Pressure — Paper 3 · IGCSE Physics 0625

5 (a) Fig. 5.1 shows a tank used for evaporating salt solution to produce crystals. For Examiner’s Use evaporating tank steam in salt solution steam out Fig. 5.1 Suggest two ways of increasing the rate of evaporation of the water from the solution. Changes may be made to the apparatus, but the rate of steam supply must stay constant. You may assume the temperature of the salt solution remains constant. 1. … … 2. … … [2] (b) A manufacturer of liquid-in-glass thermometers changes the design in order to meet new requirements. Describe the changes that could be made to (i) give the thermometer a greater range, … [1] (ii) make the thermometer more sensitive. … [1] (c) A toilet flush is operated by the compression of air. The air inside the flush has a pressure of 1.0 × 105 Pa and a volume of 150 cm3. When the flush is operated the volume is reduced to 50 cm3. The temperature of the air remains constant during this process. Calculate the new pressure of the air inside the flush. pressure = … [2]

6 marks

Mark scheme: 5 (a) increase surface area of tank B1 blow air over surface/put in windy place B1 2 (b) (i) capillary tube longer or liquid with lower expansivity B1 (ii) capillary tube thinner/finer or liquid with higher expansivity or bigger bulb B1 2 (c) p1v1 = p2v2 or 1 x 105 x 150 = p2 x50 C1 p2 = 3 x 105 (Pa) A1 2 [6]

This question in 0625/31 May/June 2006

2 Fig. 2.1 shows a steam safety valve. When the pressure gets too high, the steam lifts the weight W and allows steam to escape. 0.2 m pivot W force of steam Fig. 2.1 (a) Explain, in terms of moments of forces, how the valve works. … … … … [2] (b) The moment of weight W about the pivot is 12 N m. The perpendicular distance of the line of action of the force of the steam on the valve from the pivot is 0.2 m. The area of the piston is 0.0003 m2. Calculate (i) the minimum steam force needed for the steam to escape, force = … [2] (ii) the minimum steam pressure for the steam to escape. pressure = … [2] [Total: 6]

6 marks

Mark scheme: 2 (a) moment of W down/anticlockwise, moment of steam opposite C1 when moment of steam > moment of W, steam escapes OR when clockwise moment > anticlockwise moment, steam escapes A1 [2] (b) (i) 12 = 0.2 F C1 F = 60 N c.a.o. allow 60–61 for ans if working for 60 N shown A1 [2] (ii) (P =) F/A or 60/0.0003 e.c.f. C1 2 × 105 Pa or 200 000 Pa e.c.f. (accept N/m2) OR 20 N/cm2 A1 [2] [Total: 6] IGCSE – May/June 2007 0625 03

This question in 0625/31 May/June 2007

6 (a) A man squeezes a pin between his thumb and finger, as shown in Fig. 6.1. For Examiner’s Use finger pinhead pin thumb Fig. 6.1 The finger exerts a force of 84 N on the pinhead. The pinhead has an area of 6.0 × 10–5 m2. (i) Calculate the pressure exerted by the finger on the pinhead. pressure = … [2] (ii) State the value of the force exerted by the pin on the thumb. … [1] (iii) Explain why the pin causes more pain in the man’s thumb than in his finger. … … [2] (b) The density of the water in a swimming pool is 1000 kg / m3. The pool is 3 m deep. For Examiner’s (i) Calculate the pressure of the water at the bottom of the pool. Use pressure = … [2] (ii) Another pool has the same depth of water, but has twice the area. State the pressure of the water at the bottom of this pool. pressure = … [1] [Total: 8]

8 marks

Mark scheme: 6 (a) (i) P = F /A in any form, letters, words or numbers C1 1.4 × 106 Pa accept N/m2 A1 (ii) 84 N OR 84.0 N B1 (iii) same force over (much) smaller area B1 (much) bigger pressure B1 (b) (i) P = hdg in any form, letters, words or numbers C1 3 × 104 Pa OR 30 000 Pa OR 30 kPa accept N/m2 A1 (ii) his (i) B1 [8]

This question in 0625/31 May/June 2009

6 A vertical cylinder has a smooth well-fitting piston in it. Weights can be added to or removed For from a tray on the top of the piston. Examiner’s Use (a) Weights are added to the tray, as shown in Fig. 6.1. weights piston air cylinder Fig. 6.1 (i) State what happens to the pressure of the air in the cylinder as a result of adding these weights. … [1] (ii) The initial pressure of the trapped air is 1.05 × 105 Pa. When the weights are added, the volume of the air decreases from 860 cm3 to 645 cm3. The temperature of the air does not change. Calculate the final pressure of the trapped air. pressure = … [3] (iii) The area of the piston is 5.0 × 10–3 m2. Calculate the weight that is added to the piston. weight added = … [4] (b) The weights are kept as shown in Fig. 6.1. The temperature of the air in the cylinder is For increased. Examiner’s Use (i) State what happens to the volume of the air in the cylinder as a result of this temperature rise. … [1] (ii) State how, if at all, the pressure of the air changes as the temperature changes. … [1] (iii) State what must be done to prevent the volume change in (b)(i). … [1] (iv) The volume change in (b)(i) is prevented. State what happens to the pressure of the air in the cylinder. … [1] [Total: 12]

12 marks

Mark scheme: 6 (a) (i) increases B1 (ii) pV = const in any form C1 1.05 (× 105) × 860 (× 10–6) = p × 645 (× 10–6) C1 1.4 × 105 Pa A1 IGCSE – October/November 2009 0625 31 (iii) F = pA in any form accept weight for F C1 EITHER increase in pressure = 0.35 × 105 (Pa) C1 0.35 × 105 × 5.0 × 10–3 C1 175 N (minimum 2 s.f.) c.a.o. A1 OR 1.05 × 105 × 5.0 × 10–3 or 525 N or 1.4 × 105 × 5.0 × 10–3 or 700 N (C1) 700 – 525 N e.c.f. from (a) (ii) (C1) 175 N (minimum 2 s.f.) c.a.o. (A1) (b) (i) increases B1 (ii) no change B1 (iii) extra weight (on tray/piston) B1 (iv) increases B1 [12]

This question in 0625/31 Oct/Nov 2009

6 A vertical cylinder has a smooth well-fitting piston in it. Weights can be added to or removed For from a tray on the top of the piston. Examiner’s Use (a) Weights are added to the tray, as shown in Fig. 6.1. weights piston air cylinder Fig. 6.1 (i) State what happens to the pressure of the air in the cylinder as a result of adding these weights. … [1] (ii) The initial pressure of the trapped air is 1.05 × 105 Pa. When the weights are added, the volume of the air decreases from 860 cm3 to 645 cm3. The temperature of the air does not change. Calculate the final pressure of the trapped air. pressure = … [3] (iii) The area of the piston is 5.0 × 10–3 m2. Calculate the weight that is added to the piston. weight added = … [4] (b) The weights are kept as shown in Fig. 6.1. The temperature of the air in the cylinder is For increased. Examiner’s Use (i) State what happens to the volume of the air in the cylinder as a result of this temperature rise. … [1] (ii) State how, if at all, the pressure of the air changes as the temperature changes. … [1] (iii) State what must be done to prevent the volume change in (b)(i). … [1] (iv) The volume change in (b)(i) is prevented. State what happens to the pressure of the air in the cylinder. … [1] [Total: 12]

12 marks

Mark scheme: 6 (a) (i) increases B1 (ii) pV = const in any form C1 1.05 (× 105) × 860 (× 10–6) = p × 645 (× 10–6) C1 1.4 × 105 Pa A1 (iii) F = pA in any form accept weight for F C1 EITHER increase in pressure = 0.35 × 105 (Pa) C1 0.35 × 105 × 5.0 × 10–3 C1 175 N (minimum 2 s.f.) c.a.o. A1 OR 1.05 × 105 × 5.0 × 10–3 or 525 N or 1.4 × 105 × 5.0 × 10–3 or 700 N (C1) 700 – 525 N e.c.f. from (a) (ii) (C1) 175 N (minimum 2 s.f.) c.a.o. (A1) IGCSE – October/November 2009 0625 32 (b) (i) increases B1 (ii) no change B1 (iii) extra weight (on tray/piston) B1 (iv) increases B1 [12]

This question in 0625/32 Oct/Nov 2009

3 An ornamental garden includes a small pond, which contains a pumped system that causes For water to go up a pipe and then to run down a heap of rocks. Examiner’s Use Fig. 3.1 shows a section through this water feature. water runs down rocks pumped water rises through pipe 0.8 m rocks water inlet pump to pump Fig. 3.1 The density of water is 1000 kg / m3. A volume of 1 litre is equal to 0.001 m3. (a) Calculate the mass of 1 litre of water. mass = … [2] (b) Calculate the work done raising 1 litre of water through a height of 0.8 m. work = … [2] (c) The pump lifts 90 litres of water per minute. For Examiner’s Calculate the minimum power of the pump. Use power = … [2] (d) The pump is switched off. Immediately after the pump is switched off, what is the value of the water pressure at the bottom of the 0.8 m pipe, due to the water in the pipe? pressure = … [2] [Total: 8]

8 marks

Mark scheme: 3 (a) M = V × D in any form OR 103 × 10-3 C1 1 kg A1 (b) mgh OR his (a) × 10 × 0.8 C1 8 J (Nm) OR 7.85 J OR 7.84 J e.c.f. from (a) A1 (c) P = E/t OR (his 8 × 90) / 60 e.c.f. from (b) C1 12 W (J/s or Nm/s) OR 11.77 W OR 11.76 W A1 (d) ρgh in any form, words, letters, numbers C1 8000 Pa (N/m2) OR 7850 Pa OR 7840 Pa A1 [8]

This question in 0625/32 May/June 2010

3 An ornamental garden includes a small pond, which contains a pumped system that causes For water to go up a pipe and then to run down a heap of rocks. Examiner’s Use Fig. 3.1 shows a section through this water feature. water runs down rocks pumped water rises through pipe 0.8 m rocks water inlet pump to pump Fig. 3.1 The density of water is 1000 kg / m3. A volume of 1 litre is equal to 0.001 m3. (a) Calculate the mass of 1 litre of water. mass = … [2] (b) Calculate the work done raising 1 litre of water through a height of 0.8 m. work = … [2] (c) The pump lifts 90 litres of water per minute. For Examiner’s Calculate the minimum power of the pump. Use power = … [2] (d) The pump is switched off. Immediately after the pump is switched off, what is the value of the water pressure at the bottom of the 0.8 m pipe, due to the water in the pipe? pressure = … [2] [Total: 8]

8 marks

Mark scheme: 3 (a) M = V × D in any form OR 103 × 10-3 C1 1 kg A1 (b) mgh OR his (a) × 10 × 0.8 C1 8 J (Nm) OR 7.85 J OR 7.84 J e.c.f. from (a) A1 (c) P = E/t OR (his 8 × 90) / 60 e.c.f. from (b) C1 12 W (J/s or Nm/s) OR 11.77 W OR 11.76 W A1 (d) ρgh in any form, words, letters, numbers C1 8000 Pa (N/m2) OR 7850 Pa OR 7840 Pa A1 [8]

This question in 0625/33 May/June 2010

3 Fig. 3.1 shows a hydraulic lift in a car repair workshop. car support hydraulic fluid 4 pistons, each of area 0.02 m2 piston A, area 0.01 m2 Fig. 3.1 The hydraulic fluid transmits the pressure, caused by piston A, equally to each of the four pistons holding up the car supports. The pressure throughout the fluid is the same. A force of 1000 N on piston A is just enough to raise the car. (a) Using values from Fig. 3.1, find (i) the pressure caused by piston A on the fluid, pressure = … [2] (ii) the total upward force caused by the fluid. force = … [3] (b) The weight of each of the two car supports is 1000 N. Calculate the mass of the car. mass = … [2] [Total: 7]

7 marks

Mark scheme: 3 (a) (i) (P =) F/A in any form OR 1000/0.01 C1 100 000 Pa accept N/m2 A1 (ii) multiplication of either force or area by 4 C1 0.08 × his (i) OR 0.02 × his (i) C1 8000 N e.c.f. from (i) A1 (2000 N gets C0, C1, A1) (b) his (ii) – 2000 correctly evaluated C1 600 kg e.c.f. A1 [Total: 7] IGCSE – October/November 2010 0625 31

This question in 0625/31 Oct/Nov 2010

5 The front views of two cars are shown in Fig. 5.1, to the same scale. family car racing car Fig. 5.1 (a) Suggest which car has the greater stability, and give two reasons. car … reason 1 … … reason 2 … … [2] (b) The cars have the same weight. Study Fig. 5.1 and suggest why the stationary racing car exerts less pressure on the ground. … … [1] (c) The family car’s tyres each have an area of 0.012 m2 in contact with the ground. The weight of the car and its contents is 9600 N. Calculate the pressure exerted by the car on the ground. pressure = … [2] [Total: 5]

5 marks

Mark scheme: 5 (a) racing car + 1 correct reason M1 2nd correct reason A1 correct reasons: • wider (car) • lower (centre of mass/gravity) NOT wider tyre/surfaces o.w.t.t.e. (b) larger/wider tyres/area (of contact) ignore base area B1 (c) F/A OR 9600/0.012 OR 9600/0.048 OR 9600/(4 × 0.012) OR 800,000 C1 2 x 105 Pa OR 200 000 Pa (accept N/m2) c.a.o. A1 [Total: 5] IGCSE – October/November 2010 0625 32

This question in 0625/32 Oct/Nov 2010

3 During a period of hot weather, the atmospheric pressure on the pond in Fig. 3.1 remains constant. Water evaporates from the pond, so that the depth h decreases. force due to air pressure h Fig. 3.1 (a) Study the diagram and state, giving your reason, what happens during this hot period to (i) the force of the air on the surface of the pond, … … [1] (ii) the pressure at the bottom of the pond. … … [1] (b) On a certain day, the pond is 12 m deep. (i) Water has a density of 1000 kg / m3. Calculate the pressure at the bottom of the pond due to the water. pressure due to the water = … [2] (ii) Atmospheric pressure on that day is 1.0 × 105 Pa. Calculate the total pressure at the bottom of the pond. total pressure = … [1] (iii) A bubble of gas is released from the mud at the bottom of the pond. Its initial volume is 0.5 cm3. Ignoring any temperature differences in the water, calculate the volume of the bubble as it reaches the surface. volume = … [2] (iv) In fact, the temperature of the water is greater at the top than at the bottom of the pond. Comment on the bubble volume you have calculated in (b)(iii). … … … [1] [Total: 8]

8 marks

Mark scheme: 3 (a) (i) smaller because area smaller B1 (ii) smaller because depth/height smaller ignore less water B1 (b) (i) hρg OR 12 × 1000 × 10 C1 1.2 × 105 Pa OR 1.1772 × 105 Pa OR 1.176 × 105 Pa accept N/m2 A1 (ii) candidate’s (i) + 1.0 × 105 Pa correctly evaluated with unit (correct value 2.2 × 105) B1 (iii) p1V1 = p2V2 in any form C1 1.1 cm3 OR 0.5 × candidate’s (ii)/105 correctly evaluated A1 (iv) value in (iii) too small OR volume larger o.w.t.t.e. B1 [8]

This question in 0625/31 May/June 2011

4 (a) Complete Fig. 4.1 to show a simple mercury barometer. Insert the correct labels in the boxes. Label with the letter h the measurement required to calculate the pressure of the atmosphere. [3] Fig. 4.1 (b) The value of h taken using this barometer is 0.73 m. The density of mercury is 13 600 kg / m3. Calculate the value of the atmospheric pressure suggested by this measurement. Use g = 10 m / s2. atmospheric pressure = … [2] (c) Standard atmospheric pressure is 0.76 m of mercury. Suggest a reason why the value of h in (b) is lower than this. … … [1] [Total: 6]

6 marks

Mark scheme: 4 (a) surfaces shown at realistic levels in dish and tube AND vertical height h between levels clearly shown B1 top label: vacuum / mercury vapour B1 bottom label: mercury B1 (b) (P =) hdg OR 0.73 × 13600 × 10 C1 99280 Pa at least 2 s.f. B1 (c) one from: abnormal weather / atmospheric conditions o.w.t.t.e. air in space above mercury in tube barometer is in a high altitude location o.w.t.t.e. space above mercury is not a vacuum B1 [6] ignore atmospheric pressure varies ignore temperature

This question in 0625/31 Oct/Nov 2011

5 (a) Equal volumes of a gas held at constant pressure, a liquid and a solid undergo the same temperature rise. (i) State which of the three, solid, liquid or gas, 1. expands the most, … 2. expands the least. … (ii) Explain why the pressure of the gas must be kept constant for this comparison. … … [2] (b) Fig. 5.1 shows an alcohol thermometer. –10 0 10 20 30 40 50 60 70 °C Fig. 5.1 (i) State two properties of alcohol which make it suitable for use in a thermometer. 1. … … 2. … … [2] (ii) State two changes to the design of this thermometer which would make it more sensitive. 1. … … 2. … … [2] (c) Explain why it is an advantage for the glass surrounding the alcohol in the bulb of the thermometer to be very thin. … … [1] [Total: 7]

7 marks

Mark scheme: 5 (a) (i) most: gas least: solid both required B1 (ii) because change of pressure (also) causes volume change (in a gas) B1 NOT ‘gas can be compressed’ (b) (i) two from: expands uniformly (over required range) remains liquid over required range expands more than glass / has high expansivity / expansion has (reasonably) low specific heat capacity. has low freezing point / lower freezing point than mercury max B2 (ii) make (capillary) tube narrower (and longer) / thinner / smaller diameter B1 make bulb larger (and tube longer) B1 allow ‘bore’ for tube ignore ‘smaller’ ignore narrow thermometer IGCSE – October/November 2011 0625 31 (c) allows fast(er) flow of heat to / from alcohol OR allows fast response (to temperature change) OR because glass is a poor conductor / good insulator (so needs to be thin for fast response) OR heat transfer more efficient / faster OR glass takes up less heat B1 [7] ignore reference to sensitivity ignore ‘easier’

This question in 0625/31 Oct/Nov 2011

3 Fig. 3.1 shows a house brick of dimensions 21.0 cm × 10.0 cm × 7.00 cm. 21.0 cm 10.0 cm 7.00 cm Fig. 3.1 The brick is held under water with its largest surfaces horizontal. The density of water is 1000 kg / m3. (a) Calculate the difference in pressure between the top and the bottom surfaces of the brick. pressure difference = … [2] (b) Use your value from (a) to calculate the upward force exerted on the brick by the water. upward force = … [2] (c) The mass of the brick is 3.09 kg. Calculate the acceleration of the brick when it is released. acceleration = … [3] [Total: 7]

7 marks

Mark scheme: 3 (a) ρgh in symbols, words or numbers C1 700 Pa or N/m2 A1 [2] (b) use of F = pA C1 14.7 N ecf from (a) A1 [2] (c) (30.9 – 14.7 = )16.2 N OR evidence of calculation of resultant C1 use of a = F/m C1 5.24 m/s2 A1 [3] IGCSE – October/November 2011 0625 32

This question in 0625/32 Oct/Nov 2011

4 A soldier wears boots, each having an area of 0.016 m2 in contact with the ground. The soldier weighs 720 N. (a) (i) Write down the equation that is used to find the pressure exerted by the soldier on the ground. (ii) Calculate the pressure exerted by the soldier when he is standing to attention, with both boots on the ground. pressure = … [2] (b) The soldier is crossing a sandy desert. Explain, stating the relevant Physics, why this soldier is at an advantage over another soldier who has the same weight but smaller feet. … … … … [2] (c) The soldier’s unit is sent to a cold country, and on one occasion he has to cross a frozen lake. Suggest one way that the soldier can reduce the risk of the ice breaking under his weight. … … … [1] [Total: 5]

5 marks

Mark scheme: 4 (a) (i) (P =) F/A words or symbols B1 (ii) 22 500 Pa B1 (b) less pressure B1 less sinking B1 (c) any suggestion which involves increasing the area in contact with the ice e.g. snow shoes / skis B1 [5] IGCSE – October/November 2011 0625 33

This question in 0625/33 Oct/Nov 2011

6 (a) (i) Define pressure. For Examiner’s … [1] Use (ii) A closed box contains a gas. Explain, in terms of molecules, how the gas exerts a pressure on the walls of the box. … … … … … [3] (b) Fig. 6.1 shows a flask connected to a pump and also to a manometer containing mercury. to pump manometer 250 mm flask Fig. 6.1 The right-hand tube of the manometer is open to the atmosphere. The pump has been operated so that the mercury levels differ, as shown, by 250 mm. The density of mercury is 13 600 kg / m3. (i) Calculate the pressure, in Pa, due to the 250 mm column of mercury. pressure = … [2] (ii) The pressure of the atmosphere is 1.02 × 105 Pa. For Examiner’s Calculate the pressure of the air in the flask. Use pressure = … [1] [Total: 7]

7 marks

Mark scheme: 6 (a) (i) (pressure =) force/area OR force per unit area OR (P =) F/A with symbols explained B1 (ii) molecules collide with/hit walls/surface (of box) B1 molecule(s) exert force on wall B1 pressure is total force / force of all molecules divided by (total) area of wall B1 (b) (i) (P =) hρg OR in words OR 0.25 × 13 600 × 10 C1 34 000 Pa OR N/m2 A1 allow 1 mark for h = 250 used and 3.4 × 107 Pa obtained (ii) (P = 1.02 × 105 – 34 000) 68 000 Pa or N/m2 B1 e.c.f. from (b)(i) only if (b)(i) is less than 1.02 × 105 [Total 7]

This question in 0625/31 May/June 2013

1 Fig. 1.1 shows a side view of a large tank in a marine visitor attraction. For Examiner’s Use sea-water viewing panel M tank Fig. 1.1 (not to scale) The tank is 51 m long and 20 m wide. The sea-water in the tank is 11 m deep and has a density of 1030 kg / m3. (a) Calculate the mass of water in the tank. mass = … [3] (b) The pressure at point M, halfway down the large viewing panel, is 60 kPa more than atmospheric pressure. Calculate the depth of M below the surface of the water. depth = … [2] (c) The viewing panel is 32.8 m wide and 8.3 m high. For Examiner’s Calculate the outward force of the water on the panel. Assume that the pressure at M is Use the average pressure on the whole panel. force = … [2] [Total: 7]

7 marks

Mark scheme: 1 (a) V = W × L × D in any form words, symbols or numbers C1 use of M = ρV in any form OR ρV words, symbols or numbers C1 (M = 51 × 20 × 11 × 1030 = 11 556 600 =) 1.2 × 107 kg A1 [3] (b) p = ρg(∆)h in any form words, symbols or numbers C1 (∆h = 60 000 / (1030 × 10) =) 5.8(25) m A1 [2] (c) use of F = pA in any form or pA words, symbols or numbers C1 (F = 60 000 × 32.8 × 8.3 = 60 000 × 272.2 =) 1.6(33) × 107 N A1 [2] e.c.f. from (b) [Total: 7]

This question in 0625/32 May/June 2013

4 A large crane has a mass of 8500 kg. Fig. 4.1 shows the crane on a muddy building-site. For Examiner’s Use lifting-arm hook axle caterpillar tracks Fig. 4.1 (a) Calculate the weight of the crane. weight = … [1] (b) The crane rests on two caterpillar tracks each of which has a contact area with the ground of 3.4 m2. (i) Calculate the pressure that the crane exerts on the ground. pressure = … [2] (ii) As the crane driver walks towards the crane, he starts to sink into the mud. He lays a wide plank of wood on the mud and he walks along the plank. Explain why he does not sink into the mud when he walks along the plank. … … … [2] (c) When the crane lifts a heavy load with its hook, the load exerts a moment on the For lifting-arm about the axle. Examiner’s Use (i) Explain what is meant by moment of a force. … … [1] (ii) Despite the moment exerted on the lifting-arm, the crane remains in equilibrium. State the two conditions required for any object to be in equilibrium. 1. … 2. … [2] [Total: 8]

8 marks

Mark scheme: 4 (a) 85 000 N (accept 83 300 N) B1 (b) (i) (P = )F/A OR 85 000/3.4 OR 85 000/3.4 × 2 OR 85 000/6.8 (e.c.f. from (a)(i)) C1 1.2/1.25/1.3 × 104 Pa (e.c.f. from (a)(i)) A1 (ii) larger area M1 smaller pressure A1 (c) (i) (measure of) turning effect OR F × x B1 (ii) no resultant/net force B1 no resultant/net turning effect/moment B1 [8]

This question in 0625/33 May/June 2013

6 (a) Complete the following statements by writing appropriate words in the spaces. For Examiner’s The pressure of a gas in a sealed container is caused by the collisions of Use … with the container wall. An increase in the temperature of the gas increases the pressure because the … of the … increases. The force on the wall due to the gas is the pressure multiplied by the … of the wall. [2] (b) A mountaineer takes a plastic bottle containing some water to the top of a mountain. He removes the cap from the bottle, drinks all the water and then replaces the cap, as shown in Fig. 6.1. On returning to the base of the mountain, he finds that the bottle has collapsed to a much smaller volume, as shown in Fig. 6.2. Fig. 6.1 Fig. 6.2 (i) Explain why the bottle collapsed. … … … … [2] (ii) At the top of the mountain the atmospheric pressure was 4.8 × 104 Pa and the For volume of the bottle was 250 cm3. Examiner’s Use Calculate the volume of the bottle at the base of the mountain where the pressure of the air inside the bottle is 9.2 × 104 Pa. Assume no change of temperature. volume = … [3] [Total: 7]

7 marks

Mark scheme: 6 (a) molecules OR atoms OR particles speed OR velocity OR kinetic energy molecules OR atoms OR particles (Surface) area B2 any four correct gains 2 marks, two or three correct gains 1 mark (b) (i) (when cap is screwed on) at top of mountain: pressure of air in bottle = the low pressure of the air outside OR is less than pressure at bottom of mountain OR is low B1 (at bottom of mountain) bottle collapses because pressure outside (bottle) is greater than pressure inside B1 (ii) Boyle’s law applies OR PV = constant OR P1V1 = P2V2 C1 9.2 × 104 × V = 4.8 × 104 × 250 C1 130 cm3 A1 [Total: 7] IGCSE – October/November 2013 0625 31

This question in 0625/31 Oct/Nov 2013

5 (a) A water tank has a rectangular base of dimensions 1.5 m by 1.2 m and contains 1440 kg of water. Calculate (i) the weight of the water, weight = … [1] (ii) the pressure exerted by the water on the base of the tank. pressure = … [2] (b) Fig. 5.1 shows two water tanks P and Q of different shape. Both tanks are circular when viewed from above. The tanks each contain the same volume of water. The depth of water in both tanks is 1.4 m. 1.4 m P Q Fig. 5.1 (i) The density of water is 1000 kg / m3. The pressures exerted by the water on the base of the two tanks are equal. Calculate this pressure. pressure = … [2] (ii) Equal small volumes of water are removed from each tank. State which tank, P or Q, now has the greater water pressure on its base. Explain your answer. … … … [2] [Total: 7]

7 marks

Mark scheme: 5 (a) (i) (W = mg =1440 × 10 =) 14 400 N B1 (ii) (P =) F / A OR 14 400 / (1.5 × 1.2) C1 8000 Pa OR N / m2 A1 (b) (i) (P =) hρg OR 1.4 × 1000 × 10 C1 14 000 Pa OR N / m2 A1 (b) (ii) pressure on base of P smaller / Q greater M1 (with same volume removed) smaller decrease in depth in Q OR height in Q is greater A1 [Total: 7]

This question in 0625/31 May/June 2014

6 Fig. 6.1 shows a quantity of gas in a cylinder fitted with a piston P. P gas Fig. 6.1 (a) Describe the motion of the molecules of the gas. … … … … … [3] (b) The piston is now slowly pushed down to decrease the volume of the gas. The temperature of the gas does not change. (i) State and explain, in terms of molecules, what happens to the pressure of the gas. … … … [2] (ii) Before pushing the piston down, the pressure of the gas was 1.0 × 105 Pa. Pushing the piston down reduces the volume of the gas from 500 cm3 to 240 cm3. Calculate the final pressure of the gas. pressure = … [2] [Total: 7]

7 marks

Mark scheme: 6 (a) (molecules) move in random directions / randomly / with constant random motion / zig- zag motion / in all directions B1 (molecules) have random speeds OR a range of speeds OR move (very) fast / at (very) high speed B1 any 1 from: (molecules) collide with each other (molecules) move in straight lines between collisions (molecules) change direction in collisions (molecules) collide with walls (of cylinder) B1 (b) (i) pressure increases M1 more frequent collisions between molecules and walls OR molecules collide with walls more often / at greater rate A1 (ii) pV = constant OR p1V1 = p2V2 in any form OR 1.0 × 105 × 500 = p2 × 240 C1 2.1 × 105 Pa to 2 or more sig. figs A1 [Total: 7] IGCSE – May/June 2014 0625 31

This question in 0625/31 May/June 2014

3 (a) Fig. 3.1 shows an oil can containing only air at atmospheric pressure. can Fig. 3.1 Atmospheric pressure is 1.0 × 105 Pa. The pressure of the air in the can is reduced by means of a pump. The can collapses when the pressure of the air in the can falls to 6000 Pa. (i) Explain why the can collapses. … … [1] (ii) The surface area of face X of the can is 0.12 m2. Calculate the resultant force on face X when the can collapses. force = … [3] (b) Mercury is poured into a U-shaped glass tube. Water is then poured into one of the limbs of the tube. Oil is poured into the other limb until the surfaces of the mercury are at the same level in both limbs. Fig. 3.2 shows the result. oil water 0.32 m 0.25 m mercury Fig. 3.2 (i) State a condition that must be true in order for the mercury surfaces to be at the same level in both limbs of the tube. … [1] (ii) The height of the water column is 0.25 m. The height of the oil column is 0.32 m. The density of water is 1000 kg / m3. Calculate 1. the pressure exerted by the water on the surface of the mercury, pressure = … [2] 2. the density of the oil. density = … [2] [Total: 9]

9 marks

Mark scheme: 3 (a) (i) force / pressure greater on outside surface owtte B1 (ii) p = F / A in any form OR (F =) pA C1 = (1.0 × 105 – 6000) × 0.12 C1 11280 N to at least 2 sig. figs. A1 (b) (i) pressure of oil = pressure of water B1 (ii) 1. (p=) hρg C1 (= 0.25 × 1000 × 10 =) 2500 Pa A1 2. hρg = 2500 C1 (ρ = 2500 / (0.32 × 10) =) 781 kg / m3 to at least 2 sig. figs. A1 [Total: 9]

This question in 0625/31 Oct/Nov 2014

2 Fig. 2.1 shows a uniform, rectangular slab of concrete ABCD standing upright on the ground. The slab has height 0.60 m, width 0.30 m and mass 18 kg. A force of 40 N acts horizontally to the left at B. A B 40 N 0.60 m D C 0.30 m Fig. 2.1 (a) (i) Calculate the weight W of the concrete slab. W = … [1] (ii) The thickness of the slab is 0.040 m. Calculate the pressure exerted by the slab on the ground. pressure = … [2] (b) (i) On Fig. 2.1, draw and label an arrow to show the weight W of the slab acting at its centre of mass. [1] (ii) Calculate 1. the moment of the 40 N force about point D, moment = … 2. the moment of W about point D. moment = … [3] (iii) The ground is rough so that the slab does not slide. State and explain what happens to the slab as the horizontal force at B is gradually increased. … … … [2] [Total: 9]

9 marks

Mark scheme: 2 (a) (i) 180 N B1 (ii) (P =) F ÷ A OR 180÷(0.30 × 0.04) C1 15 000 Pa A1 (b) (i) arrow (labelled W) from / to correct centre of mass B1 (ii) 1. force × (perpendicular) distance OR 40 × 0.60 OR 180 × 0.15 in 2. C1 24 N m A1 2. 27 N m e.c.f. from (a)(i) A1 (iii) slab topples / rotates (about point D) OR corner C lifts from ground OR falls over B1 moment of force at B becomes bigger than moment of weight / W OR anticlockwise moment becomes bigger than clockwise moment OR weight/centre of mass outside base B1 [Total: 9]

This question in 0625/33 Oct/Nov 2014

4 (a) Fig. 4.1 shows a syringe containing 100 cm3 of air at atmospheric pressure. Atmospheric pressure is 1.0 × 105 Pa. piston Fig. 4.1 The open end of the syringe is sealed and the piston is pushed inwards until the air occupies a volume of 40 cm3. The temperature of the air remains constant. Calculate the new pressure of the air in the syringe. air pressure = … [2] (b) A syringe is used to transfer smokey air from above a flame to a small glass container. Extremely small solid smoke particles are suspended in the air in the container. The container is brightly illuminated from the side and viewed through a microscope. (i) The movement of the suspended smoke particles is called Brownian motion. Describe this Brownian motion. … … … [2] (ii) Explain what causes the motion of the smoke particles. … … … [2] (c) In the space below, sketch a diagram to represent the molecular structure of a solid. Show the molecules as small circles of equal sizes. [2] [Total: 8]

8 marks

Mark scheme: 4 (a) pV = constant OR p1V1 = p2V2 OR p1V1 / V2 or 1.0 × 105 × 100 ÷ 40 C1 2.5 × 105 Pa A1 (b) (i) (the particles move) randomly B1 (the particles move) slowly OR through small distances OR disappear OR zigzag OR directions change OR erratic OR straight lines between collisions B1 (ii) air molecules / particles collide with smoke particles (at high speed) B1 fast(er) air molecules OR move randomly OR many collisions B1 (c) diagram showing: molecules touching each other B1 molecules positioned in an ordered structure B1 [Total: 8]

This question in 0625/33 Oct/Nov 2014

2 A large stone block is to be part of a harbour wall. The block is supported beneath the surface of the sea by a cable from a crane. Fig. 2.1 shows the block with its top face a distance h beneath the surface of the sea. cable surface of sea h block Fig. 2.1 The force acting downwards on the top face of the block, due to the atmosphere and the depth h of water, is 3.5 × 104 N. (a) The top face of the block has an area of 0.25 m2. (i) Calculate the pressure on the top face of the block. pressure = … [2] (ii) The atmospheric pressure is 1.0 × 105 Pa. Calculate the pressure on the top face of the block due to the depth h of water. pressure = … [1] (iii) The density of sea water is 1020 kg / m3. Calculate the depth h. h = … [2] (b) Suggest two reasons why the tension force in the cable is not 3.5 × 104 N. 1. … 2. … [2] (c) The block is lowered so that it rests on the sea-bed. State what happens to the tension force in the cable. … [1] [Total: 8]

8 marks

Mark scheme: 2 (a) (i) (P =) F÷A OR 3.5 × 104 ÷ 0.25 C1 = 1.4 × 105 Pa ecf (i) A1 (ii) (1.4 × 105 – 1.0 × 105 =) 4(.0) × 104 Pa ecf (ii) B1 (iii) P = h ρ g in any form OR (h =) P÷ρ g OR 4.0 × 104 ÷ (1020 × 10) C1 = 3.9 m OR 4 m A1 (b) any 2 from: max. B2 • weight of block • upward force of water (on block) / upthrust (of water on block) • weight of cable (c) (tension force) becomes smaller or zero B1 [Total: 8]

This question in 0625/31 May/June 2015

2 Fig. 2.1 shows a cylinder containing gas compressed by the movement of a piston. compressed gas final position of piston cylinder initial position of piston Fig. 2.1 Initially the volume of the gas was 470 cm3. The piston moves up and compresses the gas to a volume of 60 cm3. The whole arrangement is left for some time until the gas cools to its original temperature. The pressure of the gas is now 800 kPa. (a) Calculate the initial pressure of the gas. pressure = … [3] (b) Explain, in terms of molecules, the effect on the pressure of the gas if it was not given time to cool to its original temperature. … … … … [3] (c) The area of the piston is 5.5 × 10–3 m2 (0.0055 m2). Calculate the force exerted by the gas on the piston when the pressure is 800 kPa. force = … [2] [Total: 8]

8 marks

Mark scheme: 2 (a) p1V1 = p2V2 in any form OR (p1 =) p2V2 ÷ V1 C1 p1 × 470 = 800 × 60 OR (p1 =) 800 × 60 ÷ 470 C1 102 OR 100 kPa A1 (b) molecules would move faster/have more KE B1 more (frequent)/harder collisions with walls/cylinder/piston B1 pressure increases B1 (c) use of p = F ÷ A in any form OR (F =) p A C1 (F =) 4400 N A1

This question in 0625/32 May/June 2015

11 Fig. 11.1 shows the main components of a cathode-ray oscilloscope. fluorescent screen plate Y1 plate X2 heater beam P anode plate plate system Y2 X1 grid Fig. 11.1 (a) (i) State the function of component P. … … [1] (ii) Tick one box to complete the sentence correctly. A cathode-ray oscilloscope contains air at about five times normal atmospheric pressure. air at about normal atmospheric pressure. air at about one fifth of normal atmospheric pressure. a vacuum. neon gas. [1] (b) Fig. 11.2 shows the front view of the screen of the cathode-ray oscilloscope. B C A Fig. 11.2 With no voltage applied between the X-plates or between the Y-plates, the spot is at A. (i) Place two ticks in each of the blank columns of the table to describe the voltages across the plates when the spot is at points B and C. The column for the spot at A has been completed as an example. spot at A spot at B spot at C plate X1 at higher voltage than plate X2 plate X1 at lower voltage than plate X2 no voltage between X-plates ✓ plate Y1 at higher voltage than plate Y2 plate Y1 at lower voltage than plate Y2 no voltage between Y-plates ✓ [3] (ii) Explain your answers for the spot at point B. … … … … [1] [Total: 6]

6 marks

Mark scheme: 11 (a) (i) (function of cathode is) to emit/produce electrons B1 (ii) 4th box: vacuum B1 (b) (i) B: box 3 no voltage between X-plates B1 B: box 4 voltage plate Y1 > voltage plate Y2 B1 C: box 2 voltage plate X1 < voltage plate X2 AND box 4 voltage plate Y1 > than voltage plate Y2 B1 (ii) no voltage between X plates so no horizontal deflection AND beam attracted upwards to higher V / Y1 OR other correct argument B1

This question in 0625/32 May/June 2015

5 Fig. 5.1 shows two men repairing a weak roof using a crawler-board. roof crawler-board Fig. 5.1 (a) Explain why use of the crawler-board prevents the men from falling through the roof. … … … … [2] (b) The crawler-board has a weight of 400 N. The total weight of the two men is 1600 N. The area of the crawler-board in contact with the roof is 0.8 m2. Calculate the pressure on the roof when the men are on the crawler-board. Include the unit. pressure = … [5] [Total: 7]

7 marks

Mark scheme: 5(a) any two from: larger area (in contact with roof) B2 weight OR force spread out lower pressure (on roof) 5(b) 400 + 1600 seen OR 2000(N) B1 P = F / A stated C1 2000 / 0.8 C1 2500 A1 N / m2 OR Pa B1 Total: 7

This question in 0625/31 May/June 2016

2 Fig. 2.1 shows the speed-time graph for a student cycling along a straight, flat road. 8 speed m / s 6 4 2 0 0 5 10 15 20 25 30 35 time / s Fig. 2.1 (a) Calculate the distance he travels in the first 10 s. distance = … m [3] (b) Fig. 2.2 shows three pairs of forces A, B and C. 60 N 20 N A backward force forward force 50 N 50 N B backward force forward force 20 N 70 N C backward force forward force Fig. 2.2 Identify which pair of forces, A, B or C, acts on the cyclist between 11 s and 16 s. Explain your choice. pair of forces … explanation … … … [3] (c) The cyclist pushes on one pedal with a force of 120 N. The area of his shoe in contact with the pedal is 16 cm2. Calculate the pressure on the pedal. Include the unit. pressure = … [4] [Total: 10]

10 marks

Mark scheme: 2(a) (distance) = area under (speed-time) graph in words, numbers or symbols OR distance = speed × time 4.4 × 10 44 (m) C1 C1 A1 2(b) C (cyclist is) accelerating (so) forward force must be greater than backward force OR there is a resultant (forward) force B1 B1 B1 2(c) P = F ÷ A 120 ÷ 16 7.5 N / cm2 C1 C1 A1 B1 Total: 10

This question in 0625/32 May/June 2016

2 Fig. 2.1 shows the speed-time graph for a student cycling along a straight, flat road. 8 speed m / s 6 4 2 0 0 5 10 15 20 25 30 35 time / s Fig. 2.1 (a) Calculate the distance he travels in the first 10 s. distance = … m [3] (b) Fig. 2.2 shows three pairs of forces A, B and C. 60 N 20 N A backward force forward force 50 N 50 N B backward force forward force 20 N 70 N C backward force forward force Fig. 2.2 Identify which pair of forces, A, B or C, acts on the cyclist between 11 s and 16 s. Explain your choice. pair of forces … explanation … … … [3] (c) The cyclist pushes on one pedal with a force of 120 N. The area of his shoe in contact with the pedal is 16 cm2. Calculate the pressure on the pedal. Include the unit. pressure = … [4] [Total: 10]

10 marks

Mark scheme: 2(a) (distance) = area under (speed-time) graph in words, numbers or symbols OR distance = speed × time 4.4 × 10 44 (m) C1 C1 A1 2(b) C (cyclist is) accelerating (so) forward force must be greater than backward force OR there is a resultant (forward) force B1 B1 B1 2(c) P = F ÷ A 120 ÷ 16 7.5 N / cm2 C1 C1 A1 B1 Total: 10

This question in 0625/33 May/June 2016

4 Fig. 4.1 shows a car parked on a road. Fig. 4.1 (a) The car has a mass of 1000 kg. Calculate the weight of the car. weight of car = … N [2] (b) (i) The combined weight of the car and its driver is 10 500 N. The area of each tyre in contact with the road is 125 cm2. Each tyre supports a quarter of the combined weight of the car and driver. Calculate the pressure that each tyre exerts on the ground. pressure = … N / cm2 [3] (ii) Later, the car is parked on some long wooden planks on a muddy field. Explain why the planks prevent the car from sinking into the mud. … … … … [2] (c) (i) A driver needs to remove a wheel from the car. To undo the wheel nuts the driver uses a wrench, as shown in Fig. 4.2. 200200200 NNN mm 0.250.25 wrench wheel nut Fig. 4.2 A force of 200 N is applied perpendicular to the wrench. The force is applied 0.25 m from the centre of a wheel nut. Calculate the moment produced by this force. Include the unit. moment = … [3] (ii) One of the wheel nuts is difficult to turn. Describe how the driver can increase the moment when using the same force of 200 N applied to the wrench. … … [1] [Total: 11]

11 marks

Mark scheme: 4(a) C1 10 000 (N) A1 4(b)(i) pressure = force/area in any form C1 (10 500 / 4) / 125 C1 21 (N/cm2) A1 4(b)(ii) (weight spread over) larger area owtte B1 pressure reduced B1 4(c)(i) moment = force × distance from pivot in any form C1 200 × 0.25 OR 50 A1 Nm B1 4(c)(ii) force applied further away from wheel nut owtte B1 Total: 11

This question in 0625/33 Oct/Nov 2016

5 The bucket at the front of a tractor is used to push fence posts down into the ground, as shown in Fig. 5.1. bucket tractor fence post Fig. 5.1 The area of each post in contact with the ground is 100 cm2. When the bucket pushes a post, the downward force from the post on the ground is 6500 N. (a) Calculate the pressure that the post exerts on the ground. pressure = … N / cm2 [3] (b) A farmer cuts the bottom of one of the posts to make it more pointed. The bucket applies the same force as before. Explain the effect this has on the pressure exerted by the post on the ground. … … … … … [2] [Total: 5]

5 marks

Mark scheme: 5(a) P = F ÷ A in any recognised form C1 6500 ÷ 100 C1 65 (N / cm2) A1 5(b) smaller area (at point) B1 greater pressure B1 Total: 5

This question in 0625/32 Feb/March 2017

5 A laboratory floor has a surface that prevents people from slipping when the floor is wet. (a) Name the force that prevents a person from slipping. … [1] (b) A stool has a round non-slip pad fitted to the bottom of each leg. (i) The stool has four legs. The area of each pad is 3 cm2. The weight of the stool is 75 N. A student sits on the stool. The weight of the student is 525 N. Calculate the pressure acting on the floor due to the student and the stool. pressure = … N / cm2 [5] (ii) The legs of the stool are made of hollow metal tubes. Fig. 5.1 shows the bottom of a stool leg with and without a pad. metal tube with pad without a pad Fig. 5.1 Explain why a stool leg without a pad does more damage to the floor. … … [2] [Total: 8]

8 marks

Mark scheme: 5(a) friction B1 5(b)(i) total area = 3 × 4 = 12 (cm2) C1 total weight = 525 + 75 N = 600(N) C1 P = F ÷ A in any form C1 600 ÷ 12 C1 50 (N / cm2) A1 5(b)(ii) less (surface) area (in contact with the ground) owtte B1 more pressure (results in more damage to the surface) B1 Total: 8

This question in 0625/31 May/June 2017

5 (a) A potato snack packet is taken onto an aeroplane. During the flight the pressure inside the aeroplane changes and the potato snack packet changes shape, as shown in Fig. 5.1. potato snack packet before flight potato snack packet during flight OO OO KK K TT ATATK CC TT OOACAC TATAAA PPSNSN OONN PP SS Fig. 5.1 Explain why the packet changes shape. Use ideas about the gas molecules inside and outside the packet in your answer. … … … … … … [2] (b) A deep-sea diver on a diving-boat experiences atmospheric pressure. When she is working underwater, she experiences an increased pressure. State two factors that affect the size of the increased pressure. 1. … 2. … [2] (c) Fig. 5.2 shows a device used for measuring atmospheric pressure. mercury Fig. 5.2 (i) State the name of the device shown in Fig. 5.2. … [1] (ii) Fig. 5.3 shows a manometer connected to a gas supply. The pressure of the gas supply is greater than atmospheric pressure. Atmospheric pressure is equal to 1033.6 cm of water. gas 4 supply 3 2 1 water cm Fig. 5.3 Determine the pressure of the gas supply. pressure = … cm of water [3] [Total: 8]

8 marks

Mark scheme: 5(a) any two from: more collide with walls more often so pressure is greater (inside bag) B2 5(b) density (of sea water) depth (of sea water) (in either order) B2 5(c)(i) barometer B1 5(c)(ii) 3.4 or 1.3 seen C1 2.1 C1 1035.7 A1 Total: 8

This question in 0625/32 May/June 2017

4 Fig. 4.1 shows a car tyre in contact with the road. road tyre Fig. 4.1 The area of tyre in contact with the road is 0.015 m2. The tyre exerts a pressure on the road of 240 kN / m2. (a) Calculate the force on the road from the tyre. force = … N [4] (b) The tyre is filled with air at high pressure. Use ideas about molecules to explain how this air exerts a pressure on the inside of the tyre. … … … … … … … [3] [Total: 7]

7 marks

Mark scheme: 4(a) B1 pressure = force ÷ area C1 transformation force = pressure × area C1 3600 (N) A1 4(b) any 3 from: molecules move about (randomly) collisions impacts with walls/surfaces (of tyre) idea of force produced (by bombarding molecules) idea of pressure as force on an area B3 Total: 7

This question in 0625/33 May/June 2017

3 Fig. 3.1 shows a glass vase used for displaying flowers. Fig. 3.1 (a) The mass of the glass is 450 g. The volume of glass in the vase is 145 cm3. (i) Calculate the density of the glass. density = … g / cm3 [3] (ii) Calculate the weight of the glass. weight = … N [3] (b) Another vase has a weight of 30 N. The area of the base in contact with a table is 80.0 cm2. Calculate the pressure this vase exerts on the table. pressure = … N / cm2 [3] [Total: 9]

9 marks

Mark scheme: 3(a)(i) D = M / V C1 450 / 145 C1 3.1 (g / cm3) A1 3(a)(ii) W = m × g in any form C1 0.45 × 10 C1 4.5 (N) A1 3(b) P = F / A in any form C1 30 / 80 C1 0.375 (N / cm2) OR 0.38 (N / cm2) A1

This question in 0625/32 Oct/Nov 2017

5 A vehicle may have tyres of type A or type B, as shown in Fig. 5.1. type A type B Fig. 5.1 (a) State and explain the type of tyre that is suitable for travelling over soft ground. … … … … [3] (b) The temperature of the air in a tyre increases. This affects the motion of the air molecules in the tyre. Describe and explain the changes. … … … … … [3] [Total: 6]

6 marks

Mark scheme: 5(a) Tyre B B1 larger / bigger surface area B1 less pressure (on ground) / weight distributed B1 5(b) molecules gain kinetic energy / move faster B1 more (frequent) / harder collisions (with tyre) B1 Increased / greater pressure (on tyre) B1

This question in 0625/33 Oct/Nov 2017

5 (a) Fig. 5.1 shows a metal can containing air. The can is sealed with a lid. lid can Fig. 5.1 The air in the can exerts a pressure of 20 000 N / m2 on the lid. The area of the can lid is 0.09 m2. Calculate the force on the lid due to the air in the can. force = … N [3] (b) The air in the can becomes warmer. State and explain what happens to the pressure of the air in the can. Use your ideas about gas molecules. … … … … … … … [3] [Total: 6]

6 marks

Mark scheme: 5(a) 1 20 000 × 0.009 1 1800 (N) 1 5(b) pressure increases 1 any two from: molecules move faster/have more ke collide harder/more often (with walls of can) (change in momentum due to) collisions impart(s) force on can walls 2

This question in 0625/33 May/June 2018

2 A student is studying elephants. Fig. 2.1 shows an elephant. Fig. 2.1 (a) The student measures the elephant and records the values, as shown in the table. Complete the table by adding a suitable unit for each measurement. Choose the units from those shown in the box. m2 kg cm mm2 g m cm2 mg mm measurements value unit mass of elephant 4000 height of elephant 3.0 average area of an elephant’s foot 0.125 [2] (b) Using information from the table in (a): (i) Calculate the weight of the elephant. weight = … N [3] (ii) Calculate the pressure the elephant exerts on the ground when it is standing on four feet. Include a unit. pressure = … [4] [Total: 9]

9 marks

Mark scheme: 2(a) mass in kg AND height in m B1 area in m2 B1 2(b)(i) W = m × g C1 4000 × 10 C1 40 000 (N) A1 2(b)(ii) P = F ÷ A in any recognisable form C1 (area = ) 0.125 × 4 = 0.50 (m2) B1 b(i) ÷ 5000 OR 40 000 ÷ 0.500 C1 80 000 N / m2 OR 80 000 Pa A1

This question in 0625/32 Oct/Nov 2018

3 A flask contains gas with a pressure lower than atmospheric pressure. Fig. 3.1 shows equipment being used to measure the pressure of the gas in the flask. atmospheric pressure A flask 280 mm B gas mercury Fig. 3.1 (a) State the name of the equipment shown in Fig. 3.1 that is used to measure the pressure of the gas. … [1] (b) The atmospheric pressure is equal to 760 mm Hg. The distance between mercury level A and mercury level B is 280 mm. Determine the pressure of the gas inside the flask. pressure = … mm Hg [2] (c) The flask is cooled. Describe the effect, if any, the cooling has on mercury level A … mercury level B … [1] [Total: 4]

4 marks

Mark scheme: 3(a) manometer B1 3(b) 760–280 C1 480 (mm Hg) A1 3(c) (level A) up (level B) down B1

This question in 0625/32 Oct/Nov 2018

4 Fig. 4.1 shows a pin. Fig. 4.2 shows a person pushing the pin into a wall. top surface of the pin Fig. 4.1 Fig. 4.2 (a) (i) The area of the top surface of the pin is 1.8 cm2. The person applies a force of 50 N. Calculate the pressure exerted on the top surface of the pin. pressure = … N / cm2 [3] (ii) The area of the top surface of the pin is 500 times larger than the area of the point. Calculate the value of the pressure exerted by the point on the wall. pressure = … N / cm2 [1] (b) Fig. 4.3 shows a simple device for measuring atmospheric pressure. space mercury column 760 mm mercury Fig. 4.3 (i) State the name given to the device shown in Fig. 4.3. … [1] (ii) State what, if anything, is in the space at the top of the tube, above the mercury column. … [1] (iii) Fig. 4.3 shows normal atmospheric pressure. Suggest a possible value for the height of the mercury column when atmospheric pressure decreases. Include the unit. reading = … [1] [Total: 7]

7 marks

Mark scheme: 4(a)(i) Pressure = force ÷ area in any form C1 50 ÷ 1.8 C1 28 (N / cm2) A1 4(a)(ii) In range 13 500 to 15 000 (N / cm2) B1 4(b)(i) (mercury) barometer B1 4(b)(ii) vacuum OR nothing B1 4(b)(iii) a value less than 760 mm (Hg ) and > 0 mm (Hg) B1

This question in 0625/31 May/June 2019

4 A student places a balloon filled with air next to a window, as shown in Fig. 4.1. The Sun warms the air in the balloon. Fig. 4.1 (a) (i) Suggest what happens to the balloon as the air in it becomes hotter than the surroundings. … [1] (ii) Use ideas about molecules to explain your answer to (a)(i). … … … … … [3] (b) The student uses a pump to inflate another balloon. Fig. 4.2 shows the student inflating a balloon. balloon pump handle Fig. 4.2 The student applies a force of 30 N to the pump handle. The force acts on an area of 12 cm2. Calculate the pressure on the pump handle. Include the unit. pressure = … [4] [Total: 8]

8 marks

Mark scheme: 4(a)(i) expand or increase in size/volume increase in pressure decrease in density B1 4(a)(ii) any 3 from: density (of air) is less molecules move faster/have more (kinetic) energy more collisions ( per second) collisions with surface OR balloon (owtte) more force (in collisions) molecules move (further) apart B3 Question Answer Marks 4(b) P = F/A in any form C1 30 ÷ 12 C1 2.5 A1 N/cm2 B1

This question in 0625/32 May/June 2019

4 Fig. 4.1 shows a tractor fitted with a device for breaking up soil in a field. tractor device heavy weight pivot point soil Fig. 4.1 (a) (i) The tractor has a heavy weight at the front. Explain why the heavy weight is needed. … … [1] (ii) Fig. 4.2 represents the weight of the device and its distance from the pivot. pivot 2.1 m 6000 N Fig. 4.2 Calculate the moment of the weight of the device about the pivot. State the unit. moment = … [4] (b) Fig. 4.3 shows a tractor fitted with narrow tyres and the same tractor fitted with wide tyres. narrow tyre wide tyre tractor fitted with same tractor fitted with narrow tyres wide tyres Fig. 4.3 (view from the front) Explain why wide tyres are more suitable for the tractor on soft soil. … … … … [3] [Total: 8]

8 marks

Mark scheme: 4(a)(i) stop the tractor tipping up/keep tractor level owtte B1 4(a)(ii) moment = force × (perp.) distance from pivot in any form C1 6000 × 2.1 C1 12 600 A1 Nm B1 4(b) Any three from: (wide tyres have) greater area (in contact with ground) pressure = force ÷ area in any form the bigger the area the smaller the pressure so tractor less likely to sink/become stuck (in soft ground) B3

This question in 0625/31 Oct/Nov 2019

4 (a) During part of a race, a skier travels a distance of 200 m in a time of 6.4 s. Calculate the average speed of the skier. average speed = … m / s [3] (b) Fig. 4.1 shows a speed–time graph for the skier in another part of the race. 20.0 Q speed m / s 15.0 P 10.0 R 5.0 S 0 0 5.0 10.0 15.0 20.0 25.0 30.0 time / s Fig. 4.1 Describe the motion of the skier at each point P, Q, R and S on the graph. P … Q … R … S … [4] (c) Skis are strapped to a skier’s feet and are longer and wider than the skier’s feet. Explain how the skis prevent the skier from sinking into soft snow. … … … [2] [Total: 9]

9 marks

Mark scheme: 4(a) C1 (s =) 200 ÷ 6.4 C1 (s =) 31 (m / s) A1 4(b) P – (constantly) accelerates (from 5 m / s) B1 Q – constant speed (of 17.5 m / s) B1 R – (non-constant) decelerates (from 17.5 m / s to rest) B1 S – at rest or stationary B1 4(c) (skis have) large (surface) area B1 (so) less pressure (on snow / ground) B1

This question in 0625/31 May/June 2020

3 Fig. 3.1 shows an archer pulling the string of a bow. string arrow fingers on string hand pushing bow archer bow Fig. 3.1 (a) The archer uses a force of 120 N. The force acts on an area of 0.5 cm2 on the archer’s fingers. Calculate the pressure on the archer’s fingers. pressure on fingers = … N / cm2 [3] (b) The archer’s other hand is pushing the bow with the same force of 120 N. This force acts on a larger area than the force in (a). State whether the pressure on this hand is greater than, the same as or less than the pressure on the fingers holding the string. … [1] (c) State the type of energy stored in the bow when the archer bends it as shown in Fig. 3.1. … [1] [Total: 5]

5 marks

Mark scheme: 3(a) P = F ÷ A in any form C1 120 ÷ 0.5 C1 240 (N / cm2) A1 3(b) Less (than) B1 3(c) elastic OR strain OR potential B1

This question in 0625/32 May/June 2020

5 Fig. 5.1 shows a steel container fitted with a liquid manometer. There is a gas in the container. sealed lid liquid manometer steel container 100 mm gas bench mercury Fig. 5.1 (a) (i) The area of the steel container in contact with the bench is 80 cm2. The total weight of the steel container and its contents is 60 N. Calculate the pressure that the steel container exerts on the bench. pressure on the bench = … N / cm2 [3] (ii) Atmospheric pressure is equal to 760 mm of mercury (mm Hg). Determine the pressure inside the container in mm Hg. pressure = … mm Hg [2] (b) The temperature of the gas inside the steel container decreases. State and explain how the pressure of the gas changes as the temperature of the gas decreases. Use your ideas about molecules in your answer. … … … … … … [3] [Total: 8]

8 marks

Mark scheme: 5(a)(i) (P =) F ÷ A in any form C1 (P =) 60 ÷ 80 C1 (P =) 0.75 (N / cm2) A1 5(a)(ii) 760(.0) + OR – 100(.0) C1 860 (mm of Hg) A1 5(b) pressure decreases B1 Any two from: (because) molecules slower speed or less kinetic energy fewer collisions OR molecules collide less often (with walls of container) collides with less force (with walls of container) B2

This question in 0625/31 Oct/Nov 2020

3 Fig. 3.1 shows a mercury barometer. X 235 mm mercury 755 mm bench glass container 50 mm Fig. 3.1 (a) (i) Determine the atmospheric pressure indicated by the barometer. Include the unit. atmospheric pressure = … unit … [2] (ii) State what is in the space labelled X above the mercury in the tube. … [1] (b) The total weight of the mercury barometer is 38 N. The area of the glass container in contact with the bench is 200 cm2. Calculate the pressure of the mercury barometer on the bench. pressure = … N / cm2 [3] [Total: 6]

6 marks

Mark scheme: 3(a)(i) 755 B1 mm Hg B1 3(a)(ii) vacuum B1 3(b) (P = ) F ÷ A C1 38 ÷ 200 C1 0.19 (N / cm2) A1

This question in 0625/32 Oct/Nov 2020

4 (a) Some vehicles have wide tyres so that they can drive over soft mud. Explain why the wide tyres enable these vehicles to drive over soft mud. … … … … [3] (b) Fig. 4.1 shows a mercury barometer. Y mercury glass tube container Fig. 4.1 (i) State what is in the space at Y above the mercury. … [1] (ii) Complete the following statement to describe the use of a mercury barometer. A mercury barometer measures … … [2] (c) The diagram in Fig. 4.2 shows a tall can containing water. The can has three identical holes 1, 2 and 3 on one side, one above the other in a vertical line, as shown in Fig. 4.2. Water is shown flowing out of hole 2 and hole 3. can 1 stream of water from hole 2 2 water stream of water 3 from hole 3 bench container Fig. 4.2 (i) State how the pressure of the water at hole 2 compares with the pressure of the water at hole 3. … [1] (ii) Draw on Fig. 4.2 to show the stream of water flowing from hole 1. [1] [Total: 8]

8 marks

Mark scheme: 4(a) any three from: pressure = force ÷ area weight or force distributed / spread or wtte (over) larger area (so) lower pressure (on the ground) (vehicle) doesn’t sink into the mud owtte 4(b)(i) vacuum B1 4(b)(ii) atmospheric / air B1 pressure B1 4(c)(i) (pressure at hole 2 is) less (than that at hole 3) ora B1 4(c)(ii) water from top hole not travelling further than from middle hole B1

This question in 0625/33 Oct/Nov 2020

5 (a) A man starts pulling his suitcase across the floor. suitcase 12 N 20 N Fig. 5.1 (not to scale) (i) Fig. 5.1 shows the horizontal forces acting on the suitcase. Calculate the resultant horizontal force on the suitcase. size of force = … N direction … [2] (ii) After a short time, the suitcase is moving at a constant speed. Suggest values for the sizes of the two horizontal forces on the suitcase when it is moving at a constant speed. pulling force = … (N) friction force = … (N) [1] (b) The total downward force of the suitcase on the ground is 150 N. The suitcase has two wheels. Each wheel has an area of 0.60 cm2 touching the ground. Calculate the pressure of the suitcase on the ground. pressure on the ground = … N / cm2 [4] [Total: 7]

7 marks

Mark scheme: 5(a)(i) 8 (N) B1 forwards B1 5(a)(ii) same non-zero values for pulling and friction force B1 5(b) (area = 2 × 0.60) = 1.2 (cm2) B1 (P =) F ÷ A C1 150 ÷ 1.2 OR 150 ÷ 0.60 C1 125 (N / cm2) A1

This question in 0625/31 May/June 2021

5 Fig. 5.1 shows a device connected to a gas cylinder. The device is used to measure the pressure of the gas inside the cylinder. gas from 7.0 cm cylinder 7.0 cm mercury Fig. 5.1 (a) (i) State the name of the device shown in Fig. 5.1. … [1] (ii) The atmospheric pressure is equal to 75 cm of mercury. Determine the pressure of the gas in the cylinder. Use information from Fig. 5.1. pressure of gas = … cm of mercury [2] (b) Fig. 5.2 shows two identical heavy stone tiles placed on soft ground. One is vertical and the other is horizontal. vertical tile horizontal tile soft ground Fig. 5.2 One of the tiles sinks into the soft ground. State and explain which tile sinks into the soft ground. … … … … … … [4] [Total: 7]

7 marks

Mark scheme: 5(a)(i) 5(a)(ii) (pressure of gas in cylinder) is 14 cm of mercury (different to atmospheric pressure) owtte C1 (75 + 14 =) 89 (cm of mercury) A1 5(b) vertical tile (is more likely to sink) B1 any three from: • (because) tiles have same weight • pressure = force ÷ area • small(er) area (in contact with ground) for vertical tile ora • (so) greater/big pressure (exerted on ground) ora B3

This question in 0625/32 May/June 2021

5 (a) Fig. 5.1 shows a tractor and a car of the same weight. Fig. 5.1 The vehicles drive over the same soft ground. Explain why the car sinks into the soft ground but the tractor does not sink. … … … … [3] (b) The car driver measures the pressure of the air in a car tyre when the air is cool. The Sun heats the air in the tyre. The driver measures the pressure of the air in the tyre when the air is warm. The pressure of the air in the warm tyre is greater. Explain the increase in the pressure of the air in the tyre. Use ideas about air molecules. … … … … [3] [Total: 6]

6 marks

Mark scheme: 5(a) P = F ÷ A in any form B1 tractor has larger area (of tyre(s) in contact with ground) / ora B1 so (tractor) pressure is less (on the ground) / ora B1 5(b) kinetic energy of molecules increases / speed of molecules increases B1 hit (tyre walls) more often / greater rate B1 hit (tyre walls) with greater force / harder B1

This question in 0625/33 May/June 2021

5 Fig. 5.1 shows a plastic bottle on a bench. The plastic bottle contains a liquid. plastic bottle liquid bench Fig. 5.1 (a) The weight of the bottle and liquid is 12 N. The area of the bottle in contact with the bench is 25 cm2. Calculate the pressure of the bottle on the bench. pressure on bench = … N / cm2 [3] (b) A student pours out all the liquid from the bottle. She then connects the bottle to a vacuum pump which removes most of the air from the bottle. Fig. 5.2 shows the bottle after most of the air is removed. to vacuum pump crushed bottle Fig. 5.2 Explain why the bottle is crushed. Use your ideas about molecules. … … … … … [4] [Total: 7]

7 marks

Mark scheme: 5(a) (P =) F ÷ A OR (pressure = ) force ÷ area in any form C1 12 ÷ 25 C1 0.48 (N / cm2) A1 5(b) Any four from: molecules in air moving at high speed / kinetic energy molecules collide with (plastic) bottle OR wall (of bottle) force of collisions (per unit area) cause pressure fewer air molecules or collisions on inside(compared to outside) (and so) greater force OR pressure on outside (than inside) B4

This question in 0625/31 Oct/Nov 2021

4 (a) A teacher wants to measure the mass of a block of metal. She also wants to measure the length, width and height of the block. Fig. 4.1 shows the block of metal. length width height Fig. 4.1 Complete each sentence using a word from the list. balance barometer protractor ruler voltmeter (i) To find the mass of the metal block, the teacher uses a … [1] (ii) To measure the length, width and height of the metal block, she uses a … [1] (b) The mass of the block is 5000 g. Calculate the weight of the block. weight = … N [3] (c) Fig. 4.2 shows another block of metal on a solid surface. 20 cm 12 cm solid surface 2.0 cm Fig. 4.2 (not to scale) (i) Calculate the area of the block of metal in contact with the solid surface. area = … cm2 [1] (ii) The weight of the block of metal in Fig. 4.2 is 60 N. Calculate the pressure of the block of metal on the solid surface. pressure = … N / cm2 [3] [Total: 9]

9 marks

Mark scheme: 4(a)(i) balance B1 4(a)(ii) ruler B1 4(b) mass = 5(.0) kg B1 (W =) m × g OR 5(.0) × 10 C1 50 (N) A1 4(c)(i) 240 (cm2) B1 4(c)(ii) (P =) F ÷ A in any form C1 60 ÷ (20 × 12) OR 60 ÷ 240 C1 0.25 (N / cm2) A1

This question in 0625/32 Oct/Nov 2021

5 A woman starts to push a trolley across the floor. Fig. 5.1 shows the horizontal forces acting on the trolley. trolley 120 N 90 N Fig. 5.1 (a) Determine the resultant horizontal force on the trolley. resultant force = … N direction of resultant force = … [3] (b) The total weight of the trolley and boxes is 900 N. The area of each wheel in contact with the ground is 8.0 cm2. The trolley has four wheels. Calculate the pressure on the ground due to the total weight of the trolley and boxes. Include the correct unit in your answer. pressure on the ground = … unit … [5] [Total: 8]

8 marks

Mark scheme: 5(a) (resultant force =) 30 (N) A2 (resultant force =) 120 – 90 (A1) to the left OR forwards B1 5(b) (pressure =) 28 A4 (pressure =) 900 ÷ 32 (C3) (total area =) (4 × 8 =) 32 (cm2) (B1) (pressure =) force ÷ area (C1) N / cm2 B1

This question in 0625/32 Feb/March 2022

3 (a) Fig. 3.1 shows a metal block and its dimensions. 12.0 cm 3.0 cm ground 2.0 cm Fig. 3.1 (not to scale) (i) Calculate the area of the metal block in contact with the ground. area = … cm2 [2] (ii) The mass of the metal block is 0.84 kg. Calculate the weight of the metal block. weight = … N [2] (b) A different metal block has a weight of 24 N. The area of this metal block in contact with the ground is 4.0 cm2. Calculate the pressure of this block on the ground. pressure = … N / cm2 [3] [Total: 7]

7 marks

Mark scheme: 3(a)(i) 24 (cm2) A2 (area in contact with ground) =length  width OR 12  2(.0) (C1) 3(a)(ii) (weight =) 8.4 (N) A2 (weight =) mass  g OR 0.84  10 (C1) 3(b) (pressure =) 6(.0) (N / cm2) A3 (pressure =) 24 ÷ 4(.0) (C2) (pressure =) force ÷ area (C1)

This question in 0625/31 May/June 2022

3 Fig. 3.1 shows a vehicle that is designed to travel on snow. snow-tracks Fig. 3.1 The vehicle has four snow-tracks. (a) Explain why the snow-tracks are better than wheels for travelling on snow. … … … [2] (b) The weight of the vehicle is 4000 N. (i) Calculate the mass of the vehicle. mass = … kg [3] (ii) The area of each snow-track in contact with the ground is 2.0 m2. Each snow-track supports a quarter of the weight of the vehicle. Calculate the pressure that each snow-track exerts on the ground. Include the unit in your answer. pressure exerted by each snow-track = … unit … [4] [Total: 9]

9 marks

Mark scheme: 3(a) any two from larger area lower pressure (on ground) does not sink in 3(b)(i) 400 A3 4000 ÷ 10 OR 4000 ÷ 9.8 (C2) (mass =) weight ÷ g OR weight ÷ 10 weight ÷ 9.8 in any form (C1) 3(b)(ii) 500 A3 1000 ÷ 2.0 OR 4000 ÷ (4  2.0) (C2) (pressure = ) force ÷ area in any form (C1) N / m2 OR Pa B1

This question in 0625/33 May/June 2022

4 (a) A student has an object with a mass of 5.0 kg. Calculate the weight of the object. weight of object = … N [2] (b) The student lifts the 5.0 kg object from the floor onto a table. He does 75 J of work on the object in lifting it onto the table. State the amount of gravitational potential energy gained by the object due to being lifted onto the table. gravitational potential energy gained by object = … J [1] (c) The weight of a table is 280 N. The table has four legs. The area of each table leg in contact with the floor is 18 cm2. Calculate the pressure of the table on the floor. Give the correct unit. pressure on the floor = … unit … [5] [Total: 8]

8 marks

Mark scheme: 4(a) (weight =) 50 (N) A2 (weight =) mass  g OR 5  10 C1 4(b) 75 (J) B1 4(c) 3.9 A4 280 / 72 C3 (P =) F / A OR (pressure =) force / area C1 (area = 4  18 =) 72 (cm2) C1 N / cm2 B1

This question in 0625/31 Oct/Nov 2022

5 An engineer measures the pressure of the gas in a gas bottle. Fig. 5.1 shows the measuring device he uses, connected to the gas bottle. mm 300 glass tube 250 from a gas bottle 200 150 100 50 mercury 0 Fig. 5.1 (a) (i) Atmospheric pressure is 756 mm of mercury. Calculate the pressure of the gas in the gas bottle. pressure of gas = … mm of mercury [3] (ii) State the name of the measuring device shown in Fig. 5.1. … [1] (b) Some gas is trapped in a cylinder fitted with a moveable piston. Fig. 5.2 shows the arrangement. gas cylinder moveable piston Fig. 5.2 (i) Describe how the gas exerts a pressure on the cylinder. Use your ideas about molecules. … … [2] (ii) The piston moves and increases the volume occupied by the gas. The temperature of the gas remains constant. Fig. 5.3 shows the new position of the piston. moveable gas piston cylinder Fig. 5.3 State and explain what happens to the pressure of the gas on the cylinder. … … [2] [Total: 8]

8 marks

Mark scheme: 5(a)(i) 880 (mm Hg) A3 (180 – 60 =) 120 (mm Hg) C2 (left tube =) 60 (mm Hg) AND (right tube =) 180 (mm Hg) seen C1 5(a)(ii) (U-tube) manometer B1 5(b)(i) any two from: B2 • molecules in air moving at high speed / kinetic energy • molecules collide with cylinder OR wall (of cylinder) OR piston • force of collisions (per unit area) cause pressure. 5(b)(ii) smaller / lower pressure (on cylinder) B1 (because) reduced rate of collisions OR fewer collisions with cylinder OR wall (of cylinder) OR piston (per unit area) B1

This question in 0625/31 Oct/Nov 2022

2 A device for measuring gas pressure is connected to a gas supply as shown in Fig. 2.1 300 250 242 connection to gas supply 200 h 150 100 82 50 mercury mm scale 0 Fig. 2.1 (a) (i) State the name of the measuring device shown in Fig. 2.1. … [1] (ii) Determine the difference h between the mercury levels shown in Fig. 2.1. h = … mm [2] (b) The atmospheric pressure is 760 mm of mercury. Determine the pressure of the gas supply. pressure of gas supply = … mm of mercury [1] (c) Suggest why this measuring device uses mercury rather than coloured water. … [1] (d) The gas supply is turned off and the device is disconnected from the gas supply. Both ends of the tube are open to the atmosphere. side A side B Fig. 2.2 On Fig. 2.2, draw and label the levels of mercury in side A and in side B of the tube. [1] [Total: 6]

6 marks

Mark scheme: 2(a)(i) (mercury / U-tube) manometer B1 2(a)(ii) 160 (mm) A2 242 – 82 (C1) 2(b) 920 (mm of mercury) B1 2(c) density (of water) too small / manometer would be too high / big owtte B1 2(d) levels at same height B1

This question in 0625/32 Oct/Nov 2022

3 Fig. 3.1 shows the distance-time graph for a cyclist. The journey has two sections, PQ and QR. Q R 250 200 150 distance / m 100 50 P 0 0 5.0 10.0 15.0 20.0 25.0 30.0 time / s Fig. 3.1 (a) (i) Calculate the speed of the cyclist in section PQ. speed = … m / s [3] (ii) Describe the motion of the cyclist in section QR on the graph. … [1] (b) Fig. 3.2 shows a bicycle fitted with wide tyres and a bicycle fitted with narrow tyres. The two bicycles have the same weight. People use bicycles fitted with wide tyres to ride over soft ground. wide tyre narrow tyre bicycle with wide tyres bicycle with narrow tyres Fig. 3.2 Explain why people use bicycles fitted with wide tyres to ride over soft ground. Use your ideas about pressure. … … … [2] [Total: 6]

6 marks

Mark scheme: 3(a)(i) (speed =) 25 (m / s) A3 (speed =) 250 ÷ 10 (C2) (speed =) gradient of d-t graph OR d ÷ t in any form (C1) 3(a)(ii) (QR –) at rest or stationary B1 3(b) any two from B2 (wide tyres have) large (contact) area (so) less pressure (on ground) so less likely to sink (into soft ground)

This question in 0625/32 Feb/March 2023

2 Fig. 2.1 shows a concrete beam resting on the ground. concrete beam 12 cm 160 cm ground Fig. 2.1 (not to scale) (a) The weight of the concrete beam is 1540 N. Calculate the pressure on the ground due to the concrete beam. pressure = … N / cm2 [4] (b) A builder starts to raise one end of the beam. He uses a force of 1030 N at a perpendicular distance of 120 cm from the pivot. Fig. 2.2 shows the arrangement. 1030 N one end of pivot the beam 120 cm ground Fig. 2.2 (not to scale) Calculate the moment of the 1030 N force about the pivot. moment = … N cm [3] (c) Describe how the builder can use a smaller force to lift the beam. … [1] (d) The builder positions the beam as shown in Fig. 2.3. 160 cm ground Fig. 2.3 (not to scale) State why the beam shown in Fig. 2.3 is less stable than the beam shown in Fig. 2.1. … [1] [Total: 9]

9 marks

Mark scheme: 2(a) (pressure =) 0.8(0) (N / cm2) A4 (pressure =) 1540  1920 OR 1540  (160  12) (C3) (pressure =) force  area (C1) (area in contact with ground =) 12  160 = 1920 (cm2) (C1) 2(b) (moment =) 120 000 (N cm) OR 1.2  105 (N cm) A3 (moment =) 1030  120 (C2) (moment =) force  (perpendicular) distance from pivot (C1) 2(c) move (lifting) force further from pivot owtte B1 2(d) centre of gravity / mass is high(er) OR idea that area of base is small(er) B1

This question in 0625/31 May/June 2023

3 A student has a battery-powered torch. Fig. 3.1 shows the torch. base of torch Fig. 3.1 (a) Fig. 3.2 shows the energy transfers when the torch is switched on. The diagram is incomplete. electrical working … energy store … energy 100 J 70 J thermal energy store … J Fig. 3.2 Show the energy transfers in the torch by completing the labels on Fig. 3.2. [3] (b) The weight of the torch is 8.5 N. The student lifts the torch a vertical distance of 0.80 m to place it on a shelf. Calculate the work done on the torch by the student. work done = … J [3] (c) The student places the torch on its base on a shelf. The area of the base of the torch is 44 cm2. The weight of the torch is 8.5 N. Calculate the pressure on the shelf due to the torch. pressure on shelf = … N / cm2 [3] [Total: 9]

9 marks

Mark scheme: 3(a) chemical (energy) B1 light (energy) B1 30 (J) B1 3(b) 6.8 (J) A3 (work done =) 8.5  0.8(0) (C2) (work done =) force  distance (moved) (C1) 3(c) 0.19 (N / cm2) A3 (P =) 8.5 ÷ 44 (C2) (P =) F ÷ A in any form (C1)

This question in 0625/32 May/June 2023

4 A tight-fitting lid keeps air inside a metal can. An airtight rubber bung holds a liquid-in-glass thermometer that is inserted through a hole in the lid, as shown in Fig. 4.1. liquid-in-glass thermometer lid metal can air Fig. 4.1 (a) (i) State what happens to the liquid in the thermometer when the air temperature rises. … [1] (ii) The temperature of the air in the can is 18 °C. Calculate the temperature of the air in kelvin. temperature = … K [2] (b) The can is placed in a refrigerator. The temperature of the air inside the can decreases. State and explain what happens to the pressure exerted by the air in the can. Use your ideas about gas particles. … … … … [3] (c) The air in another can exerts a pressure of 102 000 N / m2 on the lid. The area of the can lid is 0.0082 m2. Calculate the force on the lid due to the air in the can. force = … N [3] [Total: 9]

9 marks

Mark scheme: 4(a)(i) (liquid / it) expands B1 4(a)(ii) 273 + 18 C1 291 (K) A1 4(b) pressure decreases B1 any two from:  particles slower / less kinetic energy  collisions (with wall) less frequent  collide (with wall) with less force B2 4(c) (F =) P  A in any form C1 102 000  0.0082 C1 840 (N) A1

This question in 0625/33 May/June 2023

3 Fig. 3.1 shows a computer on the surface of a desk. computer surface of desk Fig. 3.1 (a) The weight of the computer is 48 N. The area of the computer in contact with the surface of the desk is 20 cm2. Calculate the pressure due to the computer on the surface of the desk. pressure = … N / cm2 [3] (b) A student uses a force of 12 N to tilt the computer as shown in Fig. 3.2. 12 N 32 cm pivot Fig. 3.2 Calculate the moment of the 12 N force about the pivot. moment = … N cm [3] [Total: 6]

6 marks

Mark scheme: 3(a) (P =) 2.4 (N / cm2) A3 (P =) 48 ÷ 20 (C2) (P =) F ÷ A (C1) 3(b) (moment = ) 380 (N cm) A3 (moment = ) 12  32 (C2) moment = force  (perp.) distance from pivot (C1)

This question in 0625/32 Feb/March 2024

5 Fig. 5.1 shows a metal block at room temperature on a table. metal block table Fig. 5.1 (a) Describe the arrangement, separation and motion of the particles in the metal block. … … … … [3] (b) (i) The temperature of the metal block decreases. Describe any changes in the motion and separation of the particles in the metal block. … … [2] (ii) A scientist cools the metal block until its temperature is close to absolute zero. Describe the motion of the particles in the metal block. … … [1] (c) The weight of the metal block is 26 N. The area of the metal block in contact with the table is 42 cm2. Calculate the pressure on the table due to the metal block. pressure = … N / cm2 [3] [Total: 9]

9 marks

Mark scheme: 5(a) (particles are) fixed in position / in lattice OR regular / fixed arrangement / pattern B1 can only vibrate / no translational KE B1 close / closer (than in liquids or gases) B1 5(b)(i) (particles move) closer (as temperature decreases) B1 particles vibrate slower / less OR have smaller vibrations B1 5(b)(ii) (at absolute zero particles have) least / smallest vibrations B1 5(c) (P =) 0.62 (N / cm2) A3 (P =) 26 ÷ 42 (C2) (P =) F ÷ A (C1)

This question in 0625/32 May/June 2024

2 A person pushes a pushchair. A young child rides in the pushchair. Fig. 2.1 shows horizontal forces acting on the front wheel of the pushchair. 30 N 10 N Fig. 2.1 (not to scale) (a) Calculate the resultant of the horizontal forces shown in Fig. 2.1. resultant force = … N direction = … [2] (b) (i) Another person pushes a shopping trolley with a force of 40 N. The shopping trolley moves at a constant speed along a horizontal path. Calculate the work done by the 40 N force to move the shopping trolley a distance of 50 m. work done = … J [3] (ii) The work done on the shopping trolley as it starts moving is transferred into other energy stores. State two such energy stores. 1 … 2 … [2] (c) In (a), the weight of the pushchair and child is 240 N. The total area of contact with the ground is 38 cm2. Calculate the pressure on the ground due to the pushchair and child. pressure on ground = … N / cm2 [3] [Total: 10]

10 marks

Mark scheme: 2(a) (30 – 10 =) 20 (N) B1 forwards OR in direction of 30 N force B1 2(b)(i) (work done =) 2000 (J) A3 (work done =) 40  50 (C2) (work done =) force  distance (moved in direction of force) OR (W) = F × d (C1) 2(b)(ii) internal OR thermal energy (of surroundings / tyres) B1 kinetic energy B1 2(c) (pressure =) 6.3 (N / cm2) A3 (pressure =) 240  38 (C2) (pressure =) force  area OR (p) = F  A (C1)

This question in 0625/31 Oct/Nov 2024

3 (a) A boy weighs 620 N. Calculate the mass of the boy. mass of boy = … kg [3] (b) Fig. 3.1 shows another boy sitting on a solid block of wood. block of wood ground Fig. 3.1 The total weight of the boy and the block is 1200 N. The area of the block of wood in contact with the ground is 0.16 m2. Calculate the pressure exerted on the ground. pressure = … N / m2 [3] [Total: 6]

6 marks

Mark scheme: 3(a) 63 (kg) A3 620  9.8 (C2) weight = mass  gravitation field strength OR (m = ) W  g (C1) 3(b) 7500 A3 1200  0.16 (C2) (p =) F  A (C1)

This question in 0625/33 Oct/Nov 2024

3 Fig. 3.1 shows a cross-section through the centre of a solid cone and a cross-section through the centre of a solid cuboid. The cone and the cuboid are placed on the ground as shown in Fig. 3.1. The solid cone is more stable than the solid cuboid. The symbol × indicates the position of the centre of gravity of each object. cone cuboid ground Fig. 3.1 (a) Explain why the solid cone in Fig. 3.1 is more stable than the solid cuboid. … … [1] (b) The weight of the solid cuboid is 48 N. The area of the solid cuboid in contact with the ground is 32 cm2. Calculate the pressure exerted by the solid cuboid on the ground. pressure on ground = … N / cm2 [3] (c) A horizontal force of 15 N tilts the solid cone, as shown in Fig. 3.2. 15 N 18 cm pivot Fig. 3.2 Calculate the moment of the 15 N force about the pivot. moment = … N cm [3] [Total: 7]

7 marks

Mark scheme: 3(a) (has) lower centre of gravity owtte B1 3(b) (P =) 1.5 (N / cm2) A3 (P =) 48 ÷ 32 (C2) (P =) F ÷ A (C1) 3(c) (moment =) 270 (Ncm) A3 (moment =) 15  18 (C2) moment = force  (perp.) distance (from pivot) (C1)

This question in 0625/32 Feb/March 2025

3 (a) A student determines the weight W of a metal block by using a 1.5 N load and a uniform metre ruler. She places the centre of the uniform metre ruler on a pivot. metal 0.44 m 0.21 m block load 1.5 N pivot W uniform metre ruler Fig. 3.1 (not to scale) She moves the metal block and the 1.5 N load until the uniform metre ruler balances horizontally as shown in Fig. 3.1. Calculate the weight W of the metal block. Use the principle of moments in your answer. weight W = … N [4] (b) A different metal block is lying on the ground, as shown in Fig. 3.2. 0.54 m 0.18 m ground Fig. 3.2 (not to scale) The weight of the metal block is 890 N. Calculate the pressure on the ground caused by the block in the position shown in Fig. 3.2. pressure = … N / m2 [3] [Total: 7]

7 marks

Mark scheme: 3(a) 3.1 (N) A4 1.5  0.44 = W  0.21 OR (W =) (1.5  0.44) ÷ 0.21 C3 OR (W =) 0.66 ÷ 0.21 (anticlockwise moment =) 1.5  0.44 OR 0.66 seen C1 (sum of) anticlockwise moment = (sum of) clockwise moment C1 3(b) 9200 (N / m2) A3 890 ÷ (0.54  0.18) OR 890 ÷ 0.0972 C2 (pressure =) force ÷ area C1

This question in 0625/32 May/June 2025

3 (a) A student determines the weight W of a metal block by using a 1.5 N load and a uniform metre ruler. She places the centre of the uniform metre ruler on a pivot. metal 0.44 m 0.21 m block load 1.5 N pivot W uniform metre ruler Fig. 3.1 (not to scale) She moves the metal block and the 1.5 N load until the uniform metre ruler balances horizontally as shown in Fig. 3.1. Calculate the weight W of the metal block. Use the principle of moments in your answer. weight W = … N [4] (b) A different metal block is lying on the ground, as shown in Fig. 3.2. 0.54 m 0.18 m ground Fig. 3.2 (not to scale) The weight of the metal block is 890 N. Calculate the pressure on the ground caused by the block in the position shown in Fig. 3.2. pressure = … N / m2 [3] [Total: 7]

7 marks

Mark scheme: 3(a) 3.1 (N) A4 1.5  0.44 = W  0.21 OR (W =) (1.5  0.44) ÷ 0.21 C3 OR (W =) 0.66 ÷ 0.21 (anticlockwise moment =) 1.5  0.44 OR 0.66 seen C1 (sum of) anticlockwise moment = (sum of) clockwise moment C1 3(b) 9200 (N / m2) A3 890 ÷ (0.54  0.18) OR 890 ÷ 0.0972 C2 (pressure =) force ÷ area C1

This question in 0625/33 May/June 2025

5 Fig. 5.1 shows an empty metal cylinder. metal cylinder ground Fig. 5.1 (a) Describe the arrangement, separation and motion of the metal particles. … … … … … [3] (b) The cylinder is filled with a gas. Describe how particles of the gas exert a pressure on the inside surface of the metal cylinder. … … … … … [3] (c) The weight of the metal cylinder is 420 N. The area of the metal cylinder in contact with the ground is 300 cm2. Calculate the pressure on the ground due to the metal cylinder. pressure = … N / cm2 [3] [Total: 9]

9 marks

Mark scheme: 5(a) (particles are;) fixed in position/ in lattice/regular pattern B1 (can only) vibrate / no translational KE B1 close(r than in liquids or gases) B1 5(b) any THREE from: B3 (particles/they) move at high speed OR have high/large KE move randomly (particles/they) collide with it/surface/walls (collisions) create a force (on cylinder wall) idea of P = F / A 5(c) (P = ) 1.4 (N/cm2) A3 (P = ) 420 ÷ 300 (C2) (P = ) F ÷ A (C1)

This question in 0625/31 Oct/Nov 2025

2 Fig. 2.1 shows the horizontal forces acting on an ice skater. The ice skater is moving forwards. ice skater backward force = 45 N forward force = 80 N skate edge of skate in contact with the ice Fig. 2.1 (a) Calculate the resultant horizontal force acting on the ice skater. Determine the direction of the resultant force. resultant horizontal force = … N direction = … [2] (b) The weight of the ice skater is 700 N. The area of the skate in contact with the ice is 6.2 cm2. Calculate the pressure on the surface of the ice exerted by the skate. Give your answer to two significant figures. pressure = … N / cm2 [3] (c) The weight of the ice skater is 700 N. Calculate the mass of the ice skater. Give your answer to two significant figures. mass of skater = … kg [3] [Total: 8]

8 marks

Mark scheme: 2(a) 35 (N) B1 forwards / to the right B1 2(b) 110 (N / cm2) A3 700 ÷ 6.2 C2 (pressure =) force / area C1 2(c) 71 (kg) A3 700 ÷ 9.8 C2 (mass =) weight / gravitational field strength or W / g or W / 9.8 C1

This question in 0625/32 Oct/Nov 2025