ACJC 2025 J2 H2 Physics Prelim P3 Guide

Anglo-Chinese Junior College 2025 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2025 Page 1 of 14 Qn Suggested Solutions 1(a) 2 2 R A RA PA EA t   = = = = l l l l I I ( )( )( )( ) ( )( ) 2 22 2 unit of unit of unit of kg m s m m A s m mghA t − = = Il 3 3 2unit of kg m s A −−= (b) Assume power output of phone charger ~ 10 W Typical USB charging voltage ~ 5 V 10 5 2 A P V= = = I (c) 5.02 0.038 132.105 VR = = = I RV RV   =+ I I 0.01 0.001 132.105 5.02 0.038 R =+ 3.7 4 R=  ( )132 4 RR =  

Anglo-Chinese Junior College 2025 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2025 Page 2 of 14 Qn Suggested Solutions 2(a) Coulomb's law states that the electric force acting between any two point charges is • directly proportional to the product of the charges and • inversely proportional to the square of their distance apart (b) Electric field strength (at a point in an electric field) is equal to the negative electric potential gradient (at that point) (c) For potential to be zero, one potential must be positive and the other potential must be negative OR For potential to be zero, the charges must have opposite sign For field to be zero, the fields (due to X and Y) must be in opposite directions OR For field to be zero, the charges must have the same sign The signs of the charges cannot (simultaneously) be both the same and opposite (so not possible) (d)(i) • At least six field lines drawn with the • field lines equally spaced near the plates. • Field lines are outside the sphere. • Arrows indicating the direction of the electric field is from the higher potential surface to the lower potential surface (towards the right) • 90° to the surfaces on the sphere and the plates. (ii) Charges are free to move in a conductor. In electrostatic equilibrium/for no net movement of charges, The component of electric field parallel to the surface of the sphere must be zero. Hence, the potential difference at points on the surface of the charge is zero and the potentials at K and L are equal. (iii) K 900 2 450 V V = =

Anglo-Chinese Junior College 2025 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2025 Page 3 of 14 Qn Suggested Solutions 3(a) The internal energy U of a system is the sum of a random distribution of potential and kinetic energies of the atoms/molecules/particles in the system. (b) ⅓: molecules move randomly in three dimensions (not one) so the mean square speed is any one direction is ⅓ of the mean square speed <c2>: molecules have different/a range of speeds so take average of the square of speeds (c) 2 2 2 1 3 1 3 1 3 pc Nmpc V pV Nm c = =  = 231 22 kpV Nm c E== k Using the ideal gas equation of state, 3 2 NkT E= For an ideal gas, there is no intermolecular forces of attractions so it has no microscopic potential energy. kUE= Hence, UT (d)(i) ( )( ) ( )( ) 52.0 10 0.26 8.31 273.15 20 21 mol pV nRT pVn RT = =  = + = ( )( ) 23 25 21.34 6.02 10 1.3 10 AN nN= = = (ii) ( )( )( ) 25 23 4 3 2 3 1.285 10 1.38 10 273.15 202 7.80 10 J U NkT − = =   + = (e) Since UT and for constant p, VT , therefore UV and it will be a straight line.

Anglo-Chinese Junior College 2025 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2025 Page 4 of 14 (f)  = +U Q W For a constant pressure process with an increase in temperature, negative work is done on gas. Q U W= − , where W is negative. At constant volume, no work is done on gas. QU= Hence, the heat supplied to raise the temperature by 1 kelvin ( /C Q T= ) will be higher for the constant pressure process. Cp will be higher than Cv.

Anglo-Chinese Junior College 2025 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2025 Page 5 of 14 Qn Suggested Solutions 4(a) 0mg kx= (b) Take the downward direction as positive. Resultant force with extension below equilibrium point, ( )= − + 0RF mg k x x where x0 is the extension at equilibrium. By Newton’s second law, ( )0 00 (Shown) mg k x x ma kx kx kx ma kax m − + = − − = =− since = 0mg kx OR Define upwards as positive ( )0mg k x x ma− + + = 00 Since and are opposite in direction, kx kx kx ma kax m ax kax m − + + = = =− (c)(i) From graph, T = 4.0 s 11 4.0f T== 0.25 Hz=f (ii) Comparing 2ax =− and kax m=− , ( ) 2 22 284 0.25 k m m   = = m = 11.3 kg (iii) ( ) 2 k 2 0 1 1Max 2 11.8 11.32 0.564 0.56 m s o o E mv v v − = = = 

Anglo-Chinese Junior College 2025 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2025 Page 6 of 14 (iv) 00vx = ( ) 0 0 0.564 2 0.25 0.359 0.36 m vx   = = =  (d)

Anglo-Chinese Junior College 2025 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2025 Page 7 of 14 Qn Suggested Solutions 5(a) Faraday's law states that the magnitude of the induced e.m.f. in a conductor is directly proportional to the rate of change of magnetic flux linkage experienced by the conductor. (b) Magnetic flux density in the solenoid is constant. Since the cross sectional area and number of turns of coil C remains the same, the magnetic flux linkage is constant. (c)(i) 2 2 2 10 T T      = = = 0.20 s 200 ms (Shown) T = = (ii) ( )( )( ) max 0 max 74 10 4000 4.8 0.024127 T Bn   − = = = I ( )( )( ) max max 4 cos0 71 0.024127 0.64 10 NB A −  =  = 4 max 4 1.096 10 Wb 1.1 10 Wb (Shown) − −  =  = (iii) ( ) ( ) ( ) ( ) = NBA = I        = =− =−  00 0 0 sin ( )sin sin cos t NA n t t d tdt ( )( ) max 0 410 1.1 10   − = = max 0.0034 V = (iv)

Anglo-Chinese Junior College 2025 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2025 Page 8 of 14 Qn Suggested Solutions 6(a) Any two of the following three: 1. Existence of Threshold Frequency Below a certain frequency of the incident radiation, there is no value of maximum kinetic energy, hence no electrons are emitted. Energy of radiation is quantised, and if lower than the work function, no electrons will be emitted. 2. Stopping potential is dependent on frequency but not on intensity. For the same intensity of incident light, the higher -frequency light has a larger stopping potential. Energy of radiation is quantised and increases so maximum kinetic ener