RI 2020 A-Level H2 Phys Solution

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 2020 A-Level H2 Physics Suggested Solutions Paper 1 1 B π= 34 3Vr ∆∆  ×= ×     100% 3 100%Vr Vr A graph of ∆ × 100%V V against ∆ × 100%r r will give a straight line graph through the origin. 2 D The action-reaction forces in Newton’s third law must be the same type and act on 2 different bodies. Option D: The weight of a car and the normal contact force on it are 2 forces acting on one single body i.e. the car. In addition, weight is a gravitational force which is not of the same type of force as the normal contact force. 3 D = forcestress area and = extensionstrain original length − − = × = × = = × = 2 1 2 3 force extensionunits of area under graph units of unit s of area original length Nm mm J m (since work done force displacement)m J m 4 A At = 0t , car X and car Y are at the same position. When car Y overtakes car X at =tT , the displacement of both cars from their position at = 0t is the same. Since the area under the velocity-time graph gives the displacement, ++=++ = =area under graph of car Y area under gr PQR aph o QR f car X S PS 5 C Note that = 3cbvv and the velocity of the bicycle relative to the car is given by − bcvv . 6 C Torque is a vector quantity. The statement in option C is unclear as the directions of the torques are not stated. A correct statement should read ‘The anticlockwise torque provided by the vertical forces is equal to the clockwise torque provided by the horizontal forces.’

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 7 B θ = × = × torque of couple perpendicular distance b etween the parallel forces sin F Fd 8 A When balloon is inflated with hot air, ρ = + = +120 total weight, weight of hot air weight of basket and non inflated balloonW Vg mg ρ = = 20 upthrust on inflated balloon, weight of the atmospheric air displacedairU Vg Consider the free body diagram of the inflated balloon with basket. = + 2airU WT ( ) ( )( ) ( )( ) ρρ −= −+= = −− = = 20 120 2 2 1.204 2800 0.898 2800 7002 769.104 0.77 kN (2 s.f.) airUWT Vg Vg mg g (The answer should be 0.769 kN if rounded off to 3 s.f. However, the answer given in the option is 0.770 kN. This will be the best option to choose.) 9 D ( )( ) 3rate at which useful work is done by driving force 1.6 10 22 WFv= = × ×= 63.3 10rate at which total thermal energy is produced W60 ( )( ) ( ) × = = × 3 6 1.6 10 22 % efficiency 64 % 3.3 10 60 10 B ( ) 2 2 2 8 32 2 23.85 10 27.3 24 60 60 2.73 10 m s car r T ω π π −− = =  = ×  ××× = × 11 B The minute hand takes 1 hour to make one round the clock. ππω −−= = = × × 3122 1.75 10 rad s60 60T d F F

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 12 A  − =− −−   = −+ = 12 12 12 12 12 2 2 22 2 final initial Gm m Gm mUU rr Gm m Gm m rr Gm m r Since >12 02 Gm m r , >final initialUU . Hence the increase in potential energy is 12 2 Gm m r . 13 A ( )( ) ( ) − −−×× = = = × ×× 11 31 21 22 63 6.67 10 2.0 10 5.9 10 N kg 150 10 10 GMg r 14 B =pV nRT = constantpV T From L to M, pressure is constant. Hence = constantV T . Graph of T against V is a straight-line graph through the origin. =MM NN MN pV pV TT ( )( ) ( )( )×× = 662.0 10 0.003 0.8 10 0.005 MNTT =6000 4000 MNTT TM = 1.5 TN. Graph of T against V should show that T decreases as V increases. 15 B First law of thermodynamics: ∆=+UQW Experiment 1: ∆ = +∆ + ∆ = +∆ 1 1 0UQ UQ Hence, option B is correct and option A is incorrect. Experiment 2: ( )∆ = +∆ + − ∆ = +∆ − 22 22 U QW U QW Option D is incorrect since ∆U2 < ∆U1 22W QU= +∆ −∆ Option C is incorrect.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 16 A πω = 2 T and π= 2 LT g Since L and g remain constant, T and ω remain constant. E = max. PE = max. KE ω= = 2 22 max 11 22E mv m x When max. P.E. is reduced by 4E : ω−= 22 11 1 where is the new amplitude42 EE mx x 22 2 2 1 31 1 42 2 mx mxωω = 1 3 4xx= = −= 3change in amplitude 0.1344xx x 17 C The two waves are 1 4T apart. πφπ= ×=4 2 rad2 T T 18 C distance between 2 consecutive compression points λ= λ λ = = = = − 12 1.5 Hz12 4 vf vf 19 D Since D is much greater than the slit width, the angle of diffraction is small. For single slit first minima: θλ=sinb Since θ is small, θθ λθ ≈ ≈= = sin cf c b b fb For small angles, ( )θ ≈=  = 22 2 cxD D fb Dcb fx 20 C πε= 04 QV d ( )πε πε = = =  1 00 224 24 QQVV dd

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 5 21 A πεπε = =   22 00 1 44 QQE rr Plotting a graph of E against 2 1 r , gradient = πε04 Q and y-intercept = 0. 04 gradientQ πε= × 22 B ( ) ρρρ ππ = = =22 4 2 LL LR A dd Since ρ and L are constants, =  2 1Rk d where ρ π= 4 Lk is a constant. Since X and Y are in parallel, the p.d. V across each element is the same. 2 22 22 2 V dVP V dP dd PR kk == = ⇒∝ ⇒∝  = = = = 1.0 0.816 0.821.5 xx yy dP dP 23 A circuit 1: 1 2 E Rr= +I circuit 2: 2 2 E Rr= +I ( ) 21 3 322 2 32 323 2 4 EE R r Rr Rr R r Rr R r Rr = = ++  += + += + = II 24 B Option B: By the potential divider principle, the p.d. across each R is 3.0 V. When X is at Y, the p.d. across X and Y is zero. When X is at the other side of R, p.d. across X and Y is 3.0 V. Option A: The p.d. across X and Y will vary from 3.0 V to 6.0 V. Option C: The p.d. across X and Y will always be 6.0 V. Option D: The p.d. across X and Y will vary from 3.0 V to 6.0 V.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 6 25 B The electric field strength and the electric force between the plates is constant. =upwards electric force on electron, eF ma ∆= = = ee e F Ee V ea m m dm Horizontally: − = ×= = × 2 7 5.00 10 s1.97 10 xx x x s ut st u Vertically: 2 22 7 23 19 2 2 31 7 2 1 2 1 5.00 10 2 1.97 10 1 3.00 10 1.60 10 5.00 10 2 10.0 10 9.11 10 1.97 10 1.70 10 m y e s at Veh dm − −− −− − =  ∆×∆=   ×   ×× ×= ×  ×× ×   = × 26 D ( )( )φ −−Φ= = = × × = × 5 2210 2.1 10 20 20 8.4 10 T mN NBA 27 C 2 ave rmsPR = I ave rms P P RR= =I 28 D Since a photon is emitted, >12EE . ( ) λ−=12energy of photon, hcEE ( ) λ = −12 hc EE 29 A ( )∆= ∆ 2E mc ( ) ( ) ( ) ( )( ) − − − ∆= + − + + = − −− × × = × = × 2 227 8 11 11 3 235.04 2 1.01 140.91 91.91 1.66 10 3.00 10 2.988 10 3.0 10 J U n Ba Kr nE mm m m mc 30 D Number of nuclei remaining = 5.00 × 1012 − 3.00 × 1012 = 2.00 × 1012 8 0 12 12 1.15 10 8 7 2.00 10 5.00 10 2ln 1.15 105 7.97 10 s t t N Ne e t t λ − − −× − = ×=××  = −× = ×

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 7 Paper 2 1 (a) The cyclist moves at a constant speed because the air resistance acting on him is equal in magnitude and opposite in direction to the driving force exerted by the ground on him. Hence, no net force acts on the cyclist and he moves at constant speed in a straight line i.e. constant velocity (in accordance with Newton’s first law). [1] [1] (b) ( ) 2 1 12 11.4112 11 11 22 22 1 2 1 0.01 1 0.1 1 0.02 0.1242 22 2 0.88 2 1.2 2 0.32 0.124 11.411 1.4 1 (1 s.f.) speed 11 1 m s D D D D FFc A vv cA cvF A vFc A v ρ ρ ρ ρ − = ⇒= = ∆∆∆ ∆∆=+ ++ =+ ++ = ∆= = = = ± [1] [1] [1] [1] (c) (i) 21Work done 2 DW F s c Av s ρ=⋅= for a constant force F 2311Power 22 DD dW dsc Av c Avdt dtρρ= = = [1] [1] (ii) ( )( )( )( ) 3311Power 0.88 1.2 0.32 11.411 251 W 250 W22 (If = 11 used, Power 225 W) Dc Av v ρ= = = ≈ = [1] 2 (a) Gravitational potential at a point in a gravitational field is defined as the work done per unit mass by an external force in bringing a small test mass from infinity to that point. [1] (b) ∆=−= − −   = −