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Physical Sciences Grade 11 P1 2015 Memo

MemorandumPhysical SciencesGrade 11201510 pages

Grade 11 Physical Sciences Paper 1 memorandum from the Eastern Cape November 2015, 10 pages. Read online or download the free PDF.

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Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE/GRAAD 11 NOVEMBER 2015 PHYSICAL SCIENCES P1/ FISIESE WETENSKAPPE V1 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 10 pages./ Hierdie memorandum bestaan uit 10 bladsye.
Downloaded from hlayiso.com 2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015) QUESTION/VRAAG 1 1.1 A  1.2 A  1.3 A  1.4 D 1.5 B  1.6 B  1.7 D  1.8 D  1.9 A  1.10 C  [20] QUESTION/VRAAG 2 2.1 The resultant of the forces is zero.  Die resultant van die kragte is nul.  (1) 2.2 T 20° 45° FL/FB Fg  ACCEPTED LABELS/AANVAARDE BYSKRIFTE Fw/W/Weight/mg/gravitational force/force due to gravity Fg Fw/W/gewig/mg/gravitasiekrag FT/Tension/Tension in rope T FT/Spanning/Spanning in tou FL Flegs/Force of legs on wall FB Fbene/Krag van bene op muur (3) Copyright reserved Please turn over
Downloaded from hlayiso.com (EC/NOVEMBER 2015) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 3 2.3 Fg = mg = (50)(9,8) = 490 N  SCALE/SKAAL 10 mm = 50 N Fg = 490 N = 98 mm (Accept/Aanvaar 97 – 99mm) FL  45° T  Fg = 9,8mm  20° T = 76,5 × 5 = 382,5 N  (Accept/Aanvaar 76,4 – 76,6 mm) FL = 37,0 × 5 = 185 N  (Accept/Aanvaar 184,5 – 186,5 mm) (8) NOTES/AANTEKENINGE  Mark awarded for correct label and direction./  Punt toegeken vir korrekte benoeming en pyltjie.  ANY TWO angles indicated./  ENIGE TWEE hoeke aangedui.  OR/OF CALCULATIONS instead of CONSTRUCTION/ max/maks BEREKENINGE in plaas van KONSTRUKSIES Fg = FT FT = 490 sin 45°  = 382,50 N  sin 115° sin 45° sin 115° Fr = Fg FR = 490 sin 20°  = 184,91 N  sin 20° sin 115° sin 115° [12] Kopiereg voorbehou Blaai om asseblief
Downloaded from hlayiso.com 4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015) QUESTION/VRAAG 3 3.1 fs = Fg‖ = mg sin = (60)(9,8) sin 30°  = 294 N  Any one/Enige een  (3) 3.2 s = tan θ = tan 30° = 0,58  OR/OF fs = μs FN = μs Fg = μs mg cos  294 = μs (60)(9,8) cos 30°  μs = 0,58  (3) 3.3 3.3.1 REMAINS THE SAME/BLY DIESELFDE  (1) 3.3.2 DECREASES/NEEM AF  Θ Fg will decrease,  therefore FN will decrease/fk FN  Fg sal afneem,  dus sal FN afneem /fk FN  (3) [10] QUESTION/VRAAG 4 4.1 When a resultant/net 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 ʼn resulterende/netto krag op ʼn voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel. Hierdie versnelling is direk eweredig aan die (resulterende/netto) krag  en omgekeerd eweredig aan die massa van die voorwerp.  OR/OF The net force acting on an object is equal to the rate of change of momentum  of the object (in the direction of the force). (2 or 0) Die netto krag wat op ʼn voorwerp inwerk is gelyk aan die tempo van verandering in momentum  van die voorwerp (in die rigting van die krag). (2 of 0) (2) 4.2 N/FN f/Ff  T w/Fg/mg  (4) NOTES/AANTEKENINGE  Mark awarded for label and arrow./Punt toegeken vir benoeming en pyltjie.  Any other additional force(s)/Enige ander addisionele krag(te). Max/Maks ¾  If force(s) do not make contact with body/Indien krag(te) nie met die voorwerp kontak maak nie. Max/Maks ¾ Copyright reserved Please turn over
Downloaded from hlayiso.com (EC/NOVEMBER 2015) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 5 4.3 4.3.1 OPTION/OPSIE 1 For the 6kg block./Vir 6kg blokkie: Fnet = ma T + (-ff ) = ma T – 11,76 = 6a ...........(1) For the 2kg block./Vir 2kg blokkie:  For either/vir of/6a or/of 2a Fnet = ma 2 + w + (-T) = ma 2 + (2 × 9,8) – T = 2a T = 21,6 – 2a .......(2) Substitute (2) into (1)/Stel (2) in (1): 21,6 – 2a – 11,76 = 6a a = 1,23 m.s-2  OPTION/OPSIE 2: Fnet = ma 2 + w +(-Ff )  = ma 2 + (2 × 9,8) -Ff = 8a 2 + (2 × 9,8) – 11,76 = 8a a = 1,23 m.s-2  (5) 4.3.2 POSITIVE MARKING from QUESTION 4.3.1 POSITIEWE NASIEN van VRAAG 4.3.1 T = 21,6 – 2a = 21,6 – 2(1,23)  = 19,14 N OR/OF T – 11,76 = 6a T = 11,76 + 6(1,23)  = 19,14 N (2) 4.4 INCREASES/TOENEEM  (1) [14] QUESTION/VRAAG 5 5.1 BACKWARDS/AGTERTOE  (1) 5.2 An object continues in a state of rest or uniform velocity,  unless it is acted upon by an external unbalanced force (resultant force).  ʼn Voorwerp sal in ʼn toestand van rus bly of teen ʼn konstante snelheid bly voortbeweeg,  tensy ʼn eksterne ongebalanseerde (resulterende krag) daarop inwerk.  OR/OF The tendency of an object to maintain its state of rest  or constant motion in a straight line.  Die weerstand van ʼn voorwerp wat enige verandering in sy toestand rus  of beweging teenstaan.  (2) Kopiereg voorbehou Blaai om asseblief
Downloaded from hlayiso.com 6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015) 5.3 5.3.1 EQUALS/GELYK  (1) 5.3.2 Kay will have a greater acceleration,  because her mass is smaller.  Kay sal ʼn groter versnelling het,  want haar massa is kleiner.  (2) [6] QUESTION/VRAAG 6 6.1 Action: Force of seat on astronaut (upwards)  Reaction: Force of astronaut on seat (downwards)  Aksie: Krag van sitplek op ruimtevaarder (opwaarts)  Reaksie: Krag van ruimtevaarder op sitplek (afwaarts)  (2) 6.2 FRES = m ∆v = (2,6 × 106)(160 – 0)  = 1,04 × 106 N  upwards/opwaarts  ∆t 400  Mark direction independently/Merk rigting onafhanklik (5) QUESTION/VRAAG 7 7.1 F = G m1.m2  r2 = (6,67 × 10-11)(7,35 × 1022)(6 × 1024)  (3 × 109)2 18 = 1,99 × 10 N  (4) 7.2 NEWTON’s LAW OF GRAVITATION/NEWTON se GRAVITASIEWET.  (1) 7.3 The distance between the earth and the moon changes due to the elliptical orbit.  Die afstand tussen die aarde en die maan verander as gevolg van elliptiese omwenteling.  OR/OF F α 1/r2  (1) 7.4 FEM(new) = GME1/4 MM  = FEM (1/4)(1/4)  = 1/16 FEM  (2r)2 OR/OF FEM(new) = 1/16 (1,99 × 1018)  OR/OF FEM(new) = 1,24 × 1017N  (1,24375 × 1017 N) (3) [9] Copyright reserved Please turn over
Downloaded from hlayiso.com (EC/NOVEMBER 2015) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 7 QUESTION/VRAAG 8 8.1 Angle of incidence/ i/i/sin i Invalshoek/ i/i/sin i  (1) 8.2 Type of block used/Temperature of the surroundings/Source of light/Surface on which block is placed. . Tipe blok gebruik/Temperatuur van die omgewing/Bron van die lig/Oppervlak waarop blok geplaas word. (1) 8.3 Snell’s law has been verified: * The graph is a straight line through the origin, because sin i sin r.  * Gradient = sin i = constant  sin r Snell se wet is bewys:  Die grafiek vorm ʼn reguitlyn deur die oorsprong, omdat sin i sin r.   Gradiënt = sin i = konstante  sin r (2) 8.4 Gradient/Gradiënt = n = Δ sin i = 0,77 – 0  = 1,51  Δ sin r 0,51 – 0  (4) [8] QUESTION/VRAAG 9 9.1 n sin i = n sin r  2,42 sin i  = 1 sin 90°  -1 i = sin 2,42 = 24,2°  (4) 9.2  No arrows: max Geen pyltjies: maks (2) 9.3 Total internal reflection /Totale interne weerkaatsing  (1) 9.4 Light moves from medium of higher optical density to medium of lower optical density.  Angle of incidence must be larger than critical angle.  Lig beweeg van medium van hoër optiese digtheid na medium met laer optiese digtheid.  Invalshoek moet groter wees as kritiese hoek.  (2) 9.5 Endoscope/Endoskoop  (1) [10] Kopiereg voorbehou Blaai om asseblief
Downloaded from hlayiso.com 8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015) QUESTION/VRAAG 10 10.1 Light consisting of only one wavelength.  Lig wat uit net een golflengte bestaan.  (1) 10.2 B A Broader Central band Breër sentraleband  Alternative light (bright) and dark bands. Afwisselende helder (lig) en donker bande  A: central band and bright (light) bands = CONSTUCTIVE INTERFERENCES  sentraleband en helder (lig)bande = KONSTRUKTIEWE INTERFERENSIE/ STEURINGE  B: dark bands = DESTRUCTIVE INTERFERENCES  donkerbande = DESTRUKTIEWE INTERFERENSIE/STEURINGE  (4) 10.3 10.3.1 INCREASES/NEEM TOE  (1) 10.3.2 INCREASES/NEEM TOE  Θ The narrower the opening, the greater the degree of diffraction.  Hoe smaller die opening, hoe groter die mate van diffraksie.  OR/OF Degree of diffraction 1  Width of opening Mate van diffraksie 1  Wydte van opening (2) [8] QUESTION/VRAAG 11 11.1 Force experienced per unit charge at a point in the field.  Krag ondervind per eenheidslading by ʼn punt in die veld.  OR/OF It is the force experienced by a unit positive charge placed at that point.  Dit is die krag wat deur ʼn eenheids positiewe lading geplaas by daardie punt ondervind word.  (2) Copyright reserved Please turn over
Downloaded from hlayiso.com (EC/NOVEMBER 2015) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 9 11.2 CRITERIA FOR SKETCH/KRITERIA VIR SKETS MARKS/PUNTE Correct shape / Korrekte vorm  Direction from positive to negative  Rigting van positief na negatief Fieldlines start on sphere and do not cross  Veldlyne begin op elke sfeer en kruis nie (3) 11.3 Ep = kQ = (9 × 109)(4 × 10-9)  = 3 600 N.C-1 to the right/na regs  r2 (0,10)2  Any ONE Enige EEN EQ = kQ = (9 × 109)(2 × 10-9)  = 45 000 N.C-1 to the left/na links r2 (0,02)2  Enet = 3 600 – 45 000 = 41 400 N.C-1  to the left/na links  (7) 11.4 POSITIVE MARKING FROM QUESTION 11.3 POSITIEWE NASIEN VAN VRAAG 11.3 E = F  Q 41 400 = F  1,6 × 10-19 F = 6,62 × 10-15 N  (3) [15] QUESTION/VRAAG 12 12.1 12.1.1 Φ = BA cos θ  = (0,6) r2 cos θ = (0,6)( × 0,062 ) cos 0°  = (0,6)( 0,01 ) = 6,79 × 10-3 Wb  (4) 12.1.2 POSITIVE MARKING FROM QUESTION 12.1.1 POSITIEWE NASIEN VAN VRAAG 12.1.1 = -N Δ Φ  Δt = (-1)(0 - 6,79 × 10-3)  0,04  = 0,17 V  (4) Kopiereg voorbehou Blaai om asseblief
Downloaded from hlayiso.com 10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015) 1.2.2 12.2.1 DECREASES/NEEM AF  (1) 12.2.2 INCREASES/NEEM TOE  (1) [10] QUESTION/VRAAG 13 13.1 13.1.1 P = I 2R  0,5 = I2 8  I = 0,25 A  1 =1 + 1 = 1 + 1  = 3 Rp = 4 Ω  Rp r1 r2 6 (6 + 6) 12 RT = 8 + 4  = 12 Ω VT = I RT = (0,25) (12) = 3V  (8) 13.1.2 Vp = I R  = (0,25)(4)  = 1V I = Vp = 1  = 0,17 A  R 6 (4) 13.2 13.2.1 P = V2  R 750 = 2402  R R = 76,80 Ω  OR/OF P = VI 750  = 240 I   I = 3,13 A R = V I = 240 3,13 = 76,80 Ω  (4) 13.2.2 E = Px t = 750 × 6  = 4 500 Wh = 4,5 kWh Cost/Koste = 4,5 × 1,14  = R 5,13  OR/OF Cost/Koste = (0,75)(6)  (1,14) = R5,13  (3) [19] TOTAL/TOTAAL: 150 Copyright reserved Please turn over

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