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There are fifty questions in this paper. Answer all questions. Marks will not be deducted for wrong answers. Time: 1h 45 min 1. The tensile strength of a wire is the maximum 4. What is the energy change that occurs when stress on the wire just before it breaks. What a ball rolls down an inclined plane without is the dimension of tensile strength? slipping? A ML–1T–1 C MLT–1 A Potential energy → translational kinetic B ML–1T–2 D MLT–2 energy B Potential energy → rotational kinetic energy 2. The figure shows two blocks X and Y, of C Potential energy → translational kinetic masses m and 2m respectively, placed on a energy + rotational kinetic energy smooth horizontal surface. D Potential energy → translational kinetic energy + work against friction m 2m F 5. A car of mass 1250 kg accelerates from 0 to X Y 100 km h–1 in 4.0 s. The average power of the car is A 121 kW If the blocks are accelerated by a force F, the B 341 kW force exerted on block X by block Y is 1 2 C 484 kW A —F C —F D 681 kW 3 3 B —F 1 D F 2 6. For a particle moving in a circle with a constant speed, the physical quantity which 3. The figure shows the displacement–time is always constant is graphs of two vehicles P and Q. A displacement B acceleration Displacement C linear momentum P D angular momentum Q 7. A firework in the shape of a disc can spin freely about its axis. The radius of the disc is 2.0 cm and its moment of inertia is 0 T Time 1.0 × 10–6 kg m2. When the firework is ignited, gas is ejected from the rim in the tangential Which statement describes the motion of the direction at a constant speed of 50 m s–1 two vehicles at time T ? relative to the rim. The gas is ejected at a rate A P and Q are momentarily at rest. of 1.0 × 10–5 kg s–1. The angular velocity of B The speeds of P and Q are the same. the disc after 10 s of burning is C The speed of P is greater than the speed A 10 rad s–1 of Q. B 50 rad s–1 D The accelerations of P and Q are equal C 100 rad s–1 to zero. D 200 rad s–1 1 © Majlis Peperiksaan Malaysia 2007 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 1 3/7/2008 4:48:24 PM 8. Which figure shows that coplanar forces, A C each of magnitude F acting on a disc, are in equilibrium? A F C B D F F F F B F D F F F 11. The equation of motion of an object performing simple harmonic motion is given by π x = 5 sin —t, F F F 6 where x is the displacement in metres and t is 9. The figure shows two stars X and Y of masses the time in seconds. When t = 1 s, the speed m1 and m2 respectively and separated by a of the object is distance r. A 2.3 m s–1 C 4.3 m s–1 –1 B 2.6 m s D 4.6 m s–1 r m1 P m2 12. The figure shows a mass attached to a spring X Y made to oscillate vertically by an oscillator. r − 3 Oscillator If the gravitational field strength at point P r at distance — from X is zero, the ratio of 3 m1 to m2 is Mass A 1:4 B 1:2 If the damping is slightly increased, which C 2:1 graph shows the variation of amplitude a with D 4:1 frequency f of the oscillation? A C 10. The figure shows an object in space moving a a near a planet of mass M. Increased Increased damping damping v Object f f 0 0 r B D a a Planet M M Increased Increased damping damping f f The shortest distance between the object and 0 0 the planet is r. The speed v of the object at 13. What is the phase difference between two the nearest position is 1.5 GM , where G is the —– points separated by a distance of 3.0 cm in r a wave of wavelength 5.0 cm? gravitational constant. Which figure shows A 0.6 rad C 1.9 rad the correct trajectory of the object? B 1.7 rad D 3.8 rad © Majlis Peperiksaan Malaysia 2007 2 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 2 3/7/2008 4:48:31 PM 14. The graph shows the displacement of the F particles in a rope caused by a travelling wave at a particular time. Displacement 0 r r1 r2 P Q S 0 Position Which statement is not true of the graph? A Hooke’s law is true around r1. R B The equilibrium separation is r2. Which statement is not true of the motion of C The force is an attractive force when r > r1. the particles in the rope? D The force is an attractive force when r > r2. A The speed of the particle at P is 18. When a metal wire is stretched over a limit, maximum. it undergoes plastic deformation. Which B The acceleration of the particle at Q is statement is true of plastic deformation? zero. A Stress is proportional to strain. C The energy of the particle at R is entirely B The metal is not a crystalline solid. potential energy. C The atomic planes slide over each other. D The energy of the particle at S is entirely D The atoms are displaced a little from their kinetic energy. equilibrium positions. 15. The figure shows a sound source S moving 19. A cylinder of volume 0.09 m3 contains an away from observer P towards observer Q. ideal gas at 100 °C and 8 × 104 Pa. If the mass of one mole of the gas is 0.034 kg, the P S Q r.m.s. speed of the gas molecules is A 2.71 × 102 m s–1 C 7.33 × 104 m s–1 What is the effect of the motion on the B 5.23 × 102 m s–1 D 2.73 × 105 m s–1 wavelength of the sound received by P and by Q? 20. At what temperature is the r.m.s. speed of P Q oxygen molecules the same as that of helium molecules at 300 K? A Increases Decreases [Relative molecular mass of oxygen = 32, B Increases No change relative molecular mass of helium = 4] C Decreases Increases A 38 K C 1440 K B 849 K D 2400 K D No change Decreases 21. The work done to adiabatically compress one 16. A person can hear a sound of intensity mole of a diatomic gas is 50 J. What is the 1.0 × 10–11 W m–2. When a hearing aid is used, temperature rise of the gas? the sound level increases by 35 dB. What is A 0K C 2.41 K the new intensity of the sound heard? B 1.72 K D 4.10 K A 3.50 × 10–11 W m–2 B 3.50 × 10–10 W m–2 22. Which statement is true of an adiabatic C 3.16 × 10–9 W m–2 process? D 3.16 × 10–8 W m–2 A Boyle’s law is obeyed. B The temperature is always constant. 17. The graph shows the variation of the C The internal energy always increases. interatomic force F with the separation r D No heat is transferred into or out of the between two atoms. system. 3 © Majlis Peperiksaan Malaysia 2007 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 3 3/7/2008 4:48:33 PM 23. In an experiment to measure the thermal q2 q2 A –—— C —–— conductivity of a good conductor, a long test 4πε0r 2πε0r sample is used so that q2 q2 B – –—— D – –—— A the heat loss can be neglected 4πε0r 2πε0r B the steady state is easily achieved 27. The circuit diagram shows a 10 kΩ resistor C the heat flow is parallel throughout the and a 40 kΩ resistor connected across a sample charged capacitor C. D the temperature difference between the two ends of the sample is sufficiently V large C S 24. The figure shows a uniform metal rod which 10 kΩ is not insulated. 40 kΩ θ1 X Y θ2 After switch S is closed, the potential difference V across the capacitor varies with Heat flows through the rod in a steady state, time t. The graph of ln V against t is given and the temperatures at the ends of the rod as follows. are θ1 and θ 2, where θ1 > θ 2. Which statement is true of the two positions X and Y on the In V rod as shown? A No heat is lost to the surroundings at the 2 area between X and Y. B The rate of heat loss to the surroundings at X is greater than that at Y. C The rates of heat flow at X and Y are the 0 t (s) 20 same. The capacitance of the capacitor is D The temperature gradients at X and Y are A 1.25 × 10–5 F C 1.25 × 10–3 F the same. B 2.00 × 10 F–4 D 2.50 × 10–3 F 25. The figure shows three point charges +q, +q 28. A parallel-plate capacitor is charged by a and –q at the vertices X, Y and Z respectively battery. The space between the plates is of an equilateral triangle. then completely filled with a material of dielectric constant εr while the battery remains X +q connected. What is the ratio of the energy stored after the material is inserted to the T S energy stored before the material is inserted? O A 1: ε 2 r C ε r: 1 B 1: ε r D ε2: 1r Y +q R Z 29. A lamp lights up immediately when the −q switch in a circuit is closed, although the drift velocity of electrons is small. This is The electric field strength at the centroid O because of the triangle is in the direction of A the current flowing in the circuit is large A OS C OY B the potential difference supplied in the B OT D OZ circuit is large 26. The work done to bring two point charges C the random velocity of the electrons is large +q and –q from infinity to separation r is D all the free electrons drift simultaneously in the circuit © Majlis Peperiksaan Malaysia 2007 4 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 4 3/7/2008 4:48:34 PM 30. The graph shows the variation of potential 33. The figure shows a positive charge moving difference V across a resistor with current I through a uniform magnetic field at velocity flowing through it. v in the positive z-direction. V(V) y 8 6 + x v 4 2 z 0 I (A) 0.1 0.2 0.3 0.4 If the magnetic field is in the positive x-direction, the magnetic force acting on the When the current is 0.23 A, the power supplied positive charge is in to the resistor is A the positive x-direction A 0.53 W C 2.12 W B the negative x-direction B 1.06 W D 4.23 W C the positive y-direction 31. The figure shows a circuit consisting of two D the negative y-direction batteries and three resistors. 34. Hall effect cannot be used to measure 12 V 4.0 Ω P A the charge density I1 B the magnetic flux density C the mass of charge carriers 2.0 Ω D the type of charge carriers 35. The figure shows a square coil of sides I2 Q 10 cm and resistance 2.00 Ω placed in a 8.0 V 6.0 Ω uniform magnetic field of 10.0 T directed If I1 and I2 are 2.56 A and 1.64 A respectively, the perpendicularly to the coil. potential difference between points P and Q is A 1.8 V C 9.8 V 10 cm B 4.0 V D 20 V 32. The figure shows a circuit, where XY is a 10 cm uniform wire. 100 Ω P R Q If the magnetic field is reduced to zero at a G G constant rate in 40 ms, what are the magnitude X Y and direction of the induced current? 40 cm 44 cm Magnitude Direction A 1.25 A Clockwise When the galvanometer is connected at point P and then at point Q, the balanced lengths B 1.25 A Anticlockwise are 40 cm and 44 cm respectively. The value of R is C 2.50 A Clockwise A 10 Ω C 110 Ω D 2.50 A Anticlockwise B 91 Ω D 400 Ω 5 © Majlis Peperiksaan Malaysia 2007 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 5 3/7/2008 4:48:35 PM 36. Which statement is not true of the back e.m.f. Rf A The value of — is 4. of an electric motor? Ri A The back e.m.f. opposes the applied B The voltage gain is – 4. voltage. C When the input voltage is –1 V, the output B The back e.m.f. is caused by voltage is + 4 V. electromagnetic induction. D When the input voltage is 2 V, the C The back e.m.f. helps increase the potential at point P is 2 V. magnitude of the current flowing through 40. Which figure shows the correct directions of the coil of the motor. the magnetic field B and electric field E in D The magnitude of the back e.m.f. increases relation to the direction of propagation of an if the speed of rotation of the coil of the electromagnetic wave? motor is increased. A E C E 37. An a.c. power supply with a constant r.m.s. voltage and variable frequency is connected in series to a pure inductor. Which graph shows B Direction of Direction of the variation of the r.m.s. current I with the propagation B propagation B D frequency f ? E A C Direction of I I B B propagation E Direction of propagation f f 0 0 41. Which statement is true of an electromagnetic B D spectrum? I I A Radio waves are longitudinal waves. B Only the visible spectrum can be plane- 0 f polarised. f C X-rays travel at a higher speed than 0 ultraviolet waves. D Gamma rays have a higher frequency than 38. The alternating voltage supplied to a 200 µF infrared waves. capacitor is given by V = 5 sin 300t, where V is in volts and t is in seconds. The maximum 42. The figure shows two lenses of focal current in the circuit is lengths f1 and f2 placed in contact with each A 0A C 0.30 A other. B 0.21 A D 83 A Object Image 39. The figure shows a circuit of an inverting amplifier. Rf u v +9 V Ri P Vin − Vout If the object distance is u and the image distance + 1 1 Q is v, then — + — equals −9 V u v A — 1 +— 1 C ——– 1 An input voltage of +2 V produces an output f1 f2 f1 + f2 voltage of –8 V. Which statement about the B —–— 1 1 D ——– 1 circuit is not true? f1 f2 f 1 – f2 © Majlis Peperiksaan Malaysia 2007 6 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 6 3/7/2008 4:48:37 PM 43. A diffraction grating with 2400 lines per 47. Stimulated emission occurs in the laser when centimetre is used to produce a diffraction a photon P causes the emission of a photon pattern that is recorded by a detector. The Q. Which statement is true of the photons? first-order fringe is 8.94 cm from the central A The phase of P is the same as that of fringe. If the distance between the grating and Q. the detector is 0.625 m, what is the wavelength B The frequency of P is less than that of of the electromagnetic wave used? Q. A 295 nm C 1180 nm C The wavelength of P is less than that of B 590 nm D 1190 nm Q. D The energy of P is more than that of Q. 44. An experiment is performed to show the photoelectric effect. The frequency of the 48. If the masses of the proton, neutron, light used is kept constant while the intensity electron and 12C atom are mp, mn, me and 6 is increased. Which quantity will increase? mc respectively, the mass defect ∆m of the A The momentum of the photoelectrons nucleus 12C is given by 6 B The emission rate of the photoelectrons A ∆m = 6mp + 6mn – mc C The maximum kinetic energy of the B ∆m = 6mp + 6mn – 6me – mc photoelectrons C ∆m = 6mp + 6mn + 6me – mc D The minimum de Broglie wavelength of D ∆m = 6mp + 6mn + 6me – 6mc the photoelectrons 49. A sample consists of 2 g of a radioactive 45. The figure shows the energy levels of the element. The molar mass of the element is hydrogen atom. 67 g. If the half-life of the element is 78 hours, the activity of the sample after 48 hours is n = ∞ 0 eV A 2.7 × 1016 s–1 n = 5 −0.54 eV n = 4 −0.85 eV B 2.9 × 1016 s–1 n = 3 −1.51 eV C 4.4 × 1016 s–1 n = 2 −3.39 eV D 1.0 × 1020 s–1 n=1 −13.58 eV 50. A nuclide decays by emitting x alpha particles Which transition produces radiation of and y beta particles to form an isotope of the wavelength 436 nm? original nuclide. What are the values of x and A n = 4 to n = 1 C n = 5 to n = 1 y? B n = 4 to n = 2 D n = 5 to n = 2 x y 46. In an X-ray tube, the minimum wavelength A 1 2 produced is 4.0 × 10–11 m. If the potential B 1 4 difference between the cathode and anode is C 2 1 decreased to half of the original value, the minimum wavelength becomes D 2 2 A 2.0 × 10–11 m C 8.0 × 10–11 m B 4.0 × 10–11 m D 1.6 × 10–10 m QUESTION PAPER ENDS 7 © Majlis Peperiksaan Malaysia 2007 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 7 3/7/2008 4:48:38 PM Time: 2 h 30 min Section A [40 marks] Answer all questions in this section. 1. A cylinder of mass 10 kg and radius 0.2 m rolls down an inclined plane of height 1.0 m without slipping from rest. Calculate the linear velocity of the cylinder when it reaches the ground. [4 marks] [Moment of inertia of a cylinder of mass m and radius r is — 1 mr2.] 2 2. (a) On the same axes, sketch graphs of displacement against time to show underdamped, critically damped and overdamped oscillations. Label your graphs. [3 marks] (b) What is meant by resonance in forced oscillations? [2 marks] 3. (a) In each of the following cases, state any change in the interference pattern of a two-slit arrangement. (i) One slit is covered by an opaque material. [1 mark] (ii) The distance between the slits is decreased. [1 mark] (b) Explain how interference can be used to determine the flatness of lens surface. [2 marks] 4. A cylindrical brass rod with Young’s modulus 9.7 × 1010 Pa and original diameter 10.0 mm experiences only elastic deformation when a tensile load of 200 N is applied. (a) Calculate the stress that produces the deformation. [3 marks] (b) If the original length of the rod is 0.25 m, calculate the change in the length of the rod. [2 marks] 5. A 1.5 µF capacitor and a 2.0 µF capacitor are connected in series across a 24 V source. (a) Calculate the equivalent capacitance across the source. [2 marks] (b) Calculate the charge stored on each capacitor. [2 marks] (c) If a dielectric is added to each capacitor, explain what happens to the charge stored in each capacitor. [2 marks] 6. An ideal solenoid consists of 1000 turns of wire per cm wound around an air-filled cylindrical structure. The solenoid is of 2.0 cm long and cross-sectional area 1.8 cm2. A current of 2.0 A passes through the wire. (a) Calculate the magnetic flux in the solenoid. [3 marks] (b) Calculate the self-inductance of the solenoid. [2 marks] 7. A light beam of wavelength 0.110 nm collides with an atom. After the collision, an electron is emitted with kinetic energy 180 eV. (a) Calculate the energy absorbed by the atom. [3 marks] (b) Calculate the velocity of the electron emitted. [2 marks] 8. In a nuclear reactor, a very slow moving neutron is absorbed by a stationary boron atom. The equation for the nuclear reaction is 1 10 7 4 n + 5B → 3Li + 2He 0 © Majlis Peperiksaan Malaysia 2007 8 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 8 3/7/2008 4:48:39 PM After the reaction, the speed of the helium atom is 9.10 × 106 m s–1 and the kinetic energy of the neutron is approximately zero. [mn = 1.0087 u, mB = 10.0130 u, mLi = 7.0160 u, mHe = 4.0026 u] (a) Calculate the kinetic energy of the lithium atom after the reaction. [4 marks] (b) Calculate the reaction energy. [2 marks] Section B [60 marks] Answer any four questions in this section. 9. (a) An object is projected with initial speed v at angle θ to the horizontal. Derive an equation for the path of the projectile. [3 marks] (b) The figure shows a jeep travelling at a constant speed of 10 m s–1 towards a security post. Bullet Security 15 m Jeep post 0.7 km A security guard holding his rifle 15 m above the ground shoots horizontally when the jeep is 0.7 km from the security post. The bullet of the rifle strikes the jeep. Calculate (i) the time taken by the bullet to strike the jeep [2 marks] (ii) the distance of the jeep from the security post when it is struck [2 marks] (iii) the initial speed of the bullet [2 marks] (iv) the speed and direction of the bullet when it strikes the jeep. [6 marks] 10. (a) What is meant by Doppler effect? [2 marks] –1 (b) A toy car with the siren on is approaching a wall at a speed of 2.0 m s . The siren produces a sound wave in the form y = 2.5 × 10–5 sin 2π(500t – 1.4x) where x is in metres and t is in seconds. Calculate the speed and frequency of the sound wave. [4 marks] (c) The sound wave in (b) is then reflected from the wall. A detector is fixed on the ground behind the source of the sound wave to detect any frequency changes. Determine (i) the frequencies detected by the detector [4 marks] (ii) the beat frequency [2 marks] (iii) the equation of the reflected wave [1 mark] (iv) the intensity ratio of the approaching wave to the reflected wave. [2 marks] 11. The following is a p–V diagram for a 0.2 mol ideal monatomic gas. p(Pa) A 3 × 105 C B 0 2.0 12.0 V (× 10−3 m3) The gas expands isothermally from A to B and is compressed from B to C at constant pressure. It finally undergoes a constant volume process from C to A. Calculate 9 © Majlis Peperiksaan Malaysia 2007 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 9 3/7/2008 4:48:39 PM (a) the temperatures at A, B and C [5 marks] (b) the net work done during the cycle [5 marks] (c) the net heat absorbed by the gas. [5 marks] 3 [For an ideal monatomic gas, CV = —R and Cp = CV + R] 2 12. (a) A 0.3 µF capacitor is connected to a source of alternating current with output voltage V = 240 sin 120πt (i) Calculate the reactance of the capacitor. [3 marks] (ii) Determine the r.m.s. current flowing through the capacitor. [3 marks] (b) A rectifier is a device which conducts electric current in one direction only such as a diode. (i) Describe briefly a full-wave rectifier. [4 marks] (ii) With the aid of diagrams, describe briefly the process of smoothing rectified alternating current voltages. [5 marks] 13. (a) State Bohr’s postulates for an atom. [2 marks] (b) The figure shows an electron of mass m and charge –e moving at speed v in a circular orbit of radius r round a nucleus. v −e m r +e If the force of attraction between the electron and the nucleus provides the centripetal acceleration of the electron, derive an expression for the radius of the nth orbit of the electron. [5 marks] (c) An electron in a Bohr orbit has kinetic energy 8.64 × 10 J. –20 (i) Calculate the speed of the electron. [3 marks] (ii) Determine the allowed orbit. [3 marks] (iii) Calculate the radius of the orbit. [2 marks] 14. (a) The bombardment of a beryllium nucleus by an α-particle produces a fundamental particle X, as follows. α + 9Be ⎯⎯→ 12C + X 4 6 (i) Complete the equation above by giving the proton and nucleon numbers to the α-particle and X. [2 marks] (ii) What are the α-particle and X? [2 marks] (iii) State two important properties which make X difficult to detect. [2 marks] (b) Determine the equivalent energy in MeV of a mass of 1 u. [5 marks] (c) An element of unknown atomic mass is mixed with 12C atoms in a mass spectrometer. 6 12 The radii of curvature of the tracks of the ions of the element and 6C are 26.2 cm and 22.4 cm respectively. What is possibly the element? State any assumption you make. [4 marks] QUESTION PAPER ENDS © Majlis Peperiksaan Malaysia 2007 10 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 10 3/7/2008 4:48:40 PM PAPER 1 14. A: P at the amplitude, its speed = 0. F MLT–2 1. B: [Tensile strength] = [stress] = [—] = ——– = ML–1T–2 15. A: See figure. A L2 2. C: If a = acceleration, F = (m + 2m)a = 3ma, 2F P S Q force of Y on X = force of X on Y = (2m)a = — 3 3. C: Speed = gradient of graph. When t = T, gradient of P > gradient of Q. 16. D: Intensity level = 35 dB = 10 lg ———–—I 1.0 × 10–11 4. C: Ball has translational and rotational K.E. I = 3.16 × 10–8 W m–2 1 — mv 2 17. B: F = 0 at the equilibrium separation. gain in K.E. 2 5. A: Average power = ————— = ——— = 121 kW time t 18. C: Plastic deformation, stress is not proportional to 6. D: Angular momentum = Iω = (mr 2)ω = constant. strain. dm 1 3 3RT 3(8.31)(373) 7. C: Force produced by the ejected gas, F = v —– 19. B: —Mc 2 = —RT, c = —— = ————— m s–1 dt 2 2 M 0.034 dm r = 5.23 × 102 m s–1 Torque = Fr = Iα, α = v —– — dt I 3RT1 3RT2 0.020 20. D: c1 = —— = c2 = —— = (50)(1.0 × 10–5) ——–—— rad s–2 M1 M2 1.0 × 10–6 M1 32 = 10 rad s–2 T1 = —–T2 = —(300) K = 2400 K M2 4 Angular velocity after 10 s, = αt = (10)(10) rad s–1 = 100 rad s–1 5 2W 21. C: Adiabatic, W = ∆U = —R(∆T ), ∆T = —– = 2.41 K 2 5R 8. D: Resultant force = 0, resultant torque about centre of disc = 0. 22. D: ∆Q = 0 Gm1 Gm2 23. D: Temperature gradient is small. A longer length is 9. A: EX = EY, —— = —–— (r/3)2 (2r/3)2 required to produce a sufficiently large temperature m1 1 difference. —=— m2 4 24. B: Temperature gradient at X is greater. 10. B: For object to move in circular orbit, velocity = GM —– 25. D: X +q r Escape speed = 2GM —— EY r O EZ Speed of 1.5 GM > 2GM > GM —– —— —– Y EX Z r r r +q −q Therefore trajectory is as in B. The electric field strengths EX, EY and EZ at O due to the charges at X, Y and Z are of the same magnitude π 11. A: x = 5 sin —t and in the directions shown in the figure. Therefore 6 the resultant electric field at O is in the direction dx π π OZ. Speed, v = — = 5(—) cos —t dt 6 6 π cos — = 2.3 m s–1 π 26. B: Work done = electric potential energy of the system When t = 1 s, v = 5 — 6 6 (+q)(–q) = ——–— 12. C: Increased damping: Lower peak and lower resonant 4πε0r frequency. 27. C: V = V0e–t/CR 2π t 13. D: Phase difference = —–(3.0) rad = 3.8 rad lnV = lnV0 – –— 5.0 CR 11 © Oxford Fajar Sdn Bhd (008974–T) 2008 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 11 3/7/2008 4:48:40 PM Gradient of graph = –2 = – — — 1 43. B: 20 CR 1 1 1 — = —— + —–— = —–— 5 8.94 cm R 10 kΩ 40 kΩ 40 kΩ Grating θ1 5 Capacitance, C = (10) ———— F = 1.25 × 10–3 F 40 × 103 62.5 cm 28. C: C1 = εrC0, V remains the same because the battery 1 1 remains connected. E0 = —C 0V 2, E1 = —C1V 2 = εrE 0 2 2 1 8.94 λ = d sin θ1 = ———–— —————— m = 590 nm 29. D 2400 × 102 8.942 + 62.52 0.23 30. B: When I = 0.23 A, V = ——(4 V) 44. B: When the intensity is increased, the rate of photons 0.2 incident on the cathode increases. Rate of emission 0.23 Power = IV = (0.23) ——(4) W = 1.06 W of photoelectrons increases. 0.2 31. A: Current in 2.0 Ω resistor, hc (6.63 × 10–34)(3.00 × 108) 45. D: E = — = ——————————— eV I = I1 – I2 = (2.56 – 1.64) A = 0.92 A λ (436 × 10–9)(1.60 × 10–19) VPQ = IR = (0.92)(2.0) V = 1.84 V = 2.85 eV = E5 – E2 32. A: P.d. across 40 cm = I(100) hc 46. C: λmin = — P.d. across 44 cm = I(100 + R), eV 44 hc 100 + R = —– (100) Ω, R = 10 Ω λ'min = ——— = 2λmin = 8.0 × 10–11 m 40 e(V/2) 33. C: y 47. A: Laser photons are coherent and in phase. F=vxB 48. C: Mass of C-12 atom = 6mp + 6mn + 6me 2 23 22 49. D: Initial number of atoms, N0 = — (6.02 × 10 ) = 1.80 × 10 67 + B x dN In 2 Initial activity, A 0 = – —– = λN 0 = —–– N0 z dt T—1 v 2 48 1 VH I t = 48 h = –—T — = 0.6154T — 1 34. C: e — = Bev = Be —— 78 2 2 d nAe A0 which is not in terms of the mass m of the charge Activity after 48 h, A = –—— carriers. 20.6154 1 In 2 = ——– ––— (1.80 × 1022) s–1 = 1.0 × 1020 s–1 dφ 0 – BA 20.6154 78 35. A: E = – — = – —–—— dt 40 ms 50. A: A P → Z A –4 Z 0 Q + 4 He + 2 –1 e 2 E (10.0)(0.10)2 Induced current, I = — = ——————— A = 1.25 A R (40 × 10–3)(2.00) PAPER 2 Clockwise direction: Apply Lenz’s law. Section A 36. C: Because of the back e.m.f., the current decreases. 1. Gain in translational K.E. + rotational K.E. = loss of P.E. V V 1 37. C: I = —– = —— ∝ — 1 1 XL 2πfL f —mv 2 + —Iω2 = mgh 2 2 Vmax 1 2 + — —mr 2 — 2 = mgh 1 1 v 38. C: Imax = —— = ωCVmax = (300)(200 × 10–6)(5) A = 0.30 A —mv XC 2 2 2 r 2 1 1 39. D: VP = 0 V v — + — = gh 2 4 40. C: E × B in the direction of propagation. 4 v = —gh 3 41. D 4 = —(9.81)(1.0) m s–1 1 1 1 1 1 3 42. A: — = — + — = — + — F f1 f2 u v = 3.6 m s–1 © Oxford Fajar Sdn Bhd (008974–T) 2008 12 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 12 3/7/2008 4:48:42 PM 2. (a) Displacement (b) φ = LI Critically damped 0.090 Underdamped L = —–— H = 0.045 H 2.0 Overdamped 7. (a) Energy absorbed by the atom 0 = energy of photon – K.E. of electron Time hc = — – 180 eV λ (b) • Resonance occurs when the amplitude of a (6.63 × 10–34)(3.00 × 108) = ——————————— – 180 eV forced oscillation is maximum. (0.110 × 10–9)(1.60 × 10–19) • Resonance occurs when the frequency of the = 1.11 × 104 eV driver system equals the natural frequency of 1 (b) Kinetic energy, —mv 2 = 180 eV the system that is forced into oscillation. 2 2(180)(1.60 × 10–19) Velocity of electron, v = ———–————— m s–1 3. (a) (i) The single-slit diffraction pattern as shown 9.11 × 10–31 below is obtained. = 7.95 × 106 m s–1 8. (a) Using the principle of conservation of linear (ii) The fringe separation increases. momentum, (b) Place the lens on a mirror and allow monochromatic MLiVLi + MHeVHe = 0 light to be incident on the lens. MHe If the lens surface is perfect, the interference Velocity of lithium atom, VLi = – –— VHe MLi pattern observed (Newton’s rings) would be perfect 4.0026 = – —–—– (9.10 × 106) m s–1 concentric circles. 7.0160 Monochromatic light = – 5.192 × 106 m s–1 Kinetic energy of lithium atom 1 = —MLi (VLi)2 2 Lens 1 = —(7.0160 × 1.66 × 10–27)(5.192 × 106)2 J 2 Mirror Newton s rings = 1.57 × 10–13 J F 200 4. (a) Stress = — = —————– N m–2 (b) Total mass before reaction = 1.0087 u + 10.0130 u A π(5.0 × 10–3)2 = 11.0217 u = 2.5 × 106 N m–2 Total mass after reaction = 7.0160 u + 4.0026 u (b) Young’s modulus, E = –—F/A = 11.0186 u e/l Reaction energy, Change in length, e = —Fl E = (∆m)c 2 EA = (11.0217 – 11.0186)(1.66 × 10–27)(3.00 × 108)2 J (200)(0.25) = 4.63 × 10–13 J = ——————————2 m (9.7 × 1010)π(5.0 × 10–3) = 6.6 × 10–6 m Section B 1 1 1 5. (a) — = –— + –—, C = 0.86 µF 9. (a) C 2.0 1.5 v sin θ v P (x, y) (b) Charge stored in each capacitor, Q = CV = (0.86)(24) µC y = 21 µC θ (c) With the dielectric v cos θ • Capacitance of each capacitor increases. x • Charge stored in each capacitor increases. When the object is at P, 6. (a) Magnetic flux density, B = µ0nI • horizontal displacement, x = (v cos θ)t, Magnetic flux, t = ———x φ = NBA v cos θ 1 = N(µ0nI)A • vertical displacement, y = (v sin θ)t – —gt 2 2 x 1 = (v sin θ)(—–——) – —g(—–——)2 x 1000 v cos θ 2 v cos θ = (4π × 10–7)(1000 × 2.0) ——–— (2.0) 1 × 10–2 × (1.8 × 10–4) weber gx 2 = x tan θ – ——–—— = 0.090 weber 2v cos2 θ 2 13 © Oxford Fajar Sdn Bhd (008974–T) 2008 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 13 3/7/2008 4:48:43 PM (b) (i) Consider vertical component of motion of the 360 = ———— 500 Hz bullet 360 – 2.0 u = 0, a = g, s = 15 m = 503 Hz 1 1 This frequency of 503 Hz is also detected by Using s = ut +—at 2, 15 = 0 + —(9.81)t 2 2 2 the detector. (ii) Beat frequency = f2 – f1 2(15) Time taken, t = —— s = 1.7 s = (503 – 497) Hz = 6 Hz 9.81 1 f2 503 (ii) Distance of the jeep from the security post (iii) — = — = —– = 1.4 λ2 v 360 = (700 – 1.7 × 10) m = 683 m Equation of the reflected wave is (iii) Horizontal distance travelled by the bullet, y = (2.5 × 10–5) sin 2π(503t + 1.4x) v(1.7) = 683 m (iv) Intensity is directly proportional to (amplitude)2. Speed, v = 402 m s–1 Since the approaching wave and the reflected (iv) Vertical component of the bullet’s velocity, wave have the same amplitude, vy = 0 + (9.81)(1.7) intensity of approaching wave = 16.7 m s–1 ———————————–— = 1 intensity of reflected wave Horizontal component of velocity, vx = 402 m s–1 11. (a) At A, pV = nRTA 2 2 (3 × 105)(2.0 × 10–3) Speed = v + v x y TA = ———————— K = 361 K (0.2)(8.31) = 4022 + 16.72 m s–1 At B, TB = TA = 361 K B to C, p = constant, VC ∝ TC , and VB ∝ TB = 402.3 m s–1 Direction of velocity is at an angle VC TC = –— TB vy VB 16.7 tan–1 — = tan–1 —–— vx 402.3 2.0 × 10–3 = ————— 361 K = 2.4° to the horizontal 12.0 × 10–3 = 60.2 K 10. (a) Doppler effect is the apparent change in the VB (b) From A to B, work done by the gas, W1 = nRTA ln —– frequency of sound heard by an observer when VA there is relative motion between the source of sound 12.0 × 10–3 = (0.2)(8.31)(361) ln ————— = 1075 J and the observer. 2.0 × 10–3 (b) Comparing y = (2.5 × 10–5) sin 2π(500t – 1.4x) pAVA Pressure at B, pB = —— 2π VB with y = A sin 2πft – —x , λ (3 × 105)(2.0 × 10–3) = ———————–— Pa = 5.0 × 104 Pa frequency, f = 500 Hz, λ = —– m 1 12.0 × 10–3 1.4 From B to C, work done on the gas, Speed of sound wave, v = f λ W2 = (5.0 × 104)(12.0 –2.0) × 10–3 J = 500 J 1 = (500) —– m s–1 Net work done by the gas = (1075 – 500) J = 575 J 1.4 (c) From A to B, isothermal expansion, ∆U = 0. = 357 m s–1 From ∆Q = ∆U + W, (c) (i) heat absorbed, ∆Q1 =W1 = 1075 J 5 f2 From B to C, ∆Q2 = nCp(∆T )2 CP = —R Wall uS = 2.0 m s−1 2 Detector 5 f2 Source f1 = (0.20)(—)(8.31)(60.2 – 361) J = –1250 J 2 From C to A, V = constant, ∆Q3 = nCV(∆T )3 The source is moving away from the detector. 3 Frequency detected by the detector, = (0.2) — × 8.31 (361 – 60.2) J = 750 J 2 v Net heat absorbed = (1075 – 1250 + 750) J = 575 J f1 = —–— f v + us 1 360 12. (a) (i) XC = —– (ω = 120π) = ———— 500 Hz ωC 360 + 2.0 1 = 497 Hz = ———————— Ω = 8.8 × 103 Ω (120π)(0.3 × 10–6) The source is approaching the wall. Vrms Frequency reflected by the wall (ii) Irms = –— Vrms = 240 V —– XC 2 = frequency incident on the wall, f2 240 v = ————–—— A = 0.019 A f2 = —–— f 2 × (8.8 × 103) v – us © Oxford Fajar Sdn Bhd (008974–T) 2008 14 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 14 3/7/2008 4:48:44 PM (b) (i) A full-wave rectifier that consists of four diodes mv 2 e2 •centripetal force, —— = ——— W, X, Y and Z is as shown below. r 4πε0r 2 e2 Va.c. v 2 = ———– (2) 4πε0mr e2 n 2h 2 W X From (1) and (2), m 2 ——–— r 2 = —— 0 Time 4πε0mr 4π2 A.c. supply Z Y ε0n h 2 2 Radius of the nth orbit, r = ——— πme 2 V R 1 (c) (i) Kinetic energy, —mv 2 = 8.64 × 10–20 J Full-wave rectifier 2 0 Time Full-wave rectified voltage 2(8.64 × 10–20) Speed of electron, v = ————–—– m s–1 During the first half cycle of the a.c. voltage, 9.11 × 10–31 current flows through X, R and Z. During the = 4.36 × 105 m s–1 second half cycle, current flows through Y, R (ii) Total energy of the electron in the nth orbit, and W. In both the half-cycles, current flows En = – K, the kinetic energy in the same direction through R. Therefore, 13.6 the voltage across the resistor R is full-wave Also, En = – –—– eV n2 rectified as shown. 13.6 (ii) Therefore, – –— eV = –8.64 × 10–20 J n2 W X Smoothened rectified (13.6)(1.60 × 10–19) A.c. V n = ———————— = 5 Z Y voltage 8.64 × 10–20 supply Allowed orbit n = 5 h R 0 (iii) From (mvr ) = n –— Time 2π nh Capacitor C Radius of orbit, r = ——– 2πmv The rectified voltage is smoothened by 5(6.63 × 10–34) = ———————————— m connecting a capacitor C in parallel with the 2π(9.11 × 10–31)(4.36 × 105) load R as shown in the circuit. During the first = 1.33 × 10–9 m half cycle, the capacitor C is charged until the peak voltage. Current stops flowing from the 14. (a) (i) α + 9Be → 12C + 1X 4 2 4 6 0 a.c. supply, and the capacitor discharges until the potential difference across the capacitor is (ii) α-particle: Nucleus of helium. 1 less than the voltage of the a.c. supply. Current 0X: Neutron again flows from the a.c. supply until the (iii) X or neutron capacitor is charged to the peak voltage again. • is not charged. The process is then repeated. The voltage V • does not cause ionisation. across the load R varies with time as shown (b) Energy equivalent of 1 u, E = mc 2 in the graph. = (1.66 × 10–27)(3.00 × 108)2 J (1.66 × 10–27)(3.00 × 108)2 13. (a) Bohr’s postulates = ———————–——— MeV (1.60 × 10–19)106 • An electron can only orbit the nucleus in discrete = 934 MeV allowed orbits such that mv 2 h (c) —– = qvB angular momentum of the electron = n —– r 2π h is Planck constant, and n = 1, 2, 3, … qB m = (–—)r • When an electron drops from a higher energy v level E2 to a lower energy level E1, the difference Assumption: Ions of the element and ions of C-12 in energy of the electron is radiated as a photon have the same charge, then m ∝ r. of frequency f. Hence mass number, A ∝ r E2 – E1 = hf For X, AX ∝ 26.2 cm (b) From the first postulate of Bohr, For C-12, 12 ∝ 22.4 cm h 26.2 • angular momentum, (mvr ) = n —– Mass number, AX = ——(12) = 14 2π 22.4 n 2h 2 m 2v 2r 2 = —— (1) 4π2 The element is nitrogen, N. 15 © Oxford Fajar Sdn Bhd (008974–T) 2008 http://edu.joshuatly.com http://www.joshuatly.com Ace Ahead STPM Physic vol 1 4th.indd 15 3/7/2008 4:48:45 PM