Physical Sciences Grade 12 Paper 1 Gauteng Mock 2016 Memo

Downloaded from hlayiso.com PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2016 MEMORANDUM PHYSICAL SCIENCES: PHYSICS P1 (10841) FISIESE WETENSKAPPE V1 (10841) 13 pages / 13 bladsye

Downloaded from hlayiso.com 4 10841/16 GAUTENG DEPARTMENT OF EDUCATION GAUTENG DEPARTEMENT VAN ONDERWYS PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN PROOFREAD DATE ___________ PHYSICAL SCIENCES: PHYSICS SIGNATURE _________________ (First Paper) PROOFREAD DATE ___________ FISIESE WETENSKAPPE: FISIKA SIGNATURE _________________ (Eerste vraestel) MEMORANDUM QUESTION / VRAAG 1 1.1 D  (2) 1.2 B  (2) 1.3 A  (2) 1.4 B  (2) 1.5 C  (2) 1.6 A  (2) 1.7 A  (2) 1.8 B  (2) 1.9 D  (2) 1.10 C  (2) [20] QUESTION / VRAAG 2 2.1 When a resultant/net force acts on an object, the object accelerates in the direction of the (net) force at an acceleration directly proportional to the force  and inversely proportional to the mass of the object.  Wanneer ’n resultante /netto krag op ’n voorwerp uitgeoefen word, versnel die voorwerp in die rigting van die (netto) krag met ’n versnelling direk eweredig aan die grootte van die krag  en omgekeerd eweredig aan die massa van die voorwerp.  (2) 2

Downloaded from hlayiso.com 4 10841/16 2.2 180 N/F/Fapplied/applied force  Notes/Aantekeninge: • Correct arrow and label for each 180 N/F/F toegepas/toegepaste krag  force for one mark. / Korrekte pyle en benoeming vir elke krag vir een T/Force of Y on X  punt. • Comparative lengths of arrows are T/Krag Y op X  Weight/mg/w/Fg/39,2 N  not required. / Vergelykende lengte Gewig/mg/w/Fg/39,2 N  van pyle nie vereiste nie. (3) 2.3 2.3.1 For block X / Vir blok X: Note/Aantekeninge: F net = ma  • If the system approach is used to 180 - w – T = ma first calculate acceleration and 180 - (4)(9,8) – T = 4a  then acceleration is substituted to 140,8 – T = 4a ………………(i) obtain T: Max. 2 For block Y/ Vir blok Y: 4 F net = ma • Indien die stelsel benader word T - w = ma om eerstens versnelling te T - (8)(9,8) = 8a  bereken en dan die versnelling -78,4 +T = 8a……………….(ii) gebruik om T te bereken: Maks 2 281,6 – 2T = 8a 4 -78,4 + T = 8a 360 – 3T = 0 T = 120 N T = 120 N upwards / opwaarts  (4) 2.3.2 POSITIVE MARKING FROM QUESTION 2.3.1 / POSITIEWE NASIEN VAN VRAAG 2.3.1 OPTION / OPSIE 1 OPTION / OPSIE 2 -78,4 +T = 8a 140,8 – T = 4a -78,4 + 120 = 8a  140,8 – 120 = 4a  -2 a = 5,2 m∙s  a = 5,2 m∙s-2  OPTION / OPSIE 3 281,6 -2T = 8a 281,6 + 220 = 8a  a = 5,2 m∙s-2  (2) [11] QUESTION / VRAAG 3 3.1 The object had an upward velocity when it was released.  / Die voorwerp het ’n opwaartse snelheid toe dit losgelaat is. OR/ OF The object continues with its state of motion in a straight line. / Die voorwerp beweeg voort in dieselfde rigting van beweging in ’n reguit lyn. OR/ OF The object has inertia. / die voorwerp het inersie. (1) 3

Downloaded from hlayiso.com 4 10841/16 3.2 3.2.1 Upwards / opwaarts (1) 3.2.2 Downwards / afwaarts  (1) 3.3 3.3.1 Upward positive / opwaarts Downward positive/ afwaarts positief positief vf = vi + a ∆ t  vf = vi + a ∆ t  0 = 110 - 9,8Δt  0 = - 110 + 9,8Δt  ∆ t = 11,22 s  ∆ t = 11,22 s  (3) 3.3.2 OPTION 1 Upward positive / opwaarts positief Upward negative / opwaarts negatief Δy = v i Δt + 21 a∆t 2  Δy = v i Δt + 21 a∆t 2  -550  = 110Δt + ½(-9,8)Δt2  550  = -110Δt + ½(9,8)Δt2  4,9Δt2 -110 Δt – 550 = 0 4,9Δt2 -110 Δt – 550 = 0 − ( −110 ) ± ( −110 )2 − 4( 4,9)( −550 ) Δt = 26,66 s or – 4,21 s ∆t = Δt = 26,66 s  2( 4,9) Δt = 26,66 s or – 4,21 s Δt = 26,66 s  OPTION 2 POSITIVE MARKING FROM QUESTION 3.3.1 / POSITIEWE NASIEN VAN VRAAG 3.3.1. Time(point Q to R): 11,22 s Time(point Q to R): 11,22 s Time(point R to Q): 11,22 s Time(point R to Q): 11,22 s Time(point Q to ground): Time(point Q to ground): Upward positive / opwaarts positief Upward negative / opwaarts negatief Δy = v i Δt + 21 a∆t 2  Δy = v i Δt + 21 a∆t 2  -550  = (-110)Δt + ½(-9,8)Δt2  550  = (110)Δt + ½(9,8)Δt2  4,9 Δt2 + 110 Δt – 550 = 0 4,9 Δt2 + 110 Δt – 550 = 0 − 110 ± (110 )2 − 4( 4,9)( −550 ) − 110 ± (110 )2 − 4( 4,9)( −550 ) ∆t = ∆t = 2( 4,9) 2( 4,9) Δt = 4,21 s Δt = 4,21 s Total time = 11,22 + 11,22 + 4,21 Total time = 11,22 + 11,22 + 4,21 = 26,65 s  = 26,65 s  (4) 4

Downloaded from hlayiso.com 4 10841/16 OPTION 3 POSITIVE MAKING FROM QUESTION 3.3.1 / POSITIEWE NASIEN VAN VRAAG 3.3.1. Upward positive / opwaarts positief Upward negative / Opwaarts negatief Height reached above point Q: v f = v i + 2a∆y 2 Height reached above point Q: 2 2 v f = v i + 2a∆y 2 2 2 (0)2 = (110)2 + 2(-9,8) ∆y 2 ∴ ∆y 2 = 617,35 m (0)2 = (-110)2 + 2(9,8) ∆y 2 ∴ ∆y 2 = 617,35 m Displacement from point R to ground: ∆y = 550 + 617,35 = 1 167,35 m Displacement from point R to Δy = v i Δt + 21 a∆t  2 ground: ∆y = 550 + 617,35 = 1 167,35 m -1 167,35  = (0)Δt + ½(-9,8)Δt  2 Δt = 15,435 s Δy = v i Δt + 21 a∆t 2  1 167,35 = (0)Δt + ½(9,8)Δt2 Total time = 11,22 + 15,435 Δt = 15,435 s = 26,66 s  Total time = 11,22 + 15,435 = 26,66 s  3.4 POSITIVE MARKING FROM QUESTION 3.3 / POSITIEWE NASIEN VAN VRAAG 3.3. Downward as positive / afwaarts as positief: Marking criteria/nasienriglyne Positive slope / positiewe helling  Graph starts at v = -110 m·s-1 v (m·s-1)  Grafiek begin by v = -110 m·s-1 Graph intercepts x axis at 11,22 s  Grafiek sny x axis by 11,22 s Graph ends at 26,66 s  0 11,22 26,66  t(s) Grafiek eindig by 26,66 s -110 Upward as positive / opwaarts as positief: 110 Marking criteria 0 11,22 26,66  t(s) Negative slope / negatiewe helling  Graph starts at v = 110 m·s-1 v (m·s-1)  Grafiek begin by v = 110 m·s-1 Graph intercepts x axis at 11,22 s  Grafiek sny x axis by 11,22 s Graph ends at 26,66 s  Grafiek eindig by 26,66 s (4) [14] 5

Downloaded from hlayiso.com 4 10841/16 QUESTION / VRAAG 4 4.1 The total (linear) momentum of a closed system remains constant/is conserved.  (2 marks or zero) Die totale (liniêre) momentum van ’n geslote sisteem bly konstant/bly behoue.  (2 punte or niks) (2) 4.2 Σ pi = Σ pf  m 1 v 2 + m 2 v 2 = (m 1 + m 2 )v c (68 x 20) + (12 x 0) = (68 + 12)v c  ∴ v c = 17 m·s-1  (4) 4.3 POSITIVE MARKING FROM QUESTION 4.2 / POSITIEWE NASIEN VAN VRAAG 4.2. OPTION / OPSIE 1 (E p + E k ) i = (E p + E k ) f  (mgh + ½mv c 2) i = (mgh + ½ mv c 2) f 0 + ½(80)(17)2  = (80)(9,8)h + 0  ∴ h = 14,75 m Distance up the incline, d/ Afstand opwaarts teen skuinsvlak h sin θ = ∆x 14,75 sin25° =  ∆x ∴ ∆ x = 34,89 m  OPTION / OPSIE 2 W net = ΔE k  W w = E kf -E ki 1 mgsin25°cos180° = m(v 2f –v i2 ) 2 1 (80)(9,8)sin25° ∆ xcos180° = (80)(02 - 172)  2 Δx = 34,89 m  (5) 4.4 Decreases / verminder  Friction is a non-conservative force/ opposes motion/removes kinetic energy from the system.  Wrywing is ‘n nie-konserwatiewe krag /opponeer beweging / verwyder kinetiese energie van die sisteem.  (2) [13] 6

Downloaded from hlayiso.com 4 10841/16 QUESTION / VRAAG 5 5.1 Zero/0 (J)  (1) 5.2 FN/N/Normal Force of the engine/F/FA / Normaal  Krag van enjin /F/FA   Frictional force/f/Ff Wrywingskrag /f/Ff  weight/w/mg/Fg/6 860 N gewig/w/mg/Fg/6 860 N  (4) 5.3 5.3.1 The net/total work done on an object is equal to the change in the object’s kinetic energy.  (2 marks or zero.) Die netto/totale werk verrig op ’n voorwerp is gelyk aan die verandering in die voorwerp se kinetiese energie.  (2 punte or niks.) (2) 5.3.2 Marking guidelines / nasienriglyne • Formula/ Formule: W net = ∆E k or W nc = ∆E p + ∆E k  • Formula / Formule: W = F ∆ xcos θ  • Substitution to calculate / Vervanging in formulef = µ k mgcos30°  • Substitute to calculate / Vervanging om te bereken: W f  • Substitute to calculate / Vervanging om te bereken: W w OR ΔE p  • Substitute to calculate / Vervanging om te bereken: W F  • Substitute to calculate / Vervanging om te bereken: ∆E k  • Final answer / Finale antwoord: 11,59 m·s-1  OPTION / OPSIE 1: W net = ∆E k  W N + W f + W w + W F = ∆E k 0 + f ∆ xcos 180° + w ∆ xcos 180° + F ∆ xcos 0°  = E kf - E ki 1 µ k mgcos30° ∆ xcos 180° + mg ∆ xsin30°180° + F ∆ xcos 0° = m(v f 2 –v i 2) 2 0,32(700)(9,8)cos30°(70)cos 180° + (700)(9,8)sin30°(70)cos180°+ (6000)(70)cos 0° = 1 (700)(v 2f – 0 )  2 5 5 5 2 -1,33 x 10 – 2,4 x 10 + 4,2 x 10 = 350v f v f = 11,59 m·s-1  OPTION / OPSIE 2: W nc = ΔE p + ΔE k  W f + W F = mgΔh + ½ m(v f 2 – v i 2) f ∆ xcos 180° + F ∆ xcos 0°  = mgΔh + ½ m(v f 2 – v i 2) 0,32(700)(9,8)cos30°(70)cos 180°  + (6000)(70)cos 0° = (700)(9,8)(70sin30° – 0)  + ½ (700(v f 2 – 0)  v f = 11,59m∙s-1  (8) [15] 7

Downloaded from hlayiso.com 4 10841/16 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.  Dit is die skynbare verandering in frekwensie van ’n bron wanneer daar ’n relatiewe verandering in beweging tussen die bron en die waarnemer is.  (2) 6.1.2 Away / Weg (1) 6.1.3 v ± vL fL = fs  v ± vS  340 73 = 75  340 + v s v s = 9,32 m ⋅ s −1  (4) 6.2 • The absorption spectrum observed for elements on a distant star compared to spectrum of elements on the earth or sun.  • Absorption lines in spectrum from star shifted to red end of the spectrum/ red shifted.  • Wavelengths in the absorption spectrum lines have increased.  • According to the Doppler effect the star is thus moving away from the earth.  • The absorbsiespektrum waargeneem vir elemente van ’n verafgeleë ster in vergelyking met die spektrum van elemente nader aan die aarde of die son.  • Absorbsielyne in die spektrum van ster verskuif na die rooi kant van die spektrum/ rooiverskuiwing.  • Golflengte in the absorbsiespektrumlyne vermeerder.  • Volgens die Doppler effek beweeg die ster dus verder van die aarde.  (4) 6.3 Ultrasound scans / ultraklankskandering Measuring rate of flow of blood (Blood-flow meter) / Meting van die tempo waarteen bloed vloei (bloedvloeimeter) (2) [13] 8

Downloaded from hlayiso.com 4 10841/16 QUESTION / VRAAG 7 7.1 The force of attraction that two charges at rest exert on each other is directly proportional to the product of the two charges and inversely proportional to the square of the distance between their centres.  Die aantrekkingskrag wat twee ladings in rus op mekaar uitoefen is direk eweredig aan die produk van die twee ladings  en omgekeerd eweredig aan die vierkant van die afstand tussen die middelpunte.  (2) 7.2 F12  Fnet F13  (2) 7.3 kQ1Q 2 F12 = 2  r12  (9 × 10 9 )(2 × 10 − 5 )(2 × 10 − 4 ) = (0,5)2 = 144 N, to the right / na regs kQ1Q3 F13 = 2 r13  (9 × 10 9 )(2 × 10 − 5 )(2 × 10 − 4 ) = (0,4)2  = 225 N, downwards / afwaarts F net = (225 )2 + (144 )2 = 267,13 N 225 tan θ = ∴ θ = 57,38° 144 F net = 267,13 N,  147,38° / E57,38°S / 57,38° south of east / suid van oos  (7) [11] 9

Downloaded from hlayiso.com 4 10841/16 QUESTION / VRAAG 8 8.1 12 V  (1) 8.2 8.2.1 OPTION / OPSIE 1 OPTION 2 / OPSIE 2 1 1 1 R1R 2 = +  Rp =  R p R1 R 2 R1 + R 2 1 1 (6)(10) = +  =  6 10 6 + 10 Rp = 3,75 Ω Rp = 3,75 Ω R external = 3,75 + 2  = 5, 75 Ω  R external = 3,75 + 2 = 5,75 Ω  (4) 8.2.2 POSITIVE MARKING FROM QUESTION 8.2.1 / POSITIEWE NASIEN VAN VRAAG 8.2.1. OPTION / OPSIE 1 OPTION / OPSIE 2 V 1 = IR ext  V 1 = IR ext  10 10 ∴I =  ∴I =  5,75 5,75 = 1,74 A = 1,74A ε = IR ext + Ir  V lost = Ir  12 – 10 = 1,74r  12 = (1,74)(5,75) + 1,74r  r = 1,15 Ω  ∴ r = 1,15 Ω  (5) 8.3 Increases / Verhoog  Closing S 2 effectively cuts off the 2 Ω resistor / is a short circuit. / Sluit van S 2 sny die 2 Ω weerstand effektief af  Total resistance decreases / Totale weerstand verminder.  (3) [13] QUESTION / VRAAG 9 9.1 9.1.1 AC / Alternating current / Wisselstroom  (1) 9.1.2 DC  Due to the presence of the split ring commutator, current always exits through the one brush and then through the other brush.  Agv die teenwoordigheid van die splitringkommutator, is daar altyd stroom wat deur een borsel beweeg en dan deur die ander borsel.  (2) 9.2 9.2.1 AC generator has slip rings  whilst a DC generator has a split ring commutator.  AC generator het slipringe,  terwyl die DC generator splitringkommutator het.  (2) 10

Downloaded from hlayiso.com 4 10841/16 9.2.2 An AC voltage can be stepped up. / ’n AC stroom kan verhoging ondergaan.  Increase in voltage and decrease in current. / Verhoging in potensiaalverskil en verlaging in stroom.  Loss in power/P = I2R decreases. / Verlies aan drywing/P = I2R verminder  (3) 9.3 9.3.1 OPTION / OPSIE 1 OPTION / OPSIE 2 1 cycles f= f= T time 1 3,5 =  =  0,02 0,07 = 50 Hz  = 50 Hz  (2) 9.3.2 OPTION 1 OPTION 2 V V Imax = max  Vrms = max  R 2 200 200  =  = 20 2 = 10 A = 141,42 V P ave = ½(V max I max )  Vrms = ½(200 x 10 )  Irms = = 1000 W (1 kW)  R 141,42 OPTION / OPSIE 3 = 20 V = 7,07 A Vrms = max  2 200 Pave = I rms 2 =  R 2 = (7,07)220 = 141,42 V = 999,70 W OR 2 Pave = Vrms I rms  Vrms Pave =  = (141,42)(7,07)  R = 999,84 W  (141,42)2  = 20 Accept range / Aanvaar: = 999,98 W  999,70 to 1000 W (5) [15] 11

Downloaded from hlayiso.com 4 10841/16 QUESTION / VRAAG 10 10.1 Minimum energy required to release (photo) electrons from a metal surface. Minimum energie nodig om (foto)elektrone uit ’n metaaloppervlak vry te stel.  (2) 10.2 OPTION / OPSIE 1 OPTION / OPSIE 2 W o = hf  c f=  = hc  λ λ 3 × 10 8 (6,63 × 10 −34 )(3 × 10 8 )  =  = 500 × 10 − 9 500 × 10 − 9  = 6 × 1014 Hz = 3,978 x 10-19J  Wo = hf  = (6,63 × 10 − 34 )(6 × 1014 )  = 3,98 × 10 −19 J  (5) 10.3 10.3.1 Remains the same / Bly dieselfde  (1) 10.3.2 Remains the same / Bly dieselfde  (1) 10.4 POSITIVE MARKING FROM QUESTION 10.2 / POSITIEWE NASIEN VAN VRAAG 10.2. OPTION / OPSIE 1 OPTION / OPSIE 2 E = Wo + Ek(max)  c f= hc λ = W o + E k(max) λ 3 × 10 8 =  400 × 10 − 9  (6,63 × 10 −34 )(3 × 10 8 )  = 7,5 × 1014 Hz −9 = 3,978 x 10-19 + E k(max) 400 × 10  E = hf E k(max) = 9,95 x 10-20 J  = (6,63 × 10 − 34 )(7,5 × 1014 )  = 4,97 × 10 −19 J Range/ aanvaar: 9,93 x 10-20 J to 9,95 x 10-20 J E = Wo + Ek(max)  4,97 x 10-19 = 3,98 x 10-19 + E k(max)  E k(max) = 9,93 x 10-20 J  (5) [14] 12

Downloaded from hlayiso.com 4 10841/16 QUESTION / VRAAG 11 11.1 The energy transferred to / work done on  each coulomb (of charge) / per C charge  passing through the battery. Die lading oorgedra / arbeid verrig op elke coulomb (lading) / per C lading  wat deur die battery beweeg. (2) 11.2 8 − 0,5 3, Note/Aantekening: Gradient/helling = = 0,37 (V-1) 9−0 • Accept any correct values from graph  to calculate the gradient. • Aanvaar enige korrekte waarde vanaf grafiek om helling te bereken. (3) 11.3 1  emf (1) 11.4 11.4.1 POSITIVE MARKING FROM QUESTION 11.2 / POSITIEWE NASIEN VAN VRAAG 11.2. OPTION /OPSIE 1 OPTION / OPSIE 2 1 emf = 2(0) + 2r = 0,37  emf emf = 0,26(9) + 0,26r ε (emf) = 2,70 V  0 = - 2,34 + 1,74r r = 1,34 Ω ε = 2(0) + 2r = 2(1,34) Range/Aanvaar: = 2,67 V  2,67 – 2,70 V (2) 11.4.2 POSITIVE MARKING FROM QUESTION 11.2 AND 11.3.1 / POSITIEWE NASIEN VAN VRAAG 11.2 EN 11.3.1. OPTION / OPSIE 1 OPTION / OPSIE 2 Emf = I(R + r) Any two equation using values from the 1 1 r graph/ Enige twee vergelykings waar y = mx + c: = R+ waardes van grafiek gebruik is: I emf emf r y int ercept / afsnit = emf = 2(0) + 2r  emf emf = 0,26(9) + 0,26r  r 0,5  =  0 = - 2,34 + 1,74r 2,70 r = 1,34 Ω  R = 1,35 Ω  Range/ aanvaar: 1,34 - 1,35 Ω (3) [11] GRAND TOTAL: 150 (s) only. If you have received this email in error p 13