2022 9749 H2 Physics P1 MS EJC Suggested Solutions

©EJC 2022 9749/H2 PHYSICS 2022 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2022 Paper 1 Multiple Choice Question Key Question Key Question Key 1 A 6 C 11 A 2 C 7 B 12 C 3 B 8 C 13 B 4 B 9 C 14 C 5 D 10 D 15 A 16 D 21 D 26 A 17 B 22 B 27 B 18 B 23 C 28 A 19 A 24 D 29 C 20 A 25 B 30 C 1 Since vectors are initially in opposite directions, the initial (X − Y) yields the largest magnitude, eliminate B and D 180∘ means that Y is rotated half-round only, where magnitude of (X − Y) will not return to original value. 2 took 1 sec for vertical component of velocity to be zero ( ) ( )( ) 1 1 1 50 50 m s . 9 1 m s . 0 9.81 t 1 9 81 .81 5 an tn 11 1 0a xy xx x x yy yy y x u uu u t s su t at at u u v uv u  − − − = = = =+ = − =  + == = =  = −− =

2 ©EJC 2022 9749/H2 PHYSICS 2022 3 by conservation of linear momentum ( ) ( ) ( ) ( ) ( ) ( ) 1 e 5 l 2 a 0 s . t 2 i . c 05 2.5 _____ 1 0.5 2 0 1.5 5 3 o so m s fr m 1 : P p Q Q P Q Q Q Q QP Q v v v m m m mm v vm m − + = − + − = −= = = = − 4 consider free body diagram of buoy: ( ) ( ) ( ) ( ) water submerged 3141030 0.5 200 9.8123 683 N U W T T U W V g mg    − =+ = − = − =   = 5 consider free body diagram of mass in translational equilibrium: ( ) ( )( ) ( ) ( )( ) ( ) ( ) horizontally: sin 36 25 0.06 vertically: cos 36 25 0.06 cos 36sin 36 2.06 N T kx TW W  = = =  =  = 6 mean density so assume earth is uniform sphere: ( )( ) 2 3 6 11 3 9.81 44 33 4 3 4 6.37 10 6.67 103 5512 kg m GMg g G r g r rG r MG      − − = == = =  = 7 by conservation of energy, loss of GPE = work done against air resistance: ( ) ( ) ( )( )( ) 02 0.2 0.2 loss in GPE 0.6 0.6e 0.4 9.81 0.6 1 e 0.427 J 2n mg h h mg − − =− =− − = = = T W Upthrust T W Fspring 36∘

3 ©EJC 2022 9749/H2 PHYSICS 2022 8 all points along a radius have the same angular speed (the linear speed of the point on the circumference is the max and the linear speed at the centre is zero) 9 Eliminate A and B as all geostationary satellites, regardless of their mass, has to be at a fixed distance away from centre of Earth Eliminate D, the satellite will have the same angular velocity as the point on Earth’s surface directly below them but the satellite will have far more linear velocity (see reasoning in Q8) 10 gravitational potential is a scalar sum so: ( )   P due to M due to 4M 2 10 2 4 142 GMGM GM GM dd dd   =+         +     =−          −= − −=+ 11 ideal gas so internal energy is purely KE and is directly proportional to thermodynamic temperature: ( ) new new old old 2 new new 2 oldold new new old old 1 KE KE 160 2 80 273.15 73.15350 388 m s c c c T c T T T T T − = = += = = + 12 ideal gas so internal energy is purely KE and is directly proportional to thermodynamic temperature. Since temperature remains constant, total KE of both initial or final states is same. ( )( ) 2 2 total 5 1 3 31 KE22 3 10 0.012 1500 J p c p m V Nm N V c = == = = 13 half of KE converted into thermal energy 2 2 11 22 4 v c mv mc T T  = = 14 start with displacement equation and differentiate with respect to time ( ) ( ) ( )( ) ( ) ( ) 0 0 sin sin 0 3 5 .3 sin 7 2 2 5 0. 1 2 d d cos 0. 7 2 5 3 7 cos .2 x x t xt t xv t t t T      = =  =  = =   15 A because the radio need not be outputting sounds of (driving) frequency which matches that of the natural frequency of the loudspeaker

4 ©EJC 2022 9749/H2 PHYSICS 2022 16 diagram 2 shows frequency 11 5 Hz0.2f T === diagram 1 shows wavelength ( )( ) 1 0.8 m 5 0.8 4 m svf   − = = = = 17 stationary wave so XY represents half- wavelength ( )2 5 10 cm == eliminate A and C 6 divisions on time base gives 1 period ( ) ( ) 3 3 6 0.05 1 6 z3 0 10 1 10.0 3 3 5 3H T f T − − = =  ==  18 double slit experiment so 15 xx a a D →= = 9 3 3 9 3 3 3 700 0.0001754 50 4 m 10 m10 20 40 2 0 m10 10 m10 10 0.050 50 0 1 0.000 252 10 20 A B C D x x x x − − − − − − − ==  = ==  = = =  19 approach question using kinematics consider time of flight (time spent inside uniform field) ( ) ( ) 2 2 ____ _____ 2 1 2 0 1 1 2 _ y x yt t V d eVa md eV md a v F qE me su v t y v at = = = =   =    =+   +   = 20 current along wire is constant so the larger the diameter, the lower the drift velocity, eliminate C and D ( ) 2 2 2 2 4 1 Anvq nve d nve dv r ne    = =  =        = I

5 ©EJC 2022 9749/H2 PHYSICS 2022 21 non-ideal voltmeter can be regarded as its resistance in parallel with an idea voltmeter // // // 2 12 4 1 3 22 12 3 P PP Q Q PQ P Q R R R V V V RR R RR R R = = = = = == = I I 22 e.m.f. of call is 65 cm worth of p.d. ( )( ) 2e.m.f. 65 10 V 14.3 9.30 − = = 23 the forces are N3L pairs, eliminate B & D wires attract so current flowing in same direction 24 initially current is normal to B so max value expected with θ = 0, eliminate A & C sinF B L = I so cannot be straight line 25 component of flux normal to area is ( ) ( ) ( )( ) ( ) 64 8 sin 60 sin 60 10 10 sin 60 6.75 10 Wb 65 12 BB B A BA − ⊥ ⊥ − −   =     = = = =  26 regular square wave so rms 0=I I 27 magnetic flux linkage in an a.c. generator is of the form ( ) ( ) sin d cosdt N NBA t N NBA t    = = peak e.m.f. is halved, new 2VP R= is 1 4 of original original power: ( ) 2 rms 2 0 W 2 2 20 10 4 PR R = = = = I I

6 ©EJC 2022 9749/H2 PHYSICS 2022 28 electron has mass, consider: 22 2 new old old new new 3 2 2 2 9 phE m m h mE E E E E      = = = = = = 29 mass defect is difference between total mass of individual separate nucleons and mass of nucleus bismuth isotope has 83 protons and 129 neutrons pn83 129m M M M+= − 30 alpha decay reaction: ( ) 2 238 2 13 234 4 2U Th energy released rest mass rest mass of uranium o J 238.1249 234.1165 4.0026 8 s .67 f product 10 u c c  − →+    =−       = = − −