1.a.i. At the maximum height the long jumpers vertical velocity(Vv) is 0m/s. vv = 0m/s Sv = 0.86m a = -9.8m/s/s uv = ? To find u use : v2 = u2 + 2as 02 = u2 + 2x(-9.8)x0.86 u2 = 16.856(m/s)2 u = 4.1m/s Note: 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(at2) 0 = ut + 1/2(at2)....divide by t 0 = u + (at)/2 t = -2u/a t = (-2x4.1)/-9.8 t = 0.84s vH = sH/t vH = 7.8/0.84 vH = 9.3m/s b. 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). Fun = B - W Fun = ma ma = B - W B = ma + W B = 500x1.5 + 4900 B = 5650N a.iii.Before the release the upward force must balance the downward forces. B = W + T T = B - W T = 5650 - 4900 T = 750N b. 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/Trope 1 = cos25o Trope 1 = 375/cos25o Trope 1 = 413.8N 375/Trope 2 = cos25o Trope 2 = 375/cos25o Trope 2 = 413.8N 4.a.i. Use Boyle's law to solve this problem. P1 = 1.76x105Pa V1 = 750cm3 P2 = ? V2 = 900cm3 P1V1 = P2V2 P2 = P1V1/V2 P2 = 1.76x105x750/900 P2 = 1.47x105Pa a.ii. F = PA F = 1.47x105x5x10-3 F = 735N b. The gas inside the rocket exerts the forces shown in the diagram.
4.a. Ek = 1/2(mv2) Collision with polyurethane block. Ek = 1/2(0.5x0.332) Ek = 0.027J Collision with rubber band. Ek = 1/2(0.5x0.432) Ek = 0.046J b. In elastic collisions kinetic energy is conserved. In the collision with the metal spring the 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)/tc m = 0.5kg tc = 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/.4 F = 1.125N d.ii. The vehicle accelerates to the right when 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. Vsupply = Vx + Vy Vx = Vsupply - Vy Vx = 10 - 6 Vx = 4V Ix = Vx/Rx Ix = 4/1200 Ix = 0.0033A Ix = Iy = 0.0033A Vy = IyRy Ry = Vy/Iy Ry = 6/0.0033 Ry = 1800W = 1.8kW 5.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/Rp = 1/RY + 1/RZ 1/Rp = 1/1.8 + 1/4.7 1/Rp = 0.5555 + 0.2128 1/Rp = 0.7683 Rp = 1.3kW VRp = [Rp/(RX+Rp)]Vsupply VRp = [1.3/(1.3+1.2)]10 VRp = 0.52x10 VRp = 5.2V 5.b.i. RA/RB = RC/RDb.ii.
|Resistance of A/W||Resistance of B/W||Voltmeter reading/mV|
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 = 0.5x3600 Q = 1800C 6.a.ii. E = Pt P = IV =>E=ItV E = 0.5x3600x1.2 E = 2160J OR 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 = 1.4V The internal resistance is equal to the negative of the gradient of the line given. m = (y1-y2)/(x1-x2) m = (1.0-0.6)/(1.0-2.0) m = 0.4/-0.1 m = -4 r = 4W To justify the above consider: y = mx + c ...1 V = mI + c ...2 Vtpd= E - Ir ...3 Vtpd= -Ir + E ...4 From equation (3) When I = 0A :Vtpd=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 Vout. a.ii. Vout/Vin = Rfeedback/R1 Vin = -Vout(R1/Rfeedback) Vin = -(-1.75)(15/120) Vin = 0.219V b.i. The differential amplifier amplifies the difference between V1 and V2. When V1 is adjusted to equal V2 there is no difference between the input voltages and the output voltage will be zero. b.ii. Vout = Rf/R1(V2-V1) 1.5 = 220/4.7(V2-0.219) 1.5 = 46.81(V2-0.219) 0.032 = V2-0.219 V2 = 0.251V 8.a. T = 4x5ms T = 20ms f = 1/T f = 1/20x10-3 f = 50Hz b.i. The value of resistor R2 was increased. This conclusion is drawn because the rate at which the capacitor discharges has decreased. b.ii. Q = CDV DV = Vinitial-Vfinal DV = 8V - 2V = 6V Q = 2.2x10-6x6 Q = 1.32x10-5C 9.a.i. nplastic = sinqair/sinqplastic nplastic = sin40o/sin30o nplastic = 0.633/0.5 nplastic = 1.29 a.ii. b.i. qcritical = sin-1(1/n) qcritical = sin-1(1/1.8) qcritical = sin-1(0.555) qcritical = 33.7o b.ii. nborate glass = sinqair/sinqborate glass sinqborate glass = sinqair/nborate glass sinqborate glass = sin40o/1.8 sinqborate glass = 0.0.643/1.8 sinqborate glass = 0.357 qborate glass = 20.9o 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. lsodium yellow = 589nm Ephoton = hf Ephoton = hc/l Ephoton = 6.63x10-34x3x108/589x10-9 Ephoton = 3.38x10-19J The photon energy and the energy difference are equal. Edifference = 3.38x10-19J b.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 = 300x106x60 N = 18x109 b. 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 t = 3.125h c.i. c.ii. The graph indicates that the required lead thickness is 8.8mm. 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) = 16.807/7 Mean(L) = 2.401m Random Error(L) = Range/Number of readings Random Error(L) = (Max-Min)/N Random Error(L) = (2.415-2.383)/7 Random Error(L) = 0.032/7 Random Error(L) = 0.005m b. 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 is x. c. l = dsinq l = dx/L l = 1.693x10-591x10-3/2.401 l = 641.66nm The 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 = 7.06nm l = (641.66 +- 7.06)nm d. Increasing the distance between the grating and the screen reduces the percentage error of L and, more significantly, x.
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