Technical Sciences Grade 12 P1 Gauteng Mock 2025 Memo (Eng)

PREPARATORY EXAMINATION 2025 MARKING GUIDELINES TECHNICAL SCIENCE (PAPER 1) (11101) 10 pages

11101/25 QUESTION 1: MULTIPLE-CHOICE QUESTIONS 1.1 C ✓✓ (2) 1.2 A ✓✓ (2) 1.3 D ✓✓ (2) 1.4 D ✓✓ (2) 1.5 A ✓✓ (2) 1.6 B ✓✓ (2) 1.7 B ✓✓ (2) 1.8 D ✓✓ (2) 1.9 C ✓✓ (2) 1.10 A ✓✓ (2) [20] QUESTION 2 2.1 Newton Second Law of Motion When a net/resultant force is applied to an object of mass, m, it accelerates the object in the direction of the net force. The acceleration is directly proportional to the net/resultant force and inversely proportional to the mass of the object. ✓✓ (2 or 0) (2) 2.2 FN✓ NOTE: One mark for each force represented by an arrow with a fk✓ correct label. T✓ Penalise (once) for each of the F=120 N✓ following: Fg✓ • No arrows • There is no dot ACCEPTABLE LABELS: • Gap between the line and the N/FN: Normal dot Fg/w: Force due to gravity/Weight • Dotted lines are used Ff/fk/f: friction force • Additional force is included FA: Applied force/120 N • A force diagram is given. T: Tension force in the rope • If components were drawn (5) 2.3 𝑓𝑘 = 𝜇𝑘 𝑁 ✓ = (0,15)(16 x 9,8) ✓ = 23,52 N ✓ (3) 2.4 FN = Fy + Fg any one✓ FN = Fsin𝜃 + mg FN = (120 sin 30o) + (24 x 9,8) ✓ FN = 295,2 N ✓ (3) 2

11101/25 2.5 Positive marking from 2.3 and 2.4 For the 24 kg block For the 16 kg block FN FN f Fx T f T Fy Fg Fg 𝑓𝑘 = 𝜇𝑘 𝑁 = (0,15)(295,2) = 44,28 N Fnet = ma ✓ Fnet = ma Fx – T – f = ma T – f = ma Fcos 𝜃 – T – f = ma T – 23,52 = 16a (120 cos 30°) ✓ – T – 44,28 = 24 a T = 16a + 23,52 ….②✓ 103,92 – T – 44,28 = 24a✓ 59,64 – T = 24a T = -24a + 59,64 ….①✓ ① = ②✓ 16a + 23,52 = –24a + 59,64 16a + 24a = –23,52 + 59,64 40a = 36,12 a= 0,90 m.s-2 move to right ✓ (7) 2.6 Positive marking from 2.5 Option 1 Option 2 T = 16a + 23,52 OR T = –24a +59,64 T = 16(0,90) + 23,52✓ T = –24(0,90) + 59,64 T = 37,92 N ✓ T = 38,04 N (2) [22] 3

11101/25 QUESTION 3 3.1 Direction right is positive ∑p before collision = ∑p after collision ✓ any one mx vi + my vi = (mx + y )vf (1 500)(20) + (1 000)(vf) ✓ = (1 500 + 1 000)(6) ✓ 30 000 + 1 000vf = 15 000 1 000vf = –15 000 Vf = –15 m∙s-1 vf = 15 m∙s-1 to the left / east ✓ (4) 3.2 Positive marking from 3.1 Ek before collision = ½ mv2 + ½ mv2 Both formulae correct ✓ = ½ (1 500)(20) + ½ (1 000)(–15)2 2 = 412 500 J ✓ Ek after collision = ½ mx+yv2 = ½ (1 500 + 1 000)(6)2 = 45 000 J ✓ Kinetic energy is not conserved (energy has been lost), therefor the collision is Inelastic. ✓ (4) 3.3 The product of the net force and the contact time. ✓✓ (2) 3.4 Neither ✓ Impulse x = Impulse y (1) [11] QUESTION 4 4.1 Pav = Fvav ✓ Pav = (120)(5) ✓ Pav = 600 W ✓ (3) 4.2 4.2.1 At point A ✓ (1) 4.2.2 EMA = EMB Ep + E k = Ep + E k ✓ any one 1 1 mgh + 2 𝑚𝑣 2 = mgh + 2 𝑚𝑣 2 1 15(9,8)(12) + 0 ✓ = 0 + 2 (15)𝑣 2 ✓ v = 15,34 𝑚 ∙ 𝑠 −1 ✓ (4) 4.2.3 Law of conservation of mechanical energy. ✓ The total mechanical energy of an isolated system remains constant. ✓✓ (2 or 0) (3) [11] 4

11101/25 QUESTION 5 5.1 An elastic limit is the maximum force that can be applied to body so that the body regains its original form completely on removal of the force. ✓✓ (2 or 0) (2) 5.2 Option 1 Option 2 𝐹 𝜃= 𝐴 ✓ any one 𝐴 = 𝜋𝑟 2 𝑚𝑔 𝜃= 𝐴 = 𝜋(1 × 10−3 )2 𝜋𝑟 2 (6)(9,8)✓ 𝜃= 𝐴 = 3,142 × 10−6 𝑚2 𝜋(1× 10−3 )2 ✓ 58,8 = 18 716 621,31 Pa / 𝑁 ∙ 𝑚−2 OF ✓any one 𝜃= 3,142 × 10−6 = 1,88 x 107 Pa / 𝑁 ∙ 𝑚−2 = 18 714 194,78 Pa Option 3 𝜋𝐷2 𝐴= 4 2 2 𝜋( ) 1000 = 4 = 3,141592654 × 10−6 𝑚2 𝐹 𝜃= 𝐴 58,8 = 3,141592654×10−6 = 18 716 621,31 Pa / 𝑁 ∙ 𝑚−2 (range 18 714 194,78 – 18 716 621,31) (Note to teachers: teach the learners to only round off final answer – refer to option 2) (4) 5.3 𝜀 = 𝐿✓ ∆𝑙 OR 𝜀= 𝐿 ∆𝑙 1,5 𝑚𝑚 1,5×10−3 𝑚 𝜀= ✓ 𝜀= 4 000 𝑚𝑚 4𝑚 = 3,75 x 10-4✓ = 3,75 x 10 -4 (3) 5.4 Positive marking from 5.2 and 5.3 𝜃 𝐾 = 𝜀✓ 1,88 ×107 𝐾 = 3,75 ×10−4 ✓ 𝐾 = 4,99 × 1010 𝑃𝑎 / 𝑁 ∙ 𝑚−2 ✓ (3) [12] 5

11101/25 QUESTION 6 6.1 Pressure at a particular point as the thrust acting on the unit area around that point. (2 or 0) (2) 6.2 𝑃=𝐴 𝐹 ✓ any one 𝑚𝑔 Fg = mg 𝑃= 𝐴 = 54(9,8) (54)(9,8) ✓ = 529,2 N = 0,85 ✓ = 622,59 Pa / 𝑁 ∙ 𝑚−2 OR ✓ any one 2 −2 = 6,23 × 10 Pa / 𝑁 ∙ 𝑚 (4) 6.3 Positive marking from 6.2 The same as the output system, namely 622,59 Pa / 𝑁 ∙ 𝑚−2 ✓ (1) 6

11101/25 6.4 Positive marking from 6.2 for option 1 only Option 1 Option 2 𝐹𝑖𝑛𝑝𝑢𝑡 𝐹𝑜𝑢𝑡𝑝𝑢𝑡 𝐹1 𝐹 = 𝐹 ✓ = 𝐴2 𝐴𝑖𝑛𝑝𝑢𝑡 𝑜𝑢𝑡𝑝𝑢𝑡 𝐴1 2 𝐹𝑖𝑛𝑝𝑢𝑡 𝐹1 (54)(9,8) ✓ = 622,59✓ = 0,03 0,03 0,85 𝐹𝑖𝑛𝑝𝑢𝑡 = 18,68 𝑁✓ 𝐹1 = 18,68 𝑁 Option 3 𝐹 𝑃= 𝐴 𝐹 622,59 = 0,03 F = 18,68 N (range 18,68 – 18,69) (4) 6.5 It would double / increases ✓✓ (2) [13] QUESTION 7 7.1 7.1.1 Society of Automotive Engineers ✓ (1) 7.1.2 It indicates the viscosity of oil measured at 100 oC ✓✓ (2) 7.1.3 Negative marking if incorrect SAE was chosen SAE10 ✓ It has the smallest viscosity and will therefore flow fastest. ✓ (2) 7.2 7.2.1 It is oil that has grading for viscosity at both cold and hot conditions. ✓ (1) 7.2.2 The 30 is the viscosity of the oil at cold temperatures. ✓ The W indicates that the viscosity of value 30 is measured at cold temperatures. ✓ The 60 is the viscosity of the oil measured at 100 oC ✓ (3) 7.2.3 Negative marking if incorrect SAE was chosen 10W60 ✓ Since the viscosity of 10W will be applicable at cold temperatures ✓and it has a lower viscosity than the 30W of the other oil. ✓ (3) [12] 7

11101/25 QUESTION 8 8.1 8.1.1 Refraction ✓ (1) 8.1.2 Refraction occurs because light travels at different speeds in different mediums. ✓ When light moves from a less optically dense medium (like air) to a more optically dense medium (like water), its speed decreases, causing the light to bend towards the normal. ✓ (2) 8.2 Critical angle: the angle of incidence in the denser medium such that the refracted ray just passes through the surface of separation of the two media ✓✓ (0 or 2) OR The angle of incidence in the optically dense medium for which the angle of refraction is 90° (2) 8.3 1 – Light must be traveling from a more optically dense medium to a less optically dense medium. ✓ 2 – The angle of incidence must be greater than the critical angle. ✓ (2) 8.4 (3) 8.5 8.5.1 Dispersion ✓ (1) 8.5.2 Seven colours (7) ✓ (1) 8.5.3 Red ✓ (1) 8.5.4 Decrease ✓ (1) 8.5.5 The wavelength of light decreases ✓ when it enters the prism because the speed of light decreases ✓ due to the refractive index of the prism. Since the frequency of light remains constant✓ (3) [17] 8

11101/25 QUESTION 9 9.1 Capacitance, C, of a capacitor is the amount of charge it can store per volt. ✓✓ (2 or 0) (2) 9.2 𝑄 𝐶 = 𝑉✓ 9,5 × 10−3 1,5 × 10−6 ✓ = ✓ 𝑉 𝑉 = 6 333,33 𝑉 ✓ (4) [6] 9

11101/25 QUESTION 10 10.1 Electromotive force (EMF) is the maximum potential difference a source can provide when no current is flowing✓, while terminal potential difference (terminal voltage) is the actual voltage measured across the terminals when current is flowing ✓, which is typically lower than EMF due to the source's internal resistance. ✓ (3) 10.2 10.2.1 𝑉 𝑅= ✓ 𝐼 𝑉 6= ✓ 1,5 𝑉2 = 9𝑉 ✓ (3) 10.2.2 Positive marking from 10.2.1 only for option 1 and 3 Option 1 Option 2 𝑉 𝑅=𝐼 OR R:1:2 9 12 = 𝐼 ✓ I : 1,5A : 0,75A I = 0,75 A ✓ ∴ IT = 1,5 + 0,75 IT = 1,5 + 0,75 = 2,25 A ✓ = 2,25 A Option 3 1 1 1 = + 𝑅𝑝 𝑟 𝑟 1 1 = 6 + 12 Rp = 4Ω 𝑉 𝐼=𝑅 9 =4 = 2,25 A (3) 10.2.3 Positive marking from 10.2.2 Option 1 Option 2 𝑉 V = IR OR 𝑅=𝐼 𝑉 = (2,25)(18) ✓ 18 = 2,25 = 40,5 V ✓ V1 = 40,5 V (2) 10.2.4 Positive marking from 10.2.1 and 10.2.3 V3 = V 1 + V 2 = 40,5 + 9 ✓ = 49,5 V ✓ (2) 10

11101/25 10.2.5 Option 1 Option 2 𝑞 = 𝐼∆𝑡 ✓ OR 𝑞 = 𝐼∆𝑡 = 1,5(3 × 60) ✓ = 1,5(180) = 270 C ✓ = 270 C (3) 10.2.6 Positive marking from 10.2.5 for option 1 only Option 1 Option 2 W = VQ ✓ 𝑊 = 𝑉𝐼∆𝑡 = 9(270) ✓ = 9(1,5)(180) −1 = 2 430 J / 𝑉 ∙ 𝐶 ✓ = 2 430 J Option 3 Option 4 𝑉 2 ∆𝑡 W = I2Rt 𝑊= 𝑅 92 (180) = (1,5)2(6)(180) = 6 = 2 430 J = 2 430 J (3) [19] 11

11101/25 QUESTION 11 11.1 Step-down transformer ✓ (1) 11.2 VS = NS V𝑃 𝑁P 20 𝑁 ✓ = 𝑃✓ 220 55 220 𝑁𝑃 = 1100 𝑁𝑃 = 5 Windings ✓ (3) 11.3 Cost = P x t x price 50 W = 0,05 kW = (0,05 kW) ✓ (8 x 7h)(R1,10) ✓ = R3,08 ✓ (3) [7] TOTAL: 150 12