Higher Physics 1998 PII Solutions Solutions to SQA examination 1998 Higher Grade Physics Paper II Solutions Return to past paper index page. 1.a.i. Component of the force acting parallel to the slope(Fs) is calculated using: Fs = mgsinq Fs = 2.0x9.8xsin20o Fs = 6.7N a.ii. a.iii.To calculate the acceleration use [N2] a = Funbalanced/m Funbalanced = FFriction+Fs Funbalanced = -1.3 + (-6.7) Where vectors acting down the slope are taken as negative. Funbalanced = -8N a = -8/2.0 a = -4m/s2 Which equates to a deceleration of 4m/s2, 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/s2 t = (0-3)/-4 t = ? t = 0.75s a.v. v2 = u2 + 2as s = (v2 - u2)/2a s = (02 - 32)/(2x-4) s = -9/-8 s = 1.125m b.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 = 588/9.8 m = 60kg b. 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 labelled T and downward force, weight, represented by W. Funbalanced = 678-588 Funbalanced = 90N a = Funbalanced/m a = 90/60 a = 1.5m/s2 c. The lift is decelerating when the upward force is less than the weight. Funbalanced = 498 - 588 Funbalanced = -90N The negative indicates that the unbalanced force acts in the downwards direction. a = Funbalanced/m a = -90/60 a = -1.5m/s2 This can be thought of as an upward deceleration of 1.5m/s2 d. 3.a.i. V = d/t1 d = diameter of ball = 24mm = 0.024m t = time for ball to pass through light gate = 0.060s V = 0.024/0.060 V = 0.4m/s a.ii. Favg = change in momentum/contact time Favg = (mv-mu)/t Favg = m(v-u)/t Favg = 0.045(0.4-0)/0.005 Favg = 3.6N b.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 t1 = (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 = 0.72N Favg = (3.6 +- 0.72)N 4.a. Initial volume of gas (V1) in the container = (8-5)litres = 3 litres = 3 x10-3m3 Density of gas (r) = 1.29kg/m3 m = rV1 m = 1.293x(3x10-3m3) m = 3.87x10-3Kg b.i. Pressure before pumping (Pbefore) = 1.01x105Pa Pressure after pumping (P1) = 3.0x105Pa Volume of air before and after pumping(V1) is fixed at 3x10-3m3 Area(A) = 7.0x103m2 To calculate the force, after pumping, use: P1 = F/A F = P1A F = 3.0x105x7.0x10-3 F = 2100N b.ii. When the volume of the water is 2 litres, the volume of the gas in the container(V2) is 6 litres,or 6x10-3m3. The pressure at this time is P2. P1V1 = P2V2 P2 = P1V1/V2 P2 = 3x105x3x10-3/6x10-3 P2 = 1.5x105Pa 5.a.i. Vtpd = emf - Ir I = emf/(R+r) I = 1.5/(3+0.75) I = 1.5/3.75 I = 0.4A Vtpd = 1.5 - 0.4x0.75 Vtpd = 1.2V a.ii. Vlost = Ir or emf = Vtpd+Vlost Vlost = 0.4x0.75 Vlost = emf - Vtpd Vlost = 0.3V Vlost = 1.5 - 1.2 = 0.3V b.i. R = 1W Vlost = 2V Vtpd = emf - Vlost Vtpd = 6 - 2 Vtpd = 4V b.ii. I = Vtpd/Rvariable I = 4/1 I = 4A Vlost = Ir Vlost = emf - Vtpd r = Vlost/I Vlost = 6 - 4 = 2V r = 2/4 r = 0.5W 6.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(Iinitial) is 30mA, as read from the graph. Iinitial = 30mA = 30x10-6A V = 6V R = ? R = V/I R = 6/30x10-6 R = 2x105W b.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 = 100x10-6A t = 10s Q = Ixt Q = 100x10-6x10 Q = 100x10-5C = 1x10-3C = 1mC c.iii.To calculate the capacitance use: Q = CVcapacitor At 10s Vcapacitor = 5V Q = 1x10-3C Vcapacitor = 5V C = ? C = Q/Vcapacitor C = 1x10-3/5 C = 2x10-4F = 200mF 7.a.i. To calculate the kinetic energy(Ek) use : Ek = mv2/2 Ek = ? me = 9.11x10-31kg v = 4.2x107m/s Ek = mv2/2 Ek = 9.11x10-31x(4.2x107)2/2 Ek = 8.03x10-16J 7.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. Ek(gain) = eV V = Ek(gain)/e V = 8.03x10-16/1.6x10-19 V = 5021.9V b. 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 in inverting mode. a.ii. Gain = -Rfeedback/Rinput Rfeedback = 2x106W Rinput = 100x103W Gain = -2x106/100x103 Gain = -20 a.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. 9.a. DE32 = E3-E2 DE32 = -2.416x10-19-(-5.524x10-19) DE32 = 3.006x10-19J To calculate the wavelength of a photon of this energy use : E = hf = hc/l l = hc/E32 l = 6.63x10-34x3x108/3.006x10-19 l = 6.617x10-7m = 661.7nm This corresponds to spectral line Z. This answer can also be deduced by the fact that the photon produced by the transition between the energy levels E3 and E2has the lowest energy of the four in the diagram. This lowest energy photon will also have the longest wavelength,which is Z. b. The more photons of a particular frequency emitted the brighter the spectral line will be. This means that there must be more electron transitions between the energy levels producing the lines Y and Z. c.i. E = hf E = 6.63x10-34x7.486x1013 E = 4.96x10-20J c.ii. DE54 = E5-E4 DE54 = -0.864x10-19-(-1.360x10-19) DE54 = 4.96X10-20J The transition between energy levels E5 and E4 produces the infrared photon. 10.a.i. Angle i = 50o Angle r = 28o nglass = sin(i)/sin(r) nglass = sin 50o/sin 28o nglass = 0.766/0.496 nglass = 1.63 a.ii. qcritical = sin-1(1/n) qcritical = sin-1(1/1.63) qcritical = sin-1(0.613) qcritical = 37.8o b.i. nquartz = lair/lquartz lquartz = lair/nquartz lquartz = 510/1.55 lquartz = 329nm b.ii. As the wavelength of visible light increases the refractive index of the quartz(nquartz) decreases. This means that (1/nquartz) increases and therefore qcritical[sin-1(1/nquartz)] also increases. b.iii.Flint glass has a higher refractive index than crown glass for all wavelengths of light. This would result in the spectrum being projected onto a lower position on the screen. Additionally, because the difference in refractive index across the spectrum is greater with flint glass, the distance between the red end and the violet end of the spectrum, as appearing on the screen, would be greater. 11.a.i. To conserve atomic and mass number in the decay of Thorium to Palladium a beta particle(b, 0-1e)is emitted. a.ii. Number of neutrons(n) = mass number(A) - atomic number(Z) n = 238 - 92 n = 146 a.iii.23492U -> 23090Th + 42He b.i. Soil and rocks containing natural radioactive isotopes, and cosmic rays, contribute to background radiation. b.ii. The earths atmosphere absorbs cosmic radiation and reduces the effective dose equivalent rate at sea level. At heights above 4km there is less atmosphere to absorb the radiation which results in a higher dose equivalent rate at these rarefied altitudes. b.iii. (A)t = 7h H/t = 5mSv/h H = ? H = tx5mSv/h H = 7hx5mSv/h H(flight) = 35mSv (B)During 40 flights: H = 40x35mSv H = 1400mSv (As a result of flights) In addition to this the traveller will be exposed to sea level background radiation. Time at sea level tsea = (365x24)-(7x40) = 8480h H(at sea level) = 8480hx0.2mSv/h H(at sea level) = 1696mSv H(total) = H(at sea level) + H(flight) H(total) = 1696mSv + 1400mSv H(total) = 3096mSv = 3.096(mSv/year) END OF PAPER Return to past paper index page.