Physical Sciences P1 MG 2017 hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2017 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 © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 2 of 11 QUESTION 1 1.1 B 1.2 C 1.3 D 1.4 B 1.5 D 1.6 A 1.7 A 1.8 C 1.9 A&D 1.10 D QUESTION 2 2.1 Velocity is the rate of change of position OR the rate of displacement OR the rate of change of displacement. 2.2 AB (0 – 3 s); DE (7 – 8 s); FG (9 – 10 s) 2.3 BC (3 – 5 s); EF (8 – 9 s) 2.4 CF (5 – 9 s) 2.5 Acceleration is the rate of change of velocity. ∆v 2.6 a = slope of v-t graph OR OR v= u + at ∆t −2 − 2 a = –2 = 2 + a(4) 7−3 a =–1 a=–1 a = 1 m⋅s-2 South a = 1 m⋅s-2 South 2.7 C B D E position (m) F G 0 A 0 1 2 3 4 5 6 7 8 9 10 time (s) IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 3 of 11 QUESTION 3 3.1 v t blue red 1 3.2 3.2.1 s= ut + at 2 2 s = ( 0,25 )( 8 ) +0 s=2 m ∴ cat is 6 – 2 = 4 m from mouse 3.2.2 s (m) 6 4 8 10 t (s) (or mirror image in x-axis) s 4 3.2.3 v = slope of s – t graph OR v= = t 2 v 2 m ⋅ s −1 = v 2 m ⋅ s −1 = 3.3 = 3.3.1 while a 20 m ⋅ s−2 1 s= ut + at 2 v= u + at 2 1 s= 0 + 20(15)2 v= 0 + 20 (15 ) 2 s = 2 250 m =v 300 m ⋅ s−1 while a = g v= 2 u 2 + 2as 0= 2 3002 + 2 ( −9,8 ) s s = 4 591,84 m max height = 2 250 +4 591,84 max height = 6 841,84 m IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 4 of 11 3.3.2 Time to max height after rocket runs out of fuel v= u + at = 300 + ( −9,8 ) t 0 t = 30,61 s Time to reach the ground from max height 1 s= ut + at 2 2 1 −6 841,84 =0 + ( −9,8 ) t 2 2 t = 37,37 s total t =15 + 30,61 + 37,37 total t = 82,98 s OR Time to max height after rocket runs out of fuel u +v s= t 2 300 + 0 4 591,84 = t 2 t = 30,61 s Time to reach the ground from max height 1 s= ut + at 2 2 1 −6 841,84 =0 + ( −9,8 ) t 2 2 t = 37,37 s total t =15 + 30,61 + 37,37 total t = 82,98 s OR 1 s= ut + at 2 2 1 −2 250= 300t + ( −9,8 ) t 2 2 t = 67,98 s total t = 15 +67,98 total t = 82,98 s 3.3.3 as it hits the ground IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 5 of 11 QUESTION 4 4.1 4.1.1 Newton's second law. 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 Newton's second law. The net force acting on an object is equal to the rate of change of momentum. 4.1.2 acceleration 4.1.3 Graph to show acceleration vs force 1,2 1,0 P 0,8 a (m⋅s-2) 0,6 0,4 0,2 0 0 50 100 150 200 250 300 350 Graph – on answer sheet Force (mN) Heading y-axis title and unit 1 y-axis scale (plotted points > graph paper) 2 plotted points line of best fit (6) IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 6 of 11 ∆y 4.1.4 Gradient = ∆x values from y -axis Gradient = (values must be from LOBF on graph) values from x -axis Gradient = 3,54 × 10 −3 m ⋅ s−2 ⋅ mN−1 or g-1 ( allow 3,19 × 10 − 3,89 × 10 ) −3 −3 OR Gradient = 3,54 m ⋅ s −2 ⋅ N−1 or kg-1 (allow 3,19 – 3,89) 1 4.1.5 Fnet =ma ; F − Friction =ma ; a = F m 1 1 = 3,54 × 10 −3 OR = 3,54 m m m = 282 g m = 0,282 kg 4.1.6 Sketch line parallel to line of best fit and x intercept 50 mN (line labelled P on graph). Force, F 4.2 4.2.1 Normal Weight = 4.2.2 w  plane mg sin35° w  plane = 150 ( 9,8 ) sin35° w  plane = 843,16 N 4.2.3 Ff + F cos 40° =843,16 Ff + 100 cos 40 ° = 843,16 Ff = 766,55 N 4.2.4 ⊥: FN + F sin = 40° mg cos35° : mg sin35°= Ff + F cos 40° = µg sin35 ° µ FN + F cos 40° = µg sin35° µ ( µg cos35° − F sin 40° ) + F cos 40° µg ( sin35° − µ cos35 = ° ) F (cos 40° − µ sin 40°) 150 ( 9,8 )( sin35° − 0,7 cos35 = ° ) F (cos 40° − 0,7 sin 40°) F = 0,79 N IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 7 of 11 QUESTION 5 5.1 5.1.1 Frictional force is the force that opposes the motion of an object. 1 5.1.2 EK = mv 2 2 1 EK = ( 2 ) (1,5)2 2 EK = 2,25 J 5.1.3 W = Fs W = ( 26 )( 0,7 ) W = 18,2 J 5.1.4 The work done by a net force on an object is equal to the change in the kinetic energy of the object. 5.1.5 W = ∆Ek OR Fnet = ma for both equations 1 −18,2 = 2,25 − ( 2 ) v12 coe −26 = 2a 2 =v 1 4,52 m ⋅ s −1 a = −13 m⋅s-2 v=2 u 2 + 2as 1,52 = u 2 + 2 ( −13 ) ( 0,7 ) u = 4,52 m⋅s-1 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 crate: mv 2 = mgh 2 1 (1,2 ) v 2 = (1,2 )( 9,8 )( 0,65 ) 2 =v 3,57 m ⋅ s −1 5.2.3 The total (linear) momentum of an isolated system remains constant (is conserved). 5.2.4 ( ptotal )before = ( ptotal )after 0,4v b += 0 ( 0,4 )( −0,36 ) + 1,2 ( 3,57 ) =v b 10,35 m ⋅ s −1 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 8 of 11 QUESTION 6 6.1 6.1.1 Every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Gm1m2 6.1.2 F1 = r2 F1 = ( ) ( 6,7 × 10 −11 ( 700 ) 5,8 × 1024 ) ( 7,4 × 10 ) 2 6 F1 = 4 967,49 N 1  G  m1  m2 6.1.3 F2 =  2  (1,8r )2 0,5 F2 = F1 (1,8)2 F2 = 0,15 F1 1 6.2 6.2.1 s= ut + at 2 2 1 0,6= 0 + a(3,3)2 2 = a 0,11 m ⋅ s −2 kqQ 6.2.2 ma = r2 ( 9 × 10 )(1× 10 )Q 9 −9 ( )( 0,020 0,11 = ) ( 0,6 ) 2 = 8,8 × 10−5 C Q F kQ OR Fnet = ma E= E= for all equations q r2 Fnet = ( 0,020 )( 0,11) E= 0,0022 ( 9 × 10 )Q 2,2 × 10 = 6 9 1× 10−9 (0,6)2 Fnet = 0,0022 N = 2,2 × 106 E = 8,8 × 10 −5 C Q 6.2.3 Acceleration is not constant. Electric field and hence force depends on distance from charge Q. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 9 of 11 QUESTION 7 7.1 7.1.1 The current through a conductor is directly proportional to the potential difference across the conductor at constant temperature (or constant resistance). 7.1.2 V = RI 12 = 8I I = 1,5 A 1 1 1 1 7.1.3 = + + RP R1 R2 R3 1 1 1 1 = + + RP 4 8 12 RP = 2,18 Ω 7.1.4 V = RI OR 4Ω : I = 3 A 12 = 2,18I 8 Ω : I = 1,5 A I = 5,5 A 12 Ω : I = 1 A I = 5,5 A 7.1.5 = V emf − Ir = 12 16,5 − 5,5r r = 0,82 Ω 7.2 7.2.1 Power is the rate at which work is done. OR the rate at which energy is transferred. V2 7.2.2 P = R 2202 P= 50 P = 968 W 7.2.3 cost = kW × time × unit cost 80 = ( 0,968 ) t (1,24 ) t = 66,65 hours (66 hours 39 min; 3998,93 min; 239936 s) IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 10 of 11 QUESTION 8 8.1 8.1.1 Current direction Concentric circles Magnetic field direction 8.1.2 into page 8.1.3 use a.c. or current continuously changes direction ∴ force on current carrying conductorchanges direction ∴ vibrates 8.2 8.2.1 mechanical energy → electrical energy 8.2.2 yes, there are slip rings or no split rings 8.2.3 The emf induced is directly proportional to the rate of change of magnetic flux (flux linkage) 8.2.4 A E emf B D OR negative of graph t C shape max emf for A B and D at zero ∆φ 8.2.5 at point C, = maximum ∆t ∴emf is a maximum OR ∆φ at point C, = maximum ∆t Polarity of C is opposite as coil has rotated 180° relative to A IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 11 of 11 QUESTION 9 9.1 3 9.2 ∆E = hf ( (13,6 − 3,4 ) 1,6 ×10−19 = ) 6,6 ×10−34 f (10,2 ) (1,6 ×10 ) = −19 6,6 ×10−34 f =f 2,47 × 1015 Hz hc 9.3 9.3.1 E = OR c = f λ (for both formulae) λ ( 6,6 ×10 )( 3 ×10 ) −34 8 E= 655 ×10 −9 ( 3 ×108= f 655 ×10−9 ) E 3,02 ×10−19 J = =f 4,58 ×1014 Hz E = 1,89 eV E = hf E= (6,6 ×10−34 )( 4,58 ×10 ) 14 E 3,02 ×10−19 J = E = 1,89 eV 9.3.2 n = 3 → n = 2 Direction -0,85 eV n=4 -1,51 eV n=3 -3,40 eV n=2 -13,60 eV n = 1 (ground state) 9.4 A free electron has energy of zero. Electrons will gain energy as they move up a level OR A free electron has energy of zero so an electron releases energy as it moves to a lower level Total: 200 marks IEB Copyright © 2017