Physical Sciences Grade 10 Paper 1 November Mock Exam 2 2026 Memo

Copyright reserved Please turn over Mock hlayiso mock exams Practice Series: Grade 10 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 10 PHYSICAL SCIENCES: PHYSICS (P1) NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 100 These marking guidelines consist of 7 pages.

Physical Sciences/P1 2 NSC – Marking Guidelines Hlayiso/Mock 2026 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. 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. 7. Vectors: A direction must be given wherever a vector quantity (displacement, velocity, acceleration) is asked for; candidates may choose either direction as positive if applied consistently. Copyright reserved Please turn over

Physical Sciences/P1 3 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 1 1.1 D ✓✓ (2) 1.2 C ✓✓ (2) 1.3 C ✓✓ (2) 1.4 A ✓✓ (2) 1.5 A ✓✓ (2) 1.6 B ✓✓ (2) 1.7 B ✓✓ (2) 1.8 A ✓✓ (2) [16] Notes for teachers: 1.2: Ek,P = 1 2mv2; Ek,Q = 1 2(2m)(1 2v)2 = 1 4mv2, so the ratio is 2. 1.3: moving west gives negative velocity; slowing down means the acceleration is opposite to the velocity, i.e. positive. 1.4: λ = v/f and T = 1/f, so both are halved. 1.7: the parallel pair has resistance 1 2R, so the 12 V divides 2 : 1 (8 V and 4 V). QUESTION 2 2.1 A wave in which the particles of the medium vibrate parallel to the direction of motion of the wave. ✓✓ (2) 2.2 A: compression ✓ B: rarefaction ✓ (2) 2.3 The distance from the centre of a compression (A) to the centre of the next rarefaction (B) is half a wavelength. ✓ λ = 2×0,2 = 0,4 m ✓ (2) 2.4 f = 12 3 = 4 Hz ✓ v = fλ ✓ = (4)(0,4) = 1,6 m·s–1 ✓ Accept: T = 3 12 = 0,25 s; v = λ T = 0,4 0,25 = 1,6 m·s–1. POSITIVE MARKING FROM 2.3. (3) 2.5 v = ∆x ∆t ✓ 1,6 = 2,4 ∆t ✓ ∆t = 1,5 s ✓ OR: 2,4 m = 6 wavelengths; 6×T = 6×0,25 = 1,5 s. POSITIVE MARKING FROM 2.4. (3) [12] Copyright reserved Please turn over

Physical Sciences/P1 4 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 3 3.1.1 Light travels much faster (3× 108 m·s–1) than sound (340 m·s–1). ✓ (1) 3.1.2 v = ∆x ∆t ✓ 340 = ∆x 4,2 ✓ ∆x = 1428 m ✓ (3) 3.2.1 Electromagnetic waves are produced by accelerating (oscillating) charges. ✓ The oscillating charge produces electric and magnetic fields that oscillate at right angles to each other (and to the direction of motion of the wave). ✓ (2) 3.2.2 E = hf ✓ E = (6,63×10–34)(2,45×109) ✓ E = 1,62×10–24 J ✓ (3) 3.2.3 c = fλ 3×108 = (2,45×109)λ ✓ λ = 0,12 m (0,122 m) ✓ OR: E = h c λ ⇒λ = (6,63×10–34)(3×108) 1,62×10–24 = 0,12 m. POSITIVE MARKING FROM 3.2.2. (2) [11] QUESTION 4 4.1 Every charge in the universe consists of an integer multiple of the charge of one electron (e = 1,6×10–19 C). ✓✓ (2) 4.2 Q = QP +QQ 2 +2,5×10–9 = (–3×10–9)+QQ 2 ✓ QQ = +8×10–9 C (+8 nC) ✓ (2) 4.3 P LOST electrons. ✓ Change in charge on P: ∆Q = (+2,5)–(–3) = +5,5 nC ✓ n = Q e = 5,5×10–9 1,6×10–19 ✓ n = 3,44×1010 electrons ✓ Accept ∆Q for Q: 2,5–8 = –5,5 nC. POSITIVE MARKING FROM 4.2. (4) [8] Copyright reserved Please turn over

Physical Sciences/P1 5 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 5 5.1 The energy transferred per unit (coulomb of) charge between two points in a circuit (OR the work done per unit charge). ✓✓ (2) 5.2 V15Ω= IR = (0,4)(15) = 6 V ✓ Ix = 1,0–0,4 = 0,6 A ✓ R = V I ✓ = 6 0,6 Rx = 10 Ω✓ OR: Rp = 6 1,0 = 6 Ω; 1 6 = 1 15 + 1 Rx ; Rx = 10 Ω. (4) 5.3 V1 = IR = (1,0)(5) = 5 V ✓ emf = V1 +Vp = 5+6 = 11 V ✓ emf of each cell = 11 5 = 2,2 V ✓ OR: Rtotal = 5 + 6 = 11 Ω; emf = IR = (1)(11) = 11 V; each cell 2,2 V. POSITIVE MARKING FROM 5.2. (3) 5.4 W = VQ = VI∆t ✓ W = (6)(0,6)(90) ✓ W = 324 J ✓ OR: Q = I∆t = (0,6)(90) = 54 C; W = VQ = (6)(54) = 324 J. POSITIVE MARKING FROM 5.2. (3) 5.5 DECREASES ✓ The total resistance of the circuit INCREASES (from 11 Ωto 20 Ω), so the current through the 5 Ωresistor DECREASES and V1 = IR decreases (5 V to 2,75 V). ✓ (2) [14] Copyright reserved Please turn over

Physical Sciences/P1 6 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 6 6.1 The rate of change of position (displacement per unit time) ✓✓ Accept: speed in a specified direction. (2) 6.2 ∆x = v∆t = (12)(6) ✓ ∆x = 72 m ✓ (2) 6.3 vf = vi +a∆t ✓ vf = 0+(1,5)(6) ✓ vf = 9 m·s–1 east ✓ (3) 6.4 Car: ∆x = vi∆t+ 1 2a∆t2 = 0+ 1 2(1,5)(6)2 = 27 m ✓ Position of car from the starting point of the motorbike: 40+27 = 67 m ✓ Motorbike: 72 m > 67 m, so the MOTORBIKE is in front, by 5 m. ✓ Accept ∆x = (vf+vi 2 ) ∆t = ( 9+0 2 ) (6) = 27 m. POSITIVE MARKING FROM 6.2 AND 6.3. (3) 6.5 t (s) v (m·s–1) motorbike car 12 8 Marking criteria Motorbike: horizontal line at 12 m·s–1, labelled ✓ Car: straight line through the origin with positive gradient, labelled ✓ The two graphs intersect where the car reaches 12 m·s–1 ✓ Values 12 m·s–1 (velocity axis) and 8 s (time axis) shown ✓ Time: vf = vi +a∆t: 12 = 0+1,5∆t, ∆t = 8 s. (4) [14] Copyright reserved Please turn over

Physical Sciences/P1 7 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 7 7.1 At t = 9 s ✓ The velocity changes from POSITIVE to NEGATIVE (the graph crosses the time axis), so the boat starts moving west. ✓ (2) 7.2 a = ∆v ∆t ✓ = –3–6 10–7 ✓ a = –3 m·s–2, i.e. 3 m·s–2 west ✓ (3) 7.3 ∆x = area between the graph and the time axis ✓ Area above the axis (0–9 s): 1 2(3)(6)+(4)(6)+ 1 2(2)(6) = 39 m ✓ Area below the axis (9–12 s): 1 2(1)(–3)+(2)(–3) = –7,5 m ✓ ∆x = 39–7,5 = 31,5 m east ✓ OR area above the axis as trapezium 1 2(9 + 4)(6) = 39 m and below the axis as trapezium 1 2(3+2)(3) = 7,5 m. (4) 7.4 GREATER THAN ✓ The boat changes direction (moves back west), so the total distance travelled is greater than the magnitude of the displacement (in the same time). ✓ (2) 7.5 vave = ∆x ∆t ✓ = 31,5 12 ✓ vave = 2,63 m·s–1 east ✓ Accept 2,62 or 2,63 m·s–1 east (exact value 2,625). POSITIVE MARKING FROM 7.3. (3) [14] QUESTION 8 8.1 The energy an object possesses as a result of its motion. ✓✓ (2) 8.2 EM = EP +Ek = mgh+ 1 2mv2 ✓ EM = (70)(9,8)(12) ✓+ 1 2(70)(6)2 ✓ EM = 8232+1260 = 9492 J ✓ (4) 8.3 EM(P) = EM(Q) 9492 = (70)(9,8)(2)+ 1 2(70)v2 ✓✓ v = 15,23 m·s–1 ✓ One mark for EP at Q, one mark for Ek at Q equated to 9 492 J. POSITIVE MARKING FROM 8.2. (3) 8.4 EP needed at R = mgh = (70)(9,8)(15) = 10290 J ✓ 10290 J > 9492 J, so the biker cannot reach R. ✓ OR: maximum height h = 9492 (70)(9,8) = 13,84 m < 15 m. POSITIVE MARKING FROM 8.2. (2) [11] TOTAL: 100 Copyright reserved