Physical Sciences Grade 12 P1 NSC November 2026 Memo (Eng)

Copyright reserved Please turn over Unofficial memo hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: PHYSICS (P1) NOVEMBER 2026 — UNOFFICIAL MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 21 pages.

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 2 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 IMPORTANT: These are unofficial marking guidelines prepared by Hlayiso for revision. They are not the DBE’s official memorandum; the official marking guidelines may differ in mark allocation and accepted alternatives. GENERAL MARKING GUIDELINES • Calculations: a tick (✓) shows one mark. Marks are given for the formula, the substitution and the final answer WITH the unit. A formula on its own (no substitution) earns no mark; after a wrong formula, no further marks are given for that calculation. • Units: penalise a missing or wrong unit in a final answer once per question only. Accept answers rounded to a minimum of TWO decimal places; ranges are given where rounding affects the answer. • Positive marking: an incorrect answer carried into a later calculation is marked positively. • Negative marking: if the choice in a “choose and explain” item (e.g. GREATER THAN/SMALLER THAN) is wrong, no marks are given for the explanation. • Definitions and laws: deduct 1 mark for each underlined key word/phrase that is omitted or used in the wrong context. “(2 or 0)” means no part marks. • Diagrams and graphs: marked against the “Marking criteria” box. A force earns its mark only with BOTH an arrow and a label. • Sign convention: candidates may take either direction as positive if they apply it consistently. Both versions are shown where the signs change. READING OF THE QUESTION PAPER • Drawn up from photographs of pages 2–18 of the question paper and the three data sheets (g = 9,8 m·s–2, k = 9,0×109 N·m2·C–2, h = 6,63×10–34 J·s, c = 3,0×108 m·s–1, me = 9,11×10–31 kg). • Answers that depend on our reading of a diagram, or on how the DBE has marked a similar item before, carry a short Hlayiso note. Hlayiso notes are for learners and teachers, not for marking. Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 3 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 1 1.1 B ✓✓ (2) 1.2 C ✓✓ (2) 1.3 D ✓✓ (2) 1.4 C ✓✓ (2) 1.5 B ✓✓ (2) 1.6 C ✓✓ (2) 1.7 B ✓✓ (2) 1.8 A ✓✓ (2) 1.9 D ✓✓ (2) 1.10 A ✓✓ (2) [20] Hlayiso notes (why each answer is correct): 1.1: Newton’s Second Law: a ∝Fnet (constant mass) and a ∝1 m (constant net force). 1.2: F = GMm r2 with equal masses, so FA FB = (2x)2 x2 = 4, i.e. 4 : 1. 1.3: When it is released, the stone has the balloon’s upward velocity, so (i) is false. It first rises and slows down to zero at its highest point, so (ii) is false, and its maximum height is above 50 m, so (iii) is true. 1.4: A constant force gives a constant rate of change of momentum (a straight line). The ball reverses direction, so its momentum changes sign during the contact time. Only graph C shows both. 1.5: Mechanical energy is conserved from B to C: Ek,C = Ek,B +mgh = E+mgh. 1.6: Graph A is steeper than B (fL > fs), so the source approaches and fL = v v–vs fs, with gradient v v–vs > 1. Graph C (gradient < 1) is for a receding source, with gradient v v+vs . 1.7: E = kQ r2 does not depend on the test charge q, so the field stays E (only the force on q doubles). 1.8: The battery has no internal resistance, so the total resistance doubles and the current halves. P = I2R, so P1 = ( I 2)2R = 1 4P. 1.9: The current in AB (next to X) flows upwards from A to B. With X = North, the field points to the right, and the motor rule (⃗F = I⃗L ×⃗B) gives a force INTO the page. A motor converts electrical energy to mechanical energy. With X = South the force would be out of the page, which no option pairs with “electrical to mechanical”. Only D is consistent. 1.10: One photon interacts with one electron, so photon energies do not add up. Each photon has only 1 2W0 < W0, so no electron is ejected, however many photons arrive (A). B, C and D all assume an electron is ejected. Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 4 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 2 2.1 N ✓ w ✓ T ✓ F ✓ 15◦ Accepted labels w Fg / Fw / weight / mg / 19,6 N / gravitational force / FEarth on B N FN / normal (force) / Fnormal T FT / tension / Fstring F FA / 20 N / applied force Notes • Mark awarded for label AND arrow. Do not penalise for the length of the arrows (not drawn to scale). • F must act at an angle above the horizontal (to the right); T acts to the left (towards block A). • Block B is on the frictionless section RS. Any additional force (e.g. friction): Max 3 4. • If everything is correct but there are no arrows: Max 3 4. • If force(s) do not make contact with the dot: Max 3 4. • Accept a force diagram (forces drawn on a block) instead of a dot. (4) 2.2 If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The force that opposes the motion of an object and which acts parallel to the surface (with which the object is in contact). ✓✓ Accept: the force that opposes the motion of a moving object relative to a surface (kinetic frictional force). (2) Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 5 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 2.3 Marking criteria hlayiso.com • Formula Fnet = ma (any one) ✓ • Correct equation for block A: T–11 = 5a ✓ • Correct equation for block B (20cos15◦– T = 2a) OR for the system (20cos15◦–11 = 7a) ✓ • Final answer: T = 16,94 N (range: 16,94 – 16,95 N) ✓ • Conclusion: T > 16 N, therefore the string breaks ✓ Direction of motion (to the right) as positive. OPTION 1 (EACH BLOCK SEPARATELY) Block A: Fnet = ma ✓ T+(–11) = 5a ✓. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) Block B: Fcos15◦+(–T) = 2a 20cos15◦–T = 2a ✓. . . . . . . . . . . . . . . . (2) (1) + (2): 19,32–11 = 7a ⇒ a = 1,19 m·s–2 T = 11+5(1,19) = 16,94 N ✓ 16,94 N > 16 N, therefore the string will break. ✓ OPTION 2 (SYSTEM, THEN ONE BLOCK) Fnet = ma ✓ System: 20cos15◦–11 = (5+2)a ✓ ⇒ a = 1,19 m·s–2 Block A: T–11 = 5(1,19) ✓ OR Block B: 20cos15◦–T = 2(1,19) ✓ T = 16,94 N ✓ > 16 N, so the string breaks. ✓ OPTION 3 (ASSUME THE STRING HOLDS AT 16 N) Fnet = ma ✓ Block A: 16–11 = 5aA ✓ aA = 1,00 m·s–2 Block B: 20cos15◦–16 = 2aB ✓ aB = 1,66 m·s–2✓ B would accelerate faster than A, which a taut inextensible string cannot allow, so the tension must be greater than 16 N: the string breaks. ✓ NOTE: If 20 N is used as the horizontal force (no cos15◦component): Max 3 5. (5) 2.4 Block A: continues to move to the right with a decreasing velocity (it slows down; friction is the only horizontal force on it). ✓ Block B: moves to the right with an increasing velocity (it accelerates; Fcos15◦ is now unopposed). ✓ Accept: A decelerates; B accelerates (at a greater rate than before). (2) [13] Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 6 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 3 3.1 An object upon which the only force acting is the force of gravity. ✓✓(2 or 0) Accept: an object that experiences only the gravitational force (weight). Hlayiso note: DBE Nov 2025 gave 0/2 when a learner defined a projectile instead of free fall. Expect the same strictness the other way round: the definition must describe an object, not a motion. (2) 3.2 Equal in magnitude (same speed) ✓but opposite in direction (upwards at x, downwards at t). ✓ (2) 3.3.1 UPWARDS AS POSITIVE v2 f = v2 i +2a∆y ✓ 02 = (20)2 +2(–9,8)∆y ✓ ∆y = 20,41 m ✓ DOWNWARDS AS POSITIVE v2 f = v2 i +2a∆y ✓ 02 = (–20)2 +2(9,8)∆y ✓ ∆y = –20,41 m; height = 20,41 m ✓ OPTION 2 vf = vi + a∆t: 0 = 20 + (–9,8)∆t, ∆t = 2,04 s; ∆y = vi∆t + 1 2a∆t2 ✓= (20)(2,04) + 1 2(–9,8)(2,04)2 ✓= 20,41 m ✓ Range: 20,40 – 20,41 m. NOTE: If energy principles are used: 1 3 (final answer only). (3) 3.3.2 Marking criteria hlayiso.com • Time from h to the maximum height: 1 2(1,5) = 0,75 s ✓ • Formula ✓ • Correct substitution ✓ • Subtraction from the maximum height / correct method to reach h ✓ • Final answer: 17,65 m (range: 17,64 – 17,65 m) ✓ POSITIVE MARKING FROM QUESTION 3.3.1 The motion is symmetrical: the ball spends 1 2(1,5) = 0,75 s rising from h to the top and 0,75 s falling back to h. ✓ OPTION 1 (FROM THE TOP DOWN TO h) UPWARDS POSITIVE: ∆y = vi∆t + 1 2a∆t2 ✓ = (0)(0,75) + 1 2(–9,8)(0,75)2 ✓ = –2,76 m h = 20,41–2,76 ✓ h = 17,65 m ✓ (DOWNWARDS POSITIVE: ∆y = 0 + 1 2(9,8)(0,75)2 = 2,76 m; h = 20,41 – 2,76 = 17,65 m.) OPTION 2 (VELOCITY AT h) vf = vi +a∆t = 0+(9,8)(0,75) = 7,35 m·s–1 (speed at h) ✓ v2 f = v2 i +2a∆y ✓ (7,35)2 = (20)2 +2(–9,8)∆y ✓ h = 17,65 m ✓ OPTION 3 (TIME x) x = 2,04–0,75 = 1,29 s ✓ ∆y = vi∆t+ 1 2a∆t2 ✓ = (20)(1,29)+ 1 2(–9,8)(1,29)2 ✓ h = 17,65 m ✓ (5) Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 7 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 3.3.3 POSITIVE MARKING FROM QUESTIONS 3.3.1 AND 3.3.2 OPTION 1 vf = vi +a∆t ✓ 0 = 20+(–9,8)∆t ✓ ∆ttop = 2,04 s t = 2,04+0,75 = 2,79 s ✓ OPTION 2 ∆y = vi∆t+ 1 2a∆t2 ✓ 17,65 = 20∆t+ 1 2(–9,8)∆t2 ✓ ∆t = 1,29 s or 2,79 s t = 2,79 s (the second time the ball is at h) ✓ OPTION 3 t = x+1,5 s = 1,29+1,5 ✓✓= 2,79 s ✓ (x from equations of motion) Range: 2,79 – 2,80 s. (3) 3.4 POSITIVE MARKING FROM QUESTIONS 3.3.1 – 3.3.3 Marking criteria hlayiso.com • Graph A: straight line starting at v = 20 m·s–1 at t = 0 that crosses the time axis (at 2,04 s) ✓ • Graph B starts at v = 0 at t = 2,79 s (the time ball B was dropped) ✓ • Graphs A and B are parallel (same gradient) ✓ • Both graphs end with a negative velocity; B ends after A with a smaller final speed (A: –20 m·s–1 at 4,08 s; B: –18,60 m·s–1 at 4,69 s) ✓ NB: If the graphs are not labelled A and B, deduct 1 mark. Downwards as positive: mirror image (accept). UPWARDS AS POSITIVE t (s) Velocity (m·s–1) 0 A B 20 2,04 2,79 –20 4,08 4,69 –18,60 DOWNWARDS AS POSITIVE t (s) Velocity (m·s–1) 0 A B –20 2,04 2,79 20 4,08 4,69 18,60 Hlayiso note: The end points are computed from the data: A lands after 2(2,04) = 4,08 s at 20 m·s–1; B falls 17,65 m from rest in 1,90 s, so it lands at 2,79 + 1,90 = 4,69 s at 18,60 m·s–1. These values are not required on the graph. (4) [19] Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 8 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 4 4.1 If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The total (linear) momentum in an isolated system remains constant (is conserved). ✓✓ Accept for 1 mark: in an isolated system the total momentum before a collision (explosion) is equal to the total momentum after it. IF: closed system: –1 mark. (2) 4.2 RIGHT AS POSITIVE. From the graph: after separation vP = –3 m·s–1 and vQ = 4 m·s–1. Σpi = Σpf OR (mP +mQ)vi = mPvfP +mQvfQ ✓ (any one) (1,5+2)vi ✓= (1,5)(–3)+(2)(4) ✓ vi = 1,00 m·s–1✓ LEFT AS POSITIVE: (1,5 + 2)vi = (1,5)(3) + (2)(–4), so vi = –1 m·s–1, magnitude 1,00 m·s–1. NOTE: 1 mark for the left-hand side, 1 mark for the right-hand side. Accept pbefore = pafter with the same substitutions. (4) 4.3 GREATER THAN ✓ The spring exerts forces of equal magnitude on P and Q (Newton’s Third Law), and P has the smaller mass, so a = F m is greater for P. ✓ OR: in the same time interval P’s velocity changes by 4 m·s–1 (1 to –3) and Q’s by only 3 m·s–1 (1 to 4), so the graph for P is steeper. (2) 4.4 POSITIVE MARKING FROM QUESTION 4.2 OPTION 1 (TROLLEY Q) Fnet∆t = ∆p = m(vf –vi) ✓ (any one) (10)∆t = (2)(4–1) ✓ ∆t = 0,6 s ∴ t = 3+0,6 = 3,60 s ✓ OPTION 2 (TROLLEY P) Fnet∆t = m(vf –vi) ✓ (–10)∆t = (1,5)(–3–1) ✓ ∆t = 0,6 s; t = 3,60 s ✓ OPTION 3 a = Fnet m = 10 2 = 5 m·s–2; vf = vi +a∆t ✓: 4 = 1+5∆t ✓ ∆t = 0,6 s; t = 3,60 s ✓ NOTE: If only ∆t = 0,6 s is given (not added to 3 s): Max 2 3. (3) [11] Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 9 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 5 5.1 If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. A force is non-conservative if the work done by the force on an object (moving between two points) depends on the path taken. ✓✓ OR: A force is non-conservative if the work it does in moving an object around a closed path is non-zero. ✓✓ IF the word “work” is omitted: 0 marks. (2) 5.2.1 Marking criteria hlayiso.com • Formula Wnc = ∆Ek +∆Ep OR Wnet = ∆Ek ✓ • Correct substitution for the work done by friction: 55∆xcos180◦✓ • Correct substitution of ∆Ek and ∆Ep (or Wg) ✓ • Final answer: 9,09 m ✓ The block moves down: ∆Ep = –2500 J and ∆Ek = +2000 J. OPTION 1 Wnc = ∆Ek +∆Ep ✓ fk∆xcosθ = ∆Ek +∆Ep (55)∆xcos180◦✓= 2000+(–2500) ✓ ∆x = 9,09 m ✓ OPTION 2 Wnet = ∆Ek ✓ Wg +Wf = ∆Ek (Wg = –∆Ep = 2500 J) 2500 ✓+(55)∆xcos180◦✓= 2000 ∆x = 9,09 m ✓ (4) Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 10 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 5.2.2 POSITIVE MARKING FROM QUESTION 5.2.1 Marking criteria hlayiso.com • Formula ∆Ep = mg∆h / Wg = mg∆xcosθ ✓ • Correct substitution of 2 500 J and (40)(9,8) ✓ • Correct trigonometric ratio (sinθ = h ∆x) / use of the distance from 5.2.1 ✓ • Final answer: θ = 44,55◦(range: 44,55◦– 44,58◦) ✓ OPTION 1 ∆Ep = mg∆h ✓ –2500 = (40)(9,8)(0–h) ✓ h = 6,38 m sinθ = 6,38 9,09 ✓ θ = 44,55◦✓ OPTION 2 Wg = –∆Ep: mg∆xcos(90◦–θ) = 2500 ✓ (40)(9,8)(9,09)cos(90◦– θ) = 2500 ✓✓ θ = 44,56◦✓ (accept sinθ for cos(90◦–θ)) OPTION 3 Wnet = ∆Ek ✓ [(40)(9,8)sinθ –55](9,09) ✓✓= 2000 θ = 44,55◦✓ (4) 5.3 SMALLER THAN ✓ The acceleration down the slope (a = gsinθ – µkgcosθ) does not depend on the mass, so both blocks reach Y with the same speed. Wnet = ∆Ek = 1 2mv2 f is therefore smaller for the smaller mass. ✓ OR: Fnet = ma with the same a, so the net force on the 20 kg block is half as large over the same displacement. Accept: net work is directly proportional to the mass. Hlayiso note: For the 20 kg block, Wnet = 1000 J (half of 2 000 J). (2) [12] Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 11 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 6 6.1.1 If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The 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 sound propagation. ✓✓ OR: An (apparent) change in (observed/detected) frequency (pitch) as a result of the relative motion between a sound source and an observer (listener). ✓✓ (2) 6.1.2 In front (towards the listener) Behind direction of motion of the ambulance Marking criteria hlayiso.com • Wavefronts closer together in front of the ambulance (in the direction of motion, towards the listener) ✓ • Wavefronts further apart behind the ambulance ✓ The dot (ambulance) must be off-centre, displaced towards the front. Concentric circles: 0/2. (2) Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 12 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 6.1.3 Marking criteria hlayiso.com • Formula ✓ • Police car: 340 340–(v+5)fP ✓ • Ambulance: 340 340–vfA ✓ • Equating the two detected frequencies ✓ • Using fP fA = 0,984 (or fP = 0,984fA) ✓ • Final answer: 27,50 m·s–1✓ Both vehicles move TOWARDS the stationary listener (vL = 0; subtract vs in the denominator). fL = v±vL v±vs fs OR fL = v v–vs fs ✓ Police car: fL = 340 340–(v+5)fP ✓ Ambulance: fL = 340 340–vfA ✓ Equal detected frequencies: 340 335–vfP = 340 340–vfA ✓ fP fA = 335–v 340–v = 0,984 ✓ 335–v = 334,56–0,984v ⇒ 0,016v = 0,44 v = 27,50 m·s–1✓ NOTE: If (v+vs) is used in the denominators (sources moving away): Max 3 6 (formula, equating, ratio). (6) 6.2.1 Absorption (line) spectra ✓ (1) 6.2.2 Red shift ✓ Hlayiso note: In the photo every line of Spectrum 2 sits further to the right (the red, longer-wavelength end) than the matching line of Spectrum 1: at about 25 %, 72 % and 85 % of the width, against 18 %, 65 % and 79 %. (1) 6.2.3 Away from (the Earth) ✓ (1) [13] Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 13 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 7 7.1 If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is directly proportional to the product of the (magnitudes of the) charges and inversely proportional to the square of the distance (r) between them. ✓✓ (2) 7.2 w / Fg FE T 5◦ 90◦ 85◦ Marking criteria hlayiso.com • w (Fg) vertical and FE horizontal (away from A, to the right), correctly labelled ✓ • T completes a closed triangle (arrows head to tail) ✓ • Any TWO correct angles: 5◦ (between T and w), 90◦(between w and FE), 85◦(between FE and T) ✓ Not drawn to scale. Accepted labels: T / FT / tension; w / Fg / mg / weight; FE / FAB / electrostatic (Coulomb) force. (3) 7.3 EQUAL TO 5◦✓ The electrostatic forces that A and B exert on each other are equal in magnitude (Newton’s Third Law) AND the spheres have equal masses (equal weights), so their force triangles are identical. ✓ (2) 7.4 If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The (electrostatic) force experienced per unit positive charge placed at that point. ✓✓ NOTE (1 mark only): an electric field is a region of space in which an electric charge experiences a force. (2) 7.5 RIGHT ✓ Hlayiso note: Both spheres are positive and X lies to the right of both, so both fields at X point away from the charges, to the right. (1) Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 14 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 7.6 Marking criteria hlayiso.com • Formula E = kQ r2 ✓ • Substitution for EB (distance 2r) ✓ • Substitution for EA (distance 3r) ✓ • Adding: Enet = EA +EB ✓ • Equating to 1,05×104 N·C–1✓ • Final answer: r = 0,05 m ✓ RIGHT AS POSITIVE E = kQ r2 ✓ EB = (9×109)(9×10–9) (2r)2 ✓= 20,25 r2 to the right EA = (9×109)(6×10–9) (3r)2 ✓= 6 r2 to the right Enet = EA +EB ✓ 1,05×104 = 6 r2 + 20,25 r2 = 26,25 r2 ✓ r2 = 2,5×10–3 r = 0,05 m ✓ NOTE: If EA and EB are subtracted: Max 4 6. Wrong distances (e.g. r and 2r): Max 4 6. (6) [16] Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 15 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 8 8.1.1 OPTION 1 1 Rp = 1 R1 + 1 R2 : 1 0,72 = 1 R + 1 R ✓ RL1 = 1,44 Ω✓ OPTION 2 Rp = R1R2 R1 +R2 : 0,72 = R2 2R ✓ RL1 = 1,44 Ω✓ Accept: identical bulbs in parallel, so R = 2Rp = 2(0,72) ✓= 1,44 Ω✓. (2) 8.1.2 POSITIVE MARKING FROM QUESTION 8.1.1 OPTION 1 P = I2R ✓ 25 = I2(1,44) ✓ I = 4,17 A ✓ OPTION 2 P = V2 R : 25 = V2 1,44, V = 6,00 V P = VI ✓ 25 = 6I ✓ I = 4,17 A ✓ Range: 4,16 – 4,17 A. (3) 8.1.3 POSITIVE MARKING FROM QUESTIONS 8.1.1 AND 8.1.2 Marking criteria hlayiso.com • Total current: I = 2(4,17) = 8,33 A ✓ • Formula ε = I(R+r) OR ε = Vext +Ir ✓ • Correct substitution of 7 V and the total current ✓ • Correct addition of resistances (0,72+R+0,02) OR of potential differences ✓ • Final answer: 0,10 Ω(range: 0,099 – 0,10 Ω) ✓ OPTION 1 Itotal = 2(4,17) = 8,33 A ✓ ε = I(R+r) ✓ 7 = 8,33 ✓(0,72+Rrh +0,02) ✓ Rrh = 0,10 Ω✓ OPTION 2 Itotal = 8,33 A ✓; V∥= IR = (4,17)(1,44) = 6,00 V; Vlost = Ir = (8,33)(0,02) = 0,17 V ε = V∥+Vrh +Ir ✓: 7 = 6,00+Vrh +0,17 ✓✓ Vrh = 0,83 V Rrh = V I = 0,83 8,33 = 0,10 Ω✓ (5) Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 16 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 8.1.4 GREATER THAN ✓ • The resistance of the L2 branch increases, so the resistance of the parallel combination increases. ✓ • The total (external) resistance of the circuit increases. ✓ • The total current (through the rheostat and battery) decreases. ✓ • The potential difference across the rheostat (IRrh) and the lost volts (Ir) decrease. ✓ • The emf is constant, so the potential difference across the parallel part (across L1) increases, and with RL1 constant the current through L1 (ammeter) increases. ✓ (6) 8.1.5 BRIGHTER THAN ✓ (1) 8.2.1 Higher than (battery B) ✓ Hlayiso note: Battery A has the larger internal resistance, so more of its energy is converted to heat inside it (DBE Nov 2020 8.2: “energy is converted to heat in the battery due to the internal resistance”). Strictly, the heat per second in a battery is I2r = ε2r (R+r)2 , which grows with r only while r is smaller than the heater’s resistance R. That is the normal case for a heater, so “higher” is the expected answer. (1) 8.2.2 Lower than (heater H2) ✓ The internal resistance of battery A is greater, so the total resistance of circuit A is greater and the current through H1 is smaller. ✓ The heaters have the same resistance, so less electrical energy is converted to heat in H1 per second (P = I2R is smaller). ✓ OR: more volts are lost inside battery A (Ir larger), so the potential difference across H1 is smaller and P = V2 R is smaller. NOTE: No marks for a calculation (“WITHOUT A CALCULATION”). (3) [21] Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 17 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 9 9.1 Electromagnetic induction ✓ Accept: Faraday’s Law (of electromagnetic induction). (1) 9.2 Vrms = Vmax √ 2 ✓ = 180 √ 2 ✓ = 127,28 V ✓ (3) 9.3 POSITIVE MARKING FROM QUESTION 9.2 NO ✓ OPTION 1 (CURRENT) The speakers are in series: Rtotal = 8+8 = 16 Ω✓ Irms = Vrms R ✓= 127,28 16 ✓= 7,95 A 7,95 A < 11 A: the current through each speaker is less than the 11 A needed, so neither speaker gives its maximum sound output. ✓ OPTION 2 (POWER) Needed: Pave = I2rmsR = (11)2(8) = 968 W per speaker (1 936 W for both) ✓ Available: Pave = V2rms R = (127,28)2 16 ✓✓= 1012,50 W in total (506,25 W per speaker) 1012,50 W < 1936 W (506,25 W < 968 W per speaker), so the power is not enough. ✓ OPTION 3 (POTENTIAL DIFFERENCE) Needed: Vrms = IrmsR = (11)(16) ✓✓= 176 V ✓; 127,28 V < 176 V, so not enough. ✓ NOTE: The 11 A rating of an AC device is an rms value. Comparing the PEAK current (180 16 = 11,25 A) with 11 A and answering YES is incorrect: Max 1 5 (total resistance). (5) Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 18 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 9.4 Time (s) Induced voltage (V) 0 360 180 –180 –360 0,5 1 new graph (2 Hz) original (1 Hz) Marking criteria hlayiso.com • Two complete cycles in 1 s (period 0,5 s) ✓ • Maximum induced voltage of 360 V (and –360 V) indicated ✓ • Correct shape: sine graph starting at 0 and going positive first, like the original ✓ Hlayiso note: The emf is proportional to the speed of rotation (εmax = NBAω), so doubling the speed doubles both the frequency (2 Hz) and the peak voltage (2×180 = 360 V). (3) [12] Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 19 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 10 10.1 If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The process whereby electrons are ejected from a metal surface when light (of suitable frequency) is incident on that surface. ✓✓ (2) 10.2 Light has a particle nature (light consists of photons/quanta). ✓ Accept: light energy is quantised. (1) 10.3 Marking criteria hlayiso.com • Dependent (current) and independent (intensity of the light) variables correctly identified ✓ • A question about the relationship between the independent and dependent variables ✓ Example: What is the relationship between the intensity of the incident light and the current in the circuit (photocurrent)? (2) 10.4.1 REMAINS THE SAME ✓ Accept: DECREASES (fewer photons per second at the same intensity) Hlayiso note: This is the answer the DBE is most likely to accept. DBE treats the photocurrent as depending on the intensity of the light only (Feb/March 2016 11.5: “the photocurrent is directly proportional to the intensity”). It has twice rejected “the current increases when the frequency increases” (Nov 2014 1.10 statement (iii); Nov 2020 1.10 option C). Strict physics gives a different answer: at the SAME intensity (energy per unit area per second), higher-frequency photons each carry more energy, so FEWER photons and photoelectrons arrive per second and the gradient DECREASES. We did not find a DBE memo that settles this exact item, so check the official memo, which may also accept DECREASES. (1) 10.4.2 REMAINS THE SAME ✓ The number of photoelectrons ejected per second (the current) depends only on the number of incident photons per second (the intensity), which is unchanged; each photon ejects one electron. A lower work function only increases the (maximum) kinetic energy of the ejected electrons. ✓ (2) Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 20 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 10.5 Marking criteria hlayiso.com • Formula E = W0 +Ek(max) (any form) ✓ • Substitution of hc λ : (6,63×10–34)(3×108) 390×10–9 ✓ • Substitution of W0 = 3,78×10–19 J ✓ • Substitution into 1 2mv2max with m = 9,11×10–31 kg ✓ • Final answer: 5,38×105 m·s–1 (range: 5,38 – 5,39×105 m·s–1) ✓ OPTION 1 E = hc λ = (6,63×10–34)(3×108) 390×10–9 ✓= 5,10×10–19 J E = W0 + 1 2mv2max ✓ 5,10×10–19 = 3,78×10–19 ✓+1 2(9,11×10–31)v2max ✓ vmax = 5,38×105 m·s–1✓ OPTION 2 hc λ = W0 + 1 2mv2max ✓ (6,63×10–34)(3×108) 390×10–9 ✓= 3,78×10–19 ✓+1 2(9,11×10–31)v2max ✓ vmax = 5,38×105 m·s–1✓ (Ek(max) = 1,32×10–19 J.) (5) [13] TOTAL: 150 Copyright reserved Please turn over

hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com h hlayiso.com hlayiso.com hlayiso.com hlayiso.com hlayiso.com hla hla hla hla h Physical Sciences/P1 21 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 HOW DID OUR PREDICTION DO? The real November 2026 P1 compared with the Hlayiso 2026 Physical Sciences P1 Prediction Q Topic (marks) Real paper asked Our closest prediction item(s) Verdict Covered 1 MCQ (20) N2 proportionalities; gravitation ratio; stone from rising balloon; p–t graph; E + mgh; Doppler fL–fs graph; E vs test charge; series power; motor; photon < W0 Gravitation ratio, identical-resistor power, projectile symmetry, impulse graph, N2/N3, energy graph. No Doppler graph, field or motor MCQ / 10/20 2 Newton (13) FBD with F at 15◦(4); define friction (2); prove string breaks (5); motion after break (2) Angled force on a horizontal surface, FBD, a for EACH block separately. Defined N2L, not friction  9/13 3 Projectiles (19) Define projectile (2); v at x vs t (2); max height from 20 m·s–1 (3); h from “1,5 s above h” (5); t (3); v–t graphs, two balls (4) Free fall definition; same 20 m·s–1 launch and 20,41 m; time to top; two-ball v–t graphs. No “time above a level” item  14/19 4 Momentum (11) Conservation principle (2); vi from a v–t graph (4); a of P vs Q (2); F∆t gives t (3) Principle; collision read from a v–t graph; Newton III force; impulse with contact time  11/11 5 Work–energy (12) Non-conservative force (2); distance from Wnc (4); θ (4); Wnet vs mass (2) Same definition (even-year call); crate down a rough incline with Wnc  10/12 6 Doppler, spectra (13) Doppler (2); wavefront drawing (2); two approaching sources, speed from a ratio (6); absorption, red shift, away (3) Definition; moving listener (two equations, eliminate fs); wavefront explanation, no drawing; red-shift MCQ; absorption spectrum / 9/13 7 Electrostatics (16) Coulomb (2); vector triangle (3); θ vs 5◦(2); define E at a point (2); direction (1); unknown r from net E (6) Coulomb’s law (our strongest call); net E at a point. We drilled 2D forces at 90◦, not an equilibrium triangle / 9/16 8 Circuits (21) R from parallel (2); I from 25 W (3); rheostat from emf (5); ammeter change, explain (6); brightness (1); battery and heater temperatures (4) Rated bulb, branch currents, ε = I(R+r), “switch opened, explain” chain, heat in the battery  21/21 9 AC (12) Principle (1); Vrms (3); speakers, enough power? (5); sketch at double speed (3) rms and average-power calculation; emf–time sketch. No “name the principle”  11/12 10 Photoelectric (13) Define effect (2); particle nature (1); investigative question (2); gradient changes (3); vmax (5) vmax from λ (even-year call); “one photon, one electron” reasoning. We defined the work function instead / 7/13 Marks of the real paper covered by our prediction 111/150 A learner who had practised our prediction paper would have met about 111 of the 150 marks (74%) as familiar question types. The ten-question structure and topic order held, Q4 and Q8 were covered almost item for item, and Q3 used our 20 m·s–1 launch speed. Most losses were definitions, drawings (vector triangle, wavefronts), Q1 and Q10. What we got right • Even-year calls that landed: Coulomb’s law, non-conservative force, vmax of photoelectrons, and the “due” gravitation-ratio MCQ. • An applied force at an angle on a horizontal surface, solved block by block (Q2). • Two-ball vertical projectile with v–t graphs; conservation of momentum read from a v–t graph. • Rough incline with Wnc; heat in a battery; the 6-mark “resistance up, current down, lost volts down” chain. • Red shift and an absorption spectrum; an AC sketch and rms/power work. What we missed • No pulley and no moving listener: the paper used a string-breaking test and two sources approaching one listener. • An equilibrium vector triangle (Q7.2), not the 2D forces at right angles we drilled. • Definitions we did not set: frictional force, projectile, electric field at a point, photoelectric effect. • The investigative question and the “gradient of I vs intensity” items in Q10. • MCQs on a Doppler fL–fs graph, a motor and E vs test charge. Our emf, rms and work-function definitions were not asked. Rating: 7/10. The structure, topic order and most calculation types were called correctly, and Q4, Q5 and Q8 were very close to the real paper (111/150 covered). Several definitions, the vector triangle, the wavefront drawing and the Q10 concept items were missed. Copyright reserved