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Page 1 of 17 CCC HOH FUK TONG COLLEGE Mock Examination, 2004 - 2005 Physics Paper I Secondary: Seven Date: 28/02/2005 Time allowed: 3 hours (8:30 - 11:30 a.m.) Marks: 122 Name : ______________________________ Number : ______ 1. Answer ALL questions. 2. Write your answers in the spaces provided in the question paper. you should show all the main steps in your working. 3. Assume: 1. (a) In calculations velocity of light in air = 3 108 m s-1 acceleration due to gravity = 10 m s-2 Figure 1(a) shows a skier descending the ramp of a ski jump. Figure 1(b) shows a graph of the distance travelled along the ramp against time, from the time the decent starts until the skier leaves the end of the ramp. Figure 1 (a) 110 100 90 80 70 60 50 40 30 20 10 0 distance / m Figure 1 (b) 0 2 4 6 8 time /s 10 Page 2 of 17 The skier of mass 80 kg (including equipment) skies down the ramp and leaves it horizontally. The skier gains 55% of the available gravitational potential energy as kinetic energy when descending the ramp. (i) Use Figure 1(b) to find the speed at which the skier leaves the ramp. Show your working. (2) Draw tangent to graph at the end of run. slope = speed = 23 m s-1 (ii) Assume uniform acceleration when descending the ramp, estimate its value (2) By v = u + at 23 = 0 + a(8.8) a = 2.61 m s-2 (iii) Determine the height of the ramp. (2) P.E. lost 55% = K.E. gained m 10 h 0.55 = ½ m (23)2 h = 48 m (b) Figure 1(a) shows the path taken by the skier after leaving the ramp. Assuming that there was no lift or drag due to the air during this jump, calculate (i) the time for which the skier was in flight; (2) By s = ut + ½at2 80 = 0(t) + ½(10) t2 t=4s (ii) the horizontal distance jumped by the skier before landing. (2) By s = ut = 23 (4) = 92 m (c) The landing surface is suggested to be inclined at an angle to the horizontal. Explain briefly. (2) Skier takes longer time to reduce (vertical) momentum compared with time when surface is flat. mv By F = t , force experienced by the skier is reduced. Page 3 of 17 2. Figure 2(a) shows the displacement-time graphs for two mass-spring systems X and Y that are performing simple harmonic motion. 0.3 displacement / m 0.2 0.1 X Y 0 1.0 0.5 1.5 time / s -0.1 -0.2 -0.3 (a) Figure 2(a) Deduce an equation for the motion of system X. (2) Let x = A sint 2 2 = T = 1.2 = 5.24 rad s-1 Hence x = 0.2 sin(5.24 t) m (b) The springs used in X and Y have the same spring constant. ratio of the (i) the mass used in Y to that in X m Since T (given: same spring constant) k Find the (2) From graph, they have the same period. Hence, the have the same mass, i.e. mY : mX = 1 : 1 (ii) the maximum energy stored in Y to that in X. (2) 1 1 EY : EX = 2 kY (AY)2 : 2 kX (AX)2 (same k) = (AY)2 : (AX)2 = (0.1)2 : (0.2)2 = 1 : 4 (c) Figure 2(b) shows how the potential energy of X varies with displacment. (i) Determine the spring constant of the spring used and the mass of X. (3) 1 By PE = 2 kX (AX)2 1 0.16 = 2 kX (0.2)2 kX = 8 N m-1 By T = 2 m m 1.2 = 2 k 8 Page 4 of 17 m = 0.29 kg (ii) (iii) Draw on it how the kinetic energy of X varies with its diplacement. Label this graph A. (1) Draw on it how the kinetic energy of Y varies with its diplacement. Label this graph B. (2) potential energy / J 0.2 0.15 0.1 0.05 0 -0.2 -0.1 0 0.1 0.2 displacement / m energy / J 3. (a) A small loudspeaker placed in open space emits sound energy uniformly in all directions with a total power of 5 10-5 W. (i) What is the sound intensity at a distance of 2 m from the loudspeaker? (2) P 510-5 I= = = 9.95 10-7 W m-2 4 r2 4 (2)2 (ii) If the loudspeaker were immersed in water, what change would be produced on the sound intensity at the same distance of 2 m from it? Explain briefly. (2) The sound intensity would be smaller (1) due to absorption of wave energy by water. (1) (b) In the figure below, the curves AB and EF represent the threshold of hearing and threshold of pain for the average human ear respectively. CD represents curve of equal loudness. Page 5 of 17 threshold of pain equal loudness threshold of hearing (i) Deduce the lowest intensity and lowest intensity level of sound that can be heard at a frequency of 0.25 kHz. (2) From graph AB, log10(I) = -9 I = 10-9 W m-2 (1) 10-9 h = 10 log ( 10-12 ) = 30 dB (1) (ii) What is the difference in intensity level between two sounds of frequencies 1 kHz and 0.25 kHz which would give the same loudness level corresponding to curve CD? (3) At 1 kHz, log10(I1) = -10 I1 = 10-10 At 0.25 kHz, log10(I2) = -7 I2 = 10-7 10-7 10-10 h2 – h1 = 10 log10( 10-12 ) - 10 log10( 10-12 ) = 10 log10(105) - 10 log10(102) = 50 – 20 = 30 dB (iii) A person stands in front of a small sound source of frequency 0.25 kHz so that the loudness level is the same as that corresponding to curve CD, i.e. log10(I) = -7. He cannot hear the sound if he walks a further distance of 13.5 m away from the source. Estimate the initial distance of the person from the source. (3) D d source 13.5 m I1(10-7) 1 10-7 D2 Since I r2 10-9 = d2 I2(10-9) Page 6 of 17 D = 10d d + 13.5 = 10d 9d = 13.5 d = 1.5 m 4. A source emits red and blue light, parallel beam of lights is produced when the lights pass through a collimator. It is then incident normally on a diffraction grating placed on a spectrometer turntable. (a) In the space below, draw a ray diagram showing the path of the rays passing through the spectrometer, when the spectrum of the source is observed. The diagram should include the collimator and the telescope of the spectrometer. (2) diffraction grating telescope source (b) collimator cross wire Explain why the spectrometer telescope must be adjusted to receive parallel lights in order to view a pure spectrum of the source. (1) It is adjusted so that all images will be brought into focus on the cross-wire. (c) The source, collimator and telescope are set initially in a straight line and the telescope is turned in one direction only. For the first order, what colour of the spectral lines will be seen first? Explain briefly. (2) Apply a sin = m For the same order, as the wavelength of blue light is smaller, it will be seen with a smaller diffraction angle, i.e. blue light will be seen first. (d) At a diffraction angle of 54.8o, a blue line and a red line coincide. If the grating has 600 rulings per mm and the wavelength of the blue light is 454 nm, (i) Find the order of blue light at that diffraction angle. (2) By a sin = m 1 sin54.8o = m (45410-9) 600103 m=3 (ii) Hence, estimate the order number and wavelength of red light. State the reason of your estimation. (3) At the same diffraction angle, the order number of red light must be smaller than that of blue as its wavelength Page 7 of 17 is longer, i.e. order of red light must be 1 or 2. When two lights coincide, mb b = mr r 3 454 nm = mr r If mr = 1, If mr = 2, (e) r = 1364 nm which is not reasonable. r = 681 nm which is reasonable. The same diffraction grating is used to view a monochromatic light source placed 50 cm away from the grating. The observer holds the grating close to his eye, and observes an image on either side of the source. Calculate the wavelength of the light if the images are 37.8 cm apart. (2) 18.9 cm 50 cm 18.9 tan = 50 = 20.7o a sin = 1 sin20.7o = = 589 nm 600103 5. Two conducting spheres A and B as shown below, are of equal mass and equal charge Q. They are suspended by two fine threads of equal length l. 1 Take = 9 109 N m2 C-2 . 4 π εo l l A B x In equilibrium, the threads connecting the spheres make equal angle with the vertical and the centers of the spheres are x apart. Page 8 of 17 (a) Draw and label the forces acting on sphere A. electrostatic force F (2) tension T weight mg (b) In terms of and tension T, write down an expression for (i) the electrostatic force acting on sphere A (4) F = T sin (½) (ii) the weight of sphere A mg = T cos (½) Hence, deduce that the separation x can be expressed as Q2 l x= 2 π εo m g What is the value of n? F= n (Hint: tan sin) 1 Q2 4 o x2 From (b)(i) and b(ii), F = mg tan mg sin = mg Hence, (x/2) l 1 Q2 (x/2) 2 = mg l 4 o x 1 Q2 l 3 1 and n = Solving gives x = 3 2 π εo m g (c) (3) If m = 2.0 10-2 kg, l = 0.1 m and x = 0.02 m, find the value of the charge Q. (2) 2 π εo From (b), Q = 1 3 2 mgx l 1 2 π (8.85 10 -12 ) (2 .0 10 - 2 )(10)(0.02 ) 3 2 = 0.1 = 2.9 10-8 C (d) (i) (2) If the spheres carry the same sign but with different amount of charges, will the thread be inclined at an equal angle to the vertical when the system is in equilibrium? (2) If spheres carry different amount of charges, the electrostatic forces acting on both spheres are still the same. Therefore the threads will incline at equal angle Page 9 of 17 to the vertical as before. (2) (ii) If the spheres carry the same amount of charges but different masses, will the thread be inclined at an equal angle to the vertical when the system is in equilibrium? Explain (2) If spheres have different masses, the tensions in the threads will be different. Thus the threads will not incline at equal angle to the vertical. (2) 6. The primary coil of a transformer is connected to a 3.0 V d.c. supply and a switch S1. The secondary coil is connected to a resistor and a switch S2 . S1 S2 3.0 V (a) Figure below is a graph showing the growth of current in the primary coil when switch S1 is closed and switch S2 is left open. current / A time / ms (i) Use the graph to determine the initial rate of growth of current in the primary coil. (2) dI dt (at t = 0) 0.3 = 150 10-3 = 2.0 A s-1 (2) (1.8 – 2.2 A s-1) (ii) Calculate the inductance of the primary coil. dI E = L dt (2) Page 10 of 17 3=L2 L = 1.5 H (2) (1.3 – 1.7 H) (b) Now consider the situation when switch S2 is closed before switch S1 is closed. Draw the variation of the current in the circuit connected to the secondary coil when switch S1 in the primary circuit is closed. Assume that the primary current varies as shown above. (2) current finite Imax at t = 0 becoming asymptotic at ~ 700 ms (2) 0 (c) 100 200 300 400 500 600 700 time / ms A transformer is not perfectly efficient when used with alternating current. State and explain TWO sources of energy loss and explain how they can be eliminated. (2) eddy current losses: use laminated core hysteresis losses: use soft iron as core (any 2) heat loses: use copper as wires (d) (i) Complete the circuit diagram below to show how you would use a capacitor and diodes to rectify and smooth the output from a transformer. (2) a.c. supply (ii) output Figure below shows the rectified output voltage when a current is drawn from the secondary of the transformer. Draw the output voltage you would get if a larger capacitor were used. (2) voltage time Page 11 of 17 7. In Figure 7, a uniform magnetic field B pointing into the paper and a uniform electric field E pointing upwards are applied on the left-hand side and right-hand side of y-axis respectively. Suppose that a charged particle of mass m and charge q enters the magnetic field with an initial velocity Vo at point P. After leaving the magnetic field, it will move into the electric field and leave it at point S. Neglect the gravitational effect. y E B Vo S P (a) What is the sign of the charges carried by the charged particle? paths of the charged particle in the two different fields. x State the (2) By Fleming’s left hand rule, it can be found that the charged particle carries negative charges.(1) In the uniform magnetic field, the path is part of a circle. (½) In the electric field, the path is a parabola. (½) (b) Deduce an expression, in terms of m, q , E and B, for the time taken for the charged particle to travel through. (i) the magnetic field and (3) In the magnetic field , the charged particle undergoes a circular motion with radius R. The centripetal force acting on the charged particle is the magnetic force. F= m Vo2 = q Vo B R qBR (1) m Since the charged particle describes a quarter of circle in the magnetic field only, the time taken for the charged particle to move in the magnetic field is given by 2πR 1 Tm= V 4 (1) o πm Tm = 2 q B (1) ∴Vo= (ii) the electric field (3) When the charged particle enters the electric field, its horizontal component of velocity remains uncharged, whereas its vertical component of velocity is accelerated. Page 12 of 17 By Newton’s second law of motion, the acceleration of the charged particle is given by F= ma = q E qE ∴a = m (1) For the vertical displacement, we have 1 R = 2 a te2 (te is the time taken for the charged particle to move in the electric field) (1) te2 te = (c) 2R 2mR = a = qE = m q 2 Vo BE m Vo qB qE 2m 2 m2 Vo = q2 E B (1) Suppose that the velocity of the charged particle at point S is Vs, deduce down an expression for the total work done on the charged particle by the fields in terms of m, Vo and Vs. (2) Since the motion of the charged particle is perpendicular to the magnetic force, there is no work done on the particle by the magnetic field. (1) The work done on the charged particle comes from the electric field such that it is equal to the change of kinetic energy. 1 W.D.=△K.E. = 2 m(Vs2 - Vo2) (1) (d) From the result in part (c), express Vs in terms of Vo, q, E and B. Electric work done = ∫Fe dy = Fe ∫dy = Fe R = q E R 1 m Vo 2 2 m(V s - Vo ) = q E R = q E 2 qB 1 Vo 2 2 (V s - Vo ) = E 2 B Vo Vs2 = 2E B + Vo2 Vs = Vo2 2 E Vo B (3) (3) Page 13 of 17 8. Figures below show two characteristics of an n-p-n transistor. after saturation. IB / A Take VCE = 0.2 V IC / mA 3.6 0 IB / A VBE / V 0.8 0 30 The following circuit is connected to investigate the input/output voltage characteristics of the transistor. 6V IC RL=2.2 k Vout IB Vin RB=15 k 0V (a) Find the values of IC, IB and Vin when the transistor is just saturated. (3) 6 - 0.2 = 2.64 mA (1) 2.2103 3010-6 -3 IB = (2.6410 ) = 22 A (1) 3.610-3 Vin = IBRB + VBE = (2210-6) (15103)+ 0.8 =1.13V (1) IC = (b) Draw the input/output voltage characteristics of the transistor on the following diagram, inserting the results in part (a) and their corresponding values of Vout. (2) 6 5 4 3 2 1 0 0 1 2 3 4 5 6 Page 14 of 17 (c) Find the voltage amplification from the graph. G= (d) (2) Vout 6 - 0.2 = 0.8 - 1.13 = -17.6 Vin The voltage amplification of the above transistor circuit depends on the current amplification factor . Unfortunately, varies greatly even among the same model of transistors. The following circuit design overcomes this drawback. 6V IC RC Vout Vin IE RE 0V (i) Express IC in terms of Vout and RC. (1) Ic = (6 – Vout)/RC (ii) Express IE in terms of Vin and VBE. (1) IE = (Vin –VBE)/RE (iii) Deduce the voltage amplification of the circuit in terms of RC and RE. State the assumption you need. (3) IC IE before saturation (1) (6 – Vout)/RC = (Vin –VBE)/RE Vout RC = - R (2) Vin E Page 15 of 17 9. A mass of dry air rises above a patch of warm ground. As it rises its volume and pressure change adiabatically according the graph shown below. presure / kPa 100 90 80 70 60 50 40 30 20 10 volume / 103 m3 0 0 (a) (i) 1 2 3 4 5 6 7 8 9 Explain why the temperature of the air falls as the air rises. (2) Air does work as it expands. Energy comes from the gas’s internal energy. (ii) Use the graph to determine the change in internal energy of the air for the adiabatic change shown. (3) U = Q + W = W = P dV = area under the P-V graph 1 = 2 (100103 + 75103) [(8.5 – 7) 103] = 1.3 108 J (iii) The mean relative molecular mass of the air is 28 g and the density of the air at ground level is 1.2 kg m-3. How many moles of air are there? (2) At ground level, volume = 7000 m3 mass = 7000 1.2 = 8400 kg 8400 5 number of moles = -3 = 3 10 moles 2810 (iv) Calculate the temperature change of the air as it rises. (molar gas constant = 8.31 J K-1 mol-1) (2) P1 V1 = n R T1 and P2 V2 = n R T2 P2 V2 - P1 V1 = n R (T2 - T1) (75 103)(8.5103) - (100103)(7103) = (3105)8.31 T T = -25 K Page 16 of 17 (b) Another mass of air containing water vapour rises by the same amount. Some of the vapour condenses into water droplets. State and explain how the temperature change differs from that of the dry air. (3) On condensation, latent heat of vaporisation is released. The energy goes into the air. The air becomes higher in temperature or less temperature drop. 10. One nuclear power station produces an average useful output power of 800 MW. The mass of uranium 235 in the fuel rods is reduced to 70% of its original value in a four-year period. The average yield of each fission reaction is 200 MeV and 35% of this is converted into useful power. Charge of an electron, e Avogadro constant, NA (a) = 1.610-19 C = 6.01023 mol-1 Calculate the average power generated by fission. (1) 100 power generated = 800 MW 35 = 2.286 109 W (b) Calculate the average change in the mass for each fission. (2) Energy per fission = (200106) (1.610-19) = 3.210-11 J E 3.210-11 -28 m = c2 = kg 8 2 = 3.5610 (310 ) (c) Calculate the total number of fission reaction that occurs in the four-year period. (3) Energy per fission = (200106) (1.610-19) = 3.210-11 J 2.286 109 Number of fission in 1 second = = 7.1251019 3.210-11 Number of fission in 4 yr = 7.1251019 (4365243600) = 8.991027 (d) Hence, deduce the mass of uranium that was originally present in the reactor core. (3) Number of molecules under fission in 4 yr = 8.991027 8.991027 Number of moles = = 1.5 104 6.01023 Mass = (1.5104) (235 10-3) = 3521 kg 100 Original mass = 3521 kg 30 =11740 kg Page 17 of 17 (e) In such a reactor about 5% of the reactions produce the strontium isotope 90 38Sr . This nuclide is radioactive with a half-life of 28 y. (i) Calculate the initial rate at which strontium is being formed in the reactor. (1) initial rate = 7.1251019 0.05 = 3.5631018 s-1 (ii) Estimate the number of strontium atoms after one day and its activity. (3) Number of strontium atoms after one day = 3.5631018 (2460 60) = 3.0781023 Decay constant k = Activity ln 2 = 7.8510-10 s-1 28365246060 =kN END OF PAPER