Pure Physics Nov 2020 TYS ANSWERS

Nov 2020_Suggested ANSWERS_as at July 20231 Page 1 of 6 – Nov 2020 Catholic High School | O-Level Physics NOT IN SYLLABUS: 6091 Nov 2020 P1: - Suggested Answers P2: - Paper 1 [40 marks] 1 C 11 A 21 D 31 B 2 C 12 D 22 A 32 C 3 B 13 D 23 D 33 C 4 C 14 A 24 B 34 B 5 D 15 C 25 C 35 C 6 B 16 D 26 B 36 C 7 A 17 A 27 C 37 B 8 D 18 B 28 A 38 C 9 D 19 D 29 B 39 B 10 A 20 C 30 C 40 C *Q. 4: C Students have to recognise that air resistance affects the behavior of the ball. (A and B are incorrect.) *Q. 7: A Students have to recognise that both balls being copper indicates that they have the same density. (B is incorrect.) *Q. 9: D Question was asking for the force exerted by the shoulder on stick, not F. (C is incorrect.) *Q. 11: A Students have to read the question carefully to obtain the correct information for density and height of liquid column instead of following the usual symbols. (B and C are incorrect.) *Q. 14: A To solve this question, we have to find the GPE gained and the distance moved by the box along the slope. The distance moved by the box along the slope has to be solved using Pythagoras’ Theorem. (D is incorrect.) *Q. 22: A Those who chose C did not understand that “moves backwards and forward s between two positions” means that the point is oscillating between two extreme ends whose distance apart is twice the amplitude. (C is incorrect.) *Q. 24: B Students may have mistakenly thought that when the object moves towards the lens, the image will also move toward the lens. (C and D are incorrect.) *Q. 28: A Students may have thought that the position of the earth connection affects the movement of electrons through the earth connection. (C is incorrect.)

Nov 2020_Suggested ANSWERS_as at July 20231 Page 2 of 6 – Nov 2020 *Q. 29: B The key difficulty for this question was to determine the rate of flow of electrons that is related to but different from the rate of flow of electric charge. Students have to use the information about current to determine the answer. (A, C and D are incorrect.) *Q. 30: C To solve this question, students have to add the e .m.f. of cells in the same configuration and then subtract those with the reverse configuration. (A and B are incorrect.) *Q. 34: B Students have to recall that fuses are designed to prevent excessive current from flowing through the appliance as it will melt. (D is incorrect.) 5

Nov 2020_Suggested ANSWERS_as at July 20231 Page 3 of 6 – Nov 2020 Paper 2 [80 marks] 1 a Vector: Displacement, acceleration, force, moment of a force (any one) Scalar: Distance, time, energy (any one) 1 bi Velocity of the point will change from 0.24 m / s in one direction to 0.24 m / s in the opposite direction and then back to 0.24 m / s in the initial direction. There is no net change in velocity at the end of one complete rotation even though its direction is constantly changing (despite magnitude remaining constant). 1 1 bii Radius = D / 2 = C / 2π = (velocity x time) / 2π = (0.24 x 30 x 60) / 2π → convert minutes to seconds = 69 m (least sf) 1 1 2 a Acceleration 1 bi 0.0 s to 0.6 s: Straight line starting from P to 4 m / s At 0. 6 s: Vertical straight line till -3 m / s 0.6 s to 0.8 s: Gradient of graph same as graph from 0.0 s to 0.6 s 1 1 1 bii “x” at 0.2 s 1 3 a The sum of the weight of man and the force by jetpack on man, both acting downwards on man, is balanced by force by gas from jetpack acting upwards on man. Hence, there is no resultant force acting on him. 1 bi Mass is a measure of the amount of substance in an object, while weight is the amount of gravitational force acting on it. 1 1 bii Total mass = 75 + Wjetpack /g = 75 + 160/10 = 91 kg 1 1 biii By Newton’s 2nd Law, Fnet = ma Fjetpack – Wtotal = mtotal a Fjetpack= 91 x 10 + 91 x 0.20 = 930 N (3 sf) 1 1 4 a Work done is the product of the force applied by the racket on the shuttlecock and the distance moved in the direction of the force. 1 1 b As the shuttlecock slowed down as it reaches Y, kinetic energy (KE) decreased. As the height of the shuttlecock also decreased, grav itational potential energy (GPE) also decreased. Some of the original KE and GPE is converted to thermal energy due to work done against air resistance and sound energy of air. 1 1 1 c E = ½ mv2 v = √(2E/m) = √(2 x 0.36 / 0.0050) = 12 m / s 1 1

Nov 2020_Suggested ANSWERS_as at July 20231 Page 4 of 6 – Nov 2020 5 a Thicker connecting wires have a larger cross-sectional area and hence a smaller resistance than the filament wire. As the same current passes through both types of wires, the heating effect (given by P=I2 R) is smaller for connecting wires, so less electrical energy is converted to thermal energy and they do not glow. 1 1 b Thermal energy from the filament is transferred to the gas near it. The gas is heated, expands, becomes less dense and rises, transferring thermal energy away from the filament to the top. The cooler, denser gas near the top si nks to replace the warm gas and in turn gets heated up by the filament. The process repeats, setting up convection currents in the lamp. 1 1 1 c Glass molecules are close to each other and vibrating constantly about their fixed position s due to strong inter -molecular forces . The heated gas molecules collide with the glass molecules at the inner surface of the lamp, causing them to vibrate more vigorously, transferring thermal energy to the air outside the lamp through collisions between glass molecules. 1 6 ai name of component wavelength / m Ultraviolet radiation 1 x 10-7 Infra-red radiation 1 x 10-5 1 1 aii X-rays / gamma rays 1 aiii X/Gamma rays are high energy rays that are focused on brain tumours to kill cancerous cells in radiation therapy. OR Different frequency range of X-rays are used to produce images of different shades of contents in a luggage, so as to search for illegal items in airports. 1 bi They are transverse waves. OR They are able to undergo reflection and refraction. 1 bii f = v / λ = 3.0 x 108 / 2.0 x 10-2 = 1.50 x 1010 Hz 1 1 7 ai Electrons are transferred from the screen to the cloth, so there is a net excess of positive charges on the screen. 1 aii When the dust particles are near the screen, electrons in the dust particles are attracted to the side closer to the screen, as unlike charges attract. This leaves the side of the dust particles further from the scre en to be positively charged. As the negatively charged region in the dust parti cles are nearer to the screen, the force of attraction is larger than the force of repulsion with the positively charged region in the dust , resulting i n a net attractive force between the dust particles and the screen. 1 1 1 bi As both rods are positively charged, the glass rod experiences a force of repulsion downwards, increasing the balance reading. 1 bii The charge on the plastic rod is able to exert a force on another charged object (ie. the glass rod) without any contact. 1 8 a Correct direction Near circular loops around wire Peanut shape field further out from wire 1 1 1 6

Nov 2020_Suggested ANSWERS_as at July 20231 Page 5 of 6 – Nov 2020 bi Reff = V / I = 12 / 6.0 = 2.0 Ω R + (RX // RY) = Reff = 2.0 R + (1 / 1 + 1 / 1)-1 = 2.0 R = 1.5 Ω 1 1 1 RX // RY = (1 / 1 + 1 / 1)-1 = 0.5 Ω VXY = RXY I = 0.5 x 6.0 = 3.0 V VR = 12 – 3.0 = 9 V R = VR / I = 9 / 6 = 1.5 Ω 1 1 1 bii A circuit breaker can be reset quickly by flipping the switch, unlike a fuse which has to be replaced when melted (which takes a longer time). 1 9 a switch S1 switch S2 point