Prelim H2P2 MS

1 Anderson Serangoon Junior College 2021 H2 Physics Prelim Mark Scheme Paper 2 (80 marks) 1a similarity: lines are radial / greater separation of lines with increased distance from the sphere / lines are perpendicular to the surface of sphere difference: gravitational field lines directed towards sphere and electric field lines directed away from sphere B1 B1 1bi The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. B1 B1 1bii (gravitational) potential is taken to be zero at infinity (gravitational) force is attractive work done by the external agent on the point mass moving it from infinity is negative B1 B1 B1 1ci M1 A1 1cii  magnitude of force decreases non-linearly with distance  decreasing gradient, never infinite nor zero B1 B1 1ciii It represents the increase in potential energy of the satellite. B1 2a total volume of molecules is negligible compared with volume occupied by the gas B1 2b pV = NkT 4.60 × 105 × 2.40 × 10–2 = N × 1.38 × 10–23 × (273 + 23) C1 9749/02/ASRJC/2021PRELIM [Turn Over force distance planet’s surface final position of satellite Penalise [1] if curve is drawn at distances less than planet’s surface. No penalty if curve is drawn beyond final position.

2 N = 2.7 × 1024 volume of one molecule = (4 / 3)r3 (= 1.41 × 10–29 m3) volume of all molecules = 2.7 × 1024 × 1.41 × 10–29 = 4 × 10 –5 m3 (1 s.f.; as this is an estimation question the uncertainty of final answer cannot be more precise than that of the given data.) or volume of one molecule = d3 (= 2.7 × 10–29 m3) volume of all molecules = 2.7 × 10–29 × 2.7 × 1024 = 7 × 10 –5 m3 (1 s.f.; as this is an estimation question the uncertainty of final answer cannot be more precise than that of the given data.) C1 A1 2c Since volume of all atoms (4 × 10 –5 m3 or 7 × 10 –5 m3) is 3 orders of magnitude less than volume occupied by the gas (2.4 × 10–2 m3), so assumption in (a) is justified. B1 2d work done on gas (P → Q) : 0 increase in internal energy (P → Q) : (+)97.0 J increase in internal energy (Q → R) : –42.5 J work done on gas (R→P) W = p ∆V = 2.10 × 105 × (1125 – 950) × 10–6 = 36.8 J increase in internal energy (R→P) since total change in internal energy is zero, 97 + (–42.5) + ∆U = 0 → ∆U = –54.5 J thermal energy supplied (R→P) Q = ∆U – W = −54.5 – 36.8 = –91.3 J A1 A1 A1 A1 A1 3a amplitude = 0.020 m f = 1/T = 1/0.60 = 1.7 Hz a = (–)2x and [ = 2f or  = 2/ T] = (2/0.60)2  2.0  10–2 = 2.2 m s–2 [accept –ve a] E C1 C1 A1 3b Resultant between upthrust and weight Upthrust increases with depth of immersion A B1 B1 3c wave starting with a peak at (0,6), and same period peak height decreasing successively exponentially wrt time. A B1 B1 B1 4a Polarisation is where the oscillations in a wave are confined to one direction only in a E B1 9749/02/ASRJC/2021PRELIM

3 plane norm