,65 =  2,46 2,46 Vint = 1,11 V Vext = 13,89 V Current through/Stroom deur Current through/Stroom deur 7𝛀 & 𝟔𝛀 4𝛀 & 𝟔𝛀 V V R= R= I I 13,89 13,89 13 =  10 =  I I I = 1,07 A I = 1,39 A Potential difference across/ Potential difference across/ Potensiaalverskil oor 6𝛀 Potensiaalverskil oor 6𝛀 V(B) V(A) 𝑅6𝛺 = R= I I 𝑉 𝑉 6=  6 =  1,39 1,07 V = 8,34V V = 6,42 V V2 = V𝐵 − V𝐴 V2 = 8,34 − 6,42 V2 = 1,92 𝑉 (7) 8.2 Increases/Verhoog  (1) 8.3 Total resistance increases and total current decreases.  V lost decreases  Totale weerstand neem toe en totale stroom neem af. V neem af (2) 8.4 Cost/𝐾𝑜𝑠𝑡𝑒 = Energy/𝐸𝑛𝑒𝑟𝑔𝑖𝑒 × Tarrif/𝑇𝑎𝑟𝑖𝑒𝑓  250 = (kW × 8 × 30) × 2,55  250 = 612 KW kW = 0,4084967 P = 0,4084967 × 1000 P = 408,4967 W 408,4967 Number of bulbs/𝐴𝑎𝑛𝑡𝑎𝑙 𝑔𝑙𝑜𝑒𝑖𝑙𝑎𝑚𝑝𝑒 =  = 6 bulbs/𝑔𝑙𝑜𝑒𝑖𝑙𝑎𝑚𝑝𝑒  (5) 60 [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne QUESTION/VRAAG 9 9.1.1 X  (1) 9.1.2 Change the direction of the magnetic field (turn magnet upside down).  Change the direction of the current (turn the battery around). Verander die rigting van die magneetveld (draai magneet onderstebo). Verander die rigting van die stroom (draai die battery om). (1) 9.1.3 Any ONE/Enige EEN Increase the strength of the magnetic field (use stronger magnet).  Increase the current in the conductor.  Increase the thickness of the magnet to increase the length of the magnetic field. Verhoog die sterkte van die magneetveld (gebruik sterker magneet). Verhoog die stroom in die geleier. Verhoog die dikte van die magneet om die lengte van die magnetiese veld te vergroot. (2) 9.2.1 Marking criteria/Nasienkriteria: If any of the underlined keywords/phrases in the correct context is omitted deduct 1 mark. Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks weggelaat word, trek 1 punt af. The AC potential difference which dissipates/produces the same amount of energy as an equivalent DC potential difference.  Die WS-potensiaalverskil wat dieselfde hoeveelheid energie dissipeer/produseer as 'n ekwivalente GS-potensiaalverskil. (2) Note: If rms current defined 0⁄2 Let wel: Indien rms stroom gedefinieer is 0⁄2 9.2.2 AC-slip rings/WS-sleepringe  DC- commutator (split ring)/GS-kommutator (splitring)  (2) Vmax/𝑚𝑎𝑘𝑠 9.2.3 Vrms =  √2 460 =  √2 = 325,27 V  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne 9.2.4 POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 9.2.2 OPTION/OPSIE 1 OPTION/OPSIE 2 Vmax V R = I rms  Imax = rms R 325,27 460 = 40  40 = I  rms = 11,5 A Irms = 8,13 A I Irms = max  √2 11,5 =  √2 = 8,13 A  (4) [15] QUESTION/VRAAG 10 10.1 The minimum energy that an electron in the metal needs to bemitted from the metal surface.  Die minimum energie wat 'n elektron in die metaal nodig het om vanaf die metaaloppervlak vrygestel te word. (2 or/of 0) (2) 10.2 Marking criteria/Nasienkriteria: • Correct formula for work/Korrekte formule vir arbeid  • Correct substitutions/Korrekte substitusies  • Correct final answer/Korrekte finale antwoord  OPTION/OPSIE 1 E = W0 + EK hc any one/enige een = W0 + EK λ (6,63 × 10−34 )(3 × 108 )  = (3,36 × 10−19 ) + EK (400 × 10−9 ) EK = 1,61 × 10−19 J OPTION/OPSIE 2 hc E= λ (6,63 × 10−34 )(3 × 108 ) =  (400 × 10−9 ) = 4,97 × 10−4 J E = W0 + EK  4,97 × 10−4 J = (3,36 × 10−19 ) + EK EK = 1,61 × 10−19 J (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 19 FS/VS/September 2024 Grade 12 Prep. Exam./Voorb/Eksam. Marking Guideline/Nasienriglyne 10.3 Remains the same/Bly dieselfde  (1) 10.4 EK = ½mv 2 1,61 × 10−19 = ½(1,6 × 10−19 ) v 2  v = 1,42 m ∙ s −1 ∆p = m∆v = (1,6 × 10−19 )(1,42 − 0)  = 2,27 kg ∙ m ∙ s−1  (4) 10.5 W = dEq F = Eq 1,61 × 10−19  = d(4,6)(1,6 × 10−19 ) W and/𝑒𝑛 F = d = 0,22 m d W = Eq (both = F) d ∴ W = dEq (4) [16] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ## Related documents - [PHYSICAL SCIENCES P2 QP SEPT 2024 Sesotho](https://www.hlayiso.com/library/physical-sciences-p2-qp-sept-2024-sesotho/112a3991-34d4-454a-92c4-012482040fb9) — Physical Sciences · QP · 2024 · Sesotho - [kzn=G12 Physical Sciences P1 Sep 2024 KZN QP Scanned by Farhad Akoo](https://www.hlayiso.com/library/kzn-g12-physical-sciences-p1-sep-2024-kzn-qp-scanned-by-farhad-akoo/21919713-0b36-4903-ba81-155d000a162d) — Physical Sciences · QP · 2024 - [mp=P. 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