2021 9749 H2 Physics 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 v x = 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 . mg += submergedVgρ+ 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 ρ ρ = = = = = = = Notes: ball is only half-submerged 7 Always check the axis on graphs. Area under force-extension graph gives elastic potential energy stored. E xtra 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 Fv Fv F vt FvE F vt E = = = = = = = = = = ×× = = = = × 9 magnetic force provides centripetal force 2Bqv mr Bq r ω= ω( ) mr= 2ω Bq m ω= 10 field strength is numerically equal to potential gradient at that point ( ) ( )( )( ) -2 J w.d. d6 0 6 m sd 10 2 06 25 30 g. r Fd mg d .. . φ= = ≈ = = = = 11 consider distance from Earth’s centre ( ) ( ) 1 33 - 610 236000 6400 24 0 60 3100 m 0 s 1 Evr R h ωω π = = + = +  ××  × × =

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 t hermal 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 λ φ πλ φπ π π λ λ π − − − = × × = × × × = = = ∆∆ = ∆∆= = = 17 Malus Law for Intensity: 2 0cos θ=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 b etween 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. . λ λ −×= × ≈ ≈ × = 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 πε= = 2 1QQ ( ) 2 old new 2 old ne 3 w old 1 new 2 2 N20 12 1 1 0 r r rFF r Q . Q . =             = =    21 Identify quantity related to gradient G: gradient R A G A ρ ρ = = 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 m 2 26 by faraday’s law ( ) ( ) ( ) ( ) 2 2 d d 18 00 1 3000 28 V06 0N BA BANtt r . N t t EN B . . π π ∆ =−  Φ −≈− =− = = ∆∆ = ∆

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 23EEE hc →→→ = + hc λ = 440 nm hc λ + 99 590 nm 1 m 440 10 590 10 2 1 0 1 n 5 λ λ −− −   ×× =  + = 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 Em mc m m m uc . β − = = × − −− = 440 nm 590 nm

6 ©EJC 2022 9749/H2 PHYSICS 2021 Paper 2 Structured Questions General Notes: markers noted that higher quality responses (i) did not miss out on related elaborations, (ii) reflected common sensical logic checks e.g. whether the gravitational field strength in Q4 was of a reasonable value and (iii) demonstrated good presentation. In the same year, markers of the H3 paper commented that strong presentation meant (i) writing out working clearly, (ii) manipulating equations via algebraic quantities for as long as reasonable and (iii) being sensitive towards s.f. considerations. Qns Marks 1(a) [dxn] moves downslope [magnitude] with speed that increases at a constant rate B1 1(b)(i) gradient of displacement-time graphs gives velocity 12 12 1 v 4 elocit 0 03 0 6 01 2 04 y 1 26 m s 6 s t t . s .. .. − − −≈ − ≈ = − B1 B1

7 ©EJC 2022 9749/H2 PHYSICS 2021 Qns Marks 1(b)(ii) 2 2 2 2 2 6 15 m s OR 1 26 0 0 200 1 25 63 3 2 m s0 200 2 12 6 0 2 01 0 v a. v u at . a . . ua . .a . a s . − − = = = + = × + = + = = + Note: The formula is chosen such that v calculated in the previous part is used, as required by question. M1 A1

8 ©EJC 2022 9749/H2 PHYSICS 2021 Qns Marks 2(a) 2 2 3.20 = 2(0.62) = 8.26 J 2pKE = m C1 C1 A1 2(b) Taking downwards as positive, 1.80 3.20 5.00 N s 0.68 0.53 0.15 s By N2L : 5.00 = 33.3 Nnet 0.15 net 33.3+0.62(9.81) 39.4 N pp p fi t pF t F N-W N ∆= − = −− =− ∆= − = ∆−= = −∆ = = = Average force will be 39.4 N, upwards. Note: commonly made error was missing out on change in direction C1 C1 M1 A1 2(c) 2 upon bouncing 2 (-1.8) 2(0.62) 2.61 J (2.61)Fraction of energy left after each bounce 0.316(8.26) Let be the number of bo