# Physical Sciences Grade 12 Paper 1 NSC June 2018 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the NSC June Exam 2018, 20 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sciences-p1-may-june-2018-memo-afr-eng/99d93638-0b94-4505-be49-9023e77ec651 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2018 - Exam period: NSC June Exam - Paper: 1 - Curriculum: CAPS - Pages: 20 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 Paper 1 NSC June 2018 Memo](https://www.hlayiso.com/library/physical-sciences-p1-may-june-2018-memo-afr-eng/99d93638-0b94-4505-be49-9023e77ec651) ## About this document Physical Sciences · Grade 12 · NSC June Exam · 2018. Memorandum, 20 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 SENIOR CERTIFICATE EXAMINATIONS SENIORSERTIFIKAAT-EKSAMEN PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) 2018 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 20 pages. Hierdie nasienriglyne bestaan uit 20 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/2018 SCE/SSE – Memorandum QUESTION 1/VRAAG 1 1.1 C (2) 1.2 A (2) 1.3 D (2) 1.4 A (2) 1.5 C (2) 1.6 A (2) 1.7 B (2) 1.8 B (2) 1.9 D (2) 1.10 A (2) [20] Copyright reserved/Kopiereg voorbehou ### Page 3 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 The rate of change of velocity.  Die tempo van verandering van snelheid. Accept/Aanvaar Change in velocity per unit time Verandering in snelheid per eenheidstyd (2) NOTE:/LET WEL If any one of the underlined key words in the correct context is omitted deduct 1 mark. Indien enige van die onderstreepte woorde in die korrekte konteks uitgelaat is trek 1 punt af. 2.2 OPTION 1/OPSIE 1 Δy = v i Δt + ½ aΔt2 0,5 = (0)(3) + ½ (a)(32)  a = 0,11 m·s-2 (3) OPTION 2/OPSIE 2  v + vf   Δy =  i  Δt  2  1 mark for either of the two/  0 + vf  1 punt vir enige van die twee 0,5 =  (3)  2  v f = 0,333 m∙s-1 v f = v i + aΔt 0,33 = 0 + a(3) a = 0,11 m∙s-2 (3) OPTION 3/OPSIE 3 v f = v i + aΔt = 0 + 3a 1 mark for either of the two/ v f = = 3a…………..(i) 1 punt vir enige van die twee v f 2 = v i 2 + 2aΔy = 02 + 2a(0,5) v f = a ………….(ii) 9a2 = a  ∴a = 0,11 m·s-2 (3) 2.3 POSITIVE MARKING FROM 2.2 / POSITIEWE NASIEN VANAF 2.2 OPTION 1/OPSIE 1 For the 3 kg mass/vir die 3 kg massa: F net = ma (mg – T)/(mg + T) = ma  (3)(9,8) – T = (3)(0,11)  T = 29,07 N  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2 For the 3 kg mass/vir die 3 kg massa:  v + vf  Δy =  i  Δt  2   0 + vf  0,5 =  (3)  2  v f = 0,333 m∙s-1 OR/OF POSITIVE MARKING FROM 2.2/POSITIEWE NASIEN VANAF 2.2 v f = v i + aΔt = 0 + (0,11)(3) = 0,33 m∙s-1 W net = ∆E K  W w + W T = ½ mv f 2 - ½ mv i 2 mg∆xcosθ + T∆xcosθ = ½ mv f 2 - ½ mv i 2 (3)(9,8)(0,5)cos0o + T(0,5)cos180o = ½ (3)(0,332 – 02) 14,7 – 0,5T = -0,16 T = 29,72 N  (3) OPTION 3/OPSIE 3 W nc = ΔE k + ΔE p  TΔxcos1800 = ½ mv f 2 – ½ mv i 2 + mgh f - mgh i T(0,5)cos180o = ½ (3)(0)2 – ½ (3)(0,33)2 + (3)(9,8)(0) – (3)(9,8)(0,5)  T = 29,72 N (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne 2.4 N ACCEPT/AANVAAR N f T f T w w Accepted labels/Aanvaarde benoemings F g / F w / weight / mg / gravitational force w F g / F w / gewig / mg / gravitasiekrag  f Friction/ F f / f k / 27 N/ wrywing / F w  Normal (force) / F normaL / F N / F normaal N /F reaction / F reasie  T F T /tension/spanning  Notes/Aantekeninge • Mark awarded for label and arrow / Punt toegeken vir benoeming en pyltjie • Do not penalise for length of arrows since drawing is not to scale. /Moenie vir die lengte van die pyltjies penaliseer nie aangesien die tekening nie volgens skaal is nie • Any other additional force(s) / Enige ander addisionele krag(te) Max/Maks 3 4 • If force(s) do not make contact with body / Indien krag(te) nie met die voorwerp kontak maak nie: Max/Maks: 3 4 (4) 2.5 POSITIVE MARKING FROM 2.2 AND 2.3/POSITIEWE NASIEN VANAF 2.2 EN 2.3 OR/OF For/ Vir P For/ Vir P F net = ma F net = ma   T- f = ma T- f = ma 29,07 - 27 = m(0,11)  29,72 - 27 = m(0,11)  m = 18,82 kg  (Range: 18,60 – 18,82) m = 24,73 kg  (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 Motion under the influence of gravity/weight/gravitational force only.  Beweging slegs onder die invloed van gravitasie/gewig/gravitasiekrag. OR/OF Motion in which the only force considered is gravitational.  Beweging waar die enigste krag wat in ag geneem word, gravitasie is. (2) NOTE:/LET WEL If ONLY is omitted minus 1 mark. Indien SLEGS uitgelaat is, trek 1 punt af. 3.2 OPTION 1/OPSIE 1 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF Δy = v i Δt + ½ aΔt2 = (0)(1) + ½ (-9,8)(12)  = - 4,9 m Height/hoogte = 2Δy = (2)(4,9) = 9,8 m DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF Δy = v i Δt + ½ aΔt2 = (0)(1) + ½ (9,8)(12)  = 4,9 m Height/hoogte = 2Δy = (2)(4,9) = 9,8 m OPTION 2/OPSIE 2 UPWARD POSITIVE/OPWAARTS DOWNWARD POSITIVE/AFWAARTS POSITIEF POSITIEF v f = v i + aΔt v f = v i + aΔt = 0 + (-9,8)(1) = 0 + (9,8)(1) = -9,8 m·s-1 = 9,8 m·s-1  v + vf   v + vf  Δx =  i  Δt  Δx =  i  Δt   2   2   0 + (-9,8)   0 + (9,8)   =  (1)  =  (1)  2   2  = - 4,9 m = 4,9 m Height/hoogte = (2)(4,9) Height/hoogte = (2)(4,9) = 9,8 m = 9,8 m Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 UPWARD POSITIVE/OPWAARTS DOWNWARD POSITIVE/AFWAARTS POSITIEF POSITIEF v f = v i + aΔt v f = v i + aΔt = 0 + (-9,8)(1) = 0 + (9,8)(1) = -9,8 m·s-1 = 9,8 m·s-1 v f 2 = v i 2 + 2aΔy v f 2 = v i 2 + 2aΔy (-9,8)2 = 0 + (2)(-9,8)Δy  (9,8)2 = 0 + (2)(9,8)Δy  Δy = - 4,9 m Δy = 4,9 m Height/hoogte = 2Δy Height/hoogte = 2Δy = (2)(4,9) = (2)(4,9) Height/hoogte = 9,8 m Height/hoogte = 9,8 m OPTION 4/OPSIE 4 UPWARD POSITIVE/OPWAARTS DOWNWARD POSITIVE/AFWAARTS POSITIEF POSITIEF v f = v i + aΔt v f = v i + aΔt = 0 + -9,8(1) = 0 + 9,8(1) v f = -9,8 m∙s-1 = 9,8 m∙s-1 E (mech/meg)Top/Bo = E (mech/meg at/by B)  (E p +E k ) Top = (E p +E k ) at B (mgh + ½ mv2) Top/Bo = (mgh + ½ mv2) at/by B (9,8)(h) + 0 = (9,8) ½ (h)+ (½ )(9,82)  h = 9,8 m  OPTION 5/OPSIE 5 UPWARD POSITIVE/OPWAARTS DOWNWARD POSITIVE/AFWAARTS POSITIEF POSITIEF v f = v i + aΔt v f = v i + aΔt = 0 + -9,8(1) = 0 + 9,8(1) -1 v f = -9,8 m∙s = 9,8 m∙s-1 W net = ∆K mgΔxcosθ = ½ mv f 2 - ½ mv i 2 (9,8) ½h cos0o = ½ (9,82 -0)  h = 9,8 m OPTION 6/OPSIE 6 UPWARD POSITIVE/OPWAARTS DOWNWARD POSITIVE/AFWAARTS POSITIEF POSITIEF v f = v i + aΔt v f = v i + aΔt = 0 + -9,8(1) = 0 + 9,8(1) v f = -9,8 m∙s-1 v f = 9,8 m∙s-1 W nc = ∆K + ∆U 0 = (½mv f 2 - ½ mv i 2) + (mg½h - mgh) (9,8)( ½h) = (½)(9,82 -0)  h = 9,8 m (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne 3.3 OPTION 1/OPSIE 1 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF v f 2 = v i 2 + 2aΔy = 0 + (2)(-9,8)(-9,8)  v f = 13,86 m∙s-1  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF v f 2 = v i 2 + 2aΔy = 0 + (2)(9,8)(9,8)  v f = 13,86 m∙s-1  OR/OF FROM POINT B/ VANAF PUNT B UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF v f 2 = v i 2 + 2aΔy = (-9,8)2 + (2)(-9,8)(-4,9)  v f = -13,86 m∙s-1 Magnitude = 13,86m·s-1 DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF v f 2 = v i 2 + 2aΔy = (9,8)2 + (2)(9,8)(4,9)  v f = 13,86 m∙s-1  (3) OPTION 2/OPSIE 2 Δy = v i Δt + ½ aΔt2 -9,8 = 0 + ½ (-9,8)Δt2 1 mark for either of the two/ Δt = 1,41 s 1 punt vir enige van die twee v f = v i +aΔt = 0 + (-9,8)(1,41)  = -13,82 m·s-1 Magnitude = 13,82 m·s-1  (3) 3.4 OPTION 1/OPSIE 1 POSITIVE MARKING FROM 3.3/POSITIEWE NASIEN VANAF 3.3 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF v f 2 = v i 2 + 2aΔy  0 = v i 2 + (2)(-9,8)(4,9)  v i = 9,8 m∙s-1 1 mark for any F net ∆t = m∆v  1 punt vir enige een F net ∆t = m (v f – v i ) F net (0,2)  = 0,4[9,8 –(-13,86)]  F net = 47,32 N Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF v f 2 = v i 2 + 2aΔy 0 = v i 2 + (2)(9,8)(-4,9)  v i = -9,8 m∙s-1 -1 If calculation of 9,8 m∙s is not shown 1 mark for any and it is substituted correctly award 1 F net ∆t = m∆v  1 punt vir enige een -1 mark/Indien berekening van 9,8 m∙s nie F net ∆t = m (v f – v i ) getoon is nie en dit is korrek vervang F net (0,2) = 0,4[-9,8 - (13,86)] ken 1 punt toe F net = -47,32 N F net = 47,32 N OPTION 2/OPSIE 2 POSITIVE MARKING FROM 3.3/ POSITIEWE NASIEN VANAF 3.3 E mech top/bo = E (mech ground/meg grond) 1 mark for any (E p +E k ) top = (E p +E k ) bottom/onder  1 punt vir enige een (mgh + ½ mv2) top/bo = (mgh + ½ mv2) bottom/onder (0,4) (9,8) (4,9) + 0 = ½ (0,4)v f 2  v i = 9,8 m∙s-1 -1 If calculation of 9,8 m∙s is not shown and it is substituted correctly award 1 F net ∆t = m∆v 1 mark for any -1 mark/Indien berekening van 9,8 m∙s nie F net ∆t = m(v f – v i )  1 punt vir enige een getoon is nie en dit is korrek vervang ken 1 punt toe F net (0,2)  = 0,4[9,8 - (-13,86)] F net = 47,32 N (6) OPTION 3/OPSIE 3 POSITIVE MARKING FROM 3.3/ POSITIEWE NASIEN VANAF 3.3 v f = v i + aΔt  9,8 = -13,86 + a(0,2)  a = 118,3 m·s-2 F net = ma  = (0,4)(118,3)  = 47,32 N  (6) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 E (mech top/meg bo) = E (mech bottom/meg onder) (E p + E k ) top/bo = (E p + E k ) bottom/onder 1 mark for any (mgh + ½ mv2) top/bo = (mgh + ½ mv2) bottom/onder 1 punt vir enigeen (1,5)(9,8)(2) + 0 = 0 + ½ (1,5)v2 v = 6,26 m∙s-1 (4) 4.2 In a/an closed/isolated system, the total linear momentum is conserved. /In 'n geïsoleerde/geslote sisteem bly die totale liniêre momentum behoue. (2) 4.3 POSITIVE MARKING FROM 4.1/POSITIEWE NASIEN VANAF 4.1 Σp i = Σp f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 1 mark for any m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v 1 punt vir enige een (1,5)(6,26) + 0 = (1,5 + 2)v f  v f = 2,68 m∙s-1 (4) 4.4 POSITIVE MARKING FROM 4.3/POSITIEWE NASIEN VANAF 4.3 OPTION 1/OPSIE 1 Δx = vΔt = (2,68)(3)  = 8,04 m  OPTION 2/OPSIE 2  v + vf  Δx =  i  Δt   2   (2,68 + 2,68)  =  (3)   2  = 8,04 m OPTION 3/OPSIE 3 Δx = v i Δt + ½ aΔt2 = (2,68)(3) + ½ (0)(3) 2 = 8,04 m  (Range 8,04 – 8,05) (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 T T • w w Accepted labels/Aanvaarde benoemings F g /F w /weight/mg/gravitational force/N/19,6 N w F g /F w /gewig/mg/gravitasiekrag/19,6 N T Tension/F T / F A / Spanning Notes/Aantekeninge: • Mark awarded for label and arrow/Punt toegeken vir benoeming en pyltjie • Do not penalise for length of arrows since drawing is not to scale./Moenie vir die lengte van die pyltjies penaliseer nie aangesien die tekening nie volgens skaal is nie • Any other additional force(s)/Enige ander addisionele krag(te) Max/Maks 1 2 • If force(s) do not make contact with body/Indien krag(te) nie met die voorwerp kontak maak nie: Max/Maks: 1 2 (2) 5.2 Tension/Spanning Accept/Aanvaar F applied /F toegepas (1) 5.3 W = F∆xcosθ W w = mg∆xcosθ  1 mark for any of these/1 punt vir enige van hierdie = 75(9,8) (12)cos 180o = - 8 820 J OR/OF W w = -ΔE p  = − (mgh − 0) = -(75)(9,8)(12)  = -8 820 J (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne 5.4 The work done on an object by a net force is equal to the change in the object’s kinetic energy.  Die arbeid verrig op 'n voorwerp deur 'n netto krag is gelyk aan die verandering in die voorwerp se kinetiese energie. OR/OF The net work done on an object is equal to the change in the object’s kinetic energy Die netto arbeid verrig op 'n voorwerp is gelyk aan die verandering in die voorwerp se kinetiese energie. (2) NOTE:/LET WEL If any one of the underlined key words in the correct context is omitted deduct 1 mark. Indien enige van die onderstreepte woorde in die korrekte konteks uitgelaat is trek 1 punt af. 5.5 POSITIVE MARKING FROM 5.3/POSITIEWE NASIEN VANAF 5.3 OPTION 1/OPSIE 1 W net = ∆K 1 mark for any of these 2 2  1 punt vir enige van hierdie Fnet∆xcosθ = (½ mvf - ½ mvi ) (75)(0,65)(12) cos0o  = ½(75)( v f 2 - 0)  v f = 3,95 m∙s-1 (3,949 m∙s-1)  POSITIVE MARKING FROM 5.2/ POSITIEWE NASIEN VANAF 5.2 OPTION 2/OPSIE 2 Wnet = ∆K Wnc/nk = ∆K + ∆U  1 mark for any of these/1 punt vir enige van hierdie WT + W g = ∆K Wnc/nk = (½ mvf2 - ½ mvi2 )+ (mghf - mghi) T – mg = ma 9405 = (½ (75)vf2 -0) + (75)(9,8)(12 -0) T – 75(9,8) = 75(0,65) vf = 3,95 m⋅s-1 T = 783,75 N WT = 783,75 (12) cos0o = 9405 J 9405 – (8820) = ½ (75)(vf2 – 0) vf = 3,95 m∙s-1 (3,949 m∙s-1) (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 It is the (apparent) change in frequency (or pitch) of the sound (detected by a listener) because the sound source and the listener have different velocities relative to the medium of sound propagation.  Dit is die verandering in frekwensie (of toonhoogte) van die klank (waargeneem deur 'n luisteraar) omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium van klankvoortplanting het. OR/OF An (apparent) change in(observed/detected) frequency (pitch),(wavelength)  as a result of the relative motion between a source and an observer  (listener). 'n Skynbare verandering in (waargenome) frekwensie (toonhoogte),(golflengte) as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer / luisteraar. (2) NOTE:/LET WEL If any one of the underlined key words in the correct context is omitted deduct 1 mark. Indien enige van die onderstreepte woorde in die korrekte konteks uitgelaat is trek 1 punt af. 6.2 OPTION1/OPSIE1 v ± vL v fL = fs  OR/OF fL = fs v ± vs v − vs  340  (5100) = fs 340 - 240  f s = 1 500 Hz v = fλ  340 = (1 500)λ  λ = 0,23 m  (7) OPTION 2/OPSIE 2 v ± vL v  v  fL = fs  OR/OF fL =   v ± vs v - v s  λ s     340   340   (5100) =      340 − 240   λ S   λ= 0,23 m  (7) 6.3 Greater than /Groter as (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 The two forces must be equal in magnitude  but in opposite directions /Die twee kragte moet gelyke groottes hê maar in teenoorgestelde rigtings werk. OR/OF The force experienced by Q due to P, must be equal in magnitude but opposite in direction to the force experienced by Q due to V./Die krag wat Q agv P ondervind moet gelyk in grootte maar teenoorgesteld in rigting wees met die krag wat Q agv V ondervind (2) 7.2 The magnitude of the electrostatic force exerted by one point charge on another point charge is directly proportional to the product of the (magnitudes of the) charges and inversely proportional to the square of the distance (r) between them.  Die grootte van die elektrostatiese krag uitgeoefen deur een puntlading op 'n ander puntlading is direk eweredig aan die produk van die (groottes van die) ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle. OR/OF The force of attraction or repulsion between two point charges is directly proportional to the product of the charges  and inversely proportional to the square of the distance between them.  Die aantrekkings- of afstotingskrag tussen twee puntladings is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle (2) NOTE:/LET WEL If mass is mentioned instead of charges. 0 2 Indien massa in plaas van ladings genoem word. 0 2 NOTE:/LET WEL If any one of the underlined key words in the correct context is omitted deduct 1 mark. Indien enige van die onderstreepte woorde in die korrekte konteks uitgelaat is trek 1 punt af. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne 7.3 OPTION 1/OPSIE 1 QQ F = k 12 2  r (9 × 10 9 )(Q)(5 × 10 -6 ) F PQ =  ( x )2 45×10 3 Q = x2 (9 × 10 9 )(Q)(7 × 10 -6 ) F VQ =  (1- x )2 63 × 103 Q = (1 - x )2 (F net = F PQ - F VQ = 0) 45×10 3 Q 63 × 10 3 Q =  (equating two equations) x2 (1- x )2 6,708(1 - x) = 7,937x x = 0,458 m x is 0,46 m (away from P/weg van P) (5) OPTION 2/OPSIE 2 kQ p E= 2  r kQ p kQ v E net = 2 - r r2 (E net = E p – E v = 0) (9 × 10 9 )(5 × 10 −6 ) (9 × 10 9 )(7 × 10 −6 ) 0=  -  x2 (1 − x ) 2 (9 × 10 9 )(5 × 10 −6 ) (9 × 10 9 )(7 × 10 −6 ) =  (equating two equations) x2 (1 − x ) 2 ∴x = 0,46 m (away from P/weg van P) (5) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 (2) Marks/ Criteria for sketch/Kriteria vir skets Punte Lines are directed away from the charge /  Lyne is weg vanaf die lading Lines are radial, start on sphere and do not cross./Lyne is radiaal, begin op die sfeer en  kruis nie 8.2 Q = ne = (8 x 1013)(-1,6 x10-19)  or/of (8 x 1013)(1,6 x 10-19) = -12,8 x 10-6 Net charge on the sphere = (+ 6 x 10-6 ) + (-12,8 x 10-6)  Q net = - 6,8 x 10-6 C kQ E= 2  r E= (9 ×109 )(6,8 ×10 -6 )  (0,5 )2 = 2,45 x 105 N⋅C-1 towards sphere /na die sfeer (7) [9] QUESTION 9/VRAAG 9 9.1.1 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 konstate temperatuur. OR/OF The ratio of potential difference across a conductor to the current in the conductor is constant, provided the temperature remains constant. Die verhouding van potensiaalverskil oor die geleier tot stroom in die geleier is konstant indien die temperatuur konstant bly. (2) NOTE:/LET WEL If any one of the underlined key words in the correct context is omitted deduct 1 mark. Indien enige van die onderstreepte woorde in die korrekte konteks uitgelaat is trek 1 punt af. 9.1.2 V 1 = IR  = (0,6)(4)  = 2,4 V  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne 9.1.3 POSITIVE MARKING FROM 9.1.2/ OR/OF OR/OF POSITIEWE NASIEN VANAF 9.1.2 V 6 V 4Ω = V 6Ω I 6Ω = (I) = 0,6  I 4Ω R 1 = I 6Ω R 2 R 10 2,4  I =1A (0,6)(4) = I 6Ω (6)  = I = 0,4 A I 6Ω = 0,4 A  6 6Ω = 0,4 A (2) 9.1.4 POSITIVE MARKING FROM 9.1.3 /POSITIEWE NASIEN VANAF 9.1.3 V 2 = IR = (0,4 + 0,6)(5,8)  = 5,8 V  (2) 9.1.5 POSITIVE MARKING FROM 9.1.4 AND 9.1.2/POSITIEWE NASIEN VANAF 9.1.4 EN 9.1.2 OPTION 1/ OPSIE 1 V ext = (5,8 + 2,4)  = 8,2 V V int = Ir = (1) (0,8)  = 0,8 V Emf = 0,8 + 8,2 = 9 V OPTION 2/OPSIE 2 1 1 1 R1R 2 = + R// = Rp R1 R2 R1 + R 2 1 1 (6)(4) = = 6 + 4 (6 + 4) 5 = 2,4 Ω = 12 R p = 2,4 Ω R ext = (2,4 + 5,8)  = 8,2 Ω Emf = I(R + r) = 1 (8,2 + 0,8)  =9V (3) 9.1.6 POSITIVE MARKING FROM 9.1.5 AND 9.1.3/ POSITIEWE NASIEN VANAF 9.1.5 EN 9.1.3 W = V I ∆t  W = I2R ∆t  V2 Δ t  = (0,8)(1) (15)  2 = (1) (0,8)(15)  W= R = 12 J  = 12 J 0,8 2 (15 )  = 0,8 = 12 J  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne 9.2.1 V R= I 2,8 =  0,7 =4Ω (2) 9.2.2 Increases /Neem toe Total resistance decreases,  current/power increases, motor turns faster/ Totale weerstand neem af, stroom/drywing neem toe, motor draai vinniger (3) [20] QUESTION 10/VRAAG 10 10.1 10.1.1 Split ring / commutator /Spleetring/ kommutator (1) 10.1.2 Anticlockwise /Antikloksgewys (2) 10.1.3 Electrical energy to mechanical(kinetic) energy  Elektriese energie na meganiese (kinetiese) energie (2) 10.2 10.2.1 DC generator: split ring/commutator and AC generator has slip rings GS-generator spleetringe/kommutator en WS-generator sleepringe OR/OF AC generator: slip ring and DC generator has split rings WS-generator sleepringe en GS-generator spleetringe (1) 10.2.2 Vmax Vrms = 2 320  = 2 = 226,27 V  (3) 10.2.3 OPTION 2/OPSIE 12 V Imax = max R 320 =  35 = 9,14 A I Irms = max  2 9,14 =  2 = 6,46 A Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/2018 SCE/SSE – Marking Guidelines/Nasienriglyne POSITIVE MARKING FROM 10.2.2/ POSITIEWE NASIEN VANAF 10.2.2.2. OPTION 2/OPSIE 2 2 Vrms Paverage = R 226,27 2 =  35 = 1462,80 W P average = V rms I rms  P average = I2 rms R 1 462,80 = (226,27) I rms  1 462,80 = I2 rms (35)  I rms = 6,46 A  I rms = 6,46 A  (4) POSITIVE MARKING FROM 10.2.2/ POSITIEWE NASIEN VANAF 10.2.2.2 OPTION 1/OPSIE 1 V 226,27  Irms = rms  = = 6,46 A  R 35  (4) [13] QUESTION 11/VRAAG 11 11.1 Work function (of a metal) is the minimum energy needed to eject an electron from the metal/surface Werksfunksie (van ‘n metaal) is die minimum energie benodig om 'n elektron vanaf die metaal/oppervlak vry te stel. (2) NOTE:/LET WEL If any one of the underlined key words in the correct context is omitted deduct 1 mark. Indien enige van die onderstreepte woorde in die korrekte konteks uitgelaat is trek 1 punt af. 11.2 (Maximum) kinetic energy of the ejected electrons (Maksimum) kinetiese energie van die vrygestelde elektrone. (1) 11.3 Wavelength/Frequency (of light)  Golflengte/frekwensie (van lig) (1) 11.4 Silver/Silwer - According to Photoelectric equation, hf = W o + ½ mv2 (For a given constant frequency), as the work function increases the kinetic energy decreases.  Silver has the smallest kinetic energy and hence the highest work function./ Volgens die fotoelektriese vergelyking hf = W o + ½ mv2: (Vir 'n gegewe konstante frekwensie) as die werksfunksie verhoog, verlaag die kinetiese energie. Silwer het die kleinste kinetiese energie en dus die hoogste werksfunksie. (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 20 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 20 DBE/2018 SCE/SSE – Memorandum hf = W o + ½mv2 max/maks 11.5  c h = W o + E k(max/maks) λ (6,63 × 10 -34 )(3 × 10 8 )  = W o + 9,58 × 10 -18  2 × 10 - 8 9,945 x10-18 = W o + 9,58 x 10-18 W o = 3,65 x 10-19 J (4) 11.6 REMAINS THE SAME/BLY DIESELFDE - Increasing intensity increases number of photons(per unit time) but frequency stays constant the energy of the photon is the same therefore the kinetic energy does not change. Verhoging van die intensiteit verhoog die aantal fotone(per eenheidstyd) maar die frekwensie bly konstant Foton se energie is dieselfde, daarom verander die kinetiese energie nie. 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