RI 2022 A-Level H2 Phys Solution

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 9749 H2 Physics 2022 GCE A-Level Suggested Solutions Paper 1 1 A Since X and Y are vectors, they can be positive or negative. Given: XY< initially: XY xy− = −− finally: ( )XY x y xy− = −− − = −+ xy xy−− > −+ Note: General equation ( ) ( ) 22 22cos sin 2 cosX Y x y y x xy y θθ θ− = −− + = + + As θ increases from 0° to 180° , XY− decreases. 2 C ( )100 cos 2.0 cos 100 -----(1) xxs ut ut u θ θ = = = ( )sin sin 2.0 2 sin 2.0 -----(2) yyyv u at uug ug θθ θ = + −= − = ( ) ( ) ( ) ( )1 2 2.0 9.81: tan1 100 2.0 9.81 tan 11. 1 11100 θ θ − = = = °= ° 3 B Since collision is elastic, ( ) 1 2.0 0 0.50 2.0 0.50 1.5 m s PQQP Q Q uuvv v v − −=− − = −− = −= By conservation of momentum, ( ) ( ) ( )2.0 0 0.50 1.5 2.5 1.5 1.5 3 2 3 2.5 5 2 5 P PQ QP PQ Q P PQ PQ P Q mu mu mv mv m mm mm m m +=+ += − + = = = = X Y X Y X Y vQ P Q

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 4 B Consider vertical equilibrium, u mg T T u mg = + = − ( ) ( )( ) ( ) 3 3 3 14 23 2 3 2 0.500 1030 9.81 200 9.813 683.3 683 N ww w w T V g mg r g mg r g mg ρ ρπ πρ π = − = ×− = − = − = = 5 D Consider horizontal equilibrium, sin36 sin36 ----- (1) TF T ke °= °= Consider vertical equilibrium, cos36 ----- (2)TW °= ( ) ( ) ( ) 1 sin36 2 cos36 tan36 tan36 25 0.060 tan36 2.065 2.1 N ke W ke W keW ° =° °= = ° = ° = = 6 C 22 GM G Vg rr ρ= = ( )( ) 2 2 3 11 6 3 4 3 3 9. 81 4 6.67 10 6.37 10 5512 5510 kg m gr GV gr Gr ρ π π − − = =   = × ×× = = T mg u T W F

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 7 B Maximum height after 2 oscillations, ( )0.10 2 0.20 2 0.60 0.60he e − −= = At maximum height, pendulum is momentarily at rest i.e. K.E. = 0 From initial height to maximum height after 2 oscillations, by conservation of energy, work done against air resistance = decrease in G.P.E. ( ) ( )( ) ( ) 02 0.200.40 9.81 0.60 0.60 0.4268 0.43 J mg h h e− = − = − = = 8 C Angular velocity is the rate of change of angular displacement, which is the same for every point on the disc. Hence angular velocity is independent of the distance from the centre of the disc. 9 C Gravitational force on the satellite (mass m) by the Earth (mass M) provides the centripetal force. 2 2 2 2 132 2 2 4 GMm mrr GMm mr Tr GMTr ω π π = =  =   Option A is incorrect: From the equation above, r is independent of the mass m of the satellite. Option B is incorrect: The orbital period of all geostationary satellites is 24 hours. Option D is incorrect: 2v r r vr T πω = = ⇒∝ Option C is correct: The orbits of all geostationary satellites are in the same plane as the equator of the Earth and they orbit West to East, similar to the Earth’s rotation. 10 D Gravitational potential is a scalar quantity. 4 4 22 10 PM M GM G M dd GM d φφφ= +    ×=− +−     = −

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 11 A 2 2 1 3 33 3 Nmpc V pV NkT kTc Nm Nm m = = = = 2 rmscT c T∝⇒ ∝ 21 2 1 1 160 273.15 35080 273.15 387.6 390 m s rms rms Tcc T − = += ×+ = = 12 C K.E. of 1 gas molecule, 3 2E kT= Total K.E. of gas, 3 2 TE NkT= Since temperature is constant, ET is constant and 11 2 2pV p V NkT= = . ( ) ( ) 5 2 2 11 3 3 33 1.0 10 0.010 1500 J2 2 22 TE NkT p V pV= = = = ×= 13 B decrease in K.E. = increase in thermal energy ( ) 2 2 2 50 1 100 2 50 1 1 100 2 4 mv mc vv cc θ θ  = ∆   ∆= =      14 C 0 sinxx t ω= 0 0 cos 22 cos 22 0.30cos5.0 5 .0 0.377cos1.26 0.38cos1.3 dxv dt xt xtTT t tt ωω ππ ππ = =  =     =     = = 15 A Resonance occurs for forced oscillations when the driver’s frequency is close to the natural frequency of the oscillating system. There is maximum transfer of energy and the oscillating system oscillates with maximum amplitude. Options B, C and D are examples of forced oscillations under a periodic driving force and resonance is possible. Option A is not an example of resonance. The diaphragm of the loudspeaker is vibrated (pushed or pulled) by the interaction of the electromagnetic coil attached to it and a permanent magnet. This i nteraction, which affects the amplitude of the vibrations, is dependent on how the current through the electromagnetic coil changes. The frequency of the push and pull of the magnets do not need to match the natural frequency of the diaphragm to produce large amplitudes. https://electronics.howstuffworks.com/gadgets/audio-music/vibration-speakers.htm

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 5 16 D From Diagram 1: 0.8 mλ = From Diagram 2: 0.2 sT = 10.8 4.0 m s0.2vf T λλ −= = = = 17 B From oscilloscope: ( ) 36 div 0.050 10 sT −= ×× ( ) 3 11 3.333 kHz 3.3 kHz 6 0.050 10 f T −= = = = ×× 2 antinodal distance 2 5.0 10 cmλ = × = ×= 18 B Dx a λ= Option A: ( ) ( ) 9 3 700 10 15 0.000175 15 m4.0 10 Ax − − × = = ×× Option B: ( ) ( ) 3 3 20 10 15 0.4 15 m50 10 Bx − − × = = ×× Option C: ( ) ( ) 9 3 450 10 15 0.000225 15 m2.0 10 Cx − − × = = ×× Option D: ( ) ( ) 3 3 10 10 15 0.05 15 m200 10 Dx − − × = = ×× 19 A In the uniform electric field, the proton experience s a constant electric force and acceleration that acts vertically downwards. Ey E y F ma F Eq Vea m m dm = = = = Horizontally: xxs ut y vt yt v = = = Vertically: 2 2 2 2 1 2 10 2 2 yy ys ut at Ve y eVyx dm v mdv = +   = + =     20 A 2 21 24 dAnvq nvq ned vππ= = = I 2 41v ne dπ= I Since 4 neπ I is a constant, 4K neπ= I . 2 Kv d=

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 6 21 D Equivalent resistance of P and voltmeter, 11 // 11 11 2 V PV PV VV RR RR RR −−   = += +=      // 6.0 3.0 2 6.0 3.0 6.0 2 4.03.0 PV Q V Q V Q R R R R R R = = = ×= 22 B ( ) 284 10 14.3 VwireV −= ×× ( ) 265 65 84 10 14.3 9.295 9.3 V84 84 wireEV −= × = ×× × = = 23 C By Newton’s third law, force on X by Y = − force on Y by X i.e. the force on each wire is equal in magnitude and opposite in direction Currents in the same direction will attract d ue to the interaction between the magnetic fields of the currents. 24 D ( )sin 90 cosF BL B Lθθ= °− =II The graph of F against θ is a cosine graph. 25 B ( ) ( )( ) 64 88 cos 90 60 cos30 65 10 12 10 cos30 6.754 10 6.8 10 Wb BA BA φ −− −− = °− ° = ° = ×× ° = ×= × 26 A peak current 0= I r.m.s current 2 0 0 T T= =I I 27 B ( )cos sind NBA tdE NB A tdt dt ω ωωΦ= −= − = cos 12' s in22 d NBA tdE NB A tdt dt ω ωω  Φ = −= − = 0 0 2 AE NBA RR ω= = =I 0 0 1 ' 12' 2 1 A2 NBAE RR ω = = = ×=I 2 2 0 ' 1 20 10 W22 m rmsPR R = = = ×= II

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 7 28 A ( ) 2 2 211 1 22 2 2 mv pE mv mm p Em = = = = 2 hh p Em λ = = ( ) ' '3 3229 hh h p EmEm λλ = = = = 29 C 83pn = and 212 83 129nn = −= ( )83 129pnm M MM∆= + − Mass defect is the difference in the mass of the nucleus and the total mass of its constituents (protons and neutrons). 30 C 238 234 4 2U Th He→+ Energy released is from the mass difference between the reactants and products. ( )( ) ( )( )( ) 2 2 227 8 13 13 238.1249 234.1165 4.0026 1.66 10 3.00 10 8.665 10 8.7 10 J U Th He E mc M MMc − −− = ∆ = −+ = −− × × = ×= ×

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 8 Paper 2 1 (a) Period of oscillation 0.72 s10 tT = = Since 2 mT kπ= 22 1 22 0.1204 4 9. 1385 N m0.72 mk Tππ −  = = =     Uncertainty in k 1 22 0.20.01 2 9.1385 0.599 0.6 N m (1 s.f.)7.2 mT mtk kk mT mt − ∆∆ ∆∆  ∆= + × = + ×      = + ×= =   Therefore ( ) −= ± 19.1 0.6 N mk [1] [1] [1] [1] (b)