EJC 2022 Prelim P3 Guide

©EJC 2022 9749/J2H2MYE/2022 ; EUNOIA JUNIOR COLLEGE JC2 MID-YEAR EXAMINATIONS 2022 9749 PHYSICS MARK SCHEME Paper 3 Longer Structured Questions Qns Answer Marks 1(a)(i) gravitational force of attraction per unit mass acting on (OR experienced by) a small test mass placed at that point (in the gravitational field) 1 1(a)(ii) gravitational force of attraction between two point masses is directly proportional to the product of the masses and inversely proportional to the square of separation between the masses Let m be the mass of an object distance R away from Mass M field strength is gravitational force of attraction per unit mass experienced by small test mass placed at that point 1F m m = GM m 22 GM rr = 1 1 1 1(b)(i) ( ) ( )     − =  = = =  = =  3 30 16 3 3 4 13 4 3 3 8.245 10 4volume of star 3 4 2.7 10 2 10 7.54 212 0m 3 6. m 3 .7 10 kg r m v 1 1 1(b)(ii) (words to effect of) density increase closer to centre (words to effect of) outer layers compress inner layers 1 1

2 ©EJC 2022 9749/J2H2MYE/2022 Qns Answer Marks 1(b)(iiI) By conservation of energy: ( )( ) 2 11 30 4 81 2 6.67 6.22 oss in KE gain inl 1 10 10 10 2 2. 1.75 1 GP 0 m 7 E 00 s GM G mm mv r v r − − = = − = = − −   =    1 1

3 ©EJC 2022 9749/J2H2MYE/2022 Qns Answer Marks 2(a) oscillatory motion where acceleration is directly proportional to displacement from equilibrium position and directed opposite to displacement 1 2(b)(i) 0 2 2 2 00 1 2 2 2 a 2 2 0.74 mplitude 0.7 . m m 00 1 . .0 2 5 s x xT xxvx   − = − = − =     = − = 1 1 2(b)(ii) ( ) 00 00 0.15 0.55 22 0.7 2.1374 relative to equilibrium relative to groun d m m OR sin sin 0.7 sin 0.55 sin 20.15 0.7 sin d uration 4.0 xx x x x x xx T t tt T tt   =− = ==     +  −=   +    =    = =      ( ) ( )C 8 7 s 0. a 137 48 2 c n be by G .2 −− = 1 1

4 ©EJC 2022 9749/J2H2MYE/2022 Qns Answer Marks 3(a)(i) ( ) ( ) ( ) ( ) v loss loss v l 2 1 2 v 1 os 2 s 1v 1 1 loss 2v 2 2 1 1 1 2 2 v 12 ____ 1 2 ____ 2 1 m E mL h Pt hmP tt mm tt V m m m tt P hmPL h L P P L P t V t tt L t L P = + = =    =−       =     −        =+     =+    + − − − =  −II ( )( ) ( )( ) 12 61 33 5.0 78 4.0 60 16 10 10 10 1. 2.25 10 1 J k 5 g 60 .5 60 m t −− −   − −=  − =        1 1 A0 3(a)(ii) pure substances undergo phase change at constant temperature (words to that effect) so temperature difference with surrounding kept constant 1 3(b)(i) ( )( ) 6 6 1.0 2.25 1 J2 0 0 12. 5 Q mL =  = = 1 1 3(b)(ii) ( ) ( )( ) ( )( ) on final initial 3 5 by 5 5 J 1.67 1 1 . 0 01 40 10 1.668 1 1.67 9 .0 10 1. 60 w w p V p V V − = = =−  =− − − =−  + − 1 1 3(b)(iii) ( ) 6 6 52.25 10 1. J. 67 0 2 08 10 1 U Q W=  =  + =  + − 1 3(b)(iv) l gas liq gas iquid 65 uid 62.42 10 J PE PE PE 10 3. 02.0 E 41 PE PE P 18  =  = −    = −= + 1 1

5 ©EJC 2022 9749/J2H2MYE/2022 Qns Answer Marks 3(b)(v) pure substances undergo phase change at constant temperature so total translational kinetic energy of particles remain constant large increase in volume for phase change from liquid to gas so separation between molecules instead intermolecular bonds are complete broken as potential energy between particles increases work is done against atmosphere B0 1 1 1

6 ©EJC 2022 9749/J2H2MYE/2022 Qns Answer Marks 4(a)(i) current in solenoid produces magnetic field with field lines parallel to axis of solenoid radii of copper disc cuts magnetic flux as it rotates, (by Faraday’s law) rate of change of magnetic flux linkage results in emf 1 1 1 4(a)(ii) consider a radial strip of copper: ( ) ( ) ( )( ) ( ) 2 2 2 d b 1 y faraday's law: d d 1 1 2 1 )2 d2 d 2 d BAf t N R EN t B B BAt BR Rf     == = =   =  =  1 1 4(b)(i) at null deflection, p.d. across resistor = induced e.m.f. ( ) R 0 0 same current in solenoid and resistor BUT NOT COPPER (disc/axle) VE R BAf Af f n R nA   = = = = I I 1 4(b)(ii) no electrical quantities needed so not dependent on accuracy of any voltmeter, ammeter or ohmmeter used 1 4(c) ( ) ( ) ( ) ( ) ( ) 0 0 0 __ 1 1 2 __ 2 take : BAf n Af R nAf nBAf R    = = = I ( ) AfI 0n Af ( )( ) ( )( )( ) 3 2 . 10 T 1.0 1 0.20 0.0 0 9 0 15 5 B R AF  − = = = I 1

7 ©EJC 2022 9749/J2H2MYE/2022 Qns Answer Marks 5(a) 2 max 1 2 s s h e f eV h mv Vf e = =+   = +   −   ( ) 19 14 34 when 2.4 1.6 10 5.79 10 Hz6. 3 0 10 , 6 sV f h − − =   =  ==  . ( ) when 0, 2.4 eV 2.4 V extrapolated s f e e V = =− −== − 1 cut x-axis at (5.8, 0) 1 dotted line to (0, −2.4) OR pass through (8, 0.9) 5(b) peak, supply supply, rms2 2(4.6) 6.5 V VV = == peak, diode peak, supply 6.5 V VV = = 1 Vs / V 0 0 2 4 6 8 10 f / 1014 Hz 1 − 1 − 2 − 3 5.8 − 2.4 (checking pt

8 ©EJC 2022 9749/J2H2MYE/2022 Qns Answer Marks 5(c) 1

9 ©EJC 2022 9749/J2H2MYE/2022 Qns Answer Marks 6(a)(i) time spent between plates: ( )( ) ( )( ) ( ) 1 12 5 1 5 5 2 5 19 27 2 7 22 initia 1 2 l 10 6.5 1.6 50 10 10 10 10 1 p 1 . 0 2 6 0 0 1.67 4.0 0 6.76 1 m s acce t .8 10 m 0 speed 1.54 10 s 1.84 10 m s s m s 10 Lt v qE q m md v u at F a v qE v ma a t V − − − −− ⊥ ⊥ − − − − = = + = + = +    ==  == ==   = ==  = 1 1 1 1 6(a)(ii) consider potential change from 0V equipotential line: 2 p 177 V 1 2 qV mv W V e ⊥   = = = Potential is – 177 V as protons will displace towards region of lower potential OR consider displacement from 0V equipotential line: ( )( ) ( )( ) ( ) 2 219 2 527 2 1.6 500 100 6.51.67 4.0 2. 1 V 0.01 1 2 101 10 2 01 0 0 10 250 177 cm 41 m s V s atut⊥ − − −− ⊥       =+  =  +   == = Potential is – 177 V as protons will displace towards region of lower potential 1 1 1

10 ©EJC 2022 9749/J2H2MYE/2022 Qns Answer Marks 6(b) 6(c) uniform magnetic field (where field lines) pointing upwards no deviation for particles with velocity that result in equal magnitudes of electric and magnetic force acting in opposite directions 1 1 1 