Physical Sciences Grade 12 P1 2026 Memo (Eng)

hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: PHYSICS (P1) PREDICTION EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 16 pages. Copyright reserved Please turn over

Physical Sciences/P1 2 Hlayiso/Prediction 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, deduct 1 mark (once). 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/Prediction 2026 NSC – Marking Guidelines QUESTION 1 1.1 B ✓✓ (2) 1.2 D ✓✓ (2) 1.3 C ✓✓ (2) 1.4 A ✓✓ (2) 1.5 B ✓✓ (2) 1.6 D ✓✓ (2) 1.7 A ✓✓ (2) 1.8 C ✓✓ (2) 1.9 A ✓✓ (2) 1.10 D ✓✓ (2) [20] Notes for teachers (not for marking): 1.2: w = G Mm r2 , so wX = 232 w = 43 w. 1.4: impulse = area under the graph = 21 (4)(10) = 20 N·s = m∆v, so v = 10 m·s–1 . 1.5: the source approaches, so the wavefronts are compressed: shorter wavelength, higher frequency. 1.7: after contact each sphere carries +2q; 2 F′ = k(2q)(2q) (2r)2 = kq r2 1 · k(6q)(2q) , repulsion. 1.8: the current in R is half the current in R , so P = = 12 r2 2 1 2 I 2 1 ( 2 ) R = 4 P. Copyright reserved Please turn over

Physical Sciences/P1 4 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 2 2.1 When a (non-zero) resultant/net force acts on an object, the object will accelerate in the direction of the net force with an acceleration that 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). ✓✓ (2) 2.2 N / FN (normal force) ✓ fk / f (kinetic friction) ✓ T / FT (tension) ✓ F (applied force) ✓ w / Fg (weight) ✓ Mark awarded for arrow AND label. Do not penalise relative lengths of the arrows. Any additional force: deduct 1 mark. F must point down and to the right. (5) 2.3 N = mg + F sin 30◦ = (6)(9,8) + (20) sin 30◦ = 68,80 N ✓ fk = µk N fk = (0,35)(68,80) ✓ fk = 24,08 N ✓ If the candidate uses N = mg only (fk = 20,58 N): maximum 1/3 (substitution into fk = µk N). (3) 2.4 Marking criteria • Fnet = ma (any one) ✓ • Correct equation for block A ✓ • Correct equation for block B ✓ • Final answer 1,61 m·s–2 (range: 1,60 – 1,61 m·s–2 ) ✓ Take the direction of motion as positive (A: to the right; B: downwards). Block A: Fnet = ma ✓ F cos 30◦ + T – fk = mA a 20 cos 30◦ + T – 24,08 = 6a ✓ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) Block B: mB g – T = mB a (2)(9,8) – T = 2a ✓ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2) (1) + (2): 17,32 + 19,6 – 24,08 = 8a a = 1,61 m·s–2 ✓ (Tension T = 16,39 N is not required.) POSITIVE MARKING FROM 2.3. (4) Copyright reserved Please turn over

Physical Sciences/P1 5 Hlayiso/Prediction 2026 NSC – Marking Guidelines 2.5 INCREASES ✓ The vertical (downward) component of F, F sin θ , increases, so the normal force on block A increases (fk = µk N). ✓ (2) [16] QUESTION 3 3.1 Motion during which the only force acting on the object is the gravitational force. ✓✓ (2) 3.2 UPWARDS POSITIVE DOWNWARDS POSITIVE v2f = v2i + 2a∆y ✓ v2f = v2i + 2a∆y ✓ 0 = (20)2 + 2(–9,8)∆y ✓ 0 = (–20)2 + 2(9,8)∆y ✓ ∆y = 20,41 m ✓ ∆y = –20,41 m (height 20,41 m) ✓ Also accept: ∆t = 2,04 s from vf = vi +a∆t, then ∆y = vi ∆t+ 21 a∆t2 or energy conservation. (3) 3.3 Ball Y must rise 20,41 – 8 = 12,41 m above the balcony. ✓ v2f = v2i + 2a∆y 0 = v2i + 2(–9,8)(12,41) ✓ vi = 15,59 m·s–1 ✓ (range: 15,59 – 15,60 m·s–1 ) DOWNWARDS POSITIVE: 0 = v2i + 2(9,8)(–12,41) ✓ vi = –15,59 m·s–1 , so speed = 15,59 m·s–1 ✓ POSITIVE MARKING FROM 3.2. Using 20,41 m instead of 12,41 m: maximum 1/3. (3) 3.4 OPTION 1 Ball X: vf = vi + a∆t: 0 = 20 + (–9,8)∆t ∆tX = 2,04 s ✓ Ball Y: 0 = 15,59 + (–9,8)∆t ∆tY = 1,59 s ✓ Both balls fall from rest from the SAME maximum height, so they take the same time to fall to the ground. Time interval = 2,04 – 1,59 = 0,45 s (ball Y strikes the ground first) ✓ OPTION 2 Ball X: total time = 2(2,04) = 4,08 s ✓ Ball Y: ∆y = vi ∆t + 12 a∆t2 : –8 = 15,59∆t + 12 (–9,8)∆t2 ∆t = 3,63 s ✓ Time interval = 4,08 – 3,63 = 0,45 s ✓ Range: 0,44 s – 0,46 s. POSITIVE MARKING FROM 3.3. (3) Copyright reserved Please turn over

Physical Sciences/P1 6 Hlayiso/Prediction 2026 NSC – Marking Guidelines 3.5 v (m·s–1 ) 20 15,59 2,04 t (s) 0 1,59 X Y Marking criteria • Two straight lines, each with a negative gradient and crossing the time axis ✓ • The lines are parallel (same gradient) ✓ • Initial velocities: 20 m·s–1 (X) and 15,59 m·s–1 (Y) shown ✓ • Times at maximum height: 2,04 s (X) and 1,59 s (Y) shown ✓ Downwards positive: mirror image (lines with positive gradient starting at –20 and –15,59 m·s–1 ). End points (4,08 s; 3,63 s; –20 m·s–1 ) are NOT required. POSITIVE MARKING FROM 3.3 and 3.4. (4) [15] Copyright reserved Please turn over

Physical Sciences/P1 7 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 4 4.1 The total linear momentum of an isolated system remains constant (is conserved). ✓✓ (2) 4.2 EAST positive. From the graph: viA = 3 m·s–1 and vfA = –1,2 m·s–1 . Σpi = Σpf OR mA viA + mB viB = mA vfA + mB vfB ✓ (2)(3) + (3)(–1) ✓ = (2)(–1,2) + 3vfB ✓ vfB = 1,80 m·s–1 EAST ✓ 1 mark: LHS substitution; 1 mark: RHS substitution. Final mark requires the direction. (4) 4.3 Contact time (from the graph): ∆t = 1,2 – 1,0 = 0,2 s OPTION 1 (trolley B) Fnet ∆t = ∆p = m(vf – vi ) ✓ ( ) Fnet (0,2) = 3 1,8 – (–1) ✓ Fnet = 42,00 N ✓ OPTION 2 (trolley A, Newton’s Third Law) Fnet (0,2) = 2(–1,2 – 3) ✓✓ Fnet = –42 N on A, so A exerts 42,00 N on B ✓ POSITIVE MARKING FROM 4.2 (Option 1). (3) 4.4 Ek = 21 mv2 ✓ Ek,before = 21 (2)(3)2 + 12 (3)(1)2 = 10,50 J ✓ Ek,after = 12 (2)(1,2)2 + 12 (3)(1,8)2 = 6,30 J ✓ Ek,before is not equal to Ek,after (kinetic energy is not conserved), so the collision is INELASTIC. ✓ POSITIVE MARKING FROM 4.2. (4) [13] Copyright reserved Please turn over

Physical Sciences/P1 8 Hlayiso/Prediction 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 / FN ✓ fk / f (up the slope) ✓ w / Fg ✓ The crate slides down to the left: friction is parallel to the slope, pointing UP the slope; N is perpendicular to the slope. Any additional force: deduct 1 mark. (3) 5.3 Marking criteria • fk = µk mg cos 30◦ calculated ✓ • Formula Wnc = ∆K + ∆U OR Wnet = ∆K ✓ • Work done by friction substituted ✓ • ∆K and ∆U (or Wg ) substituted ✓ • Final answer 5,77 m·s–1 (range: 5,77 – 5,78 m·s–1 ) ✓ fk = µk N = µk mg cos 30◦ = (0,25)(40)(9,8) cos 30◦ = 84,87 N ✓ OPTION 1 Wnc = ∆K + ∆U ✓ fk ∆x cos 180◦ = ( 12 mv2f – 12 mv2i ) + (mghf – mghi ) (84,87)(6) cos 180◦ ✓ = ( 12 (40)v2f – 0) + (0 – (40)(9,8)(6 sin 30◦ )) ✓ –509,22 = 20v2f – 1 176 vf = 5,77 m·s–1 ✓ OPTION 2 Wnet = ∆K ✓ Wg + Wf = ∆K (40)(9,8)(6) cos 60◦ + (84,87)(6) cos 180◦ ✓✓ = 21 (40)v2f – 0 vf = 5,77 m·s–1 ✓ (5) Copyright reserved Please turn over

Physical Sciences/P1 9 Hlayiso/Prediction 2026 NSC – Marking Guidelines 5.4 Constant speed, so Fnet = 0: Fwinch = mg sin 30◦ + fk ✓ Fwinch = (40)(9,8) sin 30◦ + 84,87 = 280,87 N ✓ Pave = Fvave Pave = (280,87)(0,8) ✓ Pave = 224,70 W ✓ (range: 224,69 – 224,70 W) POSITIVE MARKING FROM 5.3 (value of fk ). (4) [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 (listener). ✓✓ (2) 6.2 As the passenger (listener) moves towards the siren, the passenger meets (encounters) more wavefronts per second than when stationary. ✓ The number of waves received per second (the detected frequency) is therefore higher than the frequency emitted. ✓ Note: the wavelength of the sound in the air does not change; the relative speed of the waves with respect to the listener increases. (2) Copyright reserved Please turn over

Physical Sciences/P1 10 Hlayiso/Prediction 2026 NSC – Marking Guidelines 6.3 Marking criteria • Formula ✓ • (340 + vL ) in the numerator for approaching ✓ • (340 – vL ) in the numerator for moving away ✓ • Denominator 340 (vs = 0) in both equations ✓ • Eliminating fs / equating ✓ • Final answer 17,00 m·s–1 ✓ v ± vL fL = fs ✓ v ± vs ( ) 340 + vL Approaching: 892,5 = fs ✓✓ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) 340 ( ) 340 – vL Moving away: 807,5 = fs ✓ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2) 340 892,5 340 + vL (1) ÷ (2): = ✓ 807,5 340 – vL 892,5(340 – vL ) = 807,5(340 + vL ) ⇒ 1 700 vL = 28 900 vL = 17,00 m·s–1 ✓ Also accept: (1) + (2) gives fs = 850 Hz first, then substitution into (1). Candidates who treat the taxi as the SOURCE (vs in the denominator): maximum 2/6 (formula and eliminating fs ). (6) 6.4 OPTION(1 ) 340 + 17 892,5 = fs ✓ fs = 850,00 Hz ✓ 340 OPTION 2 (1) + (2): 892,5 + 807,5 = 2fs ✓ fs = 850,00 Hz ✓ POSITIVE MARKING FROM 6.3. (2) [12] Copyright reserved Please turn over

Physical Sciences/P1 11 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 7 7.1 + – Marking criteria • Correct shape (field lines curving from P to Q) ✓ • Direction of the field lines: away from P (+) and towards Q (–) ✓ • Field lines start/end on the spheres (radial at the spheres) and do not cross ✓ (3) 7.2 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) Copyright reserved Please turn over

Physical Sciences/P1 12 Hlayiso/Prediction 2026 NSC – Marking Guidelines 7.3 Marking criteria • Formula F = kQr12Q2 ✓ • Substitution for FQP ✓ • Substitution for FRP ✓ • Pythagoras (vector addition of perpendicular forces) ✓ • Final answer 1,13 N WITH direction ✓ kQ1 Q2 F= ✓ r2 (9 × 109 )(3 × 10–6 )(3 × 10–6 ) FQP = ✓ = 0,90 N EAST (attraction, towards Q) (0,3)2 (9 × 109 )(3 × 10–6 )(4 × 10–6 ) FRP = ✓ = 0,675 N SOUTH (repulsion, away (0,4)2 from R) √ Fnet = (0,90)2 + (0,675)2 ✓ Fnet = 1,13 N ✓ (1,125 N) (range: 1,12 – 1,13 N) 0,675 tan θ = θ = 36,87◦ south of east (OR 53,13◦ east of south; bearing 0,90 126,87◦ ) Final mark: magnitude AND direction. Do not penalise if the direction is described in an equivalent way. (5) 7.4 OPTION 1 F 1,125 E= ✓ E= ✓ E = 3,75 × 105 N·C–1 ✓ q 3 × 10 –6 OPTION 2 kQ (9 × 109 )(3 × 10–6 ) E = 2 : EQ = = 3,00 × 105 N·C–1 (east); ER = r (0,3)2 (9 × 109 )(4 × 10–6 ) = 2,25 × 105 N·C–1 (south) ✓✓ (0,4)2 √ Enet = (3,00 × 105 )2 + (2,25 × 105 )2 = 3,75 × 105 N·C–1 ✓ POSITIVE MARKING FROM 7.3 (Option 1). Range: 3,75 × 105 – 3,77 × 105 N·C–1 . (3) [13] Copyright reserved Please turn over

Physical Sciences/P1 13 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 8 8.1 The maximum energy provided (work done) by a battery per coulomb (unit charge) passing through it. ✓✓ (2) V2 8.2 P= ✓ R (6)2 9= ✓ R RL = 4,00 Ω ✓ Also accept: I = P V 6 V = 1,5 A and R = I = 1,5 = 4 Ω. (3) 8.3 The parallel combination has a potential difference of 6 V (the bulb works as rated). P 9 IL = = = 1,5 A ✓ V 6 V 6 6 I2 = = = 1 A and I3 = = 0,5 A ✓ R 6 12 Itotal = 1,5 + 1 + 0,5 = 3,00 A ✓ 1 1 1 1 V 6 OR: = + + , Rp = 2 Ω ✓ I = = ✓ = 3,00 A ✓ Rp 4 6 12 Rp 2 POSITIVE MARKING FROM 8.2. (3) 8.4 OPTION 1 1 1 1 1 = + + ⇒ Rp = 2 Ω ✓ Rp 4 6 12 Rext = R1 + Rp = 1,5 + 2 = 3,5 Ω ✓ ε = I(R + r) 12 = 3(3,5 + r) ✓ r = 0,50 Ω ✓ OPTION 2 VR1 = IR = (3)(1,5) = 4,5 V ✓ Vext = 4,5 + 6 = 10,5 V ✓ V 1,5 Vlost = ε – Vext = 12 – 10,5 = 1,5 V ✓ r = lost = = 0,50 Ω ✓ I 3 POSITIVE MARKING FROM 8.3. (4) 8.5 W = I2 R∆t ✓ W = (3)2 (0,5)(2 × 60) ✓ W = 540,00 J ✓ OR: W = VI∆t = (1,5)(3)(120) = 540 J. POSITIVE MARKING FROM 8.3 and 8.4. (3) Copyright reserved Please turn over

Physical Sciences/P1 14 Hlayiso/Prediction 2026 NSC – Marking Guidelines 8.6 INCREASES ✓ When S is opened, the resistance of the parallel section (and of the external circuit) increases (from 2 Ω to 2,4 Ω). ✓ The total current in the circuit decreases. ✓ The potential difference across R1 and across the internal resistance (Ir) decreases, so the potential difference across the parallel section (bulb L) 2 increases and the power of the bulb (P = VR ) increases. ✓ (With S open: I = 2,73 A and VL = 6,55 V. Not required.) (4) [19] QUESTION 9 9.1.1 Slip rings ✓ (1) 9.1.2 A to B ✓✓ Side AB moves upwards while the field points from N to S (to the right). By the right-hand (dynamo) rule, the induced current in AB flows from the front (A) to the back (B). (2) 9.1.3 emf t Marking criteria • Sinusoidal shape showing ONE complete cycle ✓ • Graph starts at a MAXIMUM emf at t = 0 (positive or negative maximum) ✓ (2) 9.1.4 Any TWO: Increase the speed (frequency) of rotation of the coil. ✓ Increase the number of turns (windings) of the coil. ✓ Use stronger magnets (increase the magnetic field strength). Increase the area of the coil. (2) 9.2.1 The rms potential difference is the DC potential difference that dissipates the same amount of energy (power) as the AC potential difference (in the same resistor). ✓✓ (2) 9.2.2 From the graph: T = 0,02 s 1 1 f= = ✓ T 0,02 f = 50,00 Hz ✓ (2) Copyright reserved Please turn over

Physical Sciences/P1 15 Hlayiso/Prediction 2026 NSC – Marking Guidelines 9.2.3 OPTION 1 Vmax 325 Vrms = √ = √ ✓ = 229,81 V ✓ 2 2 V 2 (229,81)2 Pave = rms = ✓ R 44 Pave = 1 200,29 W ✓ OPTION 2 Vrms 229,81 Vrms = 229,81 V ✓✓ Irms = = = 5,22 A Pave = Vrms Irms = R 44 (229,81)(5,22) ✓ = 1 199,61 W ✓ (1 200,29 W if Irms is not rounded) Range: 1 199,61 W – 1 200,40 W. Using Vmax instead of Vrms (P = 2 400,57 W): maximum 2/4. (4) [15] QUESTION 10 10.1.1 The minimum energy that an electron in the metal needs to be emitted from the metal surface. ✓✓ (2) 10.1.2 OPTION 1 hc E= ✓ λ (6,63 × 10–34 )(3 × 108 ) Red: E= = 2,93 × 10–19 J ✓ 680 × 10–9 (6,63 × 10–34 )(3 × 108 ) Violet: E = = 4,85 × 10–19 J ✓ 410 × 10 –9 Only the VIOLET light ejects electrons: its photon energy is greater than the work function (3,68 × 10–19 J), while the energy of a red photon is less than the work function. ✓ OPTION 2 W 3,68 × 10–19 f0 = 0 = = 5,55 × 1014 Hz ✓ h 6,63 × 10 –34 c 3 × 108 14 Hz ✓ f 3 × 108 fred = = = 4,41 × 10 violet = = 7,32 × 1014 λ 680 × 10–9 410 × 10–9 Hz ✓ Only violet: fviolet > f0 and fred < f0 . ✓ hc = 540 nm; 410 nm < 540 nm < 680 nm, so OPTION 3: threshold wavelength λ0 = W 0 only violet ejects electrons. (4) 10.1.3 E = W0 + 12 mv2max ✓ 4,85 × 10–19 = 3,68 × 10–19 + 21 (9,11 × 10–31 )v2max ✓ vmax = 5,07 × 105 m·s–1 ✓ Range: 5,06 ×105 – 5,08×105 m·s–1 . POSITIVE MARKING FROM 10.1.2. If the RED light is used: no marks for the final answer (no electrons are ejected). (3) Copyright reserved Please turn over

Physical Sciences/P1 16 Hlayiso/Prediction 2026 NSC – Marking Guidelines 10.1.4 REMAINS THE SAME ✓ The frequency (energy) of each photon is unchanged; only the number of photons per second increases, and each electron absorbs the energy of one photon. ✓ (2) 10.2.1 Absorption (line) spectrum ✓ (1) 10.2.2 The cool hydrogen atoms absorb photons with energies (frequencies) equal to the energy differences between their energy levels; the electrons move from lower to higher energy levels. ✓ Light of these specific frequencies (wavelengths) is therefore missing from the transmitted light (it is re-emitted in all directions), so dark lines appear at these positions. ✓ (2) [13] TOTAL: 150 Copyright reserved