Physical Sciences Grade 12 IEB Paper 1 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2016 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 © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 2 of 12 QUESTION 1 1.1 C 1.2 B 1.3 D 1.4 A 1.5 D 1.6 C 1.7 B 1.8 A 1.9 B 1.10 A QUESTION 2 2.1 Acceleration is the rate of change of velocity. ∆v 2.2 a = slope of v-t graph OR OR v= u + at ∆t 230, 4 − 0 a= 230, 4= 0 + a (120 ) 120 − 0 =a 1,92 m ⋅ s –2 Up =a 1,92 m ⋅ s –2 Up 2.3 Velocity is up and decreasing in magnitude OR slowing down. Acceleration is down and constant in magnitude. 2.4 297,64 s (-1 for no unit) 2.5 Speed is the rate of change of distance. 2.6 v= 2 u 2 + 2as v 2= 0 + 2(9,8)(55165,34 – 6000) = v 981,65 m ⋅ s −1 2.7 v= 2 u 2 + 2as = 1 981,652 + 2a (6000) 𝒂𝒂 = −𝟖𝟖𝟖𝟖, 𝟑𝟑𝟑𝟑 m⋅s-2 2.8 –1 if Thrust ≤ weight NB: label for weight may be 𝑤𝑤 or 𝐹𝐹𝑔𝑔 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 3 of 12 2.9 Fnet = ma Thrust – (5800)(9,8) = 5800(80,30) Thrust = 522 580,0 N up (if rounded off in Question 2.7) OR Thrust = 522 599,17 N up (if not rounded off in Question 2.7) OR 𝑚𝑚(𝑣𝑣 − 𝑢𝑢) 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝑡𝑡 5 800(1 − 981,65) 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = 12,21 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = 465 828,83 𝑇𝑇 − 5800(9,8) = 465 828,83 𝑻𝑻 = 𝟓𝟓𝟓𝟓𝟓𝟓 𝟓𝟓𝟓𝟓𝟓𝟓 N up OR 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 𝑠𝑠 = ∆𝐸𝐸𝐾𝐾 1 �𝑇𝑇 − 5 800(9,8)�6 000 = (5 800)(12 − 981,652 ) 2 𝑻𝑻 = 𝟓𝟓𝟓𝟓𝟓𝟓 𝟓𝟓𝟓𝟓𝟓𝟓, 𝟐𝟐 N up IEB Copyright © 2016 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 D, E (–1 per additional answer) 3.1.2 A, D, E (–1 per additional answer) 3.1.3 C (–1 per additional answer) 3.1.4 E (–1 per additional answer) 3.1.5 Child B and C at the same position. OR Child B passed Child C at point P OR Child B and C collided. 3.2 3.2.1 Distance is the length of path travelled. 1 3.2.2 s= ut + at 2 OR 𝑠𝑠 = 𝑣𝑣𝑣𝑣 2 = 6 2(t ) + 0 t=3s 1 3.2.3 s= ut + at 2 2 = s 4(3) + 0 s = 12 m 3.2.4 16 m − 12 m = 4 m away 1 3.2.5s= ut + at 2 2 1 =16 4 ( 3) + a ( 3) 2 2 = a 0,89 m ⋅ s −2 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 5 of 12 QUESTION 4 4.1 22 N (-1 no unit) 4.2 Frictional force due to a surface is the force that opposes the motion of the object and acts parallel to the surface with which the object is in contact. 4.3 22 N (same answer as 4.1) 4.4 Ff = µ N 22 = ( 0, 2 ) N N = 110 N wA + wC = 110 44 + wC = 110 wC = 66 N 4.5 When a net (resultant) force is applied to an object of mass, m, it accelerates the object in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass. OR The net force acting on an object is equal to the rate of change of momentum. 4.6 (force diagram -1) 4.7 Block B: Fnet = ma w wB − T = Ba g 22 22 − T = ( 2,3) 9,8 T = 16, 84 N 4.8 Block A: T − Ff = ma 44 16,84 − Ff = (2,3) 9,8 Ff = 6, 51 N 4.9 • Normal force is less as Block C has been removed. • Kinetic friction is less than static friction as surfaces are no longer 'stuck' or 𝜇𝜇𝐾𝐾 < 𝜇𝜇𝑆𝑆 . IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 6 of 12 QUESTION 5 5.1 5.1.1 The total (linear) momentum of an isolated system/in the absence of external forces remains constant (is conserved). 5.1.2 ( ptotal )before = ( ptotal )after 1, 2 ( 8 ) + = 0 1, 2 ( 4 ) + 0,5v v = 9,6 m ⋅ s –1 5.1.3 ∆pY = m(v − u ) ∆p= y 1, 2(4 − 8) ∆p y =−4,8 ∆p = y 4,8 kg ⋅ m ⋅ s −1 west ∆p y ∆𝑣𝑣 5.1.4 Fnet = OR 𝑎𝑎 = ∆𝑡𝑡 ∆t 4,8 4−8 Fnet = 𝑎𝑎 = (2,5−1)×10−3 = −2666,7 m.s-2 1,5 × 10−3 Fnet = 3200 N 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑚𝑚 for both formulae 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = 1,2(2 666,7) 𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 = 𝟑𝟑 𝟐𝟐𝟐𝟐𝟐𝟐 N OR For Cart Y: 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 (1,5 × 10−3 ) = 1,2(4 − 8) 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = 3 200 N For Cart X: 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣 3 200(1,5 × 10−3 ) = 0,5(𝑣𝑣𝑥𝑥 ) 𝑣𝑣𝑥𝑥 = 9,6 m.s-1 1 2 5.2 5.2.1 EK = mv 2 1 = (70)(15) 2 2 = 7 875 J 5.2.2 E p = mgh = (70)(9,8)(30) = 20 580 J 5.2.3 Wvs friction = Ff s Wvs friction = (12)(450) Wvs friction = 5 400 J IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 7 of 12 5.2.4 ( Emech = )top ( Emech )bottom + Wvs friction 1 20 580 + 7 875= (70)v 2 + 5400 2 =v 25,67 m ⋅ s −1 OR 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 𝑠𝑠 = ∆𝐸𝐸𝐾𝐾 1 (70 × 9,8 × 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 5 400)(450) = 70(𝑣𝑣 2 − 152 ) 2 𝒗𝒗 = 𝟐𝟐𝟐𝟐, 𝟔𝟔𝟔𝟔 m.s-1 OR 𝑊𝑊𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾 + ∆𝐸𝐸𝑃𝑃 −5400 = ∆𝐸𝐸𝐾𝐾 − 20580 ∆𝐸𝐸𝐾𝐾 = 15 180 J 1 70(𝑣𝑣 2 − 152 ) = 15180 2 𝒗𝒗 = 𝟐𝟐𝟐𝟐, 𝟔𝟔𝟔𝟔 m.s-1 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 8 of 12 QUESTION 6 6.1 Electric field is the force per unit (positive) charge. 6.2 6.3 Electric field is directly proportional to the charge. (two correct variables how electric field depends on charge) 6.4 Graph to show Electric Field vs Charge (Answer Sheet). Heading y-axis title and unit 1 y-axis scale (plotted points > graph paper, scale must be in sensible multiples) 2 Plotted points (all 6 points plotted within half small block) Line of best fit (with a ruler) IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 9 of 12 ∆y 6.5 Gradient = ∆x values from y -axis Gradient = (–1 if not shown on line of best fit) values from x-axis = 3,67 × 1012 (allow 3, 30 × 1012 − 4,04 × 1012 ) Gradient Gradient = 3,67 kN.C-1.nC-1 kQ 6.6 E= r2 k gradient = 2 r 9 × 109 3,67 ×10 =2 12 r r = 0,050 m IEB Copyright © 2016 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 Resistance is a material's opposition to electric current. 1 1 1 product 7.1.2 = + OR R = R R1 R2 sum 1 1 1 6 × 10 = + R = R 6 10 6 + 10 R = 3,75 Ω R = 3,75 Ω RT = 4 + 3,75 RT = 7,75 Ω 7.1.3 Current is the rate of flow of charge. 7.1.4 V = RT I V = (7,75)(3) V = 23, 25 V 7.1.5 = V emf − Ir = 26 − 3r 23, 25 = r 0,92 Ω 7.1.6 Total R in circuit increases Total I decreases but= V emf − Ir ∴V increases 7.1.7 I total = 0 A = V emf − Ir OR no lost volts V = 26 V 7.2 emf= Ir + IR emf= 2r + 2(2) emf= 3r + 3(1) solve simultaneously r= 1 Ω emf = 6 V IEB Copyright © 2016 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 Out of page 8.1.2 Current Magnetic field Length of conductor in the field (any two) 8.2 8.2.1 The emf induced is directly proportional to the rate of change of magnetic flux (flux linkage). 8.2.2 • Current in primary coil produces a magnetic flux in the core • As current in primary coil is alternating current, magnetic flux in core is changing • The secondary coil picks up the changing magnetic flux • By Faraday's law, changing flux induces an emf in secondary coil. 8.2.3 Power is constant, so high voltage means low current ( P = VI ) Smaller current means less energy loss as P = I 2 R N S VS 8.2.4 = N P VP N S 765 = NP 20 NS = 38, 25 NP 8.2.5 Step-up Transformer IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 12 of 12 QUESTION 9 9.1 Work function is the minimum amount of energy needed to emit an electron from the surface of a metal. hc c 9.2 E= = OR f = and E hf λ λ E= ( 6,6 ×10 )( 3 ×10 ) −34 8 = f 3 × 108 = 1,01× 1015 −9 296 × 10−9 296 × 10 and E ( = 6,6 × 10−34 )(1,01× 1015 ) E 6,69 × 10−19 J = E 6,69 × 10−19 J = 9.3 Sodium and Aluminium (-1 per wrong answer listed) 9.4 hf= W0 + EK max 6,69 × 10−19 =3,94 × 10−19 + EK max E K= max 2,75 × 10−19 J 9.5 No, intensity increases the number of photons/electrons, and energy of electrons of photons/electrons is not affected by intensity. OR No, frequency same, energy not affected by intensity. Total: 200 marks IEB Copyright © 2016