ASRJC 2022 Prelim P2 Guide

1 9749/02/ASRJC/2022PRELIMS [Turn Over Anderson Serangoon Junior College 2022 JC2 H2 Physics Prelim Mark Scheme Paper 2 1a According to Newton’s 3rd Law, F12 = −F21 … (1) Taking the time of collision to be t, Using Newton’s 2nd Law, eqn (1) can be rewritten as, 1(2pp )tt  =− OR ()2 2 2 2 1 1 1 1m v - m u m v - m u tt =− 2 2 2 2 1 1 1 1m v m u m v m u− =− + OR 1 1 2 2 1 1 2 2m v m v m u m u+ = + B1 M1 A1 1bi By conservation of momentum, the gain in momentum of body Q must be equal to the loss in momentum of body P. Since total momentum is conserved, when Q gains momentum (positive gradient), P loses momentum (negative gradient). Hence, the gradients of the graphs must have opposite sign. OR Gradient of the graph is the net force acting on the object. Since the 2 forces are action- reaction pair, hence they must be in the opposite direction. B1 1bii When there is no net external forces acting on the system. B1 1biii Total initial momentum = 20 + 16 = 36 kg m s-1 Total final momentum = 24 + 12 = 36 kg m s-1 Total momentum is conserved. A1 1biv Force acting on body P = rate of change of momentum of body P = 3 12 - 16 (300 - 150) 10 − = −26.7 N = −27 N Magnitude of force = 27 N C1 A1 1bv Impulse = F t = 26.7 × 150 × 10-3 = 4.0 N s OR Impulse = change in momentum = 24 – 20 = 4.0 N s A1 1bvi Relative speed of approach = 116 20 3.0 m s2.0 4.0 pQuu −− = − = Relative speed of separation = 124 12 0 m s4.0 2.0 QPvv −− = − = RSS is not equal to RSA, and RSS is zero, so the collision is perfectly inelastic. B1 B1 A1 2ai Constant amplitude B1 2aii Period = 0.75 s ω = 2π / T ω = 8.4 rad s–1 C1 A1

2 9749/02/ASRJC/2022PRELIMS 2aiii Either use of gradient or v = ω y0 v = 0.168 m s–1 (allow ±0.02 for construction) C1 A1 2bi f = 1/T = 1/0.75 = 1.3 Hz B1 2bii at ½ f0 ‘pulse’ / energy is provided to mass on alternate/some oscillations so ‘pulses’ / energy build up the amplitude (don’t accept maximum amplitude) M1 A1 3ai Transverse A1 3aii =  = 12 8 4 distance = speed 4.4 10 3.0 10 1.5 10 s t A1 3aiii power received by dish aerial = intensity received  area of dish   − = =  = 2 12 2 23 output (source) power 280 4 25 280 4 (4.4 10 ) 2.9 10 W P r C1 C1 A1 3aiv The actual power is greater because the transmitting aerial is directing power towards the Earth instead of radiating uniformly in all directions as assumed in (a)(iii). A1 3bi v = f  = 0.040  1.5  = 0.060 m C1 A1 3bii path difference = (44 – 29) / 6.0 = 2.5  either waves have path difference = (n + ½)  or waves have phase difference = π so destructive interference M1 M1 A1 3ci Incident sound wave from loudspeaker travels to the water surface and is reflected. The incident and reflected waves superpose / interfere / overlap to form a stationary wave. B1 B1 3cii distance =  / 2 = 0.090 m A1 3ciii letter X shown at level B and at level A A1 4a Electric field lines are radial/normal to surface (of sphere) Electric field lines appear to originate from centre (of sphere) / intersect at the centre of the sphere when extrapolated B1 B1

3 9749/02/ASRJC/2022PRELIMS [Turn Over 4bi1 Net electric potential = 0 at x = 4.0 cm (2.0 x 10−9) / 4𝜋𝜀0(4.0 x 10−2) + Q / 4 𝜋𝜀0(8.0 x 10−2) = 0 Q = −4.0 10−9 C C1 A1 4bi2 change = (2 – (−10)) x 102 = 1200 V A1 4bii Loss in KE = Gain in EPE ½ mu2 – 0 = q∆V 1 2⁄ x 4 x 1.66 x 10−27 x u2 = 2 x 1.60 x 10−19 x 1200 v = 3.4 x 105 m s−1 Allow ecf from bi2 Deduct 1 mark if candidates use mass of proton 1.67 x 10−27 kg C1 A1 4biii Electric field strength is the least when the gradient of the V-x graph is the least, hence acceptable range is from 4.5 to 6.0 cm A1 4biv Gradient of tangent = (3.0−(−7.0)) × 102 (2.0−9.0)×10−2 = − 1.4  104 Therefore, magnitude of E = 1.4  104 V m−1 acceptable range: (1.1 to 1.7)  104 Accept calculation method (assuming x = 5.0 cm) E = EA + EB = (2.0  10−9) / 4𝜋𝜀0(5.0  10−2)2 + 4.0  10−9 / 4 𝜋𝜀0(7.0  10−2)2 = 1.45  104 V m−1 A C1 A1 C1 A1 5ai More likely (higher probability) for electrons at the next higher energy level to de-excite to fill up the vacancy in the K shell. A A1 5aii Due to the higher atomic number, the attraction between the nucleus and the inner shell electrons is stronger OR the inner shell electrons of tungsten is more stable due to the higher atomic number At low accelerating potentials, the energy of electrons is not sufficient OR At high enough accelerating potentials, the energy of electrons is sufficient D B1 B1

4 9749/02/ASRJC/2022PRELIMS to knock electrons out of the inner shells of the tungsten atom. 5b hcE e= − −− =  34 8 19 11 (6.63 10 )(3.0 10 ) (1.6 10 )(6.6 10 ) = 1.9  104 eV A C1 A1 6ai Change in mass ∆m = mass of product – mass of reactants = [(17.004507 + 1.008142) - (14.007525 + 4.003860)]u = 1.264 x 10-3u = 2.09824 x 10-30 kg Change in rest-mass energy = ∆mc2 = 2.09824 x 10-30 (3.00 x 108)2 = 1.888 x 10-13 J = 1.9 x 10-13 J A A M1 M1 A0 6aii The increase in mass as shown in calculation in (a)(i) implies that energy has been converted to mass. Hence, the reaction is only possible if the alpha particle has kinetic energy that is at least equal to the value calculated in (a)(i), which can be converted into mass. A M1 A1 6aiii1 Based on conservation of momentum, as the total initial momentum is not zero, since alpha particle is moving and nitrogen – 14 is stationary, final momentum cannot be zero and hence the product particles must have kinetic energy after the reaction. D A1 6aiii2 So, the alpha particle needs to have more kinetic energy than 1.9 x 10-30 J, so that while some of the kinetic energy is converted to mass, there is sufficient kinetic energy left to move the product particles. D A1 6bi No. of Xenon particles initially N0 = (6.02 x 10 23) (5.7 x 10 −12 /140 x 10 −3) =2.451 x 10 13 A0 = λN0 =((ln 2)/16)(2.451 x 10 13) = 1.0618 x 10 12 =1.06 x 1012 Bq 0 ln 2(60)12 16 10 10 1.06 10 7.8785 10 Bq =7.88 10 Bq tA A e e − − = = =  A M1 M1 A1 6bii As the products are also radioactive, after Xenon decayed, the products also produced activity and thus is the actual activity is higher than the activity due to Xenon alone. A A1

5 9749/02/ASRJC/2022PRELIMS [Turn Over 7ai Any one of the followings: Does not leak / longer cycle life / faster charging / higher degree of freedom in shape / higher current density / less flammable. A A1 7aii1 Any one of the followings: Electric vehicle / personal mobility device / car battery / electric scooter / solid-state solar cell or panel. A A1 7aii2 Any one of the followings: Pacemakers / radio frequency identification (RFID) / wearable devices / handphone or smart phone / smart card / implantable medical device e.g. implantable defibrillator / wireless sensor / remote control / calculator. A A1 7aiii Any one of the followings: 1. Bulk solid-state battery can store more energy than thin film solid-state battery. 2. Thin film solid-state batteries have longer cycle life than bulk solid-state battery. 3. Thin film solid -state battery can be easily shaped based on application while bulk-solid state has more standard construction shape. A B1 7bi Current per u