TECH SCIENCES P1 MEMO GR12 SEPT 2024_Afr+English_hlayiso.com_.pdf

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2024 TECHNICAL SCIENCES P1/ TEGNIESE WETENSKAPPE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 13 pages./ Hierdie nasienriglyn bestaan uit 13 bladsye. Hosted by www.ecexams.co.za.

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

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 3 QUESTION/VRAAG 2 2.1 When the brakes are applied, the truck moves slower/the truck experiences a change in its state of motion, ✓ while the man keeps on moving with the same velocity the truck was moving with. ✓ Wanneer die remme toegepas word, beweeg die trok stadiger/ die trok ervaar ʼn verandering in sy toestand van beweging, ✓ terwyl die man aanbeweeg met dieselfde snelheid wat die trok beweeg het. (2) 2.2 Newton’s first law/Newton se eerste wet ✓ (1) 2.3 2.3.1 When a net 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 force and inversely proportional to the mass of the object.) ✓✓ Wanneer ʼn netto krag op ʼn voorwerp met massa, m, inwerk, versnel die voorwerp in die rigting van die netto krag. (Die versnelling is direk eweredig aan die netto krag en omgekeerd eweredig aan die massa van die voorwerp.) (2) 2.3.2 N✓ T✓ fk ✓ w✓ Marking criteria/Nasienkriteria Labels/ Marks/ Byskrifte Punte Correct direction and label of normal force/ Regte rigting en benoeming van die normale krag N/FN 1 Correct direction and label of the weight/ Regte rigting en benoeming van die gewig Fg/w 1 Correct direction and label of horizontal and vertical component of the applied force OR Correct direction and label of applied force/ Regte rigting en benoeming van horisontale en vertikale komponent van die toegepasde krag OF Korrekte rigting en benoeming van toegepaste krag F/Fa 1 Correct direction and label of frictional force/ Regte rigting en benoeming van wrywingskrag f/fk/Ff 1 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 4 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) 2.3.3 Fnet = 0 N + FV + w = 0 Any ONE/Enige EEN N + Fsinθ + mg = 0 N = (5)(9,8)  − (25sin 30°)  N = 36,5 N  (4) 2.3.4 Positive marking from QUESTION 2.3.3/Positiewe nasien vanaf VRAAG 2.3.3 For block X/Vir blok X Fnet = ma  FH + T + fk = ma 25cos30° – T – (0,2)(36,5) = 5a  -T + 14,35 = 5a………………………………………(equation1/vergelyking 1) For block Y / Vir blok Y Fnet = ma T + fk = ma T – (0,1)(3)(9,8)  = 3a  T – 2,94 = 3a…………………………………………(equation 2/ vergelyking 1) Add equation 1 and 2 / Tel vergelyking 1 en 2 op -T +14,35 = 5a T – 2,94 = 3a 11,49 = 8a a = 1,43m.s-2  (Accept 1,426m.s-2 / Aanvaar 1,426 m.s-2) (6) 2.3.5 Positive marking from QUESTION 2.3.4 / Positiewe nasien vanaf VRAAG 2.3.4 OPTION 1 from equation 1/ OPTION 2 from equation 2 OPSIE 1 vanaf vergelyking 1 OPSIE 2 vanaf vergelyking 2 -T +14,35 = 5a T – 2,94 = 3a T=14,35 – (5)(1,43) ✓ T = (3)(1,43) ✓ T = 7,22N ✓ T = 7,22N ✓ (Accept/Aanvaar 7,20N to 7,22N) (2) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 5 QUESTION/VRAAG 3 3.1 3.1.1 A system in which the net external force acting on the system is zero.  ʼn Sisteem waar by die netto eksterne kragte wat op die sisteem toegepas word, nul is. (2) 3.1.2 The product of an object’s mass and its velocity.  Die produk van ʼn voorwerp se massa en sy snelheid. (2) 3.2 Zero/Nul/0  (1) 3.3 p = mv = (3 x 10-3)(120)  = 0,36 kg∙m·s-1  (3) 3.4 The total linear momentum of an isolated system remains constant (is conserved) in magnitude and direction.  Die totale lineêre momentum van ʼn geïsoleerde sisteem  bly konstant (word behou) in grootte en rigting.  (2) 3.5 Positive marking from QUESTION 3.3/ Positiewe nasien vanaf VRAAG 3.3 Σpi = Σpf  0,36 + 0  = (50 x 10-3)vf  vf = 7,2 m·s-1 (No mark for answer/ Geen punt vir antwoord). (3) 3.6 1 Ki = 2 mv2 If a learner starts off with Ki = Kf, / As die leerder begin met Ki = Kf, 1 -3 2 = 2 (3 x 10 )(120 ) + 0 Max./Maks. 3/5 = 21,6 J 1 Kf = 2 mv2 1 = 2 (50 x 10-3)(7,2)  = 1,296 J Ki ≠ Kf  Therefore, INELASTIC collision./ Dus ONELASTIESE botsing  (5) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 6 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION/VRAAG 4 4.1 The product of the force applied on an object and the displacement in the direction of the force.  Die produk van die toegepaste krag op ʼn voorwerp en die verplasing in die rigting van die krag. (2) 4.2 OPTION 1/ OPSIE 1 OPTION 2/ OPSIE 2 WFa = Fa.∆x.cosθ WFa = Fa.∆x.cosθ WFx = Fx.∆x.cosθ  WFx = (Fcosθ).∆x.cosθ  WFx = (F.cosθ).∆x.cosθ WFx = 15(5) cos0  WFx = (15 cos30°)(5)  cos0°  WFx = 75 J WFx = 64,95 J  NOTE: (2/4) Max./Maks. (4) 4.3 4.3.1 The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant.  Die totale meganiese energie (som van gravitasie-potensiële energie en kinetiese energie) in ʼn geïsoleerde stelsel bly konstant.  (2) 4.3.2 Em(A) = Em(B) (EP + EK)A = (EP + EK)B  2 (mgh + ½mv )A = (mgh + ½mv )B 2 (1,5)(9,8)(12) + ½(1,5)(0)  = (1,5)(9,8)(7,5) + ½(1,5)𝑣2  v = 9,39 m ∙ s−1  (4) 4.3.3 The force that opposes the motion of a moving object (relative to a surface)  and acts parallel to the surface.  Die krag wat teenstand bied teen die rigting van ʼn bewegende voorwerp (relatief aan die oppervlakte) en wat parallel met die oppervlakte toegepas word. OR The force parallel to the surface that opposes the motion of a moving object  and acts in the direction opposite to the motion of the object.  Die krag parallel aan die oppervlak wat die beweging van ʼn bewegende voorwerp teenstaan en in die rigting teenoor die beweging van die voorwerp inwerk. OR The force acting between the two surfaces in contact when an object is moving.  Die krag wat toegepas word tussen die twee oppervlaktes in kontak met mekaar wanneer die voorwerp beweeg. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 7 4.3.4 No / Nee Negative marking / Negatiewe nasien The system is isolated. / Die sisteem is geïsoleerd. (2) 4.3.5 Wfk = fk∆x Cos θ  = (4,41)(2)cos180°  = − 8,82 J  (3) 4.3.6 Positive marking from QUESTION 4.3.5 / Positiewe nasien vanaf VRAAG 4.3.5 W P = ∆t  8,82 =  5 = 1,76 J  (3) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 8 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION/VRAAG 5 5.1 5.1.1 A property of the body by virtue of the body to regain its original shape and size when the deforming force is removed. /  Die eienskap van die liggaam waardeur die liggaam in staat is om sy oorspronklike vorm en grootte te herstel  wanneer die vervormingskrag verwyder word.  (2) 5.1.2 A perfectly elastic body regain it’s original shape and size completely when the deforming force is removed.  A perfectly plastic body does not regain it’s original shape and size when the deforming force is removed.   ʼn Volkome elastiese liggaam is ʼn liggaam wat sy oorspronklike vorm en grootte volkome herwin wanneer die vervormingskrag verwyder word.  ʼn Volkome plastiese liggaam is ʼn liggaam wat nie ʼn neiging toon om sy oorspronklike grootte en vorm te verkry wanneer die vervormingskrag verwyder word nie.  (2) 5.2 5.2.1 F σ = A  12 000 = 9 500  =1,26 m2  (3) 5.2.2 ∆𝑙 𝜀=  𝐿 0,12 = 1,5  = 0,08  Accept/Aanvaar (0,1) (3) 5.3 5.3.1 The property of the fluid to oppose relative motion  between the two adjacent layers. /  Die eienskap van die vloeistof om relatiewe beweging  tussen die twee aangrensende vlakke te opponeer.  OR/OF Viscosity is the internal property of a fluid  that offers resistance to flow. / Viskositiet is die interne eienskap van ʼn vloeistof  wat weerstand bied aan vloei.  (2) 5.3.2 Decrease / Verminder  (1) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 9 QUESTION/VRAAG 6 6.1 6.1.1 When light falls on a plane surface it is so reflected that the angle of reflection is equal to the angle of incidence.  The incident ray, the reflected ray and normal ALL lie on the same plane.   Wanneer lig op ʼn platvlak val, is dit so geweerkaats dat die weerkaatsingshoek gelyk is aan die invalshoek.  Die invalstraal, weerkaatste straal en normaalstraal lê almal op dieselfde vlak.  (2) 6.1.2 R – Normal / Normaal  Q – Reflected ray / Weerkaatste straal  (2) 6.2 6.2.1 The angle of incident in the dense medium such that the refracted ray just passes through the surface of separation of the two media.  Die invalshoek in die digter medium sodat die gebreekte straal net deur die oppervlak wat die twee media skei, gaan. OR/OF The angle of incidence in the optically dense medium for which the angle of refraction is 90°.  Die invalshoek in die optiese digter medium waarvan die gebreekte hoek 90° is. (2) 6.2.2 Total internal reflection / Totale interne weerkaatsing  (1) 6.2.3 Medium 1  (1) 6.2.4 i3  (1) 6.3 CRITERIA FOR MARKING/KRITERIA VIR NASIEN Object (20 mm) at 2F/ Voorwerp(20 mm) by 2F  1 Parallel line to principle axis / Parallel lyn na die hoofas  1 Diagonal ray through the optic centre/ diagonaal/skuinstraal deur die optiese middel  1 Image at F2 same size as object./ Beeld by F2 dieselfde groote as voorwerp  1 (4) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 10 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION/VRAAG 7 7.1 A changing magnetic and electric field mutually perpendicular to each other and the direction of propagation of the wave./ ʼn Verandering van magnetiese en elektriese velde onderling loodreg op mekaar en die rigting van die voortplanting van die golf.  (2) 7.2 They have high frequency./ Hulle het hoë frekwensie  (1) 7.3 7.3.1 Ultraviolet rays (UV)/ Ultraviolet-strale  (1) 7.3.2 Gamma rays / Gamma-strale  (1) 7.4 Quantum (packets) of energy./ Kwantum van energie  (1) 7.5 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 c=f𝜆 𝑐 E=h  3,0 x 108  = f(5,10x10-11)  𝜆 f = 5,88 x 1018 Hz 3,0 × 108 = (6,63 x 10-34) x  E = hf  5,10 × 10-11 = 6,63 x 10-34  x 5,88 x 1018 = 3,9 x 10-15 J  = 3,9 x 10-15J  (5) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 11 QUESTION/VRAAG 8 8.1 It is a device for storing electrical charge.  Dit is ʼn toestel wat elektriese lading stoor. (2) 8.2 • Surface area of the plates. / Oppervlakte van die plate.  • Distance between the plates./ Afstand tuseen die plate.  • Type of dielectric material. / Tipe diëlektriek materiaal.  (3) 8.3 Area = 2,0 cm2 (1/100)2 = 2,0 x 10-4 m2 εο Α C= d  (8,85 × 10-12 )((2,0 × 10-4 )✓ = 2,0 × 10-3 ✓ = 8,85×10-13 F Q C= V -12 -13 4,0 × 10 8,85 × 10 =  V V = 4,5 V  (6) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 12 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2024) QUESTION/VRAAG 9 9.1 The rate  at which (electrical) energy is converted (in an electric circuit)./ Die tempo waarteen (elektriese) energie omgeskakel word (in ʼn elektriese stroombaan.) (2) 9.2.1 𝑃 = 𝐼2𝑅  36 = 𝐼2(4)  𝐼=3𝐴 (3) 9.2.2 Positive marking from QUESTION 9.2.1/ Positiewe nasien vanaf VRAAG 9.2.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 V P = VI  V2 R= P=  I  R V 36 = V x 3  4=  3 V = 12 v  V2 36 =  4 V = 12 v  V = 12 v  (3) 9.2.3 Positive marking from QUESTION 9.2.2 / Positiewe nasien vanaf VRAAG 9.2.2 V R= I  12 16  = I  I = 0,75 A  (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 13 QUESTION/VRAAG 10 10.1 The direction of the induced emf in the coil opposes the effect that produces it.  Die rigting van die geïnduseerde emk in die spoel teen die effek wat dit produseer. (2) 10.2 ∅ = BA  = (0,4)(2,29 𝑥 10−3)  = 9,16 x 10−4 Wb  (3) 10.3 Positive marking from QUESTION 10.1/ Positiewe nasien vanaf VRAAG 10.1  0 –9,16 × 10-4 = –(75)  0,05 = 1,37 𝑉  (3) [8] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.