Physical Sciences Grade 12 Paper 1 KZN Mock 2025 Memo

} KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE 4 PHYSICAL SCIENCES P11 (PHYSICS) C PREPARATORY EXAMINATION : MARKING GUIDELINES 4 SEPTEMBER 2025 Seenee rere reese settee eee eens MARKS: 150 These marking guidelines consist of 9 pages. Copyright reserved Please turn over

Physical Sciences P1 September 2025 Preparatory Examination QUESTION 1 1.1 Dyvv h Dyv 1.3 Byy 1.4 Dvv 1.5 AVY 1.6 Cvv LF Byv 1.8 AVY 1.9 AVY 1.10 Cvv QUESTION 2 2 22 2.3.1 T Accept force diagram: -F+——_@——” -«—_ "| ! Accepted labels F | Fafapplied force/50 NY f_| (kinetic) friction/Fi/fky” _T | Fr/Tension/Ftension ~ Notes e Mark awarded for label and arrow. e Do not penalise for length of arrows since drawing is not to scale. e Any other additional force(s): Max 7% e If everything correct, but no arrows: Max % When a net force is applied on an object, the object will accelerate in the direction of the force and the acceleration is directly proportional to the forceY and inversely proportional to the mass of the object.v Accept Newton’s second law in terms of momentum fk = UkKNY = 0,3{(6)(9,8) +(3)(9,.8)]” Copyright Reserved Please Turn Over (2) (2) (2) (2) (2) (2) (2) (2) (2) (2) [20] (3) (2)

Physical Sciences P1 2.3.2 2.4.1 24.2 = 26,46 Nv’ September 2025 Preparatory Examination | POSITIVE MARKING FROM Q 2.3.1 LEFT AS NEGATIVE Block X Fret = ma v -—F+T+f=ma -50+ T+ 2646 V= Gave (1) Block Y Fnet = ma Og Solving equation (1) and (2): -50 +3a + 26,46 = -6av a= 2,62 ms? v POSITIVE MARKING FROM Q 2.3.1 LEFT AS POSITIVE Block X Fret = ma v F-T-f=ma 50 - T - 26,46 V= 6a... A) Block Y Fret = ma Ts 88 4 nono (2) ee Solving equation (1) and (2): +50 -3a - 26,46 = +6av a=2,62 ms? v Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses “and inversely proportional to the square of the distance between their centres.” If charges are mentioned ; OPTION 1 Foplanet = YF earth G(2Me)(m) /_ >G(Me)(m) ey x2 ~ R2 x = 2RV OPTION 2 Qplanet M = YeQeEarth v 1 G(2Me) ,_ 37 S(Me) / x2 Re x = 2RY Copyright Reserved Please Turn Over (3) (6) (2) (4) [20]

Physical Sciences P1 September 2025 Preparatory Examination QUESTION 3 3.1 0,6 mv 3.2.1. [DOWNWARD AS POSITIVE DOWNWARD AS NEGATIVE 3.2.2 3.2.3 3.3 vie = v,2 + 2aAyv = (1,5)? + 2(9,8)(2) v w=6,44m-s' v vi? = vi2 + 2aAy = (-1,5)2 + 2(-9,8)(-2) V w=6,44 mst v OPTION 1 POSITIVE MARKING FROM Q 3.2.1 DOWNWARD AS POSITIVE vi=vit alt ¥ 6,44 = 1,5 + (9,8)At v At = 0,50 s¥ OPTION 1 POSITIVE MARKING FROM Q 3.2.1 DOWNARD AS NEGATIVE ve = vi t+ alt Vv -6,44 = -1,5 + (-9,8)At Vv At = 0,50 sv OPTION 2 POSITIVE MARKING FROM Q 3.2.1 DOWNWARD AS POSITIVE OPTION 2 POSITIVE MARKING FROM Q 3.2.1 DOWNWARD AS NEGATIVE - (it™ _ (ity +6, -1,5 + (-6,44 [2=(-S)vae p= (Nv at At = 0,50 At=0,50s Vv OPTION 3 OPTION 3 DOWNWARD AS POSITIVE Ay = viAt+ adv [2=1,5 At +3(9,8))vAt At= 0,50 s ¥ DOWNWARD AS NEGATIVE Ay = v,At+ sadtv [-2=-1,5 At +3(-9,8))VAt At=0,50s¥_ OPTION 1 DOWNWARD AS POSITIVE ve = vi2 + 2aly ¥ 02V = vi + 2(9,8)(-0,8) ” Vi = 3,96 m:s*, upwards Vv OPTION 1 DOWNWARD AS NEGATIVE vie =v? + 2ahy ¥ 0?v = vit + 2(-9,8)(0,8) “ Vi = 3,96 ms‘, upwards Vv POSITIVE MARKING FROM QUESTION 3.2.1 & QUESTION 3.2.3 Fret At = m(vr— vi) ¥ Fnet(0,2) ¥ = 0,5(3,96 — (-6,44)) ¥ Fret = 26 N, upwards Vv Copyright Reserved Please Turn Over (1) (3) (4)

Physical Sciences P1 September 2025 Preparatory Examination 3.4 [ POSITIVE MARKING FROM QUESTION 3.1 TO QUESTION 3.3 Criteria « First part of starts at v = 1,5 m-s’ and ends at 6,44 m-s! Vv « Second part starts at v = 3,96 m:s* until v= 0 m-s? ¥ « Shape (parallel lines) v « Both axes labelled v DOWNWARD AS POSITIVE » 6,44 b £ > 1,5 t(s) -3,96+--------------- DOWNWARD AS NEGATIVE a 3,96 ‘e — > 15 t(s) -6,44 (4) [19] Copyright Reserved Please Turn Over

Physical Sciences P1 QUESTION 4 41 Easiv 4.2 | Newton’s Third Law (of Motion) “ (Mark independently) September 2025 Preparatory Examination When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A. Vv v 4.3 OPTION 1 OPTION 1 TO THE LEFT AS POSITIVE TO THE LEFT AS NEGATIVE 2pi = Lpr v 2pi = Xpr } v mtvit = Mive + Mave2 anyone mitvit = Maver + Mavee Any one Ov = 6Ovn + (4)(3) ¥ Ov = 60Ovi + (4)(-3) ¥ vi = -0,20 vr = 0,20 ms! ¥ * Vwe= 0,20 mst v OPTION 2 Apboy =- APobject mb(Vor —Vbi) = = Mo(Vor= Voi) PY Any one 60(Vvor - 0)” = -4(3-0)” Vor = -0,20 <. Vor = 0,20 mst 4.4.1 Increases Vv 4.4.2 Increases Vv 4.4.3 ApobjectincreasesY, which causes Appboy to increasev Copyright Reserved Please Turn Over (1) (3) (4) (1) (1) (2) [12]

Physical Sciences P1 September 2025 Preparatory Examination QUESTION 5 5.1 N Accept force diagram: - N f f w Ww Accepted labels Fg/ Fw/ weight/588 N/gravitational forcev” F./applied force’. (kinetic) friction/Fefky” Fn /Normal/Fnormat “ Notes Mark awarded for label and arrow. ¢ Do not penalise for length of arrows since drawing is not to scale. ¢ Any other additional force(s): Max a Z\>| n= , : 3 e If everything correct, but no arrows: Max/Maks % (4) 5.2 Work done on an object by a constant force F is F Ax cos 6 , where F is the magnitude of the force, Ax the magnitude of the displacement and 0 the angle between the force and the displacement. v OR The work done on an object is the product of the force and the displacement of the object in the direction of the displacement. vv (2) 5.3 [OPTION1 Whet = AK We + Wr = Yem(ve? — vi?) } v Any one (80)(2)cos© + (8)(2)cos180° = [¥4(7)(1,52 - 02)] © = 81,42 °V OPTION 2 We = AK + AU We + Wr = ¥em(ve2 — v2) + mg(hr— hi) a aiden (80)(2)cos© + (8)(2)cos180° = [¥4(7)(1,52 - 02) + 0] ¥ 8 = 81,42 °v (4) 5.4‘ Increasesv Vv (2) [12] Copyright Reserved Please Turn Over

Physical Sciences P1 September 2025 Preparatory Examination QUESTION 6 6.1 Frequency” (1) 6.2 The sound source (bat) and the listener (bird watcher) have different velocities (2) relative to the medium of sound propagation. vv 6.3 0,4 mv (1) 6.4.1 v=fxdAv 340 =f x 0.4v f = 850 Hzv (3) 6.4.2 POSITIVE MARKING FROM QUESTION 6.4.1 _ Viv, i= viv, ‘ y Y Any one fi=—— -f, V—Vs, _ 340 850v = = v(875) ¥ Vs = 10 m-s"' (away from the batwatcher) v (5) [12] QUESTION 7 7.1 Q= aren 2 (9x10*)+(-Sx10°) y @uC)+C5HC) 5 Accept Q= : = 2x108Cv = 2uC Vv (2) had Criteria for sketch Marks Correct direction of field lines v Shape of electric field v Field lines not crossing each other / v Field lines start from sphere / No field lines inside the spheres (3) Copyright Reserved Please Turn Over

Physical Sciences P1 September 2025 Preparatory Examination 7.3 Fr = S12 v a (9x10°)(2x10°)(1x10) | (9x10°)(2x10°)(1«10°) = (0,2)? = 5 v = 0,45 N to the left (0,1) = 1,8 N to the right Fret = 1,8 + (-0,45) ¥ = 1,35.N to the right v (5) 7.4 Electrostatic force experienced per unit positive charge placed (at that point). vv (2) 7.5 POSITIVE MARKING FROM QUESTION 7.3 OPTION 1 = 1,35x10° N-C"' to the right ¥ (3) OPTION 2 kQ E=> _(9.0x10")(2 x 10°) (0,1)° = 1,80 x 10° N:C* to the right “ nl _ 6.0% 10° 2x 10°) 7 (0,2)° = 4,50 x 105 N-C" to the left v Enet = 1,80 x 10®N— 4,50 x 108 = 1,35x10° N-C"' to the right (3) [15] Copyright Reserved Please Turn Over

Physical Sciences P1 September 2025 Preparatory Examination QUESTION 8 8.1 12V¥ B21 Vy, R Sy 8 = 0,75A V 8.2.2 OPTION 1 OPTION 2 | 1 4 Rix Ro — eee ae = v Rp R,; Ra” Rp Ry+ Ro -',1, _14«12 ) 14 12 ~ 44412 Rp =6,46Q v =6,450V 8.2.3. POSITIVE MARKING FROM QUESTION 8.1 & 8.2.2 €=(Rtr) v 12v = 0,75(6,46 + 18 + r) r= 1,54 QV 8.2.4 OPTION 1 POSITIVE MARKING FROM QUESTION 8.2.1 & 8.2.2 OPTION 2 POSITIVE MARKING FROM QUESTION 8.2.1 Vi =IR = (0.75)(6,46) ¥ =4,845V ly20 = Ms wi _ 4,845, ~~ 42— = 0,404 Av Viost = Ir = (0,75)(1,54) = 1,155 V Vian = 12 — 1,155 — 6Y = 4,845 V haa = RY 4,845 liq = | =0,404A Vv 8.3 Cost = P x At x tariff 5 =3,5v x 2v x 265v 60 =R7,73v Copyright Reserved Please Turn Over (1) (3) (4) (4) (4) [19]

Physical Sciences P1 QUESTION 9 9.1 Split ring (commutator) ¥ Accept: Commutator 9.2 Anticlockwise vv 9.3 Electrical to mechanicalv 9.4.1 Vmax Vime = v rms V2 _ 311 v2 = 220 Vv September 2025 Preparatory Examination 9.4.2 POSITIVE MARKING FROM QUESTION 9.4.1 p.. = Vins ave ~ “R 2 _ 220 J 30 = 1 613, 33 W OPTION 1 OPTION 2 Pave = ImsVims ¥ Pave = l?rmsR V 1 613,33 = Irs (220) v 1 613,33 = I?rms(30) ¥ Ims = 7,33 A ¥ Ims = 7,33 Av OPTION 3 Reavis v ne v v 30= a lms = 7,33 AY Copyright Reserved Please Turn Over (1) (2) (4) [11]

Physical Sciences P1 September 2025 Preparatory Examination QUESTION 10 10.1 Minimum frequency of light needed to emit electrons from a certain metal surface. vv (2) 10.2 4 vv 7,2x 10° Hz (2) 10.3. OPTION4 POSITIVE MARKING FROM QUESTION 10.2 Eximax)= hf - hfo ~ X = (6,63x 10) 1x 10°) v- (6,63x 10) 7,2x 10") v = 1,86x 10? Jv (4) OPTION 2 - v Gradient < Ek(max)2 - Ek(max)1 r ff, X-0 10 x 10" — 7,2 x 10" = 719 vw x = 1,86x 10% J (4) v 6,63 x 10% = 10.4 Remains the samev Intensity does not affect kinetic energy of the photonsY OR Only frequency affects Ex(max) The frequency remains the same. (2) [10] TOTAL: 150 Copyright Reserved