NYGH 2017-S3MYE-Adv Physics -Ans

2017 NYGH S3 Advanced Physics MYE Answers to Section A Note: Advanced physics questions (MCQ): 6, 9, 15, 19 and 20 1 D 6 D 11 D 16 B 2 A 7 C 12 B 17 A 3 B 8 C 13 C 18 C 4 D 9 B 14 C 19 A 5 C 10 B 15 B 20 C 6. With air resistance, the net force & hence the acceleration of the object decreases to zero. The velocity would increase to a maximum constant value (terminal velocity). 9. Let C be at the surface of the viscous medium. A to C: u = 0, s = 2.0 m, a = 10 m s -2, t1? s = ½ at2  t1 = 0.632 s v = u + at  v = 6.32 m s-1 C to B: constant v = 6.32 m s-1, s = 1.8 m, t2? s = vt  t2 = 0.285 s 15. Consider both blocks moving together (single freebody). F net = ma  54 – 6 = (6.0 + 2.0) a  a = 6.0 m s-2 Consider 6.0 kg mass alone. Fnet = ma = 6.0 a 19. r = 1.50 x 1011 m T = 1 year = 365 days, convert to seconds angular velocity ω = 2π/T = 2π / (365 x 24 x 60 x 60) Speed v = rω 20. F = mg = 2r GMm Gravitational field strength, g = 2r GM at surface of a planet of mass M and radius r. Earth: g1, M1 and r1 Another planet: g2, M2 and r2 Using EPR approach: • Equation: g = 2r GM • Proportional: g ≈ 2r M • Ratio: 1 2 g g = 1 2 M M x 2       2 1 r r = 1 2 x 2      2 1 = ½ g2 = 2g1 Answers to Section B (same as 2017 Sec 3 Physics MYE)

2017 Mid-Year Examination Secondary 3 Advanced Physics 2 Nanyang Girls’ High School Setter: CMH, MS, AJL Physics Answers to Section C 28 (a)(i) r = radius + height = 6.4 x 106 + 600 x 103 m r = 7.0 x 106 m g = F/m = GM/r2 = 8.2 m s-2 (a)(ii) a = v2/r where a = g = 8.2 m s-2  v = √(ar) v = 7576 ≈ 7600 m s-1 (a)(iii) F = mg = 11000 x 8.2 = 90200 ≈ 90 000 N (a)(iv) Any relevant point below: At this orbit, it is further up in the Earth’s atmosphere / air is very thin / the telescope is not affected by atmospheric pollution / images obtained would be very clear Note: The difference in height has little effect on distance/field of view of outer space, since this difference in height << distance to outer space. If the telescope is located on Earth’s surface, it is impossible to ‘crash’! (b)(i) a = g = 10 m s-2 and r = 40 m a = v2/r = rω2 = r (2π/T)2  √(g/r) = 2π/T  T = 2π √(r/g) T = 2π = 12.57 ≈ 13 s (b)(ii) Any relevant point below: He may have a sense of confusion as to what is “up” or “down” / he has to walk along a “curved”/ “circular” floor, or floor is not flat / he may feel giddy or nauseous due to high speed of rotation / he is unable to to experience day and night (view through window) / he experiences different values of g at difference distances (r from centre)