Physical Sciences Grade 12 P1 Western Cape Mock 2025 Memo

Western Cape Government FOR YOU Education | CAPE WINELANDS EDUCATION DISTRICT | PHYSICAL SCIENCES PAPER 1 MARKING GUIDELINE GRADE 12 fi N COMMON TRIAL EXAMINATION SEPTEMBER 2025 MARKS: 150 TIME: 3 hours This exam paper consists of 14 pages.

QUESTION 1/ VRAAG 1 4.1 1.2 1.3 14 1.5 1.6 17 1.8 1.9 4.10 D D vv vv vv vv Vv vv Vv

QUESTION 2/ VRAAG 2 24 2.2 2.3 When a net force acts on an object, the object will accelerate in the direction of the force and the acceleration is directl roportional_to_the force and_inversel proportional to the mass of the object. vv (Deduct one mark if any of the underlined phrases are omitted.) 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. (Trek een punt af indien enige van die onderstreepte frases weggelaat word.) (2) Note: 1 mark for each arrow WITH its label, deduct 1 mark for every extra force included. Let wel: 1 punt vir elke pyl MET sy byskrif, trek 1 punt af vir elke ekstra krag wat ingesluit is. AAA (5) FOS pW = [,mgcos8, Vv mark for either (0, 2)(2)(9,8)(cos40°) v = 3,0Ndownthe slope v (3)

2.4 POSITIVE MARKING FROM QUESTION 2.3/POSITIEWE NASIEN VANAF VRAAG 2.3: 2a: F =ma v net TF, — Fg, = ma T — F, — mgsin® = ma 5kg: F =ma net F,—F,—Tcos® = ma (70 — 17,15 — Tcos20°)¥ = 5a (T — 3 — (2x9, 8xsin40’))¥ = 2a¥ (forboth2a& | 52,85 — Tcos20° = 5a ...eq2 5a) T= 2@ + 15,599 ..eq 1 Substitute equation 1 into equation 2: 52,85 — (2a + 15,599)cos20° = Sav 38,19 = (5 + 2cos20°)a a=5,55ms Vv (6) [16] 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. WY (2 or 0) 'n Voorwerp waaraan 'n beginsnelheid gegee is en wat dan slegs onder die invioed van die gravitasiekrag beweeg. (2 of 0) 3.2 Downwards as positive Upwards as positive Ay = v At + tadt® Vv Ay = vdt + tant’ Vv 2 1 2 1,8 =v,0,5)+>Q9,8)0,5)¥ - 18 =v(0,5)+>(- 9,8)(0,5)v =1 ~ val 15 ms v v =~ 1,15 mrs ' sspeed = 1,15 ms! Vv 3.3 ve = ve + 2aAy Vv vy = ve + 2ady Vv 0 =v + 209,8)(- 0,9) v O=v + 2— 9,8)(0,9)v v.= 17,64 v.= 17,64 i i

v= 4,2 ms ‘upwards Vv v= 4,2 m-s ' upwards ¥ (3) 3.4 v=, t adt Vv v =v, + adt Vv 0 =— 4,2 + 9,84 v +MARKING 0=424¢ 9,8)At Vv +MARKING At = 0,435 At = 0,43s t=0,5+ 200,43) v t=0,5 + 200,43) v t=1,36s v (4) t= 1,36s v 3.5 r" 4 4,2 a ‘ : = pits : / ty) 228 ™ op/ 1,36 : -4,2|-------- 4 v Criteria for marking/ Nasienkriteria: e Lines are correctly drawn (parallel)/ Lyne is korrek getrek (parallel) W e Initial velocity correctly indicated/ Aanvangssnelheid korrek aangedui “ e Velocity of the ball as it leaves the ground./ Snelheid van die bal soos dit die grond verlaat. ¥ e Time when the ball strikes the ground after the first bounce (0,5 s)/ Tyd wanneer die bal die grond tref na die eerste bons (0,5 s).” (4) [16]

QUESTION 4/ VRAAG 4 4.1 4.2 The total linear momentum of an isolated systemW_remains constant / is conserved Vv Die totale lineére momentum in 'n geisoleerde sisteem “bly konstant / behoue. ¥ OR Accept: In an isolated system, ¥ the total linear momentum before collision is equal to the total linear momentum after collision Vv Aanvaar: In 'n geisoleerde stelsel, Y is die totale lineére momentum voor botsing gelyk aan die totale lineére momentum na botsing. ¥ (2) Right as +: Z Pinitiasy = Z Prinay ~ OR MyVin + M2Vi2 = (M, + My)v_ (5)(4) + (2)-1)v =(6 +2) (vw “V,= 2,57 ms" Vv Left as +: E Piinitiay = Z Pinar v MiVin + MoVin = (M, + Ma)Vy (5)(-4) + (2)(1)v =(+2)(y)¥ OV =-2,57ms'¥ If no subscripts:mv + mv = mv + mv max. 3/4 4.3 POSITIVE MARKING FROM 4.2/ POSITIEF NASIEN VANAF 4.2 OPTION 1: F At =mAv vW or F At = Ap net net F (0, 3)¥ = (7) — 2,57)¥ F = —59,97N net magnitude of Fe 59,97 NW OPTION 2: 0, =, + adt = 2,57 + a(0,3)v a 59,99 N magnitude of Eni = 59,99NY Range/gebied: 59,97 N to/tot 60,20 N (4)

QUESTION 5/ VRAAG 5 (4) [10] 5.1.1. The total mechanical energy in an isolated systemv is conserved/remains constant. ¥ Die totale meganiese energie in 'n geisoleerde stelsel¥ word behoue/bly konstant. Y (2) 5.1.2 OPTION 1: E wcinitiat =E al M(final) mv,” + m(9,8)(1,3) ¥ = m(9,8)(5)¥ +0 vy = 852ms! upwards/opwaarts Vv i OPTION 2: net m(9, 8)(3, 7)cos180° ¥ = 0 — mv’ v v = 852m. s- upwards/opwaarts ¥ QPTION:3: Wre = AE, + AB, = mv? - mv) + (mgh;, - mgh)) =$m(0) - S mv,” v+ m(9,8)(5) - M(9,8)(1,3)¥ B= 8,52 ms upwards/opwaarts W i 5.1.3 DECREASE / AFNEEM v Friction will remove energy from the system.v / Wrywing sal energie uit die stelsel verwyder. OR/OF Some of the object's kinetic energy would be converted into heat and sound due to friction, leaving less energy available to convert into gravitational potential energy./ Van die voorwerp se kinetiese energie sal as gevolg van wrywing in hitte en klank omgeskakel word, wat minder energie beskikbaar laat om in gravitasiepotensiéle energie om te skakel. OR/OF

Friction opposes motion, causing the object to lose energy more quickly as it moves upward. / Wrywing staan beweging teé, wat veroorsaak dat die voorwerp vinniger energie verloor soos dit opwaarts beweeg. OR/OF With less energy available to overcome gravity, the object would not rise as high, resulting in a lower maximum height. / Met minder energie beskikbaar om gravitasiekrag te oorkom, sal die voorwerp nie so hoog styg nie, wat 'n laer maksimum hoogte tot gevolg sal hé. OR/OF Increased F.. downwards/increased acceleration downwards/loses velocity faster, resulting in a lower maximum height./ Fret afwaarts neem toe/versnelling afwaarts neem toe/verloor snelheid vinniger, wat lei tot 'n laer maksimum hoogte. (2) 5.2.1 Non-conservative/ Nie-konserwatiewe krag ¥ It is a force for which the work done in moving an object between two points depends on the path taken. ¥ Dit is 'n krag waarvoor die arbeid verrig om 'n voorwerp tussen twee punte te beweeg, afhanklik is van die roete wat gevolg word. (2) 5.2.2 W = AE, + AE LV ne K P (4, 493)" — 4, 4)(0,6)) ¥ + (1,4)(9,8)0 - 1,5) ¥ friction friction = — 14,532] friction = F -rictionA*C0S® — 14,532 = F__. (1, 8)¥cos180° v friction _ = 807NY friction w= AE.V net

adhe _ dopa We +W,, =ymv, ym, F Axcos180 + mgAxsinBbcos0 = zm, = mv; = B07NV friction > uv, = v + 2aax (3)W = (0,6) + 2a(1,8) ¥ 2 a=2,4ms_ F =mav net F = A924av F = 3,36N net Ess = H, parallel ~ F, F =F sind —F net g f 3,36 = (1,49, 8)GSv - F, 3,36 = 11,43 - F, F,=8,07N ¥ (6) [16] QUESTION 6/ VRAAG 6 6.1 The apparent change in frequency (or pitch)“ detected by a listener because the source and the listener have different velocities relative to the medium of propagation. ¥ Die skynbare verandering in frekwensie (of toonhoogte) wat deur 'n luisteraar waargeneem word omdat die bron en die luisteraar verskillende snelhede het relatief tot die voortplantingsmedium. OR/OF An apparent change in detected frequencyY as a result of the relative motion between asource and a listener. ¥ 'n Skynbare verandering in waargenome frekwensie as gevolg van die relatiewe beweging tussen 'n bron en 'n luisteraar. (2) 62 v=fa Vv 340 = f(0.013) v f = 26153,85Hz Vv (3)

viv, 63 f==f v vtv 7s 340 f= (261 53,85)¥ positive marking from Q6.2 f, = (340 + v_, )(76, 92308824) vtv, f, = viv, f 340 31500¥ = [ai |(s40 + v,,.)(76, 92308824) ¥ 115600+340v car 340—v coi i} 409, 4999398 5 139229,9795 — 409, 4999398v = 115600 + 340v_ 23629, 97953 = 749, 49993980 v= 31,53ms VY OR/OF v= 113,51km.h \_ (Note: 100km.h | = 27,78m.s ’) .. The car is speeding./Die kar oorskry die spoedgrens Varese) (6) [11] QUESTION 7/ VRAAG 7 7.1.1. 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 Die grootte van die elektrostatiese_kraq wat deur een puntlading op 'n ander puntlading uitgeoefen word, is direk eweredig aan die produk van die grootte van die ladings en omgekeerd eweredi n die kwadraat van die afstand tussen hull Deduct one mark if any of the underlined phrases are omitted./Trek een punt af indien enige van die onderstreepte frases weggelaat word. (2) 71.2 F=-G Vv If Q, is written as a negative " 9. ~9 charge, no mark for final answer. -7 — (9x10°)(3,8 x10 )Q ; ; ‘ 8,54x 10 © = ————_—+ Vv Indien Q, as 'n negatiewe lading (0,4) geskryf is, geen punt vir die finale antwoord nie Q, =4x10° Cv (3) 10

7.2.1 The electric field at a point is the electrostatic force experienced per unit positive charge placed at that point. vw Die elektriese veld by 'n punt is die elektrostatiese krag wat per eenheidspositiewe- lading wat by daardie punt geplaas is, ondervind word. (2 of 0) (2) Te Right as positive Left as positive kQ, kQ, kQ, E k 2 = +—+v E et = 2 Ba 2 ov net Be e 9. —6, 9. -6. _ (oxdo’yeext0™ Hit (9x10°}(3x10 7°) =- (fe 16x10 yy fee XSx10 J (ay ay ™ " Vv v = 3375 — 27000 = — 3375 + 27000 —1 a = 23 625 N.C to the left/ na links v = 23625N.C to the left/ na links¥ 7.2.3 POSITIVE MARKING FROM Q7.2.2/POSITIEF NASIEN VANAF V7.2.2 OPTION 1/OPSIE 1 E= = Vv OR/OF F = Eq F = (1,6 x 10 )(23 625)¥ = 3,78 x 10 °N F a= — — 3.78 x10" 1,67x10 a=2,26x10'ms Vv (4) OPTION 2/OPSIE 2 = I | #2, net ? r F = (9x10°)(3x10 (6x10) _(9x10°)(6x10*)(,6x10-") net ay’ ay v F =3,78x10 N net F =ma net 3,78 x10 = (1,67 x 10 Wa a = 2,26x10'ms Vv Marking Criteria: e Correct formula Calculate force Substitute mass in F = ma Correct answer Nasienkriteria: e Korrekte formule e Bereken krag e Vervang massa in F = ma e Korrekte antwoord 7.2.4 Greater than/Groter as v (4) (1) [16] A:

QUESTION 8/ VRAAG 8 8.1.1 8.2.1 8.2.2 The potential difference across a conductor is directly proportional to the current Y in the conductor at constant temperature. Y Deduct one mark if any of the underlined phrases are omitted. Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur. Trek een punt af indien enige van die onderstreepte frases weggelaat word. (2) Diagram 1: Diagram 2: R=ty Vi =emf —V, yo é =6~4 wid =2Vv = 3.0 v, i= 7 = 4 Vv = 0,67A v Ry = a 4 =a 4 (6) 12,6V¥ (1) POSITIVE MARKING FROM Q8.2.1/POSITIEF NASIEN VANAF V8. 2.1 OPTION 1/OPSIE 1 e=1(R+ nV 12,6¥ =1(5 + 0,08)” 1 = 2,484 = IR ext v= (2,48)(5)¥ V=12,4Vv ext 12

V R R =~ xV iv OR/OF V, = z x 1 Ri+R, total Ri+R, total Sv Se OS: Ve aa lbn 5+0,0B¥ 12, 6W = 12,4Vv (5) headlamps 8.2.3 Decrease (brightness) /Verminder (helderheid)Y¥ Rexternar decrease/ Roexsteme VerminderY |, increase/ |, vermeerderY Vinternal resistance (Of Vig.) increase OR V,,, decrease (emf remain constant)// Vinterne weerstand (OF Vyertora) Vermeerder OF V,.. verminder (as emk konstant bly) P (Power) decrease (according to P = V7/R )¥ / P (Drywing) verminder (volgens P = V7/R) (5) [19] 13

QUESTION 9/ VRAAG 9 9.1.1 AC/ Alternating current/WS/Wisselstroom YW (1) 9.1.2 DC/ Direct currentY / GS / Gelykstroom Due to the presence of the split ring/commutator. ¥ A.g.v. die teenwoordigheid van die splitring/kommutator. (2) 9.2.1 The AC generator has slip rings Y whilst a DC generator has a split ring/commutator. ¥ Die WS generator het sleepringe, terwyl die GS generator splitring/kommutator het. (2) 9.2.2 Increase in voltage and decrease in current. Y Verhoging in potensiaalverskil en verlaging in stroom. Loss in power/energy decreases. (P = I?R) ¥ Verlies aan drywing/energie verminder (P = (?R) (2) 93.1 01s ov (1) _ Vmax _ 84,8 4 _ 9.3.2 Vises 7a = 59,96V ¥ (2) w 933 Pp =m oy ave 40 = 38 cov Positive marking from Q9.3.2 R = 89,880 ¥ (3) [13] 14

QUESTION 10/ VRAAG 10 c 8 3x10 = aa Vv 284 x 10 (2,06 x 10 H2)w @) 10.2 It is the minimum energy that an electron in the metal needs to be emitted from the metal surfaceY ¥ Dit is die minimum energie benodig om 'n elektron uit die opperviak van 'n metaal vry te stel. Deduct one mark if any of the underlined phrases are omitted./Trek een punt af indien enige van die onderstreepte frases weggelaat word. (2) 10.3 POSITIVE MARKING FROM Q10.1/ POSITIEF NASIEN VANAF V10.1 E=W)+ E aie OR/OF hf = Wit 4mv’ Vv (Any one of the formulae) (6,63x10 )(1,06x10%)v = WW, + £(9,11x10"")(1,48x109 ¥ W. = 6,93 x 10° J v(Range/Gebied: 6,90 x 10°? J to/tot 6,93 x 10" J) :. The metal is zinc/ Die metaal is sink ¥ (5) 10.4.1 No effect /Geeneffek v (1) 10.4.2 No effect / Geeneffek v (1) 10.4.3 Increases / Toeneem Vv (1) [13] TOTAL/TOTAAL: 150 MARKS/PUNTE 15