Technical Sciences Grade 12 P1 Eastern Cape June 2024 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 12 JUNE 2024 TECHNICAL SCIENCES P1/ TEGNIESE WETENSKAPPE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 12 pages./ Hierdie nasienriglyn bestaan uit 12 bladsye. Hosted by www.ecexams.co.za

Downloaded from hlayiso.com 2 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE 2024) QUESTION/VRAAG 1 1.1 B ✓✓ (2) 1.2 D ✓✓ (2) 1.3 B ✓✓ (2) 1.4 A ✓✓ (2) 1.5 A ✓✓ (2) 1.6 B ✓✓ (2) 1.7 C ✓✓ (2) 1.8 C ✓✓ (2) 1.9 A ✓✓ (2) 1.10 A ✓✓ (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 3 QUESTION/VRAAG 2 2.1 2.1.1 An object continues in its state of rest or uniform (moving with constant) velocity ✓ unless it is acted upon by a net (resultant) force.✓ ʼn Voorwerp bly in rus of bly beweeg teen ʼn uniforme (konstante) snelheid ✓ tensy ʼn netto (resultante) krag daarop dit inwerk. ✓ (2) 2.1.2 Gravitational force/Gravitasiekrag ✓ (1) 2.1.3 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 Fnet = ma Fnet = ma Fnet = 0 ✓ Fnet = 0 ✓ FH + fk = 0 FH + fk =0 50cos60° + fk = 0 ✓ 50cos60° + fk = 0 ✓ fk = –25 N fk = – 25 N fk = µkFN ✓ FN = mg – FV 25 = µk[ (10× 9,8) – 50sin60°] ✓ FN = mg – Fsinθ µk = 0,46 ✓ FN = 10× 9,8 = 50sin60 ✓ = 54,69 N fk = µkFN ✓ 25 = µk 54,69 µk = 0,46 ✓ OPTION 3 / OPSIE 3 FH = – fk ✓ fk = µkFN ✓ FH = – 50cos60°✓ 25 = µk[ (10× 9,8) – 50sin60°] ✓ fk = – 25 N µk = 0,46 ✓ (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE 2024) 2.2 2.2.1 When a force is exerted on an object, the object will accelerate in the direction of the force. ✓ This acceleration is directly proportional to the force, and indirectly proportional to the mass of the object. ✓ Indien ʼn krag op ʼn voorwerp inwerk sal die voorwerp versnel in die rigting van die krag. ✓ Hierdie versnelling is direk eweredig aan die krag op die voorwerp en omgekeerd eweredig aan die massa van die voorwerp. ✓ (2) 2.2.2 FN  F2  fk  F 40 kg  Fg  MARK/ Acceptable labels/ aanvaarbare byskrifte PUNT Fg/Fw/weight/gravitational force ✓ Fg/Fw/gewig/gravitasiekrag fk / Ff/Frictional force ✓ fk / Ff/Wrywingskrag FN/N/Normal Force ✓ FN/N/Normaalkrag F2/ applied force on object 2 ✓ F2/toegepaste krag op voorwerp 2 F40 kg / applied force on 40 kg object ✓ F40 kg/toegepaste krag op 40 kg voorwerp (5) Notes/Notas ✓One mark awarded for each label and arrow. Een punt toegekem vir elke byskrif en pylpunt • Do not penalise for length of arrows. Moet nie penaliseer vir die lengte van die pyle nie • Any additional forces: Maks -1 Enige addisionele krag: Maks -1 2.2.3 F1 (𝑦) = F1 sin 𝜃 = 40 sin 35 ✓ = 22,94 N ✓ (2) 2.2.4 POSITIVE MARKING FROM 2.2.3/POSITIEWE MERK VAN 2.2.3 𝐹𝑁 = Fg + F1 (𝑦) = mg + F1 (𝑦) = (40) (9,8) + 22,94 ✓ = 414,94 N ✓ (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 5 2.2.5 POSITIVE MARKING FROM 2.2.4/POSITIEWE MERK VAN 2.2.4 𝑓𝑘 = 𝜇𝑘𝐹𝑁 = 𝜇𝑘 (𝐹𝑔 − 𝐹1(𝑦)) = 𝜇𝑘 (𝑚𝑔 − 𝐹1(𝑦)) ✓ = (0,04) (414,94) ✓ = 16,60 𝑁 ✓ left/links ✓ (4) 2.2.6 For 40 kg block/Vir 40 kg blok  𝐹𝑛𝑒𝑡 = 𝐹1 (𝑥) − 𝐹10 𝑘𝑔 − 𝑓𝑘 𝑚𝑎 = 𝐹1 cos 𝜃 − 𝐹10 𝑘𝑔 − 𝑓𝑘 40𝑎 = 40 cos 35 − 𝐹10 𝑘𝑔 − 16,60 ✓ For 10 kg block/Vir 10 kg blok F𝑛𝑒𝑡 = 𝐹40 𝑘𝑔 + 𝐹2 − 𝑓𝑘 𝑚𝑎 = 𝐹40 𝑘𝑔 + 𝐹2 − 𝑓𝑘 10𝑎 = 𝐹40 𝑘𝑔 + 20 − 2,5 ✓ 40𝑎 + 10𝑎 = 40 cos 35 − 𝐹10 𝑘𝑔 − 16,60 + 𝐹40 𝑘𝑔 + 20 − 2,5 ✓ 𝑎 = 0,67 𝑚. 𝑠−2 ✓ (5) 2.2.7 Increase/Neem toe ✓✓ (2) 2.3 2.3.1 50 N ✓to the left/na links ✓ (2) 2.3.2 John’s mass is greater than Mary’s mass. Therefore, John’s inertia is greater than Mary’s inertia. ✓ John’s force on Mary is equal to Mary’s force on John ✓ Therefore, John’s acceleration is less than Mary’s acceleration ✓ John se massa is groter as Mary se massa. Dus is John se traagheid groter as Mary se traagheid. ✓ John se krag op Mary is gelyk aan die krag van Mary op John. ✓ Dus is die versnelling van John minder as die versnelling van Mary. ✓ (3) [35] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE 2024) QUESTION/VRAAG 3 3.1 3.1.1 The product of the net force acting on an object ✓ and the time the force acts on the object./ ✓ Die produk van die netto krag wat op ʼn voorwerp inwerk✓ en die tyd wat die krag op die voorwerp uitoefen. ✓ (2) 3.1.2 OPTION 1 Away from batsman / OPSIE 1 Weg van die kolwer ∆𝑝 = 𝑚 𝑣𝑓 − 𝑣𝑖 ✓ = (0,175) (−30 − 12) ✓ = −7,35 N ∙ s ✓ = 7,35 N ∙ s ✓ away from the batsman/weg van die kolwer ✓ OPTION 2 Towards the batsman / OPSIE 1 Na die kolwer ∆𝑝 = 𝑚 𝑣𝑓 − 𝑣𝑖 ✓ = (0,175) (30 − (−12))✓ = 7,35 N ∙ s ✓ = 7,35 N ∙ s ✓ away from the batsman/weg van die kolwer ✓ (5) 3.1.3 POSITIVE MARKING FROM 3.1.2/POSITIEWE MERK VAN 3.1.2 Fnet Δt = Δp ✓ Fnet (0,05) = 7,35 ✓ Fnet Δt = 147 N ✓ OR /OF Fnet Δt = Δp ✓ Fnet (0,05) = − 7,35 ✓ Fnet Δt = −147 N ✓ (3) 3.1.4 ✓ if Δp is constant ✓ crumple zones increase the time ✓ Fnet decreases. ✓ ✓ as Δp konstant is. ✓ Frommelsones veroorsaak toename in tyd ✓ Fnet neem af. ✓ (4) 3.2 3.2.1 The total linear momentum of an isolated system ✓ remains constant (in magnitude and direction). ✓ Die totale lineêre momentum van ʼn geïsoleerde sisteem ✓ bly konstant (in grootte en rigting). ✓ (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 7 3.2.2 OPTION 1 Right + / OPSIE 1 Regs + ∑ 𝑝𝑏𝑒𝑓𝑜𝑟𝑒 = ∑ 𝑝𝑎𝑓𝑡𝑒𝑟 ✓ 𝑚1𝑣1𝑖 + 𝑚2𝑣2𝑖 = 𝑚1𝑣1𝑓 + 𝑚2𝑣2𝑓 (2) (3) + (3,5) (0) ✓ = (2) (−1) + 3,5𝑣 ✓ 𝑣 = 2,29 𝑚 ∙ 𝑠−1 ✓ 𝑣 = 2,29 𝑚 ∙ 𝑠−1; right/regs ✓ OPTION 2 Left + / OPSIE 2 Links + ∑ 𝑝 𝑏𝑒𝑓𝑜𝑟𝑒 = ∑ 𝑝 𝑎𝑓𝑡𝑒𝑟 ✓ 𝑚1𝑣1𝑖 + 𝑚2𝑣2𝑖 = 𝑚1𝑣1𝑓 + 𝑚2𝑣2𝑓 (2) (−3) + (3,5) (0) ✓ = (2) (1) + 3,5𝑣 ✓ 𝑣 = −2,29 𝑚 ∙ 𝑠−1 ✓ 𝑣 = 2,29 𝑚 ∙ 𝑠−1; right/regs ✓ (5) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE 2024) QUESTION/ VRAAG 4 4.1 4.1.1 The product of the force applied on an object ✓ and the displacement in the direction of the force. ✓ Die produk van die krag wat op ʼn voorwerp toegepas word ✓ en die verplasing in die rigting van die krag. ✓ (2) 4.1.2 𝑊𝐹 = 𝐹∆𝑥𝑐𝑜𝑠𝜃 ✓ = (50) ✓(10) 𝑐𝑜𝑠0° ✓ = 500 𝐽 ✓ (4) 4.1.3 𝑊𝑓 = 𝑓∆𝑥𝑐𝑜𝑠𝜃 = (20) ✓(10) 𝑐𝑜𝑠180°✓ = −200 𝐽 ✓ (3) 4.1.4 POSITIVE MARKING FROM 4.1.2 TO 4.1.3 FOR OPTION 1 POSITIEWE MERK VAN VRAAG 4.1.2 NA 4.1.3 VIR OPSIE 1 OPTION/OPSIE 1 𝑊𝑛𝑒𝑡 = 𝑊𝐹 + 𝑊𝑓 = 500 ✓ + (−200) ✓ = 300 𝐽 ✓ OPTION/OPSIE 2 𝐹𝑛𝑒𝑡 = 50 − 20 = 30 𝑁 𝑊𝑛𝑒𝑡 = 𝐹𝑛𝑒𝑡∆𝑥𝑐𝑜𝑠𝜃 = (30) ✓(10) 𝑐𝑜𝑠0° ✓ = 300 𝐽 ✓ (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 9 4.2 4.2.1 T Acceptable labels MARK Aanvaarbare byskrifte PUNT Fg/Fw/weight/gravitational force ✓ Fg/Fw/gewig/gravitasiekrag FT/tension/ force of rope on engine ✓ FT/spanning/ krag van tou op enjin w Notes/Notas ✓ One mark awarded for label and arrow. ✓ Een punt toegeken vir byskrif en pylpunt • Do not penalise for length of arrows. • Moet nie penaliseer vir die lengte van die pyle nie • Any additional forces: Max -1 • Enige addisionele kragte: Maks -1 1 (2) 4.2.2 𝑃𝑎𝑣𝑒 = 𝐹𝑎𝑣𝑒 𝑣 ✓ 𝑃𝑎𝑣𝑒 = 𝑚𝑔𝑣 7350 ✓ = (1 500)(9,8)𝑣 ✓ 𝑣 = 0,5 𝑚 𝑠 −1 ✓ (4) 4.2.3 𝑊𝐹 = 𝐹∆𝑥 cos 𝜃 ✓ 𝑊𝐹 = (1 500)(9,8)✓ cos 00 ✓ 𝑊𝐹 = 7350 𝐽✓ (3) 4.2.4 𝐸𝑃 = 𝑚𝑔ℎ✓ 𝐸𝑝 = (1 500)(9,8)(0,5)✓ 𝐸𝑝 = 7 350 𝐽✓ (3) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE 2024) QUESTION/VRAAG 5 5.1 Within the limit of elasticity, stress is directly proportional to strain ✓✓/ Binne die grense van elastisiteit is druk eweredig aan rekking ✓✓ (2) 5.2 5.2.1 F δ= ✓ A 3 500 δ= ✓ 5 × 10-5 δ = 7 × 107 Pa ✓ (3) 5.2.2 ∆l ε= L ✓ 12 ε = 3 350 ✓ ε = 3,582 ×10-3 ✓ (3) 5.2.3 POSITIVE MARKING FROM /POSITIEWE MERK VAN 5.2.1 and/en 5.2.2 ∂ K= ε✓ 7 × 107 K= ✓ 3,582 ×10-3 K = 1,95 × 1010 Pa ✓ (3) 5.3 Clay, putty, wax, bread dough (any TWO) ✓✓/ Klei, putty, was, brooddeeg (Enige TWEE) (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 11 QUESTION/VRAAG 6 6.1 6.1.1 It is the thrust acting on the unit area around that point. ✓✓/ Dit is die stukrag op die eenheidsoppervlakte rondom daardie punt ✓✓ (2) 6.1.2 P = F ✓ A (250)(9,8) ✓ P= 4 6 (3,15 ×10 )(10) ✓ P = 7,78 × 104 Pa ✓ (4) 6.2 6.2.1 P = ρgh ✓ P = (1 000)(9,8) ✓ (2,5 − 0,9) ✓ P = 1,57 ×104 Pa ✓ (4) 6.2.2 Less Than / Minder as ✓ From/Van P = h ✓ P α ρ ✓ if g and h are constant/Indien g en h konstant is ρpetrol < ρwater ✓ (4) 6.3 6.3.1 In a continuous liquid at equilibrium, the pressure applied at a point ✓ is transmitted equally to the other parts of the liquid. ✓ In ʼn kontinue vloeistof by ewewig is die druk wat by enige punt toegepas word ✓ eweredig na die ander dele van die vloeistof versprei. ✓ (2) 6.3.2 F1 F = A2 ✓ A1 2 F1 250 -4 ✓= ✓ 200 × 10 5 × 10-4 F1 = 10 000 N ✓ (4) 6.3.3 Liquid adapts to the form of the container. Liquid is not compressible. (Any TWO ✓✓) Liquid applies pressure in all directions. ʼn Vloeistof pas by dievorm van die houer aan. ʼn Vloeistof kan nie saamgepers word nie. (Enige TWEE ✓✓) ʼn Vloestof pas druk uit in alle rigtings. (2) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE 2024) QUESTION/VRAAG 7 7.1 The bending of light when it passes from one medium to another. ✓✓/ Die buiging van lig wanneer dit van een medium na ʼn ander beweeg. ✓✓ (2) 7.2 7.2.1 Incident ray/Invalsstraal ✓ (1) 7.2.2 Normal/Normaal ✓ (1) 7.3 The angle of incidence in the denser medium ✓such that the refracted ray just passes through the surface of separation of the two mediums. ✓ OR The angle of incidence in the optically dense medium ✓ for which the angle of refraction is 90˚. ✓ Die invalshoek in die digter medium ✓sodat die gebreekte straal net deur die oppervlakte van skeiding van die twee mediums beweeg. ✓ OF Die invalshoek in die opties digte medium ✓waarvoor die brekingshoek 90˚ is. ✓ (2) 7.4 Total Internal Reflection/Totale Interne Weerkaatsing ✓ (1) 7.5 GREATER THAN / GROTER AS ✓ (1) 7.6 The ray moves from a more dense to a less dense medium ✓and will therefore be refracted away from the normal ✓ which will increase the angle. Die straal beweeg van ʼn digter medium na ʼn minder digter medium ✓en sal dus weg van die normaal ✓ gebuig word en die hoek sal dus vergroot. (2) 7.7 object image Marking guidelines/Nasienriglyne • Object at 2F / Voorwerp by 2F ✓ • Image at 2F / Beeld by 2F ✓ • Ray goes through optic centre / Straal gaan deur optiese middelpunt ✓ • Ray parallel with main axis / Straal parallel aan die hoof-as ✓ • Image the same size as object / Beeld net so groot soos die voorwerp ✓ (5) [15] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief