EJC 2022 Prelim P1 Guide

©EJC 2022 9749/J2H2Prelim/2022 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2022 General Certificate of Education Advanced Level Higher 2 PHYSICS MARK SCHEME 9749 Aug/Sep 2022 Paper 1 Multiple Choice Question Key Question Key Question Key 1 B 6 D 11 B 2 A 7 C 12 A 3 C 8 B 13 D 4 C 9 D 14 C 5 A 10 C 15 D 16 D 21 A 26 A 17 A 22 B 27 A 18 B 23 D 28 C 19 D 24 A 29 D 20 C 25 D 30 B 1 mass of electron negligible compared to nucleons question similar to asking for nuclear density: material nucleus mas 4s V = 4 u 3  3 nucleus 27 18 3 4510 10 10 kg m r − − −= 2 0.22EtRs     =  0.22 units of units of kg m EtsR  − =  =   22 m s − ( ) ( ) 2s kg ( ) ( ) 0.2 3 0.25 m m m dimensionless1 − − −    =  = 3 constant acceleration so ( ) 2 1 0 1 22 9 81 3 202 15 7 m s gs ut at t . . u. tu u − =+ −=+ − = =+ 4 upthrust depends on weight of fluid displaced so need to displace a greater volume of fluid, increase either x or z 5 each spring supports 4 N; area under force-extension graph up till 4 N: ( )( ) 21 3 10 4 0.060 J2 −= 6 a counter example is friction between tyres exists, but does not reduce total energy when there is no slippage even as vehicle (the machine) does work (increase KE)

2 ©EJC 2022 9749/J2H2Prelim/2022 7 apply knowledge from Newton’s cradle that transfer of KE between 2 masses is 100% if masses are equal and one mass is initially stationary we model the system as separated blocks doing elastic collisions, take right +ve: consider A and B only: PCLM: relative spe 1ed 0 : 0 A A A A B B BA u m m v mmvv v + = + =− ( )10 m= 4Avm+( )( ) 1 1 10 6m 1 3 5 s 10 4 5 m s 40 0 0 A A A BA v v v vv − − = + =− + + =− = = so A will move left. B will undergo newton’s cradle style collision until E moves to hit F consider E and F only PCLM: relative spee 0 d 4 : 4 E E E E F FE Fum m m v v v v m − + = + = ( )( )44 m=( ) Evm+ ( ) 1 1 12 2.4 3 4 16 m s5 2 6.4 m s 54 5 E E E F v v v v − − =+ + = = + + == so both E and F moves to the right. that leaves B, C and D at rest. 8 sketch in order visualize friction pivot at wall F W W N L N W = = ( ) cos friction2 L +  ( ) sin F LN  = ( ) cos tan2 tan 2 FW WF W N WNN N    = − += A B C D E F W NF NW friction

3 ©EJC 2022 9749/J2H2Prelim/2022 9 tension in string 1 provides centripetal force for 3 masses so must be largest magnitude; eliminate A and C centre of mass of 3 separate masses is at centre i.e. mass 2 so tension in string 1 is equivalent to a string providing centripetal force to a mass of 3m at radius 2r: ( )( ) ( ) 2 1 2 3 1 3 32 3 2 T m r T m r T T   = = = eliminate B or via considering free bodies: T3 provides centripetal force on outer mass ( ) 2 33m r T  = vector sum of T2 and T3 provides centripetal force on centre mass ( ) ( ) ( ) ( ) ( ) 2 23 2 23 22 2 2 23 5 2m r T T T m r T m r m r mr     =− =+ =+ = vector sum of T2 and T1 provides centripetal force on inner mass ( ) ( ) 12 1 2 2 2 2 22 5 6 mr T T T mr T mr m r mr     =− =+ =+ = 10 mass on moon needs just enough KE to reach location of 0 field strength, then it will accelerate towards earth “from rest”: loss in KE gain in EPE 1 2 m = 2 0v m−= ( ) ( ) ( ) 2 6 1 smaller area 10 2280 m 2 2.6 s 2v v  − = =   = 11 satellite is now further away from earth so GPE must increase 12 kinetic energy is directly proportional to thermodynamic temperature 3 2 1 2 kT = ( ) 2 new old old new old old 3 100% 100% 1.4% 3 273.15 40.5 273.15 32.1 273.15 32.1 mv k Tm v kTv m v v TT T = −  −= +  + = = = + − 13 ( )TNkpv NkT p V= → = X has less steep gradient 14 coin loses contact when piston retracts downwards at an acceleration larger than magnitude of free fall acceleration i.e. ( ) ( ) SHM 2 0 2 0 2 2 0 9.81 1 4 1.9 Hz .07 ag xg x f f      = = = mouter T3 mcentre T2 T3 T1 T2 minner

4 ©EJC 2022 9749/J2H2Prelim/2022 15 ( )2 0.90 1.80 m == ( ) ( ) 360 1.30360 3601.80 2 2 60 x x      =  =   =  = 16 check for max observable order for violet: 2 9 sin let sin 1 10 5 1 5000 400 10 d n dn    − − = →  →=  = fourth order violet is visible 17 aim is the NOT distinguish the pixels hence use the smallest wavelength to determine the shortest pixel distance that can be distinguished. sin s br   for maximum distance between pixels smax, we consider shortest wavelength (blue):  − − −     9 3 5 0.1 470 10 0.1 4.0 10 1.2 10 m d b d d 18 connecting wire so all metallic surfaces at same potential 12 0 1 4 V V  =    1 10 1 4 q r  =   2 2 01 2 1 4 q r E E    = 1 2 1 0 1 4 q r        1 11 2 122 2 22 2 1 1 q r r rqq r rr r   ==    19 electric force is vector quantity, and the charge at X is negative so electric field strength actually points to right: by W by Z by Y 2 2 2 2 2 2 22 2 vertically : 0 cos 45 c 2 os 45 2 2 2 1 XE E E E qQ LL qQ LL qQ L L L L L q =+ = −+ = −+ =− = + 2L Q 20 resistance of R3 alone is larger than the effective resistance of R2 in parallel with R3 by potential divider rule: p.d. across R1 larger p.d. across R3 smaller total circuit resistance drops so battery outputs more power ( ) 2 total emfP R= W Z Y −Q q +Q X Eon X L

5 ©EJC 2022 9749/J2H2Prelim/2022 21 by potential divider rule: ( ) ( ) ( ) ( ) 1 sec main 12 2 sec main 12 2 2 ____ 1 ____ 3 60 20 2 20 1 : 1 2 60 EER E R E RL R R R R L R + + == = = 22 with resistor across A and B, there is complete circuit for which induced current can flow, so there is damping on oscillations. 23 consider change in magnetic flux: ( ) ( ) ( )( ) ( ) final initial initial 44 6 2 2 25 10 2.0 10 sin 2 60 3 10 BA A B B AB −−  = − =− =−    =−  by Faraday’s law, induced e.m.f. E is: ( ) ( ) ( ) 6 5 d d d d 5 C 00 2 5 3 10 8.66 1 .0 0 EN t R Q R t B R N R NQ AQ N R    − − = ==  = == = = I 24 1 period is 5 ms ( )( ) ( )( ) 2 22 2 rms 2 rms area under graph 9 2 1 2 20 A ms 20 20mean sq W 2 uare 5 4 A A 52 T PR = = = =+ = = == I I I I 25 low intensity high freq can result in photoelectric effect, eliminate A ( ) ( ) elimina2 t2 e s s hf eV h f e V − −  =  B statement is correct for min frequency so should have been max wavelength, eliminate C intensity can mean increased energy per photon or more photons 26 electron KE converted to photon: ( ) ( ) 1 lg lg g 1 l hceV V hcV e    − − = = =+ straight line with negative gradient I2 / A2 t / ms 9.0 1.0 4 6 8 9

6 ©EJC 2022 9749/J2H2Prelim/2022 27 by Heisenberg’s uncertainty principle, px h  ( )( ) 34 31 6 7 6.63 m 9.11 1 0.20 100 0.20 100 10 0.20 10 10 . 0 5 2.4 100 1 e e v hh v pm x p m − − − =   = =   =   = 28 ( ) ( )0 bgexpC C C t= − + ( ) ( ) 1 ln 2exp 2877.3 8.3 8.3 752 61.6 0 s− = − − +  = 29 energy needed in reaction ( ) ( )( ) 2 H O N 2 227 8 13 6 1.007825 16.999130 14.003074 4.002604 0.001277 1.66 1.91 10 3 10 10 J 10 eV1.19 m m m uc uc m − − = + − +  =− −  =  −  =   = 30 Isotope P could be formed from isotope Y after 2 successive alpha decays. Isotope R could be formed from isotope Y after an alpha decay followed by a beta decay (or a beta decay followed by an alpha decay).