Pure Physics Nov 2022 TYS ANSWERS

Nov 2022_Suggested ANSWERS_as at June 2023 Page 1 of 9 – Nov 2021 Catholic High School | O-Level Physics NOT IN SYLLABUS: 6091 Nov 2022 P1: - Suggested Answers P2: - Paper 1 [40 marks] 1 A 11 B 21 B 31 B 2 D 12 B 22 A 32 B 3 C 13 D 23 C 33 C 4 C 14 D 24 A 34 A 5 B 15 D 25 D 35 A 6 D 16 C 26 C 36 B 7 C 17 D 27 A 37 D 8 D 18 A 28 C 38 A 9 B 19 C 29 A 39 C 10 C 20 D 30 B 40 A *Q. 2: D Students who selected C related an acceleration to a change in speed rather than to a change in velocity. *Q. 9: B Students who selected A were able to understand that Y is larger than X and Z due to Newton’s First Law. However, they erroneously thought that the magnitudes of the two forces X and Z are directly proportional to their distances from the centre of gravity. This could be due to them trying to answer using proportions rather than considering the Principle of Moments equation. *Q. 14: D Students who selected A did not consider that the temperature of the vapour and the liquid water are equal, and the effect of temperature on the molecular kinetic energies. *Q. 17: D This question is challenging, with most students selecting C which suggested that the athlete would cool down more quickly rather than more slowly. They may have learnt that shiny surfaces are poor emitters and poor absorbers of infra-red radiation but cou ld not automatically relate this to the reflection properties. Students who selected A or B did not consider it plausible that a shiny surface could result in less radiation being reflected. *Q. 19: C Students commonly selected A. It is important to read the question carefully to distinguish between total internal energy and average internal energy. *Q. 21: B Students who selected D are unable to see the inverse relation between period and frequency based on the formula. They may also not have considered the fact that the straight line shown could have been extrapolated to negative periods or frequencies which, in the context of wave motion, are meaningless concepts.

Nov 2022_Suggested ANSWERS_as at June 2023 Page 2 of 9 – Nov 2021 *Q. 30: B Students who selected C suggested that the filament lamp has a constant resistance, which is incorrect as the lam p is an ohmic conductor with a resistance that increase with voltage/current. *Q. 34: A Students who selected the wrong options may have failed to realise that resistance of (and hence p.d. across) the thermistor (or LDR) decreases when temperature (or light intensity) increases. *Q. 36: B Students who selected A or D ay have confused the magnetic fi eld with an electric field. Students who selected C may not have understood correctly that the direction of (conventional) current used in Fleming’s Left Hand Rule is the same as the direction of movement of positive charges and opposite to the direction of movement of negative charges. *Q. 40: A Students who selected B or D ignored the reduction in current due to losses in the transformer. Students need to appreciate that for a non -ideal transformer, the output power in the secondary coil is smaller than the input power in the primary coil, and the current in the secondary coil is smaller than what it would have been for an ideal transformer. The p,d, across the secondary coil remains the same as for an ideal transformer as it is dependent onl y on the turns ratio of the transformer. 5

Nov 2022_Suggested ANSWERS_as at June 2023 Page 3 of 9 – Nov 2021 Paper 2 [80 marks] 1 a Rate of change of displacement. 1 b acceleration, displacement, weight 1 ci AB: moving upwards and accelerating upwards BC: moving upwards and accelerating downwards CD: moving downwards and accelerating downwards 2 cii 10.0 s (1 dp based on precision to half a smallest square) 1 ciii Maximum height = Area under graph from t = 0.0 s to 10.0 s = ½ (10.0)(60) = 300 m 1 1 2 a Weight = Density x Volume x Gravitational Field Strength = (900) (0.20)3 (10) = 72 N 1 1 1 bi Pressure is the force acting per unit area. 1 bii p = F / A = W / (l x l) = 72 / (0.20 x 0.20) = 1800 N/m2 1 1 biii p = hρg ρ = p/hg = 1800 / (0.12)(10) = 1500 kg/m3 1 1 3 a - W must start from CG and be vertically downwards. - U should be longer in length than W as its magnitude is the sum of the magnitudes of W and the vertical components of the two F forces. 1 1

Nov 2022_Suggested ANSWERS_as at June 2023 Page 4 of 9 – Nov 2021 b Using a ruler, length of the 0.40 N vector is 10 cm. Hence, scale of Fig. 3.2 is 1 cm : 0.04 N F = 5.8 x 0.04 = 0.23 N 1 1 ci Work done = U x d = 0.60 x 20 = 12 J 1 1 cii By Conservation of Energy, Work done by U = Gain in GPE + Gain in KE + Work done against air resistance Gain in KE is maximum if Work done against air resistance = 0 Hence, Maximum KE = Work done by U – Gain in GPE = 12 – 4.0 = 8.0 J 1 ciii [Until 2023] The balloon loses energy due to work done against air resistance. [2024 onwards] As the balloon does work by moving a distance past surrounding air molecules, some of its kinetic energy is transferred mechanically to the surrounding air to increase energy in its internal store. 1 4 a Seal a glass cell illuminated by a torchlight containing a little smoke and place it under the microscope. Bright specks of light are observed to be moving about in a constant random motion. This is because the smoke particles are randomly bombarded by air molecules which are moving at random (ie. Brownian motion). 1 1 b 1. They move constantly. 2. They move randomly in all directions. 1 1 c Very large particles generally have more mass compared to air molecules. Hence, when air molecules collide with them, they require a larger force to change their direction of motion. 1 1 5 ai Ray A emerges into the air. This is because the angle of incidence of ray A is smaller than the critical angle for light travelling from glass into air, so total internal reflection does not occur. 1 1

Nov 2022_Suggested ANSWERS_as at June 2023 Page 5 of 9 – Nov 2021 aii - Line from L must hit air-glass interface at an angle of 48° between the line and the interface. - Rays emerging into air must bend away from the normal and diverge from one another. 1 1 bi The speed of light in the cladding is larger than that in the glass fibre, because the refractive index of the cladding is less than that of glass. 1 bii As the refractive index of glass is more than that of the cladding which is in turn more than that of air, the speed of light in glass is less than that in the cladding which is in turn less than that of air. Hence, the ratio of the speed of light in air to glass (or refractive index of glass with respect to air) is larger than the ratio of the speed of light in cladding to glass (or refractive index of glass with respect to cladding). This means that the sine of the critical angle (and hence the critical angle) for light travelling from glass into the cladding is larger than that from glass into air (ie 42°). 1 1 6 a Its frequency is abo ve the upper hearing limit of the human range of audibility. 1 b The transmitter emits pulses of ultrasound into the mother’s body which are then reflected back from the layers of tissue in the mother’s body to a detector. By measuring the time interval between the sending and receiving of the pulses, the image processor determines the distance of specific layers of tissue from the surface of the mother’s body and translates this information into an image. 1 1 ci v = fλ λ = v/f