2017 RI H2 Physics P3 solutions

2017 Year 6 H2 Physics Prelim Paper 3 Mark Scheme © Raffles Institution 9749/03 1 (a) Simple harmonic motion is defined as the motion of a particle a bout a fixed point such that its acceleration is proportional to its displacement from the fixed point and is always directed towards the point. [B1] [B1] (b) 0.11 m [B1] (c)  = 2f = 2   0.40 = 2.51 = 2.5 rad s 1 (2 s.f.) [M1] [A1] (d) Graph in the correct quadrant, gradient at x-axis =  and gradient at y-axis = 0. Correct values at both intercepts (x 0 = 0.11 m and vmax = 0.28 m s–1). [B1] [B1] (e) Period T = 11 0.4f  = 2.5 s Since the lamp is above the seawall for more than half the peri od (1.25 s), the equilibrium position of the lamp is above the seawall. Time taken t for the lamp to go from its highest position to the top of the seawall t = 1.8 2 = 0.90 s x = x0 cos t = 0.11 cos (2.51  0.90) = 0.0699 m Time taken t for the lamp to go from its lowest position to the top of the seawall t = 2.5 1.8 0.35 s2   x = x0 cos t = 0.11  cos (2.51  0.35) = 0.0701 m [M1] [M1] [A1] velocity / m s1 displacement / m 0 0.11  0.276 or 0.28 t 1.8 s x 2.5 t 1.8 s x

2 © Raffles Institution 9749/03 Alternatively, Time taken t for the lamp to go from equilibrium position to the top of the seawall t = 1.8 1.25 2  = 0.275 s x = x0 sin t = 0.11 cos [2.51  (1.25 + 0.275)] = 0.0697 m Time taken t for the lamp to go from equilibrium position to the top of the seawall t = 1.8 1.25 2  = 0.275 s x = x0 sin t = 0.11 cos (2.51  0.275) = 0.0700 m [M1] [M1] [A1] 2 (a) A progressive wave is one in w hich energy is transferred from one point to another without the transfer of matter. [B1] A transverse wave is one in which the particles oscillate in a direction perpendicular to the direction of energy transfer. [B1] (b) 4.0 s (1 d.p.) [B1] (c)  = (t / T)  2 = (1.4 / 4.0)  2 OR (7/20)  2 = 2.20 = 2.2 rad (2 s.f.) Ignore  signs. OR  = (t / T)  2 = (2.6 / 4.0)  2 OR (13/20)  2 = 4.08 = 4.1 rad (2 s.f.) [M1] [A1] OR [M1] [A1] (d) intensity  (amplitude) 2 intensity  1/r (for 2D circular wavefront)  (amplitude)2  1/r  rQ AQ2 = rP AP2 From Fig. 2.2, A P = 2.0 mm and AQ = 1.0 mm,  rQ 1.02 = (150)  2.02  rQ = 600 mm [B1] [M1] [A0] (e)   = 2  (x / ) 2.20 = 2  (600  150) /   = 1286 mm = 1290 or 1300 mm (2 s.f.) (allow ecf from (c)) [M1] [A1] t 1.8 s x 1.25 t t 1.8 s x 1.25 t

3 © Raffles Institution 9749/03 [Turn over (f) Graph shows amplitude of wave decreasing with distance. Accept positive/negative sine/cosine graphs. [B1] 3 (a) (i) Coherent waves have a constant phase difference. [B1] (ii) As the two transmitters emit coherent waves, a steady/observabl e two-source interference pattern is formed along the orbit. When the waves meet in phase, a high intensity signal (or const ructive interference) is detected by the satellite. When the waves meet in anti-phase, a low/zero intensity signal (or destructive interference) is de tected by the satellite. Since the satellite orbits at a constant speed and the “fringe separation” (or distance between successive maxima) is constant, the signals ha ve a constant period (or vary periodically). [B1] either one [B1] [B1] (iii) In 1 second, the satellite travels 8.0  103 m and receives 4 sets of variations in intensities, i.e. it has moved a distance of 4x, where x is fringe separation. x = (8.0  103)/ 4.0 = 2000 m D = ax/ = (150)(2000)/(1.2) = 2.5  105 m [C1] [M1] [A1] ( b ) ( i ) Actual angle subtended by the two distant sources at the telescope: actual  s / r (small angle approximation) = (2.0  1012) / (3.0  1016) = 6.7 105 rad Rayleigh criterion: RC =  / b (where b, the “slit-width”, is the size of the telescope dish) = 1.2 / 300 = 4.0  103 rad [M1] [M1] displacement distance 0 source 1 source 2 2.0  1012 m actual satellite 3.0  1016 m

4 © Raffles Institution 9749/03 Since the angle subtended by the sources is smaller than that o f Rayleigh criterion, the telescope cannot resolve these sources. Accept circular aperture: RC  1.22  / b = 1.22 (1.2 / 300) = 4.9  103 rad [A1] (ii) Maxima of second image to lie on the first minima of the first image and maxima of first image lie on the first minima of the second image. [B1] 4 (a) (i) The electric field strength at a point is defined as the electr ic force exerted per unit positive charge (placed) at that point. [B1] (ii) Magnitude of E = Q / 40r2 = (12  10-9) / (4  8.85 10-12  (2010-3)2) = 2.70  105 = 2.7  105 N C1 (2 s.f.) [M1] [A1] (iii) 0 to 20 mm: zero (must be drawn) 20 to 80 mm: curve with curve touching 20 mm, and non-zero at 80 mm [B1] [B1] ( i v ) By the principle of conservation of energy, gain in kinetic energy = loss of electric potential energy     fi if f f KE KE KE KE 00 99 3312 66 qQ qQ 4r4r 12 10 6.0 10 110 60 10 40 1048 . 8 5 1 0 5.40 10 5.4 10 J (2 s.f.)                  [M1] [A1] ( i v ) Area under the graph between d = 40 and 60 mm. OR Area between the graph and the distance axis, between d = 40 and 60 mm. [A1] 20 E / N C1 d / mm 0 40 60 80

5 © Raffles Institution 9749/03 [Turn over ( b ) The electric field lines  cannot cross.  indicated with arrow pointing from the plate towards the spher e.  are perpendicular to the surfaces of the sphere and plate.  are symmetrical about the vertical line through the centre of the sphere.  are denser directly below the sphere and spread apart towards the edge of the plate (because the potential difference between the sphere and the plate is constant, the distance between sphere and plate is shortest below the sphere). Deduct [1] for each mistake. [B2] 5 (a) (i) B = = = 4.0 10 6 T [M1] [A0] (b) (i) Faraday’s law of electromagnetic induction states that the e.m. f. induced in a conductor is proportional to the rate of chan ge of magnetic flux linkage. [B1] (ii) At d = 20 cm from wire: B = = 7(4 10 )(2.0) 2 (0.20)    = 2.0 106 T  66 8 (2.0 10 4.0 10 )(50)(0.010 0.010) (0.40) 2.5 10 V NABE t BNA t           [C1] [M1] [A1] (c) (i) By Lenz’s law, the induced current will flow in a way to oppose the decrease in magnetic flux density. Hence, the induced current flows in the clockwise direction (to reinforce the magnetic flux density into the paper). OR [M1] [A1] OR 0 2 I d   7(4 10 )(2.0) 2 (0.10)    0 2 I d   sphere plate

6 © Raffles Institution 9749/03 As the coil is being pulled away from the wire, by Lenz’s law, the induced current in the coil flows in a way to oppose this motion. Hence, the induced current must produce a net force which acts upwards by flowing clockwise (to produce a larger attractive force on the upper side and a smaller repulsive force on the lower side). [M1] [A1] (ii) The induced current in the coil produces a magnetic force that opposes the coil’s motion (or that pulls the coil towards the wire). Therefore, an external force must be applied in the direction o f motion to maintain a constant velocity (or