2021 A Level H2 Physics P1 Soln

1 C 11 D 21 B 2 A 12 C 22 A 3 B 13 D 23 C 4 B 14 B 24 A 5 A 15 C 25 D 6 B 16 D 26 A 7 D 17 B 27 B 8 A 18 D 28 B 9 A 19 D 29 D 10 B 20 C 30 B Qn Ans Explanation 1 C Smartphone is about 100 to 200 grams. mass ≈ 0.150 kg weight = mg ≈ 0.150 × 10 = 1.5 N = 150 × 10−2 N = 150 cN 2 A Choose an equation that is mostly base quantities. For quantities that are not base quantity, replace it with a combination of base quantities that is equivalent to it. Using F B L= I ,             ( ) 2 12kg m s kg A sA m F maB LL − −−= = = =II 3 B Weight is acting vertically downwards. Air resistance is present and since the ball is moving horizontally rightwards at that instant of time, the drag force due to air resistance is acting leftwards. Hence, the resultant force is both leftwards and downwards. 4 B As the collision is elastic, the speed of approach of the objects before the collision should be equal to the speed of separation of the objects after the collision. X Y Y X X X 1.0 0 0.67 0.33 u u v v v v v vv − = − − = − =− 5 A Weight W is the force by the Earth on the brick. Hence, by Newton’s 3 rd law, the force paired with W is the force by the brick on the Earth. 6 B The upthrust on the ball is 0.25 of the weight of the ball. By Archimedes’ principle, the upthrust is equivalent to the weight of the volume of water being displaced. ( ) ( ) water displaced water ball water 3 0.25 0.25 100 1.0 25 cm U m g V g m g V g V V   == = = = Hence, volume of ball will be double of that and is 50 cm 3. Common Mistake: Some students chose A because they did not realise that the hollow ball was only half submerged. 7 D The stored elastic potential energy is the area under force-extension graph.

Qn Ans Explanation ( )1 1 0 0 stored elastic potential energy 1 2 W x W x=− 8 A Power required to go against resistive forces P Fv= ( ) 6 power required to go against resistive forcesenergy from fuel time efficiency 1000 400 2048 10 20 0.16 0.0521 kg m m =  =  = 9 A vr mv mrr Be Be Be m    = == = Angular velocity is only dependent and proportional to B Common Mistake: Some students chose B because they fail to realise that increased v results with an increased r such that period T (and  ) will be constant 10 B Since the field strength is uniform, ( )( ) 6.0 10 6.0work done 2.0 2.5 3.0 J 10 g xx mgh === = = = 11 D Period T = 24 hours (need to convert to seconds) Radius of orbit (need to convert to metres) = radius of the Earth + height of the satellite above the surface of the Earth ( ) 3122 6400 36000 10 3083 m s24 60 60v r r T  −   = = = +   =       

Qn Ans Explanation 12 C Need to change the molar mass into kg. 2 c used in formula is mean square speed. Need to square root 2 c to obtain root- mean-square (r.m.s.) speed. Multiplying pV = 1 3 Nm<c2> = nRT by a factor of 3 2 gives: ( )( ) 2 5 2 1 1 3 3 2 2 2 3 1.0 10 10 3 43 5000 0.029 498 m s r m c nRT pV pVc nM c − ==    ==  = 13 D The thermal energy supplied is used to increase the water’s temperature and change its state. ( ) 3 V 7 5 4190 70 2260 10 1.28 10 J E mc mL=  + =  +  = 14 B No heat transfer (Q = 0) or no work done on the gas (W = 0) suggests that the change in internal energy of the gas is zero. Internal energy is the sum of of a random distribution of kinetic and potential energies associated with the molecules of the system (i.e. inter-molecular kinetic and potential energies). 15 C There are only conversion of energies between potential energy and kinetic energy. When the potential energy decreases, kinetic energy increases. Hence the kinetic energy line should be mirrored horizontally to the potential energy. 16 D 8 6 14 3.0 10 0.60 10 m5.0 10 2 1.522 0.60 c f x      −= = =    =  = −  = 17 B At 45°, the intensity of light is the highest: ( ) 22 0 cos cos 90  = −  II At 90°, the intensity drops to zero. 18 D min 9 16 2 10 620 10 5.7 10 50 10 7.1 10 m b x x  − − = = = 19 D

Qn Ans Explanation 20 C Coulomb’s law of electrostatic attraction: ( ) 12 2 0 12 new 2 0 4 2 1 1.0N242 QQF r QQFF r   = = = = 21 B Resistance of each wire: 1R GL= Since the wires are parallel, 1 eff 4 4 GLR GL − == 22 A The reason for drawing the lines and axis coloured red: When the voltage across the thermistor is 3.0 V, the resistor should have 0 V as the emf is 3 V. 23 C Since the lamp is connected across the thermistor, the potential difference across the thermistor needs to be higher for the lamp to glow brighter. By potential divider rule, the resistance of the thermistor should increase and the resistance of the LDR should decrease. Hence, the thermistor should be cooler in temperature and the LDR should have more light on it. 24 A Current around the circular coil produces a magnetic field directed into the page. By Fleming’s left hand rule, the magnetic force on the short wire is upwards. 25 D      2T m BA BA   = ==

Qn Ans Explanation 26 A By Faraday’s Law, ( ) ( ) ( )( ) 2 2 0 3000 1.8 0.010 0 28.3 V0.060 N t NB R t    =  −=  − == 27 B SS PP P S S 16 1.88 16 1.88 1.880.32 0.0376 A16 NV NV I == = =  = I I 28 B 3 1 3 2 2 1 photon 3 1 3 2 2 1 1 31 1 1 1 11 252nm440 590 E E E hcE     → → → → → → − → =+ = + =  = + = 29 D Given that the two beams have the same frequency, the energy of each photon from each beam is the same. To increase intensity and hence total energy per unit time in a beam of light, there must be larger quantity of photons. 30 B ( )( )( ) 22 4 27 8 13 136.90709 136.90583 5.49 10 1.66 10 3 10 1.06 10 J E m c −− − =  = − −    =