# Physical Sciences Grade 12 Paper 1 Gauteng Mock 2018 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Gauteng Mock Exam 2018, 13 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gp-2018-paper-1-memorandum/23750489-743e-47ce-ab89-fb6303430e77 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2018 - Exam period: Gauteng Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 13 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 Paper 1 Gauteng Mock 2018 Memo](https://www.hlayiso.com/library/gp-2018-paper-1-memorandum/23750489-743e-47ce-ab89-fb6303430e77) ## About this document Physical Sciences · Grade 12 · Gauteng Mock Exam · 2018. Memorandum, 13 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com PREPARATORY EXAMINATION 2018 MARKING GUIDELINES PHYSICAL SCIENCES: PHYSICS (PAPER 1) (10841) 13 pages ### Page 2 Downloaded from hlayiso.com 10841 / 18 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION – 2018 PHYSICAL SCIENCES: PHYSICS (Paper 1) MARKING GUIDELINES QUESTION 1 1.1 B PP 1.2 C PP 1.3 B PP 1.4 D PP 1.5 B PP 1.6 D PP 1.7 B PP 1.8 B PP 1.9 D PP 1.10 C PP [20] QUESTION 2 2.1 The total mechanical energy (sum of the potential energy and kinetic energy) in a system remains constant as long as the only forces acting are conservative forces. PP (2) 2.2 At point A Emech = Ep + Ek P = mgh + 0 = 50 x 9,8 x 5 P = 2 450 J At point B Emech = Ep + Ek 2 450 = 0 + ½ mvi2P vi = 9,899 m·s-1 P (4) 2.3 The total linear momentum of a closed system remains constant (is conserved). PP (2) POSITIVE MARKING FROM 2.2 2 ### Page 3 Downloaded from hlayiso.com 10841 / 18 2.4.1 Sp(before) = Sp(after) mdvd + mwvw = (md + mw)vc 50 x 9,899 + 0 P = (50 + 60 )vc P vc = 4,4995 vc = 4,50 m·s-1 P (3) 2.4.2 Impulse, Dp = mdDv = 50 x (9,899 – 4,4995) P = 269,98 kg·m·s-1 P (2) 2..4.3 Wnet = DEk = Fnet xDxCosq Ekf - Eki = Fnet xDxCosq P 2 0 - ½ x110x 4,50 P = 60 x DxCos1800 P Dx = 18,56 m P (4) [17] QUESTION 3 3.1 “free fall” is the motion of a body when the only force acting on the body is gravity/pull due to gravity/its weight. PP (2) 3.2 vf2 = vi2 + 2aDy P vf = (102 + (2x9,8x20)) P = 22,181 m·s-1P (3) 3.3 vf = vi + aDt P 22,181 - 10 Dt = P 9,8 = 1,243 s P (3) 3.4 Ball rises 19 m P after the bounce hence v f - vi tup = vf2 = vi2 + 2aDy a 02 = vi2 + 2 ( 9,8)( -19) P 0 - ( -19,298 = 9,8 vi = 19,298 m·s-1 P = 1,969 s (3) 3.5 The decrease in the magnitudes of the velocities, hence the loss in kinetic energy shows that the collision between the ball and the ground is inelasticPand energy is lost through sound and heatP (2) 3 ### Page 4 Downloaded from hlayiso.com 10841 / 18 3.6 Down is positive Marking criteria Graph starts at 10 m·s-1 P -1 22,18 m·s P Graph ends at 22,18 m·s-1 P 1,243 s indicated P 10m·s-1 P Graph starts at -19,30 m·s-1 P P Graph ends at v =0, (time 3,212 s) P 1,243s P Positive slope for both graph sections P Accept if this line is tilted -19,30m·s-1 P Upward is positive Marking criteria 19,30 m·s-1 Graph starts at 10 m·s-1 P Graph ends at 22,18 m·s-1 P 1,243 s indicated P 1,24 s Graph starts at -19,30 m·s -1 P -10 m·s-1 Graph ends at v =0 (time 3,212 s) P Negative slope for both graph sections P -22,18 m·s-1 (6) [19] 4 ### Page 5 Downloaded from hlayiso.com 10841 / 18 QUESTION 4 4.1 The net/total work done on an object is equal to the change in the object’s kinetic energy PP (OR the work done on an object by a resultant/ net force is equal to the change in the object’s kinetic energy ) (2) 4.2 OPTION 1 OPTION 2 N/FN/force of surface on objectP N/FN/force of surface on objectP f/Ff/force of frictionP f/Ff/force of frictionP Fg///wsinqP Fg/w/force of earth on objectP Fg^/ wcosq (3) 5 ### Page 6 Downloaded from hlayiso.com 10841 / 18 4.3 4.3.1 OPTION 1 OPTION 2 vf2 = vi2 + 2aDxP Dx = viDt + ½ aDt2 P = 02 + 2 x 2 x 8 P 8 = 0xDt + 0.5x2xDt P -1 vf = 5,657 m·s Dt = 2,828 Ek = ½ mv2 P Vf = vi + aDt = 0,5 x 40 x 5,6572P = 0 + 2x2,838 P = 640,03 J P = 5,657 m·s-1 Ek = ½ mv2 P = 0,5 x 40 x 5,6572 P = 640,03 J (5) OPTION 1 OPTION 2 4.3.2 Wg = Fg//Dxcosq P h = 8 sin 300 = 4 m = (mg sin q) Dx cosq Ep = mghP = 40x9,8 x sin 300 x 8 x cos00P = 40 x 9,8 x 4P = 1 568 J P = 1568 JP (4) POSITIVE MARKING FROM 4.3.1 & 4.3.2 4.3.3 Wnet = DEKP Wg + Wf = EKf - EKi 1 568 + Wf P = 640,03 - 0 P \ Wf = - 927,97 J P (4) 4.3.4 Wf = FfDxcos q P 927,97= Ff x 8 cos 1800P Ff = -115,996 N = 115,996 N P (3) [21] 6 ### Page 7 Downloaded from hlayiso.com 10841 / 18 QUESTION 5 5.1 The Doppler effect 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. PP (2) v ± vL 284,4 5.2 For car A; FL = ( )fs, P vs = = 79 m·s-1P v ± vs 3,6 340 + 0 = ( P)1200 P 340 - 79 = 1563,218 Hz. P (5) v ± vL 5.3 For car B; FL = ( )fs, v ± vs 340 + 0 1600 P = ( P)1170P 340 - v s vs = 91,375 m·s-1 P (4) 5.4 With sound • The velocity of sound in air is comparable with speed of the ambulance. P The change in frequency will therefore be significant hence a noticeable change. P With light • The velocity of light is too high compared to the speed of the ambulance. P The change in frequency will be insignificant hence no noticeable change. P (4) [15] 7 ### Page 8 Downloaded from hlayiso.com 10841 / 18 QUESTION 6 6.1 What is the relationship between the potential difference across a resistor and the current through the resistor? PP (2) 6.2 Marking criteria Components except voltmeter P connected in series. Voltmeter connected across R. P All symbols correct. P A R V (3) 6.3 6.3.1 Potential difference (V ) P (1) 6.3.2 Current (I) P (1) 6.3.3 Temperature of the fixed resistor P (1) 6.4 Take readings quickly/ allow time between readings so that resistor coolsP (1) 6.5 6.5.1 Potential difference/V P 0 1 2 3 4 5 current/A P 0 0,29 0,6 0,9 1,22 1,48 All values paired correctly P (3) 8 ### Page 9 Downloaded from hlayiso.com 10841 / 18 6.5.2 Graph of voltage vs current for a fixed resistor 6 5 potential difference/V 4 3 2 1 0 0 0,2 0,4 0,6 0,8 1 1,2 1,4 1,6 Current/A Marking criteria Axes labelled P Scales on both axes correct P At least 3 plots correct P Straight lineP passing through the origin (4) 6.5.3 The current through the resistor is directly proportional to the potential difference across the resistor if the temperature of the resistor is kept constant. PP (2) 6.5.4 - If V vs I as drawn above, gradient represents magnitude of the resistance of the resistor. P - If I vs V is plotted, then gradient represents the reciprocal of the resistance of the resistor. (1) [19] 9 ### Page 10 Downloaded from hlayiso.com 10841 / 18 QUESTION 7 (Start on a new page.) 7.1 V Rext = P (V = terminal pd) I 12 = P 1,2 = 10 W P (3) 7.2 Option 1 Option 2 1 1 1 1 1 1 = + P R // R 4 R 6 R // = R4 + R6 4x 6 4x 6 R// = P R// = 4+6 4+6 = 2,4 W = 2,4 W V// = R// I R = RT - R//P =2,4 x 1,2 = 10 - 2,4 =2,88 W = 7,6 W P VR = 12 – 2,88 = 9,12 VR 9,12 R = = I 1,2 Option 3 = 7,6 W e = I(Rext + r) P = 1,2 (2,4 + R 1,5 ) P 13,8 = 1,2R + 4,68P R = 7,6 WP (4) 10 ### Page 11 Downloaded from hlayiso.com 10841 / 18 7.3 Option 1 Option 2 Vr = Ir = 1,2x1,5 = 1,8 V P V// = IR// = (1,2)(2,4) = 2,88 V P \ emf = VT + Vr VR = IR = (1,2)(7,6) = 9,12 V P = 12 + 1,8 P Vr = 1,8 V = 13,8 V P emf = V// + VR + Vr = 2,88 + 9,12 + 1,88 = 13,8 V P (max = 3 marks) (3) 7.4 Increase P (1) 7.5 • Total resistance decreases P • Current increasesP (2) [13] 11 ### Page 12 Downloaded from hlayiso.com 10841 / 18 QUESTION 8 8.1 DC (generator) P (1) 8.2 split ring commutator P (1) 8.3 a to b P (1) 8.4 rms current is that a.c. current that would have the same effect as its dc V equivalent PP OR (Irms = max ) 2 (2) 8.5 Vmax 8 Vrms = = = 5,657 V P 2 2 Vrms Irms = P R 5,657 = P 5 = 1,131 A P (4) 8.6.1 Marking guideline emax - Starts max P - All quarters positive P 0 - 1 cycle P time/s (3) 8.6.2 No P (1) 8.6.3 induced emf drives current through the conductors P such that emf and current are in phase, P therefore shape of graphs is the same. (2) [15] 12 ### Page 13 Downloaded from hlayiso.com 10841 / 18 QUESTION 9 9.1 The photoelectric effect is a phenomenon whereby electrons are ejectedP from a metal surface when light (EM radiation) with a frequency equal to or greater than the threshold frequency of the metal is shone on it. P (2) 9.2 Work function is the minimum energy that an electron in the metal needs to be emitted from the metal surface. PP (2) 9.3 c E = hfo = h P lo 3x108 = 6,63 x 10-34 Px P 600x10 -9 = 3,315 x 10-19 J P (4) 9.4 Yes P (1) 9.5 Frequency of blue light is much higher than that of orange light Pmaking blue light photons carry more than sufficient energy to eject photoelectrons from the metal surfaceP OR Since W0 is constant the additional energy is transferred to the photoelectrons as kinetic energy. Therefore the ammeter will register a reading. 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