Technical Sciences Grade 12 P1 NSC November Mock 2026 Memo

hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 TECHNICAL SCIENCES P1 MOCK EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 11 pages. Copyright reserved Please turn over

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

Technical Sciences/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1 When a net 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. ✓✓ OR The net force acting on an object is equal to the rate of change of momentum of the object. ✓✓ (2) 2.2 N / FN ✓ FA / 300 N ✓ fk / friction ✓ Fg / w / weight ✓ Accept: FA : applied force / FT / tension; fk : f / Ff / frictional force; N: normal force; Fg : w / 784 N. Do not penalise for relative lengths of arrows. (4) 2.3 Fx = 300 cos 25◦ = 271,89 N ✓ Fnet = ma ✓ Fx – fk = ma 271,89 – fk ✓ = (80)(0,6) ✓ fk = 223,89 N ✓ (5) 2.4 N + F sin θ = Fg OR N = mg – F sin θ ✓ N = (80)(9,8) – 300 sin 25◦ ✓ N = 784 – 126,79 = 657,21 N ✓ (3) 2.5 POSITIVE MARKING FROM QUESTIONS 2.3 AND 2.4 fk = µk N ✓ 223,89 = µk (657,21) ✓ µk = 0,34 ✓ (Range: 0,34 – 0,35. No unit.) (3) 2.6 INCREASES ✓ (1) 2.7 The vertical (upward) component of the applied force is now zero, so the normal force increases. ✓ Since fk = µk N / friction is directly proportional to the normal force, the frictional force increases. ✓ (2) [20] Copyright reserved Please turn over

Technical Sciences/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 3.1 The product of the mass of an object and its velocity. ✓✓ (2) 3.2 p = mv ✓ p = (1 200)(20) ✓ p = 24 000 kg·m·s–1 east ✓ (3) 3.3 East is positive ∑ pi = ∑ pf OR m1 vi1 + m2 vi2 = (m1 + m2 )vf ✓ (1 200)(20) + (1 800)(0) ✓ = (1 200 + 1 800)vf ✓ vf = 8 m·s–1 east ✓ (4) 3.4 POSITIVE MARKING FROM QUESTION 3.3 Ek,before = 12 mv2 = 12 (1 200)(20)2 = 240 000 J ✓ Ek,after = 12 (3 000)(8)2 = 96 000 J ✓ Ek,before ̸= Ek,after (kinetic energy is not conserved) ✓ INELASTIC ✓ Any one of the formula marks may be awarded for Ek = 12 mv2 . (4) 3.5 POSITIVE MARKING FROM QUESTION 3.3 Fnet ∆t = ∆p = m(vf – vi ) ✓ Fnet (0,15) = 1 200(8 – 20) ✓ Fnet = –96 000 N Magnitude: 96 000 N ✓ (3) 3.6 Any TWO: Airbags ✓ Seat belts ✓ Crumple zones Collapsible steering column Padded dashboard Head rests (2) [18] Copyright reserved Please turn over

Technical Sciences/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 The product of the force (component) in the direction of the displacement and the magnitude of the displacement. ✓✓ OR Work is done when a force causes an object to move in the direction of the force. (2) 4.1.2 W = F∆x cos θ ✓ W = (120)(15) cos 40◦ ✓ W = 1 378,88 J ✓ (3) 4.1.3 Zero / 0 J ✓ The lawnmower moves at constant velocity, so the net force on it is zero / there is no change in its kinetic energy. ✓ (2) 4.2.1 The total mechanical energy in an isolated (closed) system remains constant / is conserved. ✓✓ (2) 4.2.2 Ep = mgh ✓ Ep = (15)(9,8)(3,5) ✓ Ep = 514,50 J ✓ (3) 4.2.3 POSITIVE MARKING FROM QUESTION 4.2.2 (Ep + Ek )top = (Ep + Ek )bottom ✓ 514,50 + 0 ✓ = 0 + 21 (15)v2 ✓ v = 8,28 m·s–1 ✓ (4) 4.3 Pave = Fvave ✓ Pave = (500)(9,8)(0,8) ✓ (F = mg since velocity is constant) Pave = 3 920 W ✓ (3) 4.4 POSITIVE MARKING FROM QUESTION 4.3 3 920 = 5,25 hp ✓ 746 (1) [20] Copyright reserved Please turn over

Technical Sciences/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 The (internal restoring) force per unit (cross-sectional) area. ✓✓ (2) F 5.2 σ= ✓ A 20 000 σ= ✓ 5,0 × 10–5 σ = 4,0 × 108 Pa ✓ (3) ∆L 5.3 ε= ✓ L 8,0 × 10–3 ε= ✓ 4,0 ε = 2,0 × 10–3 (0,002; no unit) ✓ (3) 5.4 POSITIVE MARKING FROM QUESTIONS 5.2 AND 5.3 σ K= ✓ ε 4,0 × 108 K= ✓ 2,0 × 10–3 K = 2,0 × 1011 Pa ✓ (3) 5.5 B✓ (1) 5.6 The resistance of a fluid to flow. ✓✓ OR A measure of the internal friction in a fluid. (2) 5.7 DECREASES ✓ (1) [15] QUESTION 6 6.1 The pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid, as well as to the walls of the container. ✓✓ (2) 6.2 F2 = mg = (1 500)(9,8) = 14 700 N ✓ A1 = π r2 = π (0,02)2 = 1,26 × 10–3 m2 and A2 = π (0,16)2 = 8,04 × 10–2 m2 ✓ F1 F2 = ✓ A1 A2 F1 14 700 = ✓ 1,26 × 10 –3 8,04 × 10–2 F1 = 229,69 N ✓ (Range: 229,69 – 230,38 N) Areas in cm2 (12,57 cm2 and 804,25 cm2 ) are also accepted, since the units cancel. (5) 6.3 P = ρ gh ✓ P = (1 025)(9,8)(12) ✓ P = 120 540 Pa (1,21 × 105 Pa) ✓ (3) [10] Copyright reserved Please turn over

Technical Sciences/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1 The change in direction of light (a wave) as it passes from one medium to another (of different optical density), due to a change in speed. ✓✓ (2) 7.2 Any TWO: The speed of light decreases. ✓ The light bends towards the normal / changes direction. ✓ The wavelength decreases. (2) 7.3 The angle of incidence in the optically denser medium for which the angle of refraction is 90◦ . ✓✓ (2) 7.4 Total internal reflection ✓ (1) 7.5 Light must travel from an optically denser medium to an optically less dense medium. ✓ The angle of incidence must be greater than the critical angle. ✓ (2) 7.6 Any ONE: Optical fibres (endoscopes / telecommunication) ✓ Binoculars Cat’s eye road reflectors Sparkle of diamonds (1) [10] Copyright reserved Please turn over

Technical Sciences/P1 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1.1 c = fλ ✓ 3,0 × 108 = (2,45 × 109 )λ ✓ λ = 0,12 m (1,22 × 10–1 m) ✓ (3) 8.1.2 E = hf ✓ E = (6,63 × 10–34 )(2,45 × 109 ) ✓ E = 1,62 × 10–24 J ✓ (3) 8.1.3 Any ONE: Infrared / visible light / ultraviolet / X-rays / gamma rays ✓ (1) 8.2.1 Convex / converging lens ✓ (1) 8.2.2 Real ✓ Inverted ✓ Diminished / smaller than the object ✓ Also accept: formed between F and 2F on the opposite side of the lens (only if a property above is missing). (3) [11] Copyright reserved Please turn over

Technical Sciences/P1 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 9.1.1 The ratio of the charge on one plate of a capacitor to the potential difference across the plates. ✓✓ OR The charge stored per volt. (2) ε A 9.1.2 C= 0 ✓ d (8,85 × 10–12 )(0,02) C= ✓ 5 × 10–4 C = 3,54 × 10–10 F ✓ (3) 9.1.3 POSITIVE MARKING FROM QUESTION 9.1.2 Q C= ✓ V Q 3,54 × 10–10 = ✓ 12 Q = 4,25 × 10–9 C ✓ (3) 1 1 1 1 1 9.2.1 = + = + ✓ Rp = 4 Ω Rp R2 R3 6 12 Rtotal = R1 + Rp = 4 + 4 ✓ Rtotal = 8 Ω ✓ (3) 9.2.2 POSITIVE MARKING FROM QUESTION 9.2.1 V I= ✓ R 12 I= ✓ 8 I = 1,5 A ✓ (3) 9.2.3 POSITIVE MARKING FROM QUESTION 9.2.2 P = I2 R ✓ P = (1,5)2 (4) ✓ P=9W✓ OR: V1 = IR1 = 6 V; P = VI = (6)(1,5) = 9 W (3) [17] Copyright reserved Please turn over

Technical Sciences/P1 11 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 10 10.1.1 Slip rings ✓ (1) 10.1.2 The magnitude of the induced emf is directly proportional to the rate of change of magnetic flux (linkage). ✓✓ (2) 10.1.3 Mechanical (kinetic) energy to electrical energy ✓ (1) Vs Ns 10.2 = ✓ Vp Np 12 Ns = ✓ 230 1 150 Ns = 60 turns ✓ (3) 10.3 POSITIVE MARKING FROM QUESTION 10.2 The secondary coil has fewer turns (60) than the primary coil (1 150) ✓, therefore the output (secondary) voltage is lower than the input (primary) voltage. ✓ (2) [9] TOTAL: 150 Copyright reserved