ACSI 2020 Y3 IP Physics Paper2

1 Anglo-Chinese School (Independent) FINAL EXAMINATION 2020 YEAR 3 INTEGRATED PROGRAMME PHYSICS PAPER 2 Tuesday 6 October 2020 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. Take g = 10 N kg-1 or 10 m s-2 Calculators are allowed for this paper. . There are 18 printed pages Marks Awarded Section Marks A B Penalty Sig. Fig. Units TOTAL SCORE Index Number Name

2 Section A [50 Marks] A1. Fig. 1.1 shows a skydiver, of mass 75 kg, falling towards the Earth at constant velocity, a long time after jumping from an aeroplane. Fig. 1.1 At time t = 0, he receives a radio signal. He opens his parachute 12 s later. Fig. 1.2 is the velocity-time graph for the skydiver. Fig. 1.2 (a) State the difference between speed and velocity. [1] .............................................................................................................................. .............................................................................................................................. 0 10 20 30 40 50 60 0 5 10 15 20 25 30 V / m s-1 time / s Earth

3 (b) The gravitational field strength g is 10 N kg-1. (i) Calculate the weight of the skydiver. [1] (ii) State the size of the air resistance acting on the skydiver between t = 0 and t = 12 s. [1] ………………………………………………………………………………… (c) For the period between t = 0 and t = 12 s, determine (i) the magnitude of the velocity of the skydiver, [1] (ii) the displacement of the skydiver. [1] (d) State (i) what happens to the air resistance as the skydiver opens his parachute. [1] .............................................................................................................................. .............................................................................................................................. (ii) the effect on the motion of the skydiver of opening the parachute. [1] .............................................................................................................................. .............................................................................................................................. (e) By t = 15 s, his parachute is fully opened. State what happened to the air resistance after t = 15 s. [1] .............................................................................................................................. ..............................................................................................................................

4 A2. (a) State Newton’s second law of motion. [1] …………………………………………………………………………………… …………………………………………………………………………………… (b) A car of mass 900 kg tows a trailer in a straight line along a horizontal road, as shown in Fig. 2.1. Fig. 2.1 The car and the trailer are connected by a horizontal tow-bar. The variation with time t of the velocity v of the car for a part of its journey is shown in Fig. 2.2. 8 9 10 11 12 13 14 15 16 0 5 10 15 20 25 v / m s-1 t / s horizontal road trailer tow bar car of mass 900 kg Fig. 2.2

5 At time t = 10 s, the resistive force acting on the car due to air resistance and friction is 520 N. The tension in the tow-bar is 420 N. For the car at time t = 10 s: (i) Use Fig. 2.2 to calculate the acceleration. [2] (ii) use your answer in (i) to calculate the resultant force acting on the car [1] (iii) show that a horizontal force of 1300 N is exerted on the car by its engine. [2]

6 A3. Fig. 3.1 shows a uniform lamina of length 3.0 m by 1.0 m and thickness 0.02 m freely suspended at pivot A and being displaced to the current position. The centre of gravity of the sheet is at point G. The density of the sheet is 690 kg m-3. Fig. 3.1 (a) Determine the mass of the sheet. [2] (b) State what is meant by center of gravity. [1] …………………………………………………………………………………… …………………………………………………………………………………… (c) The sheet is released from its current position. State the direction of the resulting moment. [1] …………………………………………………………………………………… (d) The sheet eventually stops with G directly below A. Explain why the sheet stops at this position. [1] …………………………………………………………………………………… centre of gravity A 3.0 m 1.0 m G

7 A4. A uniform beam AB of length 6.0 m is placed on a horizontal surface and then tilted at an angle of 30 ° to the horizontal, as shown in Fig. 4.1. Fig. 4.1 The beam is held in equilibrium by four forces that all act in the same plane. A force of 90 N acts perpendicular to the beam at end A. The weight W of the beam acts at its centre of gravity. A vertical force Y and a horizontal force X both act at end B of the beam. (a) State the name of force X. [1] …………………………………………………………………………………………… (b) (i) Show that the perpendicular distance measured from the line of action of force W to end B is approximately 2.6 m. [2] (ii) By taking moments about end B, calculate the weight W of the beam. [2] 6.0 m 90 N W X Y 30  A B 2.6 m

8 A5. A small coin of mass m is initially at rest. It is dropped from the top of a building of height 360 m above ground level. The coin has a speed v as it hits the ground. The gravitational field strength g is equal to 10 N kg-1. (a) Determine the speed of the coin as it hits the ground. You may ignore air resistance. [3] (b) When air resistance is negligible, a heavier coin hits the ground at the same speed as a lighter coin when they are both dropped from the same height. (i) Explain why. [1] …………………………………………………………………………………… …………………………………………………………………………………… (ii) When air resistance acts, coins of different masses do not hit the ground at the same speed when they are dropped from the same height. Explain why. [2] …………………………………………………………………………………… …………………………………………………………………………………… … ………………………………………………………………………………… …………………………………………………………………………………… … …………………………………………………………………………………

9 A6. A mercury barometer is setup at sea level as shown in Fig. 6.1. (a) Some air is trapped in region R of the tube. Using ideas about the molecules, explain how the trapped air exerts a gas pressure on the mercury below it. [2] ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… (b) If the trapped air exerts a pressure of 18 mm Hg on the mercury below it, calculate the atmospheric pressure in mm Hg and Pascal (Pa).The density of the mercury is 13 600 kg m-3. (i) in mm Hg [1] (ii) in Pa [2] test tube trapped air region R container 0.83 m 0.10 m 0.30 m X Fig.6.1

10 (c) Calcuate the total pressure at point X in Pascal. [2] (d) State the effect on the pressure in region R, using the terms