Pure Physics Nov 2013 TYS ANSWERS

Nov 2013_Suggested ANSWERS_Aug 2023 Page 1 of 8 – Nov 2013 Catholic High School | O-Level Physics NOT IN SYLLABUS: 5059 Nov 2013 P1: - Suggested Answers P2: - Paper 1 [40 marks] 1 D 11 D 21 B 31 B 2 D 12 A 22 C 32 D 3 B 13 B 23 D 33 A 4 D 14 A 24 B 34 A 5 A 15 B 25 B 35 A 6 C 16 B 26 C 36 C 7 C 17 D 27 D 37 D 8 D 18 D 28 A 38 C 9 C 19 B 29 B 39 A 10 C 20 D 30 A 40 C *Q. 2: (D) The word directly is important here. Although the calipers would give a more precise answer, they would not do so directly. In Physics, the most accurate determination of a property is not always what is needed and convenience and speed can be of equal significance in some circumstances. (A is incorrect.) *Q. 8: (D) Need to consider the effects of the gravitational field strength, air resistance and the mass of the ball being accelerated. (A is incorrect.) *Q. 12: (A) The centre of mass of the conical flask is lower than that of the beaker. (D is incorrect.) *Q. 30: (A) The force that acts on a positive charge is in the direction of the field, and the force on a negative charge is opposite in direction. (B is incorrect.) *Q. 38: (C) There is attraction between two wires carrying current in the same direction; not repulsion. Do not mix this up with the rule ‘like charges repel’. (B is incorrect.) *Q. 40: (C) This question deals with resistors in series and parallel in the context of a domestic power supply, as the question refers indirectly to the: • resistance of the transmission line ( ‘long transmission line ’ → hence considerable resistance in the transmission line), and • heaters as resistors. When resistor X is removed, p.d. across Y increases (potential divider circuit), as the resistance across Y alone is now larger (compared to the previous resistance of X and Y in parallel). As P = V2 R , with resistance of Y constant and V larger, PY increases. (B is incorrect.) Resistance of long transmission wire Resistor X Resistor Y

Nov 2013_Suggested ANSWERS_Aug 2023 Page 2 of 8 – Nov 2013 Paper 2 [80 marks] 1 a By finding the gradient of a speed-time graph. 1 b From A to C, the stone undergoes decreasing acceleration. From C to D, the stone undergoes a uniform and large deceleration. 1 1 1 c As the stone falls, the air resistance acting on it increases with increasing speed. As the weight of the stone is constant, the resultant force acting downwards on the stone is decreasing. By Newton’s Second L aw of Motion (Fresultant = ma) , the decreasing resultant force will result in a decreasing acceleration. 1 1 1 d The stone hit s the water at point C before the speed can b e constant at terminal velocity. [Note: Simply writing ‘the graph had not become a st raight line’ is insufficient.] 1 2 a Scale: 1 cm to 0.2 N Resultant force, R = 12.9 × 0.2 = 2.58 N [Actual: 2.57 N]  = 50 [Actual: 50; accept  1] Direction (between resultant force and horizontal: as given in question) = 90 to horizontal/ vertically upwards [Note: Compass directions should not be used in this question, e.g. ‘due north’, as compass directions are not mentioned in the question.] 1 dia- gram 1 1 b 2.58 N [Note: The weight of the stone will balance the resultant of the two tensions, as the stone is in equilibirum.] 1 3 a During the impact, some of the kinetic energy (K.E.) of the golf club i s transferred to the golf ball. Some of the K.E. of the golf club is also converted to sound and thermal energies of the surrounding air (energy losses). The Principle of Conservation of Energy applies during the impact, as the initial K.E. of the golf club is equal to the sum of t he final K.E. of the golf club, the K.E. of the golf ball, and sound and thermal energies of the surrounding air. [Note: Initial K.E. of golf club = Final K.E. of golf club + K.E. of golf ball + Thermal and sound energies of surrounding air.] 1 1 1 bi Increase in GPE of ball between A and B = mgh = (0.045)(10)(16) = 7.20 J (3 s.f.) 1 1 bii By Conservation of Energy, Total energy at A = Total energy at B K.E. at A = G.P.E. at B + K.E. at B = 7.20 + 2.5 = 9.70 J 1

Nov 2013_Suggested ANSWERS_Aug 2023 Page 3 of 8 – Nov 2013 4 a Conduction is the transfer of thermal energy, due to the vibration of particles and the movement of free electrons. Convection is the transfer of the rmal energy, due to the movement of fluid (liquid or gas) caused by density differences. [Note: Describe each process separately.] 1 1 b 1. Colour: The heat sink is black in colour and black surfaces are good emitters of infra-red radiation. 2. Material used : The heat sink is also made of metal s which are goo d conductors of thermal energy. 3. Shape/ surface area : The heat sink consists of fins pointing upwards, which increases the surface area of the heat sink in c ontact with the surrounding air , which in turn increases the rate of thermal e nergy transfer. 1 1 1 5 a Heat capacity is the amount of thermal energy required to raise the temperature of a body by 1°C (or 1 K). Specific heat capacity is the amount of thermal energy required to raise the temperature of a unit mass (or 1 kg) of a body by 1°C (or 1 K). [Note: Give the defin itions separately to explain the difference. Merely stating that ‘heat capacity involves mass but specific heat capacity does not’ is insufficient.] [Note: • Heat capacity, C: J/ °C • Specfic heat capacity, c: J/ (kg °C) • c = C m, or C = mc] 1 b Δθ = 90 - 20 = 70°C Total thermal energy required to raise temperature of water and cup, Q = Q of water + Q of cup = mcΔθ + CΔθ = (0.20)(4200)(70) + (80)(70) = 58 800 + 5 600 = 64 400 = 64.4 kJ (3 s.f.) 1 1 1 c Thermal energy was required for the molecules to break free from the bonds between the molecules of water. 1 6 ai aii 1 1 b n = 1 sin 𝑐  1.3 = 1 sin 𝑐 c = 50.3° (3 s.f.) 1 1

Nov 2013_Suggested ANSWERS_Aug 2023 Page 4 of 8 – Nov 2013 c 1. The higher refractive index will cause the angle of refraction at A to be smaller (the light ray is refracted more towards the normal) . This will cause the angle of incidence at B to be larger. 2. The higher refractive index will also cause the critical angle, c to be smaller. [Note: The larger angle of incidence at B is now greater than the smaller critical angle.] 1 1 7 a 2 [Note: Equal numbers of positive and negative charges at either end of conductor Y.] b When X is brought close to Y, the electrons in Y are attracted by X as unlike charges attract. These electrons move to the left of Y and thus there are mor e positive charges on the right of Y. 1 1 c The excess electrons on the left of Y repel the electrons in X to the left of X, leaving excess positive charges on the right side of X. 1 d Y becomes negatively charged as electrons will flow from Earth to neutralise the positive charges on the right of Y. [Note: The final charge on Y does not depend on whether the right or left side of Y was connected to Earth. The electrons on the left of Y are attracted to X, hence leaving the positive charges on the right of Y ‘free’.] 1 e An electric field is a region where an electric charge experiences an electric force. [Note: • Gravitational field: A region in which a mass experiences a gravitational force. • Magnetic field: A region where a magnetic object experiences a magnetic force.] 1 8 ai The magnetic field at X is uniform. 1 aii The magnetic field at Y is weaker (than at X). 1 b The N-pole is the pole that points towards the geographic north pole when a magnet is freely suspended. 1

Nov 2013_Suggested ANSWERS_Aug 2023 Page 5 of 8 – Nov 2013 c 1) Place the bar magnet inside a solenoid conne cted to an alternating current (a.c. supply). 2) Withdraw the magnet slowly in a East -West direc