RI 2017 A-Level H2 Phys Solution

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT ( )12 1 1 1 1 6 32ratio 31 6 2 vv v vv +× = = = ×× GCE A−level 2017 H2 PHYSICS 9749/1 answers 1 A 11 D 21 A 2 B 12 B 22 D 3 C 13 D 23 D 4 D 14 D 24 C 5 C 15 B 25 A 6 A 16 C 26 B 7 D 17 D 27 D 8 C 18 C 28 C 9 D 19 B 29 B 10 A 20 C 30 C Suggested detailed solutions: 1 A 2 B 2 Let , then 2 2 2 1 2 6xy A x y zA % %%%z Ax yz ∆∆ ∆∆= = + + = +× + = . 3 C Distance travelled is given by the area under the speed-time graph. OR By similar triangles, from the graph: 3 31Ratio = = 4 D The maximum velocity (terminal velocity Tv ) happens when air resistance is equal to weight: T0 60. v mg= . The acceleration a at any speed is found using Newton’s 2nd Law: ( ) 2net 3 0 9 81 0 60 120 60 7 4 ms30 . . .( )F mg . va. mm . −−−= = = = 12 21 00 2 6 0 12 0 vv vv−− = ⇒= −− 1 T 30 98 1 49 ms0 60 max ..vv . −×= = =

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT Acceleration when v = 12 ms-1 : 5 C Since the momentum reverses direction in the collision, the change in momentum is greater than p. Since the collision is inelastic, the final momentum must be less than p in magnitude (so that the final KE is less than the initial KE). So the change in momentum must be some value in between p and 2p. 6 A Action-reaction pairs satisfy the (A exerts on B) and (B exerts on A) relations. 7 D KE is converted into GPE on the way up, so it decreases; it increases on the downward trip since GPE is converted into KE. 8 C For a satellite in orbit, Total energy of satellite, Change in total energy of satellite, OR 11 24 2 66 9 11E= 2 6.67 10 6.0 10 6.9 10 1 1 2 6.5 10 7.2 10 2.1 10 J T fi GMm RR − ∆− −   × ×× ×× =−−  ×× = −× 9 D Angular velocity ω is common to every point on the disc. 2 2 24 Since and is a constant, a r, 1 8 ms2 rr ar aa ωω − = = = ∝ = = 10 A 2 30 98 1 06 0 1 2 30 7 4 ms . . . .a a. − × − ×= × ∴= 2 2 GMm mv RR = 2GMm mvR⇒= 2 22 T GMm mvE R= −= − 2 22 96 9 10 7900 7500 2 1 10 J2 T .E. × ∆= − − = −× 1GM RRφ = −∝ SX SX R R φ φ = ( ) 6 7 71 6 6 371 10 6 257 10 6 208 10 J kg6 371 10 50000 X . ...φ −×⇒ = × −× = −××+

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 11 D Note that 273 15T/K T/ C .= +  . 12 B The system is insulated from heat supply, therefore Q = 0. Since temperature increases, internal energy increase. From the first law of thermodynamics: doneonsystemUQW∆=+ Since Q =0, 0doneonsystemWU = ∆> By stirring, work is done on the system. 13 D For a good car suspension system, the oscillation should stop as quickly as possible so that the system may return to and stay at equilibrium position. A and C are light damping. B is over damping. D is critical damping. 14 D The dotted wave is lagging the solid wave by 2/π (quarter of a period), or leading it by 32 /π . 15 B When going from one medium to another, the frequency of the wave does not change (frequency is source dependent). Since speed is halved, wavelength is also halved. Intensity is proportional to square of amplitude. 16 C It is a fact that θθ < 2 1 2 , and higher order maximas are dimmer because most of the light passes straight through. 17 D Since kmv T v mg v k m f f∝ = > ∝ = > = = λ= >λ= ----------(1) and for a stationary wave to form, λ= ⇒λ= 2 2 LLn n ----------- (2) 22k m L Lf nfn km λ= = = > = ------------- (3) When f = 15 Hz, m = 400 g and n =1 15 322 4400 k L ( L)= = -----------(4) Subt (4) into (3) 22 4 3 324 Lf Lf fn km m( L) m = = = Since n is an integer, only option D gives an integer of n = 4 2I kA= ( ) 2 2 2 0 25 AI' I' . IIA= ⇒=

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT OR Fundamental mode, Higher harmonics, The other options did not yield integer values for n. 18 C Electric field strength At P, the directions of the field strengths for both charges are pointing to the right. Thus, the net field strength is 10.0 + 2.5 = 12.5 N C−1 19 B If the resistor obeys Ohm’s law, then the graph of I vs V will be a straight line passing through the origin for both + and −V. Since there is a diode present, we get only the + V side of the straight line. 20 C ==× =×× = ∴= 5 7 280 0 9 5 7 280 6 245 A input output rms rms rms P IV . , P . . I V I. 21 A ∝ LR A Since the lengths of the sections are equal, ∝ 1R A . This means that the gradient increases as the area A of cross-section decreases. 22 D Because the current in the bottom 30 Ω resistor is 4A + 2A = 6A, therefore the total p.d. across both resistors is (6 x 30) + (4 x 30) = 300 V. 23 D 14 0 6 -1 19 0 4 3 1020 8 93 10 m s0 88 1 6 10 20 perpen .F Bqv Bqv sin v .. . sin − − ×= = ⇒= = ××× 24 C Segments MN and QP cut through the magnetic field whereas MQ and NP do not. From E = BLv, L and v for MQ and NP are parallel ∴ E = 0. L and v for MN and QP are 90° to each other ∴ E = BLv. Hence emf is induced across MN and QP only. v T mg m v k m∝ ∝ ∝ ⇒= where 12 3 422 n nvL n , , , ,....f λ= = = 1 0 4000 900 42 69072 15 k.. k.×= ⇒=× 42 6907 0 9000 900 42 90 n. ..n ××= ⇒=× 2 1 E r∝ 2 1 2 0 30 1 1 00 25 N C0 60 4 q q q E . E ..E. − − − + = ⇒ = ×=

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 25 A −2Vt graph is the same as that for a sinusoidal p.d, hence = =0 224 22 rms VV , 26 B =  = = = ⇒= ≈ Ω 2 0 2 1 16 232 1 98 2 0 s p s N N V VP R. .RR 27 D Factual question 28 C − − − ×λ= = = ×× ×× 34 11 31 7 6 63 10 3 63 10 m9 11 10 2 10 h. .p. 29 B Because in an α-decay, both the numbers of protons and neutrons decrease by 2 30 C Next, apply −λ= 0 tN Ne to solve for the time t −−×= ×= × 1319 20 0 7708 101 5 10 4 8 10 .tN. . e ⇒ t = 4.5 x 1013 s OR 224 2 26 A 2 70 0 rms rms V .R . ∴= = = × I 2242 26 358 W 2 rms rmsP V.= =×=I 7 14 1 00 20 3 7 10 7 708 10 s4 8 10 .AN .. −−×=λ ⇒λ= = × × 12 1 14 2 ln 2 8 992 10 s 7 708 10T. . −= = ×× 19 20 5 0 1 5 10 1 1 4 8 10 32 2 N. N. ×= = =× 12 13 1 2 Time taken is 5 5 8 992 10 4 496 10 sT. .= × ×= × 7 14 1 00 20 3 7 10 7 708 10 s4 8 10 .AN .. −−×=λ ⇒λ= = × ×

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2017 A-level H2 Physics Paper 2 Solutions (9749/02) 1 (a) Extension x = 10.8 – 8.0 = 2.8 cm F = k x k = F / x = (0.140)(9.81) / (0.028) = 49.05 = 49 N (2 s.f.) (b) (i) x = L2 – L1 ∆x = ∆L2 + ∆L1 = 1 + 1 = 2 mm k = F / x = mg / x (g has no uncertainty) Δk Δm Δx=+kmx = 1.0 100 + 0.2 2.8 = 0.081 Percentage uncertainty = 8.1 % (ii) ∆k = 0.081 × 49.05 = 4 (1 s.f.) Force constant = 49 ± 4 (c) (i) T + U = mg (where T = tension; U = upthrust) U = mg – T = (0.140)(9.81) – (49)(0.103 – 0.080) = 0.246 = 0.25 N (shown) (ii) U = weight of fluid displaced = V ρliquid g = (m / ρblock) ρliquid g ρliquid = (U) (ρblock) / mg = (0.25) (7750) / (0.140)(9.81) =1400 kg m−3 (2 s.f.) 2 (a) (i) 7 3 5 5 2 2 2 (1.75 10 ) 0.200 10 5.49 10 5.495.49 10 24 60 60 6.36 days vr r T rT v s πω ππ = =  ×= = × = × ×= ×× =

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT (ii) 1. The probe will always see Charon. 2. The probe will always see the same face of Charon. Part (1) assumed that the direction of orbit of Charon is in the same direction as that of the rotation of Pluto about its axis. Hence, the more correct answer would be that Charon will appear at the same position relative to a fixed position on Pluto after each period. (It is not mentioned that the orbit of Charon is in the equatorial plane of Pluto. Hence, a “geostationary” answer is incorrect). Part (2) requires the assumptions of part (1) . Otherwise, there is no significance to the fact that both Charon and Pluto