Physical Sciences Grade 11 P1 2013 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) EXEMPLAR/MODEL 2013 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 11 pages. Hierdie memorandum bestaan uit 11 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum QUESTION 1/VRAAG 1 1.1 C  (2) 1.2 A  (2) 1.3 C  (2) 1.4 A  (2) 1.5 C  (2) 1.6 B  (2) 1.7 A  (2) 1.8 D  (2) 1.9 D  (2) 1.10 B  (2) [20] QUESTION 2/VRAAG 2 2.1 A single force  having the same effect as all other forces acting together.  'n Enkele krag  wat dieselfde effek het as al die ander kragte tesame.  (2) 2.2 2.2.1 Magnitude of P/Grootte van P = 22 + 42 = 4,47 (force units/krageenhede)  (2) 2.2.2 −1 tanθ = ∴ θ = 18,43°  −3 Direction/Rigting: 270° - 18,43°  = 251,57°  OR/OF −3 tanθ = ∴ θ = 71,57°  −1 Direction/Rigting: 71,57° + 180°  = 251,57°  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum 2.3 Rx = Px + Qx = 2 + (-3) = -1 (force units/krageenhede)  Ry = Py + Qy = 4 + (-1) = 3 (force units/krageenhede)  R= R 2x + R 2y = (−1) 2 +32 = 3,16 (force units/krageenhede)  (3) 2.4 3 tanθ = ∴ θ = -71,57° −1 Direction/Rigting: 270° + 71,57° = 341,57°  OR/OF −1 tanθ = ∴ θ = -18,43° 3 Direction/Rigting: 360° - 18,43° = 341,57°  (2) [12] QUESTION 3/VRAAG 3 3.1 The resultant of the forces is zero./Die resultant van die kragte is nul. (1) 3.2 Parallax error/Error in marking the position of the strings (because the observer's eye and the strings are not in a line perpendicular to the plane of the paper).  Parallaksfout/Fout in die merk van die posisie van die toutjies (omdat die waarnemer se oog en die toutjies nie in lyn loodreg tot die vlak van die papier is nie).  (1) 3.3 0,98 N  0,98 N  w (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum 3.4 CONSTRUCTION AND MEASUREMENT / KONSTRUKSIE EN METING Marking criteria for scale drawing: Kriteria vir nasien van skaaltekening: A vector of 98 mm at 45° with vertical. 45° (98 mm)  'n Vektor van 98 mm teen 45° met vertikaal. Another vector of 98 mm at 90° to first vector.  'n Ander vektor van 98 mm teen 90° met eerste vektor. (138,6 mm) A third vector drawn from head of one vector to tail of the 90° other./Derde vektor geteken van kop van een vektor na  stert van ander. Length of vertical vector in range of 137–139 mm.  (98 mm) Lengte van vertikale vektor in gebied 137–139 mm. 45° All vectors represented as arrows.  Alle vektore as pyltjies voorgestel. The unknown weight given in range 1,37–1,39 N.  Die onbekende gewig gegee in gebied 1,37–1,39 N. Mass/Massa = 141 g  (Accept answers in range/Aanvaar antwoorde in gebied: 139 g–142 g) OR / OF CALCULATION / BEREKENING ∑F y = 0 (0,98) sin45° + (0,98) sin45° = m(9,8)  m = 141,42 g  (7) [12] QUESTION 4/VRAAG 4 4.1 When a resultant force acts on an object, the object accelerates in the direction of the force. This acceleration is directly proportional to the force  and inversely proportional to the mass of the object.  Indien 'n resulterende krag op 'n voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel. Hierdie versnelling is direk eweredig aan die resulterende krag  en omgekeerd eweredig aan die massa van die voorwerp.  (2) 4.2 4.2.1 fk = µkN  = (0,13)(1)(9,8)  = 1,27 N  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum 4.2.2 For the 2 kg mass (to the right/downwards as positive): Vir die 2 kg-massa (na regs/afwaarts as positief): F net = ma  w + F T = ma (2)(9,8) + F T = 2a  F T = 2a – 19,6 For the 1 kg mass (to the right as positive): Vir die 1 kg-massa (na regs as positief): F net = ma F T + f = ma -(2a – 19,6) + (- 1,27) = 1a F T (1 kg) = -F T (2 kg) ∴ a = 6,11 m·s-2  (5) 4.3 Zero acceleration  F net on the 1 kg mass is zero.  According to Newton's second law of motion, its acceleration will be zero.  According to Newton's first law of motion, it will continue to move at constant velocity (until it reaches the edge of the surface).  Nul versnelling  F net op die 1 kg-massa is nul.  Volgens Newton se tweede bewegingswet, is sy versnelling nul.  Volgens Newton se eerste bewegingswet, sal dit teen konstante snelheid (4) beweeg (tot aan die einde van die oppervlak).  [14] QUESTION 5/VRAAG 5 5.1 N = w ⊥ = mgcosθ  = (60)(9,8)cos15° = 567,96 N  (2) 5.2 N fk   w Accepted Labels/Aanvaarde byskrifte F g/ F w /force of Earth on skier/weight/588 N/mg/gravitational force w F g/ F w /krag van Aarde op skiër/gewig/588 N/mg/gravitasiekrag F N /normal/567,96 N N F N /normaal/567,96 N Frictional force / F f fk Wrywingskrag / F f (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum 5.3 v f = v i + 2a∆x  2 2 152 = 02 + 2a(75)  ∴a = 1,5 m·s-2 F net = ma  w // + f = ma mgsinθ + f = ma (60)(9,8)sin15° + f k = (60)(1,5)  ∴ f = -62,19 N ∴ f = 62,19 N  up the incline/opwaarts teen skuinsvlak  (7) [12] QUESTION 6/VRAAG 6 6.1 Any two objects in the universe attract each other  with a force directly proportional to the product of their masses and  inversely proportional to the square of the distance between their centres.  Enige twee voorwerpe in die heelal trek mekaar aan  met 'n krag wat direk eweredig is aan die produk van hul massas en  omgekeerd eweredig aan die kwadraat van die afstand tussen hul middelpunte.  (3) 6.2 Gm1m2 F=  r2 (6,67 × 10 −11 )(5,98 × 1024 )(615 ) 5 000 =  r2 ∴ r = 7,0 x 106 m Height /Hoogte = 7,0 x 106 - 6,38 x 106  = 6,2432 x 105 m = 624,32 km  (5) 6.3 Gm1m2 F= r2 ∴ F = Gm12m2 = 1 Gm1m2 = 1 (5 000) = 2 500 N  (2r )2 2 r2 2 (2) [10] QUESTION 7/VRAAG 7 7.1 The ratio of sin i to sin r is constant.  Die verhouding van sin i tot sin r is konstant. (2) 7.2 1,4,2,3,6,5  (2 or/of 0) (2) 7.3 To ensure that the position of the glass block remains the same/glass block does not shift.  Om te verseker dat die posisie van die glasblok dieselfde bly/die glasblok nie skuif nie. (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum 7.4 Δ sin i 7.4.1 Gradient/Gradiënt =  Δ sin r 0,6 - 0  = 0,4 - 0  = 1,5  Note/Let Wel Accept values from/aanvaar waardes vanaf 1,44 to/tot 1,5. (4) 7.4.2 The refractive index of glass Die brekingsindeks van glas. (1) [10] QUESTION 8/VRAAG 8 8.1 n 1 sinθ 1 = n 2 sin θ 2  (1,44)sin30° = (1)sin θ 2  θ a = 46,05°  (4) 8.2 The glass block is optically more dense than air.  When light passes from an optically dense into an optically less dense medium, it bends away from the normal.  Die glasblok is opties digter as lug. Wanneer lig van 'n optiese digter medium na 'n opties minder digte medium beweeg, buig dit weg van die normaal af. (2) 8.3 Increases/Neem toe  (1) 8.4 The angle of incidence in the optically denser medium for which the angle of refraction in the optically less medium is 90°.  Die invalshoek in die opties digter medium waarvoor die brekingshoek in die opties minder digte medium 90° is. OR/OF The angle of incidence in the optically denser medium beyond which the light/wave will be totally internally reflected.  Die invalshoek in die opties digter medium waarbo die lig/golf totale interne weerkaatsing ondervind. (2) 8.5 Total internal reflection/Totale interne weerkaatsing  (1) 8.6 Optical fibres for communication/in endoscopes.  Optiese vesels vir kommunikasie/in endoskope.  OR/OF Binoculars/Verkykers OR/OF Periscopes/Periskope (1) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum QUESTION 9/VRAAG 9 9.1 The bending/spreading of waves as they pass through a narrow opening or around corners/obstacles. Die buiging/spreiding van golwe wanneer hulle 'n smal opening of om hoeke/versperrings beweeg. (2) 9.2 9.2.1 Greater than/Groter as  (1) 9.2.2 Dimmer than/Dowwer as  (1) 9.2.3 Greater than/Groter as  (1) 9.2.4 Every point on a wave front acts as a source of new wavelets that move forward with the same speed as the wave.  Elke punt op 'n golffront wat dien as 'n bron van nuwe golwe wat voortbeweeg teen dieselfde spoed as die golf. (2) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum QUESTION 10/VRAAG 10 10.1 The force per unit charge./Die krag per eenheid lading. (2) 10.2 Marking criteria for electric field lines Kriteria vir nasien van elektriese veldlyne Shape of field lines./Vorm van veldlyne.  -2 μC Direction of field lines (towards charge).  Rigting van veldlyne (na lading toe). (2) 10.3 Eresultant E5μC •P E-2μC kQ (9 × 109 )(2 × 10 −6 ) E 2μC =  =  r2 (0,1)2 = 1,8 x 106 N∙C-1 towards the 2 μC charge (as shown) na 2 μC lading (soos aangedui) kQ (9 × 109 )(5 × 10 −6 ) E 5μC = =  r2 (0,15)2 = 2 x 106 N∙C-1 away from the 5 μC charge (as shown) weg vanaf 5 μC lading (soos aangedui) E resultant = (1,8 × 106 )2 + (2 × 106 )2 = 2,69 x 106 N∙C-1  (7) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum QUESTION 11/VRAAG 11 11.1 The (magnitude of the) emf induced in a conductor is equal to the rate of change of magnetic flux linkage (through it).  Die (grootte van die emk) geïnduseer in 'n geleier is gelyk aan die tempo van verandering van magnetiese vloedkoppeling (daar deur).  (2) 11.2 11.2.1 ∆φ emf/emk = -N  ∆t ∆φ -15,2 = -(200)  3,2 × 10 −2 ∴ ∆φ = 2,43 x 10-3 Wb  or/of (2,43 x 10-3 V·s) (4) 11.2.2 ∆φ = (B f – B i )Acosθ  2,432 x 10-2 = (0,42 – 0,22)Acos0°  ∴ A = 0,012 m2 Area of circle/Oppervlak van sirkel = πr2 0,012 = πr2 ∴r = 6,22 x 10-2 m  (4) 11.3 15,2 V  (1) [11] QUESTION 12/VRAAG 12 12.1 Resistance = gradient of graph 12.1.1 Weerstand = helling van grafiek 4 − 0 = 25 × 10 − 3 − 0  = 160 Ω  (3) 12.1.2 Graph at T 1 /Grafiek by T 1  Steeper/larger gradient/Steiler helling  ∴ R is greater/R is groter  ∴ Temperature is higher/Temperatuur is hoër (3) 12.1.3 V is directly proportional to I at each of the temperatures T 1 and T 2 .  V is direk eweredig aan I by elk van die temperature T 1 en T 2 (1) 12.1.4 P = VI  = (2,5)(25 x 10-3)  = 0,06(2) W  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/2013 NSC/NSS – Grade 11 Exemplar/Graad 11 Model – Memorandum 12.2 12.2.1 Both switches open:/Beide skakelaars oop: R = 6 Ω + 1Ω + 2Ω = 9 Ω  V R=  I 4,5 9=  I ∴ I = 0,5 A  (4) 12.2.2 Both switches closed:/Beide skakelaars gesluit: V 6Ω = IR = (0,5)(6) = 3 V  V 2Ω = 4,5 – 3 = 1,5 V  V 1,5 I 2Ω = = = 0,75 A  R 2 I R = 0,75 – 0,5 = 0,25 A  V 3 R= =  = 12 Ω  I 0,25 (6) [20] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou