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