2024 ASRJC JC2 H1 Physics Prelim P2 MS

1 8867/02/ASRJC/2024PRELIMS [Turn Over Anderson Serangoon Junior College 2024 JC2 H1 Physics Preliminary Examination Mark Scheme Paper 2 (80 marks) 1ai Any two of time, temperature, current, (luminous intensity) B2 1aii Any derived quantity, e.g. energy, force, power, velocity, acceleration, pressure, density, etc B1 1bi Percentage uncertainty = 2 + (3x2) = 8% A1 1bii 2 2 2 (4 1.50) 2.48 9.63 m s g     Absolute uncertainty = 0.08 x 9.63 = 0.8 m s–2 g = 9.6 ± 0.8 m s-2 C1 C1 A1 2a As the sky-diver picks up speed, air resistance increases, the resultant force decreases and hence the acceleration decreases. B1 B1 2b Since the sky-diver starts from rest, there is no air resistance initially, hence his initial acceleration is equal to 9.81 m s–2. M1 A1 2c Before 24.0 s, speed increases with decreasing rate After 24.0 s, falling with constant velocity B1 B1 2d Find the area under the graph by using trapezium rule/counting squares B1 B1 2e Correct shape start with zero gradient and ends with constant gradient from about t = 24.0 s M1 A1

2 8867/02/ASRJC/2024PRELIMS 3a Resultant/net force (in any direction) on the object must be zero and Resultant/net moment / torque on the object about any point / axis must be zero. B1 B1 3b Taking moments about end A, (W × 0.25) + (12 × 0.35) = (17 sin 50° × 0.50) W = 9.246 = 9.2 N C1 A1 3c Consider vertical equilibrium, taking upwards as positive Sum of forces in vertical direction = 0 Fy + 17 sin 50° − 9.2 − 12 = 0 Fy = 8.177 N Consider horizontal equilibrium, taking rightwards as positive Sum of forces in horizontal direction = 0 17 cos 50° − Fx = 0 Fx = 10.93 N F = 2 2(8.177 10.93 ) = 13.7N C1 C1 A1 3d By taking the moments about end A, the moment due to the force by the block on the beam decreases, the tension in the string decreases. When the tension in the string deceases at the same angle, by considering horizontal equilibrium, the horizontal component of the force exerted on the beam by the hinge decreases. M1 A1

3 8867/02/ASRJC/2024PRELIMS [Turn Over 4a If R is inversely proportional to θ then product of R and θ is a constant. R / Ω θ / °C Rθ / Ω °C 1700 60 102000 800 100 80000 280 140 39200 As can be seen from the table, Rθ is not a constant in the range 50 °C to 150 °C. C1 A1 4bi R / Ω θ / °C T-1 / 10-3 K-1 ln(R / Ω) 200 160 2.31 5.30 Correct values for R, T-1 and ln R. A1 4bii Plot point accurately to half small square. A1 4ci  gE kTR Ae 2 M1

4 8867/02/ASRJC/2024PRELIMS     gg EER A A kT k T 1ln ln ( )( ) ln 2 2 As seen from relationship, when ln R is plotted against 1/T, a straight line should be obtained with gradient = 2 gE k and y-intercept = lnA. Since a straight line is seen in Fig. 7.3, it supports the proposal. A1 4cii Gradient = Eg/2k gE k2 =     3 5.95 4.90 4038(2.48 2.22) 10 Eg = 2(1.38  10−23 )(4038) = 1.11 × 10-19 J M1 M1 A1 4d Parallel combination of resistance 11 1( )90 120 51.43     Using potential divider, 51.43 9.051.43 243 1.57 V outV     C1 C1 A1 4e For metal, as temperature increases, the increase in lattice-electron collisions is more significant than the increase in charge carriers, thus the resistance of metal increases with temperature. A1 4f Correct shape – constant resistance then increasing (allow straight line) Include at least one set of correct resistance value, e.g. 7.5 Ω, 12.5 Ω or any correct value. B1 B1 5a electric field strength at a point is defined as the electric force exerted per unit positive charge placed at that point. OR A1 R / Ω V / V 7.5 0 2 0 12.5 4.5

5 8867/02/ASRJC/2024PRELIMS [Turn Over electric field strength at a point is defined as electric force per unit positive charge acting on a small stationary charge placed at that point. 5b F = qE ma = qE 19 4 31 1.6 10 4.0 10 9.11 ) ( 10 qEa m      = 7.0 × 1015 m s-2 C1 A1 5ci magnetic force on ion in path B provides for centripetal force By N2L, Bqv = m v2 r m = rBq v = 12.3 2 ×10ష2 × 640×10ష3 × 1.6×10ష19 9.6×104 = 6.56 × 10−26 kg = 6.56×10ష26 1.66×10ష27 = 40 u (or 39.5 u) B1 C1 A1 5cii Since the ions are of the same isotope, they all have the same mass regardless of the paths undertaken. Using the equation in answer to (b)(i), the radius of the path is inversely proportional to q (or state equation for r) Hence, the ions in path A have thrice the charge compared to ions in path B. B1 B1 B1 6a two nuclei of low nucleon number join / combine together to form one larger nucleus with release of energy. M1 A1 6b Energy released = Binding energy of products – Binding energy of reactants binding energy of Z = [(1.25 + 1.81) × 10-10 ] – 2.94 × 10-11 ( = 2.77 × 10-10 J) nucleon number of Z = 93 + 139 + 2 – 1 (= 233) Binding energy per nucleon of Z = (2.77 × 10-10) / (233 × 1.60 × 10-13) = 7.43 MeV C1 C1 C1 A1 6ci sketch: line with negative gradient starting at (0, 1.0 N0) and extending to t = 30 days exponential curve, extending from t = 0 to t = 30 days, with gradient of steadily decreasing magnitude line passing through (0, 1.0 N0), (10, 0.5 N0) and (20, 0.25 N0) B1 B1 B1 6cii Short range in tissue / high ionisation energy. B1 7ai By Newton’s 2nd Law, rotating blades pushes air and causes it to undergo a rate of change in momentum downwards giving rise to a downward force. From Newton’s 3rd Law, the air exerts an upward force of the same magnitude on the blades/helicopter. B1 B1

6 8867/02/ASRJC/2024PRELIMS When this upward force equals the weight of the helicopter, resultant force (of helicopter) is zero (and hence it can remain stationary vertically) B1 7aii1 Area = πr2 = π(0.70)2 = 1.539 m2 Volume of air per second = 1.539 x 4.0 = 6.156 m3 s-1 Mass per second = volume per second x density = 6.156 x 1.2 = 7.387 kg s-1 = 7.4 kg s-1 C1 C1 M1 A0 7aii2 Rate of change of momentum = dm/dt x velocity = 7.4 x 4.0 = 29.6 ≈ 30 N C1 A1 7aiii Fnet = 0 Mg – Force on helicopter by air = 0 Mg = Force on helicopter by air = 29.6 N M = 29.6 / 9.81 = 3.0 kg C1 A1 7bi Drag forces/ air resistance/ resistive forces increases with speed of the car At higher constant speed, there is greater work done against drag forces/ air resistance/ resistive forces hence, higher power is required B1 B1 A0 7bii v = 60  (1000 / 3600) = 16.67 m s-1 Effective power = Fdriving  v 22  103 = Fdriving  16.67 Fdriving = 1320 N Since the car is at constant speed, total resistive force = Fdriving = 1320 N C1 A1 B1 7biii 1. Total momentum before = total momentum after m x 60 – 2m x 60 = (m + 2m) V V = – 20 km h-1 Speed = 20 km h-1 C1 A1 2. During collision, force on car and truck is the same (by Newton’s 3rd Law) but since car has smaller mass, acceleration of car is greater than the acceleration of the truck. (For the same mass), force by seatbelt on car driver is greater. B1 B1 B1 8ai Horizontal component of tension / spring force provides centripetal acceleration Weight of sphere is (now) equal to the vert