13099 [Turn over 6 A door performs simple harmonic motion when it is displaced and released at time t = 0. The motion of the door is damped, enabling the door to eventually come to rest in a closed position. The amount of damping can be varied. Sketch graphs to show how the displacement d of the door varies with time t in each of the following situations. The time period of the door oscillating is T. (i) The door experiences light damping. T 2T t d 0 0 [3] (ii) The door is overdamped. T 2T t d 0 0 [2] (iii) The door experiences critical damping. T 2T t d 0 0 [2] 13099 7 A spring with a spring constant k of 209 N m-1 is suspended from a fixed point and a 650 g mass is attached to its lower end. The mass is raised slightly and the position of the bottom of the mass is noted to be at the 0.270 m mark on a half-metre rule. The mass is released and at the same time a stopwatch is started. The mass oscillates between the 0.270 m and 0.310 m marks on the half-metre rule as shown in Fig. 7.1. spring 650 g half-metre rule 0.270 m 0.310 m fixed point Fig. 7.1 (a) Calculate the period of oscillation for the spring mass system. Period of oscillation =                     s [2] 13099 [Turn over (b) Calculate the maximum acceleration of the mass. Maximum acceleration =                     m s-2 [4] (c) Calculate the first time the oscillating mass passes the 0.295 m mark on the half-metre rule. Time =                     s [3] 13099 BLANK PAGE DO NOT WRITE ON THIS PAGE 13099 [Turn over 8 (a) (i) The relationship between volume, pressure and temperature of an ideal gas can be described using gas laws. State the law which describes the relationship shown in Fig. 8.1 where p is the pressure and V is the volume. pV V Fig. 8.1 [2] (ii) Draw another line on Fig. 8.1 to show how the graph changes if the temperature of the gas is lower. [1] 13099 A car tyre is inflated to a pressure of 210 kPa above the atmospheric pressure of 101 kPa. The temperature of the air in the tyre is 13.0 °C. The car is taken for a long drive after which the tyre pressure increases by 15.0 kPa. Assume that the volume of the tyre remains constant. (b) Calculate the temperature of the air in the tyre at the end of the drive. Temperature                     °C [5] (c) (i) Calculate the average kinetic energy of one of the air molecules in the tyre before the long drive. Average kinetic energy of one molecule =              J [2] 13099 (ii) If the volume of the air in the tyre is 1.03 × 10−2 m3, calculate the total kinetic energy of all the air molecules in the tyre. Total kinetic energy =                     J [5] THIS IS THE END OF THE QUESTION PAPER Permission to reproduce all copyright material has been applied for. 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For Examiner’s use only Question Number Marks 1 2 3 4 5 6 7 8 Total Marks Examiner Number DO NOT WRITE ON THIS PAGE APH11/6 262817 DATA AND FORMULAE SHEET Physics [APH11/APH21] ADVANCED General Certifi cate of Education Assessment Units A2 1 and A2 2 12159.03 12159.03 2 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 ε0 = 8.85 × 10–12 F m–1 1 4πε0 = 8.99 × 109 F–1 m elementary charge e = 1.60 × 10–19 C the Planck constant h = 6.63 × 10–34 J s (unifi 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 the Hubble constant H0 ≈ 2.4 × 10–18 s–1 12159.03 3 Useful formulae The following equations may be useful in answering some of the questions in the examination: Mechanics conservation of energy 1 2 mv 2 – 1 2 mu 2 = Fs for a constant force Hooke’s Law F = kx (spring constant k) strain energy E = 1 2 Fx = 1 2 kx 2 Uniform circular motion centripetal Force F = mv 2 r Simple harmonic motion displacement x = A cos ωt simple pendulum T = 2π loaded spiral spring T = 2π Waves two-source interference λ = ay d diffraction grating d sin θ = n λ l g m k 12159.03 4 Thermal physics average kinetic energy of a molecule 1 2 m Gc2H = 3 2 kT kinetic theory pV = 1 3 Nm Gc2H thermal energy Q = mc∆θ Capacitors capacitors in series 1 C = 1 C1 + 1 C2 + 1 C3 capacitors in parallel C = C1 + C2 + C3 time constant τ = RC capacitor discharge Q = Q0e –t CR or V = V0e –t CR or I = I0e –t CR Light lens formula 1 u + 1 v = 1 f Electricity terminal potential difference V = E – Ir (e.m.f., E; Internal Resistance, r) potential divider Vout = R1Vin R1 + R2 a.c. generator E = BANωsinωt 12159.03 5 Nuclear Physics nuclear radius r = r0A 1 3 radioactive decay A = –λN, A = A0e–λt half-life t 1 2 = 0.693 λ Particles and photons Einstein’s equation 1 2 mvmax2 = hf – hf0 de Broglie equation λ= h p Astronomy red shift z = Δλ λ recession speed z = v c Hubble’s law v = H0 d