Physical Sciences Grade 12 Gauteng Mock 2017 Memo

Downloaded from hlayiso.com PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2017 MARKING GUIDELINES / NASIEN RIGLYNE PHYSICAL SCIENCES: PHYSICS (FIRST PAPER) (10841) FISIESE WETENSKAPPE: FISIKA (EERSTE VRAESTEL) (10841) 12 pages / bladsye

Downloaded from hlayiso.com 10841 / 17 GAUTENG DEPARTMENT OF EDUCATION GAUTENGSE DEPARTEMENT VAN ONDERWYS PREPARATORY EXAMINATION – 2017 VOORBEREIDENDE EKSAMEN – 2017 PHYSICAL SCIENCES: PHYSICS / FISIESE WETENSKAPPE: FISIKA (First Paper) (Eerste Vraestel) MEMORANDUM QUESTION / VRAAG 1 1.1 A  1.2 C  1.3 C  1.4 B  1.5 A  1.6 C  1.7 B  1.8 B  1.9 B  1.10 D  [20] Down is positive for the entire question. QUESTION / VRAAG 2 Neem af as positief vir die hele vraag. 2.1 Velocity with which the tennis ball strikes the balcony: Snelheid waarmee die tennisbal die balkon tref: Option / Opsie 1 Option / Opsie 2 vf2 = vi2 + 2a∆y  Em(top / bo) = Em(balcony / balkon) vf = 0 + 2x9,8x12  ½mv2 + mgh = ½mv2 + mgh  ½v2 + 0 = 0 + gh = 15,336 m·s-1 ½ v2 = 9,8 x 12  = 15,34 m·s-1 v = 15,34 m·s-1  (3) 2

Downloaded from hlayiso.com 10841 / 17 2.2.1 Energy that the ball has at bouncing / Energie wat die bal het wanneer dit bons Ek1 = ½ mvf2 = 0,5 x 0, 12 x 15,3362  = 14,112 J (2) POSITIVE MARKING FROM 2.2.1 / POSITIEWE NASIEN VANAF 2.2.1 2.2.2 Energy left on ball after bouncing / Energie oor op die bal na die bons Ek2 = ( 14,112 – 2 ) = 12,112 J (tick is for subtraction)/ (punt is vir die aftrekking) Option 1 Option 2 Initial velocity after bouncing / Relative to the balcony Aanvanklike snelheid na die bons Em = Epi + Eki = Epf + Ekf ½mv2 = 12,112 0 + 12,112 = mgh + 0 2 x 12,112 12,112 = 0,12(-9,8)h v =  h =0,30 m 0,12 = 14,208 m·s-1 v = 14,21 m·s-1 vf2 = vi2 + 2a∆y  0 = (-14,21)2 + 2(9,8) ∆y  ∆y = -10,30 m (6) 2.3 Height of balcony above the ground / Hoogte van die balkon bokant die grond ∆y = vi ∆t + ½ a∆t2  =( -14,208 x 3,16) + (1/2 x 9,8 x 3,162)  = 3,999 m = 4,03 m  Balcony is 4,03 m above the ground. / Balkon is 4,03 m bokant die grond. (3) 2.4 Mark if following is shown / Merk indien die volgende gewys word: • Velocity ball hits the balcony  / snelheid waarmee bal die balkon tref • Velocity ball leaves balcony  / snelheid waarmee die bal die balkon verlaat • Time of flight of ball to reach balcony (t1)  / tyd wat bal neem om balkon te bereik • Time ball reaches max height after bounce (t2)  / tyd wat bal neem om maks hoogte te bereik na die bons v(m·s-1)  15,34   t2  t1 4,72 t(s) -14,11 (4) [17] 3

Downloaded from hlayiso.com 10841 / 17 QUESTION / VRAAG 3 3.1 N Fnet w (3) 120 x 1 000 3.2 120 kmh-1 = = 33,33m·s-1  (1) 3 600 3.3 Work-energy theorem: The net work done on an object is equal to the change in the objects kinetic energy.  Arbeid energie beginsel: Die netto arbeid verrig op ʼn voorwerp is gelyk aan die verandering in die voorwerp se kinetiese energie. (2) 3.4 3.4.1 Wnet = Ekf - Eki  = 0 - ½ mv2 = 0 - ½ (1000) x 33,332  = - 555 444,45 J Fnet∆xcosθ = 555 444,45  Fnet(0,8)cos180 = 555 444,45 Fnet = - 694 305,56 N Fnet = 694 305,56 N to the left / away from the wall/opposite direction of initial motion Na links / weg van die muur / teenoorgestelde rigting van die aanvanklike beweging (5) 3.4.2 Time lapse / Tydperk Fnet ∆t = ∆p = m(vf - vi )  1 000 (0 − 33,33) ∆t =  − 694 305,56 = 0,05 s  (4) 3.4.3 • Crumbling increases stopping time / frommeling vergroot stop-tyd • thus reducing Fnet  / dus verminder Fnet ∆p • Fnet =  ∆t (3) [18] 4

Downloaded from hlayiso.com 10841 / 17 QUESTION / VRAAG 4 4.1 Option 1 / Opsie 1 Option 2 / Opsie 2 N  FA/Applied fk /friction wrywing toegepas  T N fk /friction wrywing FA/Applied force / toegepaste krag F// / Fgsinθ F⊥/Fgcosθ w/Fg/weight/gewig  (4) Accepted labels/Aanvaarde benoemings Fg/Fw/weight/mg/gravitational force w Fg/Fw/gewig/mg/gravitasiekrag Ffriction/Ff/friction/fk f Fwrywing/Fw/wrywing/fk FN/Fnormal/normal force N FN/Fnormaal/normaalkrag Deduct 1 mark for any additional force. / Trek een punt af vir enige ekstra krag Mark is given for both arrow and label. / Punt word toegeken vir beide ʼn pyl en byskrif. 4.2 0N Box slides at constant velocity / ∆Ek = 0 / acceleration = 0  Boks gly teen konstante snelheid / versnelling = 0 (2) 4.3 A non-conservative force is that force for which the work done in moving an object between two points depends on the path taken. ʼn Nie-konserwatiewe krag is daardie krag waarvan die arbeid verrig afhang van die padlengte gevolg tussen twee punte. (2) 5

Downloaded from hlayiso.com 10841 / 17 1,2 4.4 θ = sin-1  = 11,5370 6 Option 1 / Opsie 1 Wnett = ∆Ek  WA + W f + W N + W ⊥ + W// = 0  WA + Ff∆ x Cos180 + 0 + 0 +100 x 9,8Sin 11,537x 6Cos0 = 0 WA + 60 x 6Cos180 + 100 x 9,8Sin 11,537 x 6Cos0 = 0 WA = -816 J Work done by man = 816 J  Arbeid verrig2deur man Option Option 2 / Opsie 2 Fnett = FA + f + F// + N + F⊥ = FA + f + F// 0 = FA – 60 + 100 x 9,8 x Sin 11,537 = FA + 136 FA = -136 N ∴ WA = FA x ∆ x = - 136 x 6 = - 816 J ∴ work done by man is 816 J Arbeid verrig deur man Option 3 / Opsie 3 Wnc = ∆Ek + ∆Ep Wman + W f = 0 + mg(hf – hi) Wman + 60 x 6 x (-1) = 100 x 9.8 (0-1,2) Wman = -1176 + 360 = - 816 J ∴ work done by man is 816 J Arbeid verrig deur man (6) 6

Downloaded from hlayiso.com 10841 / 17 4.5 WA = F∆ x cosθ  = - 816 = F x 6 cos 180 = - 816  FA = 136 N Magnitude of force applied by man is 136 N. Grootte van krag toegepas deur man is 136 N (4) [18] QUESTION / VRAAG 5 5.1 All charge is made up of multiples of a basic charge that has a value of 1,6 x 10 -19 C  Alle ladings word opgemaak uit veelvulde van ʼn basiese lading wat die waarde het van 1,6 x 10-19 C. (1) Q + QB 5.2 5.2.1 New charge / nuwe lading = A 2 9 x 10 −6 − 3 x 10 −6 =  2 = +3µC (2) 5.2.2 For sphere A: from +9µC to +3µC is a loss of 6µC For sphere B: from -3µC to +3µC is a gain of 6µC Vir sfeer A vanaf +9µC tot +3µC is ʼn verlies van 6µC Vir sfeer B vanaf -3µC tot +3µC is ʼn toename van 6µC electrons transferred / elektrone oorgedra (− )Qlost/gained / verlies / wins = e− = (− )6 x 10 −6  (− )1,6 x 10 −19 = 3,75 x 1013 electrons (Answer must be positive.) Elektrone (Antwoord moet positief wees.) (3) 5.2.3 B to A  / B tot A (1) 5.3 Coulomb’s Law: The force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges.  Coulomb se Wet: die krag tussen twee ladings is direk eweredig aan die produkte van hul ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen die ladings. (2) 5.4 kQ 2Q 3 9 x 10 9 x 9 x 10 −6 x 3 x 10 −6  FA = = r2,3 2 0,02 2  = 607,5 N, attractive  (3) 7

Downloaded from hlayiso.com 10841 / 17 5.5 Enet = EQA at x + EQB at x kQ kQB = 2 + 2  rA rB 9 x 10 9 x 9 x 10 −6 9 x 10 9 x 3 x 10 −6 =  +  (0,04)2 0,16 2 = 50 625 000 + 1 054 687,5 = 5,168 x 107 NC-1 (4) [16] QUESTION / VRAAG 6 6.1 6.1.1 It is the apparent change in frequency of a source when there is relative motion between the source and the observer.  Die verandering in frekwensie (of toonhoogte) van die klank waargeneem deur ʼn luisteraar omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium waarin die klank voortgeplant word, het. (2) 6.1.2 v ± vL fL = ( )fs  v ± vs fL vL = (v + vs) fs  360  = (340 + 0) x    340  = 17,90 ms-1  3600  = 17,895 x    1000  = 64,42 km·h-1  (5) 6.2 source / bron   3 2 2A 1 1A Since the driver is moving in the same direction as the waves, the wavefronts will pass him at delayed intervals, thus making the wavelengths longer hence hearing a lower frequency. Aangesien die bestuurder beweeg in dieselfde rigting as die golwe, sal die golffronte verby beweeg teen vertraagde intervalley wat dus maak dat die golflengtes langer word en hy dus ʼn laer frekwensie hoor. (4) 6.3 Speed trap cameras, Doppler flow meter for measuring the speed of blood, ultra sound foetus scan  Spoedvang kameras, Dopplervloei meter vir die meting van die spoed van bloed, ultraklank fetus skandering (2) [13] 8

Downloaded from hlayiso.com 10841 / 17 QUESTION / VRAAG 7 7.1 7.1.1 Potential difference across R  / potensiaalverskil oor R (1) 7.1.2 Current flowing through R  / stroom vloei deur R (1) 7.2 ε = 2,52 + 0,3r ……i -ε = 0,71 + 1,6r .. ……ii 0 = 1,81 - 1,3r (add – any other method) / (voeg by – enige ander metode) r = 1,39 Ω  (3) 7.3 7.3.1 2,9 V  (2) 7.3.2 2,1 A  (2) 7.3.3 Gradient of graph / gradient van die grafiek (Accept any viable calculation of gradient.) 0 − 2,9 =  (Aanvaar enige sinvolle berekening van gradiënt.) 2,1 − 0 = -1,38 Ω r = 1,38 Ω (2) 7.4 The graph is an average line through the points/learners drew line of best fit ∴ gives an average value for r,whilst calculation from individual plots does not take into account instrument reading errors, and any other small errors that arise. max. Die grafiek is ʼn gemiddelde lyn deur die punte/leerders teken ʼn beste paslyn ∴ gee ʼn gemiddelde waarde vir r, terwyl berekeninge vanaf individuele punte nie enige instrumentele aflees foute in ag neem nie en enige ander sklein foute wat mag opduik. (1) [12] 9

Downloaded from hlayiso.com 10841 / 17 QUESTION / VRAAG 8 1 1 1 1 1 8.1 For the parallel network = + = +  R// = 2,4 Ω R R1 R 2 6 4 Option 1/ Opsie 1 Option 2/ Opsie 2 `````````````` Rcircuit = R// + Rseries  ε = IRext + Icr  = 2,4 + 2 + 1 = I (Rext + r) = 5,4 Ω 10 = I (2,4+2 +1)  ε  I= I = 1,85 A  R circuit / stroombaan 10  I= 5.4 I = 1,85 A  (5) 8.2 For reading of V1 Option 1/ Opsie 1 Option 2 / Opsie 2 = ε - Vr Rext = R// + Rseries / serie V1 = 2,4 + 2 = ε - Ic r  = 4,4 Ω = 10 - (1,852 x 1) V1 = Icircuit / stroombaan x Rext = 8,15 V  = 1,852 x 4,4 ( = 8,15 V (2) 8.3 Increases  When switch S is opened, circuit resistance increases leading to a decrease in I.  Fewer volts lost at “r” hence V1 increases. Toename Wanneer skakelaar S oopgemaak word, verhoog die weerstand van die stroombaan wat lei tot ʼn afname in I. Minder verlore volts by r dus sal V1 vermeerder. (3) [10] 10

Downloaded from hlayiso.com 10841 / 17 QUESTION / VRAAG 9 9.1 AC  / WS Generator has sliprings  / Generator het sleepringe (2) 9.2 Criteria for marking / Kriteria vir nasien -Wave starts at +Vmax or -Vmin  +Vmax -two cycles shown  - golwe begin by +Vmaks of -Vmin - Twee siklusse word gewys - Vmax (2) 9.3 a to b  / a tot b (1) 9.4 9.4.1 It is the DC current which dissipates the same amount of energy as AC.  Dit is die GS wat dieselfde hoeveelheid energie lewer as die WS. (2) 9.4.2 Option 1 Option 2 6 Vmax / maks 6 Imax = = = 0,6 A Vmax / maks 2  = 4,243 V R 10 Vrms = = 2 I max / maks Irms = Vrms 2 Irms =  R 0,6 = 4,243 2 =  10 = 0,424 A = 0,424 A (4) 11

Downloaded from hlayiso.com 10841 / 17 9.4.3 Option 1 / Opsie 1 Option 2 / Opsie 2 Imax Vmax Pave = IrmsVrms Pave =  2 = 0,424 x 4,243 0,6x 6 = 1,8 W =  2 = 1,8 W  Option 3/ Opsie 3 Option 4 / Opsie 4 Pave = Irms2R Pave = Vrms2/R = 0,4242x10 4,242 = = 1,8 W 10 = 1,8 W (3) [14] QUESTION / VRAAG 10 10.1 ∆E = hf = E3 - E1  6,63 x 10-34 f = - 5,44 x 10-19 - (-2,18 x 10-18)  f = 2,468 x 1015 Hz = 2,47 x 1015 Hz  (5) 10.2 3 (1) 10.3.1 The minimum energy that an electron in the metal needs, to be emitted from a certain metal surface.  Die minimum energie wat ʼn elektron in die metaal benodig, om vrygestel te word vanuit ʼn sekere metaaloppervlakte. (2) 10.3.2 hf = W 0 + ½ mv2  6.63x10-34 x 2,468x1015 = 2 x 10-19 + 0.5 x 9,11 x 10-31 v2 v = 1,78 x 106 m·s-1  (4) [12] TOTAL / TOTAAL : 150 12