Pure Physics Nov 2012 TYS ANSWERS

Nov 2012_Suggested ANSWERS_Aug 2023 Page 1 of 6 – Nov 2012 Catholic High School | O-Level Physics NOT IN SYLLABUS: 5059 Nov 2012 P1: - Suggested Answers P2: - Paper 1 [40 marks] 1 A 11 A 21 A 31 A 2 A 12 D 22 D 32 D 3 A 13 D 23 B 33 B 4 B 14 B 24 C 34 A 5 C 15 B 25 B 35 C 6 B 16 B 26 C 36 D 7 B 17 D 27 B 37 B 8 A 18 B 28 D 38 C 9 B 19 C 29 D 39 D 10 C 20 A 30 A 40 A *Q. 6: (B) The weight of the rocket must not be neglected. In F = ma, F is the resultant force. Thus F is not the thrust alone but F = thrust - weight. (D is incorrect.) *Q. 13: (D) There is kinetic energy at the top of the hill. (A is incorrect.) *Q. 22: (D) The image cannot be formed at the mirror surface unless the object is at the mirror surface. (B is incorrect.) *Q. 23: (B) There is partial reflection at the boundary. (A and D are incorrect.) *Q. 24: (C) The ray from O enters the lens normal to the surface PQ because O is the centre of the circle. Bending of the ray cannot occur at surface PQ, so D cannot be correct. (D is incorrect.) *Q. 34: (A) The flat part of the curve in B suggests that the resistance has become constant, which is inconsistent with as it is stated that ‘the resistance of the thermistor decreases as its temperature increases’ in the question. (B is incorrect.) *Q. 40: (A) The question asked for the current in the primary coil. (C is incorrect.) Paper 2 [80 marks] 1 a a = v - u t = 25−20 10−6 = 5 4 = 1.25 m s-2 (3 s.f.) 2 1 b The shaded area under the speed -time graph represents the distance travelled by the car, when it had a constant deceleration between 10 – 16 s. 1 1 ci The average speed of the car is the total distance travelled divided by the total time taken. [Note: Not ‘total distance per unit time.’!] 1 * cii The speed of the car between 6 – 16 s is always higher than 20 m s -1. 1

Nov 2012_Suggested ANSWERS_Aug 2023 Page 2 of 6 – Nov 2012 Between 0 – 20 s, the speed of the car is 20 m s-1 for half the journey. [Or: The time between 6 – 16 s is half the time taken during the journey. However the distance travelled is more than half.] [Or (longer method): Average speed between 6 – 16 s = Total distance Total time = 1 2(4)(20+25)+ 1 2(6)(25+20) 10 = 225 10 = 22.5 = 23 m s-1 (2 s.f.) Average speed between 0 – 20 s = Total distance Total time = 225+(10)(20) 20 = 425 20 = 21.25 = 21 m s-1 (2 s.f.) < 23 m s-1] [Note: Need to show comparison between 6 – 16 s and 0 – 20 s.] 2 a Chemical potential energy of swimmer to kinetic energy of swimmer. 1 b K.E. = ½mv2 = ½(60)(0.80)2 = 19.2 J (3 s.f.) 1 1 * ci The swimmer does work on the water by exerting a force backwards on the water, pushing the water backwards. [Note: Need to demonstrate ‘work done’, by showing the idea of ‘a force applied’ and ‘a distance moved’ → work done by swimmer or on the water.] 1 cii The work done by the swimmer on the water eventually becomes the thermal energy of the water. [Note: Not ‘kinetic energy of water’.] 1 3 a EM waves are able to travel through a vacuum. [OR: EM waves travel at the speed of 3.0 × 108 m s-1 in a vacuum.] 1 bi Visible light and infra-red radiation. 1 bii 1. As there is l ess current , th ere is less infra-red radiation emitted ( ie filament is less hot). 2. The radiation emitted has a lower frequency/ higher wavelength. [Or: Only the visible red part of the EM spectrum is emitted.] 1 1 ci Metal is a good conductor of heat and hence heat is conducted from the metal filament to the metal support through transfer of kinetic energy during collisions between filament molecules and support molecules. 1 cii When the nitrogen gas is heated by the metal filament, it expands, becomes less dense and rises. The cooler, denser nitrogen gas sinks to replace the warmer gas, forming convection currents. 1 1 4 a  = m V Mass of (cold or hot) air = V = (0.0012)(30) = 0.036 g Density of cold air = m V = 0.036 20 = 0.00180 g cm-3 (3 s.f.) 1 1 bi The molecules in the cylinder are in constant, random motion and collide with the walls of the cylinder, thus exerting a force per unit area. 1 bii The molecules of the cold air are moving slower at a lower temperature and 1

Nov 2012_Suggested ANSWERS_Aug 2023 Page 3 of 6 – Nov 2012 collide less frequently and less forcefully with the walls of the cylinder and piston combined. The molecules exert a smaller force but also on a smaller area of the walls of the cylinders and piston combined (as the total volume has decreased), to create the same pressure. [Or (focusing on the force exerted on the piston only): The molecules of the cold air are moving slower at a lower temperature and hence, the force of collisions will decrease on the fixed area of the piston. But the frequency of collisions increases on the fixed area of the piston, to compensate for the lower force of collisions when the temperature decreases, to create the same pressure.] 1 c As the cylinder is cooled and the pressure of the air inside decreases, the larger atmospheric pressure outside pushes the piston to move inwards. 1 5 a 95 s 1 * b The liquid is at a higher temperature (70 to 100 C) compared to the surrounding air (20 C) throughout the experiment. 1 c 1. Thermal energy is lost to the laboratory as latent heat of fusion. 2. Intermolecular bonds are being formed. 3. There is a change in state from liquid to solid/ the kinetic energy of the particles remains constant. 2 Any 2 d [Note: Beaker of water is kept at a temperature of 90C.] 1 6 a 3 [NOTE: 1. Image to be formed at 4.0 cm, ± 1 small square. 2. Dotted lines for virtual image.] b The image is formed by dotted lines, which are not light rays but which appear to come from the object. 1

Nov 2012_Suggested ANSWERS_Aug 2023 Page 4 of 6 – Nov 2012 c 1. The image becomes less enlarged. 2. The image moves closer to the lens. 2 7 a Total resistance = ( 1 R1 + 1 R2 )-1 = ( 1 30+30 + 1 10+30)-1 = ( 1 60 + 1 40)-1 = 24.0  (3 s.f.) 1 1 b V = IR  I = V R = 6 24.0 = 0.250 A (3 s.f.) 1 1 ci The e.m.f. of (or p.d. across) the battery is shared equally between the two resistors in series as they have the same resistance. 1 cii Potential difference between C and D = R1 R1+ R2 × 6.0 = 30 30 + 10 × 6.0 = 4.50 V (3 s.f.) 1 * ciii Potential difference between A & C = Potential at C - Potential at A = 4.50 - 3.0 = 1.50 V [NOTE: p.d. = higher potential - lower potential] 1 8 ai When the switch is closed, the current in the coil generates a magnetic field which interacts with the magneti c field of the magnet to produce a n unbalanced resultant magnetic field. This causes an upward force to be produced on the side of the coil near the North pole, and a downwards force on the side of the coil near the South pole (using Fleming’s Left Hand Rule). 1 1 aii When the coil passes the vertical position, the direction of the current in the coil reverses due to the split ring commutator. This causes the force on the side of the coil near the North pole to be in the same upwards direction (and for the force on the side of the coil near the South pole to be in the same downwards direction) , and the coil rotates clockwise. 1 1 bi 1 bii 1. The amplitude (maximum moment) on Fig. 8.2 increases. 2. The period on Fig. 8.2 decreases. [OR: The frequency on Fig. 8.2 increases.] 1 1 9 ai Similarity: The speeds of water waves in both deep and shallow water increase with wavelengths (up to wavelengths of 400 m). [OR: The speeds of water waves in both deep and shallow water are similar (up to wavelengths of 40 m).] 1 aii Difference: The speed of water waves increases constantly with wavelength in deep water, but the speed of water waves stop increasing and become constant from wavelengths of 400 m onwar