ACSI 2019 Y3 IP Physics Paper2 Answers

1 END-OF-YEAR EXAMINATION 2019 YEAR 3 INTEGRATED PROGRAMME PHYSICS PAPER 2 Mark Scheme Monday 7 October 2019 1 hour 45 minutes INSTRUCTIONS TO STUDENTS Write your index number in the box provided on the top right corner of this page. Do not open this booklet until you are told to do so. Section A Answer all questions in the spaces provided in the paper. Section B Answer all questions in the spaces provided in the paper. INFORMATION FOR STUDENTS Candidates are reminded that all quantitative answers should include appropriate units. Candidates are advised to show their answers in a clear and orderly manner as more marks are awarded for sound use of physics than for correct answers. The number of marks is given in brackets [ ] at the end of each question or part question. Calculators are allowed for this paper. Take g = 10 N/kg. There are 18 pages. Section A [50 Marks] Marks Awarded Section Marks A B Penalty Sig. Fig. Units TOTAL SCORE Index Number

2 A1 A 6.0 kg parcel is dropped off vertically from a helicopter 800 m above the ground. Its parachute is activated some time after the fall. The speed -time graph for the first 25 seconds of its journey is shown in Fig. A1.1. Fig. A1.1 (a) Explain what is meant by “The Earth’s gravitational field strength has a magnitude of 10 N kg-1”. [1] There is a force of 10 N acting on per unit mass of 1 kg. (b) Explain, in terms of forces, why the parcel’s velocity increased immediately after being dropped. [2] The weight of the parcel is larger than the air resistance acting on it. There is a net force in the downwards direction and hence the parcel’s velocity increases immediately. (c) State the time at which the parcel first achieves terminal velocity. [1] 10 s speed / m s-1 time / s

3 (d) Draw a labelled diagram to show all the forces that act on the parcel at t = 15 s. Indicate the values of these forces on your diagram. [2] Correct direction of arrows, correct values indicated. eeoo (e) As the parcel falls, the Earth exerts a downward force on the parcel. Name the reaction force. [1] (upwards) Force by parcel on Earth. Weight = 600 N Air Resistance = 600 N

4 A2 A ball of weight W is suspended by two ropes as shown in Fig. A2.1 (not drawn to scale). The tension in the ropes are 870 N and 500 N. Fig. A2.1 (a) What is the resultant force acting on the weight W? [1] 0 N (b) By means of a scaled diagram, determine the weight W. State the scale used. [4] Scale : scale: 1.0 cm rep 10 N [A1] Correct diagram with arrows W = 1000 N (accept from 950 N to 1050 N) 870 N 500 N 60  30 

5 (c) State the effect on the tension of each rope when the ropes are placed closer to each other. The weight W remains unchanged. [1] The tension decreases.

6 A3 Jonathan applies a vertical force F on a pedal of his bicycle, as shown in Fig. A3.1. Fig. A3.1 (a) As he travels along, the pedal moves through a circle of radius 8.0 cm. For the pedal in the position as shown in Fig. A3.1, the line of action of the force, F is 5.0 cm from the pivot. Calculate the moment of force, F, about the pivot when F = 120 N. [2] Moment of force F = 120 x 5.0 = 600 N cm or 6.0 Nm (b) Subsequently, the pedal now moves from position A to position B as shown in Fig. A3.1. Explain why the same force of 120 N applied at A will have a different effect from the same force applied at B. Justify your answer with calculations. [3] The perpendicular distance of the line of action of the vertical downward force F from the pivot changes (from 0.0 cm at A to a maximum of 8.0 cm at B). The moment produced will change from 0 N cm to 120 x 8.0 = 960 N cm. (c) Johnathan and Tony are cycling at the same speed. The combined mass of Johnathan and his bicycle is 100 kg while the combined mass of Tony and his bicycle is 120 kg. Both decided to stop for a break. Who do you think will be more difficult to stop? Why? [2] Tony will be more difficult to stop as he has a larger inertia. F = 120N

7 A4 Fig. A4.1 shows a slide in a playground. A boy of mass 35.0 kg climbs up from the side AB in 10.0 seconds and slides down from B to C. Then he decelerates uniformly in the horizontal section CD to a complete stop. The vertical height of the slide is 5.000 m. The horizontal section CD is 0.500 m above ground level. Fig. A4.1 (a) Find the average power developed by the boy when he climbs up the slide. [2] Gain in gravitational potential energy = mgh = 35.0 x 10 x 5.000 = 1750 J Average power = 1750 / 10.0 = 175 W (b) What is the loss in gravitational potential energy of the boy when he moves from point B to point C? [1] Loss in gravitational potential energy = mgh = 35.0 x 10 x (5.000 – 0.500) = 1575 J or 1580 J (c) If the work done against friction when he moves from B to C is 1000 J, what is the kinetic energy possessed by the boy at point C? [2] Along BC, By conservation of energy, Gravitational energy at B = gravitational potential energy at C + kinetic energy at C + work done to overcome friction Loss in gravitational potential energy = KE at C + 1000 1580 = KE at C +1000 KE at C = 580 J

8 (d) State the energy transformation in the horizontal section CD. [1] Energy is converted to work done against friction. (e) The length of the horizontal section CD is 5.000 m. Find the magnitude of friction, assuming that friction is constant. [2] Work done against friction = 580 J f x 5.000 = 580 f =116 N A5 Fig. A5.1 shows two vessels, X and Y, connected by a narrow tube and kept in water baths of temperatures 323 K and 363 K respectively. Compared to vessel Y, vessel X has a larger volume. The vessels contain the same type of gas. Fig. A5.1 (a) Discuss whether the following statements are true or false. (i) After some time, the gases in both vessels X and Y would reach the same equilibrium temperature. [2] False. The gases contained inside the respective vessels will be in thermal equilibrium with the water baths that they are immersed in and not with each other. Vessel X Vessel Y 323 K 363 K

9 (ii) After a long time, the gas pressure in vessel X is smaller than the gas pressure in vessel Y. [2] False. Gas particles will move from a region of high pressure to a region of low pressure until the gas pressures in both vessels are the same. (Note: Pressure Y will be greater than X actually for line 1 substantiation.) (b) Vessel Y is removed from the water bath and is then connected to a U -tube manometer as shown in Fig. A5.2 below. Fig. A5.2 (i) Given that the density of oil is 850 kg m-3 and atmospheric pressure is 1.0 × 105 Pa, calculate how much greater than atmospheric pressure is the pressure of the gas in vessel Y? [2] Py = P atm + hpg Py – P atm = hpg = (0.16-0.06)(850)(10) = 850 Pa

10 (ii) If the oil inside the manometer is replaced by water, would the level of water in the vertical arm of the manometer which is exposed to atmospheric pressure increase or decrease in height? Explain your answer. [2] The density of water is more than density of oil. Since the other end of the manometer is connected to the same gas supply, if water replaces oil, the level of water in the vertical arm that is exposed to atmospheric pressure will decrease. A6 Fig. A6.1 shows a column of gas trapped in an air-tight container. A tight-fitting piston prevent