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1.a. Squaring the equation v = at gives: v^{2}= a^{2}t^{2}equation 1The equation s = 1/2(at^{2}) can be rearranged to give: t^{2}= 2s/a which can be substituted intoequation 1v^{2}= a^{2}(2s/a) =>vb.i. F^{2}= 2as_{unbalanced}= F_{thrust}- F_{friction}F_{unbalanced}= 3150 - 450 F_{unbalanced}= 2700N a = F_{unbalanced}/m a = 2700/1000a = 2.7m/s/sb.ii. To calculate the length of the track use : v^{2}= u^{2}+ 2as s = (v^{2}- u^{2})/2a s = (33^{2}- 0^{2})/2x2.7s = 201.67mc. Use the cosine rule to calculate the resultant velocity,or, draw an accurate scale diagram. a^{2}= b^{2}+ c^{2}-2bcCosA a^{2}= 36^{2}+ 12^{2}-2x36x12Cos140 a^{2}= 1296 + 144 - 864Cos140 a^{2}= 1296 + 144 - 864x-0.766 a^{2}= 2101.9 a = 45.8m/s Use the sine rule to calculate angle C. a/sinA = c/sinC sinC = csinA/a sinC = 12sin140^{o}/45.8 sinC = 0.168 angle C = 9.7^{o}Resultant velocity = 45.8m/s with a bearing of 350.32.a.i. V^{o}_{horizontal}= Vcos36^{o}V_{horizontal}= 41.7cos36^{o}Va.ii. V_{horizontal}= 33.74m/s_{vertical}= Vsin36^{o}V_{vertical}= 41.7sin36^{o}Vb. The time for the ball to reach Q(t_{vertical}= 24.5m/s_{total}) can be calculated by summing the time to travel from O to P(t_{1}) and the time to travel from P to Q(t_{2}). t_{total}= t_{1}+ t_{2}When P is reached the vertical component of the velocity is 0m/s. This velocity is the final velocity(v) of the first part of the journey. The initial component of the vertical velocity(u) for this part of the journey is 24.5m/s. v = 0m/s v = u +at_{1}u = 24.5m/s t_{1}= (v-u)/a a = -9.8m/s/s t_{1}= (0-24.5)/-9.8 t_{1}= ? t_{1}= 2.5s When falling from P to Q take the initial velocity(u) as 0m/s for this part of the journey. u = 0m/s s = ut_{2}+ 1/2(at_{2}^{2}) s = -19.6m s = 1/2(at_{2}^{2}) a = -9.8m/s t_{2}^{2}= 2s/a t_{2}= ? t_{2}^{2}= 2x-19.6/-9.8 t_{2}^{2}= 4 t_{2}= 2s t_{total}= t_{1}+ t_{2}t_{total}= 2.5 + 2t(As required) c. S_{total}= 4.5s_{horizontal}= V_{horizontal}xt_{total}S_{horizontal}= 33.74x4.5S3.a. Use the law of conservation of momentum to solve this problem._{horizontal}= 151.8m/sP_{before}= P_{after}Before collisionP_{before}= m_{P}u_{P}+ m_{Q}u_{Q}P_{before}= 0.2x0.5 + 0.3x0 P_{before}= 0.1kgm/sAfter collisionP_{after}= m_{P}v_{P}+ m_{Q}v_{Q}0.1 = 0.2v_{P}+ 0.3x0.4 v_{P}= (0.1 - 0.3x0.4)/0.2vThe negative sign indicates the direction is to the left. b. DP_{P}= -0.1m/s_{p}= P_{p}(after) - P_{p}(before) P_{p}(after) = m_{P}v_{P}P_{p}(after) = 0.2x-0.1 P_{p}(after) = -0.02kgm/s P_{p}(before) = m_{P}u_{P}P_{p}(before) = 0.2x0.5 P_{p}(before) = 0.1kgm/s DP_{p}= -0.02-0.1DPc.i. F_{p}= -0.12kgm/s_{average}t = m_{P}(u_{P}- v_{P}) F_{average}= m_{P}(u_{P}- v_{P})/t F_{average}= 0.2[(0.5-(-0.1)]/0.06FNote: the direction of this force is to the left. c.ii. 4.a. The assumption in this experiment is that the gas in the container is the same as the water temperature. To facilitate this the can must be fully immersed to allow the gas and water temperature to equilibrate. b. P_{average}= 2N_{1}= 100kPa T_{1}= (17+273)K = 290K P_{2}= ? T_{2}= (75+273)K = 348K P_{1}/T_{1}= P_{2}/T_{2}P_{2}= P_{1}T_{2}/T_{1}P_{2}= 100x348/290Pc. P = F/A F = PA F = 120x10_{2}= 120kPa^{3}x0.001F = 120Nd. The mass and the volume of gas are fixed in this experiment,therefore, the density(mass/volume) must also remain constant. 5.a. A battery emf of 12V will supply 12J of energy to each coulomb of charge passing through the cell. b.i. P = V^{2}/R R = V^{2}/P R = 12^{2}/48R = 3Wb.ii. When S_{1}is closed the circuit can be treated as a simple series circuit with the internal resistor(r) in series with the headlamp resistance(R_{h}). R_{total}= R_{h}+ r R_{total}= 3 + 0.05 R_{total}= 3.05W I = emf/R_{total}I = 12/3.05 I = 3.93A V_{headlamp}= IR_{h}V_{headlamp}= 3.93x3Vc.i. The bulb and the starter motor are in parallel. The resistance of this network R_{headlamp}= 11.8V (as required)_{p}is calculated as shown below: 1/R_{p}= 1/R_{bulb}+ 1/R_{motor}1/R_{p}= 1/3 + 1/0.12 1/R_{p}= 8.66... R_{p}= 0.1154W R_{total}= R_{p}+ r R_{total}= 0.115 + 0.05Rc.ii. I = emf/R_{total}= 0.165W_{total}I = 12/0.165I = 72.7A6.a.i. Gain = V_{output}/V_{input}Gain = -9x30mv/5x2mV Gain = -270mV/10mvGain = 27a.ii. V_{peak}= 1.414V_{rms}V_{rms}= V_{peak}/1.414Va.iii.Gain = V_{rms}= 190.9mV_{output}/V_{input}=-R_{f}/R_{1}To produce a gain of magnitude 27 choose: R_{f}= 270kW R_{1}= 10kW Other values are acceptable as long as the ratio of R_{f}to R_{1}is 27. b.i. The bridge is balanced when the digital voltmeter reads 0V. This means the potential at X and Y are equal and there is no potential difference between these points. b.ii. V_{output}= (-R_{f}/R_{-})(V_{-}-V_{+}) (V_{-}-V_{+}) = -V_{output}(R_{-}/R_{f}) (V_{-}-V_{+}) = -(-0.18)(100x10^{3}/1x10^{6}) (V_{-}-V_{+}) = 0.18(0.1) (V_{-}-V_{+}) = 0.018VVb.iii.At 20_{Y}-V_{X}= 0.018V^{o}C the potential at Y(V_{Y}) falls to a value less than it was when the bridge was balanced at 23^{o}C. As V_{Y}is less than V_{X}=> V_{Y}-V_{X}is negative. => V_{output}=(-R_{f}/R_{-})(V_{Y}-V_{X}) V_{output}= positive value. Note: If the resistance of the thermistor varies linearly with temperature in range 20^{o}C to 26^{o}C the output voltage at 20^{o}C will be+0.18V. 7.a.i. All the supply voltage is across the resistor at the instant the switch is closed. At this instant V_{supply}= V_{resistor}. I = 100x10^{-6}R = 150,000W V_{resistor}= ? V_{resistor}= IR_{resistor}V_{resistor}= 100x10^{-6}x150000 V_{resistor}= 15VVa.ii. The current in the circuit must first be calculated when the voltage across the resistor is 6V. I = ? V_{supply}= V_{resistor}= 15V_{resistor}= 6V R = 150,000W I = V/R I = 6/150000I = 40x10From the graph read the time when the current is 40mA.^{-6}A = 40mATime = 40sa.iii.V_{supply}= V_{resistor}+ V_{capacitor}V_{capacitor}= V_{supply}- V_{resistor}V_{capacitor}= 15 - 6Vb.i.A Q = CV C = 16mF = 16x10_{capacitor}= 9V^{-6}V = 6kV = 6000V (This is the voltage across the capacitor when fully charged) Q = ? Q = 16x10^{-6}x 6000Q = 0.096Cb.i.B E = 1/2(QV) E = 1/2(0.096X6000)E = 288Jb.ii. I = Q/t I = 0.096/2x10^{-3}I = 48A8.a.i. A maxima occurs when two waves interfere constructively. This happens when waves are in phase. A minima occurs when two waves interfere destructively. This happens when the two waves are 180^{o}out of phase. a.ii. The minima is produced when the path difference YQ-XQ is equal to 3/2l. YQ-XQ = (5.2-4.0)m = 1.2m YQ-XQ = 3/2l l=(2x1.2)/3l = 0.8mb.i. The speed of sound is constant in air, therefore increasing the frequency will decrease the wavelength. If the frequency is increased by a factor of 5 the wavelength will decrease by a factor of 5.Wavelength at 1000Hz = 1.2/5 = 0.24mb.ii. If the path difference is fixed at 1.2m this will represent a whole number of wavelengths for certain frequencies and produce maxima, however, for other frequencies this will represent a whole number of half wavelengths and produce minima. For this reason a series of maxima and minima are produced. 9.a. n_{plastic}= sinq_{air}/sinq_{plastic}n_{plastic}= sin15^{o}/sin10^{o}(Note that it is the angle between the ray and the normal that must be used.)nb.i. q_{plastic}= 1.49_{crit}= sin^{-1}(1/n_{glass}) q_{crit}= sin^{-1}(1/1.44) q_{crit}= sin^{-1}(0.694)qb.ii. Ray Q will be refracted and partially reflected. Ray P will be totally internally reflected because the incident angle of 45_{crit}= 44.0^{o}^{o}is greater than the critical angle. 10.a.i. a.ii. Individual electrons in the metal atom absorb the energy of a single photon, the energy of which is dependent on the frequency of the photon. The photon energy can be calculated using the equation E = hf. Below a certain frequency the energy absorbed by an electronin the metal, from the photon, is below that required to escape from the metal atom, thus the current in the circuit will be zero. b.i. The threshold frequency will provide electrons with just enough energy to overcome the work function of the metal and eject electrons with no excess kinetic energy. Thus to estimate the work function from the graph the line must be extrapolated until it cuts the x-axis of the graph. X-axis intercept = 6.7x10^{14}Hz This is the threshold frequency. b.ii. The work function is calculated using: E_{work function}= hf_{threshold}E_{work function}= 6.63x10^{-34}x6.6x10^{14}E_{work function}= 4.4x10^{-19}J This is closest to the work function of the metalcalcium11.a. Alpha particles have a positive charge. It was therefore concluded that electrostatic repulsion from a like charge, of very large mass, was required to produce a large angle deflection. However, as large angle deflections were very rare, and most alpha particles were undeviated from their original path, it led Rutherford to conclude that most of the atom was empty space with the large positive mass concentrated at the centre of the atom. b.i. The total dose equivalent received is calculated using the equation: H =QD. H_{gamma}= Q_{gamma}D_{gamma}H_{gamma}= 1x200 H_{gamma}= 200mSv H_{neutrons}= Q_{neutrons}D_{neutrons}H_{neutrons}= 3x100 H_{neutrons}= 300mSv H_{total}= H_{gamma}+ H_{neutrons}H_{total}= 200 + 300Hb.ii. Three half value thicknesses are required._{total}= 500mSv3t_{1/2}= 3x8 = 24mm## END OF QUESTION PAPER