Technical Sciences Grade 12 P1 2026 Memo (Eng)

hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 TECHNICAL SCIENCES P1 PREDICTION EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 12 pages. Copyright reserved Please turn over

Technical Sciences/P1 2 Hlayiso/Prediction 2026 NSC – Marking Guidelines GENERAL MARKING GUIDELINES 1. Calculations: Marks are awarded for the correct formula, correct substitution and the correct final answer WITH the correct unit, as indicated by the ticks (✓). No mark is awarded for a formula that is not followed by a substitution. If an incorrect formula is used, no further marks are awarded for that calculation. 2. Units: The final-answer mark is only awarded if the correct unit is given. Penalise an omitted or incorrect unit only ONCE per question. 3. Rounding: Accept final answers rounded to a minimum of TWO decimal places. Small differences caused by rounding of intermediate values are accepted. 4. Positive marking: Where an incorrect answer from a previous question is used correctly in a later calculation, full marks are awarded for the later calculation. 5. Negative marking: If a candidate must explain or motivate a choice (e.g. INCREASES or DECREASES) and the choice is incorrect, no marks are awarded for the explanation. 6. Definitions/laws: Where key words are underlined, deduct 1 mark for EACH key word or phrase that is omitted. Accept equivalent wording that conveys the same meaning. 7. Free-body diagrams: ONE mark for each force represented by an arrow with a correct label. Penalise ONCE for: no arrows, no dot, gaps between the arrows and the dot, dotted lines, or additional (non-applicable) forces. 8. Sign convention: Candidates may choose either direction as positive, provided it is applied consistently. Copyright reserved Please turn over

Technical Sciences/P1 3 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 1 1.1 A ✓✓ (2) 1.2 B ✓✓ (2) 1.3 C ✓✓ (2) 1.4 D ✓✓ (2) 1.5 B ✓✓ (2) 1.6 C ✓✓ (2) 1.7 D ✓✓ (2) 1.8 A ✓✓ (2) 1.9 D ✓✓ (2) 1.10 C ✓✓ (2) [20] Copyright reserved Please turn over

Technical Sciences/P1 4 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 2 2.1 An object continues in a state of rest or uniform (constant) velocity unless it is acted upon by a net (resultant) force. ✓✓ (2) 2.2 N / FN / normal force ✓ fk / friction ✓ T / FT / tension ✓ Fg / w / weight ✓ Accept: Fg = 588 N; N = 588 N; fk = 147 N. Do not penalise for relative lengths of arrows. Mark awarded for label and arrow. Deduct 1 mark if an applied force F is drawn on crate A. (4) 2.3 OPTION 1 (system) fk = µk N ✓ fk = (0,25)(60 + 40)(9,8) ✓ = 245 N Fnet = 0 (constant velocity) ∴ F – fk = 0 ✓ F = 245 N ✓ OPTION 2 (each crate) fA = (0,25)(60)(9,8) = 147 N and fB = (0,25)(40)(9,8) = 98 N ✓✓ F = fA + fB ✓ = 245 N ✓ (4) 2.4 When a net (resultant) force acts on an object, the object will accelerate in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass of the object. ✓✓ Mark 2 or 0: if any key phrase is omitted, 0 marks. OR: 1 mark for ‘accelerates in the direction of the net force’; 1 mark for ‘directly proportional to Fnet and inversely proportional to mass’. (2) 2.5 POSITIVE MARKING FROM QUESTION 2.3 Fnet = ma ✓ 375 – 245 = (60 + 40)a ✓ a = 1,30 m·s–2 ✓ OR: crate A: T – 147 = 60a; crate B: 375 – T – 98 = 40a; add the equations: 130 = 100a. (3) Copyright reserved Please turn over

Technical Sciences/P1 5 Hlayiso/Prediction 2026 NSC – Marking Guidelines 2.6 POSITIVE MARKING FROM QUESTION 2.5 Crate A: Fnet = ma T – fA = mA a ✓ T – 147 = (60)(1,30) ✓ T = 225 N ✓ OR Crate B: F – T – fB = mB a ✓ 375 – T – 98 = (40)(1,30) ✓ T = 225 N ✓ (3) 2.7 INCREASES ✓ (1) 2.8 NEGATIVE MARKING FROM QUESTION 2.7 The tension (opposing force) on crate B falls away, so the net force on crate B increases. ✓ The mass that is accelerated decreases (40 kg instead of 100 kg) / acceleration is directly proportional to the net force and inversely proportional to the mass. ✓ (Calculation: aB = 375–98 40 = 6,93 m·s–2 > 1,30 m·s–2 , accept as explanation.) (2) [21] Copyright reserved Please turn over

Technical Sciences/P1 6 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 3 3.1 The total linear momentum of an isolated system remains constant (is conserved) (in magnitude and direction). ✓✓ (2) 3.2 Take east as positive: Σpi = Σpf OR mP vPi + mQ vQi = mP vPf + mQ vQf ✓ (180)(4) + (220)(–1,5) ✓ = (180)(–1) ✓ +(220)vQf 390 = –180 + 220 vQf vQf = 2,59 m·s–1 ✓ east ✓ If west is taken as positive: vQf = –2,59 m·s–1 , i.e. 2,59 m·s–1 east. (5) 3.3 POSITIVE MARKING FROM QUESTION 3.2 Ek before = 21 mP v2P + 12 mQ v2Q = 12 (180)(4)2 + 21 (220)(1,5)2 ✓ = 1 440 + 247,5 = 1 687,5 J ✓ Ek after = 21 (180)(1)2 + 12 (220)(2,59)2 ✓ = 90 + 737,89 = 827,89 J ✓ (Range: 827,89 – 828,41 J) Ek before ̸= Ek after / kinetic energy is not conserved, therefore the collision is INELASTIC. ✓ (5) 3.4.1 Impulse is the product of the net force acting on an object and the time the net force acts on the object. ✓✓ (2) 3.4.2 Take east as positive: Fnet ∆t = ∆p = mvf – mvi ✓ Fnet (0,25) ✓ = (180)(–1,2) – (180)(3) ✓ Fnet = –3 024 N Magnitude: 3 024 N ✓ OR: ∆p = –756 kg·m·s–1 ; Fnet = ∆p 756 ∆t = 0,25 = 3 024 N (4) [18] Copyright reserved Please turn over

Technical Sciences/P1 7 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 W = F∆x cos θ ✓ W = (320)(15) cos 30◦ ✓ W = 4 156,92 J ✓ (3) 4.1.2 POSITIVE MARKING FROM QUESTION 4.1.1 Wf = f∆x cos 180◦ = (210)(15)(–1) = –3 150 J ✓ Wnet = WF + Wf (WN = Wg = 0) ✓ Wnet = 4 156,92 – 3 150 = 1 006,92 J ✓ OR: Fnet = 320 cos 30◦ – 210 ✓ = 67,13 N; Wnet = Fnet ∆x = (67,13)(15) ✓ = 1 006,92 J ✓ (3) 4.1.3 The normal force is perpendicular (at 90◦ ) to the displacement / there is no displacement in the direction of the normal force. ✓ W = F∆x cos 90◦ and cos 90◦ = 0, so W = 0. ✓ (2) 4.2.1 The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. ✓✓ (2) 4.2.2 Ep = mgh ✓ Ep = (450)(9,8)(2,8) ✓ Ep = 12 348 J ✓ (3) 4.2.3 POSITIVE MARKING FROM QUESTION 4.2.2 (Ep + Ek )A = (Ep + Ek )B OR MEA = MEB ✓ 12 348 + 0 ✓ = 0 + 12 (450)v2 ✓ v = 7,41 m·s–1 ✓ OR: Ep at A = Ek at B ✓; 12 348 ✓ = 12 (450)v2 ✓; v = 7,41 m·s–1 ✓ (4) 4.3 OPTION 1 W mgh P= = ✓ ∆t ∆t (450)(9,8)(9) P= ✓ = 2 646 W 15 2 646 P= = 3,55 hp ✓ 746 OPTION 2 9 = 0,6 m·s–1 ; P v = 15 ave = Fvave ✓ P = (450)(9,8)(0,6) ✓ = 2 646 W = 3,55 hp ✓ (3) [20] Copyright reserved Please turn over

Technical Sciences/P1 8 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 5 5.1 A body which regains its original shape and size completely when the deforming force is removed. ✓✓ (2) 5.2.1 A = π r2 = π (8 × 10–3 )2 = 2,01 × 10–4 m2 ✓ F σ= ✓ A 24 × 103 σ= ✓ 2,01 × 10–4 σ = 1,19 × 108 Pa ✓ (Range: 1,19 – 1,20× 108 Pa) (4) ∆L 5.2.2 ε= ✓ L 1,8 × 10–3 ε= ✓ 3,0 ε = 6,0 × 10–4 (no unit) ✓ (3) 5.2.3 POSITIVE MARKING FROM QUESTIONS 5.2.1 AND 5.2.2 σ K= ✓ ε 1,19 × 108 K= = 1,98 × 1011 Pa ✓ (Range: 1,98 – 1,99× 1011 Pa) 6,0 × 10–4 The rod is made of STEEL. ✓ (Award the last mark only if it is consistent with the calculated value of K.) (3) [12] Copyright reserved Please turn over

Technical Sciences/P1 9 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 6 6.1 In a continuous liquid at equilibrium, the pressure applied at a point is transmitted equally to all other parts of the liquid. ✓✓ Accept: The pressure applied to an enclosed liquid is transmitted undiminished/equally to every part of the liquid (and the walls of the container). (2) 6.2 A1 = π r21 = π (0,015)2 = 7,07 × 10–4 m2 A2 = π r22 = π (0,105)2 = 3,46 × 10–2 m2 ✓ (both areas) F1 F2 = ✓ A1 A2 180 F2 = ✓ 7,07 × 10 –4 3,46 × 10–2 F2 = 8 820 N ✓ (Range: 8 800 – 8 820 N) ( )2 ( )2 OR: F2 = F1 dd2 ✓✓ = 180 21 3 ✓ = 8 820 N ✓ (4) 1 6.3 DECREASES ✓ (1) 6.4 Any ONE: hydraulic brakes of a car; hydraulic car jack/car lift; dentist’s or barber’s chair; bulldozer/excavator arm; power steering; metal press. (Not the plywood press given in the question.) ✓ (1) [8] Copyright reserved Please turn over

Technical Sciences/P1 10 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 7 7.1.1 Refraction is the bending of light when it passes from one medium to another (of different optical density). ✓✓ (2) 7.1.2 Normal ✓ (1) 7.1.3 Angle of refraction ✓ (1) 7.1.4 Any TWO: The speed of light decreases. ✓ The ray bends towards the normal / the direction of the ray changes. ✓ The wavelength decreases. (The frequency remains the same: do not award a mark for this.) (2) 7.2.1 The angle of incidence in the optically denser medium for which the angle of refraction is 90◦ . ✓✓ (2) 7.2.2 Total internal reflection ✓ (1) 7.2.3 The light travels from an optically denser medium (glass) to a less dense medium (air). ✓ The angle of incidence (65◦ ) is greater than the critical angle (41,1◦ ). ✓ (2) 7.3 Dispersion (of light) ✓ (1) [12] QUESTION 8 8.1.1 c = fλ ✓ 3 × 108 = f(254 × 10–9 ) ✓ f = 1,18 × 1015 Hz ✓ (3) 8.1.2 POSITIVE MARKING FROM QUESTION 8.1.1 E = hf ✓ E = (6,63 × 10–34 )(1,18 × 1015 ) ✓ E = 7,82 × 10–19 J ✓ (Range: 7,82 – 7,83× 10–19 J) 8 OR: E = h λc = (6,63 × 10–34 ) 254×10 3×10 –9 = 7,83 × 10 –19 J (3) 8.2.1 Convex (converging) lens ✓ (1) 8.2.2 Virtual ✓ Upright / erect ✓ Larger than the object / enlarged / magnified ✓ (Accept: on the same side of the lens as the object, as one of the three properties.) (3) [10] Copyright reserved Please turn over

Technical Sciences/P1 11 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 9 9.1.1 Capacitance is the amount of charge a capacitor can store per volt (ratio of the charge on a plate to the potential difference across the plates). ✓✓ (2) ε A 9.1.2 C= 0 ✓ d (8,85 × 10–12 )A 4,425 × 10–10 = ✓ 0,50 × 10–3 A = 0,025 m2 (2,5× 10–2 m2 ) ✓ (3) Q 9.1.3 C= ✓ V Q 4,425 × 10–10 = ✓ 24 Q = 1,06 × 10–8 C ✓ (3) 9.1.4 POSITIVE MARKING FROM QUESTION 9.1.3 C ∝ d1 : halving d doubles C, so C = 8,85 × 10–10 F ✓ Q = CV = (8,85 × 10–10 )(24) = 2,12 × 10–8 C ✓ OR: V stays 24 V, so Q doubles ✓: Q = 2(1,06 × 10–8 ) = 2,12 × 10–8 C ✓ (2) 9.2.1 P = VI ✓ 2 000 = (230)I ✓ I = 8,70 A ✓ (3) V2 V 9.2.2 R= OR R = ✓ P I (230) 2 230 R= OR R = ✓ 2 000 8,70 R = 26,45 Ω ✓ (Range: 26,44 – 26,45 Ω) (3) 9.2.3 E = P∆t = (2 kW)(3 × 30 h) ✓ E = 180 kWh ✓ Cost = 180 × R3,40 = R612,00 ✓ OR: W = Pt = (2 000)(3 × 30 × 3 600) = 6,48 × 108 J = 180 kWh (3) [19] Copyright reserved Please turn over

Technical Sciences/P1 12 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 10 10.1.1 A device that converts electrical energy into mechanical energy. ✓✓ (2) 10.1.2 Split ring (commutator) ✓ (1) 10.1.3 It reverses the direction of the current in the coil every half-turn ✓ so that the coil keeps rotating in the same direction. ✓ (2) 10.2.1 STEP-DOWN ✓ The secondary (output) voltage (230 V) is lower than the primary (input) voltage (11 kV) / the secondary coil has fewer turns than the primary coil. ✓ (2) Vs Ns 10.2.2 = ✓ Vp Np 230 Ns = ✓ 11 000 9 900 Ns = 207 turns ✓ (3) [10] TOTAL: 150 Copyright reserved