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1.a.i. Component of the force acting parallel to the slope(F_{s}) is calculated using: F_{s}= mgsinq F_{s}= 2.0x9.8xsin20^{o}Fa.ii. a.iii.To calculate the acceleration use [N2] a = F_{s}= 6.7N_{unbalanced}/m F_{unbalanced}= F_{Friction}+F_{s}F_{unbalanced}= -1.3 + (-6.7) Where vectors acting down the slope are taken as negative. F_{unbalanced}= -8N a = -8/2.0 a = -4m/s^{2}Which equates to a deceleration of 4m/s^{2}, as required. a.iv. At its furthest point up the slope the trolley will be at rest. v = 0m/s v = u + at u = 3m/s t = (v-u)/a a = -4m/s^{2}t = (0-3)/-4 t = ?t = 0.75sa.v. v^{2}= u^{2}+ 2as s = (v^{2}- u^{2})/2a s = (0^{2}- 3^{2})/(2x-4) s = -9/-8s = 1.125mb.i. b.ii. The force vectors in the above diagram are acting in opposite directions. This will produce a smaller unbalanced force than that calculated in part a.iii. This means that the magnitude of the acceleration will be less when the trolley moves down the slope. 2.a. The weight of the student is equal to the scale reading when the lift is moving at a steady speed. This weight is used to calculate the mass of the student. w = mg m = w/g m = 588/9.8m = 60kgb. The boy is accelerating at the same rate as the lift. The unbalanced force producing this acceleration is equal to the scale reading when accelerating minus the steady state scale reading. The diagram below illustrates this with the scale reading represented by the the upward force labelledTand downward force, weight, represented byW. F_{unbalanced}= 678-588 F_{unbalanced}= 90N a = F_{unbalanced}/m a = 90/60a = 1.5m/sc. The lift is decelerating when the upward force is less than the weight. F^{2}_{unbalanced}= 498 - 588 F_{unbalanced}= -90N The negative indicates that the unbalanced force acts in the downwards direction. a = F_{unbalanced}/m a = -90/60 a = -1.5m/s^{2}This can be thought of as an upward deceleration of 1.5m/sd. 3.a.i. V = d/t^{2}_{1}d = diameter of ball = 24mm = 0.024m t = time for ball to pass through light gate = 0.060s V = 0.024/0.060V = 0.4m/sa.ii. F_{avg}= change in momentum/contact time F_{avg}= (mv-mu)/t F_{avg}= m(v-u)/t F_{avg}= 0.045(0.4-0)/0.005Fb.i. Percentage error in mass = (0.01/45)x100 = 0.02% Percentage error in contact time = (0.001/0.005)x100 = 20% Percentage error in t_{avg}= 3.6N_{1}= (0.001/0.060)x100 = 1.67% Percentage error in ball diameter = (1/24)x100 = 4.17% The greatest uncertainty is in the contact time measurement. b.ii. The percentage error in the result can be taken as equal to the largest individual error. 20% of 3.6N = (20/100)x3.6 = 0.72NF4.a. Initial volume of gas (V_{avg}= (3.6 +- 0.72)N_{1}) in the container = (8-5)litres = 3 litres = 3 x10^{-3}m^{3}Density of gas (r) = 1.29kg/m^{3}m = rV_{1}m = 1.293x(3x10^{-3}m^{3})m = 3.87x10b.i. Pressure before pumping (P^{-3}Kg_{before}) = 1.01x10^{5}Pa Pressure after pumping (P_{1}) = 3.0x10^{5}Pa Volume of air before and after pumping(V_{1}) is fixed at 3x10^{-3}m^{3}Area(A) = 7.0x10^{3}m^{2}To calculate the force, after pumping, use: P_{1}= F/A F = P_{1}A F = 3.0x10^{5}x7.0x10^{-3}F = 2100Nb.ii. When the volume of the water is 2 litres, the volume of the gas in the container(V_{2}) is 6 litres,or 6x10^{-3}m^{3}. The pressure at this time is P_{2}. P_{1}V_{1}= P_{2}V_{2}P_{2}= P_{1}V_{1}/V_{2}P_{2}= 3x10^{5}x3x10^{-3}/6x10^{-3}P5.a.i. V_{2}= 1.5x10^{5}Pa_{tpd}= emf - Ir I = emf/(R+r) I = 1.5/(3+0.75) I = 1.5/3.75 I = 0.4A V_{tpd}= 1.5 - 0.4x0.75Va.ii. V_{tpd}= 1.2V_{lost}= Ir or emf = V_{tpd}+V_{lost}V_{lost}= 0.4x0.75 V_{lost}= emf - V_{tpd}VV_{lost}= 0.3V_{lost}= 1.5 - 1.2 = 0.3V b.i. R = 1W V_{lost}= 2V V_{tpd}= emf - V_{lost}V_{tpd}= 6 - 2Vb.ii. I = V_{tpd}= 4V_{tpd}/R_{variable}I = 4/1 I = 4A V_{lost}= Ir V_{lost}= emf - V_{tpd}r = V_{lost}/I V_{lost}= 6 - 4 = 2V r = 2/4r = 0.5W6.a. A capacitor of value 5mF will store 5 coulombs of charge per volt across it. b.i. Initially all the supply voltage is across the resistor in the circuit. The current in the circuit at this point(I_{initial}) is 30mA, as read from the graph. I_{initial}= 30mA = 30x10^{-6}A V = 6V R = ? R = V/I R = 6/30x10^{-6}R = 2x10b.ii. Halving the value of the resistor will double the initial charging current. Lowering the resistance will also decrease the charging time. c.i. The resistance of the variable resistor must be decreased to keep the charging current constant. c.ii. To calculate the total charge transferred use: Q = It The values for I and t are read from the graph. Q = ? I = 100mA = 100x10^{5}W^{-6}A t = 10s Q = Ixt Q = 100x10^{-6}x10Q = 100x10c.iii.To calculate the capacitance use: Q = CV^{-5}C = 1x10^{-3}C = 1mC_{capacitor}At 10s V_{capacitor}= 5V Q = 1x10^{-3}C V_{capacitor}= 5V C = ? C = Q/V_{capacitor}C = 1x10^{-3}/5C = 2x107.a.i. To calculate the kinetic energy(E^{-4}F = 200mF_{k}) use : E_{k}= mv^{2}/2 E_{k}= ? m_{e}= 9.11x10^{-31}kg v = 4.2x10^{7}m/s E_{k}= mv^{2}/2 E_{k}= 9.11x10^{-31}x(4.2x10^{7})^{2}/2E7.a.ii. The gain in the kinetic energy of the electron, with charge e, is a result of it being accelerated by a potential difference V. E_{k}= 8.03x10^{-16}J_{k}(gain) = eV V = E_{k}(gain)/e V = 8.03x10^{-16}/1.6x10^{-19}V = 5021.9Vb. Plate P can be made more positive to attract the electron. This will move the spot up vertically towards X. Making Q more positive will attract the electron and move it horizontally towards X.If plate Q is made twice as positive as plate P the combined effect will be to move the spot to position X. 8.a.i. The amplifier is working ininverting mode. a.ii. Gain = -R_{feedback}/R_{input}R_{feedback}= 2x10^{6}W R_{input}= 100x10^{3}W Gain = -2x10^{6}/100x10^{3}Gain = -20a.iii. b.i. Between 2ms and 8ms the gain of the amplifier is said to be saturated. This is due to the fact that the output voltage from the amplifier cannot be greater than that supplied to it. b.ii. To produce saturation you can:

- increase the size of the feedback resistor;
- decrease the size of the input resistor;
- increase the size of the input voltage.