Pearson Edexcel GCSE Physics Paper 1, Higher Tier (1PH0/1H)Paper 1 — every question, answered

Ray diagram B shows total internal reflection at the air and glass boundary. This is because total internal reflection only happens when light travels from a denser material into a less dense material, so the ray must be going from glass into air, and it must reflect back into the glass at the same angle it hit the boundary at, rather than crossing over and bending into the air.

Reading the heights straight off Figure 1, the image height is about 18mm and the object height is about 7mm, so magnification = 18 / 7 = 2.6.

One way to increase the magnification would be to use a different lens with a shorter focal length. A lens with more curved surfaces has a shorter focal length and a higher power, so it bends the light rays more strongly, which produces a bigger virtual image for the same object position.

For the focal length, I substitute a = 20 and b = 40 into 1/f = (a + b) / (a x b), which gives 1/f = 60 / 800 = 0.075. Flipping this gives f = 1 / 0.075 = 13.3cm, which rounds to 13cm.

Conservation of energy in a closed system means that when there are energy transfers, the total amount of energy stays the same, it does not increase or decrease. To compare sand and sawdust as insulators fairly, I would set up two identical small copper cans of hot water, each placed inside a larger copper can, packing sand around one small can and sawdust around the other, with a thermometer in each water sample and a stop clock timing how the temperature falls. To make it a fair test I must control the mass of water used, the mass or volume of insulating material packed around each can, the starting temperature of the water, and the time interval or temperature change I compare between the two set ups.

Reading Figure 4, as the density of expanded polystyrene increases, its thermal conductivity decreases, but not at a steady rate: the graph falls steeply at low densities and then the rate of decrease slows down, until the curve levels off and thermal conductivity becomes almost constant at higher densities.

Driving slower in wet weather is safer because wet roads give less grip between the tyres and the road, so braking distance increases in the wet compared to dry conditions, meaning a slower speed keeps the overall stopping distance shorter and reduces the risk of skidding or losing control.

To compare 31 m/s with 130 km/h, I convert 130 km/h into metres per second: 130 000 m divided by 3600 s gives about 36.1 m/s. Since 31 m/s is less than 36.1 m/s, this confirms 31 m/s is indeed a lower speed than 130 km/h. For the reaction time, distance travelled = speed multiplied by time, so 46 = 31 multiplied by t, giving t = 46/31 = 1.483 seconds, which rounds to 1.5 seconds to two significant figures.

Gamma radiation is the type that travels from the kidney to the scanner, since gamma can pass out through body tissue, unlike alpha or beta radiation which are absorbed inside the body before they could reach an external detector. A half-life of about 6 minutes is not suitable because the isotope would decay away too quickly to still be giving off enough radiation for readings to be taken throughout the whole 30 minute scan. A half-life of about 6 days is not suitable because the isotope would stay radioactive inside the patient's body for far longer than necessary, giving an unnecessarily high radiation dose long after the scan itself has finished.

Starting at 128 Bq at time zero, one half-life of 17s later the activity halves to 64 Bq, so I plot a point at 17 seconds, 64 Bq. After another 17s, at 34s, the activity halves again to 32 Bq, giving a point at 34 seconds, 32 Bq. After a third half-life, at 51s, the activity halves once more to 16 Bq, giving a point at 51 seconds, 16 Bq, and I would join these points with a smoothly decaying curve rather than straight lines.

Two ways of reducing the technician's radiation risk are using shielding, such as standing behind a lead barrier or lead screen, and limiting the time spent close to the radioactive source, for example by leaving the room during the scan or using tongs to keep the source at a distance rather than holding it directly.

Pushing the control rods further into the reactor slows the chain reaction because the rods absorb neutrons, so fewer neutrons are left available to go on and cause further fission reactions in the uranium fuel. For the percentage calculation, I divide the slow neutron speed by the fast neutron speed and multiply by 100: (4.0 x 10^3 divided by 3.0 x 10^7) multiplied by 100 = 0.013%, showing the slow neutrons move only a tiny fraction as fast as the fast ones. Energy is transferred from the nuclear reaction as kinetic energy of the fission fragments, which is then transferred as thermal energy to the coolant flowing through the reactor, ready to be carried on to boil water and drive a turbine.

Rearranging energy released = change in mass multiplied by the speed of light squared gives change in mass = energy released divided by the speed of light squared, so change in mass = 1.49 x 10^-10 divided by (3.00 x 10^8)^2 = 1.49 x 10^-10 divided by 9 x 10^16, which evaluates to about 1.66 x 10^-27 kg, an extremely small mass converted straight into the energy released.

A single uranium fission only releases a tiny amount of energy, but each fission also releases two or three free neutrons, and it is these neutrons that let the reactor produce large amounts of energy overall. If a released neutron goes on to hit another uranium nucleus, it can cause that nucleus to split as well, releasing more neutrons in turn, and this becomes a self sustaining chain reaction where the number of fissions happening every second keeps building, so instead of one isolated fission the reactor has millions of fissions happening continuously, and it is this huge number of fissions, not the size of any single one, that produces enough energy to supply thousands of homes.

Left completely uncontrolled, this chain reaction would keep accelerating, so the reactor uses control rods to manage exactly how many fissions happen each second. The control rods are made of a material that absorbs neutrons, so when they are pushed further into the reactor core they capture more of the free neutrons released by fission before those neutrons can reach another uranium nucleus. This directly reduces the number of fissions taking place, which slows the rate of energy release, and pulling the control rods back out allows more neutrons through to cause fissions, increasing the rate of energy release again. This is exactly how the reactor is able to track changing demand for electricity, since operators can raise or lower the control rods to increase or decrease power output as needed.

The reactor also contains a moderator, usually water or graphite, which plays a different role from the control rods. Neutrons released directly from fission travel too fast to reliably cause another fission, so the moderator slows these fast neutrons down through collisions, and slower neutrons are far more likely to be captured by a uranium nucleus and cause it to split. Without the moderator the chain reaction would not sustain itself efficiently even with the control rods withdrawn, so the moderator and the control rods work together, one making fission likely, the other limiting how often it is allowed to happen.

Acceleration = change in velocity divided by time, so a = (82 minus 0) divided by 36 = 2.3 m/s^2, which is about 2 m/s^2 as the question asks me to show. Using v squared minus u squared = 2 times a times x, I rearrange to x = (82 squared minus 0 squared) divided by (2 multiplied by 2.3), giving a distance of about 1500 m. The actual runway is always longer than this calculated distance because it needs a safety margin, for example in case the take-off has to be aborted or the aircraft is carrying a heavier load or worse weather conditions on another flight.

Using v squared minus u squared = 2ax with the stone's acceleration equal to gravity, 10 m/s^2, I substitute 17 squared minus 0 squared = 2 multiplied by 10 multiplied by distance, giving distance = 289 divided by 20 = 14.45 m, which rounds to 14.5 m.

The lift is at rest, meaning its velocity is zero, for 3.0 s during the first 16 s. I get this by reading the length of the flat section of the line that sits exactly on the time axis, where the graph shows no rise or fall in velocity at all.

Using the graph in Figure 8, the maximum velocity during the first 16 s is 4.6 m/s. I read this straight off the height of the second, higher flat section of the line, which sits above the first plateau.

For the acceleration during the first 1.4 s, I read the data point where the line reaches 4 m/s at 1.4 s, starting from (0, 0) where the lift begins moving. Acceleration is the gradient of a velocity time graph, so acceleration equals change in velocity divided by time, which is 4 divided by 1.4, giving 2.9 m/s squared.

To find the distance travelled in the first 6.0 s, I worked out the area under the graph up to that point. I split the shape into a rising triangle, a flat rectangle and a falling triangle: half of 1.4 multiplied by 4, plus 3.6 multiplied by 4, plus half of 1 multiplied by 4. That gives 2.8 plus 14.4 plus 2.0, which is 19.2 m, rounding to 19 m.

At 18 s the lift starts to move downwards. I would sketch a straight line starting exactly at the point (18, 0) on the graph and sloping downward from there into negative velocity values, showing the speed building up gradually rather than dropping instantly. I would not draw a vertical line down at 18 s, and I would not extend this new line back to before 18 s, since the lift is still shown at rest right up until that point.

Kinetic energy = half multiplied by mass multiplied by velocity squared, so KE = 0.5 multiplied by 3.6 x 10^5 multiplied by 71 squared, which evaluates to about 9.1 x 10^8 J. As the aircraft comes to a stop, this kinetic energy is transferred to the surroundings mechanically through friction in the brakes and tyres, and thermally as heating due to air resistance.

Kinetic energy = half multiplied by mass multiplied by velocity squared, so KE = 0.5 multiplied by 0.42 multiplied by 12 squared, which evaluates to 30 J.

Both red shift and cosmic microwave background radiation are explained by the Big Bang theory, since the Steady State theory cannot account for the cosmic microwave background at all, so the correct row of the table links Big Bang to both pieces of evidence. Using z = (wavelength from galaxy minus wavelength on Earth) divided by wavelength on Earth, z = (6.72 x 10^-7 minus 6.56 x 10^-7) divided by 6.56 x 10^-7, which evaluates to about 0.024, close to the 0.025 the question asks me to show. Using recession velocity = z multiplied by speed of light, v = 0.024 multiplied by 3.00 x 10^8, giving a recession velocity of about 7.2 x 10^6 m/s.

The more distant galaxy has an even longer measured wavelength, 6.92 x 10^-7 m, giving a bigger red shift value, which means that galaxy is moving away from Earth at an even higher recession velocity. Since more distant galaxies show larger red shifts and higher recession velocities, this shows that the further away a galaxy is, the faster it is moving away from us, which is exactly the pattern you would expect if the whole Universe is expanding in every direction. Some telescopes are placed on satellites in orbit because, above the Earth's atmosphere, there is no interference from dust, clouds or light pollution, and satellites can detect wavelengths such as infrared, ultraviolet and X-rays that are absorbed by the Earth's atmosphere before they would ever reach a ground based telescope.

The wavelength of light detected on Earth from distant galaxies is longer, red shifted, than the wavelength that was actually emitted, and this stretching of the wavelength shows that the galaxies emitting that light are moving away from Earth, which is only possible if the space between galaxies is itself expanding. Using the table of typical radiation wavelengths, cosmic microwave background radiation has a typical wavelength of about 1.0 mm, since it falls within the microwave part of the electromagnetic spectrum.

A nebula collapses because of the gravitational attraction pulling its particles of gas and dust together. As the nebula keeps collapsing, the increasing gravity raises the temperature and pressure at the centre until it becomes hot enough for nuclear fusion to begin, turning hydrogen into helium. Once fusion starts, the outward push from the energy released by fusion balances the inward pull of gravity, and it is this balance that stops the star collapsing any further, keeping it stable as a main sequence star.

The unit of momentum is kg m/s, since momentum is mass multiplied by velocity. To investigate conservation of momentum, I would first weigh both trolleys, then use a light gate connected to a data logger to measure the velocity of trolley A as it travels down the runway on its own before the collision. The trolleys are fitted with sticky pads so that when trolley A hits trolley B they stick together, and a second light gate further down the runway measures the combined velocity of both trolleys straight after the collision. I would then calculate the total momentum before the collision, mass of A multiplied by its velocity, and the total momentum after, the combined mass multiplied by the combined velocity, and check whether these two values are equal. I would repeat this several times and take an average to reduce the effect of timing errors.

The runway is set at a slope to compensate for friction, so that trolley A travels at a constant speed before reaching trolley B rather than gradually slowing down, which would make the momentum comparison inaccurate.

Momentum = mass multiplied by velocity, so momentum = 6.6 x 10^-26 multiplied by 480, which evaluates to about 3.2 x 10^-23 kg m/s.

Taking the direction the ball is hit back in as positive, F = (mass multiplied by final velocity minus mass multiplied by initial velocity) divided by time = (0.075 multiplied by 15 minus 0.075 multiplied by negative 8.2) divided by 0.012, which evaluates to about 145 N, since the ball's velocity has reversed direction the two momentum terms add together rather than cancel out.

Since the ball's momentum reverses direction, the change in momentum is the sum of the two magnitudes rather than their difference: change in momentum = 0.80 plus 0.60 = 1.4 kg m/s. Using force = change in momentum divided by time, with 70 ms converted to 0.070 s, force = 1.4 divided by 0.070 = 20 N.

Newton's third law applies here because the force the racket exerts on the ball and the force the ball exerts back on the racket are equal in size and opposite in direction, and they act on two different objects, the racket and the ball, rather than both acting on the same object, with both forces being the same type of contact force.

First I would set up the runway on a slight slope and let the trolley run down it on its own with no weights attached, adjusting the angle of the slope until the trolley travels at a constant speed. This step compensates for friction, so that any acceleration I measure later is caused only by the force I add, not by friction slowing the trolley down. I would then attach a length of string over a pulley at the end of the runway, with a weight hanger hanging from the other end, and place some slotted weights both on the hanger and on the trolley itself, since I need a way of changing the accelerating force without changing the total mass of the whole system.

To measure the acceleration, I would attach a card of known length to the trolley and set up two light gates connected to a data logger at fixed points along the runway, which record the time the card takes to pass through each gate and the time between the two gates. From these readings the data logger can calculate the trolley's velocity at each gate and the time taken to travel between them, and I can use acceleration = change in velocity divided by time to find the acceleration for that run.

For each reading I would move one or more slotted weights from the trolley onto the hanger, one at a time, which increases the accelerating force while keeping the total mass of the trolley and weights system exactly the same, and repeat the timing for each new force. I would repeat every measurement at least twice more and take an average of the acceleration at each force, to reduce the effect of random timing errors, then finally plot a graph of force against acceleration for all my results.

To estimate how many gold atoms fit across the foil, I divide the foil thickness by the diameter of one atom: 4.0 x 10^-7 divided by 0.15 nm, which is 4.0 x 10^-7 divided by 1.5 x 10^-10, giving about 2700 atoms. Reading the graph, at 5 degrees roughly 10 to the power 6.5 particles were detected and at 100 degrees only about 10 squared were detected, giving a ratio of around 10 to the power 4.5 to 1, somewhere between 10 000 to 1 and 100 000 to 1.

This huge difference in scattering is the key evidence for the nuclear model of the atom. Most alpha particles pass straight through the gold foil with little or no deflection, which shows that atoms are mostly empty space. A small number are scattered through small angles, showing there must be something inside the atom that can repel a positively charged alpha particle. A very small number, matching that low count at 100 degrees, are scattered through large angles or even bounce straight back, which can only happen if they hit something small, dense and positively charged concentrated at the centre of the atom, the nucleus.

To model this scattering, the students could roll marbles down the slope towards the fixed circular weight on the sheet of paper, and record where each marble ends up or measure the angle each marble deflects through, comparing the pattern to the real alpha scattering data. One real limitation of this marble model is that the marble carries no electrical charge, so it cannot represent the electrostatic repulsion between the positive alpha particle and the positive nucleus that actually causes the deflection in the real experiment.

The potential danger from waves in the electromagnetic spectrum increases as frequency increases, since higher frequency waves such as ultraviolet, X-rays and gamma rays carry more energy and become ionising, unlike the low frequency, low energy waves such as radio waves and microwaves.

For the microwave oven's wavelength, using wave speed = frequency multiplied by wavelength, wavelength = speed divided by frequency = 3.00 x 10^8 divided by 2.45 x 10^9, which evaluates to about 0.12 m. For the total energy supplied, using efficiency = useful energy transferred divided by total energy supplied, total energy supplied = 42 000 divided by 0.55, which evaluates to about 76 000 J.

X-rays sit at the high frequency, short wavelength end of the electromagnetic spectrum, which makes them highly ionising and high in energy. Because of this, hospitals use X-rays to image broken bones and damaged lungs, since X-rays pass through soft tissue but are absorbed more by dense bone, producing a shadow picture on a detector. The same high energy lets X-rays treat cancer directly in radiotherapy, and airport security scanners use them to see through luggage. Radio waves sit at the opposite end of the spectrum, with low frequency, long wavelength and low energy, so they are not ionising. This makes them safe for everyday use in broadcasting television and radio, in mobile phone communication, in satellite transmissions and in radar.

The key difference the question is really testing is how these two radiations are actually produced, not just where they sit in the spectrum. X-rays are emitted when electrons inside an atom absorb energy and jump up to a higher energy level, then almost immediately fall back down to a lower energy level. As an electron drops back down it has to lose a large amount of energy, and because atoms can only lose energy in fixed amounts, that lost energy is released as a single high energy photon, which is an X-ray. Radio waves are produced completely differently, through electrons oscillating in an electrical circuit rather than jumping between fixed energy levels inside an atom. An alternating current makes electrons flow back and forth up and down a transmitting aerial, and this oscillating movement of charge generates radio waves in the space around the aerial, with the frequency of the radio wave produced matching the frequency at which the electrons oscillate in the circuit.

So although X-rays and radio waves are both electromagnetic waves that travel at the speed of light and transfer energy without needing a medium, they sit at opposite extremes of frequency and energy, they are used for very different purposes because of that energy difference, and critically they come from two entirely different electron behaviours: energy level transitions inside atoms for X-rays, against oscillating current in a circuit for radio waves.

Microwaves transfer energy by increasing the kinetic energy of vibration of water molecules, which is how a microwave oven heats food. Infrared radiation transfers energy by heating the skin, which is how a thermal imaging camera or an electric heater works, and can also cause skin burns at high enough intensity.

The human ear detects roughly 20 Hz to 20 kHz. Two reasons there are limits to this range are the fixed size and stiffness of the parts of the inner ear, such as the cochlea, which are only built to respond to a limited range of vibration frequencies, and the fact that the eardrum itself is not sensitive enough to vibrate in response to frequencies far outside this range.

In Figure 13b, an extra reflected signal, Q, arrives back at the emitter in a shorter time than the signal reflected from the base of the bar, R, and a shorter time means a shorter distance travelled, so signal Q must be reflecting off a crack partway through the bar rather than the far end. Signal R has a smaller amplitude than signal P in the crack free bar because some of the original pulse's energy is already reflected or absorbed by the crack, so less of the signal reaches the base of the bar to be reflected back as R.

Rearranging v = f multiplied by lambda for frequency gives f = v divided by lambda = 330 divided by 0.011, which evaluates to 30 000 Hz. For the distance, the pulse has to travel to the prey and back again in 18 ms, so the total distance travelled is speed multiplied by time = 330 multiplied by 0.018 = 5.94 m, and the distance to the prey itself is half of this, about 3.0 m.

The time for sound to travel 120 m is distance divided by speed = 120 divided by 330 = 0.364 s. M's reaction time as a percentage of this is (0.23 divided by 0.364) multiplied by 100, which evaluates to about 63%. This shows the technicians' measurement would be improved by increasing the distance between L and M, since a longer travel time makes the fixed 0.23 s reaction time a much smaller proportion of the overall measurement, reducing its effect on the result.

The refraction of the sound wave is different from the light wave because refraction is caused by a change in speed at the boundary, and sound actually speeds up when it travels from air into water, while light slows down when it travels from air into water, so the two waves bend in opposite ways relative to the normal even though both are refracting.

At region A, both P and S waves are detected. Since S waves are transverse and can only travel through solids, the fact that S waves reach A at all proves that the material directly beneath the crust, the mantle, must behave as a solid. The paths of both waves also curve as they travel deeper into the Earth, which means the waves are being refracted, and a wave only refracts when it moves into a material where its speed changes. Since the paths bend gradually rather than sharply, the mantle's density must increase gradually with depth rather than changing all at once.

At region B, neither P nor S waves are detected at all, creating a shadow zone. S waves cannot travel through a liquid, so their total absence confirms that whatever lies at the centre of the Earth blocking them must be liquid rather than solid. P waves can travel through both liquids and solids, so their absence at B cannot be explained by the same reason. Instead, P waves must be refracted so strongly at the boundary between the solid mantle and this liquid layer that they bend away from region B entirely, missing it, which is only possible if there is a sudden, large change in density at that boundary.

At region C, only P waves are detected, and they arrive at slightly different times from each other rather than as one clean signal. This can only happen if the P waves have travelled through more than one distinct material on their way to C, each with a different density and therefore a different wave speed. Since the outer part of the core has already been shown to be liquid, the fact that some P waves arrive faster than others at C suggests part of their journey passed through a denser, solid region right at the very centre of the Earth, meaning the Earth has both a liquid outer core and a solid inner core.

Distance is the scalar quantity. Acceleration, force and weight all need a direction as well as a size to describe them fully, so they are vectors, but distance only has a magnitude, which is what makes it a scalar.

To investigate how the falling speed depends on the number of cupcake cases in the stack, I would use the metre rule to measure a fixed height and drop the stack from that same height each time. I would use the stop clock to measure the time of fall for that stack, then work out the average speed using speed equals distance divided by time. I would then repeat the whole process using stacks with a different number of nested cupcake cases, and for each stack size I would repeat the drop several times and average the time to reduce the effect of my reaction time on the stopwatch. Finally I would plot a graph of falling speed against the number of cupcake cases in the stack to show the pattern.

The weight of the stack is found using weight equals mass multiplied by gravitational field strength, so W = 0.005 multiplied by 10, which gives 0.05 N.

On Figure 4 I draw a single straight vertical arrow starting at the cupcake case and pointing directly upward, the same rough length as the downward weight arrow already shown: uparrow. This arrow shows the force due to air resistance acting on the cupcake case.

The acceleration of the cupcake case when it is falling at a constant velocity is zero, since the upward air resistance force is now equal in size to the downward weight force, so the resultant force on it is zero.

Rearranging change in GPE = mass multiplied by gravitational field strength multiplied by change in height gives change in height = change in GPE divided by (mass multiplied by g), so delta h = 11 divided by (0.42 multiplied by 10), which evaluates to about 2.6 m.

When the ball hits the wall, energy in the kinetic energy store of the ball is transferred to another store, either the elastic potential energy store as the ball briefly compresses, or the thermal energy store of the ball, wall and surroundings as some energy is dissipated as heat and sound on impact.

Another isotope of americium must have the same proton number, 95, but a different mass number, so 245 over 95 Am is the correct answer, since the other options change the proton number and so are not isotopes of americium at all. Alpha particles can travel about 5 cm in air at normal atmospheric pressure before being stopped.

Completing the decay equation, an alpha particle has a mass number of 4 and proton number of 2, so subtracting these from americium's 241 and 95 gives neptunium as 237 over 93 Np, balancing both the mass numbers, 241 equals 4 plus 237, and the proton numbers, 95 equals 2 plus 93, on both sides. When a positron is emitted from a nucleus, a proton inside the nucleus changes into a neutron, releasing the positron in the process.

The ray does not change direction entering at point X because it enters along the normal to the curved edge of the block, and light travelling along a normal always passes straight through a boundary without bending, even though it does still change speed. To find the critical angle from the graph, I would extend or extrapolate the curve until it reaches an angle of refraction, r, of 90 degrees, then read off the corresponding value of angle i at that point, since the critical angle is defined as the angle of incidence that produces a 90 degree angle of refraction.