MI 23M1PrelimAS (H2 PHYSICS P1)

H2 Physics PU3 Preliminary Examination Paper 1 Answers 1 A 6 B 11 C 16 B 21 C 26 B 2 C 7 B 12 A 17 D 22 A 27 C 3 B 8 C 13 A 18 C 23 A 28 B 4 B 9 D 14 B 19 D 24 D 29 D 5 D 10 D 15 A 20 C 25 B 30 C 1 A X = P – R = P + (–R) Hence vector R needs to change to the opposite direction before adding to P 2 C The object dropped from a great height will be experiencing a decreasing acceleration due to to increase in air resistance acting on it. As a result, the velocity will be increasing at a decreasing rate until it reaches a constant value. 3 B 4 B Since no net external force acts on the system of 2 particles, final momentum = initial momentum Considering vertical momentum, pX sinθ + (-pY sinα) = 0 pX sinθ = pY sinα X

5 D Let L be the distance from hip joint pivot to centre of mass. Applying principle of moment at the hip joint pivot, 6 B Upthrust The spring balance will read = 8.0 1.6 = 6.4 N 7 B By conservation of energy, Loss in GPE = Gain in KE + W.D by resistive force (600)(80 – h)(9.81) = [½ (600)(12.0)2 – 0] + (200)(1500) h = 22 m 8 C The resultant of 160 N and 120 N should provide the horizontal centripetal force. Resultant force = (1602 – 1202) =105.8 N Resultant force = ma 105.8 = (120/9.81) a  a = 8.65 ms-2 9 D g = 𝐺𝑀 𝑟2 = GM 1 𝑟2 The gradient is “GM”, where M is the mass of the planet. 10 D v = rw = r 2𝜋 𝑇 = (36000 + 6400) × 103 × 2𝜋 24×60×60 = 3100 m s-1 11 C total kinetic energy = 3 2 𝑛𝑅𝑇 = 3 2 𝑝𝑉 = 3 2 (1.0 × 10−5 × 0.010) = 1500 𝐽 Total KE is the same for the same T. 12 A m<c2> = kT 🡪 crms = = 🡪 = T’ = 188.8 K = 84.4 C

13 A Total heat supplied by the potatoes = heat for calorimeter + heat for water = C + mc = 200*12 + (2000)*(4.18)*12 = 102720 J Hence heat supplied per gram of potato = Q/m = 102720/75 =1369.6 J per gram 14 B amax = ------- (1) where xo = 0.30 cm -------- (2) Tshm = twice the period of Ek – t graph [ Remember this!] = 2 × 0.20 s = 0.4 s amax = ( 2𝜋 0.40)2(0.30 × 10−2) = 0.74 𝑚 𝑠−2 15 A With increased damping, the initial amplitudes are the same, the maximum amplitude is reduced and resonance occurs at a frequency slightly lower than natural frequency. 16 B Using Malus’ Law, Ip = Io cos2  = 2.0 (cos2 60o) = 0.5 W m-2 17 D The wave is reflected from the surface of the water and then interferes with the incident wave to form the stationary wave. 18 C 𝑥 = 𝜆𝐷 𝑎 ⇒ 𝜆 = 𝑎𝑥 𝐷 = (0.1 × 10−3)(8.0 × 10−3) 2.0 = 4.0 × 10−7 𝑚 At the second order dark fringe, the path difference will be 1.5 times of the wavelength, 6.0 × 10−7 𝑚 19 D At point A, the resultant electric field is pointing towards the left. At points B and C, the resultant electric field is pointing towards the right.

At point D, the electric field due to is pointing towards the right and that due to is pointing towards the left. Considering the magnitude of f