2025 AMKSS EOY 3E SCI PHY P2 MS

ANG MO KIO SECONDARY SCHOOL EOY 2025 SECONDARY THREE EXPRESS SC PHYSICS ANSWER SCHEME Paper 2 No. Answer Marks Total Marks Remarks Section A (55 marks) 1(a) prefix symbol value kilo k 103 giga G 109 micro μ 10−6 mega M 106 1 1 1 4 1(b) time density energy pressure 1 2(a) Average diameter of one golden sphere = (4.5−1) 5 = 0.70 cm (2 s.f.) 1 1 4 2(b) He should be using a digital micrometer screw gauge as it is more precise or it can measure up to 0.01 mm or 0.001 cm. Or He can also use a digital calipers as it is more precise or can measure up to 0.01 cm. 1 1 3(a) Friction exerted by road Air resistance 1 1 10 3(b)(i) The speed of the car is increasing [1] at a decreasing rate.[1] 2 3(b)(ii) The speed of the car is constant. 1 3(c)(i) As speed increases, air resistance increases. Hence, resultant force decreases. 1

3(c)(ii) t = 10 to 14 s Since the resultant force is zero, the maximum value of resistive force is equal to the driving force. 1 1 3(d) resultant driving resistiveF F F =− 2500 700=− 1800= N From the graph, when the resultant force is 1800 N, the time is at 2.8 s. 1 1 4(a) Pressure is the force acting per unit area in contact. 1 10 4(b)(i) Force exerted by each wheel = Pressure  Area in contact with ground = 199 kPa  0.04 m2 = 199  1000  0.04 = 7 960 N = 7.96 kN 1 1 Formula/ Working 4(b)(ii) Total weight = 7.96 kN  6 = 47.76 kN = 47.8 kN (3 s.f.) 1 1 With ecf With ecf and 3 sf 4(c)(i) Pressure = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑎𝑟𝑒𝑎 𝑖𝑛 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 = 850 5 × 10−4 = 1 700 000 N/m2 or 1 700 000 Pa 1 1 With unit 4(c)(ii) Pressure exerted by his boots = 1 700 000 N/m2 = 1700 kPa. Pressure exerted by vehicle = 6  199 kPa = 1194 kPa. Pressure exerted by his boots on the ground is greater than that from the vehicle. The farmer is more likely to sink into the soft marshy ground. 1 1 4(c)(iii) • Place two wooden pranks on the soft ground to increase contact area; • Increase the contact area of the sole of his boots. 1

5(a) • perpendicular distance from line of action of force to pivot/point X • correct perpendicular distance drawn 1 10 5(b)(i) The ladder is not uniform/ mass of the ladder not evenly distributed. 1 5(b)(ii) For an object to be in equilibrium, the sum of clockwise moments is equal to the sum of anticlockwise moments about the same pivot. 1 1 5(b)(iii) M = F x d = 90 x 1.0 = 90 Nm (clockwise) 1 1 5(b)(iv) Moment due to the bucket = 100 x 1.4 = 140 Nm (anticlockwise) Since anticlockwise moment > clockwise moment, the worker needs to apply a downward force to create a clockwise moment. Clockwise moment (M) + 90 = 140 M = 50 Nm F x d = 50 F = 50 / 0.60 = 83 N or 83.3 N Direction of force = downward 1 1 1 Allow ecf fr 6c(i) 5(b)(v) Position the shoulder nearer to bucket. Or Position the hand further away from the pivot 1 6(a)(i) At 81 °C, substance X particles are packed very closely together and are packed regularly. At 880 °C, substance X particles are very far apart and packed randomly. 1 1 7 6(a)(ii) At 81 °C, the particles are vibrating about their fixed positions. At 880 °C, the molecules are moving about randomly at high speeds 1 1 6(b)(i) Z, X, Y 1 6(b)(ii) Density of Y = mass / volume = 23.8 / (75 – 50) = 0.952 = 0.95 g / cm3 1 1 7 (a) Conduction occurs in solid, liquid and gas Convection occurs only in liquid and gas 1 10 F d

7 (b)(i) Sensor A rises more in temperature in the same time period. 1 7 (b)(ii) Heated water near the heater is less dense and it rises. The cooler surrounding water is denser and it sinks. This sets up a convection current. 1 1 7(b)(iii) Convection currents only occur above the heater Sensor B is not heated as water is a poor conductor (heat takes a long time to be conducted downwards from A to B. 1 1 7 (c) When the inside of the wall is heated, atoms gain kinetic energy and vibrates vigorously. They collide with the neighbouring particles and kinetic energy is transferred from the hot end to the cold end (from inside to outside). In metal, free electrons gains kinetic energy and move at greater speed. They diffuse into the cooler end and transfer kinetic energy when they collide with the atoms. 1 1 7 (d) Silver / white Poor emitter of infrared radiation thus it reduces rate of heat transfer from the hot container to surrounding. 1 1 No. Answer Marks Total Marks Remarks Section B (10 marks) 8(a)(i) a=(v-u)/t v= 0 + 10 (2.5) = 25 m/s (2 s.f.) 1 1 10 8(a)(ii) Straight line, positive gradient Labeled 25 m/s, 2.5 s 1 1 8(a)(iii) Distance = Area under graph = 0.5 x 25 x 2.5 = 31.25 m (or 3 s.f.) 1 v / m/s time / s 25 0 2.5

8(b) Air resistance reduces the speed and the distance travelled. OR Some energy was transferred mechanically to the internal store of the man, and lost in the form of sound waves. 1 8(c)(i) The downward force (weight) was equal to upward force (elastic tension), such that resultant force is zero/ there is no net force, therefore acceleration becomes zero 1 1 8(c)(ii) In this mechanical pathway, energy is transferred from an elastic potential store to a kinetic store, and then to the gravitational potential store as he rises 1 1 9(a) Energy cannot be created or destroyed. Energy can be transferred from one store to another. The total energy of an isolated system is constant. 1 1 10 9(b) Energy is transferred from the gravitational potential store at Point P to the kinetic store of the carriage and passengers at Point Q 1 9(c)(i) 𝐺𝑃𝐸 = 3500 × 10 × 30 = 1 050 000 J or 1.05 c 106 J 1 1 9(c)(ii) 𝑝𝑜𝑤𝑒𝑟 = 1050000 120 = 8750 𝑊 1 1 9(d)(i) 𝑔𝑎𝑖𝑛 𝑖𝑛 𝐾𝐸 = 𝑙𝑜𝑠𝑠 𝑖𝑛 𝐺𝑃𝐸 1 2 × 3500 × 𝑣2 = 1050 000 𝑣 = 24.5 m/s 1 1 Allow ecf fr 9c(i) 9(d)(ii) There is energy loss/transfer to the thermal store of surroundings due to air resistance/friction. 1