Technical Sciences Grade 10 P1 November Mock 2026 Memo (Eng)

hlayiso mock exams Practice Series: Grade 10 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 10 TECHNICAL SCIENCES P1 NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 9 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 and the choice is incorrect, no marks are awarded for the explanation. 6. Definitions: 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, gaps between the arrows and the dot, dotted lines, or additional (non-applicable) forces. 8. Vectors: Where a direction is asked for (displacement, velocity, acceleration, resultant force), the final mark is only awarded if the direction is given. Copyright reserved Please turn over

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

Technical Sciences/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 450 2.1.1 ✓ = 0,45 kg ✓ 1 000 (2) 2.1.2 15 × 60 ✓ = 900 s ✓ (2) 2.1.3 5,6 × 10–3 ✓ A ✓ (1 mark for 5,6; 1 mark for the correct power of ten) (2) 2.2 A physical quantity that has both magnitude and direction. ✓✓ (2) 2.3.1 600 + 250 ✓ = 850 m ✓ (2) 2.3.2 East positive: ∆x = 600 + (–250) ✓ ∆x = 350 m ✓ east ✓ (3) 2.3.3 600 m east 250 m west resultant: 350 m east Criteria: 600 m vector to the east with label ✓; 250 m vector to the west, drawn head-to-tail ✓; resultant drawn from tail of first to head of last vector ✓; resultant labelled 350 m east ✓. (4) distance 2.3.4 speed = ✓ ∆t 850 speed = ✓ 170 speed = 5 m·s–1 ✓ (3) 2.3.5 POSITIVE MARKING FROM QUESTION 2.3.2 ∆x 350 v= = ✓ ∆t 170 v = 2,06 m·s–1 ✓ east ✓ (3) 1 2.4.1 ∆t = ✓ = 0,02 s ✓ 50 (2) 2.4.2 CONSTANT (velocity) ✓ (1) 2.4.3 POSITIVE MARKING FROM QUESTION 2.4.1 ∆t = 10 × 0,02 = 0,2 s ✓ (10 spaces between 11 dots) ∆x v= ✓ ∆t 0,20 v= ✓ 0,2 v = 1 m·s–1 ✓ If 11 intervals are used (∆t = 0,22 s, v = 0,91 m·s–1 ): max 3/4. (4) Copyright reserved Please turn over

Technical Sciences/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines ∆v 2.5 a= ✓ ∆t 15 – 0 a= ✓ 6 a = 2,5 m·s–2 east ✓ (3) [33] QUESTION 3 3.1 A push or a pull that one object exerts on another object. ✓✓ (2) 3.2 Fg = mg ✓ Fg = (40)(9,8) ✓ Fg = 392 N (downwards) ✓ (3) 3.3 Tension / applied force ✓ (1) 3.4 N / FN / normal force ✓ f / friction ✓ T / FA / applied force ✓ Fg / w / weight ✓ Do not penalise for relative lengths of arrows. Accept a box drawn instead of a dot. (4) 3.5 East positive: Fres = 180 + (–120) ✓ Fres = 60 N ✓ east ✓ (3) 3.6 NO ✓ The resultant (net) force on the crate is NOT zero (60 N east). ✓ (2) [15] Copyright reserved Please turn over

Technical Sciences/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1 The turning effect of a force; it is the product of the force and the perpendicular distance from the pivot (fulcrum) to the line of action of the force. ✓✓ (2) 4.2 τ = F × r⊥ ✓ τ = (80)(0,3) ✓ τ = 24 N·m ✓ (3) 4.3 The perpendicular distance from the nut (pivot) is greater. ✓ The same force produces a larger moment (torque) / a smaller force is needed for the same moment. ✓ (2) 4.4.1 For a body in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point. ✓✓ (2) 4.4.2 Take moments about L: Σ clockwise moments = Σ anticlockwise moments ✓ (400)(2) + (600)(5) + (200)(7) ✓ = R(8) ✓ 5 200 = 8R R = 650 N ✓ Take moments about R: L(8) ✓ = (400)(6) + (600)(3) + (200)(1) ✓ L = 550 N ✓ OR, after R has been calculated: L = 1 200 – 650 ✓✓ = 550 N ✓ (7) 4.4.3 POSITIVE MARKING FROM QUESTION 4.4.2 Upward forces: 550 + 650 = 1 200 N ✓ Downward forces: 400 + 600 + 200 = 1 200 N ✓ ∴ upward forces = downward forces (2) 4.5.1 The ratio of the load to the effort (load divided by effort). ✓✓ (2) load 4.5.2 MA = ✓ effort 900 MA = ✓ 150 MA = 6 ✓ (no unit) (3) 4.5.3 First-class lever (class 1) ✓ (1) 4.5.4 Any ONE: Scissors / pliers / see-saw / claw hammer pulling a nail / tin-lid lever / balance ✓ (1) [25] Copyright reserved Please turn over

Technical Sciences/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 The energy that an object possesses because of its position (height) relative to a reference point (the ground). ✓✓ (2) 5.2 Ep = mgh ✓ Ep = (250)(9,8)(18) ✓ Ep = 44 100 J ✓ (accept 4,41 × 104 J / 44,1 kJ) (3) 5.3 The energy that an object possesses because of its motion. ✓✓ (2) 5.4 Ek = 12 mv2 ✓ Ek = 12 (0,16)(30)2 ✓ Ek = 72 J ✓ (3) 5.5.1 The sum of the gravitational potential energy and the kinetic energy of an object. ✓✓ (2) 5.5.2 EM = Ep + Ek ✓ EM = (2)(9,8)(4) ✓ + 21 (2)(6)2 ✓ EM = 78,4 + 36 = 114,4 J ✓ (4) [16] QUESTION 6 6.1 MORE THAN ✓ (1) 6.2 Sphere B is negatively charged / has an excess of electrons. ✓ (1) 6.3 The net (total) charge in an isolated (closed) system remains constant (is conserved). ✓✓ (2) Q1 + Q2 6.4 Q= ✓ 2 (+7 × 10–9 ) + (–3 × 10–9 ) Q= ✓ 2 Q = +2 × 10–9 C (on each sphere) ✓ (3) 6.5 POSITIVE MARKING FROM QUESTION 6.4 ∆Q n= ✓ e (+2 × 10–9 ) – (–3 × 10–9 ) –9 –9 n= ✓ (or 7×10 –2×10 –19 for sphere A) 1,6 × 10 –19 1,6×10 n = 3,13 × 1010 electrons ✓ (accept 3,125 × 1010 ) (3) 6.6 FROM B TO A ✓ (1) 6.7 REPEL ✓ Both spheres are positively charged; like charges repel. ✓ (2) [13] Copyright reserved Please turn over

Technical Sciences/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1 The rate of flow of charge. ✓✓ Accept: the amount of charge that passes a point in a conductor per second. (2) Q 7.2 I= ✓ ∆t 18 I= ✓ 60 I = 0,3 A ✓ If 1 is used for ∆t (18 A): max 1/3. (3) 7.3 The opposition that a material (conductor) offers to the flow of electric current. ✓✓ (2) 7.4 Any THREE ✓✓✓: Length of the conductor Cross-sectional area (thickness) of the conductor Type of material (resistivity) Temperature of the conductor (3) 7.5 S A 3 cells bulb V R Criteria: Battery of three cells, switch, bulb and resistor ALL in series ✓; ammeter in series ✓; voltmeter in parallel across the bulb ONLY ✓. Correct symbols must be used. (3) [13] Copyright reserved Please turn over

Technical Sciences/P1 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 The maximum energy provided (work done) by a battery per coulomb (unit) of charge passing through it. ✓✓ Accept: the potential difference across the terminals of a battery when NO current flows. (2) 1 1 1 8.2 = + ✓ Rp R2 R3 1 1 1 = + ✓ Rp 6 3 Rp = 2 Ω ✓ (3) 8.3 POSITIVE MARKING FROM QUESTION 8.2 RT = R1 + Rp = 4 + 2 ✓ RT = 6 Ω ✓ (2) 8.4 POSITIVE MARKING FROM QUESTION 8.3 V R= ✓ I 12 6= ✓ I I=2A✓ (3) 8.5 POSITIVE MARKING FROM QUESTION 8.4 V = IR ✓ V1 = (2)(4) ✓ V1 = 8 V ✓ (3) 8.6 POSITIVE MARKING FROM QUESTION 8.5 VT = V1 + V2 12 = 8 + V2 ✓ V2 = 4 V ✓ OR: V2 = IRp = (2)(2) ✓ = 4 V ✓ (2) [15] TOTAL: 150 Copyright reserved