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GCE JUN 2010 : A2 2 Fields and their Applications - Revised

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Sp N ec e i w ca tio n Centre Number 71 Candidate Number ADVANCED General Certificate of Education 2010 Physics Fields and their Applications AY221 Assessment Unit A2 2 [AY221] WEDNESDAY 9 JUNE, MORNING TIME 1 hour 30 minutes. INSTRUCTIONS TO CANDIDATES Write your Centre Number and Candidate Number in the spaces provided at the top of this page. Answer all questions. Write your answers in the spaces provided in this question paper. INFORMATION FOR CANDIDATES The total mark for this paper is 90. Quality of written communication will be assessed in question 5(a). Figures in brackets printed down the right-hand side of pages indicate the marks awarded to each question. Your attention is drawn to the Data and Formulae Sheet which is inside this question paper. You may use an electronic calculator. Question 9 contributes to the synoptic assessment required of the specification. Candidates should allow approximately 15 minutes to complete this question. For Examiner s use only Question Marks Number 1 2 3 4 5 6 7 8 9 Total Marks 5517 BLANK PAGE 5517 2 [Turn over 1 (a) State Newton s law of universal gravitation. Examiner Only Marks Remark _______________________________________________________ _______________________________________________________ _____________________________________________________ [2] (b) The International Space Station of mass ms, orbits the Earth at a height h above the Earth s surface. Take the Earth s mass as me and radius re. (i) State an equation for the gravitational force F which exists between the International Space Station and the Earth. [1] (ii) Calculate the value of the Earth s gravitational field strength g at a height 350 km above the Earth s surface. Take the mass of the Earth to be 6.0 1024 kg and the mean radius of the Earth to be 6400 km. g = _________ N kg 1 [3] (iii) Explain why an astronaut in the International Space Station experiences apparent weightlessness, even though the Earth s gravitational field acts on him. ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ _________________________________________________ [3] 5517 3 [Turn over 2 (a) State Coulomb s Law for the force F between two point charges. Examiner Only Marks Remark _______________________________________________________ _______________________________________________________ _____________________________________________________ [2] (b) A point charge q1 of charge 5 C is suspended by a thread of length 0.60 m from a point A and hangs in position B as shown in Fig 2.1. A 0.60 m q1 B Fig. 2.1 A second point charge q2 of charge 3 C is brought to, and held at, point D. Point D is a horizontal distance of 0.6 m and a vertical distance of 0.3 m below position A, as shown in Fig. 2.2. Point charge q1 will move under the influence of the second charge q2 to a new equilibrium position at point C as shown in Fig. 2.2. 0.60 m A 0.30 m P q1 0.60 m q2 C x B q1 D Path taken by q1 Diagram not drawn accurately Fig. 2.2 5517 4 [Turn over (i) Complete the force diagram (free-body diagram) in Fig. 2.3 for the point charge in the equilibrium position C by drawing another arrow. What is the name of the force represented by the arrow you have drawn? Tension q1 Electrostatic attraction [1] Fig. 2.3 Examiner Only Marks Remark (ii) Show that the distance x from C to D, as labelled in Fig. 2.2, is 0.08 m. [2] (iii) Calculate the electrostatic force acting on charge q1 due to charge q2 when q1 is in position C. Force = ________________ N [2] (iv) 1. The thread hangs at an angle to the horizontal, see Fig. 2.2. Show that = 30 . [1] 2. Hence, calculate the tension in the string. Tension = ________________ N 5517 [2] 5 [Turn over 3 Fig 3.1 shows a circuit which can be used to charge a capacitor C of capacitance 6 F. S1 Examiner Only Marks Remark S2 C V 12.0 V R Fig. 3.1 The switch S1 is closed at time t = 0. (a) On Fig 3.2, show how the voltage,V, recorded by the voltmeter varies with time t during the charging of the capacitor. Label the final value for V on the y-axis. V/V t/s 0 0 [2] Fig. 3.2 5517 6 [Turn over (b) Explain how the movement of charge carriers in the circuit, after the switch is closed, can explain the shape of the graph you have drawn in Fig. 3.2. Examiner Only Marks Remark _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _____________________________________________________ [3] (c) The switch is now moved to position S2 and the capacitor discharges through resistor R. After 48 seconds, the voltage has fallen from 12.0 V to 1.64 V. Use these data to calculate the size of resistance R. R = _______ M 5517 [4] 7 [Turn over 4 (a) (i) State Faraday s law of electromagnetic induction. Examiner Only Marks Remark ___________________________________________________ ___________________________________________________ _________________________________________________ [2] (ii) Describe an experiment in which Faraday s law of electromagnetic induction can be demonstrated. Your answer should include 1. a labelled diagram of the apparatus 2. an explanation as to how the results or observations demonstrate Faraday s law. ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ _________________________________________________ [5] 5517 8 [Turn over (b) A loop of wire is placed in the magnetic field produced by an electromagnet. The loop of wire has a resistance of 2.6 and an area of 4.0 10 3 m2. Examiner Only Marks Remark The electromagnet, when switched on, takes 0.8 s to reach its maximum flux density of 600 T. Assuming all the field links with the loop, and the loop s area is perpendicular to the field, calculate the average current that flows in the wire in the 0.8 s after the electromagnet is switched on. Current = ___________ A 5517 [3] 9 [Turn over 5 Fig. 5.1 shows a diagram of a transformer. Examiner Only Marks Remark Iron core Secondary coil Primary coil Fig. 5.1 (a) Describe how the transformer works. _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _____________________________________________________ [3] Quality of written communication [2] (b) A transformer steps down voltage from 240 V to 19 V. The maximum current drawn from the transformer is 3.3 A. Assuming the transformer is 100% efficient, calculate the current drawn from the supply. Current = ______________ A 5517 [2] 10 [Turn over (c) In practice, a transformer is not 100% efficient. Describe how the current drawn from the supply, in (b), would need to be changed, if at all, if the current drawn from the transformer is to remain 3.3 A. Examiner Only Marks Remark _______________________________________________________ _______________________________________________________ _____________________________________________________ [1] (d) State one source of energy loss in a transformer and explain how it can be reduced. _______________________________________________________ _______________________________________________________ _____________________________________________________ [2] 5517 11 [Turn over 6 A beam of helium ions of mass m, charge q and travelling at a speed of v, enters a region EFGH at right angles as shown in Fig. 6.1. The region EFGH is a square of side 8 cm and a uniform magnetic field of flux density B acts vertically out of the plane of the page throughout this region. Examiner Only Marks Remark G F 8 cm E H 4 cm Ions Fig. 6.1 (a) (i) Explain why the helium ions follow a circular path when in the region EFGH. ___________________________________________________ ___________________________________________________ ___________________________________________________ _________________________________________________ [2] (ii) Show that the radius r of the path taken by the ions while in the mv region EFGH is given by r = . Bq [2] 5517 12 [Turn over (iii) If the ions have a mass of 6.6 10 27 kg, a charge of 3.2 10 19 C and are travelling at a speed of 1.55 106 m s 1, calculate the radius of the path of the ions if the magnetic field has a flux density of 0.80 T. Give your answer to the nearest cm. r = ____________ cm Examiner Only Marks Remark [2] (iv) On Fig. 6.1, sketch the path taken by the ions within region EFGH and show the direction they travel when they leave the region EFGH. Label this path P. [2] (b) The value of the magnetic flux density is increased to 1.60 T. (i) Sketch the new path taken by the ions. Label this path Q. [1] (ii) Explain why the ions take this new path. ___________________________________________________ ___________________________________________________ _________________________________________________ [1] 5517 13 [Turn over 7 (a) Explain what is meant by antimatter. Examiner Only Marks Remark _______________________________________________________ _______________________________________________________ _____________________________________________________ [2] (b) At the CERN laboratory, antiprotons have been formed using protons accelerated in a synchrotron. The protons are smashed into an iridium rod. The antiprotons produced are separated off using magnets in a vacuum. (i) Draw and label a diagram to show the structure of a synchrotron. [2] (ii) Explain how the synchrotron causes a proton to achieve the kinetic energy needed to form antiprotons during the collision with the iridium. ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ _________________________________________________ [2] 5517 14 [Turn over (c) The annihilation of a positron occurs when it meets an electron. This is represented by the equation Examiner Only Marks Remark 0 e + 1e & 2 0 +1 where is a photon. (i) Explain why two photons are produced. ___________________________________________________ _________________________________________________ [1] (ii) Calculate the energy E of each photon produced. E = ____________ J 5517 [2] 15 [Turn over 8 (a) (i) Some physicists consider the gauge bosons to be fundamental particles. How do we define a fundamental particle? Examiner Only Marks Remark ___________________________________________________ _________________________________________________ [1] (ii) Give one example of a fundamental particle other than a gauge boson. _______________ [1] (b) (i) Two electrons approaching each other do not collide, but exert forces on one another without coming into contact. Explain the role of the GAUGE BOSON in this interaction. ___________________________________________________ _________________________________________________ [2] (ii) Complete Table 8.1 by naming the appropriate gauge boson for each of the fundamental forces. Table 8.1 Force Gauge Boson Strong Electromagnetic Weak Gravitational [2] (c) (i) Name the two types of particle that are classified as hadrons. 1. ____________________ 2. ____________________ [1] (ii) In what way are these hadrons different? ___________________________________________________ ___________________________________________________ _________________________________________________ [1] 5517 16 [Turn over 9 (a) (i) When a golf ball is hit by the head of a golf club, kinetic energy is transferred from the golf club head to the ball on impact. Assuming that all the kinetic energy is transferred from the club head to the ball and that the club head is brought to rest when it hits the ball, determine the velocity of the ball after impact. The mass of the golf ball is 46 g and that of the golf club head is 190 g. The velocity of the club head immediately before impact is 44.7 m s 1. Velocity = ______________ m s 1 Examiner Only Marks Remark [3] (ii) In practice, measurements using high speed cameras have shown that the ball velocity in this situation is 63.5 m s 1. Explain why there is such a difference between the theoretical value obtained in (a)(i) and the practical value obtained through high speed photography. ___________________________________________________ ___________________________________________________ _________________________________________________ [2] (b) The Coefficient of Restitution (CoR), as defined below in Equation 9.1, provides a numerical measure of the club head ball interaction. (velocity of ball after impact) CoR = (velocity of club head before impact) Equation 9.1 (i) Calculate the CoR for this situation. Use the measured value for velocity from (a)(ii). CoR = ______________ [1] (ii) Explain why a high CoR is desirable. ___________________________________________________ ___________________________________________________ _________________________________________________ [1] 5517 17 [Turn over (c) The CoR of the club head is related to its natural frequency of vibration. Explain the meaning of the term in bold. Examiner Only Marks Remark _______________________________________________________ _______________________________________________________ _____________________________________________________ [1] (d) It is found that the lower the natural frequency of vibration, the higher the CoR. The natural frequency of vibration of the head is given by the relationship (thickness)3 Young Modulus Natural frequency = Area hardness (i) Table 9.1 below shows the relationship between hardness and the Young Modulus for four materials whose club heads are of the same thickness and area. Which material would give the highest CoR? Table 9.1 Material Hardness/units Young Modulus/units A 14.7 1.36 B 10.5 8.44 C 67.8 2.77 D 56.6 7.34 Material __________ [1] (ii) Explain your choice. ___________________________________________________ ___________________________________________________ _________________________________________________ [1] 5517 18 [Turn over (e) The golf ball is struck and leaves the tee with a velocity of 63.5 m s 1 at an angle of 30o to the horizontal. Calculate the horizontal distance travelled by the golf ball when it first hits the ground. Distance = _____________ m Examiner Only Marks Remark [5] THIS IS THE END OF THE QUESTION PAPER 5517 19 [Turn over 1847-049-1 5517 GCE Physics Data and Formulae Sheet for A2 1 and A2 2 Values of constants speed of light in a vacuum c = 3.00 108 m s 1 permittivity of a vacuum 0 = 8.85 10 12 F m 1 1 = 8.99 109 F 1 m 4 0 elementary charge e = 1.60 10 19 C the Planck constant h = 6.63 10 34 J s (uni ed) atomic mass unit 1 u = 1.66 10 27 kg mass of electron me = 9.11 10 31 kg mass of proton mp = 1.67 10 27 kg molar gas constant R = 8.31 J K 1 mol 1 the Avogadro constant NA = 6.02 1023 mol 1 the Boltzmann constant k = 1.38 10 23 J K 1 gravitational constant G = 6.67 10 11 N m2 kg 2 acceleration of free fall on the Earth s surface g = 9.81 m s 2 electron volt 1 eV = 1.60 10 19 J *AY221INS* 5517.02 AY221INS The following equations may be useful in answering some of the questions in the examination: Mechanics Conservation of energy 1 1 mv 2 mu 2 = Fs 2 2 Hooke s Law F = kx (spring constant k) for a constant force Simple harmonic motion Displacement x = A cos t Sound intensity level/dB = 10 lg10 Two-source interference = Sound I I0 Waves ay d Thermal physics Average kinetic energy of a molecule 1 3 m c2 = kT 2 2 Kinetic theory 1 pV = Nm c2 3 Thermal energy Q = mc Capacitors Capacitors in series 1 = 1 + 1 + 1 Capacitors in parallel Time constant 5517.02 C = C1 + C2 + C3 = RC 2 [Turn over Light Lens formula Magnification 1 = vf v m=u u + Electricity Terminal potential difference Potential divider V = E Ir (E.m.f. E; Internal Resistance r) R1Vin Vout = R +R Particles and photons Radioactive decay A = N A = A0e t Half-life t1 = 0. 693 de Broglie equation = h p 2 The nucleus Nuclear radius 5517.02 1 r = r0 A3 3 5517.02 [Turn over

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Additional Info : Gce Physics June 2010 Assessment Unit A2 2, Fields and their Applications - Revised
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