Technical Sciences Grade 11 P1 November Mock 2026 Memo (Eng)

hlayiso mock exams Practice Series: Grade 11 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 11 TECHNICAL SCIENCES P1 NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 10 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, 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 C ✓✓ (2) 1.2 B ✓✓ (2) 1.3 D ✓✓ (2) 1.4 A ✓✓ (2) 1.5 C ✓✓ (2) 1.6 B ✓✓ (2) 1.7 D ✓✓ (2) 1.8 A ✓✓ (2) 1.9 B ✓✓ (2) 1.10 D ✓✓ (2) [20] QUESTION 2 2.1 A single vector that has the same effect as two or more vectors acting together (at a point). ✓✓ (2) 2.2 VECTOR ✓ (1) 2.3.1 F3x = F3 cos 30◦ = 50 cos 30◦ ✓ F3x = 43,30 N ✓ (2) 2.3.2 F3y = F3 sin 30◦ = 50 sin 30◦ ✓ F3y = 25,00 N ✓ (2) 2.4 POSITIVE MARKING FROM QUESTION 2.3 Right (east) and up (north) positive: Rx = 60 + (–43,30) ✓ = 16,70 N (to the right) Ry = 35 + (–25) ✓ = 10,00 N (upwards) R2 = R2x + R2y ✓ R2 = (16,70)2 + (10,00)2 ✓ R = 19,46 N ✓ (5) 2.5 POSITIVE MARKING FROM QUESTION 2.4 Ry 10,00 tan θ = = ✓ Rx 16,70 θ = 30,92◦ (above the horizontal, to the right / east) ✓ 10 or cos θ = 16,70 . Range: 30,90◦ – 30,95◦ . Accept sin θ = 19,46 19,46 (2) [14] Copyright reserved Please turn over

Technical Sciences/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 3.1 The gravitational force that the Earth exerts on an object. ✓✓ (2) 3.2 Fg = mg ✓ Fg = (1 500)(9,8) ✓ Fg = 14 700 N ✓ (3) 3.3 NON-CONTACT force ✓ (1) 3.4 Any ONE: Magnetic force / electrostatic (electric) force ✓ (1) 3.5.1 8 + 6 = 14 km ✓ (1) 3.5.2 ∆x2 = 82 + 62 ✓ ∆x = 10 km ✓ Accept 10 000 m. Direction not required. (2) [10] Copyright reserved Please turn over

Technical Sciences/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1 The force that opposes the (tendency of) motion of an object and acts parallel to the surface in contact. ✓✓ (2) 4.2.1 Normal force / total mass of the block ✓ (1) 4.2.2 Frictional force (maximum static / kinetic) ✓ (1) 4.2.3 Any ONE: Type (nature) of surfaces in contact / same block / same table / horizontal surface ✓ (1) 4.3 N / FN / normal force ✓ fk / friction ✓ F / FA / applied force ✓ Fg / w / weight ✓ Accept FT / tension for the applied force. Do not penalise for relative lengths of arrows. (4) 4.4 N = Fg = mg ✓ N = (2,0)(9,8) = 19,6 N ✓ (2) 4.5 fmax s = µs N ✓ 6,86 = µs (19,6) ✓ µs = 0,35 ✓ (no unit) (3) 4.6 fk = µk N ✓ 4,90 = µk (19,6) ✓ µk = 0,25 ✓ (no unit) (3) 4.7 For each mass, the kinetic frictional force is SMALLER than the maximum static frictional force ✓ / µk < µs . A smaller applied force is therefore needed to keep the block moving than to start it moving. ✓ (2) 4.8 REMAINS THE SAME ✓ (1) 4.9 Any TWO: Walking without slipping / grip of car tyres on the road / car brakes or brake pads / belt drives and pulleys / striking a match / nails and screws holding in wood / holding objects ✓✓ (2) [22] Copyright reserved Please turn over

Technical Sciences/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 A wave in which the particles of the medium vibrate perpendicular (at right angles) to the direction of propagation of the wave. ✓✓ (2) 5.2 The maximum displacement of a particle from its rest (equilibrium) position. ✓✓ (2) 5.3.1 12 cm / 0,12 m ✓ (1) 5.3.2 0,8 m ✓ (1) 5.4.1 Any ONE pair: A and C / A and F / C and F / B and E ✓ (1) 5.4.2 Any ONE pair: A and B / A and E / C and B / C and E / F and B / F and E ✓ Also accept any pair that includes D with a crest or trough. (1) 5.5 2 (two) ✓ (1) 2 5.6 T= = 0,4 s ✓ 5 1 1 f= = ✓ T 0,4 f = 2,5 Hz ✓ OR: f = numbertime of waves = 5 ✓✓ = 2,5 Hz ✓ 2 (3) 5.7 POSITIVE MARKING FROM QUESTIONS 5.3.2 AND 5.6 v = fλ ✓ v = (2,5)(0,8) ✓ v = 2 m·s–1 ✓ (3) 5.8 DECREASES ✓ (1) 5.9 The speed of the wave remains constant (same medium / same rope). ✓ Since v = fλ , the wavelength is inversely proportional to the frequency (wavelength is halved). ✓ (2) [18] Copyright reserved Please turn over

Technical Sciences/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 A wave in which the particles of the medium vibrate parallel to the direction of propagation of the wave. ✓✓ (2) 6.2.1 Compression ✓ (1) 6.2.2 Rarefaction ✓ (1) 1 6.3.1 T= ✓ f 1 T= ✓ 235 T = 4,26 × 10–3 s ✓ (3) 6.3.2 v = fλ ✓ 340 = (235)λ ✓ λ = 1,45 m ✓ (3) 6.4 Sound waves with frequencies higher than 20 kHz / above the upper limit of the human hearing range. ✓✓ (2) 0,12 6.5 Time for the pulse to reach the seabed: ∆t = = 0,06 s ✓ 2 ∆x v= ✓ ∆t ∆x 1 500 = ✓ 0,06 ∆x = 90 m ✓ OR: 2∆x = v∆t = (1 500)(0,12) = 180 m ✓✓✓; depth = 90 m ✓. If 180 m is given as the depth: max 2/4. (4) 6.6 Amplitude ✓ (1) [17] Copyright reserved Please turn over

Technical Sciences/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1 The region (in space) around a magnet in which another magnet or a ferromagnetic material will experience a (magnetic) force. ✓✓ (2) 7.2.1 N N Criteria: Correct shape – lines bend away from each other (no lines joining the two north poles) ✓; Direction of arrows away from both north poles ✓; Field lines do not cross / touch ✓. (Null point not required.) (3) 7.2.2 REPEL ✓ (1) 7.3 Any TWO ✓✓: Field lines never cross (intersect) each other. Field lines are closest together where the field is strongest (at the poles). Outside a magnet, field lines point from the north pole to the south pole. Field lines form closed loops (inside the magnet from S to N). (2) 7.4.1 SOUTH (magnetic pole) ✓ (1) 7.4.2 It deflects charged particles from the Sun (solar wind) / cosmic rays away from the Earth’s surface. ✓ Accept: protects the atmosphere (ozone layer) from being stripped away by the solar wind. (1) 7.5 Any TWO: Iron / nickel / cobalt / steel / alloys of these metals ✓✓ (2) [12] Copyright reserved Please turn over

Technical Sciences/P1 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 The net (total) charge in an isolated (closed) system remains constant (is conserved). ✓✓ (2) Q1 + Q2 (+6 × 10–9 ) + (–2 × 10–9 ) 8.2 Q= = ✓ 2 2 Q = +2 × 10–9 C (on each sphere) ✓ (2) 8.3 POSITIVE MARKING FROM QUESTION 8.2 ∆Q n= ✓ e (+2 × 10–9 ) – (–2 × 10–9 ) 6×10–9 –2×10–9 for P) n= ✓ (or 1,6 × 10–19 1,6×10–19 n = 2,5 × 1010 electrons ✓ (3) 8.4 FROM Q TO P ✓ (1) 8.5 POSITIVE MARKING FROM QUESTION 8.2 kQ1 Q2 F= ✓ r2 (9 × 109 )(2 × 10–9 )(2 × 10–9 ) F= ✓✓ (0,3)2 F = 4 × 10–7 N ✓ 1 mark for substitution of both charges, 1 mark for (0,3)2 . (4) 8.6 POSITIVE MARKING FROM QUESTION 8.5 F E= ✓ q 4 × 10–7 E= ✓ 2 × 10–9 E = 200 N·C–1 ✓ (3) [15] Copyright reserved Please turn over

Technical Sciences/P1 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 9.1 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. ✓✓ (2) 9.2.1 Rs = R1 + R2 = 5 + 12 ✓ Rs = 17 Ω ✓ V 12 R= ✓ 17 = I I I = 0,71 A ✓ (4) 9.2.2 POSITIVE MARKING FROM QUESTION 9.2.1 V R= ✓ I V 5= 1 ✓ 0,71 V1 = 3,53 V ✓ (Accept 3,55 V when 0,71 A is used.) (3) R 2 R3 (12)(6) 1 1 1 9.3.1 Rp = = ✓ OR = + ✓ R2 + R3 12 + 6 Rp 12 6 Rp = 4 Ω ✓ RT = R1 + Rp = 5 + 4 ✓ RT = 9 Ω ✓ (4) 9.3.2 POSITIVE MARKING FROM QUESTION 9.3.1 V R= ✓ I 12 9= ✓ I I = 1,33 A ✓ (3) 9.3.3 POSITIVE MARKING FROM QUESTIONS 9.3.1 AND 9.3.2 Vp = IRp ✓ Vp = (1,33)(4) ✓ V2 = 5,33 V ✓ (Accept 5,32 V) OR: V1 = (1,33)(5) = 6,67 V ✓; V2 = 12 – 6,67 ✓ = 5,33 V ✓ (3) 9.3.4 POSITIVE MARKING FROM QUESTION 9.3.3 V R= ✓ I 5,33 6= ✓ I3 I3 = 0,89 A ✓ (3) [22] TOTAL: 150 Copyright reserved