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1.a. The ball is in free fall for the first 0.5s. u = 0m/s v = -4.5m/s t = 0.5s a = ? a = (v-u)/t a = (-4.5-0)/0.5a = -9.0m/s/sb. DEk = Ek_{after}-Ek_{before}DEk = 1/2(mv^{2}-mu^{2}) 2DEk = m(v^{2}-u^{2}) 2DEk/(v^{2}-u^{2}) = m m = 2DEk/(v^{2}-u^{2}) m = 2x-1.7/(4^{2}-[-4.5^{2}]) m = -3.4/(16-20.25)m = 0.8kgc. DP = mv-mu DP = m(v-u) DP = 0.8(4-[-4.5])DP = 6.8kgm/sd. The time that the ball is in contact with the ground(t_{c}) is 0.5s. F = DP/t_{c}F = 6.8/0.5F = 13.6N2.a. w = ? m = 15000 g = 1.6N/kg w = mg w = 15000x1.6 w = 24000N The upward force(F_{up}) is equal in magnitude to the weight and acts in the opposite direction. There is no unbalanced force which means the craft must be moving at a constant speed and be moving in a straight line. b.i. F_{unbalanced}= F_{up}- w F_{unbalanced}= 25500 - 24000 F_{unbalanced}= 1500N a = F_{unbalanced}/m a = 1500/15000 a = 0.1m/s/s upwards This means that the craft has an upward deceleration of -0.1m/s/s. b.ii. a = 0.1m/s/s t = 18s u = -2.0m/s v = ? v = u + at v = -2.0 + 0.1x18 v = -0.2m/s b.iii.v^{2}= u^{2}+ 2as s = (v^{2}-u^{2})/2a s = ([-0.2]^{2}-[-2]^{2})/2x0.1 s = (0.04-4)/0.2 s = -3.96/0.2 s = -19.8m This displacement is equal to that which the craft has descended from. This means the craft must have been19.8mabove the surface of the moon to start with. 3.a. The law of conservation of linear momentum states that :Pb.i._{before}= P_{after}Before collisionP_{before}= m_{A}u_{A}+ m_{B}u_{B}P_{before}= 1400xu_{A}+ 1000x8 P_{before}= 1400u_{A}+ 8000After collisionP_{after}= m_{A}v_{A}+ m_{B}v_{B}v_{A}=v_{B}=v P_{after}= (m_{A}+ m_{B})v P_{after}= (1400 + 1000)15 P_{after}= 36000kgm/s Equate: P_{before}and P_{after}1400u_{A}+ 8000 = 36000kgm/s 1400u_{A}= (36000-8000)/1400 u_{A}= 20m/s b.ii. Consider the instant after the collision to be the starting point for this calculation. u = 15m/s v = 0m/s s = 20m a = ? v^{2}= u^{2}+ 2as a = (v^{2}-u^{2})/2s a = (0^{2}-15^{2})/2x20 a = -225/40 a = -5.625m/s/s F_{friction}= F_{un}F_{un}= ma F_{un}= 2400x-5.625FThe negative sign indicates that the force acts to the LHS. OR The kinetic energy lost by the car is equal to the work done by the frictional force over the 20m stopping distance. Ek_{un}= -13,500N_{lost}= E_{work}Ek_{lost}= Ek_{before}- Ek_{after}Ek_{lost}= (mv^{2}- mu^{2})/2 Ek_{lost}= m(v^{2}- u^{2})/2 Ek_{lost}= 2400(15^{2}- 0^{2})/2 Ek_{lost}= 27000J E_{work}= Fd F = E_{work}/d F = 27000/20F = 13500Nb.iii.Most of the kinetic energy will be converted into heat energy. 4.a. P_{1}= 400kPa V_{1}= 1000cm^{3}P_{2}= 250kPa V_{2}= ? P_{1}V_{1}= P_{2}V_{2}V_{2}= P_{1}V_{1}/P_{2}V_{2}= 400x1000/250Vb. The pressure increases. The volume remains constant. The temperature increases. c. The starting pressure is that indicated by point B on the graph and the final pressure is that indicated by point C. Over this period the volume is constant. T_{2}= 1600cm^{3}_{1}= 300K P_{1}= 200kPa T_{2}= ? P_{2}= 500kPa P_{1}/T_{1}= P_{2}/T_{2}T_{2}= P_{2}(T_{1}/P_{1}) T_{2}= 500(300/200)T5.a.i. Q = CV C = Q/V C = 24x10_{2}= 750K^{-6}/1.5 C = 16x10^{-6}FC = 16mFa.ii. Mean = total/N Mean(C) = (16+18+20+16+15)/5Mean(C) = 17mFRandom error = (max-min)/N Random error(C) = (20-15)/5Random error(C) = +-1mFC = (17+-1)mFa.iii.Measure the voltage across the capacitor directly by connecting the voltmeter across it. Additionally, to prevent leakage through the voltmeter it would be better to use an oscilloscope for this purpose. b. The effective resistance of the capacitor in the circuit decreases as the frequency of the supply increase. As the resistance decreases the current in the circuit increases and the lamp glows more brightly. 6.a. The electromotive force is the energy supplied by the power supply to each coulomb of charge. b.i. P = I^{2}R I^{2}= P/R I = (P/R)^{1/2}I = (8.0/0.32)^{1/2}I = 25^{1/2}I = 5Ab.ii. V = IR V = 5x0.32 V = 1.6V b.iii.V_{r}= emf - V_{heater}V_{r}= 2.0 - 1.6 V_{r}= 0.4V V_{r}= Ir r = V_{r}/I r = 0.4/5r = 0.08Wc. Two heaters in parallel will reduce the resistance of the circuit. The current in the heater elements will increase which has the effect of increasing the rate at which the water is heated. The resistance of the two heaters in parallel can be calculated as shown below: R_{p}= R_{heater}/2 R_{p}= 0.32/2 R_{p}= 0.16W R_{total}= R_{p}+ r R_{total}= 0.16 + 0.08 R_{total}= 0.24W I = emf/R_{total}I = 2.0/0.24 I = 8.33A V_{r}= Ir V_{r}= 8.33x0.08 V_{r}= 0.67V V_{heater}= emf - V_{r}V_{heater}= 2.0 - 0.67 V_{heater}= 1.33V This is the voltage across each heater element. I_{heater}= V_{heater}/R_{heater}I_{heater}= 1.33/0.32 I_{heater}= 4.16A The power developed in each heater can be calculated using: P = I^{2}R P = 4.16^{2}x0.32 P = 5.5W The two heating elements develop a total power of 11W. This means that the water is heated about 1.4 times faster. 7.a.i. The amplifier is working ininverting mode. a.ii. Gain = -R_{feedback}/R_{1}Gain = -1x10^{6}/2x10^{3}Gain = 500a.iii. a.iv. The LED will pulse ON/OFF. The LED will be on when V_{out}is positive. The LED will be off when V_{out}is zero. b.i. The op-amp is working indifferential mode. The gain equation for the op-amp in this mode is:Vb.ii. Use the equation in b.ii. to calculate V_{out}= (R_{f}/R_{1})(V_{2}-V_{1})_{out}. Take the voltages V_{1}and V_{2}from the graphs given.

Time interval/ms |
V_{2}/mV |
V_{1}/mV |
V_{out}/V |

0-50 | 0 | 0 | 0 |

50-100 | -6 | -6 | 0 |

100-150 | +6 | 0 | +3 |

150-200 | 0 | -6 | +3 |

200-250 | 0 | 0 | 0 |

250-300 | -6 | -6 | 0 |

300-350 | +6 | 0 | +3 |

350-400 | 0 | -6 | +3 |

8.a. A ray of sunlight is made up of many different wavelengths that is perceived as white light. However, each individual wavelength represents a particular colour of light. The diffraction grating diffracts the different wavelengths in sunlight by different amounts: the longer wavelengths being diffracted the most and the shorter wavelengths the least. At position X on the screen the short wavelength violet light has interfered constructively to produce a band of violet light. At position Y on the screen the long wavelength red light has interfered constructively to produce a band of red light. Between X and Y the other colours of the visible spectrum produce bands of indigo, blue, green, yellow and orange light. b.i. dsinq = nl d = 1/N = 1/6x10^{5}= 1.667x10^{-6}m q = ? n = 1 l = 410nm = 410x10^{-9}m sinq = nl/d sinq = 1x410x10^{-9}/1.667x10^{-6}sinq = 0.246 q = sin^{-1}0.246q = 14.2b.ii. d = 1.667x10^{o}^{-6}m q = 14.2^{o}+ 9^{o}= 23.2^{o}n = 1 l = ? l = dsinq/n l = 1.667x10^{-6}sin(23.2^{o})/1l = 656.7x10c. The diffraction grating will produce many spectra symmetrical about the q = 0^{-9}m = 656.7nm^{o}line. The prism produces one spectrum by refraction. The relative positions of each colour of light produced by refraction will be the reverse of that produced by the diffraction grating, as blue light is refracted more than red light. 9.a. Monochromatic means the laser light has a single wavelength/frequency. Coherent means the waves are in phase/step. b.i. DE = hf where DE = E_{2}-E_{2}f = E_{2}-E_{2}/h f = -4.67x10^{-20}-(-6.55x10^{-20})/6.63x10^{-34}f = 452.5nmb.ii. The wavelength of the emitted radiation corresponds to the blue end of the spectrum. c. The beam is intense because the energy is spread over a small area and a large number of photons per second are emitted from the laser. 10.a. Light incident on the pn-junction photodiode will increase the number of electron hole pairs and consequently increase the conductivity of the diode. b. The diode is operating inphotoconductive mode. c. I_{1}d_{1}^{2}= I_{2}d_{2}^{2}I_{2}= I_{1}d_{1}^{2}/d_{2}^{2}I_{2}= 3.0x1^{2}/0.75^{2}I11.a.i. a.ii. Take the background count to be 20cpm. Add this to the corrected count rate and plot as shown below. b._{2}= 5.33mATime/years Corrected Count Rate/cpm0 520 5.25 260 11.5 130 16.75 65 21 32.5After 21 years the count rate is 32.5cpm.## END OF QUESTION PAPER