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2018 Scholarship Physics (93103) Evidence Statement

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2018 Scholarship Physics (93103) Evidence Statement Scholarship Physics (93103) 2018 — page 1 of 5 Assessment Schedule – 2018 Scholarship Physics (93103) Evidence Statement 1–4 5–6 7–8 Q Evidence Below Schol Scholarship Outstanding ONE The restoring force must be directly proportional to the Thorough (a)(i) negative of the displacement. understanding of these (a)(ii) 2π applications ω = of physics. T 2π ×150 v = Aω = = 15.7079 m s−1 max 60 OR (b) Total energy = GPEhalf way + KEhalf way Partially (Partially) Correct correct correct mathematical 1 2 Total energy is found by mv = mgh and that gives 12.58 m mathematical mathematical solution to the 2 solution to the solution to the given 2 2 2 Mg × 12.58 = Mgh + ½ M (ω (150 – 75 )) given given problems. Gives h = 3.15 m problems. problems. (c)(i) AND / OR AND / OR AND Partial Reasonably Thorough understanding thorough understanding The profile is higher on each side (points A and C) with the of these understanding of these shallowest region at point B. applications of these applications of physics. applications of physics. The profile is higher at A and C because the wagon, moving of physics. slowly at these points of the motion, spends more time at these locations. The sideways motion of the falling sand can be ignored if the distance fallen to the track is regarded as negligible. (ii) Everything is constant apart from the total energy and force of the system. (d) R2 = (R – 12.58)2 + 1502 1502 +12.582 R = = 901 m 25.16 150 Angle the wagon moves through = sin–1 ⎛ ⎞ = 9.6º ⎝⎜ 901⎠⎟ This is close to the small angle limit so the motion should be acceptable to be treated as SHM. Scholarship Physics (93103) 2018 — page 2 of 5 1–4 5–6 7–8 Q Evidence Below Schol Scholarship Outstanding TWO The sources must have: Thorough (a) 1) the same wavelength understanding of these 2) fixed phase difference applications 3) a separation, d, greater than the wavelength of physics. 4) comparable amplitudes. OR (b) By realising that a rotation of ϕ for the slits will produce the standard interference pattern. The zero of intensity will be λ Partially (Partially) Correct when = d sinϕ . correct mathematical 2 correct mathematical mathematical solution to the Answer = 0.18134 degrees solution to the solution to the given given given problems. (c) At this point there is no path difference so the only factor is problems. problems. the reflection phase change, and since this is 180 degrees, the two waves cancel, leading to a dark fringe. AND / OR AND / OR AND (d) t θ = Partial Reasonably Thorough x understanding thorough understanding 2t n = of these understanding of these λ applications of these applications n 2θ of physics. applications of physics = x λ of physics. Answer approx. 1270 lines per m. or 1300 to 2 sig fig. (e) The lines become so close together that they are unresolvable. Scholarship Physics (93103) 2018 — page 3 of 5 1–4 5–6 7–8 Q Evidence Below Schol Scholarship Outstanding THREE 6.67 ×10−11 × 9.11×10−31 × 9.11×10−31 Thorough (a) Fg = 2 understanding R of these 9 −19 −19 8.98 ×10 ×1.60 ×10 ×1.60 ×10 applications of Fe = 2 R physics. F g = 2.40 ×10−43 Fe OR (b) Two in parallel (capacitor and inductor), one in series Partially (Partially) Correct (resistor). correct correct mathematical Evidence: mathematical mathematical solution to the solution to the given Finite current at a zero and high frequency implies resistor solution to the given problems. in series. given problems. problems. Can’t have the capacitor in series (at low frequency – current would be zero). AND / OR Can’t have inductor in series (at high frequency – current AND / OR AND would be zero). Zero current at finite frequency implies infinite impedance Partial Reasonably Thorough – this can happen with parallel branch containing an understanding thorough understanding inductor and a capacitor. of these understanding of these applications of of these applications of applications of (c) Take voltage to be V physics. physics. physics. V I4 = r + 4 V I = 1 r +1 4V 2 V 2 I 2 R = 16 = = (r + 4)2 (r +1)2 V 2 16(r2 2r 1) and = + + V 2 = 4(r2 + 8r +16) r2 + 8r +16 = 4r2 + 8r + 4 3r2 = 12 r = 2 Ω V 2 = 144 V = 12 volts (d) The electron associated with a single proton (forming a hydrogen atom) has a restricted set of possible energy values. We say the potential energy held by the electron is quantised because when the electron changes from large PE to less PE, the energy change is released as an electromagnetic photon. These photons always have precise values, forming the hydrogen emission spectrum. Scholarship Physics (93103) 2018 — page 4 of 5 1–4 5–6 7–8 Q Evidence Below Schol Scholarship Outstanding FOUR This system of two springs in series is equivalent to a single Thorough (a) spring, of spring constant k. understanding of these For spring 1, from Hooke’s Law applications of F = k x 1 1 physics. where x is the deformation of spring. 1 Similarly if x is the deformation of spring 2 we have 2 OR F = k x 2 2 Total deformation of the system Partially correct (Partially) Correct F F mathematical correct mathematical x1+x2 = + mathematical solution to the k1 k2 solution to the given problems. solution to the given problems. ⎛ 1 ⎞ given problems. ⇒ x1 + x2 = F ⎜ ⎟ ⎝ k1 + k2 ⎠ AND / OR AND AND / OR Rewriting and comparing with Hooke’s law we get: 1 1 1 = + Partial Thorough k k1 k2 understanding Reasonably understanding of these thorough of these applications of understanding applications of (b) Fo=kΔx2 from physics. of these physics. 1 1 1 applications of Using part a – this rearranged gives the above = + physics. k k1 k2 expression. (c) Different amplitudes, both masses will move in negative phase with each other because the centre of mass will stay at a constant position – overall motion is SHM. (d) Energy conservation and momentum conservation. N k = 3.3 using the expression in (a). effective m Using the numbers provided (F = 2N and k =3.3) implies that Δx2 = 0.6 m. That extension produces the store of energy 1 = k x2 – that energy will be shared between the masses. 2 eff When they are at max velocity (no stored elastic energy) – all energy will be kinetic therefore 1 1 1 m v 2 + m v 2 = kx2 = 0.6 J and m v = m v due to 2 1 1 2 2 2 2 1 1 2 2 conservation of momentum (or the fact that the centre of mass doesn’t move). 3 This gives v = v substituting into the above energy equation 1 2 2 –1 and solving gives v2 = 0.4 m s (e) The period of oscillation will alter and increase as there is more mass in system. The amplitude will be unchanged (since no Ek at end points). The maximum speed will decrease as overall E is constant. The centre of mass motion will be unchanged. Scholarship Physics (93103) 2018 — page 5 of 5 1–4 5–6 7–8 Q Evidence Below Schol Scholarship Outstanding FIVE Material must be an insulator (have no “free” electrons), Thorough (Partially) Correct (a) and its atoms / molecules must be easily polarised (distinct understanding correct mathematical charge separation shown in an imposed electric field). of these mathematical solution to the applications solution to the given of physics. given problems. (b)(i) 1 Q2 1 Q2 problems. Work needed = ΔE = EF − E1 = − 2 CF 2 ε rCF 1 Q2 (ε −1) = r OR AND 2 ε C r F Q = C V F F Partially AND / OR Thorough 1 2 ⎛ ε −1⎞ correct understanding So work needed = C V r F ⎜ ⎟ mathematical of these 2 ⎝ ε r ⎠ Reasonably solution to the thorough applications given understanding of physics. (ii) With the charge constant and the capacitance decreased, problems. of these the voltage (potential energy per coulomb) must have applications increased. AND / OR of physics. (c) Because the voltage must remain constant across the capacitor, the stored charge must reduce, as the Partial capacitance is reduced by the dielectric withdrawal. The understanding excess charge is returned to the battery and work must be of these done (by the withdrawing dielectric) to send the charge applications back into the battery. of physics. (d)(i) The capacitance has increased, the charge is constant, therefore voltage is reduced. (ii) Because the edge field of the capacitor has a component acting perpendicular to the plates which will attract material (electrostatically upwards). Cut Scores Scholarship Outstanding Scholarship 21 – 30 31 – 40 .
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