EJC 2023 J2 H2 PRELIM P3 MS

©EJC 2023 9749/J2H2PRELIM/2023 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2023 9749 PHYSICS MARK SCHEME Paper 3 – Longer Structured Qns Answer Marks 1a Correct positions of max and min intensities, and gradient at 0 and 180 should be approx zero [B1] Max intensity of Io/4 [B1] Working: Io = Io cos2 cos2(90−) = Io cos2 sin2 = Io/4 sin22 [note: sin2 = 2sin cos] 1bi d = 10×10-6  7 = 1.43 × 10−6 m M1 A1 1bii For number of images on each side, sin = 45 nmax =d sin45/ = 1.43 × 10−6 × sin45/ 600×10−9 = 1.68  1 Total number of images = 1 + 1 + 1 = 3 C1 A1 1biii For the first order maxima, 𝑑 sin 𝜃 = 𝜆. Most of the line spacing in the grating is uniform such that the rays converge at  to give a bright image. Some of the line spacing is smaller than the standard spacing , giving rise to some part of the image being formed at a larger angle between  and . B1 B1 180 135 90 45 0

2 ©EJC 2023 9749/J2H2PRELIM/2023 1ci The stationary wave is formed when the incident wave is reflected at the sea wall such that the incident and reflected waves are of the same type, amplitude, frequency, wavelength and speed the incident and reflected wave travelling in opposite directions then overlap and superpose. B1 B1 1cii For the increase in amplitude at wall, the incident and reflected waves must have amplitudes of the same sign / crest meets with crest. OR Constructive interference The reflected wave must have the same phase as the incident wave, hence no phase change occurred during reflection. M1 A1 1ciii 1.75 = 0.78  = 0.78/1.75 v = f = 720 × 0.78/1.75 = 321 C1 A1

3 ©EJC 2023 9749/J2H2PRELIM/2023 Qns Answer Marks 2(ai) The increase in internal energy of a system is the sum of external work done on the system and the heat supplied to the system. B1 B1 2(aii) When roasting, heat is supplied to the potato, thus toQ is positive. onW is negligible since there is little change in volume of the potato) / onW is positive since cooked potato becomes smaller in volume Therefore, increaseΔU is positive. OR onW is negative but magnitude of Q is larger than magnitude of W Therefore, increaseΔU is positive. ΔU=ΔKE+ΔPE ΔPE is negligible as there is no change in state of the potato, therefore, ΔKE is positive. OR ΔPE is positive as water from the potato is changed to gaseous state but ΔKE is also positive. Since KE is directly proportional to (thermodynamic) temperature, the temperature of the potato increases. B1 B1 B1 2(b)(i) ΔU=Q+W Constant volume, therefore W = 0 J ΔU=Q = -55 J B1 A1 2(b)(ii) Process A to B takes place at constant temperature, therefore UA = UB => UAB = 0 For cyclic process, UABCA = 0 UAB + UBC + UCA = 0