# Physical Sciences Grade 12 P1 Limpopo November 2014 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Limpopo November 2014, 10 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/ldoe-physical-science-september-2014-grade-12-paper-1-final-memo/63828ecc-9e16-47b6-8e7b-5ce3b24288fa - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2014 - Exam period: Limpopo November - Paper: 1 - Curriculum: CAPS - Pages: 10 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 P1 Limpopo November 2014 Memo](https://www.hlayiso.com/library/ldoe-physical-science-september-2014-grade-12-paper-1-final-memo/63828ecc-9e16-47b6-8e7b-5ce3b24288fa) ## About this document Physical Sciences · Grade 12 · Limpopo November · 2014. Memorandum, 10 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September2014 NSC NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (Vr1) PREPARATORY EXAMINATIONS SEPTEMBER 2014 MEMORANDUM This memorandum consists of 10 pages. / Hierdie memorandum bestaan uit 10 bladsye Copyright reserved Please turn over ### Page 2 Q 11 1.2 1.3 14 15 1.6 17 1.8 1.9 11 0 C¥v Downloaded from hlayiso.com UESTIOW/ Vraag 1 BYv (2) Avy (2) Avv (2) Cvy (2) BYv (2) Cvy (2) AvY (2) Avv (2) Avv (2) (2) [20] QUESTIOW/ Vraag 2 2.1 2.2 2.3 When a net force is applied to an object, the object accelerates 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. VW // Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, versnel die voorwerp in die rigting van die krag teen 'n versnelling direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp. (2) 2N Vv v 3kg --——> 20N Ty 3) 2.3.1 Consider the 1 kg block: 1N —— tho x+——>T Frot=ma=T+f % la=T-1Vv Tsatto oe equation/vergelyking 1 Consider the 3 kg block: FR= =mMa= Fapplied + T + Fy 8a=20-T-2 Yo Lee equation/ vergelyking 2 Substituting for T: 3a =20-(a+1)-2% 4a=17 a=4,25 ms“, (5) OR/ OF Fret = ma ¥ (F-f)=(M+m)a v v v ### Page 3 Downloaded from hlayiso.com 20 – (2 + 1) = 4a a = 17/4 = 4,25 ms–2 (5) 2.3.2 T = a + 1 = 4,25 + 1 = 5,25 N (2) [12] QUESTION/ VRAAG 3 3.1 8m  (1) 3.2 0,8 (s)  (1) 3.3 (Take downward motion as NEGATIVE. (Other option: take downwards as positive)) v f = v i + a ∆t  0 = vi + (–9,8) (0,4) vi = 3,92 m.s–1 , up   + marking (3) 3.3 3.4 + marking OPTION/OPSIE 1 3.3 OPTION/OPSIE 2  Any one ∆y = v i ∆t + 21 a∆t 2  ∆ U + ∆K = 0 2 2  = (3,92) (0,4) + ½ (–9,8) (0,4)2 ½ mvi + mghi = ½ mvf + mghf   = 0,784 m  The maximum height = 8 + 0,78 ½xm(3,92)2 + mx9,8x8 = 0 + mx9,8xh = 8,78 m  h = 8,78 m  (½xm(3,92)2 + 0 = 0 + mx9,8xh h = 0,78 h tot = 8,78 m ) + marking OPTION 3 3.3 OPTION 4 + marking 3.3 v f = v i + 2a∆y   v + vi   2 2 ∆y =  f  ∆t (0)2 = (3,92)2 + 2 (–9,8) ∆y   2  ∆y = 0,784 m   = ½ (3,92 + 0) (0,4) The maximum height = 8 + 0,78 = 0,78 m  = 8,78 m The maximum height = 8 + 0,78 = 8,78 m (4) 3.5 from maximum height downwards/ vanaf maksimum hoogte afwaarts 2 2 v f = v i + 2a∆y   + marking = (0)2 + 2 (–9,8) (–8,78) vf = 13,12 m.s–1 from 3.4 ### Page 4 Downloaded from hlayiso.com OR / of from the balcony upwards/ vanaf balkon opwaarts v f = v i + 2a∆y  2 2  = (0)2 + 2 (–9,8) (–8) vf = 13,12 m.s–1  (3) 3.6  Initial velocity/aanvanklike 3,92  snelheid  final velocity/finale snelheid  Shape/vorm v  (m.s–1) 0,4 t (s) –13,12  (3) [15] QUESTION/ VRAAG 4 4.1 N  FApplied/man 100 N   f  Weight/ w/FG/gewig (4) OR/OF N   FApplied/man 100 N  Ff  Weight/ w/F /gewig ### Page 5 Downloaded from hlayiso.com  4.2 OPTION 1 Work Done by Gravity = F ∆x Cosθ  90+20 0 W G = (10 x 9,8) (5) Cos 110 = –167,59 J  OPTION 2    Work done = ∆Ep =mgh-0 = mg ∆x Sin 200 = 10 x 9.8 x 5 x Sin 200 = 167,59 J (3) 4.3.1 OPTION 1 Σ Fy = 0: N + FG ┴ = 0  FN – mg Cos 200 = 0 FN = (10) (9,8) Cos 200  180 = 92,1 N   fk = µ FN = (0,4) (92,1) = 36,8 N ∴ Work Done by friction = fk ∆x Cos θ W = (36,8) (5) Cos 1800  = – 184 J  OPTION 2 Fk = µkN  = 0,4 mg Cos θ  = 0,4 x 10 x 9,8 Cos 200  = 36,8  ∴ Work Done by friction = fk ∆x Cos θ W = (36,8) (5) Cos 1800  = – 184 J  (6) 4..3.2 WF = F ∆x Cosθ =(100) (5) Cos00 = 500 N  (1) 4.5 Work Energy theorem states that, the net/total workdone on an object is equal to the change in the object's kinetic energy OR the work done on an object by a resultant/net force is equal to the change in the object's kinetic energy.  (2) + marking 4.2, 4.6 WNET = ∆ EK  4.3.1, 4.3.2, 4.4    ### Page 6 Downloaded from hlayiso.com W G + Wf + W N + WF = ½ mvf2 – ½ mvi2 (–168) + (–184) + (0) + (500) = ½ (10) vf2 – ½ (10) ((1,5)2 vf = 5,64 m.s–1 (4) [20] QUESTION/ VRAAG 5 5.1 Inelastic Energy lost due to sound and heat or  The two locomotives move together Onelasties Lokomotiewe beweeg saam na die botsing (2) 5.2 (take right a positive) Σp before = Σp after  [accept: m1vi + m2v2i = m1v1f + m2v2f /pbefore = pafter] (6000)(4)  + (5000) (vB)  = (6000 + 5000) (3)  24000 + 5000 vB = 33000 vB = 1,8 m.s–1 to the right (6) Other option: take right as negative [8] QUESTION/ VRAAG 6 6.1 Blood flow meter / Doppler flow meter  / bloedvloeimeter/Dopplervloeimeter (1) 6.2 6.2.1 Higher pitch /hoër toonhoogte (1) 6.2.2 Lower pitch  / laer toonhoogte (1) v ± vL v + vL 6.3 Use either fL = fs  or fL = fS  v ± vs v  330 + 30 fL= x 1800  330 - 0 fL = 1 963,6 Hz  (5) 6.4  f labels   (3) t [11] QUESTION/ VRAAG 7 7.1  force experienced per unit Electric field at a point is the (electrostatic)  positive charge placed at that point. // Die elektriese veld by 'n punt is die elektrostatiese krag wat per eenheidspositiewe-lading wat by daardie punt geplaas is ondervind word (2) Curved electric field 7.2 line important. on outside Geboë elektriese veld en lyne belangrik op A B buitekant ### Page 7 Downloaded from hlayiso.com 1 mark: shape of field between and outside 1 mark direction (2) kQ 7.3 E1 = 2  r = 9 x 109 (15 x 10–9)  (2 x 10–2)2  = 337 500 N.C–1 to the left/ links kQ E2 = r2 = 9 x 109 (Qx)  (8 x 10–2)2 = 1,40625 x 1012 x Qx to the left / links ENET = E1 + E2 3,943 x 105 = 337 500 + 1,40625 x1012 QX  QX = +4,04 x 10–8 C  (7) 7.4 decrease /  afneem (1) [12] QUESTION/VRAAG 8 V2 8.1 P =  R 13,5 = (18)2  R R = 24 Ω  OR P = VI   IR = P/V = 13,5/18 = 0,75 A V = IR R = V/IR = 18/0,75 = 24 Ω  (3) ### Page 8 8.2 8.3 8.4 8.5 Downloaded from hlayiso.com Vv R= T v 24 = 18 y | =0,75A R =v I, 12 =18 y Iho lo =1,5A% v total = 1,5 + 0,75 = 2,25 A= loo (5) L L i =_ = s-i-v OR Rp a 1 i + i, Rp i2° 24 Rp = 80% V =IRp 18 =1(8Y 1=2,25A ¥ (5) v Internal resistance is the opposition to the follow of charge within a cell (or an ammeter) ¥/ Interne weerstand is die teenstand teen die vloei van stroom binne-in ‘n sel (of ‘n ammeter) (2) When the switch is closed, the pd across the 10 Q resistor is Vio=IRiow =2,25x10v = 22,5 Vv (3) The pd across the external resistors is given by: Vi = Vet = Vp + Vio = 18+ 22,5=405V ¥ When the switch is open the total pd (emf) is: Emf = Viotal = 45,9 V «. The ‘lost volts’ is: Vi = Vrota— Ver = 45,9- 40,5 =5,4 V ¥ a v I = 54 2,25 =240 % (5) ### Page 9 Downloaded from hlayiso.com OR VTOT = E = ITOT (Rext + Rint)  45,9  = (2,25) (RP + 10 + r) = (2,25) (8 + 10 + r)  RP = 8Ω  r = 2,4 Ω 8.6 Increase / Toeneem (1) [19] QUESTION/ VRAAG 9 9.1 Ohm's Law states that: the potential difference across a conductor is  directly proportional  to the current in the conductor at constant temperature./ Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur (2) 9.3 Inverse of resistance or 1/R  / omgekeerde van weerstand (1) 9.2 Y  Y is greater The inverse of the gradient of graph  Gradient of graph Y smaller. Thus 1/R smaller. Hence, R is greater. Y; Y se gradient kleiner dus is die omgekeerde van R kleiner, dus R groter (2) 9.4 RX can be found by finding the gradient of the graph for X gradient = ∆ I ∆V m = 0,4 – 0,2  (or other correct values from graph) 8–4  (of ander korrekte waardes vanaf grafiek) = 0,05 Rx = 1 0,05  = 20 Ω  (4) [9] QUESTION/VRAAG 10 10.1 Mechanical energy is converted to electrical energy. / meganiese energie na elektriese energie (1) 10.2 10.2.1 AC generator / WS generator (1) 10.2.2 3  (slip rings/ sleepringe) (1) 10.2.3 anti-clockwise / anti-kloksgewys (1)  10.2.4 The (rate of) change in magnetic flux/ magnetic field linkage is at a maximum  // die tempo van verandering van magnetise fluks is maksimum . (2) ### Page 10 Downloaded from hlayiso.com Vmax 330 10.3 Vrms = =  2 2 = 233,36 V Prms = VrmsIrms  = 233,36 x 15  = 3500,25 W  (5) [11] QUESTION/VRAAG 11 11.1 Cathode  / katode (1) 11.2 Threshold frequency  /dumpelfrekwensie (1)  11.3 Work function is the minimum quantity of energy which is required to remove an electron from the surface of a given solid, usually a metal.  die minimum energie benodig om 'n elektron uit die oppervlak van 'n metaal vry te stel. (2)  11.4 E= W 0 = h f0  = (6,63 x 10–34) (5 x 1014) = 3,315 X 10–19 J  (3) 1  11.5 hf = W0 + mv 2  2   (6,63 x 10 ) (f1) = 3,315 x 10–19 + 11 x 10–19 –34 f1 = 2,15 x 1015 Hz  (5) 11.6 same / dieselfde (1) [13] Grand total: 150 ## Related documents - [PHYS SC P1 MEMO ENGAFR](https://www.hlayiso.com/library/phys-sc-p1-memo-engafr/da9b9ae0-718d-442a-95bc-aa0bfbf308f3) — Physical Sciences · Memo · 2014 - [2014AFR gau physical sciences p1 memo](https://www.hlayiso.com/library/2014afr-gau-physical-sciences-p1-memo/db77042f-0bdc-4e06-8f6b-ce760a7e4148) — Physical Sciences · Memo · 2014 - [2014ENG gau physical sciences p1 memo](https://www.hlayiso.com/library/2014eng-gau-physical-sciences-p1-memo/d8eca56a-f776-4202-84cd-3e2e7e299ae3) — Physical Sciences · Memo · 2014 - [Physical Sciences P1 Nov 2014 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2014-memo-afr-eng/44924678-c33b-42e1-8937-fcfda16ea7b3) — Physical Sciences · Memo · 2014 · Eng - [wkp kwod physc p1 sept 2014 eng memo](https://www.hlayiso.com/library/wkp-kwod-physc-p1-sept-2014-eng-memo/bb06a52a-4a49-454d-ae6e-5128be28cc24) — Physical Sciences · Memo · 2014 · Eng - [ldoe physc p2 sept 2014 p2 english](https://www.hlayiso.com/library/ldoe-physc-p2-sept-2014-p2-english/152643de-d636-4e6d-9fe5-b0a286e6a46c) — Physical Sciences · QP · 2014 · Eng ## More from Grade 12 Physical Sciences - [PHYSICAL SCIENCE P1 QP JCT 2025 AFR D watermark](https://www.hlayiso.com/library/physical-science-p1-qp-jct-2025-afr-d-watermark/b9429ac5-acbb-4fae-989d-af9e1478a96b) — QP · 2025 · Afr - [PHYSICAL SC P2 QP GR12 SEPTEMBER 2025 AFR1 watermark](https://www.hlayiso.com/library/physical-sc-p2-qp-gr12-september-2025-afr1-watermark/37cff2db-7c1a-4ec8-8ac5-07ec5bcf87a9) — QP · 2025 - [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 KZN](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-kzn/6d225dc8-24c6-4c28-a181-df906d5a2c0e) — QP · 2025 - [Physical Sciences Physics Grade 12 PS01 QP March 2025 Limpopo](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-ps01-qp-march-2025-limpopo/482ee9af-e88a-4818-9120-11ab9e045a1c) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-limpopo/b65e9832-a6be-4312-8f5a-57425d647121) — QP · 2025 ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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