# Physical Sciences Grade 12 P1 Mpumalanga Prelim 2024 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Mpumalanga Prelim Exam 2024, 14 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mp-physical-sciences-p1-mg/59a9a23a-20b7-43b7-9cfe-951a693e6c25 - Updated: 2026-10-29 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2024 - Exam period: Mpumalanga Prelim Exam - Paper: 1 - Curriculum: CAPS - Pages: 14 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 Mpumalanga Prelim 2024 Memo](https://www.hlayiso.com/library/mp-physical-sciences-p1-mg/59a9a23a-20b7-43b7-9cfe-951a693e6c25) ## About this document Physical Sciences · Grade 12 · Mpumalanga Prelim Exam · 2024. Memorandum. 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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 2024 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consists of 13 pages Hierdie nasienriglyne bestaan uit 13 bladsye Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 2 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne QUESTION 1 / VRAAG 1 1.1 C  (2) 1.2 D  (2) 1.3 C  (2) 1.4 B  (2) 1.5 A  (2) 1.6 D  (2) 1.7 B  (2) 1.8 B  (2) 1.9 A  (2) 1.10 A (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 3 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne QUESTION 2 / VRAAG 2 2.1 Marking criteria/Nasienkriteria If any one 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 a net force acts on an object, the object will accelerate in the direction of the force and the acceleration is directly proportional to the resultant/net 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 resultante/netto krag en omgekeerd eweredig aan die massa van die voorwerp. OR/OF The resultant/net force acting on an object is equal to the rate of change of momentum of the object. (2 or 0) Die resulterende/netto krag wat op ‘n voorwerp inwerk is gelyk aan die tempo van verandering van momentum. (2 of 0) (2) 2.2 DOWN THE INCLINE / AFWAARTS TEEN DIE HELLING  (1) 2. 3 T N f w (4) Accepted labels / Aanvaarde benoemings w Fg / Fw / force of earth on block / weight / mg / gravitational force / VALUE / gravitasiekrag T Tension in rope / FT / spankrag F Ffriction / Ff / frictional force / wrywing N Normal force / FN / Normaalkrag Notes/Aantekeninge:  Mark awarded for label and arrow.  Any additional forces: deduct 1 mark: max 3⁄4  No labels: deduct 1 mark: max 3⁄4  If everything correct, but no arrows: deduct 1 mark: 3⁄4  Force(s) not touching object: deduct 1 mark: max 3⁄4  Ignore relative sizes of the vectors Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 4 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne 2.4 Marking criteria/Nasienkriteria:  Formula for block P or block Q   Substitution of Fnet for block P   Substitusion of Fnet for block Q   5a OR 2a   Answer  For Block P: For Block Q: Fnet = ma Fnet = ma Any one  Fgǁ + (- T) + (-f) = ma T - Fg = ma (5)(9,8)sin30o –T – 2,5  = 5a  T – (2)(9,8) = 2a 22 -T = 5a T - 19,6 = 2a T = -5a + 22 T = 2a + 19,6 ∴ -5a + 22 = 2a + 19,6 7a = 2,4 a = 0,34 m∙s-2  (5) 2.5 GREATER THAN / GROTER AS  Fnet increases – no more tension pulling back Fnet neem toe – geen meer spankrag wat terug trek nie (2) [14] QUESTION 3 / VRAAG 3 3.1 An object which has been given an initial velocity and then it moves under the influence of the gravitational force only.   (2 or 0) 'n Voorwerp waaraan 'n beginsnelheid gegee is en wat dan slegs onder die invloed van die gravitasiekrag beweeg. (2 or 0) (2) 3.2.1 Upwards positive/ Downwards positive/ Opwaarts positief: Afwaarts positief: vf = vi + aΔt  vf = vi + aΔt  -30  = 30 + (-9,8)Δt  30  = -30 + (9,8)Δt  Δt = 6,12s  Δt = 6,12s  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 5 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne 3.2.2 POSITIVE MARKING FROM QUESTION 3.2.1 POSITIEWE NASIEN VANAF VRAAG 3.2.1 vf = vi + aΔt 0 = -30 + (9,8)Δt Δt = 3,06s Upwards Positive/ Downwards Positive/ Opwaarts positief Afwaarts positief Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  Δy = (30)(3,06) + ½(-9,8)(3,06)2  Δy = (-30)(3,06) + ½(9,8)(3,06)2  Δy = 45,92 m Δy = -45,92 m Δy = viΔt + ½aΔt2 Δy = viΔt + ½aΔt2 Δy = (30)(2,06) + ½(-9,8)(2,06)2  Δy = (-30)(2,06) + ½(9,8)(2,06)2  Δy = 41,01 m Δy = -41,01 m displacement in the last second displacement in the last second = 45,92 – 41,01 = 4,91 m  = 45,92 – 41,01 = 4,91 m  (4) 3.3 Upwards positive / Downwards positive/ Opwaarts positief Afwaarts positief Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  -50 = vi (4,12) + ½(-9,8)(4,12)2  50 = vi (4,12) + ½(9,8)(4,12)2  vi = 8,05 m∙s-1 vi = -8,05 m∙s-1 vi = 8,05 m∙s-1 (4) 3.4 POSITIVE MARKING FROM/POSITIEWE MERK VANAF 3.2.1 & 3.2.2 & 3.3 UPWARDS POSITIVE/OPWAARTS POSITIEF Criteria/Kriteria Marks/Punte Correct shape of A Korrekte vorm van A  Correct shape of Graph B parallel to A above A  Korrekte vorm van Grafiek parallel met A bo A Time at which both A and B reach the ground (6,12 s)  Tyd wat beide A en B die grond bereik (6,12 s) Time for A to reach the maximum height (3,06 s) shown  Tyd vir A om maksimum hoogte te bereik (3,06 s) aangedui (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 6 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne DOWNWARDS POSITIVE/AFWAARTS POSITIEF [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 7 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne QUESTION 4 / VRAAG 4 4.1 vf2 = vi2 + 2aΔy  vf2 = 0 + 2(9,8)(15)  vf = 17,15 m∙s-1  (3) 4.2. Impulse is the product of the resultant/net force  acting on an object and the time the net force acts on the object.  Impuls is die produk van die resulterende/netto krag  wat op 'n voorwerp inwerk en die tyd wat die netto krag op die voorwerp inwerk.  (2) 4.3.1 POSITIVE MARKING FROM QUESTION 4.1 Fnet ∙Δt = mvf – mvi  Fnet (0,5) = (1,5)( 0 – 17,15)  Fnet = - 51,45 N Fnet = 51,45 N  upwards/opwaarts (3) 4.3.2 POSITIVE MARKING FROM QUESTION 4.3.1 Fnet = Fmud + Fg  - 51,45 = Fmud + (1,5)(9,8)  Fmud = 66,15 N  (3) 4.4 DECREASE/VERMINDER  Δp = constant value  Fnet is inversely proportional to Δt, therefore the longer the time of contact, the smaller the net force if the change in momentum stays the same.  (3) [14] QUESTION 5 / VRAAG 5 5.1 Marking criteria/Nasienkriteria If any one 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. A force for which the work done in moving an object between two points depends on the path taken 'n Krag waarvoor die arbeid verrig om 'n voorwerp tussen twee punte te beweeg, afhanklik is van die roete wat gevolg word. (2) 5.2 EkT + EpT = EkP + EpP ½(5)(0)2 + (5)(9,8)(5)  = ½(5)v2 + (5)(9,8)(0)  245 = 2,5 v2 v = 9,90 m∙s-1 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 8 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne 5.3 N w (2) Accepted labels / Aanvaarde benoemings w Fg / Fw / force of earth on block / weight / mg / gravitational force / N Normal force / FN 5.4 Velocity stays the same there is no friction working in on the object. (1) 5.5 Marking criteria/Nasienkriteria If any one 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 work done on an object is equal to the change in kinetic energy of the object.  Die netto arbeid verrig op 'n voorwerp is gelyk aan die verandering in kinetiese energie van die voorwerp OF die arbeid verrig op die voorwerp deur 'n netto krag is gelyk aan die verandering in kinetiese energie van die voorwerp. (2) 5.6 POSITIVE MARKING FROM QUESTION 5.2 OPSIE 1/OPTION 1 Wnet = ΔEk Any one  FnetΔx cosΘ= [½mvf2 – ½mvi2 ] (18 + (5)(9,8)sinθ)Δxcos180°= [½(5) 02 – ½(5) (9,90)2]  3 -18x - 49( )x = [½(5) 02 – ½(5) (9,90)2] 𝑥 -18x – 49(3) = -245,025 Δx = 5,445 m 𝑡 sin θ = 𝑠 3 sin θ =  5,445 θ = 33,43o  (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 9 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne OPSIE 2/OPTION 2 Wnc = ΔEk + ΔEp Any one  fΔx cosΘ= [½mvf2 – ½mvi2 ]+ [mghf – mghi] 18Δxcos180°= [½(5) 02 – ½(5) (9,90)2] +[(5)(9,8(3) – (5)(9,8)(0)] -18Δx = -245 + 147 Δx = 5,44 m 𝑡 sin θ = 𝑠 3 sin θ =  5,44 θ = 33,47o  [17] UESTION 6 / VRAAG 6 6.1 Marking criteria/Nasienkriteria If any one 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/pitch/wavelength 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 van frekwensie/toonhoogte/golflengte van die klank waargeneem deur ‘n luisteraar omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium van klankvoorplanting het. OR/OF An (apparent) change in observed frequency/pitch/wavelength as the result of the relative motion between a source and an observer/listener.  ‘n (Skynbare) verandering in waargenome frekwensie/toonhoogte/golflengte as gevolg van die relatiewe beweging tussen die bron en ‘n waarnemer/luisteraar. (2) 6.2 2 1 f= or 4×10−5 1×10−4 = 40 000 Hz   (2) 6.3 Waves in front of source are more compact/ wavelength decreases  Dolphin is moving towards the recorder.  (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 10 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne 6.4 POSITIVE MARKING FROM QUESTION 6.2 v±vL v fL = f  OR fL = f v±vS S v-vS S 1500 + 0 40 000 =  39 000  1500 - vS vS = 37,5 m.s-1  (5) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 11 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne QUESTION 7 / VRAAG 7 7.1 Marking criteria/Nasienriglyne: -1 mark for each of the 3 key words omitted in the correct context. -1 punt vir elk van die 3 sleutelwoorde weggelaat in die korrekte konteks. The electric field is an area in space where an electric charge experiences a force.   Die elektriese veld is ‘n gebied in die ruimte waarin ‘n elektriese lading ‘n krag ondervind (2) 7.2 𝐸𝑛𝑒𝑡𝑡 = 𝐸1 + 𝐸2 𝑘𝑄 𝑘𝑄 𝑘𝑄 𝐸𝑛𝑒𝑡 = 𝑟 2 − 𝑟 2  (Mark for/Punt vir E = 𝑟 2 ) (9 × 109 )(4 × 10−6 ) (9 × 109 )(3 × 10−6 ) =  −   (0,3)2 (0,1)2 = 2,3 x 106 N. C-1  right/ regs/east/oos (6) 7.3 -0,5 x 10-6 C  (2) 7.4 POSITIVE MARKING FROM QUESTION 7.3 kQ1 Q2 kQ1 Q2 F=  F = r2 r2 (9×109 )(0,5×10-6 ) Q3 (9×109 )(0,5×10-6 )(0,5×10-6 ) F (Q1 on 3) =  F (Q1 on 2) =  2 (0,2)2 (0,25) = 72000 x Q3 N = 0,05625 N  Fnet = √(F1,3 )2 + (F1,2 )2 5,4 x10-1  = √(72000 Q3 )2 + (0,05625)2 Q3 = 7,459 x 10-6 C  (6) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 12 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne QUESTION 8 / VRAAG 8 8.1.1 6V  (1) 8.1.2 0,2 A  (1) 8.2.1 POSITIVE MARKING FROM Q 8.1.1 AND 8.1.2 V R =  I 3 =  0,2 = 15 Ω (3) 8.2.2 POSITIVE MARKING FROM Q8.1.2 (I=0,1A + Answer Q8.1.2) OPTION 1 1 1 1 = + R r1 r2 Any one  1 1 = + 45 90 R = 30 Ω Re = 30 + 60 = 90 Ω ε = I(R+r)  = (0,3)(90 + 2)  = 27,6 V  OPTION 2 Vp = IR Vbulb = IR  Vi = Ir = (0,3)(30)  = (0,3)(60) = (0,3)(2) =9V = 18 V  = 0,6V ε = Vp + Vbulb + Vi = 9 + 18 + 0,6 = 27,6 V  (5) 8.2.3 POSITIVE MARKING FROM Q8.1.2 + Q8.2.2 OPTION 1 OPTION 3 P = I2R  V2 = (0,3)2(60)  P=  R = 5,4 W  182 =  60 OPTION 2 = 5,4 W  P = VI  = (18)(0,3)  = 5,4W  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 13 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne 8.3 0V (1) 8.4 Total resistance increases  ∴Total current decreases  ∴’lost volts’ (Vi) DECREASE  because Vi∝I (r = constant) (3) [17] QUESTION 9 /VRAAG 9 9.1.1 Anti-clockwise (1) 9.1.2 Electrical energy to kinetic(mechanical) energy Elektriese enegie na kinetiese(meganiese) energie (1) 9.1.3 Marking criteria for graph: Correct shape Korrekte vorm  Two complete cycle/Twee volledige  siklus (2) 9.2.1 The AC potential difference/voltage which dissipates the same amount of energy as DC potential potential difference.  Die WS potensiaalverskil/spanning wat dieselfde hoeveelheid energie verbruik as GS potensiaalverskil/spanning. OR/OF (The rms value of AC is) the DC potential difference/voltage which dissipates the same amount of energy as AC potential difference/voltage.  Dit is die GS potensiaalverskil/spanning wat dieselfde hoeveelheid energie verbruik as WS potensiaalverskil/spanning. (2) 9.2.2 OPTION 1 OPTION 2 Pave= Vrms·Irms  Pave= Vrms·Irms  I = 230 . max 2100 = 230 Irms  √2 I Irms =9,13 A 2100  = 230 max  √2 I Imax = 12,9 𝐴  Irms = max √2 Imax 9,13 =  √2 Imax = 12,91 A (4) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Physical Sciences/Fisiese Wetenskappe/P1 14 MDE/MDO/September 2024 NSC/NSS –Marking guidelines/Nasienriglyne QUESTION 10 / VRAAG 10 10.1 3,6 x 10-19 J  (1) 10.2 The maximum kinetic energy is inversely proportional to the wavelength. (2) 10.3 Marking criteria/Nasienkriteria If any one 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 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.. (2) 10.4 hc Wo =  λ0 (6,63×10-34 )(3×108 ) =  4,9 ×10-7 = 4,06 × 10-19 J  (3) 10.5 POSITIVE MARKING FROM 10.4 E = Wo+ Ek(max) hc = Wo + ½mv2 𝜆 (6,63×10-34 )(3×108 )  = (4,06×10-19 ) + ½(9,11×10-31 ) (2,6 x106 )2  𝜆 𝜆 = 5,7 x10 -8 m  (5) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ## Related documents - [mp=P. 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