Technical Sciences Grade 12 P1 Eastern Cape Mock 2025 Memo

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2025 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.

2 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) QUESTION/VRAAG 1 1.1 C ✓✓ (2) 1.2 C ✓✓ (2) 1.3 C ✓✓ (2) 1.4 A ✓✓ (2) 1.5 A ✓✓ (2) 1.6 B ✓✓ (2) 1.7 D ✓✓ (2) 1.8 B ✓✓ (2) 1.9 D ✓✓ (2) 1.10 A ✓✓ (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 3 QUESTION/VRAAG 2 2.1 OPTION1/ OPSIE 1 OPTION 2/ OPSIE 2 FN✓ FN✓ Fy FA✓ ✓ Ff✓ Ff ✓ Fx Fg✓ Fg✓ (4) NOTES/NOTAS: Acceptable labels/ Allocate mark for arrow and label. / Aanvaabare byskrifte: Gee punt vir pyltjie en byskrif. Normal force/Normale Krag: N/FN/ Penalise once if/Penaliseer slegs: Fnormal No arrows/ Geen pyltjies Applied force/ Toegepasde krag: F/FA/ Gaps between the line and the arrow./ FNtombi/150N Spasies tussen die lyn en die pyltjie Gravitational force(weight)/Gravitasie Dotted lines are used./ Krag(gewig): Fg/w/Fgravity Stippellyne is gebruik Vertical component of FNtombi/ Vertikale Additional force is included./ komponent van FNtombi : Fv/Fy Additionele kragte is ingesluit Horizontal component of FNtombi/ Vertikale component van FNtombi :FH/Fx Frictional force/ Wrywings krag: f/Ff 2.2 APPLY NEGATIVE MARKING / NEGATIEWE NASIEN WORD TOEGEPAS Newton’s First law.  An object continues in a state of rest or uniform (moving with constant) velocity  unless it is acted upon by a net (resultant) force.✓ Newton se Eerste Wet ʼn Liggaam sal in sy toestand van beweging of in rus of beweging teen uniforme (konstante) snelheid volhard, totdat ʼn netto (ongebalanseerde of resulterende krag) daarop inwerk. (3) 2.3 It will remain at rest because the net force acting on it is zero.  OR It will remain at rest as long as the horizontal component (FX) of the pulling force is equal to the static friction. Dit sal in rus bly omdat die nettokrag wat daarop inwerk nul is. OF Dit sal in rus bly solank die horisontale komponent (Fx) van die trekkrag gelyk is aan die statiese wrywing. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

4 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) 2.4 2.4.1 Tension is a (pulling) force acting in a string or rope.  Force applied by Ntombi (FNtombi)/F150/150 N   Spanning is ʼn (trek) krag wat in ʼn string of tou inwerk. Krag toegepas deur Ntombi (FNtombi)/ F150 / 150 N (3) 2.4.2 Decreases/Verminder  (2) 2.4.3 Ff ✓ Fx/Ntombi ✓ FA/Mncedisi ✓ (3) NOTES/ NOTAS: Acceptable labels/ Allocate mark for arrow and label. / Aanvaarbare byskrifte: Een punt vir pyltjie en byskrif Applied force /Toegepasde Penalise once if/ Penaliseer slegs vir: krag:Fx /FA/Ntombi/FA/Mncedisi No arrows / Geen pyltjies Horizontal component of Gaps between the line and the arrow. / FNtombi/ Spasies tussen die lyn en pyltjie. Horisontale component van Dotted lines are used. / FNtombi: FH/Fx Stippellyne word gebruik. Frictional force / Additional force is included. / Wrywingskrag: f/Ff Additionele krag ingesluit. Vertical force included./ Vertikale krag ingesluit. 2.4.4 East to be positive/ Oos as positief Fnet = Fx + FMncedisi + fk  Fnet = Fx/Ntombi Cos 60° + FMncedisi + fk 105 = 150  Cos 60° + 80  + fk fk = -50 N fk = μk N fk = μk (Fg – FNtombi Sin 60°)  fk = μk (mg – FNtombi Sin 60°) 50 = μk (200 x 9,8 – 150 sin 60°)  50 = μk (1960 – 129,904) μk = 0,027/0,03 (6) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 5 QUESTION/VRAAG 3 3.1 Momentum is the product of an object’s mass and its velocity. / Momentum is die produk van ʼn voorwerp se massa en sy snelheid. (2) 3.2 𝑣𝑖 𝑚𝑖𝑛𝑖𝑏𝑢𝑠 = 120 × 1000  3600 = 33,33 m.s-1, east/oos  (2) 3.3 p = mcar vi car  = 1 100 x 16,67  = 18 337 kg.m.s-1, west/wes  (3) 3.4 3.4.1 An isolated system is one on which the net external force acting on the system is zero.  ʼn Geïsoleerde stelsel is een waarop die netto eksterne krag wat op die stelsel inwerk nul is. (2) 3.4.2 POSITIVE MARKING FROM 3.2/ POSITIEWE NASIEN VAN VRAAG 3.2 1 1 Σ𝐸𝑘𝑖 = 2 𝑚𝑚𝑖𝑛𝑖𝑏𝑢𝑠 𝑣𝑖2 𝑚𝑖𝑛𝑖𝑏𝑢𝑠 + 2 𝑚𝑐𝑎𝑟/motor 𝑣𝑖2 𝑐𝑎𝑟/𝑚𝑜𝑡𝑜𝑟   1 1 = 2 (1500)(33,33)2 + 2 (1100)(−16,67)2   = 985 255,65J 1 1 ΣE𝑘𝑓 = 2 𝑚𝑚𝑖𝑛𝑖𝑏𝑢𝑠 𝑣𝑓2 + 2 𝑚𝑐𝑎𝑟/𝑚𝑜𝑡𝑜𝑟 𝑣𝑓2  𝑚𝑖𝑛𝑖𝑏𝑢𝑠 𝑐𝑎𝑟/ 𝑚𝑜𝑡𝑜𝑟 1 1 = 2 (1500)(20,3)2 + 2 (1100)(5,32)2   = 324 633,82 J ∑Eki ≠ ∑Ekf /(Kinetic energy is not conserved/Kinetiese energie is nie behoue)  Therefore, the collision is inelastic. / Daarom is die botsing onelasties  NOTE: if ∑Eki = ∑Ekf is used (4/5 max) NOTAS: as ∑Eki = ∑Ekf gebruik word (4/5 mak) (5) 3.5 3.5.1 Inversely proportional./Omgekeerd eweredig   OR 1 Fnet = ∆𝑡 OR When the contact time increases/decrease, the net force decreases/ increase. / Wanneer die kontaktyd toeneem/afneem, verminder/vermeerder die netto krag. NOTE: Give full mark for mathematical expression LET WEL: Gee volle punt vir wiskundige uitdrukking (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

6 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) 3.5.2 Equal to/ Gelyk aan  (1) 3.5.3 • Impulse remains constant.  • Airbags increase the contact time during the crash.  • The longer the contact time, the smaller the force  exerted by the occupants on the vehicle and the lesser is the extent of injuries.  /  • Impuls bly konstant. • Lugsakke verhoog die kontaktyd tydens die ongeluk.  • Hoe langer die kontaktyd, hoe kleiner die krag  wat die insittendes op die voertuig uitoefen en die omvang van beserings is minder.  (3) 3.5.4 OPTION 1/ OPSIE 1 OPTION 2/ OPSIE 2 Direction towards the tree be Direction towards the tree positive./ Rigting na die boom positief be negative. / Rigting na die wees. boom negatief wees. Fnet∆t = ∆p Fnet∆t = ∆p  Fnet∆t = m(vf – vi) Fnet∆t = m(vf – vi) - 57 500∆t  = 1 150(0 – 15)  57 500∆t = 1 150{0 – (-15)}  ∆t = 0,30 s  ∆t = 0,30 s  (4) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 7 QUESTION/VRAAG 4 4.1 4.1.1 The sum of gravitational potential energy and kinetic energy of an isolated system remains constant.  OR The total mechanical energy in an isolated system remains constant. Die som van gravitasie potensiële energie en kinetiese energie van ʼn geïsoleerde sisteem bly konstant. OF Die totale meganiese energie in ʼn geïsoleerde stelsel bly konstant. (2) 4.1.2 ME = EP+ EK  = mgh + ½mv2 = (0,55)(9,8)( 8) + 0  = 43,12 J  (3) 4.1.3 POSITIVE MARKING FROM 4.1.2 / POSITIEWE NASIEN VAN 4.1.2  ME (3,5m) = ME (8m)  mgh + ½mv2 = ME (8m) (0,55)(9,8)(3,5)  + ½(0,55)v2  = 43,12  v = 9,39 m∙s-1  (5) 4.1.4 43,12 J  Mechanical energy remains constant,  and Ep is zero.  Meganiese energie bly konstant, en Ep is nul. (3) 4.2 4.2.1 Fg = mg = (200)(9,8)  = 1 960 N  (2) 4.2.2 POSITIVE MARKING FROM 4.2.1/POSITIEWE NASIEN VAN 4.2.1 W = F∆y cosθ  W = F∆x cosθ (Accept/ Aanvaar) = (1 960)  (3) cos0°  = 5 880 J  (4) 4.2.3 POSITIVE MARKING FROM 4.2.1 OR 4.2.2 / POSITIEWE NASIEN VAN 4.2.1 OF 4.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 ∆𝑥 v = ∆𝑡 Pave = Fvave  3 = 1960 x 4  4 = ∆𝑡 = 7840 W  ∆𝑡 = 0,75s W P=  t 5880 = 0,75  = 7840 W  (3) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

8 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) QUESTION/VRAAG 5 5.1 A force which changes the shape and size of a body.  ʼn Krag wat die vorm en grootte van ʼn liggaam verander. (2) 5.2 F =  A (10)(9,8)  =  𝜋(0.001)2 = 31194368,85 Pa (ACCEPT/ANVAAR: 3,12 x 107Pa) (4) 5.3 Δ ε=  L 3 × 10−4 =  1.8 -4 = 1.67 × 10  (3) 5.4 A body that regains its original shape and size when the deforming force is removed.  Any correct examples such as steel, springs, etc.  ʼn Liggaam wat sy oorspronklike vorm en grootte herwin wanneer die vervormende krag verwyder word. Enige korrekte voorbeelde soos staal, vere, ens. (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 9 QUESTION/VRAAG 6 6.1 6.1.1 Refraction/Breking  (1) 6.1.2 The bending of light when it passes from one medium to another.  Die buiging van lig wanneer dit van een medium na ʼn ander medium beweeg. (2) 6.2 6.2.1 Behind the object/Agter die voorwerp  ACCEPT: Between the lens and F. AANVAAR: Tussen die lens en F. (1) 6.2.2 Virtual (Cannot be projected on the screen) Upright (Same orientation as object) Any 3 Magnified (Enlarged/Larger than the object.) Behind the object (On the same side as the object) Virtueel (Kan nie op die skerm geprojekteer word nie) Regop (Dieselfde oriëntasie as voorwerp) Vergroot (Vergroot/Groter as die voorwerp.) Enige 3 Agter die voorwerp (Aan dieselfde kant as die voorwerp) (3) 6.2.3 Telescope/Teleskoop  Binoculars/Verkyker  ACCEPT: Any correct answers / AANVAAR: Enige korrekte antwoorde (2) 6.3 6.3.1 At infinity (ACCEPT: No image formed) By oneindigheid (AANVAAR: Geen beeld word gevorm) (1) 6.3.2 This is because the object is placed exactly at the focal point, where the lens cannot converge the light rays to form a real image.  OR The line passing through the (optical) centre and the line passing through the focal point are parallel/never meet. Dit is omdat die voorwerp presies by die fokuspunt geplaas is, waar die lens nie die ligstrale kan konvergeer om ʼn werklike beeld te vorm nie. OF Die lyn wat deur die (optiese) middelpunt gaan en die lyn wat deur die fokuspunt gaan, is parallel/ontmoet mekaar nooit nie. (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

10 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) QUESTION/VRAAG 7 7.1 7.1.1 An electromagnetic wave as a changing magnetic and electric field mutually perpendicular to each other and the direction of propagation of the wave.  ʼn Elektromagnetiese golf as ʼn veranderende magnetiese en elektriese veld onderling loodreg op mekaar en die voortplantingsrigting van die golf. (2) 7.1.2 𝐶 E = h.𝜆  3𝑥108 1,989 x 10-17  = 6,63 x 10-34. 𝜆  𝜆 = 0,00000001 𝑚/1,0 × 10−8 𝑚  (4) 7.1.3 Any 2  • Radiotherapy/Treatment of cancer.  • Radiographs/Internal images of objects/people.  (E.g. to examine broken bones.) • Used in airport and other security systems Enige 2 • Radioterapie/Behandeling van kanker. • Radiografieë/Interne beelde van voorwerpe/mense. (Byvoorbeeld om gebreekte bene te ondersoek.) • Word in lughawe- en ander sekuriteitstelsels gebruik. (2)  [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 11 QUESTION/VRAAG 8 8.1 8.1.1 Capacitance is the amount of charge it can store per volt.  Kapasitansie is die hoeveelheid lading wat dit per volt kan stoor. (2) 8.1.2 Farad  (1) 8.1.3 (3) 8.2 Any 3  • Decrease the distance between the plates. • Increase the area of the plates. • Use a dielectric material with a higher dielectric constant. • Increase number of plates. Enige 3 • Verminder die afstand tussen die plate. • Vergroot die oppervlakte van die plate. • Gebruik ʼn diëlektriese materiaal met ʼn hoër diëlektriese konstante. • Vermeerder die aantal plate (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

12 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2025) QUESTION/VRAAG 9 9.1 The potential difference across a conductor is directly proportional to the current in the conductor  at constant temperature.  Die potensiaalverskil oor ʼn geleier is direk eweredig aan die stroom in die geleier  by konstante temperatuur.  (2) 9.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 1 1 1 𝑅1 × 𝑅2 = + 𝑅𝑃 = 𝑅𝑃 𝑅1 𝑅2 𝑅1 + 𝑅2 1 1 1 = +  55 × 80 𝑅𝑃 55 80 = 55 + 80 𝑅𝑃 = 32,5926 + 20 ∴ 𝑅𝑇 = 52,5926 𝛺 𝑅𝑃 = 32,5926 + 20 ∴ 𝑅𝑇 = 52,5926 𝛺 𝑉 𝑅=𝐼  𝑉 𝑅=𝐼 12 𝐼 = 52,5926  12 𝐼 = 52,5926  𝐼 = 0,23𝐴  𝐼 = 0,23𝐴  (4) 9.3 𝑊 = 𝐼 2 𝑅∆𝑡  = (0.23)2 × 20 × 120  = 126,96𝐽  (4) 9.4 Cost = energy used x tariff/Koste = energie gebruik x tarief  7,5 = 1,5 × ( 60 × 30) × 30  = R168,75  (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2025) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 13 QUESTION/VRAAG 10 10.1 Faraday’s law of electromagnetic induction. /  Faraday se wet van elektromagnetiese induksie. (1) 10.2 Faraday’s law states that when the magnetic flux linked with the coil changes, an emf is induced in the coil.  The magnitude of the induced emf is directly proportional to the rate of change of the magnetic flux.  Faraday se wet staaf wanneer die magnetiesevloed wat met die spoel verbind is, verander, ʼn emk in die spoel geïnduseer word. Die grootte van die geïnduseerde emk is direk eweredig aan die tempo van verandering van die magnetiesevloed. (2) 10.3 ∆∅ 𝜀 = −𝑁 ∆𝑡  ∆∅ 20 = −11 0,5  ∆∅ = −0,909𝑊𝑏  (3) [6] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief