NJC 2022 Prelim P3 Guide

[Turn over National Junior College 2022 Preliminary Examination 9749 H2 Physics Paper 3 Suggested Solution Section A 1 (a) EK positive and decreasing, not touching x-axis B1 EP (roughly, but not too far off) symmetrical to EK B1 ET clearly shown (visibly or written in text) on the x-axis B1 (b) (i) Kinetic energy lost equals the gain in gravitational potential energy when object of mass m moves from the surface of Earth to infinity ½ mv2 – 0 = 0 – (–GMm / R) B1 v2 = 2GM / R (ii) gravitational field strength at the surface of planet g = GM / R2 equals the acceleration of free fall at the surface of planet B1 from (i), v2 = 2 (GM / R2) × R M1 v2 = 2gR A0 (c) (i) 3/2 kT = ½ m v2 = mgR C1 T = 2mgR / 3k = (2 × 6.6 x 10-27 × 9.81 × 6.4 × 106) / (3 × 1.38 × 1023) C1 T = 2.0 × 104 K A1 (ii) temperature where all substances have minimum internal energy B1 energy r EK EP ET

2 2 (a) sum of random distribution of kinetic and potential energies of the molecules of the gas B1 no intermolecular forces for ideal gas M1 so no potential energy A1 therefore, internal energy equals to total kinetic energy of the molecules only (b) (i) higher temperature increases r.m.s. speed and change in momentum of each molecule colliding with wall of container M1 increase volume reduces frequency of collision M1 force is product of frequency of collision and change in momentum which is constant when increase in change in momentum is balanced by reduction in frequency of collision A1 (ii) increase in internal energy = 3 2 k(460 − 280) × 1.2 × 6.02 × 1023 (= 2692 J) C1 heat supplied = 1.2 × 0.032 × c × (460 − 280) (= 6.912c) C1 heat supplied = increase in internal energy – work done on gas 6.912c = 2692 + 1300 (correct use of first law) C1 c = 580 J kg–1 K–1 or 578 J kg–1 K–1 A1 (c) constant volume so no work done B1 less thermal energy needed for same temperature rise, so smaller specific heat capacity B1

[Turn over 3 3 (a) series of distinct coloured lines against dark background B1 (b) photon emitted when electron changes from higher to lower energy B1 energy of photon proportional to frequency M1 each line represents photon of specific / distinct energy A1 only transition between discrete energy levels leads to specific energies (c) Δ𝐸 = 1.5 − 0.85 (= 0.65 eV or 1.04 × 10–19 J) C1 wavelength = ℎ𝑐 Δ𝐸 = 6.63×10−34×3×108 1.04×10−19 = 1.91 × 10−6 m M1 wavelength is longer than visible light / infrared, so not visible A1 (d) (i) waves passing slits spread B1 waves from each slit overlap M1 with phase difference 360° / path difference λ A1 (ii) angle of λ1 maximum = tan−1 147−73.5 240 (= 17.027o or 0.29718 rad) C1 sin 𝜃2 = sin 17.027° × 654×10−9 488×10−9 (⇒ 𝜃2 = 23.106o or 0.40328 rad) C1 𝑥−73.5