2014 H2 Physics MS EJC Suggested

©EJC 2020 9646/H2 PHYSICS 2014 EUNOIA JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 PHYSICS SUGGESTED SOLUTIONS 9646 October/November 2014 Paper 1 Multiple Choice Question Key Question Key 1 C 21 D 2 A 22 C 3 A 23 B 4 D 24 D 5 D 25 A 6 C 26 D 7 C 27 D 8 D 28 B 9 C 29 A 10 D 30 D 11 B 31 D 12 B 32 C 13 B 33 D 14 C 34 B 15 A 35 D 16 B 36 C 17 A 37 A 18 C 38 D 19 A 39 A 20 D 40 C Notes Q4: need to solve simultaneous equations, one equation with unknown time and 0.25 of distance, d, from start to ground, the other equation with required time and d Q9: calculate power output of the train first Q22: distance between 2 nodes, i.e. 2 minimum points on voltage graph is half a wavelength, not one wavelength Q25: directions of field at all 4 points are the same even though their magnitude is different Q28: current is given by charge x frequency, ef = eω/2π

©EJC 2020 9646/H2 PHYSICS 2014 Paper 2 Structured Questions Qns Marks 1(a)(i) magnitude = 60.0 N angle = 90° B1 B1 1(a)(ii) vector sum of forces = 0 along horizontal considering magnitudes: ( ) ( ) ( ) 22 cos 200 cos 30 sin 200 sin 30 sin 40 0 40 0 sin 40 cos 100 horizontally: 100 3 N vertically: 60 0 N / (20 79)) magnitude of 40 100 3 20 79 178 N tan 3 13 or 347 Y Y Y. x arcsin( . Y Y . Y θ θ θ θ θ θθ = +  = °= °= = + = = = − = − −= = −° ° Notes Do not miss out the negative signs in your working and answer. M1 A1 A1 Y S 60.0 N θ Y S 60.0 N

3 ©EJC 2020 9646/H2 PHYSICS 2014 [Turn over Qns Marks 1(b) [magnitude] as long as magnitude of X is above zero [direction] and X remains directed at 30° anticlockwise to the horizontal, there will be a horizontal component of force cosX θ acting on S to the right since vector sum of forces on S must be zero for S to be in equilibrium, Y must provide a horizontal component force acting on S to the left Y is along same direction as rope B, so must be at an angle to the left and cannot be parallel to weight of S Notes Cannot just describe general conditions for equilibrium rather than target it towards the context of this question, i.e. must focus on forces acting on object S. B1 B1 A0 2 (out of syllabus) 3(a) current decreases as resistance of R and total circuit resistance increases drop in p.d. across internal resistance decreases ( )int resistanceV r= I terminal p.d. increases as it is the difference between electromotive force and p.d. drop across internal resistance ( )terminal e.m.f. rV = − I Notes Remember that the terminal potential difference is not the potential difference across the internal resistance. Also do not simply quote the potential divider formula without explaining in detail. M1 M1 A1

4 ©EJC 2020 9646/H2 PHYSICS 2014 Qns Marks 3(b)(i) ( ) ( ) total terminal 55 02 5 4 35 77 5 e.m.fcurrent in circuit 0 645 A e.m.f. 5 0 25 4 84 75 V 5 7 R . .. . r . . V . = = = = = − = − ++ = I I or ( ) ( ) internal internal total internal internal total terminal internal 0 25 02 by potential divider rule: e.m.f. e.m.f. 5 0 161 V e 5 4 35 0 255 0 25 .m.f. 5 4V 43 8 5 4 Vr R rV R . .. VV . . . .. = ++ = = − + = = + = = − C1 C1 A1 C1 C1 A1 3(b)(ii) same current passing through ( ) external total terminal 5 efficiency 100 100e.m.f 4 84 100 96 8 P % V %. % .% P . = × = × = ×= I I 3(c)(i) PJ is balanced length so PL C(e.m.f 12.) VV. == A1 3(c)(ii) ( )( ) PQ PQ 0 645 3 5 2 26 V V iR .. . = = = OR PQ PQ total PQ by potential divider rule: e. 355 m.f. V2 2602 5 4 35 VR R .V. .. = = ++ = ( ) ( ) C PQ PQ by potential divider rule: e.m.f. 1 12 2 0 531 m 26 LV . . . = = = l l

5 ©EJC 2020 9646/H2 PHYSICS 2014 [Turn over Qns Marks 3(c)(iii) p.d. across PJ increases and is larger than the p.d. across cell C a net p.d. is exerted opposite to the polarity of cell C and results in a current flow Notes Do not just mention that p.d. across PJ changes without linking to current flow. 4(a)(i) [flux linkage] the magnetic flux linkage is the product of magnetic flux density normal to the cross sectional area and varies sinusoidally when the coil spins around PQ [Faraday’s Law] sides of coil that are parallel to PQ cut the magnetic flux lines when rotating , therefore produce sinusoidal induced e.m.f. that is directly proportional to the rate of change of magnetic flux linkage [min e.m.f.] when the cross sectional area is normal to the flux lines, there is minimum rate of change of magnetic flux linkage so magnitude of induced e.m.f. is zero [max e.m.f.] when the cross sectional area is parallel to the flux lines, there is maximum rate of the coils cutting flux lines so magnitude of induced e.m.f. is maximal Notes Note that the maximum e.m.f. is where the rate of change of flux linkage is greatest, not when the flux linkage is greatest. 4(a)(ii)1. peak-to-peak voltage: 6.8 cm maximum induced e.m.f. = ( )0 050 0 17 V2 68 ... = 4(a)(ii)2. length of trace representing 2 complete oscillations: 10 cm frequency: ( ) 3 1 25 Hz 80 11 2 0 1 0T . − = = × 4(b) ( ) ( ) ( )( ) ( ) ( )( ) ( )( ) 0 0 3 0 17 120 1 3 2 2 2 10 25 0 0605 T NBA NBA f B f EN E N . A . . ωω π π π− Φ= = = = = × = = M1 M1 A1 5(a) work done per unit mass in bringing small test mass from infinity to that point B1

6 ©EJC 2020 9646/H2 PHYSICS 2014 Qns Marks 5(b)(i) using point ( ) 8810 4 126 0,. −×× ( )( ) 88 11 27 10 102 64 6 67 10 1 92 10 kg GM r r GM . . . φ φ − =− = ×× = × = × − −− Notes Do not miss out negative sign. 5(b)(ii) ( ) ( ) moon moon moon 2 222 moon moon moon moon moon total 2 2 moon moon moon moon 2 2 moon moon moon 22 GPE 11 1KE 22 2 GPE + KE 1 2 1 2 8 2 0 2 9 2 31 mm mm m mm m GM r vr . r T E GM rrT GM rrT φ πω π π −  = =  =  = −+    = −+  = = =    =   × −( )( ) ( ) 11 27 2 28 85 31 10 1 92 10 1 1010 2 10 1 37 10 6 67 24 224 2 53 J 21 . . . . . . π − ×× ××× =−   +   × 5(c) initial distance from S can be regarded as infinity where potential is zero ( ) ( ) ( )( ) 2 final initial final initial p 19 6 27 71 KE 1 2 1 6 10 10 10 1 40 10 m 2 2 10 0 16 s 2 7 eV V eV V m qV mv v . . . . − − − = − = ∆ = ×× ≈ × = − − × Notes Cannot use kinematics equations to solve as those are only for constant acceleration (with constant force).

7 ©EJC 2020 9646/H2 PHYSICS 2014 [Turn over Qns Marks 5(d) [magnitude] similarity: gradient of both graphs approaches zero as distance increases. the m agnitude of field strength decreases to near zero with distance. [direction] difference: the gradient of gravitational potential graph is positive while the gradient of the electric potential graph is negative. the direction of the field strength is towards lower potential, so gravitational potential graph shows an attractive potential while the electric potential graph shows a repulsive potential. Notes Remember to state and explain for both the similarity and difference. 6(a) the rate of change of A decreases with time (A decreases at a decreasing rate) the rate at which the oscillating system loses energy as work done against resistive forces decreases with time Notes The gradient of graph is used to infer that rate of change of A decreases with time, not to explain it. Need to give physical significance and not use the gradient of the graph to explain this. 6(b)(i) m / kg 11 / kgm − 2 / 10 mA − ln (A / m) 0.200 5.00 1.8 -4.0

8 ©EJC 2020 9646/H2 PHYSICS 2014 Qns Marks 6(b)(ii) 6(b)(iii) ( )12 12 2 55gradien 5 25 0 92t 0 293 ..y y x. x . − −−= =− − −= − 6(b)(iv) 2 1 0 0 1ln ln 2 b m t A Ae AA m bt 