Physical 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 PHYSICAL SCIENCES: PHYSICS (P1) NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 10 pages. Copyright reserved Please turn over

Physical 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 (✓). A formula is not awarded a mark if it 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. Do not penalise units in intermediate steps. 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 to a previous question is used correctly in a later calculation, the later calculation is marked positively. 5. Negative marking: If a candidate must choose (e.g. INCREASES/DECREASES) and then explain, 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 or used in an incorrect context. Accept equivalent wording that conveys the same meaning. 7. Free-body diagrams: A mark is awarded for each force with a correct arrow AND label. Deduct 1 mark for any additional (incorrect) force. If forces do not touch the dot/object, deduct 1 mark. 8. Sign convention: Candidates may choose either direction as positive provided it is applied consistently. Copyright reserved Please turn over

Physical 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 B ✓✓ (2) 1.6 C ✓✓ (2) 1.7 A ✓✓ (2) 1.8 C ✓✓ (2) 1.9 D ✓✓ (2) 1.10 A ✓✓ (2) [20] QUESTION 2 2.1 A single vector which has the same effect as the original vectors acting together. ✓✓ (2) 2.2 Fx = 50 cos 30◦ ✓ = 43,30 N ✓ Fy = 50 sin 30◦ ✓ = 25,00 N ✓ (4) 2.3 Take the positive x- and y-directions as positive. Rx = 43,30 + (–35) = 8,30 N ✓ Ry = 25 + (–20) = 5,00 N ✓ R2 = R2x + R2y ✓ R2 = (8,30)2 + (5)2 R = 9,69 N ✓ Ry 5 tan θ = = ✓ Rx 8,30 θ = 31,06◦ above the positive x-axis (anticlockwise from the positive x-axis) ✓ POSITIVE MARKING FROM 2.2. Range for R: 9,68 N – 9,70 N. (6) 2.4 9,69 N ✓ at 211,06◦ from the positive x-axis, i.e. 31,06◦ below the negative x-axis (opposite to the resultant) ✓ POSITIVE MARKING FROM 2.3. (2) [14] Copyright reserved Please turn over

Physical Sciences/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 3.1 The force that opposes the tendency of motion of a stationary object relative to a surface. ✓✓ (2) 3.2 N / FN ✓ fs / f / Ff ✓ w / Fg ✓ Normal force perpendicular to the plank; friction parallel to the plank, UP the incline. Mark awarded for arrow AND label. Any additional force: deduct 1 mark. (3) 3.3 fs(max) = mg sin θ (forces parallel to the plank are balanced) ✓ fs(max) = (5)(9,8) sin 28◦ ✓ fs(max) = 23,00 N ✓ (3) 3.4 N = mg cos θ = (5)(9,8) cos 28◦ = 43,26 N ✓ fs(max) 23,00 µs = ✓ = N 43,26 µs = 0,53 ✓ Accept: µs = tan 28◦ = 0,53 (full marks). POSITIVE MARKING FROM 3.3. (3) 3.5 Take DOWN the plank as positive. Fnet = ma ✓ mg sin θ – µk mg cos θ = ma (5)(9,8) sin 35◦ ✓ –(0,4)(5)(9,8) cos 35◦ ✓ = 5a 28,11 – 16,06 = 5a a = 2,41 m·s–2 (down the plank) ✓ (4) 3.6 EQUAL TO ✓ µs = tan θ is independent of the mass of the crate: both the maximum static friction and the parallel component of the weight are directly proportional to the mass. ✓ (2) [17] Copyright reserved Please turn over

Physical Sciences/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1 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. ✓✓ (2) 4.2 N / FN ✓ fk / f / Ff ✓ T / FT ✓ w / Fg ✓ (4) 4.3 fk = µk N = µk mA g fk = (0,25)(8)(9,8) ✓ fk = 19,60 N ✓ (2) 4.4 Block A (to the right positive): Fnet = ma ✓ T – fk = mA a ⇒ T – 19,6 = 8a ✓ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) Block B (downwards positive): mB g – T = mB a ⇒ (4)(9,8) ✓ –T = 4a ✓ . . . . . . . . . . . . . . . . . . . . . . . . .(2) (1) + (2): 39,2 – 19,6 = 12a a = 1,63 m·s–2 ✓ POSITIVE MARKING FROM 4.3. (5) 4.5 OPTION 1 (block B) mB g – T = mB a (4)(9,8) – T = (4)(1,63) ✓✓ T = 32,67 N ✓ OPTION 2 (block A) T – 19,6 = (8)(1,63) ✓✓ T = 32,67 N ✓ Range: 32,64 N – 32,68 N. POSITIVE MARKING FROM 4.4. (3) 4.6 DECREASES ✓ The normal force on A increases, so the kinetic frictional force on A increases (19,6 N → 24,5 N) and the net force on the system decreases. ✓ F At the same time the total mass of the system increases, so a = net decreases. m ✓ (3) [19] Copyright reserved Please turn over

Physical Sciences/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A. ✓✓ (2) 5.2 The force that the learner exerts on the scale ✓, directed downwards. ✓ (2) 5.3 Upwards positive: Fnet = ma N – mg = ma ✓ N – (60)(9,8) = (60)(1,5) ✓ N = 678 N (reading on the scale = 678 N) ✓ (3) 5.4 588 N ✓ (1) [8] QUESTION 6 6.1 Every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. ✓✓ (2) GM 6.2 g= ✓ r2 (6,67 × 10–11 )(6,42 × 1023 ) g= ✓ (3,39 × 106 )2 g = 3,73 m·s–2 ✓ (3) 6.3 w = mg = (900)(3,73) ✓ w = 3 357 N ✓ (range: 3 353,54 N – 3 357 N) GMm Also accept F = . POSITIVE MARKING FROM 6.2. r2 (2) GMm 6.4 F= ✓ r2 (6,67 × 10–11 )(6,42 × 1023 )(900) F= ✓ (3,39 × 106 + 2,0 × 106 )2 F = 1 326,56 N ✓ If the height is not added to the radius: max. 1/3. (3) 6.5 EQUAL TO ✓ (1) [11] Copyright reserved Please turn over

Physical Sciences/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1.1 The net charge of an isolated system remains constant during any physical process. ✓✓ (2) Q1 + Q2 –5 + 9 7.1.2 Qnew = = ✓ 2 2 Qnew = +2 nC (+2 × 10–9 C) ✓ (2) ∆Q 7.1.3 n= ✓ e +2 × 10–9 – (–5 × 10–9 ) n= ✓ 1,6 × 10–19 n = 4,38 × 1010 electrons ✓ Electrons were transferred from P TO Q. ✓ POSITIVE MARKING FROM 7.1.2. (4) 7.2.1 The magnitude of the electrostatic force exerted by one point charge on another point charge is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. ✓✓ (2) 7.2.2 Marking criteria kQ1 Q2 • Formula F = ✓ r2 • Substitution for F12 ✓ • Substitution for F32 ✓ • Correct vector addition (subtracting the forces) ✓ • Final answer ✓ • Direction ✓ kQ1 Q2 F= ✓ r2 (9 × 109 )(2 × 10–6 )(3 × 10–6 ) F12 = ✓ = 0,60 N, to the LEFT (attraction towards (0,3)2 Q1 ) (9 × 109 )(4 × 10–6 )(3 × 10–6 ) F32 = ✓ = 2,70 N, to the RIGHT (attraction (0,2)2 towards Q3 ) Right positive: Fnet = 2,70 + (–0,60) ✓ Fnet = 2,10 N ✓ to the right (towards Q3 ) ✓ (6) [16] Copyright reserved Please turn over

Physical Sciences/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 The (electrostatic) force per unit positive charge placed at that point. ✓✓ (2) 8.2 – Marking criteria Radial field lines (evenly spaced, not touching/crossing) ✓ Arrows pointing INWARDS towards the charge ✓ (2) 8.3 Marking criteria kQ • Formula E = 2 ✓ r • Substitution for EA ✓ • Substitution for EB ✓ • Pythagoras ✓ • Magnitude ✓ • Trig ratio ✓ • Direction ✓ kQ E= 2 ✓ r (9 × 109 )(8 × 10–9 ) EA = 2 ✓ = 1 800 N·C–1 , to the RIGHT (away from A) (0,2) (9 × 109 )(6 × 10–9 ) EB = 2 ✓ = 2 400 N·C–1 , DOWNWARDS (towards B) (0,15) Enet = (1 800)2 + (2 400)2 ✓ 2 Enet = 3 000 N·C–1 ✓ 2 400 tan θ = ✓ θ = 53,13◦ 1 800 Direction: 53,13◦ BELOW the horizontal, to the right of P (i.e. 306,87◦ from the positive x-axis) ✓ Accept equivalent descriptions, e.g. 36,87◦ to the right of the vertical (downward) line PB. (7) 8.4 F = qE = (1,6 × 10–19 )(3 000) ✓ F = 4,80 × 10–16 N ✓ POSITIVE MARKING FROM 8.3. (2) [13] Copyright reserved Please turn over

Physical Sciences/P1 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 9.1.1 (current out of the page) Marking criteria Concentric circles around the conductor ✓ ANTICLOCKWISE direction ✓ (2) 9.1.2 (Right-hand) grip rule / right-hand rule ✓ (1) 9.2.1 The magnitude of the induced emf across the ends of a conductor is directly proportional to the rate of change in the magnetic flux linkage with the conductor. ✓✓ (2) 9.2.2 ∆Φ = ∆BA cos θ ✓ ∆Φ = (0,8 – 0,2)(0,03) cos 0◦ ✓ ∆Φ = 0,018 Wb (1,8 × 10–2 Wb) ✓ θ is the angle between the field and the NORMAL to the coil; here θ = 0◦ . (3) ∆Φ 9.2.3 ε = –N ✓ ∆t 0,018 ε = –(200) ✓ 0,15 |ε | = 24,00 V ✓ POSITIVE MARKING FROM 9.2.2. (3) 9.2.4 Any TWO: Increase the number of turns of the coil. ✓ Increase the area of the coil. ✓ Make the change in field take place in a shorter time (change the field faster). (2) [13] Copyright reserved Please turn over

Physical Sciences/P1 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 10 10.1.1 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. ✓✓ (2) 1 1 1 1 1 10.1.2 = + = + ✓ Rp = 2 Ω ✓ Rp R2 R3 6 3 Rtotal = R1 + Rp = 4 + 2 = 6 Ω ✓ R2 R3 Accept Rp = . R2 + R3 (3) V 12 10.1.3 I = ✓ = ✓ R 6 I = 2,00 A ✓ POSITIVE MARKING FROM 10.1.2. (3) 10.1.4 OPTION 1 Vp = IRp = (2)(2) = 4 V ✓ Vp 4 I2 = = ✓ R2 6 I2 = 0,67 A ✓ OPTION 2 4 V1 = IR1 = (2)(4) = 8 V; Vp = 12 – 8 = 4 V ✓ I2 = ✓ = 0,67 A ✓ 6 Also accept the current-divider ratio: I2 = 93 × 2. POSITIVE MARKING FROM 10.1.3. (3) 10.1.5 DECREASES ✓ When S is opened, no current flows through R3 and the resistance of the parallel section increases (from 2 Ω to 6 Ω). ✓ The total resistance of the circuit increases (from 6 Ω to 10 Ω). ✓ V The emf remains constant, so the current I = decreases (from 2 A to 1,2 A). R ✓ (4) 10.2 W = P∆t = (2 kW)(3 h × 30) ✓✓ W = 180 kW·h Cost = 180 × R2,85 ✓ Cost = R513,00 ✓ If calculated in joules: W = (2 000)(3 × 3 600 × 30) = 6,48 × 108 J; 6,48 × 108 ÷ 3,6 × 106 = 180 kW·h. Full marks if correctly converted. (4) [19] TOTAL: 150 Copyright reserved