Technical Sciences Grade 12 P1 GP Prelim 2023 Memo (English)

Downloaded from hlayiso.com : AMENDMENT TO MARKING GUIDELINES PREPARATORY EXAMINATION / VOORBEREIDENDE EKSAMEN 2023 FOR ATTENTION / VIR AANDAG: THE CHIEF INVIGILATOR / DIE HOOF TOESIGHOUER SUBJECT / VAK TECHNICAL SCIENCES/TEGNIESE WETENSKAPPE SUBJECT CODE / VAKKODE 11101 PAPER / VRAESTEL 1 DATE OF EXAMINATION / DATUM VAN EKSAMEN 22 SEPTEMBER 2023 The TECHNICAL SCIENCES/TEGNIESE WETENSKAPPE (Paper 1 / Vraestel 1) written on 22 September 2023 has reference. It has come to our attention that there was an anomaly on the data sheet which impacted on questions 6.1.3, 6.1.4 and 6.1.6. To ensure that your candidates are not disadvantaged and prejudiced in way, you are advised to please ask Educators to ignore the above-mentioned questions. In other words, the paper must be marked out of a total of 140 instead of 150 and then the learners’ marks must be converted back to a mark out of 150. E.g., Should a learner have attained 80/140, then that mark is recalculated as 86/150. Use the a b formula: 140 x 100 = b. Then, 100 x 150 = c c is the mark that is entered into SASAMS out of 150. DIRECTOR: EXAMINATIONS MANAGEMENT 22 September 2023

Downloaded from hlayiso.com 11101/23 PREPARATORY EXAMINATION 2023 MARKING GUIDELINES TECHNICAL SCIENCES (PAPER 1) (11101) 9 pages 3

Downloaded from hlayiso.com 11101/23 QUESTION 1: MULTIPLE-CHOICE QUESTIONS 1.1 C ✓✓ (2) 1.2 A ✓✓ (2) 1.3 B ✓✓ (2) 1.4 C ✓✓ (2) 1.5 A ✓✓ (2) 1.6 D ✓✓ (2) 1.7 C ✓✓ (2) 1.8 B ✓✓ (2) 1.9 D ✓✓ (2) 1.10 B ✓✓ (2) [20] QUESTION 2: MATCHING ITEMS 2.1 D✓ (1) 2.2 H✓ (1) 2.3 A✓ (1) 2.4 F✓ (1) 2.5 B✓ (1) 2.6 G✓ (1) 2.7 C✓ (1) 2.8 E✓ (1) [8] 2

Downloaded from hlayiso.com 11101/23 QUESTION 3 3.1 3.1.1 The force parallel to the surface that opposes the motion of an object and acts in the direction opposite to the motion the object. ✓✓ (2) 3.1.2 FA must not be equal or same length with Ff (-1) Symbols Name Mark allocation Ff, f, FK Force of friction ✓ Fg, W, Gravitational force ✓ FA, F Applied force ✓ FN, N Normal force ✓ (4) 3.1.3 Fnet = ma ✓ FA – Ff – = ma FA – 750 ✓ = 960 x 2 ✓ FA = 1 920 + 750 = 2670 N to the right ✓ (4) 3.1.4 Fg = 960 x 9,8 ✓ = 9408 N ✓ (2) 3

Downloaded from hlayiso.com 11101/23 3.2 3.2.1 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) 3.2.2 5 forces ✓ or 5 ✓ (1) 3.2.3 Fnet = ma FH + FDx – Ff = ma ✓ FH + FD cos Ɵ – Ff = ma 120 + 180 cos 15 – 100 ✓= 70a ✓ 193,87 = 70a a = 2,77 m·s–2 to the right ✓ (4) 3.2.4 DECREASE ✓ (1) [20] QUESTION 4 4.1 4.1.1 It is the product of the net force and the time interval for which the net force acts on the body. ✓✓ (2) 4.1.2 1 kg = 1 000 g 1 kg : 103 g 1 kg = 1 000 g x : 150 g 1000 1 000 OR 150 1 g = 0,001 kg x= 3 10 150 g = 0,15 kg ✓✓ = 150  10-3 kg ✓✓ (2) 4.1.3 DECREASE ✓ (1) 4.1.4 Option 1 Option 2 Fnet Δt = mvf – mvi = Δp ✓ Fnet Δt = mvf – mvi = Δp ✓ = 0,15(32 – (–50)) ✓ = 0,15(-32 – (+50)) ✓ 0,012 ✓ 0,012 ✓ = 1 025 N towards the wall ✓ = -1 025 N away from the wall ✓ (4) 4.2 4.2.1 From Fnet = Δp Δt The air bag causes the delay of impact by increasing the time of impact and decreasing the net force. ✓✓ (2) 4.2.2 Increase in velocity will increase the impulse since the impulse is directly proportional to the velocity. ✓✓ (2) [13] 4

Downloaded from hlayiso.com 11101/23 QUESTION 5 5.1 Work done is the product of the applied force on an object and the displacement in the direction of the force. ✓✓ (2) 5.2 W = F Δx cos Ɵ ✓ = 5 x 3 cos 0° ✓ = 15 J ✓ (3) 5.3 5.3.1 The principle of conservation of mechanical energy: The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. ✓✓ (2) 5.3.2 EK = ½mv2 ✓ = ½ x 2 x 1,52 ✓ = 2,25 J ✓ (3) 5.3.3 EP = mgh ✓ = 2 x 9,8 x 0,5 ✓ = 9,8 J ✓ (3) 5.3.4 positive marking from 5.3.2 and 5.3.3 ME top = ME bottom ✓ (½mv2 + mgh)top = (½mv2 + mgh)bottom ½ x 2 x 1,52 + 2 x 9,8 x 0,5 ✓= ½ x 2 x v2 ✓+ 2 x 9,8 x 0 ✓ 2,25 + 9,8 = v2 + 0 v2 = 12,05 v = 3,47 m·s–1 ✓ (5) [18] QUESTION 6 6.1 6.1.1 Deforming force is a force that changes the shape and size of a body. ✓✓ (2) 6.1.2 Area = πr2 ✓ = 3,14 x 0,032 ✓ = 0,0028 m2 ✓ (2,83 x 10-3 m2) (3) 6.1.3 σ = F ✓ (do not mark) Due to Data sheet A = 6 000 ✓ 0,0028 = 2 142 857,14 = 2,14 x 106 N.m–2 ✓ (3) 6.1.4 ε = ΔL ✓ (do not mark) Due to Data sheet l = 0,01 ✓ 3 = 0,0033✓ (3,33 x 10-3) (3) 5

Downloaded from hlayiso.com 11101/23 6.1.5 Hooke’s law states that, within the limit of elasticity, stress is directly proportional to the strain. ✓✓ (2) 6.1.6 K = σ ✓ (do not mark) Due to Data sheet ε = 2,14 x 106 ✓ 0,0033 ✓ = 648 484 848.5 = 6,48 x 108 Pa or N.m–2 ✓ (4) 6.2 6.2.1 Viscosity is the property of the fluid to oppose relative motion between the two adjacent layers. ✓✓ (2) 6.2.2 5W-40 ✓ (1) 6.2.3 As the temperature decreases, the viscosity increase. ✓ (1) 6.2.4 5W-20 ✓ (1) 6.2.5 • To reduce wear on the moving parts • Cleans the moving parts • Reduces corrosion • Cools the engine • Improves sealing (Any ONE) ✓ (1) Mark this question out of 13 marks [13] QUESTION 7 7.1 Capacitance of a capacitor is the amount of charge it can store per volt. ✓✓ (2) 7.2 C = Q V Any two pairs of coordinates can be used to = 10 – 5 ✓ OR calculate capacitance (gradient may be used/ any 4 – 2✓ value may be used from the graph) = 2,5 C.V–1 (F)✓ (3) 7.3 The charge is directly proportional to potential difference. OR As the charge increases, the potential difference increases. ✓✓ (2) 7.4 • Filter circuits in power supplies • Separation of frequencies between the woofer (base) speaker and tweeter (high pitch) speaker/electronic noise filtering • Power factor correction/improvement in electrical transmission systems (Any TWO) ✓✓ (2) [9] 6

Downloaded from hlayiso.com 11101/23 QUESTION 8 8.1 Ohm's law states that: The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. ✓✓ (2) 8.2 8.2.1 1 = 1+ 1 ✓ Req R1 R2 = 1 + 1✓ 4 2 =3 4 Req = 1,33 Ω Rtotal = Req + Rs ✓ = 1,33 + 3 = 4,33 Ω ✓ (4) 8.2.2 RT = VT ✓ IT 4,33 = 12 ✓ IT IT = 12 4,33 = 2,77 A ✓ (3) 8.2.3 Req = V V = IR I = (2,77) (3) 1,33 ✓ = V ✓ = 8,31 2,77 12 – 8,31 V = 1,33 x 2,77 = 3,69 V = 3,68 V ✓ (3) 8.2.4 Option 1 12 V – 3,68 V = 8, 32 V ✓ (positive marking from previous questions) P = VI ✓ = 8,32 x 2,77 ✓ = 23,05 W ✓ Option 2 P = V2 / R =8,322 / 3 = 23,07 W Option 3 P = I2 x R (4) =2,772 x 3 = 23,02 W 8.3 INCREASE ✓ NEGATIVE MARKING (2) 7

Downloaded from hlayiso.com 11101/23 The effective resistance decreases✓ [18] 8

Downloaded from hlayiso.com 11101/23 QUESTION 9 9.1 Normal Incident ray Reflection ray ɵi ɵr Mirror Angle of Incident ✓ Reflected beam ✓ (If the arrow is not Normal ✓ shown – 1 mark) Incident beam ✓ Angle of reflection ✓ (5) 9.2 Critical angle: The critical angle is measured when the incident ray passes from an optically more dense to an optically less dense medium, such that the refracted ray just passes through the surface of the two media. ✓✓ OR The critical angle is the angle incidence where the angle of refraction is 90°. The refracted beam then travels along the boundary between the two media.✓✓ (2) 9.3 Dispersion: The phenomenon whereby (white) light breaks up into its component colours. ✓✓ (Award the marks in Afrikaans) (2) 9.4 When light travels from air to water, it moves from a less dense to an optically ✓ denser medium. The colour of the light is determined by frequency ✓ and the colour of light does not change, but the speed of the light and the wavelength decrease. ✓ (3) [12] 9

Downloaded from hlayiso.com 11101/23 QUESTION 10 10.1 Electromagnetic wave: A change of a magnetic and an electric field perpendicular to each other and to the direction of propagation of the wave ✓✓ (2) 10.2 • EM waves move at a constant speed. • EM waves do not need a medium. • They have properties of waves. • EM waves have particle properties. • EM waves are transverse waves. (Any 4) ✓✓✓✓ (4) 10.3 X-rays/Gamma rays ✓ Infrared for Physiotherapy. UV rays to sterilize equipment. Marker may use own discretion on equipment or places within context. (1) 10.4 X-rays are used to ascertain if there are bone fractures. Gamma rays are used to treat cancer. Marker may use own discretion on equipment context. (Look at 10.3’s answer) ✓✓ (2) [9] TOTAL: 140 10