ASRJC 2024 JC2 H2 Physics Prelim P1 MS

1 9749/01/ASRJC/2024PRELIM [Turn Over Anderson Serangoon Junior College 2024 JC2 H2 Physics Preliminary Examination Mark Scheme Paper 1 (30 marks) 1 D 2 2 3 intensity units of I I − − = = = E At kg ms m ms kg s 2 A Z = X + (–Y) as shown by the vector triangle below. This is equivalent to option A. 3 C Constant speed up to t1 means s increases at a constant rate, hence a straight line with positive gradient. As speed decreases at constant rate, distance travelled increases at a decreasing rate (decreasing gradient) and reaches a constant (zero gradient) when speed is zero. 4 D The vertical component of acceleration is the acceleration of free fall which is a constant in the absence of air resistance. 5 C Action-reaction must be of the same type. Weight is the gravitational force on the man due to the Earth. 6 D Consider vertical equilibrium of the object in water, taking upwards as positive P + Vρwater g – W = 0 water WP Vg −= X −Y Z

2 9749/01/ASRJC/2024PRELIM Consider vertical equilibrium of the object in oil, taking upwards as positive Q + Vρoil g – W = 0 oil WQ Vg −= oil water WQ WP   −= − 7 A By Conservation of Energy, loss in KE = Gain in GPE. ➔ ∆E = ∆mgh = mg∆h Hence E varies linearly with height, i.e. a straight line. Since y is vertical displacement, at maximum height (largest y value) E = 0. Hence the answer is A. 8 B Total work done by man = Work done against friction + gain in GPE (since KE is constant) Work done against friction = 1500 – mgh = 1500 – 5.0 x 9.81 x 12 = 911. 4 J Average friction = work done against friction / distance travelled along plane = 911.4 / ( 12 sin 30 ) = 37.9 ≈ 38 N 9 B Period = 2 / (/2) = 4 s. After 6s, the marble would have completed 1.5 cycles, so directly opposite P, so displacement = 2r = 2 x 0.80 m = 1.6 m. 10 C Let m be the mass of the satellite and M be the mass of Earth. GPE of orbiting satellite = GMm r− and KE = 2 GMm r When satellite moves closer to the surface of Earth, r decrease. From the equations, GPE becomes more negative, so GPE decreases. KE increases, hence orbital speed increases. 11 B According to definition for thermodynamic temperature scale. 12 D As the temperature has stabilised, the rate of increase of internal energy (ΔU) = 0.

3 9749/01/ASRJC/2024PRELIM [Turn Over The filament is hotter than its surroundings, loses heat (Q). Thus, rate of heating the filament is negative. Positive work (W) is done on the filament by electric current (recall the potential difference across a component in a circuit as the work done to drive a unit charge through the component). 13 A Option A implies