# Physical Sciences Grade 12 Paper 1 NSC June 2017 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the NSC June Exam 2017, 23 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sciences-p1-may-june-2017-memo-afr-eng/ef24ecaa-5af8-4262-bea3-b96eddcae564 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2017 - Exam period: NSC June Exam - Paper: 1 - Curriculum: CAPS - Pages: 23 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 2017 Memo](https://www.hlayiso.com/library/physical-sciences-p1-may-june-2017-memo-afr-eng/ef24ecaa-5af8-4262-bea3-b96eddcae564) ## About this document Physical Sciences · Grade 12 · NSC June Exam · 2017. Memorandum, 23 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 EXAMINATION SENIORSERTIFIKAAT-EKSAMEN PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) 2017 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 23 pages. Hierdie nasienriglyne bestaan uit 23 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/2017 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 1.1 D  (2) 1.2 B  (2) 1.3 D  (2) 1.4 C (2) 1.5 D  (2) 1.6 B  (2) 1.7 C  (2) 1.8 C  (2) 1.9 B  (2) 1.10 B  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1.1 ACCEPT/AANVAAR N N N F f F Fy f ● f ● Fx w 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 /3 N/wrywing/F w  N Normal (force)/F normaL /F N /F normaal /F reaction/reaksie  F F A /F applied/toegepas  Notes/Aantekeninge • Mark awarded for label and arrow/Punt toegeken vir benoeming en pyltjie • Deduct 1 mark if arrow(s) is (are) missing/Trek 1 punt af indien pylpunt(e) nie aangeheg is nie • 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) 2.1.2 fk = μkN  (can use F N for N in equation/ kan F N vir N gebruik in vergelyking) 3 = (0,2)N  N = 15 N  (3) 2.1.3 POSITIVE MARKING FROM QUESTION 2.1.2 POSITIEWE NASIEN VANAF VRAAG 2.1.2 F net = ma N + F vert - w = 0 any of the three/enige van die drie N + F vert = w Fsin20o = (2)(9,8) – 15  F = 13,45 N (4) 2.1.4 POSITIVE MARKING FROM QUESTION 2.1.3 POSITIEWE NASIEN VANAF VRAAG 2.1.3 F net = ma Fcos 20o – f = ma any of the two/enige van die twee 13,45cos20o – 3 = 2a a = 4,82 m.s-2  (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/2017 SCE/SSE – Marking Guidelines/Nasienriglyne 2.2 2.2.1 Any two particles (objects) in the universe will attract each other with a force which is directly proportional to the product of the masses and inversely proportional to the square of the distance between them (their centres).  Enige twee deeltjies (voorwerpe) in die heelal sal mekaar aantrek met 'n krag wat direk eweredig is aan die produk van hul massas en omgekeerd eweredig is aan die kwadraat van die afstand tussen hul (middelpunte). (2) 2.2.2 Increases/Neem toe Gravitational force is inversely proportional to the square of the distance - between the (centres of the) objects Gravitasiekrag is omgekeerd eweredig aan die kwadraat van die afstand tussen die voorwerpe se middelpunte OR/OF 1 F α 2 r (2) [18] QUESTION/VRAAG 3 3.1 The only force acting on the ball is the gravitational force.  Die enigste krag wat op die bal inwerk is die gravitasiekrag. OR/OF The only force acting on the ball is its weight. Die enigste krag wat op die bal inwerk is sy gewig. ACCEPT/AANVAAR The only force acting on the ball is gravity. Die enigste krag wat op die bal inwerk is gravitasie. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne 3.2.1 OPTION 1/OPSIE 1 UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF Δy = v i Δt + ½aΔt2 Δy = v i Δt + ½ aΔt2 = (10)(3) + ½(-9,8)(32)  = (-10)(3) + ½ (9,8)(32)  = -14,10 = 14,10 m  Height of building = 14,10 m Height of building = 14,10 m Hoogte van gebou = 14,10 m Hoogte van gebou = 14,10 m OPTION 2/OPSIE 2 UPWARD AS POSITIVE UPWARD AS NEGATIVE OPWAARTS AS POSITIEF OPWAARTS AS NEGATIEF For maximum height For maximum height Vir maksimum hoogte Vir maksimum hoogte v f = v i + a∆t v f = v i + a∆t 0 = 10 + (-9,8)∆t 0 = -10 + (9,8)∆t ∆t = 1,02 s ∆t = 1,02 s Time taken from point A to ground Time taken from point A to ground Tyd geneem vanaf punt A tot grond Tyd geneem vanaf punt A tot grond = 3 - 2(1,02) = 0,96 s = 3 - 2(1,02) = 0,96 s Δy = v i Δt + ½ aΔt2 Δy = v i Δt + ½ aΔt2 = (-10)(0,96) + ½ (- 9,8)(0,962) = (10)(0,96) + ½ ( 9,8)(0,962)  = -14,1184 = 14,1184 m  Height of building = 14,12 m Height of building = 14,12 m Hoogte van gebou = 14,12 m Hoogte van gebou = 14,12 m OPTION 3/OPSIE 3 UPWARD AS POSITIVE UPWARD AS NEGATIVE OPWAARTS AS POSITIEF OPWAARTS AS NEGATIEF v f = v i + a∆t v f = v i + a∆t = 10 + (-9,8)(3) = -10 + (9,8)(3) = -19,40 = 19,40 v f 2 = v i 2 + 2 a∆y v f 2 = v i 2 + 2 a∆y (-19,4)2 = (10)2 + 2(-9,8)∆y (19,4)2 = (-10)2 + 2(9,8)∆y ∆y =-14,10 m ∆y = 14,10 m  Height of building = 14,10 m Height of building = 14,10 m Hoogte van gebou = 14,10 m Hoogte van gebou = 14,10 m Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 UPWARD AS POSITIVE UPWARD AS NEGATIVE OPWAARTS AS POSITIEF OPWAARTS AS NEGATIEF v f = v i + a∆t v f = v i + a∆t = (-10) + (-9,8)(0,96) = 10 + (9,8)(0,96) = -19,408 = 19,408 v f 2 = v i 2 + 2 a∆y v f 2 = v i 2 + 2 a∆y (-19,408)2 = (10)2 + 2(-9,8)∆y (19,408)2 = (-10)2 + 2(9,8)∆y ∆y =14,12 m ∆y = 14,12 m  Height of building = 14,12 m Height of building = 14,12 m Hoogte van gebou = 14,12 m Hoogte van gebou = 14,12 m OPTION 5/OPSIE 5 W net = ∆Ek = ∆K mg∆xcos0o = ½ m(v f 2 – v i 2) any one/enige een (9,8)∆x = ½ (19,4082 – 102)  ∆x = 14,12 m v f = v i + a∆t Height of building = 14,12 m = 10 + (9,8)(3) Hoogte van gebou = 14,12 m = 19,40 OPTION 6/OPSIE 6 E mechA = E mechB (E k + E p ) A = (E k + E p ) B any one/enige een ½ mv2 + mgh = ½ mv2 + 0 ½(10)2 + (9,8)h = ½(19,40)2  v f = v i + a∆t Height of building = 14,12 m = 10 + (9,8)(3) Hoogte van gebou = 14,12 m = 19,40 OPTION 7/OPSIE 7 W nc = ∆E p + ∆E k  vf = vi + a∆t 0 = mg(h f – h i ) + ½m(v f 2 - v i 2) = 10 + (9,8)(3) 0 = m(9,8)(0 – h i ) + ½m(19,4082 – 102)  = 19,40 Height of building = 14,12 m Hoogte van gebou = 14,12 m OPTION 8/OPSIE 8 UPWARD AS POSITIVE UPWARD AS NEGATIVE OPWAARTS AS POSITIEF OPWAARTS AS NEGATIEF v f = v i + a∆t v f = v i + a∆t = (10) + (-9,8)(3) = 10 + (9,8)(3) = -19,40 m∙s-1 = 19,40 m∙s-1 vi + v f vi + v f Δy = Δt  Δy = Δt  2 2 (10 - 19,40) (-10 + 19,40) Δy = 3 Δy = 3 2 2 Height of building = 14,12 m ∆y = 14,12 m Hoogte van gebou = 14,12 m Height of building = 14,12 m Hoogte van gebou = 14,12 m (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne 3.2.2 OPTION 1/OPSIE 1 UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF v f = v i + aΔt v f = v i + aΔt = (10) + (-9,8)(3)  = (-10) + (9,8)(3)  = -19,40 m∙s-1 = 19,40 m∙s-1  Speed = 19,40 m∙s-1 Speed = 19,40 m∙s-1 Spoed = 19,40 m∙s-1 Spoed = 19,40 m∙s-1 OPTION 2/OPSIE 2 UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF v f 2 = v i 2 + 2aΔx  v f 2 = v i 2 + 2aΔx  = (-10)2 + 2(-9,8)(-14,1) = (10)2 + 2(9,8)(14,1)  v f = 19,4 m∙s-1  v f = 19,4 m∙s-1  POSITIVE MARKING FROM QUESTION 3.2.1 POSITIEWE NASIEN VANAF VRAAG 3.2.1 OPTION 3/OPSIE 3 UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF vi + v f vi + v f Δy = Δt  Δy = Δt  2 2 (10 + v f ) (-10 + v f ) - 14,12 = 3 + 14,12 = 3 2 2 v f = 19,41m∙s-1 v f = 19,41m∙s-1 Speed = 19,41 m∙s-1 Speed = 19,41 m∙s-1 Spoed = 19,41 m∙s-1 Spoed = 19,41 m∙s-1 (3) 3.2.3 OPTION 1/OPSIE 1 UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF v f 2= v i 2 + 2aΔy v f 2= v i 2 + 2aΔy 0 = v i 2+ (2)(-9,8)(8)  0 = v i 2+ (2)(9,8)(-8) v i = 12,52 m∙s-1 v i = -12,52 m∙s-1 Speed = 12,52 m∙s-1 Speed = 12,52 m∙s-1  Spoed = 12,52 m∙s-1 Spoed = 12,52 m∙s-1 OPTION 2/OPSIE 2 UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF Δx = v i Δt + ½aΔt2 Δx = v i Δt + ½aΔt2 -8 = 0 + ½ (-9,8)Δt2 8 = 0 + ½ (9,8)Δt2 Δt = 1,28 s Δt = 1,28 s v f = v i + aΔt  v f = v i + aΔt  0 = v i + (-9,8)(1,28) 0 = v i + (9,8)(1,28) v i = 12,52 m∙s-1 v i = -12,52 m∙s-1 Speed = 12,52 m∙s-1 Speed = 12,52 m∙s-1  Spoed = 12,52 m∙s-1 Spoed = 12,52 m∙s-1 (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/2017 SCE/SSE – Marking Guidelines/Nasienriglyne 3.3 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF 12,52 10 Velocity/Snelheid (m·s-1) CHECKLIST FOR MARKING RIGLYNE VIR NASIEN 0 Each line correctly drawn with  3 3,2 t(s) calculated velocities. 1 x 2 Elke lyn korrek geteken met berekende waardes 1 x 2 Mark for lines being parallel  Twee parallelle lyne korrek geteken -19,41 Times 3 s and 3,2 s correctly shown  Tye 3 s en 3,2 s korrek getoon 3.3 DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF 19,41 CHECKLIST FOR MARKING RIGLYNE VIR NASIEN Velocity/Snelheid (m·s-1) Each line correctly drawn with  calculated velocities. 1 x 2 3,2 Elke lyn korrek geteken met 0 berekende waardes 1 x 2 3 t(s) Mark for lines being parallel  Twee parallelle lyne korrek geteken Times 3 s and 3,2 s correctly shown  -10 Tye 3 s en 3,2 s korrek getoon -12,52 (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 4.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 2/OPSIE 2 Δx Δx Δx v= v= v= Δt Δt Δt 0, 2 0, 4 0, 6 = = = 0,4 0,8 1, 2 -1 -1 = 0,5 m∙s = 0,5 m∙s = 0,5 m∙s-1  Formula/Formule  Correct substitution in all three equations/Korrekte vervanging in al drie vergelykings Correct answer/korrekte antwoord • If only two calculations used with correct answer, award 2/3 Indien slegs twee berekenings met korrekte antwoord gebruik word, ken 2/3 toe • If unit(s) omitted lose 1 mark./Indien eenhede weggelaat is verbeur 1 punt OPTION 2/OPSIE 2 Trolley moves with equal displacements in equal time (intervals). (Full marks or zero) Trollie beweeg met gelyke verplasings in gelyke tydintervalle (Vol punte of nul) (3) 4.2 The total linear momentum of a closed (isolated) system remains constant (is conserved). Die totale lineêre momentum in 'n geslote (geïsoleerde) sisteem bly konstant (is behoue). OR/OF In an isolated system, the total linear momentum before collision is equal to the total linear momentum after collision In 'n geslote (geïsoleerde) sisteem is die totale lineêre momentum voor botsing gelyk aan die totale lineêre momentum na botsing. (2) NOTE:/LET WEL If the words “closed (isolated)” or “total” omitted, award 1/2. Indien die woorde “geslote (geïsoleerde)” of totaal weggelaat is, ken 1/2 toe Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne 4.3 POSITIVE MARKING FROM QUESTION 4.1/POSITIEWE NASIEN VANAF VRAAG 4.1 OPTION 1/OPSIE 1 Σp i = Σp f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Any one/Enige een (3,5)(0,5) = (3,5 + 6)v f  Fnet = ma v f = v 6kg = 0,184 m∙s-1 (v f - v i ) (0,184 -0) m = (6) Δt 0,05 For trolley B:/Vir trollie B: = 2,21 N F net ∆t = ∆p = m∆v F net (0,5) = 6(0,184 -0)  F net = 2,21 N (2,24 N) OPTION 2/OPSIE 2 Σp i = Σp f Any one/Enige een m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (3,5)(0,5)  = (3,5 + 6)v f  v f = v 6kg = 0,184 m∙s-1 For trolley B:/Trollie B: For trolley A:/Trollie A: F net ∆t = ∆p = m∆v F net ∆t = ∆p = m∆v F net (0,5) = 6(0,184 -0)  F net (0,5) = 3,5(0,184 -0,5)  F net = 2,21 N (2,24 N)  F net = - 2,21 N (2,24 N) ∴ Magnitude of the average net force experienced by trolley B = 2,21 N (2,24 N) ∴Grootte van die gemiddelde netto krag deur trollie B ondervind = 2,21 N (2,24 N) (6) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 Σp i = Σp f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Any one/Enige een (3,5)(0,5)  = (3,5 + 6)v f  v f = v 6kg = 0,184 m∙s-1 For trolley B:/Trollie B: For trolley A:/Trollie A: v f = v i + aΔt v f = v i + aΔt 0,184 = 0 +a(0,5) 0,184 = 0,5 +a(0,5) a = 0,368 ms-1 a = - 0,632 ms-1 F net = ma F net = ma = (6)(0,368) = (3,5)(-0,632) = 2,21 N = - 2,21 N ∴ Magnitude of the average net force experienced by trolley B = 2,21 N (2,24 N) ∴Grootte van die gemiddelde netto krag deur trollie B ondervind = 2,21 N (2,24 N) (6) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1 ACCEPT/AANVAAR f N f N ● f N ● W┴ w// 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 /50 N/wrywing/F w  N Normal force/F NORMAL /F NOR /F NORMAAL  Notes/Aantekeninge • Mark awarded for label and arrow/Punt toegeken vir benoeming en pyltjie • Deduct 1 mark if arrow(s) is (are) missing/ Trek 1 punt af indien pylpunt(e) nie aangeheg is nie • 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 nieAny other additional force(s)/Enige ander addisionele krag(te) Max/Maks 2 • If force(s) do not make contact with body/Indien krag(te) nie met die voorwerp kontak maak nie: Max/Maks: 3 4 • Award 1 mark if both resolved components of w are correct/Ken 1 punt toe indien beide komponente van w korrek is. 5.2 The net/total work done on an object equals the change in the object's kinetic energy.  Die netto/totale arbeid op 'n voorwerp verrig is gelyk aan die verandering in die voorwerp se kinetiese energie OR/OF The work done by the net force equals the change in the object's kinetic energy.  Die arbeid verrig deur die netto krag is gelyk aan die voorwerp se verandering in kinetiese energie. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne 5.3 OPTION 1/OPSIE 1 W net = ∆E K f∆xcosθ + F g ∆xcosθ = ½mv f 2 – ½mv i 2 Any one/Enige een (50)(25cos180o) + (60)(9,8) (25cos70o)  = ½(60)(152 – v i 2)  -1 250 + 5 027,696 = 6 750 – 30v i 2 v i = 9,95(4) m.s-1 OPTION 2/OPSIE 2 W net = ∆E K f∆xcosθ + F g|| ∆xcosθ = ½mv f 2 – ½mv i 2 Any one/Enige een (50)(25cos180o ) + (60)(9,8sin20o)(25cos0o)  = ½(60)(152 – v i 2)  -1 250 + 5 027,696 = 6 750 – 30v i 2 v i = 9,95(4) m.s-1 OPTION 3/OPSIE 3 W nc = ∆ E K + ∆ E P f∆xcosθ = ½(mv f 2 - mv i 2) + (mgh Q – mgh P ) Any one/Enige een E mechP + E mechQ + W nc = 0 (50)(25cos180o) = ½(60)(152 – v i 2)  + (60)(9,8)(-25sin 20o)  -1 250 = 6 750 – 30 v i 2 – 5 027,696 v i = 9,95 m.s-1 (5) OPTION 4/OPSIE 4 F net = 201,11 – 50 = 151,11 N W net = ΔE k  F net Δxcosθ = ½(mv f 2 - mv i 2) (151,11)(25)cos0o = ½ (152 – v i 2) v i = 9,95 m∙s-1  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne 5.4 POSITIVE MARKING FROM QUESTION 5.3 POSITIEWE NASIEN VANAF VRAAG 5.3 OPTION 1/OPSIE 1 Pave / gemid = Fv ave / gemid (9,95 + 15)  = 50 2 = 623,75 W  OPTION 2/OPSIE 2 W (v i + v f ) P= ∆x = Δt ∆t 2 Any one/Enige een f Δx cos θ (9,954 + 15) Pf = 25 = Δt  Δt 2 [50 (25 cos 180 o )] ∆ t = 2,004 s = 2,004 = - 623,75 W (- 625 W) F Δx cos θ Pboy/seun =  P boy/seun = 623,75 W ( 625 W)  Δt [50 (25 cos 0 o )]  = 2,004 = 623,75 W NOTE/LET WEL: Candidates can substitute -1 250 from QUESTION 5.3 directly into the equation Kandidate kan – 1 250 vanaf VRAAG 5.3 direk in die vergelyking vervang (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 6.1.1 It is the 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. OR/OF An apparent change in frequency (pitch), (wavelength) as a result of the relative motion between a source and an observer (listener). Die skynbare verandering in frekwensie (toonhoogte) as gevolg van die relatiewe beweging tussen die bron en die waarnemer. (2) 6.1.2 v ± vL v fL = f s OR/OF fL = fs  v ± vs v − vs  (340 + 0) 365 =  x 330  (340 - v s ) v s = 32,60 m·s-1  (5) 6.2 According to the Doppler Effect if the star moves away from the observer a lower frequency/longer wavelength is detected. This lower frequency/ longer wavelength corresponds to the red end of the spectrum Volgens die Doppler-effek as die ster weg vanaf die waarnemer beweeg word 'n laer frekwensie/langer golflengte waargeneem. Hierdie laer frekwensie/ langer golflengte stem ooreen met die rooi ent van die spektrum. (3) [10] QUESTION/VRAAG 7 7.1 The magnitude of the electrostatic force exerted by one point charge (Q 1 ) on another point charge (Q 2 ) 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 punt lading (Q 1 ) op 'n ander puntlading (Q 2 ) is direk eweredig aan die produk van die (groottes van die) ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne 7.2.1 Negative/Negatief (2) 7.2.2 Q1Q3 QQ F= k 2  (Accept/Aanvaar F = k 1 2 2 ) r d (9 × 10 9 )Q1(2 × 10 -6 ) 0,012 =  FQ23 FQ13 (2,5)2 Fnet -6 Q 1 = 4,17 x 10 C  F net = F Q13 + F Q23  (9 × 10 9 )( Q 2 ) (2 × 10 - 6 ) - 0,3 = 0,012 -  12 OR/OF (9 × 10 9 )( Q 2 ) (2 × 10 - 6 ) 0,3 = - 0,012 +  12 Q 2 = 1,73 x10-5 C  Do not penalise for the nature of the charges. Moenie vir die aard van die ladings penaliseer nie. (7) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 8.1 Electric field is a region of space in which an electric charge experiences a force.  Elektriese veld is 'n gebied in die ruimte waarin 'n elektriese lading 'n krag ondervind. (2) 8.2 Marks/ Criteria for sketch/Kriteria vir skets Punte Correct shape as shown.  Korrekte vorm soos getoon. Direction away from positive  Rigting weg van positief . Field lines start on spheres and do not cross for correct diagram.  Veldlyne begin op elke sfeer en kruis nie. (3) 8.3 kQ E PA =  r2 = ( )( 9 ×10 9 5 ×10 - 6 ) (1, 25 )2 = 2,88 x 104 N∙C-1 to the right/na regs kQ E PB = r2 = ( )( 9 ×10 9 5 ×10 - 6 ) (0,75 ) 2 = 8,00 x 104 N∙C-1 to the left/na links E net = E PA + E PB = 2,88 x 104 + (-8,00 x 104) Vector addition/Vektoroptelling = 5,12 x 104 N∙C-1  (5) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION/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 deur die geleier by konstante temperatuur. NOTE/LET WEL If constant temperature is omitted -1 mark Indien konstante temperatuur weggelaat word -1 punt (2) 9.1.2 Graph of potential difference versus current Grafiek van potensiaalverskil teenoor stroom ● Potential difference/Potensiaalverskil (V) 5,0 ● ● 4,0 ● 3,0 ● 2,0 ● 1,0 ● 0 0 1,0 2,0 3,0 4,0 5,0 Current/Stroom(A) Straight line passing through 4 or five points. Straight line with intercepts on both axes. Reguitlyn deur 4 of 5 punte Reguitlyn met afsnitte op beide asse (2) 9.1.3 5,5 V (accept any value from /aanvaar enige waarde vanaf 5,4 V to/tot 5,6 V based on graph drawn/op die grafiek gebaseer.) NOTE /LET WEL : The value must be the y-intercept./ Die waarde moet die y-afsnit wees. (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne 9.1.4 ΔV y -y Slope/helling = or/of 2 1 ΔI x 2 - x1 5,5 - 0  = = - 1,2 0 - 4,6  Internal resistance/Interne weerstand (r) = 1,2 Ω NOTE/LET WEL: Any correct pair of coordinates chosen from the line drawn Enige korrekte paar koördinate vanaf die getekende lyne gekies For the equation ε = I(R + r) or ε = V ext + Ir marks are awarded only if the correct I and V values are used from the graph Vir die vergelyking ε = I(R + r) of ε = V ext + Ir sal punte slegs toegeken word indien die korrekte waardes van I and V vanaf die grafiek gebruik is. (3) 9.2.1 V R= I V any of these/enige van hierdie I= R V = IR 21,84 = I tot (8)  I tot = 2,73 A (3) 9.2.2 OPTION 1/OPSIE 1 R 30 × R 20 1 1 1 R|| = = + R 30 + R 20 R // R 30 R 20 1 1 1 = +  30 × 20 R|| 30 20 R|| =  50 = 12 Ω R || = 12 Ω (2) 9.2.3 POSITIVE MARKING FROM QUESTION 9.2.1 AND 9.2.2 POSITIEWE NASIEN VANAF VRAAG 9.2.1 EN 9.2.2 OPTION 1/OPSIE 1 R tot = (8 + 12 + r) (for the addition/vir optelling) = (20 + r) ℇ = I(R + r)  60 = (2,73)(20 + r)  ∴ r = 1,98 Ω Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 20 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 20 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne POSITIVE MARKING FROM QUESTION 9.2.1 POSITIEWE NASIEN VANAF VRAAG 9.2.1 OPTION 2/OPSIE 2 V || = I tot R || = (2,73)(12)  = 32,76 V ∴V terminal = (32,76 + 21,84)  for addition/vir optelling = 54,6 V “V lost ” = 60 - 54,6 = 5,4 V ℇ = Vlost + V|| + V8 60 = (V”lost” + 32,76 + 21,84)  V R= V”lost” = 5,4 V I V any of these/enige van hierdie I= R V = IR 5,4 = 2,73 r r = 1,98 Ω NOTE/LET WEL: No penalisation for omitted subscripts Geen penalisering vir weggelate onderskrifte nie (4) 9.2.4 POSITIVE MARKING FROM 9.2 1 AND 9.2.2 POSITIEWE NASIEN VANAF VRAAG 9.2.1 EN 9.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 2 V 2 W = I R∆t W = VI∆t W= Δt  = (2,73)2 (20)(0,2)  = (54,6)(2,73)(0,2)  R = 29,81 J = 29,81 J  W= (54,6 ) 2 (0,2)  20 = 29,81 J (3) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 21 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 21 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne QUESTION/VRAAG10 10.1.1 R: armature/Coil(s)/Spoel T: (Carbon/Koolstof) brushes/borsels X: Slip rings/sleepringe NOTE/LETWEL: Answers must be in that order if R, T and X are omitted. Antwoorde moet in hierdie orde wees as R, T en X weggelaat (3) 10.1.2 Faraday's Law/Faraday se wet  (1) 10.2.1 15 V (1) 10.2.2 OPTION 1/OPSIE 1 V rms = I rms R 15 Irms =  45 = 0,333 A any one/enige een I max I rms = 2 I max =(0,333) 2  = 0,47 A OPTION 2/OPSIE 2 V Vrms = max 2 V max = (15) 2  any one/enige een = 21,213V V max = I max R 21, 213  I max = 45 = 0,47 A OPTION 3/OPSIE 3 I max= I rms 2 Vrms = 2 any one/enige een R V 2 rms = R 15 = 2  45 = 0,47 A  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 22 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 22 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 V2 15 2 Pave = rms = = 5W R 45 P ave = V rms Irms any one/enige een 5 = (15)Irms  I rms = 0,33 A I Irms = max 2 I 0,33 = max  2 I max = 0,47 A  (4) [9] QUESTION/VRAAG 11 11.1 It is the process whereby electrons are ejected from a (metal) surface when light of suitable frequency is incident on that surface.  Dit is die proses waarby elektrone vanaf die oppervlak van 'n metaal vrygestel word wanneer lig van geskikte frekwensie daarop skyn. (2) 11.2 INCREASE /TOENEEM  Increase in intensity means that (for the same frequency) the number of - photons incident per unit time increase.  Therefore the number of electrons ejected per unit time increases.  Thus current increases. Toename in intensiteit beteken dat (vir dieselfde frekwensie) die aantal fotone per eenheid tyd neem toe. Dus sal die aantal elektrone per eenheid tyd vrygestel, toeneem. (3) 11.3 OPTION 1/OPSIE 1 E = W 0 + E k(max) hf = hf 0 + E k(max) Any one/Enige een hf = hf 0 + ½ mv2 E = W 0 + ½ mv2 -34 14 (6,63 x10 -34 )(3 ×10 8 ) (6,63 x 10 x 5,9 x 10 )  = + 2,9 x 10-19  λ0 19,89 × 10-26 39,117 x 10-20 – 2,9 x 10-19 = λo λ o = 1,97 x 10-6 m Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 23 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 23 DBE/2017 SCE/SSE – Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2 E = W o + E k(max) hf = hf o + E k(max) Any one/Enige een hf = hf o + ½ mv2 E = W o + ½ mv2 (6,63 x 10-34 x 5,9 x 1014)  = (6,63 x 10-34)f 0 + 2,9 x 10-19  f o = 1, 52 x 1014 Hz c = f oλo 3 x 108 = (1,52 x 1014) λ 0  λ o = 1,97 x 10-6 m (5) OPTION 3/OPSIE 3 E = W o + E k(max) hf = hf o + E k(max) Any one/Enige een hf = hf o + ½ mv2 E = W o + ½ mv2 (6,63 x 10-34 x 5,9 x 1014)  = W 0 + 2,9 x 10-19  W o = 1,01 x 10-19 J W o = hf o 1,01 x 10-19 = (6,63 x 10-34)f o f o = 1, 52 x 1014 Hz c = f oλo 3 x 108 = (1,52 x 1014) λ 0  λ o = 1,97 x 10-6 m 11.4 From the photo-electric equation, for a constant work function,  the maximum kinetic energy of the photoelectrons is proportional to the energy of the photons. Vanaf die foto-elektriese vergelyking is die maksimum kinetiese energie van die foto-elektrone eweredig aan die energie van die fotone vir 'n konstante arbeidsfunksie. 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