Physical Sciences Grade 12 P1 Limpopo November 2015 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (Vr1) SEPTEMBER 2015 MEMORANDUM MARKS: 150 TIME: 3 hours This memorandum consists of 10 pages.

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2015 NSC QUESTION / VRAAG 1 1.1 B (2) 1.2 A (2) 1.3 D (2) 1.4 C (2) 1.5 B (2) 1.6 D (2) 1.7 A (2) 1.8 C (2) 1.9 A (2) 1.10 C (2) [20] QUESTION / VRAAG 2 2.1 When a net force (Fnet) is applied to an object (of mass, m) it accelerates in the direction of the (net) force. The acceleration (a) is directly proportional to the (net) force and inversely proportional to the mass of the object. // Wanneer ‘n netto krag op ‘n voorwerp toegepas word, sal die voorwerp versnel. Die versnelling is direl eweredig aan die krag en omgekeerd eweredig aan die massa (2) 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).  // Die netto krag wat op ‘n voorwerp inwerk is gelyk aan die tempo van verandering van momentum van die voorwerp (2 or 0) 2.2 f   5 kg F (3) T  f (friction / fw) T (tension/spankrag) and F (40 N) 2.3 Let the acceleration of the 3 kg mass and trolley be a and let T represent tension. Take right as positive (+) 2.3.1 Consider the 5 kg trolley: T 40 N 5 kg 2N Fnet = ma  40 + (–T) + (–f) = ma 40 – T – 2= 5a  T = 38 – 5a ……….. equation 1 2

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2015 NSC Similarly, consider the 3 kg suspended block: T 3 kg Fg Fnet = ma T + Fg = 3a T – (3 x 9,8)= 3a T = 3a + 29,4 ……….. equation 2 (NB: one mark for 38 – 5a OR 3a + 29,4) Since T is the same:  38 – 5a 3a + 29,4 = 8a = 8,6 a = 1,075 m.s–2  (7) 2.3.2 From eq 1: T = 38 – 5a   = 38 – 5 x 1,075 = 32,625 m.s–2 (2) [14] QUESTION / VRAAG 3  3.1 displacement of the brick/height of scaffolding/final height above the ground // verplasing van die baksteen/ hoogte van steier bokant grond/ (1)  3.2 1,4 – 0,6 = 0,8 s  Motion is parabolic, therefore tx = 1,4 + 0,8 = 2,2 s OR 2,8 – 0,6 = 2,2 s  (2)  3.3 1,4 s (1) 3

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2015 NSC 3.4 3.4.1 Take downward motion as NEGATIVE. (Other option: take downwards as positive) v f  vi  a t  0 = vi + (–9,8) (1,4) vi = –13,72 m.s–1 = 13,72 m.s–1 , upwards  /opwaarts (4) 3.4.2 y  v i t  21 at 2 = (13,72)(2,2)+ ½ (–9,8) (2,2)2 = 6,47 m OR/OF y  v i t  21 at 2  = 0 + ½ (–9,8) (0,8)2  = 3,136 m   y1 = 9,6 – 3,136 = 6,46 m (4) 3.5   9,8 a t (s) (m.s–2) a OR (m.s–2) t (s)   –9,8 (2) [14] QUESTION / VRAAG 4 4.1 T Tension/Fcable/Fapplied/spankrag/FKabel   f  (3) Fg Weight/Gravitational force//gewig 4

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2015 NSC   4.2 Work done = Fapplied ∆x Cos Θ = (800) (12) (1) = 9600 J  (3) 4.3 Work Energy theorem states that, the net/total work done on an object is  to the change in the object's kinetic energyOR the work done on equal  to the change in the object's an object by a resultant/net force is equal kinetic energy.  // Die netto arbeid verrig is gelyk aan die verandering in kinetiese energie/ die arbeid verrig op ‘n voorwerp deur ‘n netto krag is gelyk aan die verandering in kinetiese energie (2) 4.4 W NET =  EK  WG + WA + W f = ½ mvf2 – ½ mvi2  120 x 9,8 x 12 Cos 180 + (9600)+ W f = ½ (120) (52) – ½ (120) 02 W f = 1500 – 9600 + 14112 = 6012 J  OR/ OF WNC = EK + EP  WA + Wf = (½ mvf2 – ½ mvi2) + (mghf – mghi) 9600 + W = (½ (120)(52) – ½ (120) 0) + (120 x 9,8 x 12 – 0) f Wf = 6012 J  (4)   4.5 tension in the cable OR air friction//spankrag OF lugweerstand (1) [13] QUESTION / VRAAG 5  5.1 impulse EQUALS change in momentum //gelyk aan (1)  5.2 Yes. The system is isolated / closed with no external forces acting on person and ball. / Ja dis ‘n geslote/geisoleerde sisteem/geen eksterne kragte werk in op die person of die bal nie (2) 5.3 OPTION 1 OPTION 2 Take direction towards player as Take direction towards player as negative / na speler is - positive / na speler is+   Fnet∆t = m∆v = mvf – mvi  Fnet∆t = m∆v = mvf – mvi  F (0,02) = (0,5) ((9)– (–6)) = 7,5  – (6)) = –7,5 F (0,02) = (0,5) ((–9) F = 375 N  F = –375 N = 375 N   (4) 5.4 Greater. The change in momentum for the ball will be the same but the time of contact will be smaller. // Groter Verandering in momentum dieselfde maar kontaktyd kleiner (3) [10] 5

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2015 NSC QUESTION / VRAAG 6 6.1 6.1.1 The Doppler Effect is the change in the observed frequency (or pitch) of the 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 die waargenome frekwensie/ toonhoogte/ klank waargeneem deur die luisteraar omdat die klankbron en die luisteraar verskillende snelhede het relatief tot die medium van klank voortplanting OR/ OF The change in the (observed) frequency when there is relative motion between the source and the observer.  // die verandering in die waargenome frekwensie wanneer daar relatiewe beweging is tussen die bron en die waarnemer (2) v  vL 6.1.2 Movement towards: fL  fs  v  vs  1560  0 50 x 1000 = x fs 1560 - v s  1560fs vs = 1560 –  ……equation 1 50000 v  vL Movement away: fL  fs v  vs 1560  0 49 x 1000 = x fs  1560  v s 1560fs vs = – 1560 ……equation 2 49000 eq 1  = eq 2 1560fs 1560fs 1560 – = – 1560 50000 49000 fs = 49494,95 Hz  (8) 1560fs 6.1.3 vs = 1560 – 50000 1560x 49494,95  vs = 1560 – 50000 = 15,76 m.s–1  (2) 6.2  off light and absorption Emission spectra occurs when a light source gives spectra occurs when white light is observed through a cold gas  // Emissiespektra vorm wanneer ‘n ligbron lig afgee terwyl absorbsiespektra vorm wanneer wit lig deur ‘n koue gas waargeneem word (2) [14] 6

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2015 NSC QUESTION / VRAAG 7 7.1  Direction/ rigting Q1 Q2  Pattern/ vorm  field lines do not touch / veldlyne raak nie (3) 7.2 Charges can neither be created nor destroyed but transferred from one body to another.   // ladings kan nie geskep of vernietig word nie maar slegs van een voorwerp na ‘n ander oorgedra word The net charge of an isolated system remains constant during any physical process.   // die netto lading in ‘n geslote sisteem bly constant tydens enige fisiese proses (2) 7.3 same position  / dieselfde posisie (1) kQ 7.4 E  r2 = 9 x 109 (7,5 x 10–9)  (10 x 10–3)2  = 675000 N.C–1 to the right or 6,75 x 105 N.C (3) 7.5   7.5.1 Q1 = –7,5 + (–12,5) = –20 n C   Q2 = –7,5 – (–12,5) = + 5 n C (4) 7.5.2 Positive marking from QUESTION 7.5.1 kQ1Q 2 F  r2  9 x 109 x 20 x 10 -9 5 x10 -9 F (30 x 10 -2 ) 2  F = 0,001 N (attraction)  (4)  7.5.3 No // nee (1) [18] 7

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2015 NSC QUESTION / VRAAG 8   8.1 Yes. V2 will be in series to the circuit / no current in the circuit / no energy use in the circuit / V2 is effectively across the battery – no current in the circuit. // Ja V2 sal in serie verbind wees in die stroombaan/ daar is geen stroom in die stroombaan/ geen energie verbruik in die stroombaan nie (2) 8.2 8.2.1 Zero or 0 V  // nul (1) 8.2.2 OPTION 1 OPTION 2  V1Ω = ITotalR = (2,5) (1) = 2,5V V 5 I6    A R 6  V// = 7,5 – 2,5 = 5 V    IR = 2,5 - V// 5 R //    2   I tot 2,5 R= = = 3Ω     R = 3Ω (7) 8.2.3 Internal resistance is the opposition to the flow of charge within a cell  // die teenstand teen die vloei van lading binne-in die battery (2) 8.2.4 OPTION 1 OPTION 2 Vlost = Vtot – V1 = 10 – 7,5 = 2,5 V  V 10 R //  //  4 Vlost = Ir  I tot 2,5 I = (2,5) = 1 Ω   Rext = 2 + 1 = 3 Therefore, r = 1  2,5 OPTION 3 emf = I(R +r)  10 = 2,5(3 + r)  r=1Ω (3) 8.3 Decrease  //afneem (1) [16] 8

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2015 NSC QUESTION / VRAAG 9   9.1 Temperature. Temperature affects resistance // temperatuur omdat temperatuur die weerstand beinvloed (2) 9.2 emf or total potential difference // emk of totale potensiaalverskil (1) 9.3 9.3.1 7V  (allow 6,9 – 7,1) (1) 9.3.2 r can be found by finding the gradient of the graph gradient = ∆ V ∆I = 6–0  0,05 – 0,3  = 1 –0,05  = – 20 Rint = 20 Ω  (other correct values from the graph can be used for the calculation) (4) [8] QUESTION / VRAAG 10 10.1 10.1.1 (split – ring) commutator  //(splitring) kommutator (1) 10.1.2 A – North  /Noord B – South  / Suid (2) I max 10,6  10.2 10.2.1 I rms    7,495A  (3) 2 2 10.2.2 Pave = VrmsIrms  V 300 = max x7,495  x7,495  2 2 = 1589,93 W  (3) 10.2.3 maximum  // maksimum (1) [10] 9

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2015 NSC QUESTION / VRAAG 11  11.1 threshold frequency // drumpelfrekwensie (1) 11.2   (6,63 x 10 x 3 x 108) = 3,5 x 10–19 + Ek(max) –34 429 x 10–9  Ek(max)= (6,63 x 10–34 x 3 x 108) – 3,5 x 10–19 429 x 10–9 = 4,64 x 10–19 – 3,5 x 10–19 = 1,14 x 10–19 J  (5) 11.3.1 same  // dieselfde (1) 11.3.2 same  // dieselfde (1) 11.3.3 increase (or double) //verhoog/ verdubbel (1) 11.4 q = I  t  = 3.2 x 10-7 x 4 = 12,8 x 10-7 C    = = 8 x 1012 electrons (4) [13] Grand total: 150 10