2018 RI H2 Physics P3 solutions

© Raffles Institution 9749/03   2018 Raffles Institution Preliminary Examinations – H2 Physics Paper 3 – Solutions Section A 1 ( a ) ( i ) (The graph is a curve. This implies there is non-constant acceleration due to effect of air resistance. Only at t = 0 is the acceleration equal to g) Draw tangent at t = 0 acceleration = gradient 24.00 1.74 m s2.30  (ii) The Moon’s surface has no atmosphere (very negligible compared to Earth). There is no air resistance outside the container. Read off from tangent (as this is the v-t graph for no air resistance): v = 1.4 m s-1, t = 0.80 s or use equation: v = u + at (substitute value of a from (a)(i), v = 1.4, u = 0) (b) (i) At terminal velocity, resistive force F reaches its maximum value.  max 3 0.0027 1.74 4.70 10 N Fm g     (ii)  3 31 is terminal velocity on Moon)( 1.40 4.70 10 3.36 10 kg s TT vkv mg k k      0.0 1.0 2.0 3.0 4.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 P (a)(i) Q

2    © Raffles Institution 9749/03 (iii)  '' 3' '1 is terminal velocity on Earth)( 3.36 10 0.0027 9.81 7.89 m s TT T T kv mg v v v      ( i v ) 1 . Total mass is 2.5 x initial mass.  3 1 3 2.5 4.70 10 3.50 m s3.36 10 T T kv mg mgv k         Initial gradient is the same, and graph always above the origin al (but below the tangent) – B1 Curve reaches terminal velocity 3.5 m s–1 – B1 (time it reaches terminal velocity is later – not marking pt) 2 . Starts from 4.0 m s–1 Curve must show speed decreasing to vT – B1       2 (a) (i) The gravitational field strength at a point in space is defined as the gravitational force experienced per unit mass at that point. (ii) Gravitational field strength is a vector since it is defined us ing gravitational force which is a vector. (iii) Newton’s law of gravitation states that two point masses attrac t each other with a force that is directly proportional to the product of their mas ses and inversely proportional to the square of the distance between them.  Since gravitational field strength g is defined as the gravitational force acting per unit mass: (b) (i) 22 Sun Jupiter 2 230 7 27 11 6 1.99 10 7.14 10 1.90 10 7.79 10 8.80 10 Sun Jupitersun Sun Jupiter SunJupiter Jupiter GM rrgM gM r GM r              (ii) Since the gravitation field strength due to the Sun on the surf ace of Jupiter is only ~106 (< 0.001%) that of the gravitational field strength on the surface of Jupiter due to its mass, it can be neglected. (c) (i) For a moon in circular orbit about its planet, the centripetal force is provided by the gravitational force acting on the moon by the planet: GF GMm GMg m rm r22 

3    © Raffles Institution 9749/03 2 2 2 32 3 2 23 2 2 23 22 , 2 4 4 4 pm m p p p p p GM M MRR Since T we have T GM R R T Rearranging, we have: TR GM Since M is the mass of the planet, is a constant GM T KR where K GM               (ii) Hence, the ratio of the orbital period of Io: Europa: Ganymede is 1: 2: 4     3 (a) Since the test-tube is in equilibrium,    0netFM m gA H g       Mm g A H g  MmH A  (b) (i) Agy (ii)  Agy M m a  Agay Mm   (c) By comparing with 2ax  , 2 Ag Mm    2 2 T Mm Ag       4 0.012 0.0252 1000 6.0 10 9.81 0.498 0.50 s       EE E II I GG G II I R TR R TR R TR R TR R 23 33 38 9 33 39 8 Since T 6.71 10 2.01 2 4.22 10 1.07 10 4.04 4 4.22 10                     

4    © Raffles Institution 9749/03 (d) As the test-tube oscillates, it experiences drag force exerted by the water . This results in light damping and energy is gradually lost as heat. (e) (i) (ii) The amplitude of oscillation is small because the frequency of the driving force (the waves) is too low (0.30 Hz) compared to the natural frequency o f the test-tube (2 Hz). The amplitude of the oscillations can be increased by adding ba ll bearings to the test-tube to decrease the natural frequency so that it is close r to the frequency of the driving force.   4 (a) (i) The principle of superposition states that when two or more waves of the same kind meet at a point in space, the resultant displacement at that po int is equal to the vector sum of the displacements of the individual waves at that point. (ii) According to Huygen’s principle, all the points on the wavefronts that pass through the single slit are individual sources of circular wavelets whi ch interfere with one another by the principle of superposition to form the interference pattern. (b) (i) For single slit diffraction, sinbm  Positions of minima,  9 3 3 600 10sin 2.0 10 0.30 10mmmb        where 32.0 10b   (ii) limiting angle of resolution for the single slit, 3 min 2.0 10 radb    3 31.0 10 4.0 10 rad0.25 s r     Since min , the interference patterns due to the two point sources of li ght can be resolved. y t I b b b b b b

5    © Raffles Institution 9749/03 (c) (i) Position of first minima of the single slit diffraction envelope is given by: sin ----- (1)b   where b is the slit width Positions of maxima of the double slit interference pattern is given by: sin ----- (2)an   where a is the slit separation Where the first minima of the single slit diffraction envelope coincide with a double slit maxima,  is the same in equations (1) and (2). (2) 1.2: 4(1) 0.30 an b  Hence the 4th orders will be missing from the double slit interference pattern. Number of maxima in the central region / within the diffraction envelope = 3 + 3 + 1 = 7 (3 maxima on either side of the principle axis and the central maximum) (ii)  3 3 3 3 12 10 1tan 2 12 10 2 12 10 22 . 0 1 0 3.0 m D D           

6    © Raffles Institution 9749/03 5 (a) Faraday's law of electromagnetic induction states that the indu ced e.m.f. is proportional to the rate of change of magnetic flux linkage. (b) (i) When magnetic field increases in the positive direction, induced current decreases in the negative direction. When magnetic field reaches maximum, induced current becomes zero. As magnetic field decreases in the positive direc tion, induced current increases in the positive direction until it reaches a minimum value when B = 0. This corresponds to Lenz’s law as induced current flows in a di rection to produce effects that opposes the change producing it. Hence sensitive ammeter deflects in opposite directions. (ii) 1. Accept t = 115, 230, 345 or 450  103 s 2 . Using (85  10 -3, 0.40  10-2) and (145  10-3, -0.40  10-2),   22 33 max 0.40 10 0.40 10d 0.1333dt 145 10 85 10 B        4 max max max d 4.2 10 400 0.1333 d5 . 0 E AN Bi RR t   3 max 4.5 10 Ai  3 .  232 max 4.479 10 5.0 22 2 oPi RP    55.0 10 WP  B / 102 T t / ms 0.8 0.4 0.4 0 100 200 300 400 500 0.8 0.2 0.6 0.2 0.6

7    © Raffles Institution 9749/03 (iii)       ( i v )  t / ms 0 P / W 500 400 300 200 100 1.00  104 0.45  102 0 100 200 300 400 500 0.45 102 i / A t / ms 0.40  102 0.40  102

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