EJC 2022 Prelim P2 Guide

©EJC 2020 9749/J2H2JCT/2020 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2022 General Certificate of Education Advanced Level Higher 2 PHYSICS MARK SCHEME 9749 August/September 2022 Paper 2 Structured Questions Qns Answer Marks 1(a) Forces must be labelled forces clearly in full – Weight, Reaction Force, Tension Lines of action forces (including intersection of the 3 lines) must be clearly shown. 1 (b) Let T be the tension in the spring. Taking moments about A, Sum of clockwise moment = Sum of anticlockwise moment mg (0.20 cos30°) = T (0.40) 1.2(9.81) (0.20 cos30°) = T (0.40) T = 5.10 N 1 1

2 ©EJC 2022 9749/J2H2Prelim/2022 Qns Answer Marks (c) Let Rx and Ry be the horizontal and vertical components of the reaction force R. Resolving forces horizontally, Rx = T cos 60° = 5.10 cos 60° = 2.55 N Resolving forces vertically, Ry + Tsin 60° = W Ry + 5.10 sin 60°= (1.2)(9.81) Ry = 7.36 N 22 XY 22 magnitude of reaction force 2.55 7.36 7.79 N RR R+ =+ = = 1 1 1 (c) increase (stiffer spring so a smaller extension needed for the same elastic force, resulting in larger angle) 1 Ry Rx

3 ©EJC 2022 9749/J2H2Prelim/2022 [Turn over Qns Answer Marks 2(a)(i) Total momentum of an isolated system of interacting bodies before and after collision remains constant if no net external force acts on the system. 1 (a)(ii) Presence of external force like weight / force from the ground 1 (b) By Conservation of Energy, loss of gravitational potential energy = gain in kinetic energy m 1 2gh m= ( )( ) 2 1 2 2 9.81 2 2 . . 0 6 6 m s v v gh −= = = 1 (c) collision is elastic, total KE before and after bounce conserved ½ (0.8)(6.26)2 = ½ (0.3)(v2) + ½ (0.5)(3.2)2 v = 9.35 m s-1 1 1

4 ©EJC 2022 9749/J2H2Prelim/2022 Qns Answer Marks 3(a)(i) Since the direction of field strength is directed upwards at point O, the horizontal component of the electric field strength due to the 2 charges at O are equal in magnitude and opposite in direction. oo12 22 o 1 o 2 sin 30 sin 6044 sin 60 1.73 sin 30 1.7 oo qq rr q q   = == = 1 3(a)(ii) correct shape, graph does not touch the y-axis or dotted line (else max 1) E = 0 at position nearer to Y Relative field strength at X larger in magnitude than at Y 1 1 3(b)(i) electric field strength is a vector quantity Since the resultant field strength at point O is directed upwards, the resultant electric field strength is the vector sum of the vertical component of the electric field strength due to the 2 charges. ( ) ( )( ) ( ) resultant 1 2 1 2 22 00 2 2 0 9 212 41 11 cos 30 cos 6044 cos 3 0 1.7 0 cos 604 1 1.7 cos 30 cos 60 4 8 200 .85 0.5 V 010 1.42 10 m EE qq rr E q r     − − − = = +     +        = +    = +   = 1 1 X Y distance

5 ©EJC 2022 9749/J2H2Prelim/2022 [Turn over Qns Answer Marks (b)(ii) electric potential is a scalar quantity total electric potential at point O is sum of electric potential at that point due to the 2 charges. ( ) ( )( )( ) 9 2 resultant 1 2 1 00 2 12 0 11 44 10 2.74 4 10 9710 V 2001.7 1 8.85 0.50 VV qq rr q r V     − − +    +       ==  =  = = +  1 (b)(iii) ( )( ) ( )( )( ) 2 final initial 16 9 V 22 w 971 ork done 2 0.00388 J 2. 0 0 1 01 43 00 0 e W q V q V V −= −  − = = − − =− =−  1 1

6 ©EJC 2022 9749/J2H2Prelim/2022 Qns Answer Marks 4(a) Force per unit current unit length of wire Carrying a wire that is normal to the magnetic field. 1 (b)(i)1. direction (dots above current, crosses below current) field strength stronger when nearer wire (higher density of dots/crosses nearer the current) 1 1 (b)(i)2. positive 1 (b)(ii) Force is always perpendicular to the velocity of the particle No displacement in the direction of force, so work done by the force is zero, no change in kinetic energy so no change in speed. 1 1 (b)(iii) Done in vacuum, so no loss of kinetic energy due to no collision with air particles, so no change in speed. OR Weight of the particle is negligible compared to the magnetic force. 1 (c) Observe that 3 cm, 4 cm and 5 cm form a right -angle triangle. So the magnetic flux density at point Z due to currents in X and Y are perpendicular to one another. The magnetic flux density at Z is the vector sum of magnetic flux densities due to X and Y. ( )( ) ( ) ( )( ) ( ) 0 22 resultant 2277 22 5 4 2 7.0 8.0 2 T 10 4 10 3.0 10 4. 2 0 10 6.15 10 XY B B BB d     −− −− − = =+        =+    =         I 1 1 X Y

7 ©EJC 2022 9749/J2H2Prelim/2022 [Turn over Qns Answer Marks 5(a)(i) At 22.5 oC, RT = 1.6 k (read from graph) Total resistance = 1 11 1600 1600 −   + = 800  1 1 (a)(ii) Using potential divider principle, = + = 800 9 0 800 1200 3 6 V V . V. 1 1 (b)(i) Using potential divider principle, 4 1200 96 5 0 AB AB R R = = 1 1 (b)(ii) =+ = 1 1 1 960 1600 2400 T T R R From graph, temperature = 10.5 °C OR 10.75 C (read to half the smallest division) OR 11 C 1 1 (c) From graph, no calibration data available between 24C to 25C (accept “at 25 C”O). Since graph is non-linear, hence cannot extrapolate to get values of temperature. Based on temperature range of 0 – 25 C, VAB has a range of 0.48 V to 3.5 V. Since only a small part of the voltage scale is used, the measurement of temperature will be less sensitive. 1 1

8 ©EJC 2022 9749/J2H2Prelim/2022 Qns Answer Marks 6(a)(i) Progressive wave is one where energy is transferred in the direction of propagation of the wave. Transverse wave is one where the direction of oscillations is normal to the direction of energy propagation. 1 1 6(a)(i) Polarisation is where oscillations of a wave to one direction only, in the plane normal to the direction of energy propagation. Since the oscillations in a longitudinal wave is parallel to the direction of energy propagation, longitudinal wave cannot be polarised. 1 1 6(b)(i) Based on the amplitudes of the electricity field strength graphs,    = = = o cos 16 9cos 56 oAA 1 1 6(b)(ii) 2 2 2 2 0 cos ( ) 9cos ( ) 0.31616 A I    = = = =  I I I 1 1

9 ©EJC 2022 9749/J2H2Prelim/2022 [Turn over Qns Answer Marks 7(a)(i) Loss in electric potential energy = gain in kinetic energy ( ) ( )( ) 1 2 6 -1 3 2 19 6 1 9 11 10 1 60 10 1 452 3 98 2 4 0 1 1 0 m s 0 e .. m . v eV v . v−− =   =  = 1 (ii) ( )( ) 31 6 24 -1 34 24 10 9 11 10 4 0 10 3 644 10 kg m s 6 63 10 3 644 10 1 82 10 m ep m v .. . h p . . .  − − − − − = =   = = =  = • 1 1 (b)(i) Spacing between the crystals is of the same order of magnitude as the de Broglie wavelength of electron (so diffraction effect is significant ). The crystal acts as a diffraction grating for the beam of electrons. Bright spots observed when constructive interference takes place. 1 1 1 (ii) As the accelerating potential increase, the de Broglie wavelengt