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2008 1 2008 nov advanced physics complete yearly solutions 2008 Nov 2008 2 3. 2008 Nov Paper 1 [Measurement] Solution MCQs Manufacturer’s specification of a digital voltmeter is: “accuracy ± 1 % with an additional uncertainty of ± 10 mV” Answer all questions. 1. The meter reads 4.072 V. [Measurement] Recorded reading (4.072 ± 1% ± 10 mV) Solution (4.072 ± 1% ± 0.010) Velocity is a vector. Time is a scalar. (4.072 ± 0.04072 ± 0.010) To define velocity with respect to time, the measurement of distance travelled must be of vectorform. (4.07 ± 0.05) V (error has only 1sf and the main reading follows the decimal place of the error) Recorded reading, together with its uncertainty: Distance travelled in unit time (4.07±0.01) V Distance travelled per unit time (4.07±0.04) V Speed in a particular direction (4.072±0.052) V Displacement per unit time (4.07±0.05) V (A) (ans) themis 2. [Measurement] (D) (ans) themis 4. [Kinematics] Solution Solution Reasonable estimate: Air resistance is negligible. The average kinetic energy of a bus (say, m 3000kg) travelling on an expressway (say, v 100km/h) is 12 mv 2 1 MJ. A metal ball is at rest and dropped over a bed of sand for 1.0 s before impact making a depth of 8.0 mm. Just before hitting the sand, the metal ball (mass m) from rest (initial speed u = 0) has a final speed the kinetic energy of a bus travelling on an expressway is 30,000 J (v) u + gt (0) + (10)(1) 10 ms-1. The average power of a domestic light is 12 – 100W. Kinetic energy before impact is 12 m 10 50m 2 the power of a domestic light is 300 W The average temperature of a hot oven is 300C. Work done against sand Ffrictions mas the temperature of a hot oven is 300 K 1 mv 2 2 where a is the deceleration of ball in sand and s is its depth of displacement. The average volume of air in car tyre is say, 1.0m (tube) 0.2m (height) 0.2m (wide) : 0.04m3 (close) the volume of air in a car tyre is 0.03 m3 50m = mas a (D) (ans) themis 50 0.0080 a 6.3 103 ms-2 (2sf) (close) The average deceleration of the ball: 6.0 102 m s-2 1.2 103 m s-2 Advanced Physics solutions - yearly themis 2008 3 1.2 104 m s-2 3.0 ms-1 5.0 ms-1 6.0 10 m s 3 -2 6.0 kg 10 kg (C) (ans) themis 5. [Kinematics] pin cork The trolleys had a head-on collision that lasted 0.20 s. After the collision, the trolleys stick together. Solution Taking to be positive, A cyclist accelerates down a hill, then constant speed, before decelerating back up another hill. Total momentum 6.05.0 10 3.0 To deduce the graph of distance s moved by the cyclist against time taken, t: Since objects stick together, When the cyclist goes down slope, for every t , there would be an increasing s , giving an upwardly increasing graph. When the cyclist is travelling at a constant speed, for every t , there would be a constant s , giving a constant increasing graph. When the cyclist goes up another slope, for every t , there would be a decreasing s , giving an downwardly decreasing graph. = 0 kgms-1 Total kinetic energy = 0 J Average force change in momentum per unit time of impact 6.0 5.0 0.20 150 N Since the cyclist is moving further away from his original position, the graph of s is ever increasing. The shape of graph: s 0 t 0 0 The average force acting on each trolley during the collision (N) 0 300 60 150 60 300 0 150 (C) (ans) t 0 s s Total momentum of the two trolleys before the collision (kgms-1) themis s 7. [Forces] Solution 0 t 0 0 t A ladder of weight W rests against a vertical wall. 0 (D) (ans) themis 6. Friction exists at each contact point prevents the ladder from slipping. [Dynamics] Solution Trolley 1: Mass = 6.0 kg, Velocity = 5.0 ms-1 Direction: to the right Trolley 2: W According the Lami’s theorem, the lines of action of the 3 co-planar forces must meet at a common point. Mass = 10 kg, Velocity = 3.0 ms-1 Direction: to the left 2008 Nov 2008 4 The diagram that shows the directions of the forces on 9. the ladder: [Work, Energy & Power] Solution A car, of mass m, has a driving force F. W In a time t, it travels a distance s and its speed increases from u to v. W Useful work done by the car engine = final energy initial energy = W = m v 2 u2 2 Fs , it has a unit equivalent to power t mv - u , it has a unit equivalent to momentum Ft , it has a unit equivalent to impulse (C) (ans) themis 8. [Forces] Solution l rubber handle m v 2 u2 2 , it has a unit equivalent to energy (D) (ans) themis wooden section 2.10 l 10. [Electromagnetic Induction] A uniform rod with a rubber section and wooden section is made to balance as shown. 1 mu2 2 The useful work done by the car engine: W 4.00 l 1 mv 2 2 4.00 l l Ww 2.10 l 2.00 l Solution A square loop of wire is placed in a region of uniform magnetic field. Wr Direction of field perpendicular to plane magnetic field Taking moment about the pivot: Ww 4.00 1.00 2.00 2.10 l Wr 2.10 0.5 l v 0.90 Ww 1.60 Wr : loop of wire W V g AL g 0.90 ALw wg 1.60 ALr r g The loop is pulled out of the field at a uniform speed v. 0.90 4.00l l r 1.60 w The ratio, The ratio, Initially when the loop is pulling, the loop experienced no change in magnetic flux linkage, i.e., no work done. density of rubber r = 2.25 density of wood w When the loop is partially out of the magnetic field, the loop experienced a change in magnetic flux linkage, i.e., work done needs to be done to oppose that change. Since the change of flux is constant, the work needs to be done is also constant. density of rubber density of wood 1.71 2.50 3.27 2.25 (B) (ans) themis Advanced Physics solutions - yearly themis 2008 5 The graph shows work done W versus the speed v: Tension in string W Fc — in its usual notations 0 mg m W W v 0 0 Tension in the string: 11.8 N W W 0.5752 1.27 9.81 13.2 N 0.600 v 0 0.5752 v2 1.27 9.81 0.600 r 12.5 N 13.7 N 0 v 0 0 v 0 13.2 N (C) (ans) (D) (ans) themis themis 13. [Gravitational Field] 11. [Motion in a Circle] Solution Solution A satellite has: The rotor of an engine is rotating at 3000 rpm (revolutions per min); radius 8.0 cm. Mass m Orbit Circular, initially at radius r1 around the Earth, then moved to a new circular orbit of radius r2 , as shown. Centripetal acceleration r 2 — in its usual notations 2 3000 2 -2 8.0 102 7900 m s 60 Mass of the Earth M Gravitational constant G The centripetal acceleration: 25 ms-2 3.1104 ms-2 7.2107 ms-2 C Earth mass M satellite mass m r1 7900 ms-2 r2 (B) (ans) themis Change in potential energy of the satellite, 12. [Motion in a Circle] m m2 1 Solution 1 1 m GM GM r2 r1 A pendulum has: Bob mass 1.27 kg Radius of oscillation 0.600 m 1 1 GMm (note: ve) r1 r2 Velocity (horizontally) 0.575 ms , at the centre of its motion when the string is vertical. -1 Increase in the potential energy of the satellite: T a W 1 1 GM r2 r1 1 1 GM r1 r2 2008 Nov 2008 6 1 1 GMm r2 r1 1 1 GMm r1 r2 a / m s-2 10 (D) (ans) 5 themis 14. [Gravitational Field] x / cm 0 -5 -10 Solution Deduce from graph, amplitude of motion 5.0 cm Earth is orbiting round the Sun. Radius of the Earth's orbit 1.50 1011 m gradient Period 365 days GMSmE — r2 — in its usual notations Centripetal force, mEr2 : r2 GMS r2 Mass of sun, MS r 32 G 2 r3 T G 2 1.50 10 365 242 60 60 11 3 6.67 10 2 10 2 2 2 5.0 10 T 2 Period of motion, T 0.44 s Amplitude (cm) Period (s) 5.0 14 10 0.44 10 14 5.0 0.44 (A) (ans) themis 11 16. [Oscillations] 2.01 1030 kg Solution The mass of the sun: 6.41029 kg 1.161033 kg 3.311033 kg 2.011030 kg An object is said to undergo lightly-damped oscillations. (B) (ans) themis 15. [Oscillations] Solution Defining equation for a particle moving in simple harmonic motion is Although the object is undergoing lightly-damped oscillations, i.e., its amplitude will reduce slowly, but its period remains unchanged. Diagram shows the displacement, y of that object against time, t: y y a 2 x t where, acceleration of the particle a , displacement x , and angular frequency How a varies with x for a particle moving in simple harmonic motion, is as shown. t y y t t (D) (ans) themis Advanced Physics solutions - yearly themis 2008 7 Original Ideal gas equation: 17. [Thermal Physics] pV nRT Solution 1 1 p V nRT — A container contains a hot vapour and is allowed to cool New Ideal gas equation: and loses heat to its surroundings. 1 pV 2n R 2T 1 p 14 V — The graph shows how its temperature changes with nRT time. The new gradient would be 14 that of the original temperature graph. P X Line represents the new gas: Q 1/p R S Y 0 0 0.4 T position of original graph 0.3 time The length of the line Q is greater than the length of the line S, implies it needs to lose more heat in liquefaction than freezing. 0.2 0.1 V 0 The part of the graph shows that the specific latent 0 heat of vaporization of the substance is greater than its (D) (ans) specific latent heat of fusion: The gradient of the graph at P is greater than the gradient at R. 1 2 3 4 themis 19. [Thermal Physics] The gradient of the graph at T is greater than the gradient at R. Solution The value of X is greater than the value of Y. An electric kettle contains: 500 g of water at 15°C Power 2.2 kW The length of the line Q is greater than the length of the line S. Specific heat capacity of water 4.2 103 J kg-1 K-1 (C) (ans) themis The temperature of the water is raised to 100°C. 18. [Thermal Physics] Recall, Solution A fixed amount of an ideal gas: Pressure p Volume V The graph shows the variation of 1 p with V at a constant temperature. Pt mc — in its usual notations Time taken, t 1/p 0.4 mc P 0.500 4200 100 15 81 s (2sf) 2200 0.3 Minimum time taken to raise the temperature of the water to 100°C: 0.2 22 s 8.1104 s 81 s (B) (ans) 0.1 V 0 0 95 s 1 2 3 themis 4 The amount of gas and the thermodynamic temperature are both doubled. 2008 Nov 2008 8 20. [Wave Motion] c E hf h — in its usual notations Recall, Solution If measured in the other units, then c E ' 1.6 1019 h ' 109 A point source produces waves is made to pass through an area that is 2.0 cm wide, as shown. 16 cm point source of waves hc E' 109 1.6 10 19 hc K 109 1.6 1019 Within this area, the intensity of the waves is I and their amplitude is A. The waves reach a second area of width 16 cm. 6.63 10 3.00 10 10 1.6 10 8 9 19 1240 (3sf) Numerical value of K: I : Area 64 I 64 64 3.1810-53 3.1810-35 A : I 64 A 8 8 1.2410-15 1.24103 (D) (ans) themis Intensity at 2nd area Amplitude at 2nd area I 8 A 4 I 64 A 8 I 256 A 16 I 64 A 4 1 ' 34 Recall, I 2.0 cm 1 — and I A2 — Area — in its usual notations 22. [Superposition] Solution Diagrams show air particles moving in a column. A stationary wave is formed. L node antinode node antinode (B) (ans) themis 21. [Quantum Physics] Solution Energy E of a photon and its wavelength is E K , where K is a constant. The first diagram shows the displacement of some particles at one instant and the second diagram shows the displacement of some particles half a cycle later. Deduce from diagrams that 3 4 L . Maximum varying pressures occur at points with maximum particle movements (anti-nodes). If E is measured in electronvolts E ' and in nanometres ' , then: Advanced Physics solutions - yearly themis 2008 9 Length L in terms of wavelength A 3 4 Position within the column that pressure change by the largest amount (D) (ans) themis 25. [Electric Fields] Solution node In an electric field, the electric potentials at points P1 and P2 in free space are V1 and V2 respectively. 3 2 node 3 2 antinode A point electric charge q is brought from P1 to P2 by an external agent. antinode Work done on the charge q V2 V1 3 4 q V1 V2 (B) (ans) themis Work done on the charge: 23. [Superposition] Solution A beam of incident monochromatic light is shined normally to grating: For maximum images, set sin 1 : 1 3.0 105 600 109 q V2 V1 q V2 V1 q V1 V2 26. [Current of Electricity] Solution Recall, d sin n — themis Diffraction grating 3.0 105 lines per metre d q V2 V1 (D) (ans) Wavelength 600 nm : nmax A wire has: Resistivity 1.3 108 m Diameter 0.50 mm Length 30 times round an insulating rod Insulating rod’s diameter 1.5 cm 5.5 5 images, maximum Total number of images produced by light transmitted through this grating: 8 9 11 wire wound round rod 30 times 5 insulating rod diameter 1.5 cm (A) (ans) themis Recall, resistance, R — A — in its usual notations 24. [Electric Fields] Solution Electric field strength is defined: : R 1.3 108 “force per unit positive charge on a small test charge” The key definition of electric field strength came from Coulomb. Hence, it is necessary for the test charge to be small, so that the test charge does not distort the electric field so that the force on the test charge is small so that the test charge does not create any forces on nearby charges 30 2 1.5 10 2 2 0.5 103 2 2 9.4 102 Resistance of the wire: 1.1 101 4.7 105 7.0 104 9.4 102 (B) (ans) themis so that Coulomb's law for point charges is obeyed 2008 Nov 2008 10 Thermistor is connected to the potential divider circuit as shown. 27. [Current of Electricity] Solution +9.0 V 1.5 k A A battery, connected to an external resistor, has: e.m.f. 6.0 V E Internal resistance 0.40 r Rv External resistor 2.90 R Recall, current i B output 1.5 k E 6.0 R r 2.90 0.40 0V 1.82 A At 0730, an output of 6.O V is required. Power dissipated P i 2R 1.82 2.90 If the output is to be set at 6.0 V, then across the other section (A), the voltage drop needed to be 3.0 V. 2 9.6 W At 0730, the resistance of the thermistor 1.5 k Power supplied to the external resistor: 1.3 W 5.3 W 12.4 W 9.6 W The current flowing through section (A): i (C) (ans) themis 3.0 V 0.001 A 1.5 k 1.5 k Similarly, the same current flows through the section (B): 6.0 V RV 1.5k0.001 A 28. [D. C. Circuits] Solution The resistance R of a thermistor varies during part of a day is shown. R/ k 4.0 Rv 4.5k Value of the variable resistor be set: 1.5 k 3.0 k 6.0 k 4.5 k (C) (ans) themis 29. [D. C. Circuits] 3.0 Solution 2.0 The circuit consists: 1.0 0.0 0500 0600 0700 0800 time 0900 of day A battery of e.m.f. 6.0 V, of negligible internal resistance; Three resistors, each of resistance R, and a variable resistor T, as shown. T X R 6.0 V +6.0 V +0.0 V V R R Y The resistance of T changes from R to 5R. Advanced Physics solutions - yearly themis 2008 11 When variable resistor, T is set to R: Potential X +3.0 V Potential Y +3.0 V Potential difference, V 0.0 V must exactly opposed in order for the beams to emerge undeflected. The arrangement that would allow the beam to pass through undeflected: When variable resistor, T is set to 5R: Potential X +1.0 V Potential Y +3.0 V Potential difference, V 2.0 V Change in the reading of the high-resistance voltmeter 2.0 V 0.0 V 2.0 V electrons protons 4V 5V magnetic field Change in the reading of the high-resistance voltmeter: zero 2V (B) (ans) themis electrons 30. [Electric Fields] protons Solution (B) (ans) themis A charged particle, in vacuum, travels in a straight line, enters a uniform field. The particle then travels in a curved path that is not the arc of a circle. 32. [Electromagnetic Induction] Solution Only electric field influences a charged particle in a non-circular way. A coil has area A and n turns. For the charged particle to go on a subsequent nonparallel path, the field direction has to be non-parallel. A uniform magnetic field of flux density B acts at an angle to the plane of the coil, as shown. Type of field Initial direction of the particle compared to the field Electric parallel magnetic parallel magnetic perpendicular electric perpendicular pivot coil area A (B) (ans) themis 31. [Electromagnetism] Solution magnetic flux density B Different particle beams are made to enter a region between two metal plates in which there are uniform electric and magnetic fields. By applying Fleming’s left hand rule and electric field direction, the forces acting on the particle beams 2008 Nov 2008 12 Initial flux linkage i BAn sin 34. [Alternating Currents] The coil is rotated by an angle of . Solution Final flux linkage f 0 The graph of an alternating current with a square waveform is shown. Change in flux linkage BAn sin current Change in magnetic flux linkage when the coil rotates so that the angle is reduced to zero: BAn cos 2BAn cos 2BAn sin square wave-form Io BAn sin time (B) (ans) - Io themis Its peak value for the current is lo . 33. [Electromagnetic Induction] Consider a complete period, T. Solution current The magnetic flux linking a coil, with respect to time is as shown. Io2 Io flux 0 time - Io t 0 2t time Squaring the wave to give the Io2 waveform. The induced e.m.f. of a coil, varies with the rate of change of flux linkage, i.e., the gradient of the fluxtime graph. Further increasing or decreasing flux linkage induces differently signed e.m.f.. 0 2t time 0 0 t 0 2t time Io 2 Io 2 Io (D) (ans) themis 35. [Quantum Physics] e.m.f. e.m.f. Io2 Io2 Io Root-mean-square value for the current: t Io2 T Io2 . T Irms e.m.f. 0 Io2 Take the square root of this average value, The corresponding e.m.f. induced in the coil, with respect to time: e.m.f. Take its average over its period, Solution 0 0 t 2t time 0 0 t 2t time A beam of electrons is made to pass through a thin carbon film, velocity, v, as shown. (C) (ans) themis alternating regions of bright and dark zones electron beam graphite plate Advanced Physics solutions - yearly fluorescent screen themis 2008 13 The electrons produce a pattern of concentric circles on the fluorescent screen. Graph shows the probability of finding an electron at each position: According to de Broglie’s principle of wave-particle duality, the electrons exhibited wave-like property, i.e., diffraction. From mv h probability — in its usual notations When mv 0 From d sin n — in its usual notations When distance probability More fringes and diameters of circles decrease. 0 Cause of the pattern Changes to pattern when the velocity is increased diffraction diameters of circles increase refraction diameters of circles increase refraction diameters of circles decrease diffraction distance probability 0 distance probability diameters of circles decrease 0 (B) (ans) themis distance 36. [Quantum Physics] (B) (ans) themis Solution Sketch shows how the wave function of an electron varies with position. 37. [Lasers and Semiconductors] Solution The symbols and represent the majority carriers in the p-type and n-type sides of a p-n junction. 0 distance As interpreted in standard text. We usually interpret the absolute square of the wavefunction r ,t as the probability density for the particle to be found at each point in space. In other words, r ,t d 3r is the probability, at time t, of 2 finding the particle in the infinitesimal region of volume d 3r surrounding the position, r . 2008 Nov 2008 14 Pair of diagrams illustrates how a p-n junction acts as a rectifier: voltage source p n conventional current (D) (ans) themis 39. [Nuclear Physics] p n no current Solution A detector has: voltage source p n conventional current p n no current A radioactive source has: voltage source p n conventional current p n Ionising radiation Background count rate 24 counts per minute Reading 532 counts per minute True initial reading 532 24 508 no current After 2 half-lives, the final true reading voltage source p n conventional current p n no current 508 127 4 We now have to add back the background count, the final meter reading 127 24 151 Reading after two half-lives of the source: (D) (ans) themis 127 133 157 151 (C) (ans) 38. [Lasers and Semiconductors] themis Solution 40. [Nuclear Physics] Solution In a helium-neon laser: Helium atoms collide with neon atoms and excite them It produces a population inversion which allows stimulated emission. A nucleus of bohrium yx Bh decays to mendelevium 255 101Md by a sequence of three -particle emissions. bohrium yx Bh dubnium The spontaneous infra-red emission is low energy. lawrencium The stimulated emission of red light is high energy. The total excitation energy should equal its total emissions. Rewriting, Neon energy level diagram that correctly shows the excitation of the neon atoms by the helium atoms, the spontaneous infra-red emission from the neon, and the stimulated emission of red light: spontaneous emission stimulated emission excitation stimulated emission spontan. emission excitation mendelevium x y Bh 255 101Md 255 101Md 3 — : x 255 3 4 267 y 101 32 107 no. of neutrons 267 107 160 No. of neutrons in a nucleus of yx Bh : 267 261 154 160 (C) (ans) themis spontaneous emission excitation spontaneous emission stimulated emission Advanced Physics solutions - yearly excitation stimulated emission themis 2008 15 [2008N P1 MCQ Key] Q. Key Q. Key Q. Key Q. Key 1. A 11. B 21. D 31. B 2. D 12. C 22. B 32. B 3. D 13. D 23. A 33. C 4. C 14. B 24. D 34. D 5. D 15. A 25. D 35. B 6. C 16. D 26. B 36. B 7. C 17. C 27. C 37. D 8. B 18. D 28. C 38. D 9. D 19. B 29. B 39. C 10. D 20. B 30. B 40. C themis 2008 Nov 2008 16 (b) Upon impact, the speed of the stone is reduced from 34 ms-1 to 2.0 ms-1 in a time of 0.95 s. 2008 Nov Paper 2 Using momentum considerations, Questions Impulse, Faverage t mv u — in its notations Answer all questions. 1. [Kinematics] Average force acted on the stone during this time, Faverage Solution 4.38N (ans) [2] At the top of a cliff of height 32m, a stone of mass 130 g is thrown horizontally as shown. (c) If the stone causes a big splash on hitting the sea, using energy considerations, the high kinetic energy from the stone is not only transferred to its immediate vicinity but over a wide area. Hence, the whole water surface and column is made to resist the impact. Together with a louder sound, the stone will be slowed down in a shorter time (shorter distances) than when its impact is restricted to a small water column, where no splash is produced. (ans) [2] path of stone 32 m 0.130 2.0 34 0.95 s A sea themis Air resistance is negligible. 2. The stone hits the sea at speed of 34 ms-1 . Solution (a) At point of impact, (i) Let the velocity of impact be v. [Oscillations] vy v A vx The diagram showed a flat horizontal plate vibrating in a vertical plane. Consider its vertical descent, plate v y uy 2gs — in its usual notations 2 2 vy 2 02 2 9.81 32 vy 25.1 ms-1 (the vertical component of the vertical oscillations The plate’s the variation with displacement x of the acceleration a is as shown. velocity of the stone) (ans) [2] a / m s-2 20 (ii) Deduce from the final velocity, v 2 v y 2 v x 2 342 25.12 vx 2 10 vx 23.0 ms-1 The angle to the horizontal of the stone’s path vy 25.1 tan1 tan1 v 23.0 x -8 -6 -4 -2 0 0 2 4 6 8 x / mm -10 -20 47.5 (3sf) (ans) [2] Advanced Physics solutions - yearly themis 2008 17 (a) From the straight-line graph, the negative gradient gives a relationship i.e., a x (oppositely directed). It is also given that it is a periodic oscillation, i.e., the graph does not extend itself indefinitely. balanced by the satellite’s kinetic urge to leave the orbit. By definition, the oscillations of the plate are simple harmonic (SHM) as it is a periodic motion in which acceleration (a) is proportional to, but oppositely directed to, the displacement (x) from its equilibrium position. (ans) [3] GMm v2 — in its usual notations m R R2 GMm kinetic energy of the satellite 2R (ans) [2] 1 mv 2 2 (ii) Gravitational potential energy GMm — in its R usual notations (b) Some sand is introduced onto the plate. Amplitude of vibration of the plate is gradually increased from zero. gravitational potential energy of satellite kinetic energy of satellite At one particular amplitude, the sand is seen to lose contact with the plate. GMm R GMm 2R = 2 (QED) [1] acceleration (i) When the sand particle is resting reaction, R on the vibrating plate, the sand grain, external forces acting on the mass m particle are as shown in the free body diagram (upward positive). ma R W — weight, W When the sand particle loses contact, the reaction force, R become zero. : ma 0 W ma mg a g 9.81 m s-2 (ans) 10.0 energy 9 / 10 J 8.0 6.0 4.0 At a g (upward ), the plate is travelling downward from its top displacement position. At this point, the acceleration due to gravity g and the acceleration due to the vibrating plate is the same, hence, if the plate were to accelerate faster than g (downwards), the sand particle will not be able to “catch-up” and stay in contact, thereby losing contact with the plate. (ans) [3] (ii) From graph, when a g , the amplitude of vibration of the plate at which the sand first loses contact is at x 3.8 mm. (ans) [1] themis 3. (b) The variation with orbital radius R of the gravitational potential energy of the satellite is shown. [Gravitational Field] Solution A planet, with a mass M and radius Rp , has an orbiting satellite of mass m and the orbit is R . 2.0 0 Rp 2Rp 3Rp 4Rp R -2.0 -4.0 -6.0 -8.0 -10.0 (i) The variation with orbital radius of the kinetic energy of the satellite (The line should extend from R 1.5Rp to R 4Rp ): Relating the ratio in (a)(ii) with the actual given K.E. values in the graph. We get, K.E. of satellite 12 G.P.E. of satellite (a) In terms of M, m and R: (i) For the satellite to be in a stable orbit, it implies that the attractive gravitational force that provides the centripetal acceleration is perfectly 2008 Nov 2008 18 battery e.m.f. E 10.0 energy / 109 J 8.0 A ammeter 6.0 kinetic energy of satellite 4.0 current i resistor R V 2.0 0 internal resistance r switch voltmeter p.d. V Rp 2Rp 3Rp 4Rp R The resistance of resistor R can be varied. Take the potential difference across R to be denoted as V and the power dissipated in R as P . -2.0 -4.0 The graph thus show the variation with V of P . -6.0 5.8 P/W -8.0 5.6 -10.0 (ans) [2] 5.4 (ii) The radius of the orbit of the satellite is changed from R 4Rp to R 2Rp . 5.2 Given the mass m of the satellite is 1600kg. 5.0 3.0 From K.E. 12 mv 2 , v 4.0 4.5 (i) 5.5 V/V P is maximum at 5.62 W. Its corresponding potential difference across the resistor is 4.50 V. 2 1.25 109 1250 m s-1 : v 1600 The current in the circuit current in the resistor When R 2Rp , K.E. 5.00 109 J i : v 5.0 (a) For the maximum value of P, 2 K.E. — m When R 4Rp , K.E. 1.25 109 J 2 5.00 109 2500 m s-1 1600 5.62W P 1.25 A (3sf) (ans) [2] 4.50V V (ii) Correspondingly, Resistance in R Change in orbital speed of the satellite 2500 1250 1250 m s-1 (ans) [5] [Current of Electricity] Solution 4.50 V V 1.25 A i 3.6 (QED) [1] themis 4. 3.5 (b) Set the resistor R to 2.03 , the current in the circuit is 1.60 A. When R 2.03 and i 1.60 A , An electrical circuit (as shown) has: A variable resistor R connected A battery of e.m.f. E with internal resistance r . Advanced Physics solutions - yearly Using the data and answers to part (a), Let E be the e.m.f. of the battery, E i R r themis 2008 19 From (a), E 1.253.6 r — From (b), E 1.60 2.03 r 6. — Solution Substituting E of into , Strontium-90 is a radioactive nuclide. 1.253.6 r 1.60 2.03 r 3.6 r 1.28 2.03 r 3.6 r 2.5984 1.28r 1.0016 0.28r r 3.58 Internal resistance r of the battery 3.58 (this is a proof of maximum power theorem) (ans) [3] themis 5. [Nuclear Physics] [Lasers and Semiconductors] (a) A substance is said to be radioactive if it comprised unstable nuclei that will disintegrate into more stable configurations by the emission of alpha-particles (helium nuclei), beta-particles (electrons or positrons) and/or gamma radiation (electromagnetic waves of short-wavelengths). (ans) [2] (b) A sample of Strontium-90 has a mass of 2.40 10 8 g. The average activity of this sample during a period of 1 hour is found to be 1.26 105 Bq. Solution The electrical resistance of an intrinsic semiconductor material is found to decrease as its temperature rises. The explanation using the band theory of conduction, with a diagram, is shown below. (i) The decay constant () of a radioactive nuclide is defined as the constant of proportionality relating its activity to the number of undecayed nuclei. (ans) [2] A — in its usual notations — N : The decay constant of Strontium-90, energy of electrons, E (ii) From (i), conduction band small energy gap valence band In semiconductors, the conduction and valence bands are spaced closely enough together (1 eV) that, at room temperature, a nontrivial number of electrons is found in the conduction band. These materials have significant conductivity that is highly temperaturesensitive. 2.40 10 103 kg 90 1.66 1027 kg 7.84 1010 s-1 (3sf) (ans) [3] energy bands in semiconductors 1.26 105 Bq 8 (c) For nuclides that have relatively small decay constants, it will have a relatively large half-life. It is thus suitable to accurately measuring its changes in mass and activity to determine its decay constant; whereas if the half-life is short, its mass and activity measurements would not be too accurate. (ans) [1] themis At low temperatures, semiconductors behave like insulators. The valence electrons have no adjacent energy levels to transit into, and do not have sufficient energy to the energy gap. 7. At high temperatures, semiconductors behave like conductors. The valence electrons have sufficient energy to cross over the small energy gap to assist in the conduction. (ans) [4] A serious hazard for fire-fighters is the explosion of containers of `liquefied gas' (butane) that have been heated in a fire. When the butane suddenly burns in an explosion, the fire spreads very rapidly in the form of a spherical fireball of increasing radius that is at very high temperature. themis [Thermal Physics] [Data Response] Solution In order to study such fireballs, a series of experiments is carried out. Some butane of volume 12.5 103 m3 is put in a sealed container and is then heated until it 2008 Nov 2008 20 20 1.4 lg (R / m) explodes. The variation with time t of the radius R of the fireball is determined. The results are shown in the diagram. 1.3 1.1 16 1.2 R/m 1.0 12 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 8 1.7 lg (t / ms) (i) On the diagram, (1) At t 40 ms, R 14.6 m 4 lg R 1.16 and lg t 1.60 0 0 10 20 30 40 50 The point corresponding to time t 40ms is plotted at (1.60,1.16). (ans) [1] t / ms (a) Use the diagram, (2) The best-fit line for all the plotted points is drawn. (ans) [1] (i) Without any calculation, From t 0 2 ms, initially the rate at which the radius of the fireball increases is very high. From t 2 25 ms, the rate reduces. From t 25 ms onwards, the rate is reduced further to an approximate constant (still positive). (ans) [2] (ii) In a room of length 12 m, width 5 m and height 3 m, such an explosion would be very hazardous. From the graph, it will take less than t 25 ms for the explosion to reach every part of the room. This is a very short reaction time. The space is relatively small and enclosed, the blast and sound will add sudden pressure onto the ears and may confuse the victim subsequently. The debris created by the exploding gas will behave like bullets from a shotgun thus harming the victim quickly. (ans) [3] (b) It is thought that, for a fixed volume of butane, the radius R of the fireball varies with time t according to the expression Rn kt m , where n and m are integers and k is a constant. Some corresponding values of lg t and lg R for the data in the diagram are plotted on the graph as shown following. Advanced Physics solutions - yearly 1.4 lg (R / m) 1.3 1.2 1.1 (b)(i)1. 1.0 (b)(i)2. 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 lg (t / ms) (ii) From the best-fit line drawn in (i) part (2), The gradient 1.24 0.935 0.386 (ans) 1.79 1.00 [2] (iii) Recall, Rn kt m n lg R m lgt lg k m 1 lg R lg t lg k n n m : 0.386 0.4 n Given that n and m are integers, making m as small as possible, implies n 5 and m 2 (ans) [3] (c) The experiment is repeated using similar containers but with different volumes of butane. The results are shown in the next diagram. themis R/m 2008 21 20 volume of container / m3 16 12.5 10-3 10.0 10-3 7.5 10-3 5.0 10-3 12 2.5 10-3 8 4 0 10 0 20 30 40 50 60 t / ms Without drawing a further graph, Proof: R5 cV , where c is a constant 5lg R lgV lg c lg R 15 lgV 15 lg c — From the graph at t 40ms, R / m V / 10-3 m3 R5 / 105 R5 V / 107 14.6 12.5 6.63 5.30 14.0 10.0 5.38 5.38 13.1 7.5 4.01 5.35 12.2 5.0 2.70 5.41 10.6 2.5 1.34 5.35 average 5.36 Within limits of 1.1% error, it is found that R5 is a V constant; hence, the relationship R5 cV is true. (QED) [3] (d) (i) The equation in (c) may also be applied to other exploding gases. One suggested physical quantity on which the constant c will depend is its surrounding pressure. (ans) [1] (ii) The data were collected for butane in a container in a room. One other situation where the theory developed predicts a high level of hazard for fire-fighters is fire-fighting onboard ships where flammable gases build up in enclosed small cavities onboard the ship. (ans) [1] themis 2008 Nov 2008 22 (ii) Within limits of errors, inverse proportion is demonstrated by the plotted graph, since the plot of 1 p against V is approximated to a straight line passing through the origin, p is shown to be inversely proportional to V . (QED) [1] 2008 Nov Paper 3 Questions Section A (b) F and d are in the same direction. Answer all questions in this section. 1. The variation with displacement d of the force F applied to an object is as shown below (C). [Work, Energy and Power] 30 F /N Solution (a) At constant temperature, for a given fixed mass of gas, the variation of volume V of the pressure p is as shown (A). 20 (C) 10 p / 105 Pa 3 0 2 0 1.0 2.0 3.0 (A) 0 100 200 300 400 V / cm3 (i) Using values from (A), using the axes drawn, plot in diagram below (B) to show that p is inversely proportional to V . work / J Draw a graph showing the variation with d of the work done in (D). 1 0 4.0 d/m (D) 0 (B) 1.0 2.0 3.0 4.0 d/m Work done is area under the force displacement graph. 0 work / J 60 0 1/p / 10-5 Pa-1 3.0 40 (D) (B) 0 1.0 0.0 20 2.0 0 0 1.0 2.0 3.0 4.0 d/m (ans) [4] 100 200 300 400 V / cm3 (ans) [2] Advanced Physics solutions - yearly themis 2008 23 (c) Measurements are made of the Earth's gravitational field strength g for different distances r from the centre of the Earth. Diagram (E) shows the variation of lg r against lg g . lg (g / ms-2) 1.0 Hence, when the coil rotates, its flux linkages changes with the magnetic field resulting in an e.m.f. being generated between the ends of the coil. (ans) [2] 0.5 0.0 6.7 6.8 6.9 7.0 7.2 7.1 7.3 -0.5 7.4 lg (r / m) (E) The gradient of the graph (E) 0.73 0.85 7.5 6.7 1.98 (3sf) (ans) [3] (ii) Within limits of errors, since the plot of lg g against lg r is approximated to a straight line, lg g is shown to have a linear relationship with lg r From (i), the gradient 2 or lg g relates to 2lg r g relates to (ii) Two factors that affect the magnitude of the maximum e.m.f.: -1.0 (i) (i) Faraday's law of electromagnetic induction states that the magnitude of the induced e.m.f. in a conductor is directly proportional to the rate at which magnetic flux is cut by the conductor or flux linkage is changed in a coil. the number of turns in the coil the field magnetic field (the coil is rotated at a relatively faster speed) (ans) [2] (iii) When the coil is at its horizontal position, its direction of motion is perpendicular to the magnetic field. Its induced e.m.f. is therefore at its maximum. When the rotation of the coil reaches its vertical position, its direction of motion is parallel to the magnetic field. No e.m.f. would be induced. e.m.f / V 1 r2 +Eo The value of gradient relates g to vary inversely with the square of the distance 1 2 from the centre of the earth. (ans) [2] r A A B B B 0 T T/2 one revolution [Electromagnetism] Solution B A time/ t -Eo themis 2. A A B Voltage output-time graph Since the motion of the coil is circular, its induced e.m.f. follows a sinusoidal curve. (ans) [2] (a) A simple generator has: A coil with a large number of turns Rotates at a constant rate in a uniform magnetic field, as shown. Q C permanent magnet B D S (b) The output from a similar generator is connected to the input of an ideal transformer: primary coil 30 turns secondary coil 600 turns input e.m.f. 72 V r.m.s.. output is connected to a resistor of resistance 160 , as shown in diagram. transformer input A 30 turns 600 turns 160 Ω N P (i) The peak input e.m.f. Vrms 2 72 2 102 V (3sf) (ans) [1] 2008 Nov 2008 24 (ii) The r.m.s. value of the p.d. across the resistor Noutput Ninput Vinput (d) 600 72 30 light 1440 V (3sf) (ans) [1] unpolarised wave (iii) The r.m.s. value of the current in the resistor themis 4. P Irms R 9.00 160 Solution 13.0 kW (3sf) (ans) [1] (a) Briefly define: (v) The r.m.s. value of the current from the generator Vrms, input (i) Every minute particle (such as ions, electrons, atoms or molecules) in a body has random potential energy Ep , due to their state and 12960 72 position, and, random kinetic energy Ek , due to 180 A (3sf) (ans) [2] themis 3. [2008N P3] [Thermal Physics] 2 2 P planepolarized Polarization, also called wave polarization, is an expression of the orientation of the lines of electric flux in an electromagnetic field. (ans) [1] Vrms 1440 9.00 A (3sf) (ans) [1] R 160 (iv) The mean power dissipated in the resistor, Polaroid lens [Superposition] their motion. Collectively, the sum of these randomly distributed energies is termed the internal energy U of the body, i.e., U Ep Ek . (ans) [2] Solution Explanations of the meaning of each of the following terms as applied to waves. (ii) The first law of thermodynamics is defined as the heat q absorbed by a system either raises the internal energy U of the system and/or does work by the system on the environment wby , (a) i.e., qin U wby . — (ans) [1] A standing wave, also known as a stationary wave, as opposed to progressive wave, is a wave that remains in a constant position, i.e., their waveforms do not move. (ans) [2] (b) Diffraction is the bending and spreading of waves when they meet an obstruction or gap into regions where a shadow might be expected. (ans) [2] (c) Sources are said to be coherent, if the waves leaving them bear the same phase relationship to each other at all times. The same phase relationship implies the same frequency, and therefore the same wavelength. (ans) [2] (b) An ideal gas undergoes a cycle of changes A B C A , as shown in the diagrams below. pA cylinder area A x pressure p / 105 Pa B 7 1 A C 5 Advanced Physics solutions - yearly frictionless piston gas 20 volume V / cm3 themis 2008 25 (i) The work done by a gas during the change C A expanding against an external pressure p is given by area under p V graph or pave V wby 1 105 Pa 20 - 5 106 m3 work done on gas / J A B zero 4.2 BC 8.5 (a) (i) The electric field strength, E , or electric field intensity or electric field is defined by the ratio of the electric force on a charge at a point in free space to the magnitude of the charge placed there, i.e., increase in internal energy / J electric field strength, E CA section of cycle [Electric Fields] Solution (ii) The figure below is a table of energy changes during one cycle. heating supplied to gas / J Answer two questions in this section. 5. 1.5 J (2sf) (ans) [2] section of cycle Section B electric force charge (ans) [1] heating supplied to gas / J work done on gas / J increase in internal energy / J A B zero 4.2 4.2 BC 8.5 zero 8.5 CA 5.8 1.5 4.3 Rearranging , (ii) The diagram below shows a uniform electric field of electric field strength E where a charge q is placed at point. d q uniform field E U qin won — Process A B : Y X : U 0 4.2 U 4.2 (ans) The charge at X is moved to Y through a distance d. Process B C : Using the definition in (i), Since the charge is positive, positive work done has to be won p 0.0 0.0 (ans) W.D. FE d qE d (ans) [1] : U 8.5 0 U 8.5 (ans) (iii) The potential difference between X and Y is V. Process C A : Using the answer from (ii), Ucycle UAB UBC UCA 0 The electric potential (V) at a point in free space in an electric field is defined as the work done in bringing a unit positive charge from infinity to the point, i.e., 0 4.2 8.5 UCA UCA 4.3 (ans) : 4.3 qin 1.5 qin 5.8 (ans) [4] themis electric potential, V : VXY work done — charge work doneXY qEd V charge q Ed (ans) [2] (b) An X-ray tube has: a vacuum anode and cathode 2008 Nov 2008 26 electrons are accelerated from rest through a potential difference of 60 kV between the cathode and the anode. The current in the tube is 8.6 mA. 5.38 1016 (n) (3sf) (ans) [2] (ii) Work done on the electron qV This work done is used to accelerate the electron to a high speed – converted to pure kinetic energy, qV 12 mv 2 — in its usual notations the speed of electrons arriving at the anode, v 2 1.6 10 60 10 2qV m 9.11 1031 3 (i) By referring to the above diagram, If the conducting sphere is replaced by a point charge of the same value, beyond the boundaries of the original conducting sphere, it will produce exactly the same field pattern and strength as that of the conducting sphere at points A, B and C. Hence, to points A, B and C, it appears as if the charge is concentrated at the centre of the sphere. (ans) [1] (ii) Electric field strength at the surface of the sphere, E 1.45 108 m s-1 (3sf) (ans) [4] Power supplied by the electrons hitting the anode nqV 5.38 1016 1.6 1019 60 103 516 W (3sf) (ans) [2] (c) X-ray production: uses a negligible fraction of the power reaching the anode in (b) has to be cooled by passing a coolant through the anode. The coolant has specific heat capacity of 3500 J kg-1 K-1 and the temperature rise is 30 K. The heat that needed to be removed (iii) Rate at which the coolant must be pumped work done through the anode c 516 4.91 103 kg s-1 (ans) [3] 3500 30 1 Q — in its usual notations 4 o r 2 (iii) Work done by the electrons per second field of isolated point charge A B 8.6 10-3 A current chargeelectron 1.6 10 19 C 19 0.060 C C (i) The number of electrons passing through the tube in one second The electric field around the sphere is as shown in the diagram below. 1 0.060 106 4 o 0.102 5.39 104 N C-1 (ans) [2] (iii) Point A and B are 0.40 m and 0.50 m from the centre of the sphere respectively. Potential difference (V) for AB V for BC. VAB Q 1 1 4 o rB rA Similarly, VBC — in its usual notations Q 1 1 . 4 o rC rB Since VAB VBC , rearranging 1 1 1 1 rB rA rC rB 1 1 1 1 0.50 0.40 r 0.50 C Distance from the centre of the sphere to C, rC 0.67 m (ans) themis (d) A conducting sphere has: radius 0.10 m charge 0.060 C Advanced Physics solutions - yearly themis 2008 27 6. [Oscillations] At its equilibrium position, kx mg Solution (a) Frequency f is defined as the number of complete oscillations produced per unit time, is distinct from angular frequency which is defined as the magnitude of the vector quantity angular velocity, that is defined as the rate of change of angular displacement with respect to time. (ans) [2] (b) A spring has: mg 0.400 9.81 19.6 N m-1 e 0.200 At its lowest point of its movement, the resultant force on the load T ' W k x e mg 19.6 0.400 0.400 9.81 3.92 N (ans) [2] (ii) Angular frequency of the oscillation, An unstretched length 0.650 m Attached to a fixed point A mass 0.400 kg is attached to the spring and gently lowered until equilibrium is reached. The spring has then stretched elastically by a distance of 0.200 m. For the stretching of the spring, (i) k 7.00 rad s-1 (ans) [2] (iii) Maximum speed of the mass, v x02 x2 7.00 0.2002 1.40 m s-1 (ans) [1] The loss in gravitational potential energy of the mass mgh 0.400 9.81 0.200 (e) Complete the table below which is a list of energies of the simple harmonic motion. 0.785 J (ans) [1] gravitational potential energy / J elastic potential energy / J kinetic energy / J total energy / J gravitational potential energy / J elastic potential energy / J kinetic energy / J total energy / J lowest point 0 1.57 0 1.57 equilibrium position 0.785 0.392 0.392 1.57 highest point 1.57 0 0 1.57 (ii) The elastic potential energy gained by the spring 12 Fx 12 0.400 9.81 0.200 0.392 J (ans) [2] k 19.6 m 0.400 lowest point 0 equilibrium position highest point (c) The two answers to (b) are different because the springmass system is a rather inefficient mechanical energy storage system. It only manages to store half of the energy supplied, while the rest are lost through heat, sound and other forms of energy. (ans) [2] (d) The load on the spring is now set into simple harmonic motion of amplitude 0.200 m. (i) simple mass-spring system spring By observation, (ans) [5] T e equilibrium position T’ e+x W W 2008 Nov 2008 28 (f) Based on the axes below, sketch four graphs to show the shape of the variation with position of the four energies. Label each graph. Most of the incident alpha particles pass straight through the foil, to hit the fluorescent screen. There are few particles, however, which suffer deviations in the forward and backward directions. A very few particles even retrace their original path, backwards. energy Conclusions: lowest point equilibrium position The observation can be explained by proposing that the atom is made up of a very small, positively charged nucleus surrounded by a cloud of electrons. The atom is mainly empty space, so that most alpha particles pass through the foil with practically no deviation. If the alpha particles, however, come too close to the nucleus, the strong Coulomb repulsion between the nucleus and the positively charged alpha particle will cause the alpha particle to deviate from its original direction. The repulsive force indicates that the nucleus must be of the same nature as the alphaparticle. The retracing of a few alpha-particles resolved that the atomic nucleus must be real, massive, hard and physical. (ans) [4] highest point By conservation of energies: energy TE EPE GPE KE –xo 0 xo x (ans) [3] themis 7. [Nuclear Physics] Solution (a) In the past, experiments were performed to provide evidence for a small charged nucleus in the atom. Rutherford and his students conducted an experiment on alpha particles scattering. Alphaparticles from a radium source are made to impinge on a very thin sheet of gold foil, approximately 10-6 m thick. (b) Uranium-235 nucleus: at rest, and absorbs a slow neutron and undergoes fission. Sometimes the fission don’t emit any neutrons. One such fission: 235 1 92U 0 n 235.0439 u zirconium 235 92U x 52Te 98 y Zr neutron 1 0n 1.0087 u. uranium tellurium fluorescent screen (i) Nucleons: gold atom x 98 52Te y Zr The masses of these particles are Observations: -particles 137.9603 u 97.9197 u 235 1 x 98 x 138 (ans) 92 0 52 y Protons: y 40 (ans) [2] radium source gold foil Advanced Physics solutions - yearly alpha nuclei themis 2008 29 Speed of the zirconium nucleus, (ii) Assume all particles are initially at rest. Energy released in the fission mass excess equivalent (235.0439 u 1.0087 u 137.9603 u 97.9197 u) c 2 (0.1777 u) c 2 0.1777 1.66 1027 3.00 108 2 2.65 1011 J (ans) [4] vZr 1.29 107 m s-1 (ans) [3] (vi) Two assumptions made in the calculation in (iv) part 1.: 1. The fission takes place in isolation. 2. Newtonian laws of motion are obeyed at atomic level. (ans) [2] themis (iii) Part of the energy released, 2.3 1011 J were used to become kinetic energy of the tellurium and zirconium nuclei. The remaining energy released may be in a form of photon(s). (ans) [1] (iv) 1. Initial momentum is zero, hence final momentum must also be zero mZrvZr mTevTe (oppositely directed) the ratio speed of zirconium nucleus speed of tellurium nucleus m 137.9603 u vZr Te v Te mZr 97.9197 u 1.41 (3sf) (ans) [2] 2. The ratio kinetic energy of zirconium nucleus kinetic energy of tellurium nucleus 1m v 2 m v K.E.Zr 2 Zr Zr Zr Zr 1 2 m v K.E.Te mTe v Te 2 Te Te 1 m v Te Zr mZr v Te 2 2 v Zr 1.41 (ans) — [2] v Te (v) K.E.Zr K.E.Te 2.3 1011 — Combining and : K.E.Zr K.E.Zr 2.3 1011 1.41 K.E.Zr 1.346 1011 1m v 2 2 Zr Zr 1 2 1.346 1011 97.9197 1.66 10 v 27 2 Zr 1.346 1011 2008 Nov 2008 30 Notes: Advanced Physics solutions - yearly themis