Physical Sciences Grade 12 P1 Western Cape Mock 2018 Memo

Downloaded from hlayiso.com 1 MEMORANDUM PHYSICAL SCIENCE/FISIESE WETENSKAPPE SEPTEMBER 2018 CW PLC COMMON PAPER 1 QUESTION 1 1.1 A (2) 1.2 A (2) 1.3 C (2) 1.4 D (2) 1.5 D (2) 1.6 A (2) 1.7 B (2) 1.8 B (2) 1.9 D (2) 1.10 D (2) [20]

Downloaded from hlayiso.com 2 QUESTION 2 21. 2.1.1 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) 2.1.2 (3) 2.1.3 (a) Commented [1]: If a candidate add the masses (12 kg) directly into Fnet = ma, 2 marks maximum! -1 mark for formula - 1 mark for correct calculation of combined weight (117,6 N) Commented [2]: Accepted Option 2 for resolving two equations: 140,8 - T = 4a………(i) -78,4 + T = 8a ……..(ii) 62,4 = 12 a a = 5,2 ms-2 T = 8 a + 78,4 Commented [3]: Candidate receives full marks for (b) if = 120 N (5) Option 2 is used in (a). Candidate must show the answer in (b) Commented [4]: Yes...answer MUST be shown in (b).

Downloaded from hlayiso.com 3 2.1.3 (b) (2) 2.2 2.2.1 2 x 103 N. √ (1) 2.2.2 Option 1: Distance above Earth’s surface/ Afstand bo Aardoppervlak = Option 2: Commented [5]: I added Option 2 Commented [6]: Thanks (5) Commented [7]: I suggest that the candidate receives full marks for a final answer in meter. Commented [8]: Accepted [18] QUESTION 3

Downloaded from hlayiso.com 4 3.1 g = 9,8 m.s-1 ⩗ downwards/afwaarts ⩗ (2) 3.2 3.2.1 (3) 3.2.2 (3) 3.2.3

Downloaded from hlayiso.com 5 (4) 3.3

Downloaded from hlayiso.com 6 Correct shape/Korrekte vorm ⩗ (NOTE: The graphs are not drawn to scale. The length on the y-axis between 0 and 17 should be longer than the length between 0 and 14,51) (4) (NOTE: The graphs are not drawn to scale. The length on the y-axis between 0 and 17 should be longer than the length between 0 and 14,51) [15]

Downloaded from hlayiso.com 7 QUESTION 4 4.1 The total linear momentum in a closed system remains constant./is conserved / Die totale lineêre momentum in 'n geslote stelsel _bly konstant/bly behoue. √√ OR/OF In a closed/isolated system, the total momentum before a collision is equal to the total momentum after the collision./In 'n geslote/geïsoleerde stelsel is die totale momentum voor 'n botsing gelyk aan die totale momentum na die botsing. (2) 4.2 4.2.1 (4) 4.2.2 (4) 4.3 (1) Greater than/groter as ⩗ [11]

Downloaded from hlayiso.com 8 QUESTION 5 5.1 (3) Option 3: Commented [9]: I added option 3 Wnc = ∆Ep + ∆Ek √ Commented [10]: Thank you 0√ = mg(hf - hi) + ½m(vf2 - vi2) 0 = 3 x 9,8 (0 - 0,15)√ + ½ x 3 (vf2 - 02)√ vf = 5,42 m.s-1 √ 5.2 0 J. Nul resulterende krag/ konstante snelheid/kinetiese energie is die selfde/slegs konserwatiewe kragte is teenwoordig (2) 5.3 Die netto/totale arbeid op ‘n voorwerp verrig is gelyk aan die verandering in die kinetiese energie van die voorwerp OF Die arbeid verrig op ‘n voorwerp deur ‘n resulterende/netto krag is gelyk aan die voorwerp se verandering in kinetiese energie. (2)

Downloaded from hlayiso.com 9 5.4 (3) OR 5.5 Option 3: Wnc = ∆Ep + ∆Ek √ -30√ = mg(hf - hi) + ½m(vf2 - vi2) -30 = 3 x 9,8 (0 - h)√ + ½ x 3 (02 - 5,422)√ h = 0,48 m √ (5) [15] QUESTION 6

Downloaded from hlayiso.com 10 6.1 OR (2) 6.2 6.2.1 170 Hz (1) 6.2.2 130 Hz (1) 6.3 Commented [11]: 6.3: 3/6 if candidate take fs = 150 Hz. (fs is in fact 147,33 Hz) give the first three marks (6) 6.4 ANY ONE/ENIGE EEN Doppler flow meter/Doppler-vloeimeter Measuring foetal heartbeat/Meet van fetale hartslag Measure speed of blood flow/Meet spoed van bloedvloei Ultra sound/Ultraklank Sonar Radar (for speeding/vir jaag) (1) 6.5 The red shift occurs when the spectrum of a distant star moving away from the earth is shifted toward the red end of the spectrum. √√ Rooi verskuiwings vind plaas wanneer die spektrum van 'n vêr afgeleë ster wat vanaf (2) die aarde wegbeweeg na die rooi ent van die spektrum skuif. [13] QUESTION 7

Downloaded from hlayiso.com 11 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 (2) aan die produk van die twee ladings en✓ omgekeerd eweredig aan die vierkant van die afstand tussen die middelpunte.✓ 7.2 (3) 7.3 (5) [10]

Downloaded from hlayiso.com 12 QUESTION 8 8.1 The electric force ✓ per unit charge ✓ at that point. (2) 8.2 Mark allocation: Number of electric field lines from both spheres the same✓ (3) Correct direction of electric field lines ✓ Correct pattern✓ 8.3 OPTION 1:

Downloaded from hlayiso.com 13 OPTION 2: (6) [11]

Downloaded from hlayiso.com 14 QUESTION 9 9.1 I = V/R ✓ = 4/4 ✓ =1 A ✓ (3) 9.2 Opsie 1: ( I parallel) : One branch ( 4 + 4 = 8 Ω ) current is 1 A. The 16 Ω branch current is 1 A/2 = 0,5 A✓ (dubbel resistance means half the current ) I tot = I2 = 1 + 0,5 ✓ = 1,5 A ✓ Opsie 2: Vp = 2 x 4 = 8 V therefore V(R1) = 20 - 8 = 12 V I(R1) = V/R ✓ (3) = 12/8 ✓ = 1,5 A ✓ 9.3 1/Rp = 1/R1 + 1/R2 = 1/(4 +4) + 1/16 ✓ = 3/16 ✓ = 5,33 Ω ✓ or Rp = R1 x R2 / R1 + R2 ✓ (3) = 8 x 16 / 8 + 16 ✓ = 5,33 Ω ✓ 9.4 emf = I ( R + r) ✓ 24 = 1,5 [ (5,33 + 8) + r ] ✓ 24 = 19,995 + 1,5 r r = 2,67 Ω ✓ OR (3) V(int) = Ir. ✓ 4 = 1,5 x r. ✓ r = 2,67 ohm ✓ 9.5 With S open the total resistance increases and the current decreases. ✓

Downloaded from hlayiso.com 15 emf = IR + Ir , emf and r is constant thus Ir will decrease. Therefore V1 will increase ✓ Wanneer S oop is neem die totale weerstand van die stroombaan toe, dus sal die stroom afneem Emk = IR + Ir, emk en r bly konstant daarom sal Ir afneem, dus sal V1 (2) toeneem [14] QUESTION 10 10.1. Move the bar magnet very quickly up✓ and down inside the coil ✓ Beweeg die staafmagneet baie vinnig op (2) en af binne in die spoel. 10.2. 10.2.1 Electromagnetic induction/Elektromagnetiese induksie ✓ (1) 10.2.2 Commutator/kommutator / split ring/splitring ✓ (1) 10.3 10.3.1

Downloaded from hlayiso.com 16 (3) 10.3.2

Downloaded from hlayiso.com 17 (4) [11] QUESTION 11 11.1 11.1.1 Photoelectric effect/foto-elektriese effek✓ (1) 11.1.2 Increase/neem toe✓ An increase in intensity, will result in more electrons being emitted per second/time unit / ‘n toename in intensiteit lei tot meer elektrone wat per (2) sekonde/tydseenheid vrygestel word✓ 11.1.3 Increase in Kinetic Energy/ toename in kinetiese energie✓ Blue light has higher frequency than red light/blou lig het ‘n hoër frekwensie as rooi lig✓ Therefore the energy of blue is higher than red because E = hf/daarom is (3) die energie van blou lig hoër omdat E = hf✓ 11.2

Downloaded from hlayiso.com 18 E = W0 + Kmax ✓ (5) -20 2,95 x 10 ✓= 1 x 10-20 ✓ + 0,5 (9,11 x 10-31) v2 ✓ v = 2,07 x 105 m.s-1 ✓ [11] TOTAAL: 150