# Physical Sciences Grade 12 Paper 1 North West June 2024 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the North West June Exam 2024, 18 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-12-physical-sciences-2024-north-west-june-exam-question-paper-paper-1/cab9ec02-679e-4c95-b32d-60bae447cfa2 - Updated: 2026-10-29 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2024 - Exam period: North West June Exam - Paper: 1 - Curriculum: CAPS - Pages: 18 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 Paper 1 North West June 2024 Memo](https://www.hlayiso.com/library/grade-12-physical-sciences-2024-north-west-june-exam-question-paper-paper-1/cab9ec02-679e-4c95-b32d-60bae447cfa2) ## About this document Physical Sciences · Grade 12 · North West June Exam · 2024. Memorandum. 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 PROVINCIAL ASSESSMENT PROVINSIALE ASSESSERING GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) JUNE/JUNIE 2024 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 16 pages including a cognitive grid. Hierdie nasienriglyne bestaan uit 16 bladsye en ‘n kognitiewe tabel. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Physical Sciences P1/Fisiese Wetenskappe V1 2 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 B  (2) 1.2 D  (2) 1.3 C  (2) 1.4 C  (2) 1.5 A  (2) 1.6 A  (2) 1.7 B  (2) 1.8 C  (2) 1.9 D  (2) 1.10 B  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Physical Sciences P1/Fisiese Wetenskappe V1 3 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force acts on it.  ‘n Liggaam sal in rus of beweging teen konstante snelheid volhard tensy ‘n nie-nul resulterende/netto krag daarop inwerk. (2 or 0) (2) 2.2 FN f T Fg Accept the following symbols/aanvaar die volgende simbole N /Normal/Normal force FN  Normaal/Normaalkrag T FT/ Ft /Tension/Spanningskrag Fg  w /mg/weight/gravitational force Gewig/gravitasiekrag f Ff /Force of friction/Wrywingskrag 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 3/4 • If everything is correct, but no arrows/Indien alles korrek is, maar geen pyltjies: Max/Maks ¾ 2.3.1 FY = 0 N + Fg = 0 N – Fg = 0 any one/enige een  N = mg N = 8 × 9,8  N = 78,4 N  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Physical Sciences P1/Fisiese Wetenskappe V1 4 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne 2.3.2 Fnet = ma ✓ (Direction of motion as positive) (Rigting van beweging as positief.) Block/Blok A T + fK = ma T - fK = ma any one/enige een T - 23,52 = 8a T = 8a + 23,52……………(1) ✓ Block/Blok B F - (T + Fg) = ma T = F - (ma) - Fg any one/enige een = 192 - 16a - 16(9,8)…………..(2) ✓ Solving (1) and (2)/Los (1) en (2) op: 8a + 23,5 = 192 - 16a - 156,8 a = 0,49 m.s-2 ✓ Substitute a into (1) or (2)/Vervang a in (1) of (2) T = 8a +23,52 OR/OF T = 192 - 16a - 156,8 = 8(0,49) + 23,52 = 192 - 16(0,49) -156,8 T = 13,72 N ✓ = 13,72 N (5) 2.4 To the left.  The only force acting of the object is frictional force. (According Newton’s second law), the body will accelerate in the direction of the (net) force.  Na links. Die enigste krag wat op die voorwerp inwerk is wrywingskrag. (Volgens Newton se tweede wet), die liggaam sal versnel in die (2) rigting van die (netto) krag. 2.5 Increase  Vermeerder (1) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Physical Sciences P1/Fisiese Wetenskappe V1 5 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 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 force or the component of a force which a surface exerts on an object  with which it is in contact, and which is perpendicular to the surface.  Die krag of komponent van ‘n krag wat deur die oppervlak uitgeoefen (2) word op ‘n voorwerp waarmee dit in kontak is, is loodreg op die oppervlak. 3.2 Accept the following symbols/aanvaar die volgende simbole FN/Normal/Normal force N Normaal/Normaalkrag T FT/ Ft /Tension f  Ff /Force of friction/Wrywingskrag w Fg/mg/weight/gravitational force Fg/mg/gewig/gravitasiekrag 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 3/4 • If everything is correct, but no arrows/Indien alles korrek is, maar geen pyltjies: Max/Maks 3/4 (4) 3.3 OPTION/OPSIE 1: OPTION/OPSIE 2: Fnet = ma Fnet = ma Fnet = 0 any one/enige Fnet = 0 any one/enige een een T =Fgll + f T = Fg =(44 x 9,8 Sin 28,66˚) + 87,2  Fg = mg = 294 N  = 30(9,8)  (3) = 294 N  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Physical Sciences P1/Fisiese Wetenskappe V1 6 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne 3.4 POSITIVE MARKING FROM QUESTION 3.3 POSITIEWE NASIEN VANAF VRAAG 3.3 To the right as positive/Na regs as positief: Fnet =ma Fnet =0 any one/enige een T - Wsinθ - f = ma 294 - 44x9,8sinθ - 87,2 = 0  θ = 28,66˚  (2) [11] 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 net (or 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) Die netto/resultante krag wat op ‘n voorwerp uitgeoefen word is gelyk aan die tempo van momentsverandering in die rigting van die netto krag. 4.2.1 Ʃpi = Ʃpf mvi1 + mvi2 = mvf1 + mvf2 any one/enige een  (0,41)(15) + (79)(0)  = (0,41)(-12) + (79)vf1  vf1 = 0,14 m.s-1 west/wes (4) 4.2.2 Positive marking from Question 4.2.1/ Positiewe merk vanaf Vraag 4.2.1 FnetΔt = Δp ∆𝑝 Fnet = ∆𝑡 any one/enige een  mvf−mvi = ∆𝑡 (79)(0,14)−(79)(0) =  2 = 5,53 N east/oos (3) 4.2.3 Positive marking from Question 4.2.2/ Positiewe merk vanaf Vraag 4.2.2 Fnet = 5,53/4 N west/wes  (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Physical Sciences P1/Fisiese Wetenskappe V1 7 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 In an isolated system the total (linear) momentum is conserved/remains constant.   In 'n geïsoleerde sisteem bly die totale (lineêre) momentum behoue/konstant. (2 or 0) (2) 5.2.1 FnetΔt = Δp any one/enige een  ∆𝑝 Fnet = ∆𝑡 mvf−mvi = ∆𝑡 111 000−0 7,4 x 105 =  𝑡1 −0 t1 = 0,15 s  (4) 5.2.2 Between t = t1 and t = 0,25 s, there is a decrease in momentum. Therefore, the velocity of the ball decreases which implies that there is a force opposing the motion of the ball (i.e friction)  (2) Tussen t = t1 en t = 0,25 s, is daar ‘n afname in momentum. Daarom sal die snelheid van die bal verminder, dus kan ons aflei dat daar ‘n krag is wat beweging teenwerk (wrywing). 5.2.3 Ʃpi = Ʃpf pAi + pBi = pAf + pBf any one/enige een  100 000 + 0 = 88 800 + pBf pBf = 11 200 kg.m.s-1  but/maar p = mv 11 200 = (15 000)v  v = 0,75 m.s-1 (to the right/na regs)  (4) 5.3 According to Newton’s second law, according to momentum FnetΔt = Δp  Crumple zones causes an increase in collision time  which results in a lesser/smaller force  Volgens Newton se Tweede wet in terme van momentum FnetΔt = Δp Frommelsones veroorsaak ‘n verlengde botsingstyd wat ‘n kleiner krag tot gevolg het. (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Physical Sciences P1/Fisiese Wetenskappe V1 8 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 An object which has been given an initial velocity and then it moves under the influence of the gravitational force only.  ‘n Voorwerp met ‘n aanvanklike snelheid beweeg slegs onder die invloed van gravitasiekrag. (2 or 0) (2) 6.2.1 OPTION/OPSIE 1 OPTION/OPSIE 2 UPWARDS POSITIVE UPWARDS POSITIVE OPWAARTS POSITIEF OPWAARTS POSITIEF vf 2 = vi 2 + 2aΔy  vf = vi + aΔt (- 50)2 =(- 40)2 + 2(-9,8)Δy  - 50 = - 40 + (-9,8)Δt Δy = - 45,92 Δt = 1,02 s Δy = 45,92 m downwards/afwaarts Δy = viΔt + ½gΔt2  =(-40)(1,02) + ½(-9,8)(1,02 )2  Δy = 45,90 m  (3) 6.2.2 UPWARDS POSITIVE/OPWAARTS POSITIEF vf = vi + a∆t  -50 = -40 + (-9,8)(Δt)  Δt = 1,02 s (time to reach to ground)/(tyd om die grond te bereik) vf = vi + a∆t 0 = 24 + (-9,8)(Δt) Δt = 2,45 s  (time to reach maximum height)/(tyd om maks hoogte te bereik) t = 1,02 + 2,45 t = 3,47 s  (4) 6.2.3 Positive marking from 6.1.1/Positiewe merk vanaf 6.1.1 UPWARDS POSITIVE/OPWAARTS POSITIEF vf 2 = vi 2 + 2aΔy 0 = 242 + 2(-9,8) Δy  Δy = 556,40 m Displacement/Verplasing = - 556,40 + 45,90  = - 510,50 = 510,50 m (downwards/afwaarts)  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Physical Sciences P1/Fisiese Wetenskappe V1 9 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne 6.3 (3) [15] OUESTION 7/VRAAG 7 7.1 free fall/vry val  (1) 7.2 Newton's First Law/Newton se Eerste Wet  A body will remain in its state of rest or motion at a constant velocity unless a non-zero net/resultant force acts on it. 'n Voorwerp sal in sy posisie van rus of beweging bly teen 'n konstante snelheid tensy 'n nie-nul netto/resulterende krag daarop inwerk. (3) 7.3 7.3.1 OPTION/OPSIE 1 OPTION/OPSIE 2 (Upwards as (Downwards as positive/Opwaarts as positive/Afwaarts as positief) positief) vf2 = vi2 + 2aΔy  vf2 = vi2 + 2aΔy  0 = (16)2 + 2(-9,8)Δy  0 = (-16)2 + 2(9,8)Δy  Δy = 13,06 m Δy = - 13,06 m Height/Hoogte = 13,06 + 25  Height/Hoogte = 13,06 + 25  = 38,06 m  = 38,06 m  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Physical Sciences P1/Fisiese Wetenskappe V1 10 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne 7.3.2 OPTION/OPSIE 1 OPTION/OPSIE 2 (Whole motion upwards as (Whole motion downwards positive/Hele beweging as positive/Hele beweging opwaarts as positief) afwaarts as positief) Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  = 16(3,5) + ½(-9,8)(3,5)2 = -16(3,5) + ½(9,8)(3,5)2 = - 4,03 m = 4,03 m Height/Hoogte = 25 - 4,03  Height/Hoogte = 25 - 4,03  = 20,97 m  = 20,97 m  OPTION/OPSIE 3 OPTION/OPSIE 4 (Whole motion upwards as (Whole motion downwards positive/Hele beweging as positive/Hele beweging opwaarts as positief) afwaarts as positief) vf = vi + g∆t vf = vi + g∆t 0 = 16 − 9,8∆t 0 = -16 + 9,8∆t ∆t = 1,63 s ∆t = 1,63 s Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  = 0 + ½(-9,8)(3,5 - 1,63)2 = 0 + ½(9,8)(3,5 - 1,63)2 = - 17,13 m = 17,13 m Height/Hoogte =38,06 - 17,13 Height/Hoogte =38,06 - 17,13 = 20,93 m = 20,93 m OPTION/OPSIE 5 OPTION/OPSIE 4 (From point of projection (From point of projection upwards as positive/Vanaf downwards as punt van projeksie positive/Vanaf punt van beweging opwaarts as projeksie beweging positief) afwaarts as positief) vf = vi + g∆t vf = vi + g∆t 0 = 16 − 9,8∆t 0 = -16 + 9,8∆t ∆t = 1,63 s ∆t = 1,63 s Δy = viΔt + ½aΔt2 Δy = viΔt + ½aΔt2 = -16(0,24) + ½(-9,8)(3,5 - 3,26)2 = 16(0,24) + ½(9,8)(3,5 - 3,26)2 = - 4,12 m = 4,12 m Height/Hoogte = 25 - 4,12 Height/Hoogte = 25 - 4,12 = 20,88 m  = 20,88 m  RANGE/GEBIED: 20,88 - 20,98 m RANGE/GEBIED: 20,88 - 20,98 m (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Physical Sciences P1/Fisiese Wetenskappe V1 11 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne 7.3.3 OPTION/OPSIE 1 OPTION/OPSIE 2 (Whole motion upwards as (Whole motion downwards positive/Hele beweging as positive/Hele beweging opwaarts as positief) afwaarts as positief) ∆y = vi∆t + ½g∆t2  ∆y = vi∆t + ½g∆t2  -25 = 16Δt + ½(-9,8)Δt2 25 = -16Δt + ½(9,8)Δt2 Δt = 4,42 s  Δt = 4,42 s Note: For options 3 and 4 use any of the equations of motion containing ∆t to calculate ∆t and substitute into equation of ∆y = vi∆t + ½g∆t2. Nota: Vir opsies 3 en 4, gebruik enige van die bewegingsvergelykings wat ∆t bevat om ∆t te bereken en vervang in vergelyking van ∆y = vi∆t + ½g∆t2. OPTION/OPSIE 3 OPTION/OPSIE 4 (From maximum height (From maximum height upwards as positive/Vanaf downwards as maksimum hoogte positive/Vanaf maksimum opwaarts as positief) hoogte afwaarts as positief) vf = vi + g∆t 0 = 16 − 9,8∆t vf = vi + g∆t ∆t = 1,63 s 0 = - 16 + 9,8∆t ∆t = 1,63 s ∆y = vi∆t + ½g∆t2  - 38,06= 0Δt + ½(-9,8)Δt2 ∆y = vi∆t + ½g∆t2  Δt = 2,79s 38,06= 0Δt + ½(9,8)Δt2 Δt = 2,79s Δt = 2,79 + 1,63 = 4,42 s  Δt = 2,79 + 1,63 = 4,42 s  (4) 7.4 Positive marking from 7.3.3/Positiewe merk vanaf 7.3.3 UPWARDS POSITIVE/OPWAARTS POSITIEF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Physical Sciences P1/Fisiese Wetenskappe V1 12 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne DOWNWARDS POSITIVE/AFWAARTS POSITIEF CRITERIA FOR GRAPH/RIGLYNE VIR GRAFIEK Shape of graph/Vorm van grafiek  Initial velocity/Aanvanklike snelheid  Time to complete motion/Tyd vir volledige beweging  (3) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Physical Sciences P1/Fisiese Wetenskappe V1 13 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 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. A force for which the work done in moving an object between two points depends on the path taken.  ‘n Krag waar die werk gedoen op ‘n bewegende voorwerp tussen twee punte, afhanklik is van die pad geneem. (2) 8.2 Frictional force./Gravitasie krag  (1) 8.3 fk = µkN = µk(mgcosθ) any one/enige een  = 0,112 x (70 x 9,8cos30˚)  = 11,85 N  (down the slide/glyplank af) (3) 8.4 POSITIVE MARKING FROM QUESTION 8.3 POSITIEWE NASIEN VAN VRAAG 8.3 OPTION/OPSIE 1 Wnet = ΔEk fΔxcosθ + Fg//Δxsinθ = ½mvf2 - ½mvi2 any one/enige een  (11,85)(2,8)cos180˚+(70)(9,8)(2,8)sin30˚=½(70)vf2 - ½(70)(0,35)2 vf =7,39 m.s-1  OPTION/OPSIE 2: Wnc = ΔEk + ΔEp fΔxcosθ = (½mvf2 - ½mvi2) + (mghf + mghi) any one/enige een  (11,85)(2,8)cos180˚= [½(70)vf - ½(70)(0,35) ] + [0 - (70)(9,8)(2,8)] 2 2 vf = 7,39 m.s-1  (5) 8.5 INCREASES  The frictional force is proportional to the normal force , and since the normal force increases with an increase in the angle of the slide, the frictional force also increases. TOENEEM Die wrywingskrag is direk eweredig aan die normaalkrag, en omdat die normaalkrag vermeerder as die hoek van die glyplank vermeerder, sal wrywingskrag ook meer word. (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Physical Sciences P1/Fisiese Wetenskappe V1 14 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1 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 or 0) (2) 9.2 Accept the following symbols/aanvaar die volgende simbole air friction f/fair/lugweerstand F  FA/FT/ w Fg/mg/weight/gravitational force Fg/mg/gewig/gravitasiekrag 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 (3) 9.3 OPTION/OPSIE 1 Wnet = ΔEk WT + WW + Wf = ΔEk any one/enige een FTΔycosθ + FgΔycosθ + Wf = 0 (15 000)(50)cos0˚ + (2500)(9,8)(50)cos180˚ + Wf = 0  Wf = 475 000 J  (5) OPTION/OPSIE 2: UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF Fnet = ma F + f + w = ma 15 000 - f - (2500)(9,8) = 0 f = 9 500 N Both equations/Beide vergelykings Wf = fΔycosθ = 9500(50)cos180˚ = 475 000 J (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Physical Sciences P1/Fisiese Wetenskappe V1 15 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne 9.4 W P=  ∆t 475 000 =  (30 ×60) = 263,89 W  (3) [13] QUESTION 10/VRAAG 10 10.1 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 die frekwensie (toonhoogte) van die waargenome klank deur die luisteraar agv die klankbron en die luisteraar wat verskillende snelhede relatief tot mekaar het. OR/OF An (apparent) change in (observed/detected) frequency (pitch), as a result of the relative motion between a source and an observer (listener). 'n (Skynbare) verandering in(waargenome) frekwensie (toonhoogte), as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer/luisteraar. (2) 10.2 EQUAL TO /GELYK AAN  The driver of the taxi is at rest relative to the source of sound. Die bestuurder van die taxi beweeg nie relatief tot van die bron van klank nie. OR/OF No relative motion between the taxi driver and taxi. Geen relatiewe beweging tussen bestuurder en taxi nie. (2) 10.3 OPTION/OPSIE 1: OPTION/OPSIE 2:   340 340 383  =340+ v  400  418,5  =340 - v  400  s s vs = 15,09 m.s-1 vs = 15,03 m.s-1 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Physical Sciences P1/Fisiese Wetenskappe V1 16 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne 10.4 ANY TWO  • It is used (in flow meters) in medical science to measure the speed and direction (velocity) of blood flow. • movement of the heart of a foetus. • To find the rate of blood flow (Doppler scanning) • To see the unborn child (Ultra sound scanning) • To hear the heart of a foetus (Ultra sound scanning) • It is used in medical sonography to generate images (and sounds) of flowing blood. • To detect blood clotting (Doppler ultrasound test) ENIGE TWEE • Dit word (in vloeimeters) gebruik in mediese wetenskap om die volgende te meet: • Die spoed en rigting (snelheid) van bloedvloei. • Beweging van 'n fetus se hart • Om die tempo van bloedvloei te meet (Doppler skandering) • Om 'n ongebore baba te sien (Ultraklankskandering) • Om die hart van 'n fetus te hoor (Ultraklankskandering) • Dit word gebruik in mediese sonografie om beelde (en klanke) te vorm van vloeiende bloed. • Om vorming van bloedklonte op te spoor. (Doppler- ultraklamktoets) (2) 10.5 Doppler effect/Doppler effek  (1) 10.6 DIAGRAM 2  Star Y’s frequency is lower than star X/Ster Y se frekwensie is laer as Ster X (3) Star Y is redshifted/Ster Y ervaar rooiverskuiwing  [15] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou ### Page 17 Physical Sciences P1/Fisiese Wetenskappe V1 NW/June/Junie 2024 Grade 12 – Marking Guidelines/Nasienriglyne SUBJECT: Physical Sciences P1 ASSESSMENT TASK: GRADE 12 - June exams QUESTION ANALYSIS GRID 2024 Cognitive Levels Topic Difficulty Levels Momentum&Impulse Vertical Projectile Work, Energy and Newton's Laws Doppler effect TOTAL Moderate Difficult QUESTION Mark 1 2 3 4 Easy motion Power 1.1 2 2 2 2 2 1.2 2 2 2 2 2 1.3 2 2 2 2 2 1.4 2 2 2 2 2 1.5 2 2 2 2 2 1.6 2 2 2 2 2 1.7 2 2 2 2 2 1.8 2 2 2 2 2 1.9 2 2 2 2 2 1.10 2 2 2 2 2 Ques 1 20 2 10 8 0 4 4 2 4 6 20 10 6 4 2.1 2 2 2 2 2 2.2 4 4 4 4 4 2.3.1 3 3 3 3 3 2.3.2 5 5 5 5 5 2.4 2 2 2 2 2 2.5 1 1 1 1 1 Ques 2 17 2 7 3 5 17 0 0 0 0 17 5 5 7 3.1 2 2 2 2 2 3.2 4 4 4 4 4 3.3 3 3 3 3 3 3.4 2 2 2 2 2 Ques 3 11 2 4 5 0 11 0 0 0 0 11 2 6 3 4.1 2 2 2 2 2 4.2.1 4 4 4 4 4 4.2.2 3 3 3 3 3 4.2.3 1 1 1 1 1 Ques 4 10 2 1 7 0 0 10 0 0 0 10 3 7 0 5.1 2 2 2 2 2 5.2.1 4 4 4 4 4 5.2.2 2 2 2 2 2 5.2.3 4 4 4 4 4 5.3 3 3 3 3 3 Ques 5 15 2 5 8 0 0 15 0 0 0 15 6 9 0 6.1 2 2 2 2 2 6.2.1 3 3 3 3 3 6.2.2 4 4 4 4 4 6.2.3 3 3 3 3 3 6.3 3 3 3 3 3 Ques 6 15 2 3 10 0 0 0 15 0 0 15 2 6 7 7.1 1 1 1 1 1 7.2 3 3 3 3 3 7.3.1 4 4 4 4 4 7.3.2 5 5 5 5 5 7.3.3 4 4 4 4 4 7.4 3 3 3 3 3 Ques 7 20 1 3 16 0 0 0 20 0 0 20 4 3 13 8.1 2 2 2 2 2 8.2 1 1 1 1 1 8.3 3 3 3 3 3 8.4 5 5 5 5 5 Copyright reserved/Kopiereg voorbehou 8.5 3 3 3 3 3 Ques 8 14 2 4 3 5 0 0 0 14 0 14 8 6 0 9.1 2 2 2 2 2 ### Page 18 Ques 6 15 2 3 10 0 0 0 15 0 0 15 2 6 7 7.1 1 1 1 1 1 7.2 3 3 3 3 3 7.3.1 4 4 4 4 4 Physical 7.3.2 Sciences 5 P1/Fisiese Wetenskappe 5 V1 5 NW/June/Junie 5 2024 5 7.3.3 4 Grade 412 – Cognitive grid/Kognitiewe 4 tabel 4 4 7.4 3 3 3 3 3 Ques 7 20 1 3 16 0 0 0 20 0 0 20 4 3 13 8.1 2 2 2 2 2 8.2 1 1 1 1 1 8.3 3 3 3 3 3 8.4 5 5 5 5 5 8.5 3 3 3 3 3 Ques 8 14 2 4 3 5 0 0 0 14 0 14 8 6 0 9.1 2 2 2 2 2 9.2 3 3 3 3 3 9.3 5 5 5 5 5 9.4 3 3 3 3 3 Ques 9 13 2 6 0 5 0 0 0 13 0 13 2 6 5 10.1 2 2 2 2 2 10.2 2 2 2 2 2 10.3 5 5 5 5 5 10.4 2 2 2 2 2 10.5 1 1 1 1 1 10.4 3 3 3 3 3 Ques 10 15 3 7 5 0 0 0 0 0 15 15 5 5 5 SUMMARY QUES 1 20 2 10 8 0 4 4 2 4 6 20 10 6 4 QUES 2 17 2 7 3 5 17 0 0 0 0 17 5 5 7 QUES 3 11 2 4 5 0 11 0 0 0 0 11 2 6 3 QUES 4 10 2 1 7 0 0 10 0 0 0 10 3 7 0 QUES 5 15 2 5 8 0 0 15 0 0 0 15 6 9 0 QUES 6 15 2 3 10 0 0 0 15 0 0 15 2 6 7 QUES 7 20 1 3 16 0 0 0 20 0 0 20 4 3 13 QUES 8 14 2 4 3 5 0 0 0 14 0 14 8 6 0 QUES 9 13 2 6 0 5 0 0 0 13 0 13 2 6 5 QUES 10 15 3 7 5 0 0 0 0 0 15 15 5 5 5 Total marks 150 20 50 65 15 32 29 37 31 21 150 47 59 44 Norm marks 150 22,5 45 67,5 15 32 30 32 32 24 150 45 60 45 Total % 14,6% 30,6% 44,6% 10,0% Norm % 100 15 30 45 10 100 30 40 30 Copyright reserved/Kopiereg voorbehou ## Related documents - [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 - [Physical Sciences Physics P1 MEMO May June 2024 KZN](https://www.hlayiso.com/library/physical-sciences-physics-p1-memo-may-june-2024-kzn/49134363-7da2-4c98-ab54-c6d186d63bdc) — Physical Sciences · Memo · 2024 - [2024 MCED Sept Gr 12 P.Sciences P1 MEMO](https://www.hlayiso.com/library/2024-mced-sept-gr-12-p-sciences-p1-memo/10a9100b-159b-46b7-bf75-b22879248ca6) — Physical Sciences · Memo · 2024 - [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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