ASRJC 2024 JC2 H2 Physics Prelim P3 MS

1 9749/03/ASRJC/2024PRELIM [Turn Over Anderson Serangoon Junior College 2024 H2 Physics Preliminary Examination Mark Scheme Paper 3 (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 (Note: g must have same place value as Δg) 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 / the only force he experiences is his weight, hence his initial acceleration is equal to 9.81 m s–2. M1 A1 2c Terminal velocity is the area under the a-t graph (by using trapezium rule/counting squares) A1 2d Correct shape start with zero gradient and ends with constant gradient from about t = 24.0 s M1 A1

2 9749/03/ASRJC/2024PRELIM 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 = 22(8.177 10.93 )+ = 13.7N C1 C1 A1 3d By taking 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 9749/03/ASRJC/2024PRELIM [Turn Over 4ai reduction in energy / amplitude (of the oscillations) due to force opposing motion / resistive forces / dissipative forces B1 B1 4aii amplitude is decreasing (very) gradually / oscillations would continue for a long time / many complete oscillations hence light damping M1 A1 4bi frequency = 1 / 0.3 = 3.3 Hz A1 4bii energy = ½ mv2 and v = ωxo = ½ × 0.065 × (2π/0.3)2 × (1.5 × 10–2)2 = 0.00321 J = 3.2 mJ B1 B1 A0 4c amplitude reduces exponentially / amplitude decreases by a smal