Physical Sciences Grade 12 Paper 1 Limpopo Mock 2025 Memo

PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE PHYSICAL SCIENCES: PHYSICS (P1) PREPARATORY 2025 MARKING GUIDELINE/ MERK RIGLYNE MARKS: 150 These marking guidelines consist of 12 pages Copyright reserved please turn over

Physical Sciences/P1 QUESTION/VRAAG 1 11. Dv 1.2 BYY 1.3 AVY 140 AW 1.5 Dv¥v 16 Av 1.7 Cvv 1.8 AYY 19 BYY 1.10 BYY Copyright reserved LimpopoDoE/September 2025 (2 x 10 =20) Please turn over

Physical Sciences/P1 QUESTION/VRAAG 2 2.1 2:2 LimpopoDoE/September 2025 When a (non-zero) net force acts on an object, the object will accelerate in the direction of the (net) force at an acceleration that is directly proportional to the (net) force and inversely proportional to the mass of the object.vV Wanneer 'n (nie-nul) resulterende/netto krag op 'n voorwerp inwerk, versnel die voorwerp in die rigting van die krag teen 'n versnelling direk eweredig aan die (netto)krag en omgekeerd eweredig aan die massa van die voorwerp.v ¥ FerakesY% Fnv¥ fe ACCEPTED SYMBOLS/AANVAARDE SIMBOLE the body: -1 mark -1 mark if all arrowheads are omitted but correctly labelled. Fn Normal force/N /Fsurtace on truck Normaalkrag/ N/Fopperviak op trok fic Kinetic friction / friction /Frictional | Kinetiese force wrywing/wrywing/wrywingskrag Fg w, weight/Gravitational force/ Gewing/Gravitasiekrag/ Fearth on truck Faarde op die trok Feraxes | Force of Brakes/ Fa/Applied force | Fremme / Krag van remme/ F7/ _. Toegepaste krag Notes: Notas: e Mark awarded for label! and arrow e Punt toegeken vir pyl en e¢ Do not penalize for length of arrows pylpunt e Any additional force/s: -1 e Addisionele krag/te:-1 e If force(s) do not make contact with e As kragte nie kontak maak met kol nie: -1 e -7 punt as alle kragte teenwoordig is en benoem Copyright reserved maar geen pylpunte nie. (4) Please turn over

Physical Sciences/P1 2.3.1 2.3.2 2.4.1 2.4.2 Copyright reserved For == mgsin® (no mark for formula) = (40 000)(9,8)(sin 33° ¥ = 213 498,50 N ¥ (213 498,5017 N) W = FAxcos0 v = (10 000)(400)(cos180°) v = -4,0 x 108 J ¥ (- 4 000 000 J) POSTIVE MARKING FROM 2.3.1 The speed of the truck is calculated from the motion of A to B. The pre- amble clearly states that the brakes were still functional during this motion. Fnet = ma Fon — Ferakes ~ fy = ma mgsin@ — 10000 - pxmgcos® = ma 40000(9,8)(sin33°) — 10000 — (0,42)(40000)(9,8)cos33°“ = 40000aY OR 213 498,50 — 10000 — (0,42)(40000)(9,8)cos33 Vv =40000a_ Vv a = 1,64 m-s? (downhill) (Range: 1,63549 — 1,64) v? = v? + 2aAyv Carry over = (5,4)? + 2(1,64)(400) ¥ vi = 36,62 m-s* (downhill/down the incline) Y (Range: 36,5726-36.62) (af teen helling/afwaarts) The product of the resultant/net force acting on an object and the time the net force acts on the object. vv Die produk van die resulterende/netto krag wat op 'n voorwerp inwerk en die tyd wat die netto krag op die voorwerp inwerk VV LimpopoDoE/September 2025 (2) (3) (5) (2) Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 2.43 244 POSITIVE MARKING FROM QUESTION 2.4.1 The only horizontal force present, is the friction caused by the gravel. Therefore friction is seen as the net force. NOTE: ONLY THE IMPULSE-MOMENTUM THEORY METHOD IS ALLOWED (4) FnetAt = Ap ~ Fnet(2,5) ¥ = 40 000(0 - 36,62) v Fret = 585 161,71N (range: 585 161,71 — 585 920 N) v POSITIVE MARKING FROM QUESTIONS 2.4.1 , 2.4.3 OPTION 1 W = fAxcos68 ¥ = 585 161,71 (55) cosi80 ¥ = -3,22 x10" ¥ OPTION 2 (OPTIONS 2 and 3 Provides a different answer) Whe = AEp + AEk¥ :. Wi=0+ % (40 000)(0)? — % (40 000)(36,62)? V = - 2,675 x10’ N (Range: =2,675 x10’~ 2,68 x 10” N) ¥ OPTION 3 Whet = AE, ¥ = % (40 000)(0)* — % (40 000)(36,62)2V = - 2,675 x10’ N (Range: =2,675 x10’~ 2,68 x 10” N) ¥v (3) [25] Copyright reserved Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 3.1 3.2 3.3 3.4 QUESTION/ VRAAG 3 An object that has an initial velocity and moves under the influence of the gravitational force only. ¥¥ (2) ‘n Voorwerp met ‘n aanvangsnelheid wat slegs onder die invloed van die gravitasiekrag beweeg.v¥ v? = v? + 2aAy ¥ 0 = vj + 2(9,8)(-7) V v; = 11,71 m-s~t upward VY (3) POSITIVE MARKING FROM QUESTION 3.2 Ve = Vi tat v¥ vr = (-11,71) +9,8(2,83) ¥ (3) = 16,02 m-s downward v POSITIVE MARKING FROM QUESTION 3.3 NOTE: Candidate may not show calculation, but is credited if it was calculated and substituted 75 % of 16,02= 12,015 m-s'V Carry over mark — if initial value calculated (5) incorrectly, candidate gets mark Ve =Vi +adt for substituting correctly but not further. 14,938Y¥ =-12,015+9,8 -Atv At = 2,7495 sv (Range: 2,7495 — 2,75) Copyright reserved Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 3.5 POSITIVE MARKING 3.2,3.3,3.4 OPTION 1: AB: Ay = vjAt + SaAt? v = (-11,71)(2,83) + % (9,8)(2,83)? ¥ = 6,10 m (6,10431 m) BC: Ay = v,At+ 5 adt? = (-12,015)(2,749) + % (9,8)(2,749)? v| OPTION 2: AB v? =v? + 2ady v (16,02)? = (—11,71)* + 2(9,8)AyY Ay = 6,0977m BC v? =v} + 2ady ¥ (14,93)? = (—12,015)? + 2(9,8)Ay¥ Ay = 4,00m =4,00m OPTION 3 AB _ vety; Ayae = (“*)aev Ayan = eae) (2,83) Py, Ay = -6,09865 BC: Ayap = Co) At AYan = (ees) (2,749) v Ay = -4,015 v (5) f4a1 MARKING CRITERIA: e One mark for either formula present to calc Ay for AB or BC e Substitution to calc AB e Substitution to calc BC e Addition of AB + BC e Final answer Total height: a = 10,11 mv Copyright reserved Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 QUESTION/VRAAG 4 Afrikaans and English papers differ in mass of objects: Afrikaans = 85 kg English = 85 g NOTE: Mass conversions is imperative. Penalise when conversion is not done correctly. 4.1 Em=Ep+ Ex y Any one AFRIKAANS: =mgh + % mv? = (85)(9,8)(0,7) + = (85)(0)2 = (85 x 10°)(9,8)(0,7) + 5 (85 x 10°)(0)2¥ euanesemll “ienaiees = 83,1 J = 0,5831J ¥ (3) 4.2 POSTIVE MARKING FROM 4.1 OPTION 1 AFRIKAANS: _nal Emtop = Embottom Y anyone | 983,1= 0 +5 (85)v? = Ep +%mv2 v=3.70 ms1 0,5831 = 0 +5 (85x 10°)’ v = 3,70 m-s* rightY (3,7041 m-s") OPTION 2: Wret = AE, + AE, ¥ 0 = (Mghbottom -mghtop) + (% mv? - % mv?) 0 = [0-(85 x 10°°)(9,8)(0,7)] +[% (85 x 10°)v? -O] ¥ v= 3,70 ms right ¥ (3) 4.3. The total linear momentum of an isolated system remains constant/is conserved. VY Die totale linéere momentum van ‘n geisoleerde sisteem bly konstant/ bly behoue V¥ (2) Copyright reserved Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 44 POSITIVE MARKING 4.2 tpi = <Xper Yany one mva +mve=mva + mva (4) (85 x 10°)(3,7) ¥+0 = 0 +85 x 103 viv vi =3,7 m-s* Right ¥ AFRIKAANS: Zpi = =Ipr Y any one mva +Mve=mMva + MvB (85 )(3,7) V+0 = 0 +_(85)vi“ ve =3,7 m-s"right ¥ OPTION 2: Apa = —Apz ¥ [0 - (85 x 10°)(3,7)] = - [(85 x 10°)(va)-0] ¥ Vb = 3,7 m.s? Right” 4.5 0,7m V Same value as A started (1) [13] QUESTION/VRAAG 5 5.1. The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. Vv“ Die totale meganiese energie (som van gravitasie- potensiéle energie en kinetiese energie) in 'n geisoleerde sisteem bly konstant. VV (2) Copyright reserved Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 5.2 5.3.1 OPTION 1 Ema = Eup i y : An (Ep + Exa = (Ep + En ‘ (4410)(9,8)(23) + (2) (4410)(18)2V = (4410)(9,8)(37) + (-) (4410)v? V ve =7,04 ms! ¥ (7,04273 ms") (4) OPTION 2 Wye = AE, + AE, 0 = [(4410)(9,8)(37) — (4410)(9,8)(23)] + (5) (4410)(18)? — (6) (4410)v2|v v=7,04m.s" POSITIVE MARKING from 5.2 OPTION 1 Wac = AEp + OE, % 1 = (4410)(9,8)(1,5 — 37)¥ + (6) (4410) (25? — 7,042732)¥ (4) = - 265 482,1011 J v (- 265 397,33 J) OPTION 2 Calculate Emechanical at points B and C Emc = Ep + Ex Emc = =mv? +mgh v(Formula at either MC or MB) = 5 (4410 x 252) + (4410 x9,8x1,5) Vv = 1442 952 J Emp = 5 (4410 x7,047) + (4410 x 9,8 x 37) ¥ =1 708 349,328 J Wa = Emc — Emp = 1 442 952 — 1 708 349,328 = - 265 397,328 JV Copyright reserved Please turn over

Physical Sciences/P1 5.3.2 NO v 6.1 6.2 6.3 6.4 Friction is present / The external net force is not equal to zero There is a non-conservative force acting. Vv Neev Wrywing (is teenwoordig)/Die eksterne netto krag is nie nul nie Daar is ‘n nie-konservatiewe krag teenwoordig “ QUESTION/VRAAG 6 Doppler Effect Doppler Effek V fi, = ¢ =) fs Vv Vivs 2428,57V = (= ) 1000 v 0~-Vs5 Vs = 200,00 m-s! v (199,9999176 m-s*) DECREASESY The wavelengths are longer. ~ ACCEPT: Waves are stretched out/Wave crests are apart/wavelength increase “ a | VERLAAG Y Die golflengtes is langer ¥ AANVAAR::Golwe is uitgerek/Golffronte is verder uit mekaar ¥ Ultrasound Detection of blood clots Detection of heart problems Any ¥ Fetal blood flow Blood flow in the heart/organs Copyright reserved LimpopoDoE/September 2025 (2) [12] (1) (4) farther (2) Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 Ultraklank Bloedkionte (1) Hart probleme Fetale bloed vioei [8] Bioed vioei in hart/organe QUESTION/VRAAG 7 7.1 The electric field at a point is the electrostatic force experienced per unit positive charge placed at that point vv Die elektrostatiese krag wat per eenheid positiewe lading wat by daardie punt geplaas is, ondervind word. Vv (2) 724 8 = Fy q _ 3,5x 1076 Sf 6,2 x 10-9 (3) = 564,516 N+ C~* Right 7.2.2 Take to the right as positive: OPTION 1 _ kQe E= 2 _Use of formula v kQ_ kQ Substitution of any one ¥ Enet = 72 72 y — (OX10°)Qe _ (9x10°)(6,2x10-%) (x)? (2x)? SIMPLIFY BY MULTIPLYING WITH x* THROUGHOUT (9x10) (6,2x109) 0 = (9x10°)Qa - ri Qg = 155x107°C ¥ Copyright reserved Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 OPTION 2 kQr E= a Use of formula ¥ (4) Substitution of any one Ev atp Fate —— (9x10°)QR— (9x10°)(6,2x10~°) ~? (2x)? Qa = 1,55x10°CV NOTE: Consider to the left as positive also. 7.2.3 POSITIVE MARKING FROM 7.2.2 OPTION 1 ¥ kQ,Q2 Fa — (9x10°)(6,2x107°)(1,55x10"9 -v OX = r2 r= 0,157 —0,1572 3x = 0.1572 (Division with 3) x=0,052m v (0,0524 m) OPTION 2 kQ E = 7 — (9x10°)(1,5x10~? 7 (3x)? X =0,0157m (4) 7.3. POSITIVE MARKING FROM 7.2.2 RtoT v (NOTE: Wrong calculation in QUESTION 7.2.2 giving Qr > Qr: T to R) (1) Copyright reserved Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 7.4 8.1 8.2 8.3 8.4 POSITIVE MARKING FROM /.z.z (1,55x107) + (6,2x10~°)y me 2 (2) Qnew = 3,875 x10-°C V (3,88 x 10°C) [16] QUESTION/VRAAG 8 {t is the maximum energy provided by a battery per unit charge passing through it. VW Dit is die maksimum energie wat 'n battery lewer per eenheidslading wat daardeur vioei. (2) OPTION 1 OPTION 2 Vinternal = Ir ¥ £ = Vextemal + Vintemal V 12 —11,3 = 1(0,5 42 = 11,3 + (0,5) ¥ (3) OR 1=1,40AV 0,7 = 1(0,5)” 1=14AVY INCREASE ¥ TOENEEM (1) S2 causes a parallel circuit: Rtotai decreases ¥ ltotal increases since I « ; v OR: | is inversely proportional to R. Therefore, as Rtotai decrease, Itotal will increase. VV S2 veroorsaak dat die stroombaan verander in parallel skakeling: Riotaal neem af ~ Iotaai neem toe want I a = Y OF: (2) | is omgekeerd eweredig aan R. Daarom, wanneer Rootaai afneem, Sal Itotaai toeneem. V~ Copyright reserved Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 85 OPTION 1 e=1(R+r) 1 1 1 ¥Formula for Rp R, Ri Ry e= Vexternal +Ir Pp 4 s 1 i 1 12=11,10 +1(0,5) ¥ ie Ei, 617 x 3x I=18A Carry over x= 8,230 ¥ (8,2267 O mark¥ V, = IR Pp Pp OR 11,10 = 18R, , 1 _ Rat (R1+2+3) Rp = 6,170 (6,1667 O) Rp ~ R4xR, 1 _ Xx +3x% 7 6,17 3x? 3x = 24,6667 X = 8,2220V OPTION 2 Vinternat = Ir 0,9 = 1(0,5) v = arry y= 18 over mark¥ e=1(R+r) 12 = 1,8(RA0,5) R =6,1667 OPTION 3 Calculate Ra in the initial circuit, V=IRa ¥ 11,3 Y= 1(1,4) ¥ Ra = 8,070 ¥ Ro = Rav (5) Copyright reserved Please turn over

Physical Sciences/P1 8.6 POSITIVIE MARKING OPTION 1 V=IRY LimpopoDoE/September 2025 Rs = 24,68 (8,2267 x 3) 11,10 = 1246801) ) 7 1 = 0,4498 AY OPTION 3 V=IRVY 11,10 = (8,22) ¥ IRH = 1,3504 Trot = Tri + tra 1,8 = Ir1 + 1,3596 In = 0,45A ¥ QUESTION/VRAAG 9 OPTION 2 Ratio’s: 1. 1,8 xqVvY = 045 AV Resistors in series are potential dividers: 9.1. Mechanical (energy) to Electrical (energy) VY Meganiese (energie) na Elektriese (energie) 9.2 AC/Alternating It has slip rings ¥ WS/Wisselstroom ¥ Dit het sleep ringe V 93 ._1 ~ f me ll 50 =0,02sv Copyright reserved (3) [16] (2) (2) (2) Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 9.4.1 9.4.2 The AC potential difference which dissipates/produces the same amount of energy as an equivalent DC potential difference. VW Die WS _potensiaalverskil wat dieselfde hoeveelheid energie verbruik/oordra as ‘n ekwivalente GS potensiaalverskil. VW” (2) Vinax — Y rms v2 220(V2)¥ = Vmax Vmax = 311,13 VY (3) = o 0 » 2 0,04 = ime (s) = NOTES: Question does not require 2 cycles to start at the same position as the coil in the ACCEPT, since the instruction does not specify diagram. That the graph must start with same orientation Period (0,02 s) ¥ As the diagram. , Vmax =311,13 v CULES pga eeny Correctness: (Shape and 2 full cycles) ~ = 0 5 02 04 3) £ Time (s) ( $ [14] Copyright reserved Please turn over

Physical Sciences/P1 LimpopoDoE/September 2025 QUESTION/VRAAG 10 10.1. The minimum energy that an electron in the metal needs to be emitted from the metal surface. VY Die minimum energie benodig om 'n elektron uit die oppervilak van 'n metaal vry te stel VV (2) 10.2 Photoelectric effect V and particle nature of light “ Foto-elektriese effek V en deeltjie-aard van lig V (2) 10.3 Ejight = Wo + Excmax) Y (note that subscript (max) is essential) 1 (6,63 x10-**)(1x10!7)¥ = (3,66x10-19) + (5) (9,11x10-*)v2,4.7 Vmax = 1,20 x 107 ms ¥ (12 031 243,63 m-s*) (4) [8] GRAND TOTAL/GROOTTOTAAL: 150 Copyright reserved Please turn over