Physical Sciences Grade 12 IEB Paper 1 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2019 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 © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 2 of 12 QUESTION 1 1.1 D 1.2 C 1.3 B 1.4 B 1.5 C 1.6 A 1.7 B 1.8 D 1.9 C 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 12 m ⋅ s –1 North 2.1.3 s = area under v-t graph 1 s = (10)(12) + (25 − 10)(12) 2 s = 240 m total s average v = total t 240 average v = 25 = average v 9,6 m ⋅ s –1 2.1.4 Acceleration is the rate of change of velocity. ∆v 2.1.5 a = slope of v-t graph OR ∆t 12 − 0 a= 10 − 0 =a 1,2 m ⋅ s –2 1 2.2 Helen: s= ut + at 2 2 s = ( 0,37 ) t 1 ( 0,91)( t − 1,8 ) 2 Matthew: s 0+ = 2 1 Combined: ( 0,37 ) t ( 0,91)( t − 1,8 ) 2 = 2 t = 3,48 s IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 3 of 12 1 OR Helen: 𝑠𝑠 = 𝑢𝑢𝑢𝑢 + 2 𝑎𝑎𝑡𝑡 2 𝑠𝑠 = 0,37(𝑡𝑡 + 1,8) 1 Matthew: 𝑠𝑠 = 0 + 2 (0,91)𝑡𝑡 2 1 Combined: 0,37(𝑡𝑡 + 1,8) = 2 (0,91)𝑡𝑡 2 𝑡𝑡 = 1,68 s Helen: 𝑡𝑡 = 1,8 + 1,68 𝒕𝒕 = 𝟑𝟑, 𝟒𝟒𝟒𝟒 s 1 OR Helen: 𝑠𝑠 = 𝑢𝑢𝑢𝑢 + 2 𝑎𝑎𝑡𝑡 2 𝑠𝑠 = (0,37)(1,8) 𝑠𝑠 = 0,66 m 1 Matthew: 𝑠𝑠 = 0 + 2 (0,91)𝑡𝑡 2 1 Combined: 0,66 + 0,37𝑡𝑡 = 2 0,91𝑡𝑡 2 𝑡𝑡 = 1,68 s Helen: 𝑡𝑡 = 1,8 + 1,68 𝒕𝒕 = 𝟑𝟑, 𝟒𝟒𝟒𝟒 s IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 4 of 12 QUESTION 3 3.1 3.1.1 A is greater than B, speed = gradient of position-time graph and gradient A > gradient B. 3.1.2 Yes 4 s (accept 3 s – 5 s) 3.1.3 Object A 1 3.2 3.2.1 s= ut + at 2 2 1 =s 2 ( 4,32 ) + ( −9,8 ) (4,32)2 2 s = −82,81 m ∴ Helicopter 82,81 m above trampoline 3.2.2 v= u + at OR 𝑣𝑣 2 = 𝑢𝑢2 + 2𝑎𝑎𝑎𝑎 v = 2 + ( −9,8)(4,32) 𝑣𝑣 2 = 22 + 2(−9,8)(−82,81) v = −40,34 𝑣𝑣 = 40,34 =v 40,34 m ⋅ s –1 down = v 40,34 m ⋅ s –1 down 3.2.3 Force due to trampoline Force due to trampoline > force due to weight Weight 3.2.4 Upward as he is launched back into the air OR Upward as he experiences a net force up IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 5 of 12 QUESTION 4 4.1 4.1.1 Normal Frictional force Weight 4.1.2 Fg slope = mg sinθ 52,52 = 48 ( 9,8 ) sinθ sinθ = 0,111 θ 6, 41° = 4.1.3 FN = mg cos θ Ffk = µk FN Ffk = ( 0,12 )( 48 )( 9,8 ) cos ( 6,41) Ffk = 56,10 N 4.1.4 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. 4.1.5 Fnet = ma Fg slope + Ffk = ma 52,52 + 56,10 48a= −2 =a 2,26 m ⋅ s down slope 4.1.6 Greater from A to B Friction and component of weight in the same direction So resultant force is more OR Less from B to A As friction acting up slope, opposite to component of weight So resultant force is less IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 6 of 12 4.2 Answer Sheet 4.2.1 Point C BC parabola 4.2.2 Straight line onwards from C In same direction as before C IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 7 of 12 QUESTION 5 5.1 5.1.1 Velocity 5.1.2 Graph – on answer sheet Heading y-axis title and unit x-axis title and unit 1 scale (plotted points > graph paper) 2 plotted points line of best fit Graph to show time taken to stop vs initial velocity ∆y 5.1.3 gradient = ∆x ∆y gradient = ∆x (values must be from LOBF on graph – not data points) gradient = 0,61 s2 ⋅ m−1 (accept 0,57 – 0,66) IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 8 of 12 5.1.4 Fnet ∆t = m∆v Fnet ∆t = m ( 0 − v ) m ∆t =− v Fnet m 5.1.5 gradient = − Fnet m 0,61 = − −140 m = 85, 4 kg 5.2 5.2.1 Kinetic energy is the energy an object has as a result of the object's motion. 1 5.2.2 At bottom: EK = mv 2 2 1 EK = ( 60 ) (7,5)2 2 EK = 1 687,5 J 5.2.3 At top: EP = mgh EP = (60)(9,8)(2,8) EP = 1 646, 4 J 5.2.4 Work done vs friction = 1 687,5 – 1 646,4 work done vs friction = 41,1 J 2,8 5.2.5 sin25 = s s = 6,63 m Wfriction = Ff s 41,1 = Ff ( 6,63 ) Ff = 6,20 N IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 9 of 12 QUESTION 6 6.1 6.1.1 • pattern • direction + 6.1.2 Coulomb's law states that two-point charges in free space or air exert forces on each other. The force is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges. 6.1.3 Fup = Fdown kq1q2 = mg r2 ( )( )( 9 × 109 30 × 10 −9 100 × 10 −9 ) h2 = (1,5 × 10 ) ( 9,8 ) −3 h = 0,043 m kQ 6.2 6.2.1 E6 µC = r2 E6 µC = ( 9 × 109 )( 6 × 10−6 ) ( 0,09 ) 2 E= 6 µC 6,67 × 106 N ⋅ C−1 kQ 6.2.2 E−2 µC = r2 E−2 µC = ( 9 × 109 )( 2 × 10−6 ) ( 0,04 ) 2 E−2 µC = 1,125 × 107 N ⋅ C−1 Enet = 1,125 × 107 − 6,67 × 106 E net = 4,58 × 106 N ⋅ C−1 towards charges or right IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 10 of 12 QUESTION 7 7.1 7.1.1 The potential difference is the work done per unit positive charge. 7.1.2 15 V 7.1.3 15 V 7.1.4 = V1 emf − Ir so as switch closes, I increases, more lost volts. OR V1 emf − Vlost is lost volts, as switch closes, I ≠ 0, more lost volts. = 1 1 1 7.1.5 = + RP R1 R2 1 1 1 = + RP 6 12 RP = 4 Ω 7.1.6 V = IRT = I (1 + 4 ) 10,5 I = 2,1 A 7.1.7 = V emf − Ir = 15 − 2,1r 10,5 r = 2,14 Ω V2 7.2 7.2.1 P = R 502 100 = R R = 25 Ω 𝑉𝑉 7.2.2 P = VI OR 𝑃𝑃 = 𝐼𝐼 2 𝑅𝑅 OR 𝐼𝐼 = 𝑅𝑅 50 100 = 50 I 100 = 𝐼𝐼 2 (25) 𝐼𝐼 = 25 I =2A I =2A I =2A 7.2.3 V = RT I = 220 ( R + 25 ) 2 R = 85 Ω IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 11 of 12 QUESTION 8 8.1 8.1.1 Into the page. 8.1.2 In the same direction as B field shown half the magnitude (only if weaker) 8.2 Generator Transformer (Microphone) (Induction stove) Any 2 correct 8.3 8.3.1 solenoid N Direction Pattern I 8.3.2 Faraday's law states that the emf induced is directly proportional to the rate of change of magnetic flux (flux linkage). 8.3.3 No relative movement – rate of change of flux is zero 8.3.4 Move coil or solenoid relative to each other. Rotate the coil so effective area changes. Continuously change the current in the solenoid. [21] IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 12 of 12 QUESTION 9 9.1 Threshold frequency is the minimum frequency of incident radiation at which electrons will be emitted from a particular metal. 9.2 c = fλ 3 × 10= 8 ( f0 583 × 10 −9 ) = f0 5,15 × 1014 Hz 9.3 W0 = hf0 ( )( 6,6 × 10 −34 5,15 × 1014 W0 = ) −19 = W0 3,40 × 10 J hc 9.4 = W0 + EKmax λ ( 6,6 × 10 )( 3 × 10 ) = 3,40 × 10 −34 8 −19 + EKmax −9 450 × 10 EKmax = 1× 10−19 J 9.5 9.5.1 Higher intensity has no effect. 9.5.2 Higher intensity increases the number of ejected electrons. Total: 200 marks IEB Copyright © 2019