# Physical Sciences Grade 12 Paper 1 NSC November 2015 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the NSC November Exam 2015, 28 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sciences-p1-nov-2015-memo-afr-eng/140f25b7-f3c1-406a-a99d-6549fa7ea9e0 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2015 - Exam period: NSC November Exam - Paper: 1 - Curriculum: CAPS - Pages: 28 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 November 2015 Memo](https://www.hlayiso.com/library/physical-sciences-p1-nov-2015-memo-afr-eng/140f25b7-f3c1-406a-a99d-6549fa7ea9e0) ## About this document Physical Sciences · Grade 12 · NSC November Exam · 2015. Memorandum, 28 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 NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2015 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 28 pages. Hierdie memorandum bestaan uit 28 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/November 2015 NSC/NSS – Memorandum QUESTION 1/VRAAG 1 1.1 B (2) 1.2 D (2) 1.3 C (2) 1.4 D (2) 1.5 A (2) 1.6 A (2) 1.7 A (2) 1.8 D (2) 1.9 C (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/November 2015 NSC/NSS – Memorandum QUESTION 2/VRAAG 2 2.1.1 When body A exerts a force on body B, body B exerts a force of equal magnitude in the opposite direction on body A.  Wanneer liggaam A 'n krag uitoefen op liggaam B, oefen liggaam B 'n krag van gelyke grootte in die teenoorgestelde rigting op liggaam A uit. OR/OF If body A exerts a force on body B, then body B exerts an equal  and opposite force on body A Indien liggaam A 'n krag uitoefen op liggaam B, dan sal liggaam B 'n gelyke maar teenoorgestelde krag op liggaam A uitoefen (2) 2.1.2 For 2,5 kg block/Vir 2,5 kg blok OR/OF OR/OF T = mg F net = ma F net = ma ∴T = (2,5)(9,8) T – mg = (2,5)(0)  mg - T = (2,5)(0)  T – (2,5)(9,8) = 0 (2,5)(9,8) - T = 0 = 24,5 N T = 24,5 N  T = 24,5 N  (3) 2.1.3 POSITIVE MARKING FROM 2.1.2 POSITIEWE NASIEN VANAF 2.1.2 For mass M/Vir mass M OR/OF f s = μ s N μ s N = μ s Mg 24 , 5  24,5= (0,2)M(9,8)  ∴N= = 122,5 N M = 12,5 kg 0, 2  N = Mg = 122,5 N M(9,8) = 122,5 N M = 12,5 kg (5) 2.1.4 For the 5 kg block/Vir die 5 kg blok: fk = μkN f k = (0,15)(5)(9,8)  = 7,35 N F net = ma T – f k = ma  T – 7,35 = 5a For the 2,5 kg block/Vir die 2,5 kg blok w – T = ma (2,5)(9,8) – T = 2,5 a  17,15 = 7,5 a a = 2,29 m∙s-2 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 2015 NSC/NSS – Memorandum 2.2 m1m 2 F=G  r2 (6,67 × 10 -11 )(6,5 x 10 20 )(90)  F= (550 × 10 3 )2  = 12,90 N (12,899 N) (4) OR/OF Gm g= 2  r (6,67 × 10 -11 )(6,5 × 10 20 )  g= (550 × 10 3 )2 = 0,143...m∙s-2 w = mg = (90)(0,143..) = 12,89 N(downwards/afwaarts) (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 2015 NSC/NSS – Memorandum QUESTION 3/VRAAG 3 3.1. OPTION 1/OPSIE 1 Upwards positive/Opwaarts Downwards positive/Afwaarts positief: positief: v f = v i + a∆t v f = v i + a∆t -16 = 16 – 9,8(∆t)  16 = -16 +9,8(∆t)  ∆t = 3,27s ∆t = 3,27s (4) OPTION 2/OPSIE 2 Upwards positive/Opwaarts Downwards positive/Afwaarts positief: positief: v f = v i + a∆t v f = v i + a∆t To the top/By bopunt: To the top/By bopunt: 0 = 16 – 9,8(∆t)  0 = -16 +9,8(∆t)  ∆t = 1,63s ∆t = 1,63s Total time/Totale tyd = 1,63 x 2 Total time/Totale tyd = 1,63 x 2 = 3,26(7) s = 3,26(7) s (4) 3.1 OPTION 3/OPSIE 3 Upwards positive/Opwaarts Downwards positive/Afwaarts positief: positief: ∆y = v i ∆t + ½ a∆t2 ∆y = v i ∆t + ½ a∆t2 0 = 16∆t + ½ (-9,8) ∆t2 0 = -16∆t + ½ (9,8) ∆t2 ∆t(16 - 4,9∆t) = 0 ∆t(-16 +4,9∆t) = 0 ∆t = 0 or/of 3,27 s ∆t = 0 or/of 3,27 s Time taken/Tyd geneem = 3,27 s Time taken/Tyd geneem = 3,27 s (accept/aanvaar 3,26 s)  (accept/aanvaar 3,26 s)  (4) OPTION 4/OPSIE 4 Upwards positive/Opwaarts Downwards positive/Afwaarts positief: positief: v f 2 = v i2 + 2aΔy v f 2 = v i2 + 2aΔy At highest point/By hoogste punt At highest point/By hoogste punt 0 = 162 + 2(-9,8)∆y 0 = (-16)2 + 2(9,8)∆y ∆y = 13,06 m ∆y = 13,06 m ∆y = v i ∆t + ½a∆t2 ∆y = v i ∆t + ½a∆t2 13,06 = 16∆t – 4,9∆t2 13,06 = -16∆t + 4,9∆t2 ∆t =1,62 or 1,65 ∆t =1,62 or 1,65 Total time/Totale tyd = (1,62/1,65)x2 Total time/Totale tyd = (1,62/1,65) x 2 = 3,24 sor/of 3,3 s = 3,24 sor/of 3,3 s (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 2015 NSC/NSS – Memorandum OPTION 5/OPSIE 5 Upwards positive/Opwaarts Downwards positive/Afwaarts positief: positief: v f 2 = v i2 + 2aΔy v f 2 = v i2 + 2aΔy At highest point/By hoogste punt At highest point/By hoogste punt 0 = 162 + 2(-9,8)∆y 0 = (-16)2 + 2(9,8)∆y ∆y = 13,06 m ∆y = -13,06 m  v + vi   v + vi  Δy =  f ∆t  Δy =  f ∆t   2   2   0 + 16   0 − 16  13,06 =  ∆t  - 13,06 =  ∆t   2   2  Δt = 1,63 s Δt = 1,63 s Total time/totale tyd = 3,26 s Total time/totale tyd = 3,26 s (4) 3.1 OPTION 6 /OPSIE 6 Upwards positive/Opwaarts Downwards positive/Afwaarts positief: positief: F net ∆t = ∆p F net ∆t = ∆p mg ∆t = m (v f -v i ) mg ∆t = m (v f -v i ) -9,8∆t = (0 - 16)  9,8∆t = {0 – (-16)}  ∆t = 1,63 s ∆t = 1,63 s Total time/Totale tyd = (1,63)(2) Total time/Totale tyd = (1,63)(2) = 3,26 s = 3,26 s (4) OPTION 7 /OPSIE 7 Upwards positive/Opwaarts Downwards positive/Afwaarts positief: positief: F net ∆t = ∆p F net ∆t = ∆p mg ∆t = m (v f -v i ) mg ∆t = m (v f -v i ) -9,8∆t = [-16 – (+16)]  9,8∆t = [16 – (-16)]  ∆t = 3,26 s ∆t = 3,26 s Total time/Totale tyd = 3,26 s Total time/Totale tyd = 3,26 s  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2015 NSC/NSS – Memorandum 3.2 POSITIVE MARKING FROM 3.1./POSITIEWE NASIEN VANAF 3.1 Upwards positive/Opwaarts positief: 16 Velocity/snelheid (m∙s-1) 3,26 0 1,63 t(s) -16 (3) POSITIVE MARKING FROM 3.2./POSITIEWE NASIEN VANAF 3.2 Downwards positive/Afwaarts positief: 16 Velocity/snelheid (m∙s-1) 0 1,63 3,26 t(s) -16 Criteria for graph/Kriteria vir grafiek Marks/Punte Correct shape for line extending beyond t = 1,63 s.  Korrekte vorm vir lyn verleng verby t = 1,63 s Initial velocity correctly indicated as shown.  Beginsnelheid korrek aangedui soos getoon. Time to reach maximum height and time to return to the ground correctly shown.  Tyd om maksimum hoogte te bereik en om na die grond terug te keer. (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/November 2015 NSC/NSS – Memorandum 3.3 OPTION 1 / OPSIE 1 Upwards positive/Opwaarts positief: Take y A as height of ball A from the ground. (no penalising)/Neem y A as hoogte van bal A vanaf die grond. (geen penalisering) ∆y A = v i ∆t + ½ a∆t2  y A - 0 = 16∆t + ½(-9,8)∆t2 =16∆t – 4,9∆t2 Take y B as height of ball B from the ground./Neem y B as hoogte van bal B vanaf die grond. ∆y B = v i ∆t + ½ a∆t2 y B – 30 = (v i ∆t + ½ a∆t2) y B = 30 - [ -9(∆t -1) + ½(-9,8)(∆t – 1)2 = 34,1 +0,8∆t - 4,9 ∆t2 yA = yB ∴16∆t – 4,9∆t2 = 34,1 + 0,8∆t - 4,9∆t2 15,2∆t = 34,1 ∆t = 2,24 s  y A = 16 (2,24) - 4,9(2,24)2 = 11,25 m (6) Downwards positive/Afwaarts positief: Take y A as height of ball A from the ground.(no penalising)/Neem y A as hoogte van bal A vanaf die grond. (geen penalisering) ∆y A = v i ∆t + ½ a∆t2  y A - 0 = -16∆t + ½(9,8)∆t2 = -16∆t + 4,9∆t2 Take y B as height of ball B from the ground/Neem as hoogte van bal B vanaf die grond.. ∆y B = v i ∆t + ½ a∆t2 y B – 30 = – (v i ∆t + ½ a∆t2) y B = 30 – [ 9(∆t -1) + ½(9,8)(∆t – 1)2 = 34,1 + 0,8∆t - 4,9 ∆t2  yA = yB 16∆t – 4,9∆t2 = 34,1+ 0,8∆t – 4,9∆t2 15,2∆t = 34,1 ∆t = 2,24 s ∆y A = (- 16 (2,24) + 4,9(2,24)2) = 11,25 m (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 2015 NSC/NSS – Memorandum 3.3 OPTION 2/OPSIE 2 Upwards positive/Opwaarts positief: Δy A = v i ∆t + ½ a∆t2  = 16∆t + ½(-9,8)∆t2 =16∆t – 4,9∆t2  Distance travelled by ball A = y A = 16∆t – 4,9∆t2 Δy B = v i ∆t + ½ a∆t2 = -9(∆t -1) + ½(-9,8)(∆t – 1)2 = 0,8∆t – 4,9∆t2 + 4,1 Distance travelled by ball B = y B = 0,8∆t – 4,9∆t2 + 4,1 y A +(-y B ) = 30 16∆t – 4,9∆t2 – (0,8∆t – 4,9∆t2 + 4,1) = 30 15,2∆t = 34,1 ∆t = 2,24 s  ∴ ∆y A = v i ∆t + ½ a∆t2 y A = 16 (2,24) - 4,9(2,24)2 = 11,25 m (6) 3.3 Downwards positive/Afwaarts positief: y A = v i ∆t + ½ a∆t2  = -16∆t + ½(9,8)∆t2 = -16∆t + 4,9∆t2 y B = v i ∆t + ½ a∆t2 = 9(∆t -1) + ½(9,8)(∆t – 1)2  = - 0,8∆t + 4,9∆t2 - 4,1  (-y A ) + y B = 30 -(-16∆t + 4,9∆t2) - 0,8∆t + 4,9∆t2 - 4,1 = 30 15,2∆t = 34,1 ∆t = 2,24 s  ∴ ∆y A = v i ∆t + ½ a∆t2 ∆y A = -16 (2,24)+ 4,9(2,24)2 = - 11,25 m ∴Height of ball A/Hoogte van bal A = 11,25 m (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 2015 NSC/NSS – Memorandum OPTION 3/OPSIE 3 Upwards positive/Opwaarts positief: v f = v i + a∆t After 1 s, speed of ball A/Spoed van bal A na 1 s v f = 16 + (-9,8)(1) = 6,2 m∙s-1 Distance travelled by ball A in 1 s/Afstand deur bal A afgelê in 1 s ∆y A = v i ∆t + ½ a∆t2  = (16)(1) + ½(-9,8)12 = 11,1 m For ball A, after 1 s/Vir bal A na 1 s ∆y A = 6,2∆t – 4,9∆t2  For ball/Vir bal B, ∆y B = v i ∆t + ½ a∆t2 = -9∆t + ½(-9,8)∆t2 y A + (-y B ) = (30 - 11,1) = 18,9 6,2∆t – 4,9∆t2 – [ -9∆t + ½(-9,8)∆t2] = 18,9 15,2∆t = 18,9 ∆t = 1,24 s  The balls meet after/Die balle ontmoet na (1,24 +1) = 2,24 s ∆y A = [6,2 (1,24) – 4,9 (1,24)2] = 0,154 m Meeting point/Ontmoetingspunt = (11,1 + 0,154) = 11,25 m OR/OF Δy = (-9)(1,24) + ½ (-9,8)(1,24)2  = -18,69 m Meeting point/Ontmoetingspunt = (30 -18,69) = 11,31 m (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 2015 NSC/NSS – Memorandum Downwards positive/Afwaarts positief: v f = v i + a∆t After 1 s, speed of ball A/Spoed van bal A na 1 s v f = -16 + (9,8)(1) = -6,2 ms-1 Distance travelled by ball A in 1 s/Afstand deur bal A afgelê in 1 s ∆y A = v i ∆t + ½ a∆t2  = (-16)(1) + ½(9,8)(1)2 = - 11,1 m For ball A, after 1 s/Vir bal A na 1 s ∆y A = - 6,2∆t + 4,9∆t2 For ball/Vir bal B ∆y B = v i ∆t + ½ a∆t2 = 9∆t + ½(9,8)∆t2 - ∆y A + ∆y B = 18,9 6,2∆t - 4,9∆t2 + [ 9∆t + ½(9,8)∆t2] = 18,9 15,2∆t = 18,9 ∆t = 1,24 s The balls meet after/Die balle ontmoet na (1,24 +1) = 2,24 s ∆y A = -6,2 (1,24) + 4,9 (1,24)2 = - 0,154 m Meeting point/Ontmoetingspunt = (-11,1 - 0,154) = 11,25 m OR/OF Δy = (9)(1,24) + ½ (9,8)(1,24)2  = 18,69 m Meeting point/Ontmoetingspunt = (30 -18,69) = 11,31 m (6) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/November 2015 NSC/NSS – Memorandum QUESTION 4/VRAAG 4 4.1 OPTION 1/OPSIE 1 Take motion to the right as positive/Neem beweging na regs as positief. Σp i = Σp f (m 1 + m 2 )v i = m 1 v f1 + m 2 v f2  Any one/Enige een (m 1 +m 2 )v i = m 1 v f1 + m 2 v f2 (3 + 0,02)(0)  = (3)(-1,4) + (0,02) v f2  v f2 = 210 m∙s-1  OR/OF Take motion to the left as positive/Neem beweging na links as positief. Σp i =Σ p f (m 1 + m 2 )v i = m 1 v f1 + m 2 v f2 Any one/Enige een  (m 1 +m 2 )v i = m 1 v f1 + m 2 v f2 (3 + 0,02)(0)  = (3)(1,4) + (0,02) v f2  v f2 = -210 m∙s-1 Speed/Spoed = 210 m∙s-1  (4) OPTION 2/OPSIE 2 Take motion to the right as positive/Neem beweging na regs as positief. Δp bullet = -Δp block  m(v f – v i ) = - m(v f – v i ) (0,02)(v f – 0) = -(3)(-1,4 – 0) ∴v i = 210 m∙s-1  OR/OF Take motion to the left as positive/Neem beweging na links as positief Δp bullet = -Δp block  m(v f – v i ) = - m(v f – v i ) (0,02)(v f – 0) = -(3)(1,4 – 0) ∴v i = -210 m∙s-1 Speed/Spoed = 210 m∙s-1  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 2015 NSC/NSS – Memorandum 4.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 v f 2 = v i 2 + 2a∆x  v + vf  0 = 2102 + 2a(0,4) ∆x =  i ∆t   2  a = - 55 125 m∙s-2  210 + 0  0,4 =  ∆t  F net = ma  2  = (0,02)(-55 125) ∆t = 0,004 s (0,00381s) = -1 102,5 N F net ∆t = ∆p = m∆v Magnitude of force = 1 102,5 N (0,02)(0 - 210) Grootte van krag = 1 102,5 N F net =  (0,004 ) = -1 050 N Magnitude of force = 1 050 N Grootte van krag = 1 050 N (Accept/Aanvaar: 1102,5 N) (5) OPTION 3/OPSIE 3 v f 2 = v i 2 + 2a∆x 0 = 2102 + 2a(0,4) a = - 55 125 m∙s-2 v f = v i + a∆t 0 = 210 – (55 125)∆t ∆t = 0,004 s (0,00381 s) F net ∆t = ∆p = m∆v (0,02)(0 - 210) F net =  (0,004 ) = -1 050 N Magnitude of force = 1 050 N (Accept/Aanvaar: 1 102,5 N) Grootte van krag = 1 050 N (5) OPTION 4/OPSIE 4 W net = ∆K Any one/Enige een F net ∆xcosθ = ∆K = ½ m(v2 i - v2 f )  F net (0,4) cos180o= ½ (0,02)(02 - 2102)  F net = 1 102,5 N OR/OF W nc = ΔE p + ΔE k Any one/Enige een 2 2  F net Δxcosθ = 0 + ½ m(v i - v f ) F net (0,4) cos180o= ½ (0,02)(02 - 2102)  F net = 1 102,5 N (5) 4.3 The same as/equal Dieselfde as/gelyk (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/November 2015 NSC/NSS – Memorandum QUESTION 5/VRAAG 5 5.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 800 800 v ave = = 10,67m∙s-1 v ave =  = 10,67m∙s-1 75 75 P ave = Fv ave  ∴Distance covered in 1s = 10,67m∙ P ave = (240)(10,67) ∴W(Work done in 1 s) = FΔxcosθ =2 560,8 W (2,56 kW)  = (240)(10,67)(1) = 2 560,8 J s-1 ∴ P ave = 2 560,8 W (2,56 kW) OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 W W P=  P=  ∆t ∆t F∆x cos θ F∆x cos θ = = ∆t ∆t (240 )(800 ) cos 0 o (240 )(10,67) cos 0 o =  =  75 1 = 2 560 W = 2 560 W (3) 5.2 FD/FA FN/N ● Ff/f w/Fg Accepted labels/Aanvaarde benoemings F g / F w / weight / mg / gravitational force/2 940 N w F g / F w / gewig / mg / gravitasiekrag F friction / F f / friction /294 N /f k f F wrywing / F w / wrywing/294 N /f k F N / F normal / normal force N F N / F normaal / normaalkrag F Applied/toegepas /350 N/Average driving force FD F driving/dryfkrag /350/Gemiddelde aandrywingskrag (4) 5.3 The net/total work done on an object is equal to the change in the object's kinetic energy  Die netto/totale arbeid verrig op 'n voorwerp is geyk aan die verandering in die voorwerp se kinetiese energie. OR/OF The work done on an object by a resultant/net force is equal to the change in the object's kinetic energy.  Die arbeid verrig op 'n voorwerp deur 'n resulterende krag is gelyk aan die verandering in die voorwerp se kinetiese energie. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 2015 NSC/NSS – Memorandum 5.4 OPTION 1/OPSIE 1 W nc = ∆U + ∆K W f + W D = ∆U + ∆K f∆xcos θ + W D = mg(h f - h i ) + ½ (m)(v f 2 – v i 2 ) (f∆xcosθ + F D ∆x cos θ = mg(h f - h i ) + ½ m(v f 2 – v i 2 ) (294)(450)(cos180o) + (350)(450)cos0o= (300)(9,8)(5 - 0) + ½(300)(v f 2 – 0) v f = 8,37 m∙s-1 (6) OPTION 2/OPSIE 2 W net = ∆K W net = W D + W g + W f + W N = (F D ∆xcos θ) + (mg sinα)∆xcosθ) + (f∆xcosθ) + 0 5 W net = [350(450)](cos0)+ (300)(9,8) (450)(cos180)+ 450 294(450)(cos180o) = 157 500 – 14 700 – 132 300 = 10 500 J OR/OF 5 W net = ∆K α = sin-1 10 500 = ½ (300)(v f 2 – 0)  450 o v = 8,37 m∙s  -1 = 0,64 (6) f OPTION 3/OPSIE 3 W net = W D + W g + W f + W N = (F D ∆xcos θ) + mg∆xcosθ) + f∆xcosθ + 0 W net = (350)(450)(cos0o)+(300)(9,8)(450)cos(90 + 0,64) +294(450)(cos180o) = 157 500 – 14 777,74 – 13 2300 = 10 430,51 J OR/OF 5 W net = ∆K  α = sin-1 10 430,51 = ½ (300)(v f 2 – 0) 450 v = 8,34 m∙s-1  = 0,64o f (6) OPTION 4/OPSIE 4 F net = F D + (-mgsinα) +(-f) = 350 +[-(300)(9,8)sin0,64o] + (- 294) = 23,16 N OR/OF W net = F net ∆x cos θ Fnet = 350  - (300)(9,8)sin0,64o - 294 = (23,16)(450) cos0o = 350 - 32,84 - 294 = 10 422 J = 23,16 N W net = ∆K  10 422 = ½ (300)(v f 2 – 0) v f = 8,34 m∙s-1  (6) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 2015 NSC/NSS – Memorandum QUESTION 6/VRAAG 6 6.1.1 Frequency (of sound detected by the listener (observer)) Frekwensie van klank deur luisteraar (waarnemer) waargeneem (1) 6.1.2 The apparent change in frequency or pitch of sound (detected (by a listener) because the sound source and the listener have different velocities relative to the medium of sound propagation.  Die verandering in frekwensie (of toonhoogte) van die klank deur 'n luisteraar waargeneem omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium van klankvoortplanting het. (2) 6.1.3 Away/Weg van Detected frequency of source decreases Waargenome frekwensie van bron neem af (2) 6.1.4 OPTION 1/OPSIE 1 EXPERIMENT/EKSPERIMENT 2 v ± vL v fL = f s OR/OF fL = fs  v ± vs v + vs  v  874 = (900 )  v + 10 v = 336,15 m∙s-1 (Accept/Aanvaar : 336,15 m∙s-1 – 323,33 m∙s-1) (5) EXPERIMENT/EKSPERIMENT 3 v ± vL v fL = f s OR/OF fL = fs  v ± vs v + vs v  850 = (900 )  v + 20 v = 340 m∙s-1 (Accept/Aanvaar : 313,33 m∙s-1 – 340 m∙s-1) (5) EXPERIMENT 4/EKSPERIMENT 4 v ± vL v fL = f s OR/OF fL = fs  v ± vs v + vs  v  827 = (900 )  v + 30 v = 339,86 m∙s-1 (Accept/Aanvaar : 339,86 m∙s-1 – 345 m∙s-1) (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/November 2015 NSC/NSS – Memorandum OPTION 2/OPSIE 2 v ± vL v fL = fs OR/OF fL = fs  v ± vs v + vs Experiment/Eksperiment 2 and/en 3 874( v + 10) 850( v + 20)  = v v  both frequencies / beide frekwensies 874v + 8740 = 850v + 1700 ∴v = 344,17 m∙s-1  Experiment/Eksperiment 2 and/en 4 874( v + 10) 827( v + 30)  = v v 874v + 8740 = 827v + 24810  both frequencies / beide frekwensies ∴v = 341,91 m∙s-1  Experiment/Eksperiment 3 and/en 4 850( v + 20) 827( v + 30)  = v v 850v + 1700 = 827v + 24810  both frequencies / beide frekwensies ∴v = 339,57 m∙s-1  (5) 6.2 Away from the Earth/Weg vanaf die aarde (1) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/November 2015 NSC/NSS – Memorandum QUESTION 7/VRAAG 7 7.1 Q n=  e 0,5 × 10 -6 n=  1,6 × 10 -19 n = 3,13 x 1012 electrons/elektrone (3) 7.2 T/FT FE P w/ Fg Accepted labels/Aanvaarde benoemings F g / F w / weight / mg / gravitational force w F g / F w / gewig / mg / gravitasiekrag F T / tension T F T / spanning Electrostatic force/F C / Coulombic force/F Q /F RP/PR FE Elektrostiesekrag / Coulombkrag / F Q /F RP/PR (3) 7.3 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 wat deur een puntlading (Q 1 ) op 'n ander puntlading (Q 2 )uitgeoefen word, is direk eweredig aan die produk van die (groottes van die) ladings en omgekeerd eweredig aam die kwadraat van die afstand (r) tussen hulle. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/November 2015 NSC/NSS – Memorandum 7.4 OPTION 1/OPSIE 1 QQ FE = k 12 2  r Tsinθ/(Tcosθ) = F E 9 -6 -6  ∴ T sin7o/(Tcos83o) = (9 × 10 )(0,5 × 102 )(0,9 × 10 ) (0, 2)  ∴T = 0,83 N (Accept/Aanvaar 0,82 N) (5) OPTION 2/OPSIE 2 kQ1Q 2  FE = r2 (9 × 10 9 )(0,5 × 10 -6 )(0,9 × 10 -6 )  FE = (0,2)2  = 0,101 N T 0,101 tan 7 o = x =  TY TY T Y = 0,823 N T = TX2 + TY2 = (0,101)2 + (0,823 )2 = 0,83 N  (5) OPTION 3/OPSIE 3 kQ1Q 2 (9 × 10 9 )(0,5 × 10 −6 )(0,9 × 10 −6 )  F= = = 0,101 N r2 (0,2)2  FE T o = sin 7 sin 90 o 0,101 T o =  sin 7 sin 90 o T = 0,83 N  (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 20 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 20 DBE/November 2015 NSC/NSS – Memorandum QUESTION 8/VRAAG 8 8.1 E X = E 2 + E (-8)  kQ kQ − 8 = 22 + correct equation /korrekte vergelyking r r2 (9 × 10 9 )(2 × 10 −5 ) (9 × 10 9 )(8 × 10 −6 ) = +  (0,25)2 (0,15)2 = 2,88 x 106 + 3,2 x 106 = 6,08 x 106 N∙C-1  to the east/na oos  OR/OF Q E= k 2  r (9 × 10 9 )(2 × 10 -5 )  E2 = (0,25)2 = 2,88x 106 NC-1 to the east/na oos (9 × 10 9 )(8 × 10 -6 ) E -8 =  (0,15)2 = 3,2 x 106 N∙C-1 to the east/na oos E X = E 2 + E (-8) = (2,88 x 106 + 3,2 x 106)  = 6,08 x 106 N∙C-1  to the east/na oos (6) 8.2 OPTION 1/OPSIE 1 F E = QE = (-2 x 10-9) (6,08 x 106) = -12,16 x 10-3 N = 1,22 x 10-2 N to the west/na wes (4) OPTION 2/OPSIE 2 F (-2)Q1 = qE (2)  = (2 x 10-9) (2,88x 106 ) = 5,76 x 10-3 N to the west/na wes F (-2)Q2 = qE (8) = (2 x 10-9)(3,2 x 106) = 6,4 x 10-3 N to the west/na wes F net = 5,76 x 10-3 + 6,4 x 10-3  = 1,22 x 10-2 N to the west/na wes (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 21 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 21 DBE/November 2015 NSC/NSS – Memorandum OPTION 3/OPSIE 3 QQ F = k 12 2  r (9 × 10 9 )(2 × 10 -9 )(2 × 10 -5 ) F (-2)2 = (0,25)2 = 5,76 x 10-3 N to the west/na wes (9 × 10 9 )(2 × 10 -9 )(8 × 10 -6 ) F (-2)(-8) = (0,15)2 = 6,4 x10-3 N to the west/na wes F net = (5,76 x 10-3 + 6,4 x 10-3)  = 1,22 x 10-2 N to the west/na wes  (4) 8.3 2,44 x 10-2 N (1) [11] QUESTION 9/VRAAG 9 9.1 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. (provided temperature and all other physical conditions are constant)  Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur (mits temperatuur en alle fisiese toestande konstant bly) OR/OF . The current in a conductor is directly proportional to the potential difference across the conductor, provided temperature and all other physical conditions are constant  Die stroom in 'n geleier is direk eweredig aan die potensiaalverskil oor ‘n geleier by konstante temperatuur mits temperatuur en alle fisiese toestande konstant bly (2) 9.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 V = IR V = IR V 8 = (0,5)(8) = 4 V V 8 = (0,5)(8)  = 4 V V 8 = V 16 1 1 1 ∴ V 16 = 4 V = + R R1 R 2 V 4 I 16 = = = 0,25 A 1 1 R 16 = +  8 16 I tot// = A 1 = (0,5 + 0,25)  = 0,75 A R = 5,33 Ω 4 I tot// = 5,33 A 1 = 0,75 A (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 22 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 22 DBE/November 2015 NSC/NSS – Memorandum OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 I1R1 = I2R2 2R 8Ω = R 16Ω  (0,5)(8) = I 16 (16)  ∴ I R16 = ½ I R8  (8)(0,5) ∴ I R16 = ½ (0,5) = 0,25 A I 16 = = 0,25 A 16 A 1 = (0,5 + 0,25)  = 0,75 A I tot// = A 1 =(0,5 + 0,25) = 0,75 A (4) 9.3 OPTION 1/OPSIE 1 V = IR V 20Ω =( 0,75) (20)  = 15 V V //tot = (15 + 4)  = 19 V V R = 19 V P = VI 12 = (19)I  I R = A 2 = 0,63 A (5) OPTION 2/OPSIE 2 1 1 1 1 1 = + = +  OR/OF R R1 R 2 8 16 R1R 2 8 × 16 R // = 5,33 Ω R= =  = 5,33 Ω R1 + R 2 8 + 16 R // + R 20 = (5,33 + 20)  = 25,33Ω V //tot = I(R // + R 20 ) = (0,75)(25,33) = 19 V P = VI 12 = I(19)  I R = A 2 = 0,63 A (5) OPTION 3/OPSIE 3 V = IR V 20Ω =( 0,75) (20)  = 15 V V //tot = (15 + 4)  = 19 V V R = 19 V V2 P= R (19)2 12 = R R = 30,08 Ω P = I2R 12 = I2(30,08) I = 0,63 A (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 23 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 23 DBE/November 2015 NSC/NSS – Memorandum 9.4 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 ( ε ) = I(R + r)  V int = Ir = V terminal + V int = (0,75 + 0,63)(1) = 19 + (0,75 + 0,63)(1)  = 1,38 V = 20,38 V ε = V terminal + V int  = 19 + 1,38 = 20,38 V  (3) OPTION 3/OPSIE 3 V 19 R= = = 30,16 Ω I 0,63 1 1 1 1 1 = + = + ∴R p = 13,77 Ω R p R1 R 2 30,16 25,33 I tot = 0,63 + 0,75 = 1,38 A ε = I(R + r)  = (1,38)(13,77 + 1) = 20,38 V [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 24 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 24 DBE/November 2015 NSC/NSS – Memorandum QUESTION 10/VRAAG10 10.1.1 Move the bar magnet very quickly up and down inside the coil  Beweeg die staafmagneet baie vinnig op en af binne in die spoel. (2) 10.1.2 Electromagnetic induction/Elektromagnetiese induksie  (1) 10.1.3 Commutator/kommutator / split rings/spleetringe  (1) 10.2.1 OPTION 1/OPSIE 1 V2 2 Vrms Paverage = rms  W= ∆t  R R 220 2  220 2  = (1) = 40,33  40,33  = 1200,10 J  = 1 200,10 W (J∙s-1) (4) OPTION 2/OPSIE 2 Vrms V Irms =  Irms = rms  R R 220 220 =  =  40,33 40,33 = 5,45 A = 5,45 A 2 2 W = I rms R∆t Paverage = I rms R = (5,452)(40,33)(1) = (5,45 )(40,33)  2 = 1 197,9 J OR/OF 1 200,10 J = 1 197,9 W OR/OF 1 200,10 W (4) OPTION 3/OPSIE 3 V Vrms  Irms = rms  Irms = R R 220  220  = = 40,33 40,33 = 5,45 A = 5,45 A Paverage = Vrms I rms W = Vrms I rms ∆t = (220 )(5,45)  = (220 )(5,45)(1)  = 1 199 W or/of 1 200,10 W = 1 199 J or/of 1 200,10 J (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 25 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 25 DBE/November 2015 NSC/NSS – Memorandum 10.2.2 OPTION 1/OPSIE 1 Vmax Vrms = 2 Vmax 220 = 2 V max = 311,13 V Vmax 331,13  Imax = = R 40,33 = 7,71 A OR/OF V I P ave = max max 2 (311,13) Imax 1200,1 = 2 I max = 7,71 A (3) OPTION 2/OPSIE 2 Paverage = Vrms I rms  1200,1 = (220)I rms  I rms = 5,455 A I max = 2 (5,455) = 7,71 A (7,715 A) (3) OPTION 3/OPSIE 3 Paverage = I 2rmsR  1200,1 = I2 rms (40,33) I rms = 5,455 A I max = 2 I rms = 2 (5,455) = 7,71 A (3) OPTION 4/OPSIE 4 V rms = I rms R  220 = I rms (40,33) I rms = 5,455 A I max = 2 I rms = 2 (5,455) = 7,71 A (3) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 26 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 26 DBE/November 2015 NSC/NSS – Memorandum QUESTION 11/VRAAG 11 11.1 It is the process whereby electrons are ejected from a metal surface when light (of suitable frequency) is incident on it.  Dit is die proses waartydens elektrone vanaf ‘n metaaloppervlak vrygestel word wanneer van geskikte frekwensie daarop inval (2) 11.2 1 Graph of/Grafiek van Ek(max/maks) vs λ • 6 4 • Ek(max) x (10 -19 J) 2 • • 0 3 5 1 2 4 1 × 10 6 (m -1 ) λ -2 -4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 27 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 27 DBE/November 2015 NSC/NSS – Memorandum 11.3.1 OPTION 1/OPSIE 1 1 = 1,6 × 10 6 m-1  λ 1 fo = c  λ = (3 x 108)(1,6 x 106)  = 4,8 x 1014 Hz (Accept/Aanvaar 4,8 x 1014 Hz to/tot 5,1 x 1014 Hz) (4) OPTION 2/OPSIE 2 By extrapolation: y-intercept = -W o /Deur ekstrapolasie : y-afsnit = -W o W o = hf o  3,2 x 10-19 = (6,63 x 10-34)f o  f o = 4,8 x 1014 Hz (Accept/Aanvaar 4,8 x 1014 Hz to/tot 4,83 x 1014 Hz) (4) OPTION 3/OPSIE 3 (Points from the graph/ Punte vanaf grafiek) E = W o + E k(max) hc = hf0 + E k (max)  λ0 (6,63 × 10 −34 )(3 × 10 8 )(1,6 × 10 6 )= (6,63 × 10 −34 )f0 + 0  f o = 4,8 x 1014 Hz  OR/OF (6,63 × 10 −34 )(3 × 108 )(5 × 106 ) = (6,63 × 10−34 )f0 + 6,6 × 10 −19 f o = 4,92 x 1014 Hz OR/OF (6,63 × 10 −34 )(3 × 108 )(3,3 × 106 ) = (6,63 × 10 −34 )f0 + 3,3 × 10 −19 f o = 4,8 x 1014 Hz OR/OF (6,63 × 10 −34 )(3 × 108 )(2,5 × 106 ) = (6,63 × 10 −34 )f0 + 1,7 × 10 −19 f o = 4,94 x 1014 Hz OR/OF (6,63 × 10 −34 )(3 × 108 )(2,2 × 106 ) = (6,63 × 10 −34 )f0 + 0,7 × 10 −19 f o = 5,54 x 1014 Hz (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 28 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 28 DBE/November 2015 NSC/NSS – Memorandum 11.3.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 hc = Gradient/ Helling  W 0 = y intercept/afsnit  ∆y = 3,2 x10-19 J = ∆x 6,6 × 10 -19 Accept /Aanvaar =  3,2 x10-19 J to/tot 3,4 x10-19J) (5 - 1, 6) × 10 6 = 1,941 x10-25 (J∙m) W o = hf o gradient / helling 3,2 x 10-19 = h(4,8 x 1014)  h= c h = 6,66 x 10-34 J∙s 1, 941× 10 - 25 Accept /Aanvaar h=  3 × 10 8 6,66 x10-34 J∙s to/tot 7,08 x10-34J∙s) = 6,47 x10-34 J∙s  (4) OPTION 3/OPSIE 3 (Points from the graph OPTION 4/OPSIE 4 (Punte vanaf grafiek) hc 1 hc Wo = or / of Wo = hc = W0 + K max = 3,2 × 10 -19+ 6,6 × 10 -19  λo λo λ 9,8 × 10 -19 3,2 x 10 = h(3 x 10 )(1,6 x 106) -19 8 h= J ⋅ s = 6, 53 × 10 -34 h = 6,66 x 10-34 J∙s  (3 × 10 8 )(5 × 10 6 ) OR/OF hc = W0 + K max = 3,2 × 10 -19+ 3,3 × 10 -19  λ 6,5 × 10 -19 = 6, 57 × 10 h= -34 J ⋅ s (3 × 10 8 )(3,3 × 10 6 ) OR/OF hc = W0 + K max = 3,2 × 10 -19+ 1,7 × 10 -19 λ 4,7 × 10 -19 = 6, 27 × 10 -34 J ⋅ s h= (3 × 10 8 )(2,5 × 10 6 ) OR/OF hc = W0 + K max = 3,2 × 10 -19+ 0,7 × 10 -19  λ 3,9 × 10 -19  h= = 6, 5 × 10 -34 J ⋅ s (3 × 10 8 )(2 × 10 6 ) (4) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou ## Related documents - [MP 2015 SEPT Physics P 1 Memo](https://www.hlayiso.com/library/mp-2015-sept-physics-p-1-memo/888a8f9f-c0da-4ad1-b3ae-e2ecd133acdd) — Physical Sciences · Memo · 2015 - 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