Pure Physics Nov 2018 TYS ANSWERS

Nov 2018_Suggested ANSWERS_as at July 2023 Page 1 of 8 – Nov 2018 Catholic High School | O-Level Physics NOT IN SYLLABUS: 6091 Nov 2018 P1: - Suggested Answers P2: - Paper 1 [40 marks] 1 B 11 A 21 D 31 A 2 D 12 B 22 B 32 A 3 C 13 D 23 C 33 A 4 C 14 C 24 B 34 A 5 C 15 C 25 D 35 D 6 B 16 D 26 B 36 C 7 A 17 A 27 B 37 D 8 C 18 A 28 D 38 B 9 A 19 B 29 B 39 B 10 A 20 D 30 C 40 C *Q. 9: A This question tested the way in which the gravitational fields of the Earth and the Moon interact. It is a commonly held, but erroneous view that beyond the Earth’s atmosphere, objects no longer experience a gravitational force due to the Earth. (B is incorrect.) *Q. 12: B Both of the forces referred to in the question are one force of a Newton’s third law pair and it is necessary to be clear which of the two forces, in each case, is the one being asked for. Confusion with this led to the selection of any of the three incorrect options. *Q. 15: C The question concerns the energy transfer that is taking place as the stone slows down as it moves downwards in the water. Both the kinetic energy and the gravitational potential energy are decreasing during this time. (A and B are incorrect.) *Q. 21: D Convection in air can only transfer heat upwards. Consequently, the options that include convection could not be correct. The question states that the temperature of the air is greater than that of the ice and so there will also be some transfer of heat by radiation. Thermal radiation from gases must be considered. *Q. 30: C It is commonly understood that the resistance of the circuit decreased, and deduced from this that the current supplied by the cell and hence the current in the lamp, increased – thus an increased potential difference across the lamp was often successfully chosen. As there are two significant potential differences in the circuit – one increases whilst the other decreases – it is easy to be confused. (A is incorrect.)

Nov 2018_Suggested ANSWERS_as at July 2023 Page 2 of 8 – Nov 2018 *Q. 32: A This question tested both factual knowledge and one consequence of the fact. Most are aware that a voltmeter measures the potential difference between the two terminals of the component with which it is connected in parallel. The question makes clear that the voltmeter does not affect the current and it is wrong to interpret this to mean that the resistance of the voltmeter had to be small. (B is incorrect.) *Q. 33: A (B and C are incorrect.)

Nov 2018_Suggested ANSWERS_as at July 2023 Page 3 of 8 – Nov 2018 Paper 2 [80 marks] 1 ai The value on the main scale is 7.50 mm, and should be added to 0.48 mm on the rotating scale. 1 1 aii Action: Close the jaws of the micrometer completely. Reason: To check for and subtract away the zero error. OR: Action: Take average of diameters at different positions along cylinder. Reason: The cylinder may not be uniformly shaped. [Note: ‘Precaution’ = ‘Action’ + ‘Reason’] 1 b Diameter of cylinder = 7.98 mm = 0.798 cm Volume of cylinder = 𝜋( 𝑑 2)2𝑙 = 𝜋 ( 0.798 2 ) 2 (2.6) cm3 Density = Mass Volume = 3.5 𝜋(0.798 2 ) 2 (2.6) = 2.69 g/cm3 (3 s.f.) 1 1 1 2 a Fr = ma (2.5 × 105) – (1.2 × 105) = (1.5 × 106)a a = 0.0867 m/s2 (3 s.f.) 2 1 b The force acting forwards on the tanker due to the propellers is constant, but the drag force acting on the tanker increases with speed. Thus the resultant force acting on the tanker decreases and the acceleration decreases by Newton’s Second Law. 1 1 ci 2 cii The gradient of the displacement-time graph, which gives the velocity, is decreasing. 1 3 a Kinetic energy, K.E. = ½mv2 = ½(60)(2.5)2 = 188 J (3 s.f.) 1 1 bi There is friction between the slide and the child. Hence, the child slows down and the kinetic energy is converted to thermal energy (internal energy) of the child and the slide. 1 1 bii Internal energy is the sum of the potential and kinetic energies of all the molecules in a body. 1 biii There is no height difference between X and Y. 1 4 a The air molecules are in constant, random motion. 1 b The air molecules collide with the sides of the containers and exert a force on them. This creates a force per unit area, which is the pressure on the sides. 1 1

Nov 2018_Suggested ANSWERS_as at July 2023 Page 4 of 8 – Nov 2018 The pressure is large due to the large number of air molecules colliding with the slides. 1 5 a Due to magnetic induction, the bottom of the iron plate becomes an induced South pole and the top becomes an indcued North pole. As the f orce of attraction between unlike poles is larger than the force of repulsion between like poles , there is a net attractive force between the magnet and the iron plate. 1 1 b Due to electrostastic induction, electrons flow from the earth to the bottom of the iron plate. Unlike charges attract, so the foil moves towards the iron plate. 1 1 c Both the aluminuim foil and the iron plate are electrical conductors. Hence, electrons on the iron plate flow to neutralise the positive charges on the aluminuim foil. 1 1 6 a 2 b RP = V ÷ I = 1.5 ÷ 0.25 = 6.00  (3 s.f.) 1 1 c ( 1 𝑅𝑃 + 1 𝑅𝑅 ) −1 + RQ = 12 ( 1 6.00 + 1 18) −1 + RQ = 12 RQ = 7.50  (3 s.f.) 1 1 di The current splits to flow through resistor R and lamp P, before joining to flow through lamp Q. 1

Nov 2018_Suggested ANSWERS_as at July 2023 Page 5 of 8 – Nov 2018 dii As the current that flows through lamp Q is higher, it has a higher temperature. Thus the resistance of lamp Q increases as filament lamps are non-ohmic. 1 1 7 a 1 bi As the coil rotates, the magnetic flux linking the magnet and the coil is changing. According to Faraday’s Law of Electromagnetic Induction, an electromagnetic force (e.m.f.) is induced in the coil. 1 1 bii C 1 biii The coil is in the vertical position, with side X above side Y. 1 c The output current from the a.c. generator is constantly changing in both direction and magnitude, but the current from a battery is constant in both direction and magnitude. 1 8 a 1. The particle must be charged. 2. The particle must be moving in a direction not parallel to the magnetic field. 1 1 b 3 9 a Boiling Evaporation Occurs at particular temperature Occurs at any temperature Relatively fast Relatively slow Takes place throughout liquid Takes place only at surface liquid Bubbles are formed No bubbles are formed Temperature remains constant Temperature may change External thermal energy source No external thermal energy source 2

Nov 2018_Suggested ANSWERS_as at July 2023 Page 6 of 8 – Nov 2018 bi Q = mlv 2.3 × 106 = m(2.4 × 106) Mass of water lost by evaporation in 1 hr, m = 0.958 kg (3 s.f.) 1 1 bii Energy gained by body in one hour = 3.2 × 108 – 2.3 × 106 = 9.0 × 105 J Q = mc  = Q ÷ mc = 9.0 × 105 ÷ (70)(3500) = 3.67 C (3 s.f.) 1 1 1 biii The rise of 3.67 C will cause the average body temperature to be 37 + 3.67 = 40.67 C, which will be higher than 40 C. 1 biv Energy lost by vaporisation when the runner exhales in 1 hr = 3.2 × 106 – 2.3 × 106 = 9.0 × 105 J The whole race takes 3.5 hrs. Total energy loss in 3.5 hrs = (9.0 × 105) × 3.5 = 3.15 × 106 J Q = mlv 3.15 × 106 = m(2.4 × 106) Mass of water vapour exhaled in 3.5 hr, m = 1.31 kg (3 s.f.) 1 1 bv From (b)(i): mass of water lost by evaporation in 1 hr = 0.958 kg From (b)(iv): mass of water vapour exhaled in 3.5 hr = 1.31 kg Total mass of water lost = 1.31 + 3.5(0.958) = 4.663 kg Percentage of body mass lost by evaporation = (4.663 ÷ 70) × 100% = 6.67% (> 5%) The runner suffers from severe dehydration. 1 1 10 a Potentiometer (OR: Potential divider) 1 b 2 ci From 0 to 0.54 V, the current is zero, and the resistance of the diode is infinite/ v