Physical Sciences Grade 10 P1 November Mock 2026 Memo (Eng)

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 8 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. 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 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 B ✓✓ (2) 1.6 D ✓✓ (2) 1.7 C ✓✓ (2) 1.8 A ✓✓ (2) [16] QUESTION 2 2.1 A wave in which the particles of the medium vibrate perpendicular (at right angles) to the direction of motion of the wave. ✓✓ (2) 2.2 0,15 m ✓ (1) 2.3 A and C ✓ (OR B and D) (1) 2.4 2,5 wavelengths fit into 3,0 m ✓ 3,0 λ= = 1,2 m ✓ 2,5 (2) 10 2.5 f= = 5 Hz ✓ 2 1 1 T = ✓ = = 0,2 s ✓ f 5 2s OR: T = ✓✓ = 0,2 s ✓ 10 waves (3) 2.6 v = fλ ✓ v = (5)(1,2) ✓ v = 6 m·s–1 ✓ OR v = λ /T = 1,2/0,2 = 6 m·s–1 . POSITIVE MARKING FROM 2.4 AND 2.5. (3) [12] Copyright reserved Please turn over

Physical Sciences/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 3.1.1 Longitudinal ✓ (1) 3.1.2 Total distance travelled by the pulse: ∆x ∆x v= ✓ 340 = ✓ ∆t 0,025 ∆x = 8,5 m (to the moth and back) ✓ 8,5 Distance to moth = = 4,25 m ✓ 2 OR: ∆t one way = 0,0125 s ✓; ∆x = v∆t ✓ = (340)(0,0125) ✓ = 4,25 m ✓ (4) 3.2.1 A packet (quantum) of energy ✓ of light/electromagnetic radiation ✓. (2) c 3.2.2 E=h ✓ λ (6,63 × 10–34 )(3 × 108 ) E= ✓✓ 250 × 10–9 E = 7,96 × 10–19 J ✓ One mark for h and c substituted correctly, one mark for converting 250 nm to 250 × 10–9 m. (4) [11] QUESTION 4 4.1 The net charge of an isolated system remains constant during any physical process (e.g. when two charges make contact and are separated). ✓✓ (2) Q1 + Q2 (+6 × 10–9 ) + (–14 × 10–9 ) 4.2 Q= = ✓ 2 2 Q = –4 × 10–9 C (–4 nC) on EACH sphere ✓ (2) 4.3 Change in charge on A: ∆Q = –4 – (+6) = –10 nC ✓ Q –10 × 10–9 n= = ✓ e –1,6 × 10–19 n = 6,25 × 1010 electrons ✓ Electrons are transferred from B TO A. ✓ Accept ∆Q for B: –4 – (–14) = +10 nC. POSITIVE MARKING FROM 4.2. (4) [8] Copyright reserved Please turn over

Physical Sciences/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 The total energy supplied per coulomb of charge by the battery (OR the potential difference across the terminals of a battery when no current flows). ✓✓ (2) 1 1 1 1 1 5.2 = + = + ✓ Rp R1 R2 6 12 Rp = 4 Ω ✓ Rtotal = Rs = 4 + 4 ✓ = 8 Ω ✓ (4) V 12 5.3 R= ✓ 8= ✓ I I I = 1,5 A ✓ POSITIVE MARKING FROM 5.2. (3) 5.4 Q = I∆t ✓ Q = (1,5)(2 × 60) ✓ Q = 180 C ✓ POSITIVE MARKING FROM 5.3. (3) 5.5 DECREASES ✓ The total (external) resistance INCREASES (from 8 Ω to 10 Ω), so for the same emf the current decreases. ✓ (2) [14] Copyright reserved Please turn over

Physical Sciences/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 The rate of change of velocity. ✓✓ (2) 1 000 6.2 90 × = 25 m·s–1 ✓ 3 600 (1) 6.3 ∆x = v∆t (OR ∆x = vi ∆t + 12 a∆t2 with a = 0) ∆x = (25)(0,7) ✓ ∆x = 17,5 m ✓ (2) 6.4 Take the direction of motion as positive. v2f = v2i + 2a∆x ✓ 02 = (25)2 + 2(–5,5)∆x ✓✓ ∆x = 56,82 m ✓ One mark for vf = 0 and vi = 25; one mark for a = –5,5. POSITIVE MARKING FROM 6.2. (4) 6.5 YES Total stopping distance = 17,5 + 56,82 = 74,32 m ✓ 74,32 m < 75 m, so the taxi stops (0,68 m) before reaching the cow. ✓ POSITIVE MARKING FROM 6.3 AND 6.4. (2) 6.6 vf = vi + a∆t ✓ 0 = 25 + (–5,5)∆t ✓ ∆t = 4,55 s ✓ (3) [14] Copyright reserved Please turn over

Physical Sciences/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1 Constant velocity (of 8 m·s–1 east) OR uniform motion. ✓ (1) ∆v 8–0 7.2 a= ✓ = ✓ ∆t 4–0 a = 2 m·s–2 east ✓ (3) 7.3 ∆x = area under the graph ✓ ∆x = 12 (4)(8) + (6)(8) + 21 (4)(8) ✓✓ ∆x = 80 m east ✓ OR: area of trapezium = 12 (14 + 6)(8) ✓✓✓ = 80 m east ✓ (4) ∆x 80 7.4 vave = ✓ = ✓ ∆t 14 vave = 5,71 m·s–1 east ✓ POSITIVE MARKING FROM 7.3. (3) 7.5 a (m·s–2 ) 2 10 14 t (s) 4 –2 Marking criteria Horizontal line at +2 from 0 to 4 s ✓ Zero acceleration from 4 s to 10 s ✓ Horizontal line at –2 from 10 s to 14 s ✓ (3) [14] Copyright reserved Please turn over

Physical Sciences/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 The energy an object possesses due to its position (height) relative to a reference point (in the gravitational field). ✓✓ (2) 8.2 EP = mgh ✓ EP = (55)(9,8)(3,2) ✓ EP = 1 724,8 J ✓ (3) 8.3 The total mechanical energy of an isolated system remains constant. ✓✓ Accept: In the absence of friction/air resistance, the sum of the kinetic and gravitational potential energy of an object remains constant. (2) 8.4 EM(A) = EM(C) (EP + Ek )A = (EP + Ek )C ✓ 1 724,8 + 0 ✓ = (55)(9,8)(1,5) + 12 (55)v2 ✓ v = 5,77 m·s–1 ✓ OR: mghA + 0 = mghC + 12 mv2 ⇒ v2 = 2(9,8)(3,2 – 1,5). POSITIVE MARKING FROM 8.2. (4) [11] TOTAL: 100 Copyright reserved