ASRJC 2022 Prelim P1 Guide

1 9749/01/ASRJC/2022PRELIM [Turn Over Anderson Serangoon Junior College 2022 JC2 H2 Physics Prelim Mark Scheme Paper 1 (30 marks) 1 2 3 4 5 6 7 8 9 10 B C A C D B B B A A 11 12 13 14 15 16 17 18 19 20 C B C B A D D B B A 21 22 23 24 25 26 27 28 29 30 B C D B B C B A C C 1 B Replace J with SI base unit of energy kg m2 s–2 in the original expression J kg–1 K–1 will give m2 s–2 K–1. 2 C In the calculation of fractional uncertainty of 3X, the coefficient 3 will cancel out, also, for calculation of uncertainty, we always add and not subtract because the uncertainty should increase with more terms involved rather than decrease, so the final answer is x + 2y. 3 A All responses show a flat constant v at the initial distance travelled. For the next part where a is constant, using v2 = u2 + 2as, we see that when we plot v vs s, we get a square root curve, so A is the answer. Alternatively, 1gradient of graph = =  = dv dv dt ads dt ds v As v approaches zero, 1 v approaches infinity, so the gradient approaches infinity. 4 C Since we are using the same scales, it is expected that the initial velocity is the same as the air resistance is negligible when the velocity is small. 5 D Consider horizontal direction, the change of momentum = p cos θ – p cos θ = 0 So average horizontal force = 0 Consider vertical direction, Change in momentum = p sin θ – (– p sin θ )= 2p sin θ Average vertical force = 2 sindp p dt t = 6 B Moment by weight about bottom left corner of book = Wx/2 clockwise

2 9749/01/ASRJC/2022PRELIM To avoid rotation (i.e. achieve rotational equilibrium), the resultant moment must be zero about bottom left corner of book. Hence, student must provide Wx/2 anticlockwise 7 B rate of increase of kinetic energy = Fnet × v = mav = (250000)(0.90)(5.0) = 1.1 MW Alternatively, output force of train’s engine, F, can be found using N2L: Fnet = ma F – 15000 = 250000(0.90) F = 240000 N output power of the train’s engine = Fv = 240000 × 5.0 = 1 200 000 W However, some of this output power is used against resistive force, while the rest accelerates the train and increases the kinetic energy. Hence, rate of increase of kinetic energy = 1 200 000 – 15000(5.0) = 1.1 MW 8 B T = 15.9 days = 15.9 × 24 × 3600 s r = 1.22 × 109 m centripetal acceleration = rω2 = (1.22 × 109)(2π/(15.9 × 24 × 3600))2 = 2.55 × 10−2 m s−2 9 A Since v = rω, and ω is constant and the same for every point on the record, v is then prop ortional to r. 10 A For the molecule to escape, total energy at Earth’s surface must at least be zero (min energy at infinity). Hence, − GMm r + 1 2 mv2 = 0 ➔ v = √ 2GM RE Therefore, vnew = √ 2GM 3RE = 1 √3 √ 2GM RE = 1 √3 × 1.1 × 104 = 6.4 × 103 m s−1 Alternatively, use ratio method  1v r , therefore = originalnew original new rv vr = 41.1 10 3 new E E v R R vnew = 1 3 (1.1  104) = 6.4 103 m s−1

3 9749/01/ASRJC/2022PRELIM [Turn Over 11 C Option A: Geostationary satellites can only be found above the equator Option B: The linear speed of the geostationary satellite has to be greater as it is further away from Earth’s centre and has the same angular speed as Earth Option D: Geostationary satellites travel in the same direction as Earth’s rotation and that is from west to east. 12 B Container X: 2 2 X X X X X Y X X X X n RT n R T n RTp V n RT p V V V= → = = = Container Y: Y Y Y Y Y Y Y Y Y n RT n RTp V n RT p VV= → = = Since XYpp = , XYnn = , 1X Y n n = . 13 C Let the amount of added heat be Q. So, wwQ mc T= ……………. (1) {heat added to water} iiQ mc T= ……………. (2) {heat added to iron} Eqn. (1) = (2), wi iw cT cT =  Since cw > ci, ∆Ti > ∆Tw → the final temperature of iron is going to be higher than the water’s final temperature. 14 B Net work done by gas = enclosed area = ½  (4 - 1)  10−3  (8 - 4) 104 = 60 J Net work done on gas = - 60 J Alternatively, Net work done by gas = (1/2  (8 + 4)  104 x (4-1)  10−3) - ((4-1)  10−3 x 4 104) = 180 - 120 = 60 J Net work done on gas = - 60 J 15 A Amplitude of oscillation decreases exponentially, causing the shape of the graph to be a curve. 16 D i. The speed at point P is zero, not maximum.

4 9749/01/ASRJC/2022PRELIM ii. The displacement at any point, including point Q, varies sinusoidally with time, thus it is not always zero. within a period. iii. The energy at point R is entirely potential, not kinetic. iv. The acceleration at point S is a maximum since displacement is maximum. 2ax =− 17 D I  A2 For maximum intensity, Io  (2A + A)2 For minimum intensity, I  (2A – A)2 2 3 9 A A = = o o I I II 18 B s = r θ  == -5 -5 3 25 = 7.46×10 = 7.5×10 rad335×10 s r Minimum angle of resolution θm =   -9 -6 -3 500 10 = 2.5 10 rad200 10 λ =b Images seen are resolved since angular separation between two sources is larger than the minimum angle of resolution. 19 B In order to hold the positively-charged particle P halfway between the two plates, the upper plate must be at a lower potential with respect to the lower plate (so that electric force acts upwards which is balanced by the gravitational force acting downwards). By decreasing V and causing the positively-charged particle P to move towards the lower plate (which is at a higher potential), the EPE of the particle increases (since a positive charge’s EPE increases with electric potential). However, as the particle has fallen to a lower position, its GPE decreases. 20 A Electric field strength is given by the negative potential gradient. Consider the line joining PQ, from P till the 0 V equipotential line, for every decrease in potential of 100 V, the distance between the equipotential line gets larger and larger, hence the magnitude of the electric field strength decreases. Similarly from the centre of PQ where the 0 V equipotential line towards Q, for every decrease in potential of 100 V, the distance between the equipotential line gets smaller and smaller, hence the magnitude of the electric field strenght increases. 21 B For metal wire at constant temperature, the value of resistance should be constant. For filament lamp, the resistance increases with potential difference.

5 9749/01/ASRJC/2022PRELIM [Turn Over For diode, it requires a p.d. of around 0.3 – 0.7 V to conduct. After that, when it is forward bias, the resistance is very low as seen by the near vertical line of the graph. 22 C No current flows through voltmeter. Hence I1 = I2. Current through ammeter = I1 = I2 Voltmeter reading = R1I2 = R2I3 = E - I4R3 23 D Using LHR, for the magnetic force on the wire to be downwards, the magnetic field must be towards the right. 24 B E = Blv, where v is the horizontal component of the rod’s velocity. From P to Q, the rod is moving parallel to the B-field and hence E is zero. From Q to R, as the rod rolls down the slope, the component of its weight parallel to the slope causes the velocity to increase at a constant rate, and hence the horizontal component of the velocity increases at a constant rate. Hence E increases at a constant rate. From R onwards, the rod is moving in a projectile motion under free fall and its horizontal velocity is constant and hence, E remains constant. 25 B Vrms = 150 √2 P = Vrms 2 R 13 = 1502 2R R = 865 , hence Rtotal = 2  865 P = 2402 2865 = 33 W 26 C 5 5 240 12001 S S P V VVV =  =  = Current in the secondary circuit = 12