DHS_H1_PHY_P1_MS

Page 1 of 6 DHS Mark Scheme Syllabus Year 6 Preliminary Examinations H1 Physics 2016 8866 Paper 1 Question Number Key Question Number Key 1 A 16 C 2 B 17 D 3 C 18 C 4 A 19 D 5 C 20 A 6 D 21 B 7 D 22 A 8 D 23 A 9 A 24 B 10 B 25 D 11 A 26 B 12 A 27 C 13 A 28 D 14 C 29 B 15 C 30 D

Page 2 of 6 Paper 2 1 (a) (i) Correct lengths of vector components. B1 (ii) All horizontal components must be of the same length B1 vertical component of A should be zero B1 vertical component of B is downwards and less than that at S B1 (b) (i) correct shape of the graph (parabola), max at mid point B1 (ii) correct shape B1 at maximum height, min value of KE is not zero B1 [7] 2 (a) apply conservation of momentum 0 = (0.500 Χ 3.8) – (0.310 Χ v) C1 v = 6.1 m s -1 A1 (b) F = Δp/Δt C1 = (0.500 x 3.8)/0.25 C1 = 7.6 N A1 [5] 3 (a) mass is the property of a body resisting changes in motion / quantity of matter in a body/ measure of inertia to change in motion B1 weight is the force due to gravitational field/ force due to gravity / force due to gravitational force B1 (b) (i) Constant horizontal speed, net force in the horizontal direction is zero M1 Resistive force = F cosθ = 300 cos(30°) = 260 N A1 (ii) vertical direction  300sin(30°) + contact force = mg M1 Contact force = (50 x 9.81) – 300sin(30°) = 340 N A1 [6] 4 (a) (i) constant phase difference B1 (ii) wavelength estimate: 550 to 700 nm C1 Separation = λD/ a = (650 Χ 10 -9 Χ 2.4) / (0.86 Χ 10-3) = 1.8 mm A1 (iii) dark fringes are brighter as amplitude no longer completely cancel A1 vV vH

Page 3 of 6 bright fringes are less bright as resultant amplitude is smaller A1 (iv) shorter wavelength for blue, so separation is less A1 (b) (i) 1 6 0 c m A 1 (ii) v = fλ, = 20 x 1.6 = 32 m s -1 A1 (iii) progressive wave reflected at the fixed ends B1 incident and reflected (two) waves travelling in opposite directions interfere B1 speed is the speed of one of these waves A1 [11] 5 (a) Oscillation of the molecules of the wave is along the direction of transfer of energy of the wave. B1 (b) A = v2 = (2700)(3100)2 = 2.59 x 1010 A1 U n i t s o f A = (kg m-3)(m s-1)2 = kg m-1 s-2 = Pa A1 (c) (d)

Page 4 of 6 (e) The waves should be weaker after traveling longer distances, B1 hence direct waves should show larger amplitude than reflected waves. B1 (f) 1. SD 8, t = 0.4 s, so SD8 = (3.1)(0.4) = 1.24 km A1 2 . S X D 8, t = 0.6 s, so SXD8 = (3.1)(0.6) = 1.86 km A1 (g) Assume SX = XD 8 Using Pythagoras Theorem, depth d = 2 8 2 8 )()( ODXD  = 2 8 2 8 22          SDSXD = 22 2 24.1 2 86.1          C1 = 0