2018 9749 H2 Physics MS EJC Suggested ver Jun2022

©EJC 2021 9749/H2 PHYSICS 2018 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2018 Paper 1 Multiple Choice Question Key Question Key Question Key 1 A 6 A 11 C 2 A 7 C 12 D 3 B 8 B 13 D 4 C 9 D 14 C 5 A 10 A 15 D 16 C 21 A 26 A 17 D 22 A 27 C 18 C 23 D 28 A 19 C 24 A 29 C 20 A 25 A 30 B Notes: Candidates found Questions 1, 2, 5, 9, 11, 18, 24, 26 and 27 relatively challenging. Question 1 Percentage uncertainty in L needed to be subtracted from, rather than added to, the percentage uncertainty in g. Question 2 Distance is the total magnitudes of the areas of the triangles defined by the graph lines and the time axes. Question 9 Those who chose option C confused between angular displacement and linear displacement. Question 11 Those who chose option D assumed wrongly that the spacecraft must start from rest. Question 17 Those who chose B found the centre-to-fringe distance instead of fringe-fringe distance. This requires a doubling of the new centre-fringe distance. Question 23 Those who chose either B or C assumed wrongly that current must split equally at the junction Question 24 Those who chose option B, omitted a factor of 2 either from omitting the sin 30° factor or from missing that the magnetic force is exerted on the coil at double the distance away from the pivot as is the weight. Those who chose option D treated the coil as consisting of a single turn rather than 50. Question 27 Those who chose option D neglected to consider the effect of half-wave rectification.

©EJC 2021 9749/H2 PHYSICS 2018 Qns Marks 1(a) when a body is in rotational equilibrium, the sum of clockwise moments about any point must be equal to the sum of anticlockwise moments about the same point. B1 1(b)(i) Solving (1) and (2): 31 NAT = 13 NBT = Notes: Be careful not to get the = 3 7 B A T T values the wrong way around. A1 1(b)(ii) Taking moments about the centre of mass, T A (d – 0.20) = TB (1.00 – d – 0.20) 31 (d – 0.20) = 13 (1.00 – d – 0.20) d = 0.38 m C1 A1 Paper 2 Structured Questions TA TB mg As metal sign is in translational equilibrium, TA + TB = mg = (4.5)(9.81) TA + TB = 44.145 ---- (1) Since the perpendicular distance of T A is smaller than that of TB from the centre of mass, TA > TB therefore ---- (2) M1

3 ©EJC 2021 9749/H2 PHYSICS 2018 [Turn over Qns Marks 1(c) wind exerts a horizontal force on the sign to the right to maintain translational equilibrium of the metal sign, force that each support provides on the sign must have a horizontal component to balance the rightward force. Notes: Do not simply state that a horizontal force is needed as wind is horizontal. Remember to state explicitly that this force is to maintain translational equilibrium. Common misconception: Some may attribute the horizontal force as resulting from N3L applied to the wind force. From N3L, the sign is the one exerting an equal and opposite force on the wind, this force is not a force acting on the sign by the support. B1 Total horizontal components of the forces of the supports on sign Average force of wind on sign mg Total vertical components of the forces of the supports on sign

4 ©EJC 2021 9749/H2 PHYSICS 2018 Qns Marks 2(a) the change in length of a material is directly proportional to the force applied on it, provided that the limit of proportionality is not exceeded. B1 2(b) elastic potential energy stored = area under load-compression graph ( ) ( ) 3 J 1 85 6.82 0.29 10 −= × = M1 A1 2(c) loss in elastic PE = gain in KE and gravitational PE ( )( ) 2 32 3 1 1 2 10.29 0 (42 00 ) ( )910 42 102 2.4 m .81 0.40 s iEPE hm v mgv −− − −= − + +−= × × = Notes: Be careful not to forget the GPE too. B1 M1 A1 2(d)(i) At minimum speed, normal contact force 0N = , Hence weight provides centripetal force. Thus 2 minmvmg r= min 1 (0.20)(9.81 ) 1.4 m s v rg − = = = vmin = 1.4 m s–1 B1 M1 A1 2(d)(ii) minv rg = vmin is independent of mass and hence does not change B1 mg N P

5 ©EJC 2021 9749/H2 PHYSICS 2018 [Turn over Qns Marks 3(a) a wave spreads out after passing through a slit or around the edge of an obstacle [do not accept spreading “through” an obstacle, “bending” of waves, “splitting” of waves] B1 3(b) sin b λθ = For very small angle θ, 0.14sin tan 2.7θθ≈= 6 0.14 2.7 12 x 10 λ −= λ = 6.2 x 10–7 m M1 M1 A1 3(c) sin b λθ = , a narrower slit (smaller b) means a largerθ. more spreading of the waves. broader central maxima. less light passes through, lower overall intensity B1 B1 3(d)(i) sin b λθ = , a longer wavelength λ would mean a larger θ for the angular positions of the first minima. The central maximum would be broader. B1 B1

6 ©EJC 2021 9749/H2 PHYSICS 2018 Qns Marks 3(d)(ii) white light spectrum consists of wavelengths spanning 400 – 700 nm. sin b λθ = , amount of spreading depends on wavelength λ of the incident light. longer wavelengths at the red end spreads more as c ompared to shorter wavelengths, hence the edges are red at the central region, the different wavelengths overlap to produce white Notes: Remember to use scientific terms such as interference/ overlapping. Common misconception: There is white light at centre as there is no diffraction at the centre. B1 B1 Intensity Displacement from centre 0 Blue Red

7 ©EJC 2021 9749/H2 PHYSICS 2018 [Turn over Qns Marks 4(a) minimum frequency of electromagnetic radiation for electrons to be emitted from metal surface. at this frequency, electrons are emitted with zero kinetic energy B1 B1 4(b) 4(b)(ii) gradient is equal to the Planck constant h ,max ,maxKKhh EEff Φ+ = Φ= →− B1 ( 0f ) B1 (b)(iii) B1 4(c)(i) 2 max 1 2 hc mvλ =Φ+ 34 8 1 2 ma 9 31 x9 2.(6.63 10 )(3.00 10 ) 1(1.60 10 ) (9.11 10 )2(490 0 ) 51 v − −− − ×× ×= + ×× 1 max 51.1 10 m sv −= × M1 C2 A1 4(c)(ii)1 blue light has frequency higher than threshold frequency of europium, hence photoelectric effect occurs. electrons are emitted from the electroscope, gold leaf and rod become less charged and hence smaller electrostatic repulsion. frequency of red light is below threshold frequency of europium so no electrons are emitted B1 4(c)(ii)2 gold leaf rises due to electrostatic repulsion due to net lack of electrons. even if photoelectric effect occurs, electrons emitted from europium surface will only make the gold leaf and rod more positively charged and hence larger repulsion Notes: Remember that only electrons are free to move and do not use terms like positive charge cannot be emitted. B1 B1 (b)(iii)

8 ©EJC 2021 9749/H2 PHYSICS 2018 Qns Marks 5(a) 3 63 mass of 1 molevolume of 1 mole density 63.5 10 8960 = 7.087 10 m − − = ×= × 23 63 28 3 number density of copper atoms volume of 1 mole 6.02 10 7.087 10 m 8.49 10 m AN − − = ×= × = × each atom has 1 conduction electron, so number density of charge carriers is 8.49 × 1028 m–3. C1 C1 B1 5(b) 2 25.0 (30) 0.41 A PR= = = I I I 32 28 19 41 (0.36 10 )0.41 ( 3.4 10 )( )(1.60 10 )4 7.4 10 m s Anvq v v π − − −− =  ×= ×× = × I C1 C1 C1 A1 5(c) ( )( ) LV R nAqv nev L A ρ ρ= = =I same material, so number density of charge carriers n is the same same temperature and material, so resistivity ρ is same