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## Section A

1.D11.D2.E12.E3.A13.C4.D14.B5.A15.B6.C16.B7.E17.D8.C18.A9.B19.E10.E20.A## Section B

21.a.i. V_{horizontal}=V_{resultant}x cosq V=35xcos40^{o}V_{hor}= 26.81(m/s)

a.ii. V_{vertical}=V_{resultant}x sinq V_{ver}=35xsin40^{o}V_{ver}= 22.5(m/s)

a.iii.Maximum height reached when the vertical component of velocity (V_{ver}) is 0m/s. The vertical component of velocity calculated in part a.ii. is the initial velocity (u) for the purpose of this calculation. v_{ver}= 0(m/s) u_{ver}= 22.5(m/s) a = -9.8(m/s/s) t = ? v = u + at t = (v-u)/a t = (0-22.5)/-9.8t = 2.29(s)b. Horizontal distance is calculated by multiplying the total time of flight by the horizontal velocity, which is constant throughout the flight. Total time = [(2x2.29) + 0.48]s t_{total}= 5.06(s) V_{hor}=26.81(m/s) d = V_{hor}x t_{total}d = 26.81(m/s)x5.06(s)d = 135.66(m)22.a. DP_{s}= m_{s}(v_{s}-u_{s}) DP_{s}= 38(4.6-2.2) DP_{s}= 91.2(kgm/s) b. F_{avg}t_{c}= DP t_{c}= DP/F_{avg}t_{c}= 91.2/130tc. DP_{c}= 0.7(s)_{R}= -DP_{S}DP_{R}= -91.2(kgm/s) DP_{R}= m_{R}(v_{R}-u_{R}) -91.2 = 54(v_{R}-2.2)vd. If kinetic energy is conserved the interaction is elastic. Ek_{R}= 0.51(m/s)_{before}= 1/2M_{R}U_{R}^{2}+ 1/2M_{S}U_{S}^{2}Ek_{before}= 0.5x54x2.2^{2}+ 0.5x38X2.2^{2}Ek_{before}= 222.64(J) Ek_{after}= 1/2M_{R}V_{R}^{2}+ 1/2M_{S}V_{S}^{2}Ek_{after}= 0.5x54x0.51^{2}+ 0.5x38x4.6^{2}Ek_{after}= 409(J) The gain in kinetic energy indicates that the collision isNOT ELASTIC. The interaction is more like an explosion. 23.a.i. a.ii. P_{liquid}= rgh P_{liquid}= 1000x9.8x0.25Pa.iii.As the depth increases the total pressure acting on the air inside the tubing increases. Liquid enters the tube,compressing the air, until the total pressure acting on the air inside the tube is equal to the pressure of the air inside the tube. b. The pressure resulting from tank and water can be calculated using: P = F/A where F is the weight of the tank and water. F = w_{liquid}= 2450(Pa)_{total}= m_{total}g w_{total}= (2.7x10^{3}+300)9.8 w_{total}= 3000x9.8 w_{total}= 29400(N) A = 2.0x1.5 = 3.0m^{2}P = 29400/3 P = 9800(Pa) Atmospheric pressure is pushing down on the tank. This means that the total pressure acting on the surface that the tank is resting on is the sum of the atmospheric pressure and the pressure resulting from the weight of the tank. P_{total}= P_{tank+water}+ P_{atmospheric}P_{total}= 9800 + 1.01x10^{5}P24.a. P_{total}= 1.108x10^{5}(Pa)_{1}= 1.56x10^{5}Pa T_{1}= (27+273)K = 300k P_{2}= ? T_{2}= (77+273)K = 350k P_{1}/T_{1}= P_{2}/T_{2}P_{2}= P_{1}T_{2}/T_{1}P_{2}= 1.56x10^{5}x350/300Pb.i. P = I_{2}= 1.82x10^{5}Pa^{2}R_{element}I = V/R_{total}I = 30/(0.5+1.5) = 30/2 I = 15A P = 15^{2}x0.5P = 112.5Wb.ii. The output power of the element would beless. This is because the current in the circuit would be less because of the increase of the total series resistance of the circuit. Other explanations involving V_{tpd}are also valid. 25.a.i.Y-gain = 5volts/diva.ii.f = 1/T T = 2.5ms = 2.5x10^{-3}f = 1/2.5x10^{-3}f = 400Hzb.i. V_{RMS}= V_{Peak}/SQRT2 V_{RMS}= 12/1.414Vb.ii. E = 1/2(CV_{RMS}= 8.49V_{pk}^{2}) E = 1/2(220x10^{-6}x12^{2})E = 0.016Jb.iii. The reading on the ammeter increases because the current is directly proportional to the frequency. b.iv. The capacitor is constantly charging and discharging, as the polarity of the supply voltage changes in this ac circuit. The flow of charge during this process means that the current in the circuit does not fall to zero. In a dc circuit the capacitor charges and opposes the flow of charge from the supply. When fully charged no current flows in the circuit. 26.a. When the bridge is balanced: R_{1}/R_{2}= R_{3}/R_{4}Where R_{1}= 5500W R_{2}= R_{LDR}R_{3}= 330000W R_{4}= 150000W R_{LDR}= (R_{1}xR_{4})/R_{3}R_{LDR}= (5500x150000)/330000)Rb.i._{LDR}= 2500WMOSFET (n-channel)b.ii.V_{1}= V_{X}= 1.28V V_{2}= V_{Y}= 1.50V V_{ouput}= (R_{feedback}/R_{1})x(V_{2}-V_{1}) V_{ouput}= (22.5/1.5)x(1.5-1.28) V_{ouput}= 15x0.22Vb.iii.When water reaches the maximum level the beam of light is not totally internally reflected. This means that the light intensity incident on the LDR decreases, increasing its resistance and increasing V_{ouput}= 3.3V_{X}. This decreases the output voltage from the differential amplifier. When the output voltage falls below 2.0V the MOSFET does not conduct and the current in the solenoid falls to zero, closing the valve. 26.c. The incident angle at the glass water boundary is less than the critical angle so there is no total internal reflection. 27.a.i. The energy of the emitted photon is equal to the energy difference between levels.The longest wavelength radiation will have the lowest frequency and energy. The smallest energy gap is beween E_{4}and E_{3}a.ii. E_{photon}= E_{3}-E_{2}E_{photon}= (-2.4x10^{-19})-(-5.6x10^{-19}) E_{photon}= 3.2x10^{-19}J E_{photon}= hf f = E_{photon}/h f = 3.2x10^{-19}/6.63x10^{-34}f = 4.83x10b.i. f^{14}Hz_{air}= f_{glass}fb.ii. v_{glass}= 4.74x10^{14}Hz_{air}= f_{air}l_{air}l_{air}= v_{air}/f_{air}l_{air}= 3x10^{8}/4.74x10^{14}l_{air}= 632.9nm l_{glass}= l_{air}/n l_{glass}= 632.9/1.6l28.a.i. Maxima are produced when the path difference of the microwaves, from each gap, to a point along AB is zero or a whole number of wavelengths. This is because constructive interference occurs at these points as the waves are in phase. Minima are produced when the path difference of the microwaves, from each gap, to a point along AB is l/2, 3l/2, 5l/2, 7l/2 etc. This is because destructive interference occurs at these points as the waves are P out of phase. a.ii. PD = (766 - 682)mm PD = 84mm Third Max PD = 3l 3l = 84mm l = 84/3_{glass}= 395.6nml = 28mmb.i. Gain = -R_{feedback}/R_{input}Gain = -10kW/10kW Gain = -1 V_{output}/V_{input}= -R_{feedback}/R_{input}V_{output}= -V_{input}x(R_{feedback}/R_{input}) V_{output}(peak) = -6x(3.0x10^{3}/2.0x10^{3})Vb.ii. 29.a.i._{output}(peak) = -9V^{4}_{2}He is is an alpha particle. a.ii. A = N/t N = 7.2x10^{5}disintegrations t = 2 minutes = 120s A = 7.2x10^{5}/120A = 6000Bq = 6kBqb.i. Half value thickness(t_{1/2}) = thickness of barrier required to reduce the count rate by half. 60counts/s -> 30counts/s requires 3cm 30counts/s -> 15counts/s requires 3cmtb.ii. The bar indicates the uncertainty in the average reading. This results from the random nature of radioactive decay. c.i. H_{1/2}= 3cm_{total}= 6.4x10^{-5}SvAlphaQ_{a}= 20 H_{a}= Q_{a}xD_{a}H_{a}= 20D_{a}GammaQ_{g}= 1 D_{g}= 1.2x10^{-5}Gy H_{g}= QxD_{g}H_{g}= 1x1.2x10^{-5}Sv H_{g}= 1.2x10^{-5}Sv H_{total}= H_{a}+ H_{g}6.4x10^{-5}Sv = H_{a}+ 1.2x10^{-5}Sv H_{a}= 6.4x10^{-5}Sv - 1.2x10^{-5}Sv H_{a}= 5.2x10^{-5}Sv H_{a}= QD_{a}H_{a}= 20D_{a}20D_{a}= 5.2x10^{-5}Sv D_{a}= 5.2x10^{-5}/20D_{a}= 2.6x10^{-6}Gy = 2.6mGyReturn to past paper index page. ## END OF QUESTION PAPER