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