2025 ASRJC Physics Prelim P3 MS and ans

1 9749/03/ASRJC/2025PRELIM [Turn Over Anderson Serangoon Junior College 2025 H2 Physics Prelim P3 Mark Scheme Paper 3 (80 marks) E – Easy, A – Average, D – Difficult ECF Error carried forward SF Significant figures error M0 No A marks awarded AE Arithmetic error BOD Benefit of doubt ^ More is needed in answer POT Power of ten error CON Contradictory response XP Wrong physics TE Transcription error IR Irrelevant (part) response 1a sx = vxt vx = 24 / 1.5 = 16 m s-1 tan 28° = vY / vX or vX = v cos 28° and vY = v sin 28° vY = 16 tan 28° or vY = 16 × (sin 28° / cos 28°) so vY = 8.507 = 8.5 m s–1 Examiner’s comments: Most students were able to solve. Some workings did not show complete substitution of values to meet the demand of a “Show” question. There were still some students who did not show more s.f. before final answer which was required in a “Show” question. A C1 C1 A0 1b Taking upwards as positive vY = uY + at t = (0 – 8.5) / (– 9.81) = 0.87 s Examiner’s comments: Students did not substitute the sign of the direction correctly in their Kinematics questions which resulted in wrong answer. Some students wrongly assumed initial velocity as zero. C1 A1 1c straight line from positive vY at t = 0 to negative vY at t =1.5 s line starts at (0, 8.5) and crosses t-axis at (0.87, 0) and does not go beyond t = 1.5 s. M1 A1

2 9749/03/ASRJC/2025PRELIM Examiner’s comments: Those who drew a straight line graph often did not ensure the x-intercept was within half smallest square and extended line beyond t = 1.5 s. 1d acceleration (of freefall) is unchanged / not dependent on mass, and so no effect (on maximum height and time taken) Examiner’s comments: Many did not pinpoint acceleration as the quantity that was unchanged / independent of mass. Even though net force (= weight) was larger, the acceleration of free fall remained as g, so this showed it was independent of mass. E B1 1e Since air resistance acts downwards, net downward force is larger hence, shorter time taken Examiner’s comments: Students need to be explicit about the direction of the net force / net deceleration as downwards. Stating AR is directed opposite to the motion of the object was too generic and not contextualizing. Students need to illustrate their awareness about the net deceleration or net force is greater than before. Many students used inappropriate term such as “increases” to illustrate greater / larger. Some did not conclude for the time taken. D M1 A1 2ai Use A ρR l= =       = − − 350 4 )10x020.0( )0.2)(10x5.5( 23 8 power dissipated = I2R = (0.42)2(350) = 62 W Examiner’s comments: Most students performed the correct calculation and obtained the correct answer. A C1 A1 2aii1 Mthd 1 Common current in both wires, so nAv for tungsten= nAv for copper. 1 2 32 tungstentungsten coppercoppercopper tungsten sm24.0 )02.0)(4.3( )10x021.0)(4.1)(0.8( An vAn v − − = = = Mthd 2 Use I = nAve 1 19 23 28 tungstentungsten tungsten sm25.0 )10x60.1)(4 )10x02.0()(10x4.3( 42.0 eAnv − − − =  = = I Note that v is the drift speed and not the actual speed of the electrons. Examiner’s comments: This was commonly correct. A C1 A1

3 9749/03/ASRJC/2025PRELIM [Turn Over 2aii2 The higher speed of the electrons in tungsten means that they have a much greater kinetic energy than those in copper. As the electrons collide with the fixed atoms, energy is lost to these atoms, resulting in a rise in temperature. Examiner’s comments: This question proved challenging. Many students did not use the values of drift speed calculated from the previous part. E1: did not explain in microscopic terms; diameter/current/power is macroscopic E2: did not associate kinetic energy of an electron with its drift speed. E3: did not mention electrons transfer energy to lattice during collisions. E4: misconception, e.g: higher drift speed does not, by itself, mean electrons collide more often. The collision rate is set mainly by the material, impurities, and temperature, not by the small drift speed. D M1 A1 2bi resistance of thermistor at 0 C = 3900  using potential divider, 00.150.1x3900R R =     + R = 7800  resistance of thermistor at 30 C = 1250  using potential divider, V29.1 50.1x12507800 7800readingvoltmeter =      += OR p.d. across thermistor = 1.50 – 1.00 = 0.50 V resistance of thermistor at 0 C = 3900  common current in circuit = 3900 50.0 R 00.1 = R = 7800  resistance of thermistor at 30 C = 1250  common current, I in circuit = 9050 50.1 12507800 50.1 1250R 50.1 =+=+ voltmeter reading = IR = 1.29 V Examiner’s comments: It is advisable to use potential divider method instead of using current method which is longer. The current in the circuit is not the same when temperature changes. A C1 A1 2bii resistance of thermistor at 0 C = 3900  effective resistance of R and voltmeter, X = 7800/2 = 3900  (same as thermistor’s) voltmeter reading = p.d. across X = 1.50/2 = 0.750 V Examiner’s comments: The answer should be written to 3 s.f. because all data are given as 3 or more s.f. A C1 A1

4 9749/03/ASRJC/2025PRELIM 3ai gravitational force provides the centripetal force GMm / R2 = mv2 / R EK = ½mv2 and clear algebra leading to EK = GMm / 2R Examiner’s comments: Answers need to be clear in presentation flow, since this is a “show” question. Any symbols that are not found in the question must be defined. A B1 M1 A1 3aii ET = EK + EP = GMm / 2R – GMm / R = – GMm / 2R Examiner’s comments: The concept of total energy must be shown, as this is a “show” question. A M1 M1 A0 3b As the satellite gradually loses energy, i.e. its total energy ET decreases (i.e. more negative), its radius of orbit R decreases. As the satellite radius of orbit R decreases, its kinetic energy Ek increases, i.e. its speed v increases, and it would also move nearer or into the Earth’s atmosphere. The satellite’s increasing speed gives rise to increase of resistive forces which results in increasing rate of conversion of its energy to thermal energy, and so the satellite could eventually ‘burn up’. Examiner’s comments: The question started with “small resistive forces”. Good answers need to show how this transitioned into increasing severity of the satellite eventually “burning up”. D B1 B1 B1 3c advantage: • Continuous coverage: They orbit at the same speed as the Earth's rotation, appearing stationary over one location, providing constant monitoring of a specific area. • Ideal for communications: They are well-suited for broadcasting and communication services due to their fixed position. • Weather monitoring: They can provide continuous weather observations and imagery of a particular region. • Surveillance: Their constant visibility makes them useful for surveillance applications. Any 1 of the points above or other suitable advantages. disadvantage: • Their high altitude (about 35,786 km above Earth’s surface) leads to longer signal transmission times. • Limited polar coverage: Due to the curvature of the Earth, they have limited coverage of the polar regions. • Lower detail: Their high altitude results in lower -resolution images compared to polar-orbiting satellites. Any 1 of the points above or other suitable disadvantages. Examiner’s comments: Good answers require not only the characteristics of a geostationary satellite, but how that characteristic leads to an advantage/disadvantage. D B1