2017 9749 H2 Physics MS EJC Suggested ver Jun2022

©EJC 2021 9749/H2 PHYSICS 2017 H2 PHYSICS SUGGESTED SOLUTIONS 9749 November 2017 Paper 1 Multiple Choice Question Key Question Key Question Key 1 A 6 A 11 D 2 B 7 D 12 B 3 C 8 C 13 D 4 D 9 D 14 D 5 C 10 A 15 B 16 C 21 A 26 B 17 D 22 D 27 D 18 C 23 D 28 C 19 B 24 C 29 B 20 C 25 A 30 C Notes: Questions that candidates found more challenging were Questions 5, 12, 14, 15, 16, 17, 21, 22 and 26. Question 5 Option A - a momentum change of less than p would result in the ball continuing to move in the same direction. Option D – an elastic bounce will result in a change of momentum of 2p Question 12 Note that work is done on the system. Question 14 Those who chose option B did not realise that one full cycle corresponds to a rotation of 2π radians. Question 15 Those who chose option A did not realise that frequency is constant as a wave moves from one medium to another. Question 16 Those who chose option A did not recognise that the relationship between ϴ and order n of maxima is not proportional. Question 22 Those who chose option B did not realise that the 2 A current is added to the 4 A current to give 6 A through the bottom 30 Ω resistor. Question 26 Those who chose option C did not recall that the mean power in a resistive load is half the peak power for a sinusoidal alternating current.

©EJC 2021 9749/H2 PHYSICS 2017 Paper 2 Structured Questions Notes: Ensure that all steps in a ‘show that’ question are clearly presented and that all assumptions made are declared. As a logic check, final numerical answers should be realistic. Question Marks 1(a) 3 2 0 (14 10 0 )(9.81 1 ) (10.8 8.0 0 ) F mg x F kx k LL − − ×= =− × = = − M1 1 N49.1 m−= A1 1(b)(i) ( ) 0 00 00 1 () () () 12 100 108 80 12 100 108 80 1249.1 100 108 80 4.0 Nm 2 s.f. mg LL LL L Lkm m km L L k Lm L kk − − ∆ − ∆ +∆∆∆ ∆= += + −− = + −  ∆= +  −  = +  −  =  =  Notes: Portion in square bracket not necessary but as a reminder, if need to evaluate absolute uncertainty, give to more than 1 s.f.. M1 12100% 100%100 108 80 8.1% k k ∆ × = + × − = A1 1(b)(ii) 1 1 8.1 (49.1)100 4 N m (1 s.f.) 49 4 N m k kkk k − − ∆ × = ∆= = ± = ± Notes: When presented in this form, the absolute uncertainty is 1 s.f. and the precision of the quantity follows that of the absolute uncertainty (i.e. in this case, k must be quoted to the nearest ‘1’) B1

©EJC 2021 9749/H2 PHYSICS 2017 Question Marks 1(c)(i) magnitude of upthrust is equivalent to reduction of elastic force for (10.8 - 10.3 =) 0.5 cm of extension C1 2 0.5 cm (49)(0.5 10 2 N ) 0. 5 U kx −= = ≈× C1 A0 1(c)(ii) volume of block, ( ) 3 53 140 10 7750 1.81 10 m block block mV ρ − − ×= = × = Notes: This is not the final answer, so in your calculator/working work with more s.f.’s. M1 Upthrust is the weight of liquid displaced 3 3 140 10 (9.81)7750 kg m 0.25 1410 liquid liquid liquid Um gV g U Vg ρ ρ − −  ×  = = = = = Notes: Remember to distinguish between mass and weight. M1 A1

©EJC 2021 9749/H2 PHYSICS 2017 Question Marks 2(a)(i) 4 3 5 3 5 2 22 (1.75 10 10 ) = 5.498 100.200 10 5.498 10number of days = 6.36 days s 24 60 60 CC CC C CC C v Trr r T v πω ππ     = = ×× ×  × = = ××  × = Notes: Remember to (show how to) convert the number of seconds into days. M1 M1 A0 2(a)(ii)1. Charon hovers over the same position in the sky all the time B1 2(a)(ii)2. same side of Charon observed all the time B1 2(b) 22 11 2 62 3 11.31 (1.20 0(6.67 10 ) 10 0.60 ) 7 kg m P P P Mg G R − − ×= = × × = Notes: Be careful with the power-of-ten, and to square the radius. M1 A1 2(c)(i) Loss in Gravitational Ep of rock = Gain in EK of rock B1 2100 2 initial final final i l ni uto tial P U U KE KE GM M Mvr −= − −= − 2 PlutoGM rv = 22 PlutoGM rv r   =    M1 Since 2 Pluto P GMg r= , 2 Pv gr= M1 A0 Notes: Need to explicitly show the (i) conservation of energy and (ii) algebra and substitution clearly. 2(c)(ii) 61(0.607)(12 2 10 1 m 0 s. 20 ) 12min Pv gr −= =×= Notes: Be careful with the power-of-ten B1

©EJC 2021 9749/H2 PHYSICS 2017 Question Marks 3(a) Oscillations in one direction, in a plane normal to the direction of transfer of energy. Notes: not asking for definition of transverse wave and need to mention either “oscillation” or “vibration”. B1 B1 3(b) speed, v, is the rate of change of distance (s) with time taken (t) v s t= M1 for a progressive wave, it travels one wavelength (λ ) in a period T: v s tT λ= = M1 Since 1f T= , 1vf Tλλ= = Notes: Start with the definition of speed A0 3(c)(i) Path difference BP – AP = 22 12 1.0 km12 5.0 −=+ Number of wavelengths within path difference: 3 83 1.0 10 4(3.00 10 ) (1200 10 ) BP AP BP AP cfλ −− × = = =÷ ×÷ × M1 since waves are emitted at source in phase and meet with path difference of integer multiple of wavelengths at P, M1 waves reach in phase, undergo constructive interference, a maximum is detected Notes: It is not sufficient to just state the conditions for constructive/destructive interference. To explain, data needs to be used, such as the relationship between wavelength and the path difference. A1 3(c)(ii) resultant amplitude varies between maxima and minima with constant frequency (OR at a constant rate) B1 3(c)(iii) use polarisers at both A and B axis of polarisation is aligned perpendicularly to each other Notes: both sources need to be polarised 90deg relative to each other. B1 B1

©EJC 2021 9749/H2 PHYSICS 2017 Question Marks 3(d) assuming point source transmitters 22 22 2 1 4 13 12 AB B A PII rr Ir I r π= →∝ = =  C1 M1 = 1.2 (do not accept fraction) Notes: cannot leave as 169/144. Do not confuse A for amplitude/area. A1 4(a) at least 3 concentric circles with increasing distance from each other B1 each circle labelled with clockwise arrows Notes: best to bring your compass (alongside geometry set) for exams. B1 4(b) magnetic flux density due to long straight wire: 7 0 2 6 (4 10 )(8.5) 2 2 (19 10 ) 9.0 10 T B I d µ π π π − − − ×= × = = × Notes: Be careful when converting cm to m. M1 A1

©EJC 2021 9749/H2 PHYSICS 2017 Question Marks 4(c)(i) wire H 6 wire H 5 t T an(12 ) tan(12 ) tan(12 9.0 10 4.2 1 ) 0 B B B B − − ×= = × ° °= = ° Notes: The examiners were looking out for (at least) a vector diagram showing how the 2 magnetic fields combined to form the resultant field with the angle. M1 A1 4(c)(ii) (if more marks allocated / if more numerical accuracy required) Find the distance from wire that will result in same magnitude of flux as with Earth’s field: 7 0 H 7 5 (4 10 )(8.5) 22 (4 10 )(8.5) 0.0404 m2 (4.21 10 )d B I dd µ π ππ π π − − − ×= ×= =× = X to be marked 4.04 cm (about 1/5 distance between wire and compass) from wire at the 3 o’clock position. Magnetic flux density is a vector quantity. At X, the magnetic flux density due to wire is pointing downwards, opposite to the upward Earth’s magnetic field. The point is nearer to the wire than the compass in Fig. 4.3 so that the magnitude matches that of the Earth’s. B1 A1 Notes: Need to explain how the 2 fields cancel out (magnitude and direction).

©EJC 2021 9749/H2 PHYSICS 2017 Question Marks 5(a)(i) 22 8 2 32 2 4 4(1.7 10 )(96) (0.18 10 ) 64 R LL L A r d L d ρρ ρ π π ρ ππ − − = =   ×= = × = = Ω C1 M1 A1 5(a)(ii) Since volume = L x A, increased length causes reduced cross-sectional area, Since R L A ρ= , resistance increases. Notes: Need to consider the reducti