Physical 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 PHYSICAL SCIENCES: PHYSICS (P1) MOCK EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 14 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 (the first time it occurs). 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 (ranges are given where relevant). 4. Positive marking: Where an answer to a previous question is used in a later calculation, the later calculation is marked positively, i.e. full marks are awarded if the method and substitution are correct using the candidate’s earlier (incorrect) answer. 5. Negative marking: An incorrect answer that is based on a conceptual mistake cannot be correctly motivated. If a candidate must explain or motivate an answer (e.g. INCREASES/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 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, maximum 34 of the marks. 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 C ✓✓ (2) 1.3 D ✓✓ (2) 1.4 A ✓✓ (2) 1.5 A ✓✓ (2) 1.6 D ✓✓ (2) 1.7 B ✓✓ (2) 1.8 C ✓✓ (2) 1.9 B ✓✓ (2) 1.10 D ✓✓ (2) [20] Copyright reserved Please turn over

Physical Sciences/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1 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. ✓✓ OR The net force acting on an object is equal to the rate of change of momentum of the object (in the direction of the net force). ✓✓ If any of the underlined key phrases is omitted, deduct 1 mark. (2) 2.2 N (normal force) ✓ F (applied force) ✓ T (tension) ✓ fk (friction) ✓ w / Fg (weight) ✓ Accepted labels: w: Fg / Fw / weight / 58,8 N T: FT / tension fk : f / Ff / friction N: FN / normal force. Mark awarded for label AND arrow. Do not penalise for relative lengths of arrows. Any additional force: deduct 1 mark. Arrows not touching the dot: max 4/5. (5) 2.3 OPTION 1 N + F sin θ = mg (vertical forces balance) ✓ N + 60 sin 30◦ = (6)(9,8) ✓ N = 28,80 N ✓ OPTION 2 Fnet,y = 0: N + F sin 30◦ – w = 0 ✓ N + 30 – 58,8 = 0 ✓ N = 28,80 N ✓ (3) Copyright reserved Please turn over

Physical Sciences/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines 2.4 Marking criteria • Fnet = ma (for A OR for B) ✓ • fk = µk N for block A: 0,2(4)(9,8) = 7,84 N ✓ • Substitution for block A ✓ • Substitution for block B (using fk = 0,2 × 28,8 = 5,76 N) ✓ • Final answer: 23,18 N (range: 23,16 N – 23,20 N) ✓ Take motion to the right as positive. Block A: Fnet = ma ✓ T – fkA = mA a T – (0,2)(4)(9,8) = 4a ✓✓ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(1) Block B: F cos θ – T – fkB = mB a 60 cos 30◦ – T – (0,2)(28,8) = 6a ✓ . . . . . . . . . . . . . . . . . . . . . . . . . . . (2) (1) + (2): 51,96 – 5,76 – 7,84 = 10a ⇒ a = 3,84 m·s–2 T = 4(3,84) + 7,84 = 23,18 N ✓ POSITIVE MARKING FROM QUESTION 2.3. (5) 2.5 Fnet = ma fk = ma ⇒ µk mg = ma (0,2)(4)(9,8) = 4a ✓ a = 1,96 m·s–2 (block A decelerates) ✓ Accept: a = –1,96 m·s–2 . (2) [17] Copyright reserved Please turn over

Physical Sciences/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 3.1 Motion during which the only force acting on the object is the gravitational force (weight). ✓✓ (2) 3.2 UPWARDS POSITIVE DOWNWARDS POSITIVE vf = vi + a∆t ✓ vf = vi + a∆t ✓ 0 = 14 + (–9,8)∆t ✓ 0 = –14 + (9,8)∆t ✓ ∆t = 1,43 s ✓ ∆t = 1,43 s ✓ (3) 3.3 OPTION 1 (upwards positive) v2f = v2i + 2a∆y ✓ 0 = 142 + 2(–9,8)∆y ✓ ∆y = 10 m Height above ground = 25 + 10 = 35 m ✓ OPTION 2 ∆y = vi ∆t + 12 a∆t2 = 14(1,43) + 12 (–9,8)(1,43)2 ✓✓ = 10,00 m Height above ground = 35 m ✓ ( ) Also accept ∆y = vi +v 2 f ∆t or energy conservation. POSITIVE MARKING FROM 3.2. (3) 3.4 UPWARDS POSITIVE ∆y = vi ∆t + 12 a∆t2 ✓ Ball P (from rest at its maximum height): –35 = 0 + 12 (–9,8)∆t2P ✓ ∆tP = 2,67 s Ball Q (dropped from rest): –25 = 0 + 12 (–9,8)∆t2Q ✓ ∆tQ = 2,26 s Time interval = 2,67 – 2,26 = 0,41 s ✓ Downwards positive: same values with positive displacements. POSITIVE MARKING FROM 3.3. Range: 0,41 s – 0,42 s. (4) Copyright reserved Please turn over

Physical Sciences/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines 3.5 v (m·s–1 ) 14 1,43 0 t (s) Marking criteria • Straight line with negative gradient, crossing the time axis ✓ • Initial velocity 14 m·s–1 shown ✓ • Time 1,43 s shown where the line crosses the time axis ✓ Downwards positive: line with positive gradient from –14 m·s–1 . Final velocity (–26,19 m·s–1 at 4,10 s) is NOT required. (3) [15] Copyright reserved Please turn over

Physical Sciences/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1 The total linear momentum of an isolated system remains constant (is conserved) (in magnitude and direction). ✓✓ (2) 4.2 EAST positive: Σpi = Σpf OR mA viA + mB viB = mA vfA + mB vfB ✓ (2)(3) + (1,5)(0) = (2)(0,9) + (1,5)vfB ✓✓ vfB = 2,80 m·s–1 EAST ✓ 1 mark: LHS substitution; 1 mark: RHS substitution. Final answer mark requires direction. (4) 4.3 Ek,before = 12 mv2 = 12 (2)(3)2 = 9,00 J ✓ Ek,after = 21 (2)(0,9)2 + 12 (1,5)(2,8)2 ✓ = 0,81 + 5,88 = 6,69 J ✓ Ek,before ̸= Ek,after (kinetic energy is not conserved), therefore the collision is INELASTIC. ✓ POSITIVE MARKING FROM 4.2. (4) 4.4 OPTION 1 (trolley B) Fnet ∆t = ∆p = m(vf – vi ) ✓ Fnet (0,15) = 1,5(2,8 – 0) ✓ Fnet = 28,00 N ✓ OPTION 2 (trolley A, Newton’s Third Law) Fnet (0,15) = 2(0,9 – 3) ✓✓ F = –28 N on A, therefore A exerts 28,00 N on B ✓ POSITIVE MARKING FROM 4.2. (3) [13] Copyright reserved Please turn over

Physical Sciences/P1 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 A force for which the work done in moving an object between two points depends on the path taken. ✓✓ (2) 5.2 N✓ F✓ f / fk ✓ w / Fg ✓ F parallel to incline (up); f parallel to incline (down); N perpendicular to incline; w vertically downwards. Any additional force: deduct 1 mark. (4) 5.3 Marking criteria • Formula Wnc = ∆K + ∆U (or Wnet = ∆K) ✓ • Work done by F and by friction substituted ✓ • ∆K substituted ✓ • ∆U (or Wg ) substituted ✓ • Final answer 398,50 N (range: 398,40 N – 398,60 N) ✓ OPTION 1 Wnc = ∆K + ∆U ✓ WF + Wf = ( 12 mv2f – 12 mv2i ) + (mghf – mghi ) F(8) cos 0◦ + (90)(8) cos 180◦ ✓ = 12 (60)(4)2 – 0 ✓ +(60)(9,8)(8 sin 25◦ – 0) ✓ 8F – 720 = 480 + 1 988,00 F = 398,50 N ✓ OPTION 2 Wnet = ∆K ✓ WF + Wf + Wg = ∆K F(8) cos 0◦ + 90(8) cos 180◦ ✓ +(60)(9,8)(8) cos 115◦ ✓ = 12 (60)(4)2 – 0 ✓ F = 398,50 N ✓ (5) Copyright reserved Please turn over

Physical Sciences/P1 10 Hlayiso/Mock 2026 NSC – Marking Guidelines 5.4 OPTION 1 Pave = Fvave ✓ () 0+4 Pave = (398,50) ✓ 2 Pave = 797,00 W ✓ OPTION ( 2) ( ) vi +vf 0+4 ∆x = 2 ∆t: 8 = 2 ∆t ⇒ ∆t = 4 s W (398,50)(8) P= = ✓✓ = 797,00 W ✓ ∆t 4 POSITIVE MARKING FROM 5.3. (3) [14] QUESTION 6 6.1 It is the (apparent) change in frequency (or pitch) of the sound detected by a listener because the sound source and the listener have different velocities relative to the medium of propagation. ✓✓ OR An apparent change in observed frequency (pitch) as a result of the relative motion between a source and an observer. ✓✓ (2) 6.2 Reading 2 (920 Hz) ✓ (1) 6.3 Marking criteria • Formula ✓ • Numerator v = 340 in both equations ✓ • Denominator (340 – vs ) for approaching ✓ • Denominator (340 + vs ) for receding ✓ • Eliminating fs / equating ✓ • Final answer 28,00 m·s–1 ✓ v ± vL fL = fs ✓ v ± vs ( ) 340 Approaching: 920 = fs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) 340 – vs ( ) 340 Receding: 780 = fs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(2) ✓✓✓ 340 + vs 920(340 – vs ) 780(340 + vs ) fs = = ✓ 340 340 920(340 – vs ) = 780(340 + vs ) ⇒ 1 700 vs = 47 600 vs = 28,00 m·s–1 ✓ Note: fs = 844,24 Hz (not required). Candidates who calculate fs first and then vs : award marks accordingly. POSITIVE MARKING FROM 6.2 (if readings swapped, max 4/6). (6) Copyright reserved Please turn over

Physical Sciences/P1 11 Hlayiso/Mock 2026 NSC – Marking Guidelines 6.4 The galaxy is moving AWAY from the Earth. ✓ The light detected on Earth has a longer wavelength (lower frequency) than the light emitted, i.e. the spectral lines shift towards the red end of the spectrum. ✓ (2) [11] QUESTION 7 QX + QY (+6 × 10–9 ) + (–2 × 10–9 ) 7.1.1 Qnew = = ✓ 2 2 QX = +2 × 10–9 C (+2 nC) ✓ (2) 7.1.2 OPTION 1 (sphere Y) ∆Q (+2 × 10–9 ) – (–2 × 10–9 ) n= = ✓ qe 1,6 × 10–19 n = 2,50 × 1010 electrons ✓ Electrons are transferred from sphere Y to sphere X. ✓ OPTION 2 (sphere X) ∆Q 6 × 10–9 – 2 × 10–9 n= = ✓ = 2,50 × 1010 ✓ From Y to X ✓ qe 1,6 × 10–19 POSITIVE MARKING FROM 7.1.1. (3) 7.2.1 The (electrostatic) force experienced per unit positive charge placed at that point. ✓✓ (2) 7.2.2 Marking criteria • Formula E = kQ r2 ✓ • Substitution for E1 ✓ • Substitution for E2 ✓ • Vector addition (subtraction of magnitudes) ✓ • Final answer with direction ✓ Take RIGHT as positive. kQ E= 2 ✓ r (9 × 109 )(4 × 10–6 ) E1 = ✓ = 1,00 × 105 N·C–1 to the right (away from Q1 ) (0,6)2 (9 × 109 )(6 × 10–6 ) E2 = ✓ = 1,35 × 106 N·C–1 to the left (towards Q2 ) (0,2)2 Enet = 1,00 × 105 + (–1,35 × 106 ) ✓ Enet = 1,25 × 106 N·C–1 to the left (towards Q2 ) ✓ (5) Copyright reserved Please turn over

Physical Sciences/P1 12 Hlayiso/Mock 2026 NSC – Marking Guidelines 7.2.3 F = qE F = (1,6 × 10–19 )(1,25 × 106 ) ✓ F = 2,00 × 10–13 N ✓ to the right (away from Q2 ) ✓ Also accept: F = kQ12qe and F = kQ22qe added as vectors. POSITIVE MARKING FROM r1 r2 7.2.2. (3) [15] QUESTION 8 8.1 The maximum energy provided (work done) by a battery per coulomb (unit charge) passing through it. ✓✓ (2) 1 1 1 1 1 R2 R 3 6 × 12 8.2 = + = + ✓ OR Rp = = ✓ Rp R2 R3 6 12 R2 + R3 6 + 12 Rp = 4 Ω ✓ Rext = R1 + Rp = 3 + 4 = 7 Ω ✓ (3) 8.3 OPTION 1 V 10,5 I= = ✓ = 1,5 A ✓ R 7 ε = I(R + r) 12 = 1,5(7 + r) ✓ r = 1,00 Ω ✓ OPTION 2 10,5 V 1,5 I= = 1,5 A ✓✓ Vlost = 12 – 10,5 = 1,5 V ✓ r = lost = = 1,00 Ω ✓ 7 I 1,5 POSITIVE MARKING FROM 8.2. (4) 8.4 OPTION 1 Vp = IRp = (1,5)(4) = 6 V ✓ Vp 6 I2 = = ✓ = 1,00 A ✓ R2 6 OPTION 2 V1 = IR1 = (1,5)(3) = 4,5 V; Vp = 10,5 – 4,5 = 6 V ✓ I2 = 66 ✓ = 1,00 A ✓ Also accept the current-divider method: I2 = 12 18 (1,5) = 1 A. POSITIVE MARKING FROM 8.3. (3) 8.5 I3 = 1,5 – 1,0 = 0,5 A W = I2 R∆t ✓ W = (0,5)2 (12)(2 × 60) ✓ W = 360,00 J ✓ V2 ∆t (6)2 (120) OR W = = ✓✓ = 360,00 J ✓ R 12 POSITIVE MARKING FROM 8.4. (3) Copyright reserved Please turn over

Physical Sciences/P1 13 Hlayiso/Mock 2026 NSC – Marking Guidelines 8.6 INCREASES ✓ The resistance of the external circuit increases (from 7 Ω to 9 Ω). ✓ The (total) current in the circuit decreases. ✓ The potential difference across the internal resistance (”lost volts”, Ir) decreases, so the terminal potential difference (V = ε – Ir) increases. ✓ (4) [19] QUESTION 9 9.1.1 Slip rings ✓ (1) 9.1.2 Electromagnetic induction (Faraday’s Law) ✓ (1) 9.1.3 Mechanical (kinetic) energy to electrical energy ✓ (1) 9.1.4 ANY ONE: Increase the speed (rate) of rotation of the coil / Use a stronger magnet / Increase the number of turns (windings) on the coil / Increase the area of the coil ✓ (1) 9.2.1 The rms current is the value of the direct current that dissipates the same amount of energy (power) as the alternating current (in the same resistor). ✓✓ (2) 9.2.2 Pave = Vrms Irms 2 000 = (230)Irms ✓ Irms = 8,70 A Imax Imax Irms = √ ⇒ 8,70 = √ ✓ 2 2 Imax = 12,30 A ✓ (range: 12,30 A – 12,31 A) (3) 9.2.3 Energy = P∆t = (2 kW)(0,25 h × 30) ✓ Energy = 15 kW·h ✓ Cost = 15 × R3,50 = R52,50 ✓ OR: W = P∆t = 2 000 × (15 × 60 × 30) = 5,4 × 107 J = 15 kW·h. Cost = R52,50. (3) [12] Copyright reserved Please turn over

Physical Sciences/P1 14 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 10 10.1 The process whereby electrons are emitted (ejected) from a metal surface when light (electromagnetic radiation) of suitable frequency is incident on it. ✓✓ (2) 10.2 5,5 × 1014 Hz ✓ (1) 10.3 W0 = hf0 ✓ W0 = (6,63 × 10–34 )(5,5 × 1014 ) ✓ W0 = 3,65 × 10–19 J ✓ POSITIVE MARKING FROM 10.2. (3) 10.4 Marking criteria • Formula E = W0 + Ek(max) / hc λ = W0 + 21 mv2max ✓ • Substitution of hc λ ✓ • Substitution of W0 ✓ • Substitution of 12 mv2 ✓ • Final answer ✓ hc = W0 + 12 mv2max ✓ λ (6,63 × 10–34 )(3 × 108 ) ✓ = 3,65 × 10–19 ✓ + 21 (9,11 × 10–31 )v2max ✓ 400 × 10 –9 4,97 × 10–19 = 3,65 × 10–19 + 12 (9,11 × 10–31 )v2max vmax = 5,39 × 105 m·s–1 ✓ (range: 5,38 × 105 – 5,40 × 105 m·s–1 ) POSITIVE MARKING FROM 10.3. (5) 10.5 INCREASES ✓ A higher intensity means more photons are incident on the metal per second (per unit time). ✓ More photoelectrons are therefore emitted per second, so the current (rate of flow of charge) increases. ✓ (3) [14] TOTAL: 150 Copyright reserved