RI 9646 2014

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 2014 GCE A-Level 9646 H2 Physics Suggested Solutions Paper 1 Solutions 1 C Absolute error presented to 1 sf = 3 m s−1 Quantity presented to same precision as absolute error = 348 m s−1 Speed = (348 ± 3) m s−1 2 A 5 µm = 5000 nm (not visible ⇒ not B, C) 0.5 µm = 500 nm (visible ⇒ A or D) Radio waves are in the range of cm to m (not D) 3 A Gravitational field strength = initial acceleration = gradient of graph when object is just released from rest = 6 m s−2 4 D Let total distance travelled be x. Use s = ut + ½ at2. For initial 0.25 x: 0.25 x = 0 + ½ (9.81) t2 For final 0.75 x: x = 0 + ½ (9.81) (t + 1)2 ⇒ 2 (9.81) t2 = ½ (9.81)(t2 + 2t + 1)2 ⇒ 3t2 – 2t – 1 = 0 ⇒ t = 1 s Total = 1 + 1 = 2 s 5 D Since system total momentum is non -zero, both spheres cannot be at rest simultaneously in order to conserve momentum. 6 C ( ) ( ) ( ) 4 -2 Let the required tension be . Isolate th e wagons 3, 4, 5 and 6 and apply N2L. 4 4 4000 N 4 6.0 10 kg 0.15 m s 52 kN T T f ma T T −= −× = × × × = 7 C Work done in stretching the fibre is given by the area under the force -extension graph. When force is increased from 0 to F, the extension increases, work is done on the fibre, and this work is the area P. When the force is reduced from F to 0, the extension decreases, work is done by the fibre, and this work is the area Q. The net work done on the fibre is therefore (P – Q). Note: if the axes are swapped, work done will be area between graph and vertical axis. 8 D When the ball is falling vertically with constant speed v, the result ant force acting on it is zero. Therefore, weight W is balanced by the sum of upthrust U and viscous force kv: W U kv= + Note: if velocity is high, the viscous is kv2.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 9 C ( ) ( )( ) 45 6 Let the power developed by the locomotiv e be . If its acceleration is when travelling at speed , the equation of motion is pulling force frictional force mass 5.0 10 3.0 10 0.5010 2.0 10 P av P av P P −= ⇒−× = × ⇒= × max 6 4 max 6 -1 max 4 W The speed reaches a maximum when the acceleration decreases to zero: frictional force 0 2.0 10 5.0 10 2.0 10 40 m s5.0 10 P v v v −= ×⇒= × ×∴= = × 10 D The direction of electric field V/x points from left to right within the parallel plates. The electric force (q)(V/x) acts to the right on the charge having displacement x to the right. The work done by this electric force on the charge is +[( q)(V/x)] ( x) = +qV, causing its kinetic energy to increase by qV. Therefore, its electric potential energy decreases by qV, due to conservation of mechanical energy. 11 B A & D are equivalent statements and both are wrong. If the centripetal acceleration of the astronaut and capsule is the same, their centripetal force may not be the same because they may have different mass. 12 B The car tends to lose contact at the top of the lump if it’s speed is too high. The car just loses contact with the lump if the normal contact N just becomes zero. Applying N2L to car at top of lump: Fc = W – N ⇒ mv2/r = mg – 0 ⇒ r = v2/g = 202/9.81 = 41 m 13 B The g-field strength at the mid -point between stars X & Y must be zero since both stars have the same mass and they are equidistant from the line PQ. (B) As we move away from that mid -point, the vertical component of g due to each star continues to cancel each other, leaving only the horizontal component gx: gx = −2GM cosθ / r2, where θ is the angle each g-field makes with the line PQ. Initially, the horizontal component g x increases as we move away from the mid- point. But when r is too large, g x decreases to zero. This explains the rise and fall of option B. 14 C Applying N2L, FR = ma ⇒ FG = m (rω2) ⇒ GMm/r2 = mrω2 ⇒ M = r3(2π/T)2/G = (2.95x108)3 (2π/1.89x24x60x60)2/(6.67x10-11)= 5.70x1026 kg 15 A Definition of shm: a = −ω2x. Hence, the a -vs-t graph has the same shape as the negative x-vs-t graph.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 16 B Since the amplitude of the driver is constant, the amplitude of the new graph must be same as graph X at low frequency (B or C). However, with less damping, the new graph must above graph X at all frequencies. (So, C is wrong.) 17 A Whether it is the Celsius or Kelvin scale, the temperature difference between ice and boiling point are both 100 oC or K. 18 C Q = m c θ ⇒ dQ/dt = m c dθ/dt ⇒ P = m c (gradient) ⇒ c = P / m (gradient) ⇒ c ∝ 1/gradient, since P & m are constant Since Y has the smallest gradient, it has the largest specific heat capacity c. Note: Option A is wrong because horizontal portions represent boiling points of the 3 liquids. 19 A The total number of gas particles remains constant before and after heating. ⇒ Ntotal, i = Ntotal, f ⇒ p1i V1i / kT1i + p2i V2i / kT2i = p1f V1f / kT1f + p2f V2f / kT2f ⇒ p1f [V1i / T1i + V2i / T2i] = p1f [V1f / T1f + V2f / T2f] ⇒ (1.00x105) [0.40 + 0.20] / 300 = p1f [0.40 / 300 + 0.20 / 600] ⇒ p1f = 1.2x105 Pa 20 D Malus law: I = Io cos2θ ⇒ A = Ao cosθ From wave theory, we know that I ∝ A2. Hence, I ∝ Ao2 cos2θ, where Ao = a. 21 D Speed of wave = wavelength/period = half-wavelength/half=period = (x3-x2)/(t2-t1) 22 C Incident waves and reflected waves superpose to produce a stationary wave. Distance between 2 consecutive nodes (eg 90- 30mm) or 2 consecutive antinodes (eg 60-120mm) is equal to half-wavelength. Therefore, λ = 2x60 = 120mm. 23 B Applying the double-slit formula, x = λD/a = (6.0x10-7)(1.50)/(3.0x10-3) = 0.0003 m 24 D That is how the direction of an electric field is defined. Note: Electric field strength depends on how close or far apart the field lines are. 25 A Do by elimination method – otherwise, there are many lengths, angles & resolutions to deal with. By symmetry, E-field at S & Q must the same. Hence, options C & D are wrong. By drawing vectors, E -fields at P & R are in same direction as E -fields at S & Q. So, option B is wrong. 26 D T = 2π/ω I = Q/t = e/T = eω/2π = (1.60x10-19)(4.1x1016)/(2π) = 1.04x10-3 A

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 27 D For D, Vx = (4/6)x12 = 8V. Options B & C are in reverse bias and hence no current is flowing. 28 B In the dark LDR resistance increases. The net resistance of the parallel branch also increases (apply resistors in parallel formula). Therefore, total current I 1 decreases. Since net resistance of the parallel branch increases, pd across the parallel branch also increase, hence driving a higher current. 29 A Since the potentials at points X & Y are the same, the potential difference between them is zero and hence no current is driven through the wire. 30 D By the principle of conservation of energy, loss in KE = gain in EPE ⇒ −∆KE = ∆PE = q ∆V [since epe = qV] ⇒ (KEi – KEf) = q(E∆x) [since E=−dV/dx] ⇒ ½ m (v2 – vf2) = q(Ex) [∆V, and hence Ex, must be positive] ⇒ vf2 = v2 – 2(2e)Ex/m [q = 2e] ⇒ vf = √(v2 – 4eEx/m) 31 D Magnetic flux BAφ = . Hence 2unit of Unit of Wb munit of B A φ −= = 32 C Magnetic flux density B is independent of the coil position. Hence it remains constant. Total magnetic flux Φ of the coil is dependent on the number of turns N, the area A of coil and the component of B which is perpendicular to A. Since this component of B is smaller in the new position, Φ decreases. 33 D 750 600 9 250 750 600 1800250 750 9 27 V250 SPP P P S SP P P INV N V NVI N V = = ⇒= = ∴= × = ∴ = ×= 34 B Mean power dissipated is 2 r.m.s.PIR= Rearranging, r.m.s. 160 4.0 A10 PI R= = = 35 D The graphs show the instantaneous power dissipated in the resistor. Since Pinstant = Iinstant2 R = (Iosin ωt)2 R = Io2R sin2 ωt = (√2I)2 R sin2 ωt = 2 I2R sin2 ωt. The shape of graphs A & B are wrong because s