Physical Sciences Grade 12 IEB Paper 1 2018 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2018 PHYSICAL SCIENCES: PAPER I MARKING GUIDELINES Time: 3 hours 200 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 2 of 10 QUESTION 1 1.1 B 1.2 B 1.3 C 1.4 A 1.5 D 1.6 D 1.7 B 1.8 D 1.9 A 1.10 D QUESTION 2 2.1 2.1.1 Velocity is the rate of change of position OR the rate of displacement OR the rate of change of displacement. 2.1.2 s = area under v-t graph OR s = area under v-t graph 1 1 1 29,8 = (3,5)v + (14)v 29,8 = (17,5) v 2 2 2 –1 v = 3,41 m⋅s v = 3,41 ms–1 v 𝑚∆𝑣 2.1.3 a = slope of v-t graph OR (for both formulas) OR 𝐹𝑓 = ∆𝑡 t 0  3,41 (20)(0−3,41) a= (coe from 2.1.2) 𝐹𝑓 = 14 14 a = – 0,24 m⋅s–2 𝐹𝑓 = −4,87 N Ff = ma 𝐹𝑓 = 4,87 N South Ff = (20)(–0,24) Ff = – 4,87 Ff = 4,87 N South 2.2 2.2.1 Stage 1 v = u + at 44 = 0 + a(4) a = 11 m⋅s–2 2.2.2 Stage 1 v2 = u2 + 2as 442 = 02 + 2(11)s1 (coe from 2.2.1) s1 = 88 m Stage 2 v = u + at v2 = u2 + 2as 280 = 44 + a(8) 2802 = 442 + 2(29,5)s2 a = 29,5 m⋅s–2 s2 = 1 296 m Total distance = 88 + 1 296 Total distance = 1 384 m IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 3 of 10 QUESTION 3 3.1 3.1.1 No, velocity is increasing and friction opposes motion / friction slows objects down. 3.1.2 Yes 3.2 3.2.1 Acceleration is the rate of change of velocity. 1 3.2.2 v2 = u2 + 2as OR 𝑣 = 𝑢 + 𝑎𝑡 𝑠 = 𝑢𝑡 + 2 𝑎𝑡 2 (for both eqns) 1 342 = 02 + 2(0,21)s 34 = 0 + 0,21𝑡 𝑠 = 0 + 2 (0,21)(161,91)2 s = 2 752,37 m 𝑡 = 161,91 s 𝒔 = 𝟐𝟕𝟓𝟐, 𝟑𝟕 m 3.2.3 Weight is not the only force acting on the hailstone. OR There are upward forces (e.g. drag) acting on the hailstone. 3.2.4 v2 = u2 + 2as 02 = 342 + 2a(0,12) a = – 4 816,67 a = 4 816,67 m⋅s–2 up 3.2.5 Fnet = Fground – Fg OR Fnet = ma (0,7)(4 816,67) (coe) = Fground – (0,7)(9,8) Fnet = (0,7)(4 816,67) Fground = 3 378,53 N Fnet = 3 371,67 N Fnet = Fground – Fg Fground = 3 371,67 + 6,86 Fground = 3 378,53 N IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 4 of 10 QUESTION 4 4.1 Normal Frictional force Force B on A Force, F Weight diagrams updated (lengths of arrows changed) 4.2 Normal Frictional force Force A on B Weight 4.3 When a net force acts on an object, the object accelerates in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass of the object. OR The net force acting on an object is equal to the rate of change of momentum. 4.4 F cos 36 – (45 + fB on A) = ma OR 134 cos 36 – 45 – fB on A = 23a 4.5 When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A. 4.6 fA on B – 45 = 31 a OR Fnet = ma fA on B = fB on A = f 134 cos 36 – 2(45) = 54a f  45 a= a = 0,34 31 23  f  45  134 cos 36 – 45 – f = f – 45 = 31a 31 f = 55,56 N f = 55,56 N 4.7 The force that opposes the motion of an object. 4.8 Box A experiences the same normal force as box B even though mass of box A is smaller, due to the vertical component of F  Ff = FN greater than expected IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 5 of 10 QUESTION 5 ∆𝑠 5.1 5.1.1 v = slope of x-t graph (or 𝑣 = ∆𝑡 ) 10,2  4,8 v 52 v = 1,8 m⋅s–1 5.1.2 Newton's third law 5.1.3 (ptotal)before = (ptotal)after (0,9)(2,4) + 0 = (0,9 + m)(1,8) (coe) m = 0,3 kg 5.1.4 Momentum is a vector and perpendicular directions are conserved independently. OR Momentum is a vector and velocity of 𝑚 has no horizontal component. 5.2 5.2.1 In the absence of air resistance or any external forces, the mechanical energy of an object is constant. 1 5.2.2 hammer: mgh = mv2 OR v2 = u2 + 2as 2 1 (600)(9,8)(3,5) = (600)v2 v2 = 0 + 2(9,8)(3,5) 2 v = 8,28 m⋅s–1 v = 8,28 ms–1 5.2.3 Work done by a net force on an object is equal to the change in the kinetic energy of the object. 5.2.4 W = EK OR v2 = u2 + 2as 1 Fnet(0,16) = 0 – (600)(8,28)2 0 = 8,282 + 2a(0,16) 2 a = 214,25 ms–2 Fnet(0,16) = (–)20 567,52 Fnet = ma (both eqn) Fnet = (–)128 547 N Fnet = (600)(214,25) Fnet = 128 547 N W 𝑊 5.2.5 P = OR 𝑃= 𝑡 t mgh 𝐹𝑔 𝑠 P= 𝑃= 𝑡 t 180 =  600  9,8  3,5  180 = (5880)(3,5) t 𝑡 t = 114,33 s t = 114,33 s IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 6 of 10 QUESTION 6 6.1 Weight is the gravitational force that the earth exerts on an object (on or near its surface) while mass is the quantity of matter in a body. 6.2 Graph – on answer sheet Heading y-axis title and unit x-axis title and unit 1 scale (plotted points > graph paper) 2 plotted points (accurate and visible to within half a small square) line of best fit Graph to show weight vs mass 3,0 2,5 2,0 weight (N) 1,5 1,0 0,5 0 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 mass (kg) IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 7 of 10 y 6.3 Gradient  x values from y -axis Gradient  (values must be from LOBF on graph) values from x -axis Gradient = 3,76 N⋅kg-1 or 3,76 m⋅s–2 (accept 3,56 – 3,96) 6.4 w = mg  g = gradient g = 3,76 m⋅s–2 6.5 Mars GM 6.6 g= r2 0,58 =  6,7  10  M 11 1,19  10  2 6 M = 1,23 × 1022 kg QUESTION 7 7.1 7.1.1 Resistance is a material's opposition to the flow of electric current. 7.1.2 Emf is the total energy supplied per coulomb of charge by the cell. 7.1.3 emf = I1 (r + 2R) OR 6 = 0,6(r + 2R) 7.1.4 emf = I2( r  R ) 2 OR R 6 = 1,5(r + ) 2 7.1.5 6 = 0,6(r + 2R) 10 = r + 2R r = 10 – 2R (method) R 6 = 1,5(10 – 2R + ) 2 3 6= R 2 R=4 7.1.6 Power is the rate at which work is done. OR the rate at which energy is transferred. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 8 of 10 𝑉2 7.1.7 P = I2R OR 𝑉 = 𝑅𝐼 𝑃 = 𝑅 (for both eqns) 2 P =  1,5  32  (4) 𝑉 = (2)(1,5) 𝑃 = 4 coe  2  P = 2,25 W 𝑉=3V P = 2,25 W 7.1.8 Circuit resistance decreases, current increases. As V = emf – Ir, voltmeter reading decreases. 7.2 7.2.1 Potential difference is the work done per unit positive charge. 7.2.2 neither 7.2.3 bulb A 7.2.4 bulb IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 9 of 10 QUESTION 8 8.1 8.1.1 N Direction Parallel lines P Q S 8.1.2 P to Q 8.2 8.2.1 Lenz’s law states the induced current flows in a direction so as to set up a magnetic field to oppose the change in magnetic flux. 8.2.2 diagram b, clockwise diagram d, anticlockwise 8.2.3 Zero induced current in (c) as the rate of change of magnetic flux is zero. 8.3 8.3.1 The emf induced is directly proportional to the rate of change of magnetic flux (flux linkage). 8.3.2 4,0 3,0 emf (mV) 2,0 1,0 0,0 2 4 6 8 10 12 14 16 18 20 t (ms) −1,0 −2,0 −3,0 −4,0 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 10 of 10 QUESTION 9 9.1 9.1.1 Work function is the minimum amount of energy needed to emit an electron (from the surface of a metal) 9.1.2 f0 = 6,90 × 1014 Hz W0 = hf0 W0 = (6,6 × 10–34)(6,90 × 1014) W0 = 4,55 × 10–19 J 9.1.3 Sketch line parallel to line of best fit and x intercept 8,9 ×1014 – 9,1 × 1014 Hz hc 9.2 9.2.1 E = OR c=f  E=   6,6  1034 3  108  3 × 108 = f(6,58 × 10–7) 7 6,58  10 E = 3,0 × 10–19 J f = 4,56 × 1014 Hz E = hf (for both formula) E = (6,6 × 10–34)(4,56 × 1014) E = 3,0 × 10–19 J 1 9.2.2 Wavelength inversely proportional to energy difference (  ) E Highest E therefore smallest wavelength ∴ Line P 9.2.3 Emission lines are unique to each element. So can be used to identify substances. Total: 200 marks IEB Copyright © 2018