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1.a.i. At the maximum height the long jumpers vertical velocity(V_{v}) is 0m/s. v_{v}= 0m/s S_{v}= 0.86m a = -9.8m/s/s u_{v}= ? To find u use : v^{2}= u^{2}+ 2as 0^{2}= u^{2}+ 2x(-9.8)x0.86 u^{2}= 16.856(m/s)^{2}u = 4.1m/sNote: The positive square root is taken because the motion is in the upwards direction. a.ii. The total time in the air for the vertical jump in part a.i. will be the same as that for the actual jump if the vertical component of velocity is the same. The total displacement for the jump in part a.i. is 0m. s = 0m a = -9.8m/s/s u = 4.1m/s t = ? To solve for t use: s = ut + 1/2(at^{2}) 0 = ut + 1/2(at^{2})....divide by t 0 = u + (at)/2 t = -2u/a t = (-2x4.1)/-9.8 t = 0.84s v_{H}= s_{H}/t v_{H}= 7.8/0.84vb. With a maximum height of less than 0.86m the time in the air would be less than that 0.84s. This means that the horizontal distance must be covered in a shorter time. This can only be achieved if the horizontal velocity is greater than 9.3m/s. 2.a.i. a.ii. m = 500kg g = 9.8N/kg W = ? W = mg W = 500x9.8 W = 4900N With no rope in position there is no tension force(T). This means that the unbalance force causing the acceleration is the difference between the buoyancy force(B) and the weight(W). F_{H}= 9.3m/s_{un}= B - W F_{un}= ma ma = B - W B = ma + W B = 500x1.5 + 4900B = 5650Na.iii.Before the release the upward force must balance the downward forces. B = W + T T = B - W T = 5650 - 4900T = 750Nb. Each rope contributes half of the downward component of the tension. Downward tension from rope 1 = 375N Downward tension from rope 2 = 375N To calculate the tension in each rope use trigonometry. 375/T_{rope 1}= cos25^{o}T_{rope 1}= 375/cos25^{o}T375/T_{rope 1}= 413.8N_{rope 2}= cos25^{o}T_{rope 2}= 375/cos25^{o}T4.a.i. Use Boyle's law to solve this problem. P_{rope 2}= 413.8N_{1}= 1.76x10^{5}Pa V_{1}= 750cm^{3}P_{2}= ? V_{2}= 900cm^{3}P_{1}V_{1}= P_{2}V_{2}P_{2}= P_{1}V_{1}/V_{2}P_{2}= 1.76x10^{5}x750/900Pa.ii. F = PA F = 1.47x10_{2}= 1.47x10^{5}Pa^{5}x5x10^{-3}F = 735Nb. The gas inside the rocket exerts the forces shown in the diagram.

- The lateral forces cancel.
- The downward force accelerates the water out of the rocket.
- The upward force accelerates the rocket.

4.a. Ek = 1/2(mvb.ii.^{2}) Collision with polyurethane block. Ek = 1/2(0.5x0.33^{2})Ek = 0.027JCollision with rubber band. Ek = 1/2(0.5x0.43^{2})Ek = 0.046Jb. In elastic collisions kinetic energy is conserved. In the collision with themetal springthe least amount of kinetic energy is lost and is therefore the most like an elastic collision. c. To propel the vehicle with the same initial speed the force providing the impulse must be equal in each experiment. This could be achieved if the impulse was provided by a stretched elastic band and ensuring the band was pulled back by the same amount each time. d.i. F = (mv-mu)/t_{c}m = 0.5kg t_{c}= contact time = 0.4s u = -0.55m/s v = 0.35m/s F = 0.5[0.35-(-0.55)]/0.4 F = 0.5x0.9/.4F = 1.125Nd.ii. The vehicleaccelerates to the rightwhen in contact with the block. This is because the direction of the force causing the acceleration is towards the right and the acceleration must be in the same direction. 5.a.i. V_{supply}= V_{x}+ V_{y}V_{x}= V_{supply}- V_{y}V_{x}= 10 - 6 V_{x}= 4V I_{x}= V_{x}/R_{x}I_{x}= 4/1200 I_{x}= 0.0033A I_{x}= I_{y}= 0.0033A V_{y}= I_{y}R_{y}R_{y}= V_{y}/I_{y}R_{y}= 6/0.0033R5.a.ii. The voltage across each resistor in a potential divider circuit is proportional to the value of its resistance. When resistor Z is connected in parallel with resistor Y a parallel network, with a lower resistance than resistor Y alone, is created. The voltage across this network, and each resistor in the network, is therefore less than the voltage across Y alone. 5.a.iii.1/R_{y}= 1800W = 1.8kW_{p}= 1/R_{Y}+ 1/R_{Z}1/R_{p}= 1/1.8 + 1/4.7 1/R_{p}= 0.5555 + 0.2128 1/R_{p}= 0.7683 R_{p}= 1.3kW V_{Rp}= [R_{p}/(R_{X}+R_{p})]V_{supply}V_{Rp}= [1.3/(1.3+1.2)]10 V_{Rp}= 0.52x10V5.b.i._{Rp}= 5.2VR_{A}/R_{B}= R_{C}/R_{D}

Resistance of A/W | Resistance of B/W | Voltmeter reading/mV |

120 | 120 | 0 |

121 | 120 | -21 |

121 | 121 | 0 |

121 | 122 | +21 |

121 | 119 | -42 |

The values in the table are worked out using the fact that the voltage reading is proportional to the out of balance resistance. 6.a.i. Q = Ixt Q = ? I = 0.5A t = 1h = 3600s Q = 0.5x3600Q = 1800C6.a.ii. E = Pt P = IV =>E=ItV E = 0.5x3600x1.2E = 2160JOR E = QV (same answer) b.i. The emf is the energy the cell supplies to each coulomb of charge passing through it. b.ii. The emf can be found by projecting the graph line back until it cuts the voltage axis.emf = 1.4VThe internal resistance is equal to the negative of the gradient of the line given. m = (y_{1}-y_{2})/(x_{1}-x_{2}) m = (1.0-0.6)/(1.0-2.0) m = 0.4/-0.1 m = -4r = 4WTo justify the above consider: y = mx + c ...1 V = mI + c ...2 V_{tpd}= E - Ir ...3 V_{tpd}= -Ir + E ...4 From equation (3) When I = 0A :V_{tpd}=emf Comparing (2) and (4) m = -r 7.a.i. The light from the window will result in the solar cell producing a small voltage that is amplified to produce a non zero V_{out}. a.ii. V_{out}/V_{in}= R_{feedback}/R_{1}V_{in}= -V_{out}(R_{1}/R_{feedback}) V_{in}= -(-1.75)(15/120)Vb.i. The differential amplifier amplifies the difference between V_{in}= 0.219V_{1}and V_{2}. When V_{1}is adjusted to equal V_{2}there is no difference between the input voltages and the output voltage will be zero. b.ii. V_{out}= R_{f}/R_{1}(V_{2}-V_{1}) 1.5 = 220/4.7(V_{2}-0.219) 1.5 = 46.81(V_{2}-0.219) 0.032 = V_{2}-0.219V8.a. T = 4x5ms T = 20ms f = 1/T f = 1/20x10_{2}= 0.251V^{-3}f = 50Hzb.i. The value of resistor R_{2}was increased. This conclusion is drawn because the rate at which the capacitor discharges has decreased. b.ii. Q = CDV DV = V_{initial}-V_{final}DV = 8V - 2V = 6V Q = 2.2x10^{-6}x6Q = 1.32x109.a.i. n^{-5}C_{plastic}= sinq_{air}/sinq_{plastic}n_{plastic}= sin40^{o}/sin30^{o}n_{plastic}= 0.633/0.5na.ii. b.i. q_{plastic}= 1.29_{critical}= sin^{-1}(1/n) q_{critical}= sin^{-1}(1/1.8) q_{critical}= sin^{-1}(0.555)qb.ii. n_{critical}= 33.7^{o}_{borate glass}= sinq_{air}/sinq_{borate glass}sinq_{borate glass}= sinq_{air}/n_{borate glass}sinq_{borate glass}= sin40^{o}/1.8 sinq_{borate glass}= 0.0.643/1.8 sinq_{borate glass}= 0.357 q_{borate glass}= 20.9^{o}10.a.i. Electrons are moving from a high to low energy level within the atom. A photon of light is emitted when this happens. a.ii. l_{sodium yellow}= 589nm E_{photon}= hf E_{photon}= hc/l E_{photon}= 6.63x10^{-34}x3x10^{8}/589x10^{-9}E_{photon}= 3.38x10^{-19}J The photon energy and the energy difference are equal.Eb.i. Photons emitted from the sodium lamp and passing through the flame containing vaporised sodium will be absorbed by sodium electrons. This means that sodium light passing through the flame will be reduced in intensity and produce a dark shadow behind the flame. b.ii. There is no energy gap in cadmium with the same energy as a photon emitted from the sodium lamp. Therefore, no absorption will take place and there will be no shadow region. 11.a. The activity(A), measured in becquerels(bq), is a measure of the number of disintegrations(N) per second. A = N/t N = At N = 300x10_{difference}= 3.38x10^{-19}J^{6}x60N = 18x10b. H/t = 16mSv/h (at a distance of 1m) H = DQ H/t = (D/t)xQ 16mSv/h = (50mGy/t)x1 t = 50mGy/16mSv/h^{9}t = 3.125hc.i. c.ii. The graph indicates that the required lead thickness is8.8mm. d. The polystyrene packaging increases the distance between the porters and the source. As the intensity at a distance(d) from the source is inversely proportional to the square of the distance the dose equivalent rate for the porters will be less. 12.a. Mean(L) = Total/Number of readings Mean(L) = (2.402+2.399+2.412+2.408+2.388+2.383+2.415)/7 Mean(L) = 16.807/7Mean(L) = 2.401mRandom Error(L) = Range/Number of readings Random Error(L) = (Max-Min)/N Random Error(L) = (2.415-2.383)/7 Random Error(L) = 0.032/7Random Error(L) = 0.005mb. Percentage error in L = [Random Error(L)/Mean(L)]x100 Percentage error in L = [0.005/2.401]x100 Percentage error in L = 0.21% Percentage error in x = [Random Error(x)/Mean(x)]x100 Percentage error in x = [1/91]x100 Percentage error in x = 1.1% The measurement with the largest percentage error isx. c. l = dsinq l = dx/L l = 1.693x10^{-5}91x10^{-3}/2.401l = 641.66nmThe percentage error in l can be taken to be equal to the percentage error in x, as this is the largest individual error. 1.1% of 641.66nm = (1.1/100)x641.66 = 7.06nml = (641.66 +- 7.06)nmd. Increasing the distance between the grating and the screen reduces the percentage error of L and, more significantly, x.## END OF QUESTION PAPER