YIJC 2024 JC2 PRELIM H2 Phy P2 Soln

1 ©YIJC 9749/02/YIJC/23 2024 JC2 Prelim Exam H2 Physics Paper 2 Solution 1 (a) (i) 1Distance area under the graph (8.8)(0.90) 4.0 m2vt= − = = B1 (ii) ( ) 2 change in velocityaverage acceleration time 4.4 8.8 0.50 26 m s − = −−= =− (Circle but not penalise for missing -ve sign) C1 A1 (b) (i) The accelerations are the same as the gradients of the graph before and after the rebound are the same. (the girl is in a state of free fall before and after the rebound). B1 (ii) The area under the lines represents height and the second area is smaller than the first area, hence, the rebound height is less than the initial height. Thus the GPE of the girl at t = 0 is larger than the GPE at t = 1.85 s. OR The kinetic energy (KE) of the girl at trampoline equals to the GPE at maximum height. The speed after rebound is smaller and hence the KE becomes smaller. Thus the gravitational potential of the girl after rebound is less. M1 A1 (M1) (A1) 2 (a) Using hinge as pivot: Clockwise moment due to weight = anti-clockwise moment due to tension ( )Moment due to cos60 50 Moment due to (sin60)10 y x TT TT = = ( )   cos60 50 (sin60)10 10 5.0 50(cos60) 10(sin60) 16.8 17 cm Wx T T x x =+ =+ == Note: if either moment vertical/horizontal is not considered, max 1 mark. M1 A1

2 ©YIJC 9749/02/YIJC/23 (b) (i) Both vectors drawn and labelled 3 forces have common point B1 B1 (b) (ii) 2 2 2 2 5.0sin60 4.33 N 10 5.0cos60 7.5 N 4.33 ( 7.5) 8.7 N xx yy yy xy RT R T W R W T R R R = = = += = − = − = = + = + − = M1 M1 A0 3 (a) (i) ( ) ( ) 6 3 3 80 6.4 10 1.5 10 10 earth final GMGPE =−  +   Total energy on Earth’s surface = total energy at LEO ( ) ( ) ( ) ( )( )( ) 9 6 6 3 3 9 11 24 66 9 energy supplied 80 80 3.0 106.4 10 6.4 10 1.5 10 10 113.0 10 6.67 10 6.0 10 80 7.9 10 6.4 10 2.1 10 J i i f f earth earth f f f KE GPE KE GPE GM GM KE KE KE − + + = +  − +  = + −   +    =  +   −   = C1 C1 A1 (ii) Centripetal force is provided by gravitational force ( )( ) ( ) 2 11 24 26 6.67 10 6.0 10 80 7.9 10 510 N cG C c c FF GMmF r F F − = =  =  = (1m for both working and answer.) B1 B1 (b) (i) Gravitational potential at a point is the work done per unit mass by external agent in bringing a small test mass from infinity to that point. B1 wall hinge R x 10 N T cord 60° 10 cm 50 cm L-shaped beam

3 ©YIJC 9749/02/YIJC/23 (ii) Since dFm dr =− , the negative of the gradient of the ϕ-r graph gives the force. Positive gradient means gravitational force (vector) is negative Since gravitational force is negative when displacement is positive, it is in the opposite direction as the displacement from Earth and hence is attractive. OR Potential decreases nearer to the Earth. Gravitational force must always be in direction of decreasing potential and points towards Earth and is attractive. OR The gravitational potential is always negative, so the external force points in the opposite direction to displacement from infinity to that point / points away from the earth. Hence, the gravitational force points opposite to the external force towards the earth and is an attractive one. M1 A1 (M1) (A1) (M1) (A1) (ii) ( ) 22 27 -1 11 22 1 6.25 10 0 02 11.2 km s initial initial final final initial final KE U KE U mu m mv m u u  + = + + = + + −  = + = M1 A1 4 (a) (i)1 distance moved by wavefront/wave during one period / during one oscillation of a particle in that wave OR minimum distance between two wavefronts/crests/troughs/peaks OR minimum distance between two points which are in phase B1 (B1) (B1) (i)2 The phase difference between two particles refers to how much one particle lags or leads another with respect to a cycle. OR how one wave lags or leads another wave of the same wavelength. B1 (B1) (ii) 2 x   = OR 360 x   = B1 (b) (i) Using v T = or vf = and 1f T= From Fig. 4.1, T = 0.60 s ( )( )20 0.60 12 cm vT == = M1 A1 (b) (i) ( ) 0.20 0.60360 120 or 360 120 240 t T    ==  =  − =  o (Award M1 mark only if answers expressed in radians (2.09 rad).) M1 A1

4 ©YIJC 9749/02/YIJC/23 (ii) At t = 0.45 s, using the principle of superposition, ( )net displacement 1.00 3.00 2.00 mm = + − =− (Accept ±0.1mm) M1 A1 (iv) Intensity  (amplitude)2 ( ) 2 2 2.0 3.0 0.444 3sf QQ PP A A  ==  = I I M1 A1 (v) Two sources (waves) are said to be coherent when the phase difference is always the same OR there is a constant phase relationship at all times. B1 5 (a) Current = 2.7 – 1.5 = 1.2 A B1 (b) p.d. across XY = 12 V = 1.5 (5.0 + R) Resistance R = 3.0  B1 (c) p.d. across XZ = (1.6/2.0)12 = 9.6 V p.d. across XW = 1.5(5.0) = 7.5 V potential difference = 9.6 – 7.5 = 2.1 C1 A1 (d) When the resistance of the variable resistor is now increased, the effective resistance of the circuit increases. For the same e.m.f. supply, the power dissipated in the effective resistor decreases since power is inversely proportional to the effective resistance. Thus the power supplied by the battery decreases. OR When the resistance of the variable resistor is now increased, the effective resistance of the 5.0 ohm and variable resistor increases , so the current in the variable resistor decreases. The current in the resistance wire is unchanged since the p.d. across the wire and the resistance remain the same. So, the current in the battery decreases (same e.m.f.) the power decreases. B1 B1 B1 (B1) (B1) (B1) 6 (a) ( )( ) ( )( ) ( ) 0 max max 7 2 6 2 4 10 1.2 2 20 10 1.2 10 T 2sf IB d     − − − =  =  = C1 A1 (b) (i) ( ) ( ) ( ) ( ) 2 max max 262 7 1500 1.2 10 0.50 10 1.41 10 Wb 3sf NB A NB r  =− −  = = =   = C1 A1

5 ©YIJC 9749/02/YIJC/23 (ii) The alternating current in the cable induces an alternating (changing) magnetic field at the plastic ring / toroidal solenoid. This cuts the solenoid and thus there is an alternating (changing) magnetic flux linkage at the solenoid. Hence, by Faraday’s law of electromagnetic induction, e.m.f. is induced in the solenoid. B1 B1 (iii) The root-mean-square e.m.f. of the alternating voltage source is the equivalent to the value of steady direct current e.m.f. that dissipates the same power as the average amount of power dissipated by the alternating voltage. (subtract 1m if average is not mentioned) B1 B1 7 (a) 32 hc EE =− (identifying which two levels.) OR Uses wavelength of 658 nm 34 8 19 39 19 3 6.63 10 3.00 10 ( 3.40 1.60 10 )658 10 2.42 10 J E E − − − −    = − −   =−  C1 (C1) C1 A1 (b) (i) ( )( ) 19 6 19 15 1 Energy of each photon, 2.84 10 Power intensity area intensity area intensity area 160 2.5 10 number per unit time, 2.84 10 1.4 10 s hcE N Et N tE N t  − − − − = =  =  =  =  =  = C1 M1 A0 (ii) forcePressure area= Pressure on the mirror by the light beam is equal to the total force of impact by the photons per unit area. Change of momentum for one photon after ‘reflected’ = 2p (where p = momentum of the photon) ( ) ( ) 34 15 12 7 12 6 6 Total force exerted Comboned rate of chan ge of momentum 22 6.63 102 1.4 10 2.65 107.0 10 2.65 10Pressure 2.5 10 1.1 10 Pa N N hptt  − − − − − − =     ==           =  =    =  = C1 C1 A1

6 ©YIJC 9749/02/YIJC/23