Physical Sciences Grade 11 Paper 1 November Mock Exam 2 2026 Memo

Copyright reserved Please turn over Mock 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 12 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. 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 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 1 1.1 C ✓✓ (2) 1.2 B ✓✓ (2) 1.3 D ✓✓ (2) 1.4 B ✓✓ (2) 1.5 D ✓✓ (2) 1.6 D ✓✓ (2) 1.7 A ✓✓ (2) 1.8 C ✓✓ (2) 1.9 A ✓✓ (2) 1.10 C ✓✓ (2) [20] Notes for markers: 1.2: √ 52 +122 = 13 N. 1.5: w ∝1 r2 , so r = 3R from the centre, i.e. a height of 2R. 1.6: F′ = k(2Q1)(2Q2) ( 1 2r)2 = 16F. 1.7: n = 4,8×10–9 1,6×10–19 = 3×1010. 1.8: the induced pole opposes the magnet moving away (Lenz), so X becomes a SOUTH pole that attracts the north pole. 1.9: the external resistance decreases, so the current increases and the lost volts (Ir) increase. Copyright reserved Please turn over

Physical Sciences/P1 4 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 2 2.1 The resultant (net) force acting at O is zero. ✓✓ Accept: the forces, drawn head-to-tail, form a closed vector diagram (2 marks). (2) 2.2 w = 176,4 N T1 T2 50◦ 35◦ Marking criteria Three force vectors drawn head-to-tail forming a CLOSED triangle with correct directions ✓ All three forces labelled ✓ At least TWO correct angles shown (e.g. 50◦between w and T1, 35◦between w and T2, 95◦between T1 and T2) ✓ (3) 2.3 w = mg = (18)(9,8) = 176,4 N ✓ OPTION 1 (components) Horizontal: T1 cos40◦= T2 cos55◦✓ Vertical: T1 sin40◦+T2 sin55◦= 176,4 ✓ T2 = T1 cos40◦ cos55◦ = 1,3356T1 ✓ 0,6428T1 +(0,8192)(1,3356T1) = 176,4 ✓ T1 = 101,57 N ✓ T2 = 135,65 N ✓ OPTION 2 (sine rule on the closed triangle) T1 sin35◦= 176,4 sin95◦✓✓ T2 sin50◦= 176,4 sin95◦✓✓ T1 = 101,57 N ✓ T2 = 135,65 N ✓ The w = mg ✓line applies to BOTH options. Ranges: T1 101,5 – 101,6 N; T2 135,6 – 135,7 N. (7) 2.4 176,4 N ✓vertically UPWARDS ✓ (Equal in magnitude and opposite in direction to the weight.) POSITIVE MARKING FROM 2.3. (2) [14] Copyright reserved Please turn over

Physical Sciences/P1 5 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 3 3.1 The force that opposes the motion of a moving object relative to a surface. ✓✓ (2) 3.2 N / FN ✓ w / Fg ✓ fk / f / Ff ✓ F / FA ✓ F is directed at 40◦BELOW the horizontal, to the right. Accept F drawn as a push ending at the dot. (4) 3.3 An object continues in its state of rest or uniform motion (moving with constant velocity) unless it is acted upon by a net (resultant) force. ✓✓ (2) 3.4 Constant velocity, so Fnet = 0 in the horizontal and vertical directions. N = mg+Fsin40◦✓ fk = µkN ✓ Horizontally: Fcos40◦= fk ✓ Fcos40◦= 0,35[(25)(9,8)+Fsin40◦] ✓ 0,7660F–0,2250F = 85,75 F = 158,48 N ✓ Range: 158,4 – 158,6 N. If the vertical component of F is omitted from N (i.e. N = mg): max. 3/5. (5) 3.5 N = mg+Fsin40◦= (25)(9,8)+(158,48)sin40◦✓ N = 346,87 N ✓ Also accept N = fk µk = 158,48cos40◦ 0,35 . POSITIVE MARKING FROM 3.4. (2) 3.6 SMALLER THAN ✓ The vertical component of the applied force now acts UPWARDS, so the normal force decreases (N = mg – Fsinθ < mg). Since fk = µkN, the kinetic frictional force decreases. ✓ (2) [17] Copyright reserved Please turn over

Physical Sciences/P1 6 NSC – Marking Guidelines Hlayiso/Mock 2026 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 ✓ w / Fg ✓ T / FT ✓ fk / f / Ff ✓ N perpendicular to the incline; T up the incline; friction down the incline (opposes motion up the incline). (4) 4.3 fk = µkN = µkmAgcos30◦✓ fk = (0,2)(6)(9,8)cos30◦✓ fk = 10,18 N ✓ (3) 4.4 Block A (up the incline positive): Fnet = ma ✓ T–mAgsin30◦–fk = mAa T–(6)(9,8)sin30◦–10,18 = 6a ✓ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) Block B (downwards positive): mBg–T = mBa ⇒ (5)(9,8)–T = 5a ✓ . . . . . . . . . . . . . . . . . . . . . . . . . . . (2) (1) + (2): 49–29,4–10,18 = 11a ✓ a = 0,86 m·s–2✓ Range: 0,85 – 0,86 m·s–2. POSITIVE MARKING FROM 4.3. (5) 4.5 OPTION 1 (block B) mBg–T = mBa: (5)(9,8)–T = (5)(0,86) ✓✓ T = 44,70 N ✓ (44,72 N if the unrounded a = 0,856 m·s–2 is used) OPTION 2 (block A) T–29,4–10,18 = (6)(0,86) ✓✓ T = 44,74 N ✓ Range: 44,70 – 44,75 N. POSITIVE MARKING FROM 4.4. (3) 4.6 SMALLER THAN ✓ Block B accelerates DOWNWARDS, so the net force on B is downwards; its weight must be greater than the (upward) tension. ✓ (2) [19] Copyright reserved Please turn over

Physical Sciences/P1 7 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 5 5.1 T / FT ✓ w / Fg ✓ (2) 5.2 Upwards positive: Fnet = ma T–mg = ma ✓ T–(2500)(9,8) = (2500)(0,8) ✓ T = 26500 N (2,65×104 N) ✓ (3) 5.3 The cage slows down while moving upwards, so its acceleration is DOWNWARDS. Upwards positive: T–mg = ma T–(2500)(9,8) = (2500)(–1,2) ✓✓ T = 21500 N (2,15×104 N) ✓ One mark for the substitution of –1,2 m·s–2 (correct direction of a) and one for the rest of the substitution. If +1,2 m·s–2 is used (answer 27 500 N): max. 1/3. (3) [8] Copyright reserved Please turn over

Physical Sciences/P1 8 NSC – Marking Guidelines Hlayiso/Mock 2026 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) 6.2 F = GMm r2 ✓ F = (6,67×10–11)(2,4×1025)(1200) (9,2×106)2 ✓ w = 22695,65 N (2,27×104 N) ✓ Also accept: g = GM R2 = 18,91 m·s–2 ✓and w = mg = (1200)(18,91) ✓= 22692 N ✓. Range: 22 690 – 22 700 N. (3) 6.3 Weight on Earth: F = mg = (1200)(9,8) = 11760 N ✓ F = GMm r2 : 11760 = (6,67×10–11)(2,4×1025)(1200) r2 ✓ r = 1,28×107 m ✓ h = r–R = 1,278×107 –9,2×106 = 3,58×106 m ✓ Range for h: 3,57 – 3,59 ×106 m. Accept the ratio method: r R = √ 22695,65 11760 , r = 1,28× 107 m. POSITIVE MARKING FROM 6.2 where used. (4) 6.4 NO ✓ According to Newton’s Third Law, the force that Kepra exerts on the lander and the force that the lander exerts on Kepra are EQUAL in magnitude (and opposite in direction), regardless of the masses. ✓ (2) [11] Copyright reserved Please turn over

Physical Sciences/P1 9 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 7 7.1.1 After X and Y touch: QY = –9+0 2 = –4,5 nC ✓ After Y and Z touch: QZ = –4,5+5 2 = +0,25 nC (+2,5×10–10 C) ✓ (2) 7.1.2 n = ∆Q e = (5–0,25)×10–9 1,6×10–19 ✓ n = 2,97×1010 electrons ✓ Electrons moved from Y TO Z. ✓ Accept ∆Q calculated for Y: 0,25–(–4,5) = 4,75 nC. POSITIVE MARKING FROM 7.1.1. (3) 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 T / FT ✓ w / Fg ✓ FE / FTS / electrostatic force ✓ Tension along the string (towards the support); electrostatic force horizontal, towards T. (3) 7.2.3 OPTION 1 Vertical: Tcos12◦= mg; horizontal: Tsin12◦= FE ✓ FE = mgtan12◦= (2×10–3)(9,8)tan12◦✓ FE = 4,17×10–3 N ✓ OPTION 2 T = mg cos12◦= (2×10–3)(9,8) cos12◦ = 2,004×10–2 N ✓ FE = Tsin12◦= (2,004×10–2)sin12◦✓ = 4,17×10–3 N ✓ Range: 4,16 – 4,17 ×10–3 N. If the mass is not converted to kg: max. 2/3. (3) 7.2.4 F = kQSQT r2 : 4,17×10–3 = (9×109)(40×10–9)QT (0,05)2 ✓ QT = 2,89×10–8 C (28,9 nC) ✓ T is NEGATIVE (S is positive and is attracted). ✓ Range: 2,89 – 2,90 ×10–8 C. POSITIVE MARKING FROM 7.2.3. (3) [16] Copyright reserved Please turn over

Physical Sciences/P1 10 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 8 8.1 The (electrostatic) force per unit positive charge placed at that point. ✓✓ (2) 8.2 + – A B Marking criteria Correct shape: field lines from A to B (lines curve between the charges) ✓ Direction: arrows from A (positive) towards B (negative) ✓ Field lines do not cross and start/end on the spheres ✓ Accept: more lines ending on B than leaving A (|QB| > |QA|); do not penalise equal densities. (3) 8.3 E = kQ r2 ✓ EA = (9×109)(4×10–6) (0,15)2 ✓= 1,6×106 N·C–1 towards B (away from A) EB = (9×109)(9×10–6) (0,15)2 ✓= 3,6×106 N·C–1 towards B Enet = 1,6×106 +3,6×106 ✓ Enet = 5,2×106 N·C–1 towards B (to the right) ✓ The final mark is awarded only if the direction is given. If the two fields are subtracted: max. 3/5. (5) 8.4 At P the fields of A and B are opposite in direction and equal in magnitude: kQA x2 = kQB (x+0,3)2 ✓ (9×109)(4×10–6) x2 = (9×109)(9×10–6) (x+0,3)2 ✓ x+0,3 x = 3 2 ⇒ x = 0,6 m ✓ (3) [13] Copyright reserved Please turn over

Physical Sciences/P1 11 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 9 9.1.1 NORTH ✓ (The induced current opposes the approach of the north pole.) (1) 9.1.2 Zero (no deflection) ✓ There is no change in the magnetic flux (linkage) through the solenoid, so no emf/current is induced. ✓ (2) 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 = BAcosθ = (0,5)π(0,04)2 cos0◦✓ Φ = 2,51×10–3 Wb ✓ (2) 9.2.3 ∆Φ = BA(cos60◦–cos0◦) = (0,5)π(0,04)2(0,5–1) = –1,26×10–3 Wb ✓ ε = –N∆Φ ∆t ✓ ε = –(150)(–1,26×10–3) 0,02 ✓ ε = 9,42 V ✓ Range: 9,42 – 9,45 V. Ignore the sign of ∆Φ for the magnitude. If the final flux is taken as zero (i.e. ∆Φ = BA): max. 3/4. POSITIVE MARKING FROM 9.2.2. (4) 9.2.4 DECREASES ✓ The same change in flux linkage takes place over a longer time, so the RATE of change of flux linkage is smaller (ε ∝1 ∆t). ✓ (2) [13] Copyright reserved Please turn over

Physical Sciences/P1 12 NSC – Marking Guidelines Hlayiso/Mock 2026 QUESTION 10 10.1.1 The maximum energy provided (work done) by a battery per coulomb (unit) of charge passing through it. ✓✓ Accept: the potential difference across the terminals of a battery when no current flows (open circuit). (2) 10.1.2 R34 = R3 +R4 = 4+8 = 12 Ω 1 Rp = 1 6 + 1 12 ✓ Rp = 4 Ω✓ Rext = R1 +Rp = 3+4 = 7 Ω✓ Accept Rp = R2R34 R2 +R34 = (6)(12) 18 . If R3 and R4 are not added in series: max. 1/3. (3) 10.1.3 ε = I(R+r) ✓ 12 = I(7+0,5) ✓ I = 1,60 A ✓ If the internal resistance is ignored (I = 1,71 A): max. 1/3. POSITIVE MARKING FROM 10.1.2. (3) 10.1.4 OPTION 1 V = ε –Ir = 12–(1,6)(0,5) ✓ V = 11,20 V ✓ OPTION 2 V = IRext = (1,6)(7) ✓ V = 11,20 V ✓ POSITIVE MARKING FROM 10.1.3. (2) 10.1.5 Vp = IRp = (1,6)(4) = 6,4 V I3 = Vp R34 = 6,4 12 = 0,53 A ✓ P3 = I2R = (0,53)2(4) ✓ P3 = 1,14 W ✓ Range: 1,12 – 1,14 W. Also accept: V3 = I3R3 = 2,13 V and P = VI or P = V2 R . If Vp = 6,4 V is used across R3 alone: max. 1/3. POSITIVE MARKING FROM 10.1.3. (3) 10.1.6 INCREASES ✓ When S is opened the R2 branch is removed and the external resistance increases (from 7 Ωto 3 + 12 = 15 Ω), so the current in the battery decreases (from 1,6 A to 0,77 A). ✓ The lost volts (Ir) across the internal resistance decrease, so the terminal potential difference (V = ε –Ir) increases (to 11,61 V). ✓ (3) 10.2 P = V2 R = (230)2 24 = 2204,17 W (2,20 kW) ✓ W = P∆t = (2,204 kW)(1,5 h×30) = 99,19 kW·h ✓ Cost = 99,19×R3,15 = R312,44 ✓ Range: R312,40 – R312,50. If calculated in joules: W = (2204,17)(1,5×3600×30) = 3,57×108 J = 99,19 kW·h. Full marks if correctly converted. (3) [19] TOTAL: 150 Copyright reserved