2021 9749 H2 Physics P1 MS EJC Suggested Solutions

©EJC 2020 9749/H2 PHYSICS 2020 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2021 Paper 1 Multiple Choice Question Key Question Key Question Key 1 C 6 B 11 D 2 A 7 D 12 C 3 B 8 A 13 D 4 B 9 A 14 B 5 A 10 B 15 C 16 D 21 B 26 A 17 B 22 A 27 B 18 D 23 C 28 B 19 D 24 A 29 D 20 C 25 D 30 B 1 Estimate mass ≈ 150 g = 0.15 kg W = mg = (0.15)(9.81) ≈ 1.5 N = 150 cN Notes: centi- means divide by 100 e.g. centimetre, deci- means divide by 10 hence decimal point 2 One tesla is the uniform magnetic flux density which, acting normally to a long straight wire carrying a current of 1 ampere, causes a force per unit length of 1 N m–1 to act on the conductor. ( )( )( ) -2 -1 -2 11 N 1 A N T A 1 m sin 90 units of T kg m s kg A s m A m B =  = == = 3 air resistance acts in direction opposing relative motion so left weight acts down constantly Vector addition gives B 4 elastic collision so special result is speed of approach = speed of separation. Let vx be speed of X after collision. v - 0= 0.67 v – vx vx = 0.33 v 5 N3L states that force is of same type (gravity): eliminate C & D force acts on another body (S acts on brick so N3L-pair cannot act on brick): eliminate B & D

2 ©EJC 2022 9749/H2 PHYSICS 2021 6 forces acting on ball include tension read from newton meter, weight and upthrust 0 75 T U mg . m g += submergedVg+ mg= ( ) ( )( ) ( )( ) submerged 3 submerged total submerged total submerged 3 0 25 0 25 0 25 in cm 0 2 100 1 2 252 100 1 1 50 cm V . m .mV V . . V VV   = = = = = = = Notes: ball is only half-submerged 7 Always check the axis on graphs. Area under force-extension graph gives elastic potential energy stored. Extra potential energy is difference in “area under graph” 8 constant speed so zero net force force provided by engine has same magnitude as resistive forces ( ) ( )( )( ) ( )( ) output engine resistive resistive resistive fuel resistive fuel fuel 66 6 so work with time 50 s KE KE 0 16 0 16 0 16 10 10 0 16 10 0 0052 kg 1000 20 48 48 400 20 50 48 E t Pt .. .m t . . Pt s v P F v Fv F v t FvE F vt E = = = = = = = = = =  = = = =  9 magnetic force provides centripetal force 2Bqv mr Bq r = ( ) mr= 2 Bq m = 10 field strength is numerically equal to potential gradient at that point ( ) ( )( )( ) -2 J w.d. d6 0 6 m sd 10 2 0 6 2 5 3 0 g. r Fd mg d . . . = =  = = = = 11 consider distance from Earth’s centre ( ) ( ) 1 33 - 610 236000 6400 24 0 60 3100 m 0 s 1 Ev r R h   = = + =+      =

3 ©EJC 2022 9749/H2 PHYSICS 2021 12 assume ideal behaviour: ( )( ) ( )( ) 51 10 10 3 4 5000 8 31 289 K 1 2 pV nRT pVT nR . =    = == 2 3 2 rmsmv = ( )( ) 23 -1 3 3 1 38 10 28 9 10 93 2 500 m s rms A kT .kTv m N − −=  = =  Notes: need to change molar mass into SI units of kg. 13 thermal energy supplied goes to (i) heat liquid water up to 100 °C then (ii) boil ( ) ( ) ( )( ) ( ) vap vap 3 7 5 00 4190 100 30 2260 1 28 10 J 10 Q mc T mL m c L . . T=  + =   = − +  = +  14 the macroscopic “external” KE/PE does not affect internal energy (unchanged) 15 At max PE, KE = 0. At max KE, PE = 0. in simple(r) systems, by conservation of energy, PE + KE = constant total E 16 this works for laser light because the beam is coherent 8 14 7 6 7 3 10 5 10 6 10 m 2 2 1 5 10 6 10 5 0 rad 1 0 rad 2 v . v f f . . x x          − − − =   =   = ==  =  = = = 17 Malus Law for Intensity: 2 0cos =II where θ is relative angle X polarises incoming light vertically (0°) From 45 to 90, output intensity after rotated polarizer and before Y decreases to zero, because angle between X and rotated polarizer are 90 to each other: (B) From 90 to 135, output intensity after Y decreases to zero, because angle between rotated polarizer and Y are 90 to each other: (B) 18 sketch: ( ) 9 16 10 620 105 7 10 05 7 1 10 m s r rb s. b. .   −=     = s

4 ©EJC 2022 9749/H2 PHYSICS 2021 19 recall definition of electric field line. 20 take ratio: 12 old 2 0 2new ol 3 d 4 1F r F F QQ QQ = = 21QQ ( ) 2 old new 2 old ne 3 w old 1 new 2 2 N2 0 1 21 1 0 r r rFF r Q . Q . =            ==     21 Identify quantity related to gradient G: gradient R A G A   = = l 4 wires are parallel (current “splits up” and rejoins”) so effective resistance is 1/4 original eff 44 RGR = = l 22 thermistor and resistor have same current flowing through them At current = 0.1 A, total p.d. across them is 3.0 V, equal to emf of battery. 23 2VP R= for bulb to glow more, the p.d. across it must increase, then the effective resistance between it and the variable resistor in parallel must increase by potential divider concepts, > LDR must be brighter to decrease its resistance > thermistor must be cooler to increase its resistance and consequentially the resistance of the parallel branch comprising bulb and thermistor 24 By Right Hand Grip Rule on circular coil, B points out into plane of paper by FLHR, upwards force on short wire 25 units of BA = T m2 26 by faraday’s law ( ) ( )( ) ( ) 2 2 d d 1 8 0 01 3000 28 V 06 0NBA BANtt r . N t t EN B . .    =−   −− =− ==  = 

5 ©EJC 2022 9749/H2 PHYSICS 2021 27 ideal so no loss of energy ( ) pri sec pri pri sec sec pri sec pri sec 160 32 13 6 P P VV V V .. . = =      =  =  II II 28 visualise the 3 possible transitions. longer wavelength is lower in energy: 13 21 23E E E hc →→→ =+ hc  = 440 nm hc  + 99 590 nm 1 m 440 10 590 10 2 1 0 1 n5   −− −    =  + = 29 t n t P nhf AA hf A == = I I 30 working with mass: ( ) ( ) 2 2 rxtnt pd 13 t 1 1 10 J Ca Ba E m m c m m m uc .  − = =  − −− = 440 nm 590 nm