Physical Sciences Grade 12 P1 2017 Memo

Physical Sciences/P1 Downloaded from 1 hlayiso.com NSC – Memorandum Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA PHYSICAL SCIENCES P1 MEMORANDUM PREPARATORY EXAMINATION SEPTEMBER 2017 NATIONAL SENIOR CERTIFICATE GRADE 12 N.B. This memorandum consists of 8 pages including this page. Copyright Reserved Please Turn Over

Physical Sciences/P1 Downloaded from 2 hlayiso.com Preparatory Examination 2017 NSC – Memorandum QUESTION 1 1.1 C 1.2 C 1.3 C 1.4 B 1.5 B 1.6 D 1.7 B 1.8 C 1.9 A 1.10 D 10x 2 = [20] QUESTION 2 2.1.1 When a net force acts on an object, the object will accelerate in the direction of the net force with an acceleration that is directly proportional to the net force and inversely proportional to the mass of the object.   (2) OR Net force is equal to a rate of change in momentum. 2.1.2 𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝐹𝑜𝑟𝑐𝑒1 𝑁𝑜𝑟𝑚𝑎𝑙 𝐹𝑜𝑟𝑐𝑒  70°  𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒⬚  (4) 𝑊𝑒𝑖𝑔ℎ𝑡  Copyright Reserved Please Turn Over

Physical Sciences/P1 Downloaded from 3 hlayiso.com Preparatory Examination 2017 NSC – Memorandum 2.1.3 T2 − m2 g = 0 ⃗⃗⃗T2 = m2 g ⃗⃗⃗T2 = (5)(9,8) T2 = 49 N (2) 2.1.4 ⃗Fnet = 0 (on box mass 2kg) T1x = 49 N to the left Using trigonometry to work out T1y : T1x = tan α T1y T1x T1y = tan α 49 T1y = tan 70° T1y = 17,83 N Working with vertical forces (up as positive): ⃗Fnet,y = 0 T1y + N − Fg = 0 17,83 + N − (2)(9,8) = 0 N = 1,77 N (i.e. the magnitude of the normal force is 1,77 N) (4) 2.2 W = mg 152,28= 94g g = 1,62 m.s-2 (downwards)  (3) [15] QUESTION 3 3.1 An object upon which the only force acting is the force of gravity.  (2) 3.2 Upward is positive Downward is positive vf = vi + aΔt vf = vi + aΔt 0 = 5 +(–9,8)Δt 0 = -5 +(9,8)Δt Δt = 0,51 s Δt = 0,51 s (3) 3.3 Upward is positive Downward is positive vf = vi + aΔt vf = vi + aΔt = 8 + (-9,8) (0,51) = -8 + (9,8) (0,51) = 3,00 m.s-1, upwards = 3,00 m.s-1, upwards (3) Copyright Reserved Please Turn Over

Physical Sciences/P1 Downloaded from 4 hlayiso.com Preparatory Examination 2017 NSC – Memorandum 3.4 Ball A Ball B vf = vi + aΔt vf = vi + aΔt = 5 + (-9,8)Δt = 8 + (-9,8)Δt = 5 - 9,8Δt………..(1)  = 8 - 9,8Δt…………..(2)  But Ball A is moving downwards Solving (1) and (2) -(5 - 9,8Δt) = (8 - 9,8Δt) Δt = 0,66 s (Δt = 0,6633 s) (5) Positive marking from Question 3.4 3.5 Δy = vi Δt + ½ a Δt2  = (5)(0,66) + ½ (-9,8)(0,66)2  = 1,17 m  (using Δt = 0,6633 s, Δy = 1,16 m) (3) [16] QUESTION 4 4.1 Principle of Conservation of linear momentum.  The total linear momentum of an isolated system remains constant.  (3) 4.2 To the left is positive Total pbefore = total pafter Total pbefore = total pafter (340)(2.4) = (340)(0.8) + mZvZ (340)(-2.4) = (340)(-0.8) + mZvZ MZvZ = 544 kg.m.s-1 MZvZ = -544 kg.m.s-1 EKbefore = EKafter EKbefore = EKafter ½ (340)(2.4) = ½ (340)(0.8) + ½ mZvZ ½(340)(-2.4)2 = ½(340)(-0.8)2+½ mZvZ2 2 2 2 870.4 = ½ (mZvZ)vZ = ½ (544)vZ 870.4 = ½ (mZvZ)vZ = ½ (-544)vZ -1 vZ = 3.2 m.s (to the left) vZ = 3.2 m.s-1 (to the left) mZ = 170 kg mZ = 170 kg (8) [12] QUESTION 5 Fg = gravitational force or 5.1 FN weight Ff FN = Normal force Ff = frictional force (3) 60o Fg 5.2 Wnet = FnetΔx cosθ = (Fg||+( - Ff))Δx = (343 – 150)(120) = 23160 J  (5) Copyright Reserved Please Turn Over

Physical Sciences/P1 Downloaded from 5 hlayiso.com Preparatory Examination 2017 NSC – Memorandum 5.3 The work done by a net force on an object is equal to the change in the kinetic energy of the object. OR Net work done on an object is equal to the change in the kinetic energy of the object. (2) Mark positively from 5.2 5.4 Wnet = ΔEk = ½ mvf2 – ½ mvi2 23160 = ½ (70)vf2 - 0 vf = 25,72 m.s-1  (4) [14] QUESTION 6 6.1 Doppler Effect.  It is the change in 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.  OR Doppler Effect.  It is the change in the observed frequency of a sound wave when the source of sound is moving relative to the listener. (3) 6.2 56,56 m.s-1 (2) 6.3 v fL  fS  v - vS  340  (5) 298,84  fS 340 - 56,56  f S  250 Hz  6.4 Determine whether arteries are clogged/narrowed  Determine heartbeat of foetus (2) [11] Copyright Reserved Please Turn Over

Physical Sciences/P1 Downloaded from 6 hlayiso.com Preparatory Examination 2017 NSC – Memorandum QUESTION 7 7.1 Force experienced per unit positive charge placed at a point (2) 7.2 Accept more lines on the 5nC Checklist Marks Criteria for electric field Direction  Shape  Field lines not touching each other or entering the spheres  (3) 7.3 E3 = kQ  E5 = kQ r2 r2 = (9x109)(3x109)  = (9x109)(5x109)  (10x10-3)2 (30x10-3)2 = 270 000 N.C-1 (to right) = 50 000 N.C-1  (to left) Enet = 270 000 + (-50 000) = 220 000 NC-1 (6) Mark positively from 7.3 7.4 F = qE = (1,6x10-19)(220 000)  = 3,52 x 10-14 N (to the right)  (4) [15] Copyright Reserved Please Turn Over

Physical Sciences/P1 Downloaded from 7 hlayiso.com Preparatory Examination 2017 NSC – Memorandum QUESTION 8 8.1.1 The current through a conductor is directly proportional to the potential difference across the conductor at constant temperature (2) 1 1 1 8.1.2 = + 𝑅𝑝 𝑅1 𝑅2 1 1 1 = +  𝑅𝑝 2 6 Rp = 1,5 Ω RT = 1,5 + 3,5 = 5 Ω           (3) Mark positively from 8.1.2 8.1.3 emf = I(RT + r) 12  = 2,2 (5 + r) r = 0,45 Ω (4) OR emf - Vext = Vint  (12-11)  = (2,2)r  r = 0,45 Ω       (4) 8.1.4 V2 = IR1    OR V// = IR// = 2,2×1,5 = 3,30 V = (2,2)(3,5)    V2 = VT - V// = 11 – 3,3 = 7,70 V = 7,7 V  (3) Mark positively from 8.1.4 8.1.5 Q = I t  = 2,2 x (5 x 60)  = 660 C OR W = VIΔt W = V Q = 7,7 x 660= 5082 J  = (7,7)(2,2)  (5×60)  = 5082 J (4) OR P = I2R = (2,2)2(3,5)  = 16,94 W OR W = I2RΔt W = Pt = (16,94)(5x60)  = 5082 J = (2,2)2(3,5)  ( 5×60)  = 5082 J (4) OR 2 v W Δt   R 7,7 2  (5  60)   3,5  5082 J  (4) Copyright Reserved Please Turn Over

Physical Sciences/P1 Downloaded from 8 hlayiso.com Preparatory Examination 2017 NSC – Memorandum 8.2 Decreases (1) 8.3 E=P×t = 2 × 2 = 4 kWh C = tariff × E = 1,25 × 4 = R5,00  (3) [20] QUESTION 9 9.1 DC – polarity of emf does not change (2) 9.2 Electromagnetic induction (1) 9.3 Mechanical energy to electrical energy (1) 9.4 Replace split-ring commutator with slip rings (2) 9.5 A (1) 9.6 Friction between moving parts Electrical resistance in wires (2) 9.7 Increase magnetic field strength Increase number of turns on the coil (2) 9.8 Pave= I2rmsR OR P = I2R  1.035×106 = I2rms(490)  1.035×106 = I2(490)  Irms= 45.96 A I= 45.96 A (3) [14] QUESTION 10 10.1 E = hf = 6.63 x 10-34 x 4.65 x 1014 = 3.08 x 10-19 J E < W 0 for gold OR f < f0 for gold E > W 0 for caesium f > f0 for caesium  choose caesium  choose caesium (5) 10.2 There will be a higher current reading on the ammeter.  Each photon ejects one electron, so more electrons per second will be ejected. (2) 10.3.1 E = W 0 + EK  6.6 x 10-34 x 1.21 x 1015= 3 x 10-19 + ½ (9.1 x 10-31) v2  v = 1.05 x 106 m.s-1 (5) 10.3.2 Remain the same (1) [13] GRAND TOTAL: 150 Copyright Reserved Please Turn Over