IGCSE physics diagrams

Reading cue. The 5.0 cm and 3.0 cm are edges of the front face and the 2.0 cm runs into the page: three different edges, so no length is used twice, and 240 g is the mass of all of it.

1. aIdentify Identify which of the three edge lengths in Fig. 3.1 gives the depth of the block, and state how many faces of the block measure 5.0 cm by 3.0 cm.

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   Check answer · 2 marks

   1. depth is the 2.0 cm edge, the one drawn receding into the page (1)
   2. two faces measure 5.0 cm by 3.0 cm, the front face and the hidden back face (1)

2. bCalculate A second block is cut from the same metal and measures 5.0 cm by 3.0 cm by 4.0 cm. Calculate its mass.

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   Check answer · 3 marks

   1. volume of second block = 5.0 × 3.0 × 4.0 = 60 cm³ (1)
   2. volume of block in Fig. 3.1 = 30 cm³, so the second block has twice the volume, or 8.0 g in every cm³ used (1)
   3. mass = 480 g (1)

3. cExplain The block in Fig. 3.1 is cut in half by a single cut parallel to its 5.0 cm by 3.0 cm faces. State the volume and mass of one half, and explain what happens to the density of the metal.

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   Check answer · 4 marks

   1. volume of one half = 5.0 × 3.0 × 1.0 = 15 cm³ (1)
   2. mass of one half = 120 g (1)
   3. density is unchanged at 8.0 g/cm³ (1)
   4. mass and volume are both halved, so the mass in each cm³ is the same; density does not depend on how much material is present (1)

4. dSuggest The three edge lengths in Fig. 3.1 are all measured with the same ruler, so each carries the same uncertainty in centimetres. Suggest which of the three measurements limits the accuracy of any result calculated from them most severely, and justify your choice by comparison with the other two.

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   Check answer · 3 marks

   1. the 2.0 cm edge, the depth (1)
   2. taking the uncertainty as 0.1 cm for the sake of argument, that is 5% of 2.0 cm but about 3% of 3.0 cm and only 2% of 5.0 cm (1)
   3. the smallest length carries the largest percentage error, and because the three lengths are multiplied together that error passes into the volume (1)

Reading cue. Both distances are measured from the pivot, not from the ends of the beam, and the pivot is at the centre of a uniform beam, so the beam's own weight pulls straight down through it.

1. aIdentify Identify which of the two weights in Fig. 4.1 hangs the nearer to the pivot when the beam balances, and state the reason.

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   Check answer · 2 marks

   1. the 8.0 N weight (1)
   2. it is the larger force, so it needs the smaller distance from the pivot to give the same moment (1)

2. bCalculate Calculate the moment of the 4.0 N weight about the pivot in Fig. 4.1. Give the unit.

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   Check answer · 2 marks

   1. moment = force x perpendicular distance from pivot = 4.0 × 0.80 (1)
   2. = 3.2 N m (1)

3. cDetermine The 8.0 N weight in Fig. 4.1 is taken off and a 12 N weight is hung in its place, 0.30 m from the pivot, with the 4.0 N weight left where the figure shows it. Determine whether the beam still balances, and if it does not, state which way it turns.

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   Check answer · 3 marks

   1. anticlockwise moment of the 4.0 N weight = 4.0 × 0.80 = 3.2 N m (1)
   2. clockwise moment of the 12 N weight = 12 × 0.30 = 3.6 N m (1)
   3. the two moments are not equal, so the beam does not balance; the clockwise moment is the larger, so the beam turns clockwise, the 12 N end going down (1)

4. dExplain The beam is replaced by one of the same length that is thicker, and therefore heavier, towards its left-hand end. It rests on the same pivot at its mid-point and carries only the two weights of Fig. 4.1, the 4.0 N weight still 0.80 m from the pivot. Explain why the 8.0 N weight must now hang further from the pivot than it does in Fig. 4.1.

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   Check answer · 4 marks

   1. the centre of mass of the new beam lies to the left of the mid-point, and so to the left of the pivot (1)
   2. the weight of the beam therefore has a turning effect about the pivot, anticlockwise, where in Fig. 4.1 it acted through the pivot and had none (1)
   3. the total anticlockwise moment is now greater than 3.2 N m (1)
   4. the clockwise moment must rise to match it and the force is still 8.0 N, so the distance from the pivot must increase (1)

Reading cue. Both 1.8 m and 1.2 m run from the ground up to the ball, not from one to the other, and 1.2 m is the top of the rebound, where the ball is momentarily at rest.

1. aIdentify Identify the point on the path drawn in Fig. 5.1 at which the ball has its greatest kinetic energy, and state the height of the ball above the ground there.

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   Check answer · 2 marks

   1. just before it reaches the ground, at the bottom of the fall (1)
   2. height = 0 m, at ground level (1)

2. bDetermine Determine the speed of the ball just before it strikes the ground. Take the gravitational field strength as 9.8 N/kg and ignore air resistance.

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   Check answer · 3 marks

   1. kinetic energy gained equals gravitational potential energy lost: (1/2)mv² = mgh (1)
   2. v² = 2 × 9.8 × 1.8 = 35.28 (m/s)² (1)
   3. v = 5.9 m/s (1)

3. cCalculate Calculate the percentage of its kinetic energy at the ground that the ball still has immediately after the bounce, and show that this percentage does not depend on the mass of the ball.

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   Check answer · 3 marks

   1. energy immediately after the bounce is mgh with h = 1.2 m; energy at the ground before the bounce is mgh with h = 1.8 m (1)
   2. ratio = 1.2/1.8 = 0.67, so 67% (1)
   3. m and g appear in both expressions and cancel, so only the two marked heights decide the percentage (1)

4. dEstimate The ball is allowed to bounce a second time. Assuming the same percentage of energy is lost at every bounce, estimate the height of the ball at the top of its second rebound, and suggest one reason why the true height would be lower than your estimate.

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   Check answer · 3 marks

   1. height = 1.2 x (1.2/1.8) (1)
   2. = 0.80 m (1)
   3. reason, any one: air resistance also removes energy during the flight; the fraction lost at each bounce is not really constant, since the ball deforms differently at different speeds (1)

Reading cue. Only horizontal forces are drawn: the weight and the road's upward push are missing, and the shorter arrow carries no number, so its size cannot be read off the drawing at all.

1. aIdentify Identify what the difference in length between the two arrows in Fig. 7.1 shows about the motion of the car at the moment drawn.

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   Check answer · 2 marks

   1. there is a resultant force on the car, acting forwards (1)
   2. so the car is accelerating, gaining speed in its direction of travel (1)

2. bDetermine The car starts from rest and covers 100 m of road in the first 8.0 s. Determine the average power developed by the forward force shown in Fig. 7.1 over this time.

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   Check answer · 3 marks

   1. work done by forward force = force x distance = 4500 × 100 (1)
   2. = 4.5 × 10⁵ J (1)
   3. power = work done / time taken = 4.5 × 10⁵ / 8.0 = 5.6 × 10⁴ W, that is about 56 kW (1)

3. cDetermine The car has a speed of 24 m/s at the end of the 8.0 s. Determine the energy transferred to the surroundings by the resistive force drawn in Fig. 7.1 during that time.

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   Check answer · 3 marks

   1. kinetic energy gained = (1/2) x 1200 × 24² = 3.456 × 10⁵ J (1)
   2. energy to surroundings = work done by forward force - kinetic energy gained = 4.5 × 10⁵ - 3.456 × 10⁵ (1)
   3. = 1.044 × 10⁵ J, that is 1.0 × 10⁵ J to two significant figures (1)

4. dExplain No value is printed beside the resistive arrow in Fig. 7.1. Explain why an acceleration worked out from a starting speed of 0, a final speed of 24 m/s and a time of 8.0 s is only an average value, and describe how the resistive force behaves over that time.

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   Check answer · 4 marks

   1. 24/8.0 uses only the total change in velocity divided by the total time, so it is a mean value for the whole 8.0 s (1)
   2. the resistive force grows as the car goes faster, because air resistance increases with speed, while the forward force stays at 4500 N (1)
   3. the resultant force therefore falls during the 8.0 s, and the mass is fixed, so the acceleration falls too: greater than the mean at the start, smaller at the end (1)
   4. a constant acceleration would give an average speed of (0 + 24)/2 = 12 m/s and so 96 m in 8.0 s; the car in fact covers 100 m, which is what a larger acceleration early on gives (1)

Reading cue. In the upper panel the 2.5 m/s belongs to A alone, B at rest carrying none; in the lower panel v is the speed of the joined 2.0 kg, not of either trolley on its own.

1. aState State the momentum of trolley B before the collision, and give the reason from the upper panel of Fig. 8.1.

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   Check answer · 2 marks

   1. zero (0 kg m/s) (1)
   2. B is at rest, so its velocity is zero, and momentum is mass x velocity (1)

2. bDetermine After the collision the joined trolleys move to the right at 1.0 m/s. Determine the change in momentum of trolley A, and give its direction.

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   Check answer · 3 marks

   1. momentum of A before = 0.80 × 2.5 = 2.0 kg m/s to the right (1)
   2. momentum of A after = 0.80 × 1.0 = 0.80 kg m/s to the right (1)
   3. change = 1.2 kg m/s, directed to the left, that is opposite to A's motion (1)

3. cCalculate The two trolleys are in contact for 0.15 s. Calculate the average force that trolley B exerts on trolley A during the collision.

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   Check answer · 3 marks

   1. force = change in momentum / time taken (1)
   2. = 1.2 / 0.15 (1)
   3. = 8.0 N, acting to the left on A (1)

4. dExplain Using both panels of Fig. 8.1, explain why the momentum gained by trolley B is exactly equal to the momentum lost by trolley A, even though B is the heavier trolley and its velocity changes by less.

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   Check answer · 4 marks

   1. the force B exerts on A is equal in size and opposite in direction to the force A exerts on B (1)
   2. the two trolleys are in contact for the same length of time, so force x time is the same for both and the changes in momentum are equal and opposite (1)
   3. A's velocity changes by 1.5 m/s, giving 0.80 × 1.5 = 1.2 kg m/s; B's changes by 1.0 m/s, giving 1.2 × 1.0 = 1.2 kg m/s (1)
   4. B's larger mass is offset exactly by its smaller change in velocity, so the total momentum of the two trolleys is unchanged (1)

Reading cue. The 50.0 cm³ is water alone and the 68.0 cm³ is water with the stone in, so the quantity everything turns on is the difference between the two levels, and it is printed on neither scale.

1. aDetermine Determine the volume of the stone from the two readings in Fig. 10.1.

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   Check answer · 2 marks

   1. volume = 68.0 - 50.0 (1)
   2. = 18.0 cm³ (1)

2. bDetermine A second stone, identical to the first, is lowered into the same cylinder on a thread beside it. Determine the new reading of the water surface, and state one condition that must hold for your answer to be correct.

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   Check answer · 3 marks

   1. the second stone displaces a further 18.0 cm³ (1)
   2. new reading = 68.0 + 18.0 = 86.0 cm³ (1)
   3. condition: both stones must be completely below the surface, and the water must not reach the top of the cylinder (1)

3. cExplain The cylinder in Fig. 10.1 is graduated in divisions of 1 cm³, so each of the two levels can be judged only to within half a division. Explain why the volume of the stone is known far less precisely than either reading, supporting your answer with figures.

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   Check answer · 3 marks

   1. each reading may be out by up to 0.5 cm³, so their difference may be out by up to 1 cm³ (1)
   2. 1 cm³ in 18.0 cm³ is about 6% (1)
   3. 0.5 cm³ in 68.0 cm³ is less than 1%, so subtracting two large readings to obtain a small difference magnifies the error (1)

4. dSuggest A larger stone is to be measured, but lowering it into the cylinder drawn in Fig. 10.1 would take the water above the top graduation. Suggest a change to the method that still uses the same cylinder, and explain why the volume obtained is still correct.

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   Check answer · 3 marks

   1. pour out some water first, so that the starting level is much lower, for example 20 cm³ rather than 50.0 cm³ (1)
   2. enough water must remain to cover the stone completely once it is lowered in (1)
   3. only the difference between the two readings is used, and that difference does not depend on the starting level (1)

Reading cue. The 300 cm³ and 120 cm³ are the trapped gas columns, not the whole cylinder, and the note fixing the temperature is what allows pressure and volume to be linked at all.

1. aIdentify Identify two quantities that are the same for the trapped gas in both drawings of Fig. 2.1, and state how you know each from the figure.

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   Check answer · 2 marks

   1. the mass, or number, of gas molecules is the same, because the cylinder is closed at one end and the piston seals the other so no gas escapes (1)
   2. the temperature is the same, as stated in the note below the drawings (1)

2. bDetermine Determine the volume the trapped gas would occupy if the piston were pushed in until the pressure reached 4.0 × 10⁵ Pa at the same temperature.

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   Check answer · 3 marks

   1. pressure x volume is constant at constant temperature: 1.0 × 10⁵ x 300 = 4.0 × 10⁵ x V (1)
   2. V = 3.0 × 10⁷ / 4.0 × 10⁵ (1)
   3. V = 75 cm³ (1)

3. cExplain Explain, in terms of the molecules of the trapped gas, why the pressure in the lower drawing of Fig. 2.1 is greater than in the upper drawing.

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   Check answer · 4 marks

   1. the same number of molecules is now contained in a smaller volume (1)
   2. the temperature is unchanged, so the average speed and average kinetic energy of the molecules are unchanged (1)
   3. each molecule has a shorter distance to travel between the walls, so it strikes them more often (1)
   4. more collisions each second on each unit area of wall gives a greater average force per unit area, and so a greater pressure (1)

4. dSuggest The piston is instead pushed in very quickly from 300 cm³ to 120 cm³, so that the note below Fig. 2.1 no longer applies. Suggest how the pressure reached compares with the value for a slow compression, and explain your answer.

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   Check answer · 4 marks

   1. the pressure reached is greater than the constant-temperature value (1)
   2. the piston does work on the gas and there is no time for that energy to pass out to the surroundings (1)
   3. the temperature of the gas rises, so the molecules move faster on average (1)
   4. faster molecules strike the walls harder and more often, raising the pressure further; as the gas then cools to room temperature the pressure falls back towards the constant-temperature value (1)

Reading cue. The element lies a short way above the floor and spans only half the width, and the top is open: water below it, water beside it and the surface are three different places.

1. aState State how the density of the water immediately above the element in Fig. 5.1 changes when the element is switched on, and state why it changes.

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   Check answer · 2 marks

   1. the density decreases (1)
   2. the water expands when it is heated, so the same mass now occupies a larger volume (1)

2. bExplain The element drawn in Fig. 5.1 spans only half the width of the tank. Explain how the water in the half of the tank the element does not reach is warmed.

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   Check answer · 3 marks

   1. water heated by the element becomes less dense and rises above it (1)
   2. it spreads sideways below the surface, while cooler, denser water sinks in the far half of the tank and flows back along the floor towards the element (1)
   3. the circulating convection current carries warmed water, and the energy it holds, into every part of the tank (1)

3. cSuggest The tank in Fig. 5.1 is open at the top. Suggest two ways in which the heated water loses energy from that open surface, and suggest one change to the tank that would reduce the loss.

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   Check answer · 3 marks

   1. evaporation: the more energetic molecules escape from the surface and carry energy away, leaving the slower ones behind (1)
   2. energy passes into the air above the surface, which is warmed and carried away by convection, or is radiated from the surface (1)
   3. fit a lid or an insulating cover over the open top (1)

4. dExplain The element is switched off after a long time. A student says that the hottest water must be at the bottom of the tank, since that is where the element is. Using Fig. 5.1, explain why the student is wrong, and state where the coolest water lies.

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   Check answer · 4 marks

   1. the hottest water is at the top of the tank (1)
   2. all through the heating the warmed water above the element rose and collected there, while cooler water sank (1)
   3. the coolest water is the layer trapped between the element and the floor of the tank (1)
   4. warmed water cannot sink to reach it, and water is a poor conductor, so energy reaches that layer only very slowly (1)

Reading cue. The lamp shines in from the side through the lens while the microscope looks straight down, at right angles to the beam, and the specks in view are smoke, not air molecules.

1. aState State the purpose of the converging lens drawn between the lamp and the cell in Fig. 7.1.

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   Check answer · 2 marks

   1. it brings the light together, focusing the beam inside the cell (1)
   2. so that the smoke particles are lit brightly enough to be seen as separate specks (1)

2. bExplain Explain why the microscope in Fig. 7.1 is placed vertically above the cell rather than in line with the lamp.

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   Check answer · 3 marks

   1. the beam from the lamp crosses the cell sideways and does not travel up into the microscope (1)
   2. only light scattered from the smoke particles travels upwards into the microscope (1)
   3. the specks are then seen bright against a dark background, whereas looking along the beam would flood the view with light from the lamp (1)

3. cExplain The cell in Fig. 7.1 is warmed gently. Explain the change seen in the movement of the specks.

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   Check answer · 3 marks

   1. the specks move faster and change direction more often, so the movement looks more violent (1)
   2. at the higher temperature the air molecules have greater average kinetic energy and move faster (1)
   3. each impact on a smoke particle is therefore harder and impacts arrive more frequently (1)

4. dSuggest A student looking down the microscope says that the specks being watched are the air molecules themselves. Suggest two pieces of evidence, one from the apparatus in Fig. 7.1 and one from what is seen, that show this cannot be so.

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   Check answer · 4 marks

   1. the specks can be seen through an ordinary light microscope, so they are far larger than molecules, which are much too small to be seen in this way (1)
   2. the cell held air before the smoke was introduced and nothing was visible until the smoke was added (1)
   3. the specks move in short, sudden, random jerks rather than smoothly (1)
   4. this is what is expected of a particle being struck unequally on opposite sides by very many smaller particles that cannot themselves be seen (1)

Reading cue. The vacuum gap runs down both sides and under the base but stops at the sealed neck; the stopper and the small supports are the only solid links to the outer case.

1. aIdentify Identify the two parts drawn in Fig. 9.1 that provide a solid path for thermal energy out of the inner container, and state the property of the material chosen for them.

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   Check answer · 2 marks

   1. the stopper closing the neck and the small supports under the base (1)
   2. both are made of insulating material, that is poor conductors of thermal energy, and their contact area is kept small (1)

2. bExplain Explain how the stopper drawn in Fig. 9.1 slows the cooling of the drink, other than by being a poor conductor.

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   Check answer · 3 marks

   1. it seals the neck, so vapour and warm air above the drink cannot escape (1)
   2. evaporation removes the fastest molecules from the surface of the drink, and this cools the drink (1)
   3. with the neck closed the escaped molecules stay trapped just above the surface and many return to the liquid, so the net loss by evaporation is small, and the warm air cannot be carried away and replaced by cooler air (1)

3. cExplain The drink does not reach the stopper: Fig. 9.1 shows a space between the surface of the drink and the neck. Explain how energy still crosses this space, and suggest how filling the flask closer to the stopper would change the rate of cooling.

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   Check answer · 3 marks

   1. the warm surface of the drink emits infrared radiation across the space (1)
   2. air in the space is warmed, circulates by convection, and some liquid evaporates into it (1)
   3. filling closer to the stopper leaves a smaller space, so less evaporation and less convection are possible and the drink cools more slowly (1)

4. dSuggest The same flask is used to keep a cold drink cold on a hot day. Suggest whether the vacuum and the silvered surfaces still do useful work, and explain your answer in terms of the direction in which energy travels.

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   Check answer · 4 marks

   1. both still work (1)
   2. the energy now travels inwards, from the warm surroundings towards the colder drink (1)
   3. the vacuum contains no particles, so neither conduction nor convection can carry energy across the gap in either direction (1)
   4. the silvering on the outer of the two glass walls reflects the infrared radiation arriving from the surroundings back outwards, before it can cross the gap to the drink (1)

Reading cue. Read the 48 W as a rate, not an amount: the drawing fixes the energy delivered each second, and the two holes are drilled apart so the thermometer reads the block, not the heater.

1. aState State the energy transferred to the aluminium block by the heater shown in Fig. 10.1 in one second, and state the quantity the label 48 W measures.

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   Check answer · 2 marks

   1. 48 J
   2. 48 W is a power, the rate at which the heater transfers energy

2. bCalculate The specific heat capacity of aluminium is 900 J/(kg °C). Calculate the time for which the heater in Fig. 10.1 must run to raise the temperature of the block by 20 °C, assuming no thermal energy escapes.

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   Check answer · 2 marks

   1. E = mcΔθ = 1.0 × 900 × 20 = 18 000 J
   2. t = E/P = 18 000 / 48 = 375 s (6.25 minutes)

3. cExplain The heater and the thermometer occupy holes at opposite sides of the block. Explain why the thermometer reading rises more slowly in the first minute after switching on than it does later in the run.

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   Check answer · 3 marks

   1. energy must be conducted through the aluminium from the heater hole across to the thermometer hole
   2. this takes time, so the thermometer lags behind the temperature of the aluminium next to the heater
   3. some of the early energy also warms the heater itself and the metal immediately around it before the block is at a uniform temperature

4. dDetermine The student switches the heater off at the end of the run and finds that the block cools by 0.60 °C in the next minute. Determine the rate at which the block was losing thermal energy at the end of the run, and hence determine the percentage of the heater's 48 W that was actually raising the temperature of the block at that moment.

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   Show a hint

   The block cools at the same rate whether or not the heater is on, so the cooling run measures the loss that was happening during the heating run.

   Check answer · 4 marks

   1. energy lost in 60 s = mcΔθ = 1.0 × 900 × 0.60 = 540 J
   2. rate of loss = 540/60 = 9.0 W
   3. useful power = 48 − 9.0 = 39 W
   4. percentage = 39/48 × 100 = 81%

Reading cue. The 2.5 cm dimension runs crest to next crest, not crest to trough, and the scale line is the undisturbed water level, so amplitude is measured from that line rather than trough to crest.

1. aDetermine Determine the total horizontal distance occupied by the complete waves drawn in Fig. 2.1.

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   Check answer · 2 marks

   1. four complete wavelengths are drawn
   2. 4 × 2.5 = 10 cm

2. bDetermine The pattern drawn in Fig. 2.1 takes 0.50 s to travel a distance equal to the whole marked scale. Determine the frequency of the waves.

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   Check answer · 2 marks

   1. speed = 10 cm / 0.50 s = 20 cm/s
   2. f = v/λ = 20 / 2.5 = 8.0 Hz

3. cDescribe A small cork floats on the water at the 5.0 cm mark on the scale in Fig. 2.1. Describe how the cork moves as the waves pass, and state how far it travels along the scale.

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   Check answer · 3 marks

   1. the cork moves up and down about the undisturbed water level, at right angles to the arrow showing the direction of travel
   2. it completes one full oscillation each time a whole wave passes it
   3. it travels no distance along the scale, because the wave transfers energy along the tank without carrying the water along with it

4. dExplain A student says that doubling the frequency of the dipper will double the number of complete waves fitting into the 10 cm scale in Fig. 2.1, but will leave the speed of a crest unchanged. Explain whether the student is correct.

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   Check answer · 4 marks

   1. the speed of the water waves is fixed by the depth of the water, not by the dipper, so the crest speed is unchanged
   2. since v = fλ and v is fixed, doubling f halves the wavelength from 2.5 cm to 1.25 cm
   3. the 10 cm scale then holds 8 complete waves instead of 4
   4. the student is correct on both counts

Reading cue. The 165 m dimension line is the one-way distance to the wall, so the sound covers it twice; the figure deliberately draws no sound path for you to count.

1. aDetermine Determine the total distance travelled by the sound between the clap and the student hearing the echo.

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   Check answer · 2 marks

   1. the sound travels from the student to the wall and back again
   2. 2 × 165 = 330 m

2. bDetermine The student walks towards the wall until the echo returns 0.40 s after she claps. Taking the speed of sound in air as 330 m/s, determine her new distance from the wall.

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   Check answer · 2 marks

   1. total path = 330 × 0.40 = 132 m
   2. distance to the wall = 132/2 = 66 m

3. cDetermine Back at the position drawn in Fig. 5.1 she times the echo at 1.0 s after the clap, but her timing could be wrong by up to 0.20 s either way. Determine the largest and the smallest speeds of sound consistent with the 165 m marked in the figure.

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   Check answer · 3 marks

   1. path length = 2 × 165 = 330 m
   2. largest speed = 330/0.80 = 412.5 m/s, that is 410 m/s to 2 significant figures
   3. smallest speed = 330/1.20 = 275 m/s

4. dSuggest Suggest whether she would obtain a better value for the speed of sound by doubling the 165 m marked in Fig. 5.1 or by buying a stopwatch that reads to 0.01 s, and give reasons.

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   Check answer · 3 marks

   1. the timing error is set by her reaction time, not by how finely the stopwatch reads, so a more precise stopwatch changes almost nothing
   2. doubling the distance doubles the echo time while the reaction error stays the same size
   3. the error is then a smaller fraction of the measured time, so moving further from the wall is the better improvement

Reading cue. The 30° arc is between a wavefront and the boundary, not between a ray and a normal, and the shallow half of the tank is left blank on purpose: you must say what belongs there.

1. aState The shallow region of Fig. 7.1 is drawn empty. State how the spacing of the wavefronts there compares with the spacing of those drawn in the deep region, and state what happens to the frequency of the waves as they cross the boundary.

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   Check answer · 2 marks

   1. the wavefronts in the shallow region are closer together, because the waves travel more slowly there
   2. the frequency is unchanged

2. bDetermine The dipper makes twelve complete waves in 1.5 s, producing the twelve wavefronts drawn in Fig. 7.1. Determine the frequency of the waves, and hence their wavelength in the shallow region, where they travel at 9.6 cm/s.

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   Check answer · 2 marks

   1. f = 12/1.5 = 8.0 Hz
   2. λ = v/f = 9.6/8.0 = 1.2 cm

3. cDetermine The twelve wavefronts drawn in the deep region of Fig. 7.1 are evenly spaced, one wavelength of 1.8 cm apart. Determine the distance from the first of them to the twelfth, and determine the distance those same twelve wavefronts occupy once all of them have crossed into the shallow water.

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   Check answer · 3 marks

   1. there are 11 wavelengths between the first wavefront and the twelfth, not 12
   2. distance in the deep region = 11 × 1.8 = 19.8 cm
   3. the spacing in the shallow region is 1.2 cm, so the distance = 11 × 1.2 = 13.2 cm

4. dExplain The dipper is now vibrated twice as fast and nothing else is altered. Explain what happens to the wavelength in each region of Fig. 7.1, and explain why the wavefronts in the shallow water still make the same angle with the boundary as before.

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   Check answer · 4 marks

   1. the speed in each region is fixed by the depth of the water there, so neither speed changes
   2. λ = v/f, so doubling the frequency halves each wavelength: 0.90 cm in the deep region and 0.60 cm in the shallow region
   3. the ratio of the two wavelengths is 0.60/0.90, the same value as 1.2/1.8 before
   4. the change of direction at the boundary is set by that ratio and not by the frequency, so the angle the wavefronts make with the boundary is unchanged

Reading cue. The dashed normal is perpendicular to the fibre wall, so light running nearly along the fibre strikes that wall at a large angle close to 90°, not at a small one.

1. aState State how the two angles marked at the reflection in Fig. 8.1 compare with each other, and state the line from which each is measured.

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   Check answer · 2 marks

   1. the two angles are equal, since the angle of incidence equals the angle of reflection
   2. both are measured from the normal drawn perpendicular to the wall of the core

2. bDetermine The core drawn in Fig. 8.1 is replaced by one of refractive index 1.60. Determine the critical angle for the new core, and state whether total internal reflection will now occur for a wider or a narrower range of angles.

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   Check answer · 3 marks

   1. sin C = 1/1.60 = 0.625
   2. C = 39° (38.7°)
   3. the critical angle is smaller than for the 1.50 core, so a wider range of angles exceeds it and gives total internal reflection

3. cDetermine Light travels at 3.0 × 10⁸ m/s in air. Determine the speed of light inside the core labelled in Fig. 8.1, and the time taken for light to travel 1.0 km along a straight fibre if it runs along the axis without reflecting.

   demanding3 marks

   Check answer · 3 marks

   1. v = c/n = 3.0 × 10⁸ / 1.50 = 2.0 × 10⁸ m/s
   2. t = d/v = 1000 / (2.0 × 10⁸)
   3. t = 5.0 × 10⁻⁶ s (5.0 μs)

4. dExplain Fig. 8.1 shows the light following a zig-zag path rather than running straight along the middle of the core. Explain why a short pulse of light sent into a long fibre arrives at the far end spread out over a longer time than it was sent, and state what this places a limit on.

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   Check answer · 4 marks

   1. light following the zig-zag path travels a greater distance along the fibre than light running along the axis
   2. all of the light travels at the same speed, 2.0 × 10⁸ m/s, in the glass, so the zig-zag light arrives later than the axial light
   3. the pulse therefore arrives stretched out in time
   4. this limits how closely pulses may be sent one after another, and so how much information the fibre can carry each second

Reading cue. Both i and r are measured from the dashed normal, never from the block face, and r is the angle inside the glass, so r is always the smaller of the pair.

1. aState State how the direction of the emergent ray in Fig. 10.1 compares with that of the incident ray, and give a reason.

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   Check answer · 2 marks

   1. the emergent ray is parallel to the incident ray, though shifted sideways from it
   2. the top and bottom faces are parallel, so the ray bends away from the normal on leaving by the same angle as it bent towards the normal on entering

2. bCalculate The block in Fig. 10.1 has a refractive index of 1.50 and the angle i is set to 40°. Calculate the angle r.

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   Check answer · 2 marks

   1. sin r = sin i / n = sin 40° / 1.50 = 0.429
   2. r = 25° (25.4°)

3. cDetermine On the drawing the angle i measures 40° and the angle r measures 25°, each read to the nearest degree. Determine the largest and the smallest values of the refractive index consistent with these two readings.

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   Show a hint

   Pair the extremes the right way round: a ratio is largest when its top is largest at the same time as its bottom is smallest.

   Check answer · 3 marks

   1. largest value uses the largest i with the smallest r: sin 40.5° / sin 24.5° = 1.57
   2. smallest value uses the smallest i with the largest r: sin 39.5° / sin 25.5° = 1.48
   3. the refractive index therefore lies between 1.48 and 1.57

4. dSuggest Suggest which of the two angles marked in Fig. 10.1 does more to limit the precision of the refractive index obtained, and justify your answer.

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   Check answer · 3 marks

   1. the angle of refraction r limits it more
   2. r is the smaller angle, so the same reading error of half a degree is a larger fraction of r, and changes sin r by proportionally more than the same error changes sin i
   3. r is also not measured directly: it is drawn between the normal and a line joining the entry and exit points, so errors in the outline drawn round the block and in the marked emergent ray feed into it as well

Reading cue. Both resistors lie between the same two junction dots, so they have the same p.d. across them; it is the current, not the voltage, that divides at the left-hand dot.

1. aState State which of the two resistors in Fig. 3.1 carries the larger current, and give a reason.

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   Check answer · 2 marks

   1. the 3.0 Ω resistor
   2. both branches lie between the same two junctions and so have the same p.d. across them, and I = V/R, so the smaller resistance carries the larger current

2. bDetermine The potential difference between the two junction dots in Fig. 3.1 is 12 V. Determine the current in each resistor and the current in the single wire arriving from the left.

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   Check answer · 3 marks

   1. current in the 6.0 Ω resistor = 12/6.0 = 2.0 A
   2. current in the 3.0 Ω resistor = 12/3.0 = 4.0 A
   3. current in the wire from the left = 2.0 + 4.0 = 6.0 A

3. cDetermine Each resistor in Fig. 3.1 is rated at a maximum power of 12 W. Determine the largest potential difference that may safely be applied between the two junction dots, and the current drawn at that potential difference.

   demanding4 marks

   Check answer · 4 marks

   1. for the 3.0 Ω resistor, V = √(PR) = √(12 × 3.0) = 6.0 V
   2. for the 6.0 Ω resistor, V = √(12 × 6.0) = 8.5 V, so the 3.0 Ω resistor is the one that limits the pair
   3. largest safe p.d. = 6.0 V
   4. current drawn = 6.0/6.0 + 6.0/3.0 = 3.0 A

4. dExplain A student claims that connecting a third resistor between the same two junction dots in Fig. 3.1 must raise the total resistance, because there is then more resistance present. Explain why the student is wrong, and state the largest total resistance the arrangement could ever have once a third resistor is added there.

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   Check answer · 4 marks

   1. a resistor connected between the same two dots is in parallel with the others and gives the current an extra path
   2. for the same p.d. the total current is therefore larger, and R = V/I, so the total resistance falls
   3. the pair alone gives 1/R = 1/6.0 + 1/3.0, so R = 2.0 Ω
   4. adding a third branch can only bring the total below 2.0 Ω, so 2.0 Ω is a value the arrangement approaches but never reaches

Reading cue. The arrows leave the north pole and enter the south, so the field runs left to right; the force is perpendicular to both field and wire, so it cannot lie in the plane of the page.

1. aDescribe Describe what happens to the force on the wire in Fig. 5.1 if the current I is reversed so that it flows down the page instead of up.

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   Check answer · 2 marks

   1. the force acts in the opposite direction
   2. its size is unchanged, because neither the current nor the strength of the field has been altered

2. bDescribe The two magnets in Fig. 5.1 are now exchanged, so that a south pole faces the gap from the left and a north pole from the right, and at the same time the current is reversed. Describe the effect of these two changes together on the force on the wire.

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   Check answer · 3 marks

   1. reversing the field alone would reverse the force, and reversing the current alone would reverse it as well
   2. with both reversed the two changes cancel each other
   3. the force therefore acts in the same direction as before and has the same size

3. cExplain The wire in Fig. 5.1 is turned slowly in the plane of the page, away from the vertical position drawn, until it finally lies horizontally along the field arrows. Explain how the size of the force on the wire changes as it is turned, and state the position in which the force is largest.

   demanding4 marks

   Check answer · 4 marks

   1. the force is largest in the position drawn in Fig. 5.1, with the wire at right angles to the field arrows
   2. as the wire is turned away from that position the force becomes steadily smaller
   3. when the wire lies along the field arrows the force is zero, because no part of the current then crosses the field
   4. throughout the turning the force stays perpendicular to the page, so only its size changes

4. dSuggest Suggest three separate changes to the arrangement in Fig. 5.1, each of which would increase the size of the force on the wire, and explain why only the length of wire lying between the pole faces affects that force.

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   Check answer · 4 marks

   1. increase the current in the wire
   2. use stronger magnets, or bring the pole faces closer together, so that the field across the gap is stronger
   3. increase the length of wire lying in the field, for example by using wider pole faces or by replacing the single wire with several wires side by side carrying the current the same way
   4. outside the gap the field is very weak, so the parts of the wire beyond the pole faces experience almost no force and do not contribute

Reading cue. Take the turns from the labels, not from the loops drawn: five on the left and seven on the right, but the ratio that decides the output is 200 to 5000.

1. aIdentify Identify which coil in Fig. 6.1 has the greater number of turns, and state whether the transformer is a step-up or a step-down transformer.

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   Check answer · 2 marks

   1. the coil wound on the right-hand limb, the secondary, with 5000 turns
   2. it is a step-up transformer, because the secondary has more turns than the primary

2. bExplain The 12 V a.c. supply in Fig. 6.1 is replaced by a 12 V d.c. supply. Explain what a voltmeter connected across the output terminals would read.

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   Check answer · 3 marks

   1. the voltmeter reads zero, apart from a momentary reading as the supply is switched on or off
   2. a steady direct current produces a steady magnetic field in the iron core
   3. there is then no change of magnetic field through the secondary coil, so no e.m.f. is induced in it

3. cDetermine The output terminals in Fig. 6.1 are connected to a lamp, and the current in the primary coil is 0.50 A. Assuming the transformer is 100% efficient, determine the current in the secondary coil and the power delivered to the lamp.

   demanding3 marks

   Check answer · 3 marks

   1. input power = 12 × 0.50 = 6.0 W, and at 100% efficiency the lamp receives 6.0 W
   2. secondary current = primary current × 200/5000 = 0.50 × 0.040
   3. current in the secondary = 0.020 A (20 mA)

4. dExplain The core in Fig. 6.1 is drawn as a solid rectangle of iron. In a real transformer it is built from thin sheets separated by insulation. Explain how this changes what happens in the core, and describe one further reason why a real transformer is not 100% efficient.

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   Check answer · 4 marks

   1. the changing magnetic field induces currents in the iron of the core itself
   2. in a solid core these currents circulate freely and heat the core, so energy from the supply is wasted
   3. insulated sheets break up the paths available to these currents, so they are much smaller and less energy is wasted
   4. one further loss: the copper coils have resistance, so the current in them heats the windings (accept: not all the field from the primary passes through the secondary, or energy is wasted repeatedly magnetising the core)

Reading cue. The 40 Ω sits before the first junction dot, so it carries the whole battery current; only past that dot is the current shared between the 30 Ω and the 60 Ω.

1. aDetermine The current drawn from the battery in Fig. 7.1 is 0.20 A. Determine the potential difference across the 40 Ω resistor.

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   Check answer · 2 marks

   1. the 40 Ω resistor lies before the junction, so it carries the whole 0.20 A
   2. V = IR = 0.20 × 40 = 8.0 V

2. bDetermine Determine the power dissipated in the 40 Ω resistor in Fig. 7.1, and the total power supplied by the battery.

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   Check answer · 2 marks

   1. power in the 40 Ω resistor = I²R = 0.20² × 40 = 1.6 W
   2. total power from the battery = VI = 12 × 0.20 = 2.4 W

3. cDetermine The 40 Ω resistor in Fig. 7.1 is replaced by a variable resistor. Determine the resistance it must be set to for the potential difference across the parallel pair to be 6.0 V.

   demanding3 marks

   Check answer · 3 marks

   1. the parallel pair has a combined resistance of (30 × 60)/90 = 20 Ω, so the current is 6.0/20 = 0.30 A
   2. p.d. across the variable resistor = 12 − 6.0 = 6.0 V
   3. R = V/I = 6.0/0.30 = 20 Ω

4. dExplain The 30 Ω resistor in Fig. 7.1 is replaced by a filament lamp whose resistance rises as it warms up. Explain what happens to the potential difference across the 40 Ω resistor as the lamp warms, and state whether the current in the 60 Ω resistor rises or falls.

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   Check answer · 4 marks

   1. the resistance of the lamp rises, so the combined resistance of the parallel pair rises
   2. the total resistance of the circuit rises, so the current drawn from the battery falls
   3. the p.d. across the 40 Ω resistor is that current × 40, so it falls
   4. the parallel pair therefore takes a larger share of the 12 V, so the current in the 60 Ω resistor, equal to that p.d. divided by 60, rises

Reading cue. Read the pole from the end of the magnet nearest the coil, not the labelled far end, and note the arrow sits on the magnet: the magnet moves and the coil stays still.

1. aIdentify Identify the feature of the meter drawn in Fig. 8.1 that lets the direction of the induced current be found, and state what its needle reads before the magnet is moved.

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   Check answer · 3 marks

   1. it is a centre-zero ammeter / zero is at the middle of the scale (1)
   2. the needle can swing either side of zero, so the direction of the current is shown (1)
   3. reads zero before the magnet moves (1)

2. bState For the motion drawn by the arrow in Fig. 8.1, state which magnetic pole is produced at the end of the coil nearest the magnet, and state the effect this has on the magnet as it moves in.

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   Check answer · 2 marks

   1. the near end of the coil becomes a north pole (1)
   2. it repels the approaching north pole of the magnet / opposes the magnet's motion (1)

3. cExplain The magnet in Fig. 8.1 is now held still and the coil is moved to the left towards it, at the same speed. Explain what the ammeter shows.

   demanding4 marks

   Check answer · 4 marks

   1. the needle deflects in the same direction as before (1)
   2. by the same amount (1)
   3. only the relative movement of magnet and coil matters (1)
   4. the field through the coil changes at the same rate, so the same e.m.f. is induced (1)

4. dExplain The two wires running down from the coil in Fig. 8.1 are disconnected from the ammeter and the magnet is pushed in again at the same speed. Explain why less force is now needed to push the magnet in, and state the source of the energy that was previously measured as a current.

   top of the paper4 marks

   Check answer · 4 marks

   1. with the circuit complete an induced current flows in the coil (1)
   2. this current makes the coil into a magnet whose near pole repels the incoming north pole, so a force must be overcome (1)
   3. with the wires disconnected the circuit is broken, so no current flows and there is no opposing force (an e.m.f. is still induced) (1)
   4. the electrical energy came from the work done by the person pushing the magnet / from the magnet's kinetic energy (1)

Reading cue. Trace all three wires out of the supply box before answering: note which one the fuse sits in, and note that one wire ends on the metal case instead of joining the heating element.

1. aIdentify Identify the wire in Fig. 9.1 that the fuse has been placed in, and identify the wire that carries no current while the shower is working normally.

   recall2 marks

   Check answer · 2 marks

   1. the fuse is in the live wire (1)
   2. the earth wire carries no current in normal operation (1)

2. bShow (that) Show that the resistance of the heating element drawn between the live and neutral wires is about 6 Ω when the shower runs at its marked power on the marked supply.

   routine3 marks

   Check answer · 3 marks

   1. current I = P / V = 8500 / 230 (1)
   2. I = 37 A (1)
   3. R = V / I = 230 / 37 = 6.2 Ω, which is about 6 Ω (1)

3. cDetermine Fuses rated at 13 A, 30 A and 45 A are available. Determine which one should be fitted at the position drawn in Fig. 9.1, and give a reason for rejecting each of the other two.

   demanding3 marks

   Check answer · 3 marks

   1. the normal working current is about 37 A (1)
   2. a 13 A and a 30 A fuse would both melt during normal use, cutting off the shower (1)
   3. the 45 A fuse is chosen: it is the lowest rating above the working current, so it still melts on a fault (1)

4. dSuggest The earth wire in Fig. 9.1 has broken away at the solid dot on the metal case, but the shower still heats the water normally. Suggest why this fault is dangerous, and suggest how it could be found before anyone is hurt.

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   Check answer · 4 marks

   1. the live–neutral circuit through the element is untouched, so the shower still works and there is no sign of the fault (1)
   2. if the live wire later touches the case, the case becomes live and there is now no low-resistance path to earth (1)
   3. the current is then too small to melt the fuse, and only flows when a person touches the case, through their body (1)
   4. test the continuity between the metal case and the earth of the supply / have the appliance checked regularly by a qualified electrician (1)

Reading cue. The paper and the 5 mm aluminium are alternative absorbers put in the same place, one at a time and never stacked, so each reading tests one absorber only.

1. aState State the quantity that Fig. 2.1 deliberately keeps unchanged when the absorber is put in place, and state why it must be kept unchanged.

   routine2 marks

   Check answer · 2 marks

   1. the distance between the source and the window of the tube (1)
   2. so that any fall in count rate is caused by the absorber alone, and not by the radiation spreading out over a greater distance (1)

2. bDescribe Describe how the counter in Fig. 2.1 should be used so that the reading with the absorber can be fairly compared with the reading without it.

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   Check answer · 3 marks

   1. first record the count with the source removed, to obtain the background (1)
   2. take every count over the same measured counting time, or convert each count to a count rate (1)
   3. subtract the background from each reading before the two are compared (1)

3. cExplain With no absorber the counter records 620 counts in 100 s. With the sheet of paper in place, at the same distance, it records 611 counts in 100 s. Explain whether this shows that the paper absorbs some of the radiation.

   demanding4 marks

   Check answer · 4 marks

   1. count rates are 620 / 100 = 6.2 counts/s and 611 / 100 = 6.1 counts/s (1)
   2. radioactive decay is random, so repeated counts vary even when nothing has changed (1)
   3. the difference of about 0.1 counts/s is no larger than this random variation, so it is not evidence of absorption (1)
   4. count for much longer, or repeat the readings, to decide (1)

4. dSuggest The paper is replaced by the 5 mm sheet of aluminium, at the same distance, and the counter records 33 counts in 100 s; with the source taken right away it records 30 counts in 100 s. From these readings a student writes that the aluminium stops all the radiation from the source. Suggest why that conclusion is not safe, and suggest one change to the arrangement in Fig. 2.1 that would test it.

   top of the paper5 marks

   Check answer · 5 marks

   1. corrected count rate with the aluminium = (33 − 30) / 100 = 0.03 counts/s, which is small but not zero (1)
   2. a difference of 3 counts is smaller than the random variation in counts of this size, so the readings cannot show whether anything is getting through (1)
   3. a weak, more penetrating emission such as gamma could be present as well and would still be passing through the aluminium (1)
   4. put lead absorbers of increasing thickness at the same position and look for any further fall in the corrected count rate (1)
   5. count for far longer at each thickness, so that the random variation is small compared with the fall being looked for (1)

Reading cue. The plates, the air gap, the cell and the alarm form one series loop, so whatever happens in the gap fixes the current everywhere; the arrow shows smoke entering, not the source moving.

1. aState State the part of the series circuit drawn in Fig. 6.1 that the radiation from the source must act on before any current can flow round the loop, and state why.

   recall2 marks

   Check answer · 2 marks

   1. the air gap between the two plates (1)
   2. air is an insulator unless it is ionised, and the radiation from the source produces the ions that carry the charge across the gap (1)

2. bExplain The source in Fig. 6.1 emits α-particles. Explain why the current in the circuit falls when smoke drifts sideways into the gap, as shown by the arrow.

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   Check answer · 3 marks

   1. smoke particles absorb the α-particles before they have crossed the whole gap (1)
   2. fewer ions are produced in the air, and ions that do form attach to the smoke particles and move more slowly (1)
   3. so less charge passes between the plates each second, the current falls and the alarm circuit is triggered (1)

3. cExplain Explain why the α-source in Fig. 6.1 is fixed to the underside of the upper plate rather than mounted outside the chamber, and why the gap between the plates is only a few millimetres wide.

   demanding4 marks

   Check answer · 4 marks

   1. alpha particles travel only a few centimetres in air (1)
   2. mounted outside, they would be absorbed by the wall of the chamber and no ions would be made in the gap (1)
   3. a narrow gap means ions are produced right across it, all the way to the lower plate (1)
   4. so a steady, measurable current can be maintained between the plates (1)

4. dSuggest A manufacturer suggests replacing the α-source in Fig. 6.1 with a beta source of the same activity. Suggest the effect on the current between the plates and on how well the alarm works.

   top of the paper4 marks

   Check answer · 4 marks

   1. beta particles are much less strongly ionising than alpha particles (1)
   2. far fewer ions are produced in the gap each second, so the current is much smaller (1)
   3. beta is barely absorbed by smoke, so the current changes very little when smoke enters (1)
   4. the alarm becomes unreliable or fails to trigger, and the more penetrating beta also escapes through the casing (1)

Reading cue. The three lines are the paths of three different α-particles, not one particle bouncing about; each angle is measured from the original beam direction set by the channel in the lead.

1. aState State the purpose of the narrow channel cut through the lead block in Fig. 7.1.

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   Check answer · 2 marks

   1. the lead absorbs alpha particles emitted in all other directions (1)
   2. so a narrow beam of known direction reaches the foil, against which deflection angles can be measured (1)

2. bDescribe Describe the force that acted on the α-particle following the middle path in Fig. 7.1, the one deflected upwards through a moderate angle, and describe where in the gold atom that force acted.

   routine3 marks

   Check answer · 3 marks

   1. an electrostatic force of repulsion (1)
   2. between the positively charged α-particle and the positively charged nucleus (1)
   3. acting as the particle passed close to a nucleus, but not directly at it (1)

3. cExplain Explain how Fig. 7.1 would have to be redrawn if the positive charge of each gold atom were spread evenly throughout the whole atom, as in the earlier model of the atom.

   demanding4 marks

   Check answer · 4 marks

   1. only the straight-through path, with at most very small deflections, would be drawn (1)
   2. no path turned back through more than 90° would appear (1)
   3. charge spread through the whole atom produces a much weaker repulsion at any point in it (1)
   4. this force is far too small to reverse the motion of a fast, massive α-particle (1)

4. dSuggest The gold foil in Fig. 7.1 is replaced by an aluminium foil of the same thickness. An aluminium nucleus holds 13 protons and a gold nucleus holds 79. Suggest how the three paths drawn would change.

   top of the paper4 marks

   Check answer · 4 marks

   1. most α-particles would still pass straight through, so that path is unchanged (1)
   2. the aluminium nucleus carries a much smaller positive charge than the gold nucleus (1)
   3. so the repulsive force at the same distance of approach is much smaller (1)
   4. fewer particles are deflected through large angles, and far fewer are turned back through more than 90° (1)

Reading cue. The window-to-source gap is fixed for every reading, and the counter shows a total number of counts, not a rate, so divide by the counting time before comparing any two readings.

1. aState State why the tube in Fig. 10.1 is clamped with its window pointing straight down at the source, and state one effect on the results if the retort stand were nudged part-way through the experiment.

   recall2 marks

   Check answer · 2 marks

   1. the same fraction of the radiation emitted by the source enters the window for every reading (1)
   2. nudging the stand changes the distance or direction, so the count rate would change for a reason other than the decay of the source (1)

2. bDetermine With the source taken away but the stand and tube left exactly as drawn, the counter records 300 counts in 12.0 minutes. The source is then placed at the fixed distance and the counter records 1085 counts in 5.0 minutes. Determine the count rate due to the source alone.

   routine3 marks

   Check answer · 3 marks

   1. background rate = 300 / 12.0 = 25 counts/min (1)
   2. measured rate with source = 1085 / 5.0 = 217 counts/min (1)
   3. corrected rate = 217 − 25 = 192 counts/min (1)

3. cDetermine The corrected count rate due to the source is 192 counts/min at the start, and the half-life of the source is 30 minutes. Determine the total number of counts the counter in Fig. 10.1 will record in one minute, 90 minutes after the start.

   demanding4 marks

   Check answer · 4 marks

   1. 90 minutes is 3 half-lives (1)
   2. 192 → 96 → 48 → 24, so the source alone gives 24 counts/min (1)
   3. the background must be added back, because the tube detects it as well as the source (1)
   4. total recorded = 24 + 25 = 49 counts in that minute (1)

4. dSuggest The student repeats the whole experiment with the tube clamped twice as far above the source, everything else unchanged. Suggest how her corrected count rates and her value of the half-life would each be affected.

   top of the paper4 marks

   Check answer · 4 marks

   1. every corrected count rate would be much smaller (1)
   2. the radiation spreads out in all directions, so a smaller fraction of it enters the window (1)
   3. the background and the random variation then form a larger proportion of each reading, so the results are less reliable (1)
   4. the half-life obtained would be the same, because it is a property of the source and does not depend on how much of the radiation is detected (1)

Reading cue. The 1.1 × 10¹¹ m is measured from the centre of the star, so it is the orbit radius, not the distance travelled; 1.9 × 10⁷ s is the time for one full lap of the dashed circle.

1. aCalculate Calculate the distance the planet travels in going once round the dashed circle in Fig. 3.1.

   recall2 marks

   Check answer · 2 marks

   1. distance = 2πr with r = 1.1 × 10¹¹ m (1)
   2. distance = 6.9 × 10¹¹ m (1)

2. bDetermine Light travels at 3.0 × 10⁸ m/s. Determine, in minutes, the time light takes to travel from the star to the planet at the position drawn in Fig. 3.1.

   routine3 marks

   Check answer · 3 marks

   1. t = d / v = 1.1 × 10¹¹ / 3.0 × 10⁸ (1)
   2. t = 367 s (1)
   3. t = 6.1 minutes (1)

3. cExplain The arrow at the planet in Fig. 3.1 shows the direction in which it is travelling. Explain how the direction of the star's gravitational pull on the planet is related to that arrow, and explain why this pull changes the planet's direction without changing its speed.

   demanding4 marks

   Check answer · 4 marks

   1. the gravitational force acts from the planet towards the centre of the star (1)
   2. it is therefore at right angles to the arrow showing the direction of travel (1)
   3. so the force has no part of it along the direction of motion, and does no work on the planet (1)
   4. it changes only the direction of the velocity, so the planet keeps a constant speed while continually turning (1)

4. dExplain The orbit is drawn as a dashed circle, but the planet's path is described only as nearly circular. Explain how the planet's distance and speed would vary round a slightly squashed orbit, and explain what this means for any single speed calculated from Fig. 3.1.

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   Check answer · 4 marks

   1. the distance from the star would vary round the orbit rather than staying at 1.1 × 10¹¹ m (1)
   2. the planet moves fastest at the point of the orbit closest to the star (1)
   3. and slowest at the point furthest from the star (1)
   4. a value found from circumference ÷ period is therefore only an average speed for the whole orbit (1)

Reading cue. Both strips share one wavelength scale increasing to the right, and the labelled gap is the distance between two positions of one line — it is not itself a wavelength read off the scale.

1. aIdentify Identify which of the two strips in Fig. 6.1 was recorded in the laboratory, and state what must be true of the two lines being compared for the labelled gap to have any meaning.

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   Check answer · 2 marks

   1. the upper strip is the laboratory spectrum (1)
   2. both lines must come from the same element / be the same spectral line (1)

2. bCalculate The marked line lies at 486 nm in the laboratory strip and the same line lies at 500 nm in the galaxy strip. Calculate the increase in wavelength labelled on Fig. 6.1, and calculate it as a percentage of the laboratory wavelength.

   routine3 marks

   Check answer · 3 marks

   1. increase = 500 − 486 = 14 nm (1)
   2. percentage = 14 / 486 × 100 (1)
   3. = 2.9 % (1)

3. cExplain Light from a second galaxy shows the same line shifted twice as far to the right along the scale as the shift drawn in Fig. 6.1. Explain what this tells you about the second galaxy, in terms of both its speed and its distance.

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   Check answer · 4 marks

   1. a larger shift towards longer wavelengths means a greater speed of recession (1)
   2. the second galaxy is moving away about twice as fast (1)
   3. speed of recession is proportional to distance (1)
   4. so the second galaxy is roughly twice as far away as the one drawn (1)

4. dSuggest A star within our own Galaxy gives the same line at almost exactly the laboratory position, with a shift far too small to draw on Fig. 6.1. Suggest what this shows about the star, and suggest why such a measurement is of little use for finding a value of the Hubble constant.

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   Check answer · 4 marks

   1. the star is moving away from us very slowly, or hardly at all (1)
   2. it is extremely close compared with distant galaxies, and is held within our own Galaxy rather than being carried apart by the expansion (1)
   3. a shift that small cannot be measured accurately, so the speed found from it is very uncertain (1)
   4. dividing a very uncertain speed by a very small distance would give a value of the constant with an enormous uncertainty (1)

Reading cue. This is a plan view and not to scale: check what each dashed circle is drawn around before you name it, and note that both arrows go the same way round.

1. aIdentify Identify what each of the two dashed circles in Fig. 7.1 represents, and state the body at the centre of each one.

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   Check answer · 2 marks

   1. the large circle is the path of the Earth, centred on the Sun (1)
   2. the small circle is the path of the Moon, centred on the Earth (1)

2. bExplain The Sun is the only source of light drawn in Fig. 7.1. Explain why an observer on Earth sees the whole of the Moon's lit face when the Moon reaches the point of the small circle furthest from the Sun, and identify where on that circle the Moon is almost invisible.

   routine3 marks

   Check answer · 3 marks

   1. the Moon is seen only by the sunlight it reflects (1)
   2. at the far point the lit half of the Moon faces the Earth, so the whole lit disc is seen (1)
   3. on the side of the small circle nearest the Sun, the lit half faces away from the Earth, so almost none of it can be seen (1)

3. cDetermine The Moon takes 27 days to travel once round the small dashed circle and the Earth spins once on its axis in 24 hours, both in the directions shown by the arrows. Determine the angle the Moon moves round its orbit in one day, and hence determine the extra time the Earth must spin each day before the observer faces the Moon again.

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   Show a hint

   In one day the Earth must turn through 360° plus the extra angle the Moon has moved on round its own circle.

   Check answer · 4 marks

   1. angle moved in one day = 360 / 27 = 13.3° (1)
   2. after one complete spin the Earth must turn a further 13.3° to point at the Moon again (1)
   3. extra time = 13.3 / 360 × 24 hours (1)
   4. = 0.89 hours, about 53 minutes (1)

4. dSuggest Fig. 7.1 is drawn not to scale. Suggest two ways in which a true scale drawing would look different, and suggest why the diagram is drawn as it is.

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   Check answer · 3 marks

   1. the Moon's orbit would be very much smaller compared with the Earth's orbit (1)
   2. the Sun would be drawn very much larger than the Earth, and both would be far too small to see beside orbits of this size (1)
   3. the diagram is drawn out of scale so that the directions of travel and the relative positions of the three bodies can all be shown on one page (1)

Reading cue. Each axis is labelled 'quantity divided by a power of ten', so the numbers along it are not the measured values — convert every reading into the axis units before plotting or reading off.

1. aState A galaxy is 8.0 × 10²⁴ m away and is receding at 1.8 × 10⁷ m/s. State the number to be plotted on each axis of the grid in Fig. 10.1 for this galaxy.

   recall2 marks

   Check answer · 2 marks

   1. plotted at 8.0 on the distance axis (1)
   2. 1.8 × 10⁷ m/s = 18 × 10⁶ m/s, so plotted at 18 on the speed axis (1)

2. bExplain Explain why the straight line of best fit drawn on the grid in Fig. 10.1 must be taken through the origin, and state what it would mean if one plotted point lay well above that line.

   routine3 marks

   Check answer · 3 marks

   1. a galaxy at zero distance from us would have zero speed of recession (1)
   2. so the line must pass through the point (0, 0) (1)
   3. a point well above the line is a galaxy receding faster than the trend for its distance, from a measurement error or from its own motion within a cluster (1)

3. cDetermine A straight line of best fit through the origin is drawn on Fig. 10.1 and reaches the right-hand edge of the grid at the point d = 10, v = 22. Determine the speed of recession of a galaxy 3.5 × 10²⁴ m away, in m/s.

   demanding3 marks

   Check answer · 3 marks

   1. gradient in plotted units = 22 / 10 = 2.2 (1)
   2. reading at d = 3.5 gives v = 3.5 × 2.2 = 7.7 in plotted units (1)
   3. v = 7.7 × 10⁶ m/s (1)

4. dExplain A fifth galaxy is measured at a distance of 2.4 × 10²⁵ m, receding at 5.0 × 10⁷ m/s. Explain why it cannot be plotted on the grid in Fig. 10.1, state whether it follows the same relationship as the line of best fit in the previous part, and describe the change needed to each axis so that all five galaxies can be shown.

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   Check answer · 5 marks

   1. it would be plotted at 24 on the distance axis, which stops at 10 (1)
   2. it would be plotted at 50 on the speed axis, which stops at 25 (1)
   3. v / d = 50 / 24 = 2.1 in the plotted units, close to the gradient of 2.2, so it does follow the same relationship (1)
   4. extend the distance axis to at least 25, for example a numbered gridline every 2.5 (1)
   5. extend the speed axis to at least 50, for example a numbered gridline every 10, keeping both scales linear so the points still spread across the grid (1)