ACJC 2025 J2 H2 Physics Prelim P2 Guide

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2025 Page 1 of 10 Qn Suggested Solutions 1(a)(i) ( )( )0.140 4.0 5.4 p mv mu = − = − − 1.3 N sp =− (ii) 1.3 32.5 N0.04 pF t = = = ( )32.5 0.14 9.81 34 N F N mg N N =− =− = By Newton’s 3rd law, magnitude of the force exerted by the ball on the bar = magnitude of the force by the bar on the ball, N (b) Taking moments about B, ( ) ( )( ) ( )34 75 0.450 9.81 25 20 AF+= 133 NAF =

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2025 Page 2 of 10 Qn Suggested Solutions 2(a)(i) 2 y y y 1 2s u t a t=+ 21220sin(30 )(16.8) (9.81)(16.8)2 =+x 3230 m=x (ii) 1 xx 220cos(30 ) 190.5 m svu −= = = y y yv u a t=+ 1 y 220sin(30 ) 9.81 (16.8) 274.8 m sv −= + = 22 xyv v v=+ 22190.5 274.8=+v 1334 m sv −= OR ( ) 2 k 1initial 220 2Em= ( )( ) Ploss in gravitational 9.81 3232 E mgh m = = final kinetic energy = loss in gravitational potential energy + initial kinetic energy ( )( ) ( ) 2211 9.81 3232 22022mv m m=+ 1334 m sv −= (iii) (b) vx / t / s vx / 1m s− t/s 191 16.8 S vx / 1m s− t / s 191 16.8 R S

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2025 Page 3 of 10 Qn Suggested Solutions 3(a) (b)(i) Although the magnitude of the velocity is constant, its velocity is changing as the direction of its velocity is always changing. According to Newton’s Second Law, the astronaut must experience rate of change of momentum / an acceleration and hence a resultant force. Both the acceleration and force act towards the centre. (ii) 2 2 9.81 100 10199.81 vg r r == == 1020rm=

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2025 Page 4 of 10 Qn Suggested Solutions 4(a) A polarised wave has its vibrations/oscillations occur in a single direction in a plane / restricted to one plane (axis) and perpendicular to the direction of transfer of energy/propagation. (b) Applying Malus’ Law and assuming that the intensity of the polarised light after polaroid Q is I0, ( ) 2 0 2 0 0 cos cos 30 0.75 = = = II I I Since ( ) 2 2 00 0 0 0 cos 30 0.866 0.87 A A A A A AA A  =  = =  I I I (c) It is the spreading of the wave into its geometrical shadow when it is incident on an aperture/opening or edge of an obstacle. (d)(i) ( ) 6 sin 163.4 10 sin 2 dn  − = =  7 7 4.73 10 m 4.7 10 m  − − =  (ii) Blue/Indigo (iii) For an emerging beam to be observed on the screen, 90  and hence sin 1  . Since sin n d  = , 1n d   6 7 3.4 10 4.73 10 7.19 dn n n  − −     no. of emerging beams 7 7 1 15= + + = (iv) Without the polaroids, the incident light intensity on the diffraction increases, so the diffraction maxima become brighter. The separation between maxima depends only on the wavelength of light and the line spacing of the grating, which are unchanged, so it remains the same.

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2025 Page 5 of 10 Qn Suggested Solutions 5(a) The resistance R of a circuit component is defined as the ratio of potential difference V across the conductor to current I flowing through the conductor. (b) current in series circuit total E R E AB = = + I pd across (Shown) AA E AAB A EAB = = + = + I (c)(i) As the temperature of the thermistor is raised, its resistance will decrease. This will result in the combined resistance of the thermistor and the 4000  resistance to decrease due to their parallel circuit arrangement. As a result, based on the potential divider principle, the potential difference across the 1500  resistor will increase and hence, the voltmeter reading will increase. (ii) 1 1 1 1 1 1 4000 2700 1612 =+ =+ = TR B R R R 1500 (5)1500 1612 = + = + AVE AR V 2.41 V=V (iii) With internal resistance, the effective resistance of the circuit increases / terminal p.d. decreases. For the same change in temperature (same change in thermistor’s resistance), there is a smaller change in the voltmeter reading.

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2025 Page 6 of 10 Qn Suggested Solutions 6(a) The magnetic flux density of a magnetic field is the force per unit length acting on a straight, current-carrying conductor, carrying unit current and placed at right angles to this external magnetic field. (b)(i) The wire exerts a downward force on the magnet. By Newton’s third law, the magnet exerts an upward force on the wire. (ii) A to B (iii) ( ) ( )( )( ) 32 sin sin 90 3.7 10 4.6 8.5 10 FB B  −− = = =   Il Il 3 3 1.447 10 1.4 10 N F − − =  (iv) Assume 𝜃 is the angle of rotation from the initial position. length of wire in magnetic field cos= l ( ) ( ) sin 90cos coscos FB FB FB    = −  =  = lI lI Il The force acting on the wire is independent of the angle of rotation.

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2025 Page 7 of 10 Qn Suggested Solutions 7(a) Binding energy per nucleon is a maximum/close to the maximum at around A = 56. Products of splitting a 56 26Fe nucleus have a lower total binding energy which requires a net input of energy (b) ( )( )( ) 227 8 energy released BE of products BE of reactants 1.273 0.8405 1.910 1.66 10 3.00 10− =− = + −   113.04 10 J−= (c) ln2 28.8 yrs = ( ) 0 ln2 14428.8 0 1 32 tN N e N eN − − = = = 0 0 1 32 1 31 NNR N N N −= =− =− = OR 144no. of half-lives 28.8= = 5.00 = = 0 5 0 2 1 32 NN N 0 0 1 32 1 31 NNR N N N −= =− =− = (d)(i) 0 27 23 0.13 90 1.66 10 8.7 10 N −=  = (ii) ( ) 00 23 14 14 ln2 8.70 1028.8 365 24 60 60 6.64 10 6.6 10 Bq AN = =     =  (iii) ( )( ) 14 6 19 energy released per decay 6.64 10 0.546 10 1.60 10 PA − = =     58 WP =

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