Physical Sciences Grade 12 Free State March Test 2025 Memo

education Department of Education FREE STATE PROVINCE CONTROL TEST/ KONTROLETOETS GRADE/GRAAD 12 PHYSICAL SCIENCES MARCH 2025 CONTROLLED TEST MARKING GUIDELINE/ NASIENRIGLYN. This MARKING GUIDELINE consists of 15 pages/ Hierdie NASIENRIGLYN bestaan uit 15 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Marking Guideline/Nasienriglyn Grade/Graad 12 QUESTION 1 / VRAAG 1 1.1 12 1.3 1.4 1.5 1.6 1.7 1.8 1.8 1.10 Copyright reserved/Kopiereg voorbehou Byv Cvv Dvv Dvv Bvv Avy Byv Cvv Cv Avy FS/March/Maart 2025 [20] Please turn over/Blaai asseblief' om

FS/March/Maart 2025 Marking Guideline/Nasienriglyn Grade/Graad 12 QUESTION 2 / VRAAG 2 2.1. When a net force acts on an object, the object will accelerate in the direction of the force, at an acceleration directly proportional to the force and inversely proportional to the mass of the object.v Vv 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. ¥ N 22 Tv fv w v ACCEPTED LABELS/AANVAARBARE BYSKRIFTE w | F, /weight/gravitational force/ 39,2 N/ F iriction!/F¢/friction/fx/ wrywingskrag f N | Fy/Frormai/Normal force/Normaalkrag/Normaal T F;/Tension/Spanning NOTE: - 1mark awarded for label and arrow. - Deduct 1 mark ONCE for any additional forces. - Donot consider the length of the forces. LET WEL: - 1 punt toegeken vir etiket en pyl. - Trek 1 punt EEN keer af vir enige bykomende kragte. - Moenie die lengte van die kragte in ag neem nie. (2) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

FS/March/Maart 2025 Marking Guideline/Nasienriglyn Grade/Graad 12 2.3 Toolbox B/ Gereedskapkis B Fnet = ma —— —... T+(-f,) =ma T- KN = ma T- 0,2x4x9.8v = 4x5,852V T = 31,248 N | ——————— Anyone/ Enige eenv Toolbox A/ Gereedskapkis A Fnet = ma ————_—— F + (-T)+ (- fk) =ma F-T—pxN =ma (6) F - 31,248 — 0,3x6x9.8Y = 6x5,852V F=84NV 2.4 Force of toolbox A on earth. v v Krag van gereedskapkis A op aarde (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

FS/March/Maart 2025 Marking Guideline/Nasienriglyn Grade/Graad 12 QUESTION 3 / VRAAG 3 3.1 3.2.2 3.2.3 3.3 Momentum is the product of an object's mass and its velocity. vv Momentum is die produk van die massa en snelheid van 'n voorwerp. (2) (2 or/of 0) West / Westwards Vv ¥ Wes / Weswaarts (2) 0 (N)/Zero ¥ (1) Velocity/momentum is constant between 0 s and 10 s¥ Change in momentum/Ap = 0 ¥ «According to Fret = 2 Yi Fiat =ON Snelheid/momentum is konstant tussen 0s en 10s Verandering in momentum/Ap = 0 -.Volgens Fnet =e Fnet=0N ru OR/OF Velocity is constant between 0 s and 10 s¥ Change in velocity/Av = 0 OR Acceleration/a = 0 ¥ «According Fret = ma ¥ ,Fnet = ON Snelheid is konstant tussen 0s en 10s Verandering in snelheid/Av = 0 OF Versnelling/a = 0 -.Volgens Fnet = ma ,Fnet = 0 N OR/OF Gradient of graph between 0 s and 10s =0 ¥ Gradient of graph = 2 v :. Accoding to Fret = 2 Y , Fret= 0 Gradiént van grafiek tussen 0 s en 10s =0 Gradiént van grafiek = 2 -Volgens Fnet =2, Fnet = 0 N (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

FS/March/Maart 2025 Marking Guideline/Nasienriglyn Grade/Graad 12 3.4.1 3.4.2 3.4.3 Fret = Ap Fret = pr - pi Fret (5)¥_ = 40000 - (- 60000) v Fret = 20000 N v Any one/Enige een Fret = ma 20000 = m(4)v m = 5000 kg¥ (5) POSITIVE MARKING FROM QUESTION 3.4.1 POSITIEWE NASIEN VANAF VRAAG 3.4.1 p=mvY (- 60000) = (5000)v~ v=-12ms" v=12 ms! West/Westward/Wes/Weswaartv (3) POSITIVE MARKING FROM QUESTION 3.4.1 and 3.4.2. POSITIEWE NASIEN VANAF VRAAG 3.4.1 and 3.4.2. Ex = smv? v ¥ Ex = SMarucks Verve? od FM can Vicar)? © Ex = 3(5000)(12)? + =(2000)(20)?v > Exi = 360 000 + 400 000 > Exi = 760 000 J > Ee #D Exw « Inelastic / onefastiesY (4) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

QUESTION 4 / VRAAG 4 4.1. NOv NEE FS/March/Maart 2025 Gravitational force is NOT the only force acting on the hot-air ballon/ Acceleration of the ballon is equal to zero/ Net force acting on the hot-air ballon is equal to zero.” (2) Gravitasiekrag is NIE die enigste krag wat op die warmlugballon inwerk nie/Versnelling van die ballon is gelyk aan nul/ Netto krag wat op die warmlugballon inwerk is gelyk aan nul. 4.2 12m-s‘VupwardsY 12 m-s" opwaarts (2) 4.3 POSITIVE MARKING FROM QUESTION 4.2 POSITIEWE NASIEN VANAF VRAAG 4.2. OPTION/OPSIE 1 Up/Op +ve Vy = VitaAt Y 0 =12+(-9,8)AtY At=1,22s Vv OPTION/OPSIE 2 Up/Op -ve Vy = VitaAt Y 0 =-12+(9,8)AtY At=1,22 s ¥ (3) 4.4 POSITIVE MARKING FROM QUESTION 4.2 and 4.3. POSITIEWE NASIEN VANAF VRAAG 4.2 and 4.3. OPTION/OPSIE 1 Up/Op +ve Max. height (camera)/Maks. hoogte (kamera) Ay = viAtt Laat? v = 12(1,22)+2(-9.8)(1,227) v =7,35m Height (hot-air balloon) after 1,22 s Hoogte (warmlugballon) na 1,22 s Ay = vat + Saat? = 12(1,22) +3.(0)(1,227) v = 14,64 m :. Distance/Afstand = 14,64 —7,35 v =7,29mv Copyright reserved/Kopiereg voorbehou OPTION/OPSIE 2 Up/Op -ve Max. height (camera)/Maks. hoogte (kamera) Ay = viAtt Laat? v = -12(1,22)+4 (9.8)(1,227) v =-7,35m Height (hot-air balloon) after 1,22 s Hoogte (warmlugballon) na 1,22 s Ay = viAt + Saat? = 4 2 = -12(1,22) +5(0)(1,22°) ¥ = -14,64 m -. Distance/Afsiand = -14,64 — (-7,35) v =-7,29m =7,29mv Please turn over/Blaai asseblief om

March/Maart 2025 OPTION/OPSIE 3 Up/Op +ve Max. height (camera)/Maks. hoogte (kamera) v? = ve+2aAy ¥ 0? = 127+2(-9,8)Ay ¥ Ay =7,35m Height (hot-air balloon) after 1,22 s Hoogte (warmlugballon) na 1,22 s Ay=vjAt+ saat? = 12(1,22) +3(0)(1,22) ¥ OPTION/OPSIE 4 Up/Op -ve Max. height (camera)/Maks. hoogte (kamera) v? = ve+2ady ¥ 0* = -127+2(9,8)Ay V Ay =7,35m Height (hot-air balloon) after 1,22 s Hoogte (warmlugballon) na 1,22 s Ay = viAt + Saat? = -12(1,22) +3.(0)(1,22") v Height (hot-air balloon) after 1,22 s Hoogte (warmlugballon) na 1,22s Ay = vat + Saat? = 12(1,22) +4.(0)(1,227) v =14,64m =-14,64m -. Distance/Afstand = 14,64 —7,35 v «. Distance/Afstand = -14,64 — (-7,35) ¥ =7,29m¥ =-7,32m =7,29mv OPTION/OPSIE 5 OPTION/OPSIE 6 Up/Op +ve Up/Op -ve Max. height (camera)/Maks. hoogte || Max. height (camera)/Maks. hoogte (kamera) (kamera) ; a Ay = (“*)xat v Ay = ()xat v -12+0 Ay = (= )x1,22v Ay = (= )x1,22v Ay =7,32m Ay =-7,32m Height (hot-air balloon) after 1,22 s Hoogte (warmlugballon) na 1,22 s Ay = v,At + saat? “a 1 2 = -12(1,22) +5(0)(1,22 yy = 14,64m = -14,64m «. Distance/Afstand = 14,64 — 7,32 Vv :. Distance/Afstand = -14,64 — (-7,32) Vv =7a2m = -7,32m =7,32mv¥ Note: RANGE FINAL ANSWER (7,29 m to 7,35 m) Let wel: REEKS FINALE ANTWOORD (7,29 m tot 7,35 m) (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

4.5 March/Maart 2025 OPTION/OPSIE 1 Up/Op +ve Time taken by the camera to reach the ground/ Tyd wat die kamera neem om die grond te bereik Ay = viAt + saat v -100¥ = 12At+ 5(-9,8)At? v M=5,91s Time taken by the jogger to cover 15 m/ Tyd wat die joggie neem om 15m af te lé Ay=vAt+ Saat? 15¥=3At+ 5 (O)Ar ¥ At=5s Yes, the jogger will catch the camera Vv Ja, die joggie sal die kamera vang OPTION/OPSIE 2 Up/Op -ve Time taken by the camera to reach the ground/ Tyd wat die kamera neem om die grond te bereik Ay = viAt+ Saat? v 100¥ =-12At + 5(9,8)At? v At =5,91s Time taken by the jogger to cover 15 m/ Tyd wat die joggie neem om 15m af te lé Ay=vjAtt+ saat? 15v= 3at + 2 (Oat? v At=5s Yes, the jogger will catch the camera ¥ Ja, die joggie sal die kamera vang Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

March/Maart 2025 OPTION/OPSIE 3: POSITIVE MARKING FROM 4.3 AND 4.4 POSITIEWE NASIEN VANAF VRAAG 4.3 AND 4.4. Up/Op +ve Time taken by the camera to reach the ground from Max. height/ Tyd wat die kamera neem om die grond vanaf die Maksimum hoogte te bereik: 1 2 Ay = vjAt + past -(100+7,35)¥ = OAt+ 5(-9,8)At? v At =4,681s At (total)= 4,681+ 1,22 =5,90s Time taken by the jogger to cover 15 m/ Tyd wat die joggie neem om 15m af te lé Ay = vjAt + sary 15v= Sat + 2(O)at’ ¥ At=5s Yes, the jogger will catch the camera ¥ Ja, die joggie sal die kamera vang OPTION/OPSIE 4: POSITIVE MARKING FROM 4.3 AND 4.4 POSITIEWE NASIEN VANAF VRAAG 4.3 AND 4.4. Up/Op -ve Time taken by the camera to reach the ground from the Max. height/ Tyd wat die kamera neem om die grond vanaf die Maksimum hoogte te bereik: 1 2 Ay = vjAt + pant (100+7,35)¥ =OAt+ +(9,8)at? v At = 4,681 s At (total)= 4,681+ 1,22 =5,90s Time taken by the jogger to cover 15 m/ Tyd wat die joggie neem om 15 m af te lé Ay = viAt+ saary 15¥= Bat + 1 (O)AP v At=5s Yes, the jogger will catch the camera ¥ Ja, die joggie sal die kamera vang (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

FS/March/Maart 2025 Marking Guideline/Nasienriglyn Grade/Graad 12 4.6 POSITIVE MARKING FROM QUESTION 4.3 POSITIEWE NASIEN VANAF VRAAG 4.3 OPTION/OPSIE 1 Up/Op +ve a v(m:s") 1,22 t s) OPTION/OPSIE 2 Up/Op -ve 7 v (ms) 4,22 t(s) ¥ Criteria for graph/ Kriteria vir grafiek Marks Graph starts at/ Grafiek begin by v = 12 m-s" att=0s. v Maximum height reached/Maksimum hoogte bereik (v = 0 m-s‘') at//by t | ¥ =1,22s. v Correct shape as shown: straight line with gradient and section of the graph after max height longer than before max height./ Korrekte vorm soos getoon: reguit lyn met gradiént en snit van die grafiek na maksimum hoogte langer as voor maksimum hoogte. Copyright reserved/Kopiereg voorbehou (3) [21] Please turn over/Blaai asseblief om

QUESTION 5 / VRAAG 5 5.1.1 Wv 5.1.2) 27 5.1.3 V¥ 5.2 Hydroxyl (group) ~ Hidroksiel (groep) 5.3.1 Propan-1-ol/ 1-propanol. ¥ ~ (2 or 0) Propaan-1-of 5.3.2 6-bromo-3-ethyl-2-methylheptane/ 6-bromo-3-etiel-2-metielheptaan Marking criteria: e Correct stemi.e. heptane. ¥ e Substituents (bromo, ethyl and methyl) correctly identified. v e IUPAC name completely correct including numbering, sequence, hyphens and commas. ¥ Nasienkriteria: e Korrekte stam d.i. heptaan. ~ e Substituente (bromo, etiel en metiel) korrek geidentifiseer. “ e /UPAC-naam heeltemal korrek insluitende nommering, volgorde, koppeltekens en kommas. “ H H H | Cc Cc C oTH H | H H—C—H ae H Marking criteria: e Correct functional group (triple bond) drawn. “ e Whole structure correct.” Nasienkriteria: e Korrekte funksionele groep (drievoudige binding) getrek e Hele struktuur korrek (1) (1) (1) (1) (2) (3) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

FS/March/Maart 2025 Marking Guideline/Nasienriglyn Grade/Graad 12 5.5.1 CaHanO2 ¥ (1) 5.5.2 CHsCH2CH2CH2CH2COOH OR CH3(CH2)aCOOH OR CHs(CH2)s— FO OR CH3CH2CH2CH2CH2 —C——-OH j j Marking criteria: Nasienkriteria: e Functional group for e Funksionele groep van carboxylic acids is correct karboksielsuur is korrek (carboxyl group) “” (karboksiel groep) ¥ e Whole structure is correct e Hele struktuur is korrek (carbon with 4 bonds, all H (koelstof met 4 bindings, al H- atoms shown) “ atome word getoon) “ (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

FS/March/Maart 2025 Marking Guideline/Nasienriglyn Grade/Graad 12 QUESTION 6 / VRAAG 6 6.1 6.2 6.3 6.4 6.5 The temperature at which the solid and liquid phases of a substance are at equilibrium. v ¥ Die temperatuur waarby die vaste- en vioeistoffases van 'n stof in ewewig is. (2) (Type of) Functional group/Type of Intermolecular forces/ Homologous series.” (1) (Tipe van) Funksionele groep/Tipe intermolekulére kragte/ Homoloé reeks. SECONDARY ’Y, carbon bonded to hydroxyl group (OH’) is bonded to two other carbon atoms. “” (2) SEKONDER, koolstof gebind aan hidroksielgroep (OH-) is gebind aan twee ander kooistofatome Liquid. (1) Vioeistof Ahas london forces only”, D has london forces and dipole-dipole forces”, B has london forces, dipole-dipole forces and hydrogen bonds.” FROM Ato D to B: - Strength of intermolecular forces increases.v - Energy required to overcome the strength on intermolecular forces increases.v Therefore the melting point of D is greater than that of A but less than that of B. A het slegs Londense magte, D het Londense kragte en dipool-dipoolkragte, B het Londense kragte, dipool-dipoolkragte en waterstofbindings. VAN A tot Dna B: - Die sterkte van intermolekulére kragte neem toe. - Energie wat nodig is om die sterkte op intermolekulére kragte te oorkom, neem toe. Daarom is die smeltpunt van D groter as dié van A maar minder as dié van B. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief'om

FS/March/Maart 2025 Marking Guideline/Nasienriglyn Grade/Graad 12 OR/OF A has london forces only”, D has london forces and dipole-dipole forces”, B has london forces, dipole-dipole forces and hydrogen bonds.” FROM B to DtoA: - Strength of intermolecular forces decreases. ¥ - Energy required to overcome the strength on intermolecular forces decreases.v Therefore the melting point of D is greater than that of A but less than that of B. A het slegs Londense magte, D het Londense kragte en dipool-dipoolkragte, B het Londense kragte, dipool-dipoolkragte en waterstofbindings. VAN B tot Dna A: - Die sterkte van intermolekulére kragte neem af. - Energie wat benodig word om die krag op intermolekulére kragte te oorkom, neem af. Daarom is die smeltpunt van D groter as dié van A maar minder as dié vanB (5) [11] Grand Total: 100 marks Groottotaal: 100 punte Copyright reserved/Kopiereg voorbehou