2024 JPJC H2 Prelim P2 solutions Printed

2024/JPJC/Prelim/9749/02 Answers to 2024 JC2 H2 Preliminary Examinations Paper 2 Suggested Solutions: No. Solution Remarks 1(a) 22 22 f A vk k f A v = = I I I 2 2 3 2 3 2 2 1 3 3 S base units of W m kg m s m kg m s m m s s m k − − − − − − −= = = [1] for correct units of frequency and amplitude [1] for correct base units of power [1] for correct answer 1(b) 3 3 133 84 10 53.133 m s0.56 P v b Pv b − = = = = 3 3 1 1 11 33 11(0.05) (0.07)53.133 3 3 2.1253 m s 2 m s P v b Pv b v P b v P b v v v − − = =   =+  =+ =  [1] for correct value of v [1] for correct substitution [1] for correct answer to 1 s.f. 2(a)(i) ( )( ) ( )( ) 33 4 Applying Newton's second law, 4 10 9.81 4 10 0.32 4.05 10 N T T −  =  = [1] for correct substitution [1] for answer 2(a)(ii)1. ( ) ( ) ( ) ( ) 2 mass per unit time vol per unit time 1.3 10 408 410 Av v v v    = = =  = = [1] for correct substitution [1] for correct intermediate value 2(a)(ii)2. Applying Newton’s second law, ( ) 4 1 9.1 410 14.9 10 ms net dp dmFv dt dt vv v −  ==  = = [1] for correct substitution [1] for correct answer

2 2024/JPJC/Prelim/9749/02 2(b) For vertical equilirium, ( )( ) ( ) 3cos65 4 10 9.81 --- 1T mg = =  Horizontally, ( ) ( ) ( ) 33 33 sin65 6 10 4 10 sin65 4 10 6 10 --- 2 T ma a Ta −  = =  =  +  ( ) ( ) 3 22 2 6 10: tan651 19.5 m s 20 m s ma mg a −− += == [1] for correct substitution [1] for correct answer 3(a)(i) The moment of a force about a pivot is the product of the force and the perpendicular distance from the pivot to the line of action of the force. [1] 3(a)(ii) Let N be the number of 5.0 g mass needed for equilibrium. Taking moments about the edge of the table, anti-clockwise moments = clockwise moments ( ) ( ) ( ) ( ) ( ) ( )1.2 0.05 0.11 0.40 0.005 0.75 4.3 g g N g N =+ = Since N is a whole number, the maximum number is 4. [1] correct substitution [1] correct answer 3(b)(i) [1] for T 1.0 kg 1.2 kg 0.75 m string pulley 0.10 m table T 65° mg T air resistance

3 2024/JPJC/Prelim/9749/02 [Turn over 3(b)(ii) Taking moments about the edge of the table, anti-clockwise moments = clockwise moments ( ) ( ) ( ) ( ) ( ) ( ) ( )1.2 0.05 sin60 0.40 0.11 0.40 1.0 0.75 0.346 7.2 20.8 N 21 N g T g g T T +  = + = == [1] correct substitution [1] correct answer 4(a) Since the direction of motion of the body is always changing, its velocity is not constant and thus a resultant force acts on the body. Since the speed of the body is constant, the resultant force acts perpendicular to its velocity and thus, direction of motion, which is towards the centre of the circle. [1] [1] 4(b)(i) The net force towards centre of the circle provides the centripetal force required to move in a circular path. 2 2 cos cos , as . mvT mg r mvT mg r L L   −= = + = [1] for statement and equation [1] for r = L 4(b)(ii) The point where the force in the rod changes from tension to compression is when T = 0 N. ( ) ( )( ) 2 2 22 cos 0 cos 2.0cos 0.51 9.81 0.80 120.6 120 mvT mg L mvmg L v gL     = + = =− =− =− =− = =  [1] for T = 0 N [1] for correct substitution and answer A B L  mg T cosmg 

4 2024/JPJC/Prelim/9749/02 4(b)(iii) [1] graph is cosine and cuts at about 120° [1] correct values 4(b)(iv) The K.E. of the mass is constant but its G.P.E. is constantly changing. So as the mass is moving upwards, an external device is needed to supply energy to increase its G.P.E. As the mass moves downwards, its G.P.E. is decreasing and the external device must absorb the lost in G.P.E. [1] energy supply [1] energy absorb 5(a)(i) A longitudinal wave is one where the oscillations of the wave particles is parallel to the direction of the wave propagation / energy transfer. [1] 5(a)(ii) Sound waves are longitudinal . Since there is no component of a sound wave's oscillation that is perpendicular to its direction of motion, sound waves cannot be polarised. [1] [1] 5(b)(i) Sound waves from the speaker travel leftward through the tube and gets reflected by the piston rightwards. The incident sound waves from the speaker superpose with the reflected sound waves by the piston to form stationary waves. [1] [1] 5(b)(ii)1. [1] 5(b)(ii)2. wavelength, 330 0.75 m440 v f = = = 0.75 0.563 m 33 4 = 56.3 cm4L  = = = [1] for wavelength [1] for correct answer 5(b)(iii) Stationary waves can only form at specific lengths of the air column, such that a node and an antinode are formed at the closed and opened end of the tube respectively. [1] [1] piston tube L speaker T  0

5 2024/JPJC/Prelim/9749/02 [Turn over This can only occurs when L is odd number multiples of quarter wavelength or ( )odd number . 4L = 6(a) At V = 12 V, I = 2.5 A 0.208V=I At V = 6 V, I = 1.25 A 0.208V=I At V = 9.6 V, I = 2.0 A 0.208V=I Since ratio of I to V is constant, I is proportional to V. [1] show at least 2 values of ratio of V  or V  . [1] conclusion 6(b)(i) Resistance of resistor X = 12 2.5 = 4.8  [1] answer 6(b)(ii)1. Current in wire AB = 9.0 4.0 5.0+ = 1.0 A [1] answer 6(b)(ii)2. Current in resistor X = 9.0 4.8 2.7+ = 1.2 A [1] answer 6(b)(iii) Resistance of wire AC = 0.70 4.01.0  = 2.8  Potential difference across AC = 2.8 9.04.0 5.0 + = 2.8 V Potential difference across AD = 4.8 9.04.8 2.7 + = 5.76 V Potential difference between C and D = 5.76 − 2.8 = 2.96 V = 3.0 V [1] value of VAC [1] value of VAD [1] answer 6(b)(iv) With internal resistance, the potential difference across the power supply would be less than 9.0 V. The potential difference across AC and AD would also be less, and hence the answer in (iii) would be less. [1] 7(a) Electric field strength E = 3 40 ( 40) 40 10 V d − −−=  = 2.0 × 103 V m−1 [1] answer 7(b) Acceleration a = F eE mm= = 19 3 31 (1.6 10 )(2.0 10 ) 9.11 10 − −   = 3.51  1014 m s−2 [1] substitution [1] answer (2 or 3 s.f.) 7(c) Duration of time for electron to stay in electric field is given by t = 3 7 80 10 1.5 10 s v −=  = 5.333 × 10−9 s [1] for value of t

6 2024/JPJC/Prelim/9749/02 Vertically, using v = u + at, v = 0 + (3.51  1014)( 5.333 × 10−9) = 1.87  106 m s−1 = 1.9  106 m s−1 [1] for velocity 1.87  106 m s−1 7(d) tan  = 6 7 1.9 10 1.5 10 y x v v = → = 7.2 [1] answer 7(e)(i) Magnetic force provides centripetal force 2 31 6 4 19 22 (9.11 10 )(1.9 10 ) (1.62 10 )(1 6.68 10 6.7 1 .60 10 ) 0mm mvBev R mvR Be − − − −− = → = =  =  = [1] statement [1] substitution [1] answer 7(e)(ii) 2 31 4 19 2 2 2 (9.11 10 ) (1.62 10 )(1.60 10 ) mv mv mBev Be mR R T mT Be   − −− = → = = = → = =  = 2.21 × 10−7 s = 2.2 × 10−7 s [1] substitution [1] answer 8(a)(i) 20.8 knots = 110.6912 m s− Maximum KE of ship ( )( ) 27 9 1 3.40 10 10.69122 1.94 10 J = = [1] for correct substitution [1] for correct answer 8(a)(ii) Angular speed ( ) 1 1 115 2 60 12.0428 rad s 12.0 rad s  − − = =  [1] for correct substitution [1] for correct answer 8(b) Power required to convert water at 100 °C to steam at 100 °C every hour ( )( ) 56 8 2.15 10 2.26 10 3600 1.3497 10 W  = = Maximum efficiency 3 8 29400 10 100% 1.3497 10 21.8% =  = [1] for correct substitution for power required [1] for correct substitution for efficiency [1