AS Level Content
A student wrote · Physical Quantities and UnitsA student wrote: 'the current is 470 microA, which is 470 x 10-3 A.'
Why it loses marksThe prefix micro means x 10-6, not x 10-3. Writing the wrong factor makes the current a thousand times too large, and the same slip between milli and micro is a serious error in a dose or a capacitance.
The answer that scores470 microA = 470 x 10-6 A = 4.7 x 10-4 A. The factor for micro (mu) is 10-6, while milli (m) is 10-3, centi (c) is 10-2, nano (n) is 10-9 and pico (p) is 10-12. Convert to base units first, then substitute into the equation.
Revise this topicPhysical Quantities and Units
A student wrote · Physical Quantities and UnitsA student wrote: 'the pascal is the same as N m, because pressure is force divided by area.'
Why it loses marksDividing by area means dividing by m2, not multiplying by m. The student has also mis-stated the defining equation in terms of units, which makes the answer wrong by a factor of m3.
The answer that scoresPressure p = F / A, so 1 Pa = 1 N / 1 m2 = 1 N m-2 = 1 kg m s-2 / m2 = 1 kg m-1 s-2. The newton itself is 1 kg m s-2, and dividing by m2 reduces the power of the metre by two.
Revise this topicPhysical Quantities and Units
A student wrote · Physical Quantities and UnitsA student wrote: 'mass 60.0 +/- 0.6 g and volume 8.0 +/- 0.4 cm3 give a density of 7.5 g cm-3 with a percentage uncertainty of 5.0 - 1.0 = 4.0 per cent.'
Why it loses marksUncertainties are never subtracted. Density is a quotient, so the percentage uncertainties add, and subtracting them understates how uncertain the answer really is.
The answer that scoresThe percentage uncertainties are 0.6 / 60.0 x 100 = 1.0 per cent for the mass and 0.4 / 8.0 x 100 = 5.0 per cent for the volume. For a quotient they add, so the percentage uncertainty in the density is 1.0 + 5.0 = 6.0 per cent. The absolute uncertainty is 0.060 x 7.5 = 0.45 g cm-3, so the density is 7.5 +/- 0.5 g cm-3, quoted to the same decimal place as the value.
Revise this topicPhysical Quantities and Units
A student wrote · Physical Quantities and UnitsA student wrote: 'student A's readings of the wire diameter are accurate, because 0.58, 0.59, 0.58, 0.57 and 0.58 mm are all very close to each other.'
Why it loses marksBeing close to each other is precision, not accuracy. Accuracy is closeness to the true value, and every one of these readings sits about 0.04 mm above the accepted 0.54 mm, so the set is precise but inaccurate.
The answer that scoresThe readings are precise: their spread is only 0.02 mm, so random errors are small. They are not accurate, because their mean of 0.58 mm differs from the accepted value of 0.54 mm by 0.04 mm, which is far more than the spread. A constant shift of every reading in the same direction indicates a systematic error such as a zero error on the micrometer, which has to be measured and subtracted rather than averaged away.
Revise this topicPhysical Quantities and Units
A student wrote · KinematicsA student wrote: 'A cyclist rides one complete lap of a 400 m circular track in 40 s, so her displacement is 400 m and her average velocity is 10 m s-1.'
Why it loses marksDisplacement is measured in a straight line from the finishing point back to the starting point, and after a complete lap the cyclist is back where she began, so the displacement is zero. The 400 m is the distance travelled, which is a different quantity with a different definition.
The answer that scoresDistance travelled = 400 m, but displacement = 0 m, because the start and finish positions coincide. Average speed = 400 / 40 = 10 m s-1, while average velocity = 0 / 40 = 0 m s-1. The two quantities are only equal when the motion is in a straight line in one direction and never reverses.
Revise this topicKinematics
A student wrote · KinematicsA student wrote: 'The area under a displacement-time graph gives the velocity of the object.'
Why it loses marksThe two graphs carry different information. On a displacement-time graph it is the gradient, not the area, that gives the velocity, and the area under that graph has no physical meaning at all.
The answer that scoresOn a displacement-time graph the gradient is the velocity, and the gradient of the tangent at a point gives the instantaneous velocity. It is on a velocity-time graph that the area between the graph and the time axis gives the displacement and the gradient gives the acceleration.
Revise this topicKinematics
A student wrote · KinematicsA student wrote: 'A heavy ball dropped from a height hits the ground before a light ball dropped from the same height, because gravity pulls on the heavier ball with a bigger force.'
Why it loses marksThe bigger force acts on a proportionally bigger mass, so the acceleration is the same for both. If air resistance is negligible, every object in free fall has a = g = 9.81 m s-2, so the time of fall is identical.
The answer that scoresBoth balls have the same acceleration, because a = F / m and for the heavier ball F and m are both larger by the same factor, leaving a = g. They land together, as Commander David Scott showed on the Moon in 1971 with a hammer and a feather. In the laboratory a light, fluffy object really does fall more slowly, but that is air resistance, not gravity.
Revise this topicKinematics
A student wrote · KinematicsA student wrote: 'For a ball projected horizontally off a cliff, the horizontal velocity decreases as it falls, so the horizontal distance is less than the horizontal velocity multiplied by the time of flight.'
Why it loses marksWith air resistance negligible there is no horizontal force acting on the ball, so the horizontal velocity cannot change. Gravity acts only downwards and alters the vertical velocity alone.
The answer that scoresThe horizontal velocity stays constant at its initial value, so the horizontal distance is exactly the horizontal velocity multiplied by the time of flight. The time of flight is found from the vertical motion alone, using h = (1/2)gt^2 with an initial vertical velocity of zero.
Revise this topicKinematics
A student wrote · DynamicsA student wrote: 'A book rests on a table. Its weight and the contact force from the table are equal and opposite, so they are a Newton's third law pair.'
Why it loses marksNewton's third law pairs act on two different bodies. The weight of the book and the contact force from the table both act on the book, and they are of different types, so they are balanced forces that make the resultant force zero. They are not a third-law pair.
The answer that scoresThe weight (the gravitational pull of the Earth on the book) and the contact force (the push of the table on the book) both act on the book and simply balance. The third-law pairs are: the Earth pulls the book down while the book pulls the Earth up, and the table pushes the book up while the book pushes the table down. Each pair acts on two different bodies and is of the same type of force.
Revise this topicDynamics
A student wrote · DynamicsA student wrote: 'A skydiver at terminal velocity has a zero resultant force, which means the skydiver must be at rest.'
Why it loses marksA zero resultant force means zero acceleration, not zero velocity. Terminal velocity is a constant velocity, and a large constant velocity is perfectly consistent with no resultant force, which is exactly what Newton's first law says.
The answer that scoresAt terminal velocity the drag force has increased with speed until it equals the weight. The resultant force is zero, so by F = ma the acceleration is zero, and the skydiver continues to fall at a constant speed of about 55 m s-1. The forces have not disappeared: weight and drag still act, they cancel.
Revise this topicDynamics
A student wrote · DynamicsA student wrote: 'A ball of mass 0.058 kg travels at 12 m s-1 and then rebounds at 18 m s-1, so its change in momentum is 0.058 x (18 - 12) = 0.35 kg m s-1.'
Why it loses marksThe ball has reversed its direction, so the two velocities have opposite signs. Subtracting the speeds as if they were both positive makes the change look five times smaller than it really is, and the direction of the change in momentum is also lost.
The answer that scoresTake the original direction of travel as positive. Then the initial velocity is +12 m s-1 and the final velocity is -18 m s-1, so the change in momentum is 0.058 x (-18) - (0.058 x 12) = 0.058 x (-30) = -1.74 kg m s-1. The change in momentum is 1.7 kg m s-1 in the direction of the rebound, and dividing by the contact time of 0.012 s gives an average force of 1.74 / 0.012 = 145 N in that same direction.
Revise this topicDynamics
A student wrote · DynamicsA student wrote: 'A crate of mass 20 kg is pushed across a rough floor with a horizontal force of 60 N while a frictional force of 20 N acts on it, so its acceleration is 60 / 20 = 3.0 m s-2.'
Why it loses marksThe force in F = ma is the resultant force, not the force the student applied. Substituting the applied force ignores the friction, which acts in the opposite direction and must be subtracted before dividing by the mass.
The answer that scoresResultant force = 60 - 20 = 40 N, so a = F / m = 40 / 20 = 2.0 m s-2 in the direction of the push. The acceleration is always in the direction of the resultant force.
Revise this topicDynamics
A student wrote · Forces, Density and PressureA student wrote: 'A force of 15 N acts at the end of a spanner 0.40 m long, so the moment about the nut is 15 x 0.40 = 6.0 N m.' The force is actually applied at 60 degrees to the spanner.
Why it loses marksThe distance in the definition of a moment is the perpendicular distance from the pivot to the line of action of the force, not the distance along the spanner. Using the full 0.40 m assumes the force acts at right angles when it does not, so the answer is too large.
The answer that scoresResolve the force or use the perpendicular distance. The component perpendicular to the spanner is 15 x sin 60 degrees = 13 N, and the moment is 13 x 0.40 = 5.2 N m. Equivalently, the perpendicular distance is 0.40 x sin 60 degrees = 0.35 m, and 15 x 0.35 = 5.2 N m. The unit is N m, not J.
Revise this topicForces, Density and Pressure
A student wrote · Forces, Density and PressureA student wrote: 'A couple is a pair of forces, so it produces a resultant force as well as a turning effect.'
Why it loses marksThe two forces of a couple are equal in magnitude and opposite in direction, so they cancel exactly and the resultant force is zero. A couple cannot produce a resultant force, and it is precisely that fact which distinguishes it from the moment of a single force.
The answer that scoresA couple is a pair of equal and opposite forces whose lines of action do not coincide. The resultant force is zero, so a couple acts to produce rotation only, with no translation. Its torque is F x d, where F is the magnitude of one of the forces and d is the perpendicular distance between the two lines of action, and this torque is the same about any point.
Revise this topicForces, Density and Pressure
A student wrote · Forces, Density and PressureA student wrote: 'The upthrust on a floating object is equal to the weight of the object, so the upthrust depends on the mass of the object.'
Why it loses marksThe upthrust is equal to the weight of the fluid displaced, rho g V, and so depends on the density of the fluid and the volume displaced, not on the mass of the object. For a floating object the upthrust is numerically equal to the weight, but that is a consequence of the object being in equilibrium, not the definition of upthrust.
The answer that scoresUpthrust is F = rho g V, the weight of the fluid the object displaces, where rho is the density of the fluid and V is the displaced volume. A floating object is in equilibrium, so upthrust = weight is true for it, but a body that is fully submerged and denser than the fluid also has an upthrust, smaller than its weight, and it sinks because the resultant force is downwards.
Revise this topicForces, Density and Pressure
A student wrote · Forces, Density and PressureA student wrote: 'The pressure at the bottom of a tank of water depends on the width of the tank, because a wide tank holds more water pressing down.'
Why it loses marksIn the derivation of delta-p = rho g delta-h the base area cancels, so the pressure difference between the top and the bottom of a column of fluid depends only on the density of the fluid, g and the height of the column. The extra water in a wide tank is spread over a proportionally larger base, so the pressure is unchanged.
The answer that scoresdelta-p = rho g delta-h, so the pressure depends on the depth and the density only. Water 0.50 m deep in a wide tank and 0.50 m deep in a narrow tube both give 1.0 x 10^3 x 9.81 x 0.50 = 4.9 x 10^3 Pa at the bottom. The total pressure is this value plus atmospheric pressure at the surface, again the same for both vessels.
Revise this topicForces, Density and Pressure
A student wrote · Work, Energy and PowerA student wrote: 'When a ball bounces, some of its energy is used up and destroyed by the impact, so the total amount of energy in the world decreases.'
Why it loses marksEnergy cannot be created or destroyed, and a bounce is no exception. The ball leaves the floor with less kinetic energy, but that energy has been transferred to other stores - mainly the thermal store of the ball and the floor as they deform, plus sound - where it is spread out and no longer useful for the bounce.
The answer that scoresThe principle of conservation of energy says energy is only ever transferred between stores. Just before the impact the ball has 1/2 m v^2 in its kinetic store. After the bounce it has less, because some of that energy has been transferred to the thermal store of the ball and the floor by the work done in deforming them, and to the sound waves produced. The total energy is unchanged; only the useful, ordered store has shrunk, which is why the ball rebounds to a smaller height.
Revise this topicWork, Energy and Power
A student wrote · Work, Energy and PowerA student wrote: 'The efficiency of the motor is 85 J.'
Why it loses marksEfficiency is a ratio of two energies, or of two powers, so it is a pure number and has no unit at all. Writing 85 J confuses efficiency with an amount of energy, and a question asking for the efficiency as a percentage would score nothing for this answer.
The answer that scoresEfficiency = useful energy output / total energy input = 2040 / 2400 = 0.85, which is 85%. The per cent sign belongs after the number, or the answer can be left as the fraction 0.85. The joule is the unit of energy and of work; it is never the unit of efficiency.
Revise this topicWork, Energy and Power
A student wrote · Work, Energy and PowerA student wrote: 'A girl pulls a sledge 5.0 m across level snow with a rope at 60 degrees to the horizontal. The tension in the rope is 40 N, so the work done is 40 x 5.0 = 200 J.'
Why it loses marksOnly the component of the force along the direction of motion does work on the sledge. The vertical component pulls upwards at right angles to the motion, so there is no displacement along its line and it transfers no energy. Using the whole 40 N overcounts the work done.
The answer that scoresThe component of the tension along the direction of motion is 40 cos 60 = 20 N. Work done = F s cos theta = 20 x 5.0 = 100 J. The upward component of the tension does no work on the sledge because the sledge moves horizontally, so the correct answer is half of the one the student gave.
Revise this topicWork, Energy and Power
A student wrote · Work, Energy and PowerA student wrote: 'A car doubles its speed from 15 m s-1 to 30 m s-1, so the work needed to bring it to rest doubles.'
Why it loses marksKinetic energy depends on the square of the speed, not on the speed itself. Doubling the speed multiplies the kinetic energy, and therefore the work that must be done to remove it, by four. That is why braking distances grow so steeply with speed.
The answer that scoresE k = 1/2 m v^2. For a 1200 kg car at 15 m s-1, E k = 0.5 x 1200 x 225 = 1.35 x 10^5 J. At 30 m s-1, E k = 0.5 x 1200 x 900 = 5.4 x 10^5 J, which is four times as much. The same average braking force must therefore act over four times the distance, so the braking distance is four times greater, not twice.
Revise this topicWork, Energy and Power
A student wrote · Deformation of SolidsA student wrote, in a write-up of the Young modulus experiment: 'I measured the diameter of the wire with a metre rule and used A = 2 pi r to work out the cross-sectional area.'
Why it loses marksA metre rule cannot read the diameter of a wire accurately, because the diameter is well under a millimetre. Worse, 2 pi r is the circumference of the circle, not its area, so the value used for A is about 20 times too large for a wire of this size and the calculated Young modulus comes out far too small.
The answer that scoresMeasure the diameter with a micrometer screw gauge, at several places along the wire and in two perpendicular directions at each place, and take the mean diameter. Then find the area from A = pi d^2 / 4, which is the same as pi r^2. For d = 0.42 mm, r = 2.1 x 10^-4 m and A = pi x (2.1 x 10^-4)^2 = 1.4 x 10^-7 m2.
Revise this topicDeformation of Solids
A student wrote · Deformation of SolidsA student wrote: 'A wire of cross-sectional area 2.0 x 10^-7 m2 carries a force of 15 N, so the stress is 15 x 2.0 x 10^-7 = 3.0 x 10^-6 N m-2.'
Why it loses marksStress is force divided by area, not multiplied by it. Multiplying gives a number far too small and with the wrong unit pattern; a stress that small is characteristic of nothing real in this experiment.
The answer that scoresstress = F / A = 15 / (2.0 x 10^-7) = 7.5 x 10^7 Pa. Dividing by a small area makes the stress large, which is why a thin wire carries a much greater stress than a thick one under the same load.
Revise this topicDeformation of Solids
A student wrote · Deformation of SolidsA student wrote: 'The limit of proportionality is the point beyond which the wire is permanently stretched, so it is the same as the elastic limit.'
Why it loses marksThe two are defined differently. The limit of proportionality is a statement about the graph: beyond it the extension is no longer proportional to the force. The elastic limit is a statement about the material: beyond it the wire does not return to its original length. For a metal wire the limit of proportionality is reached slightly earlier than the elastic limit.
The answer that scoresThe limit of proportionality is where the force-extension graph stops being a straight line through the origin and begins to curve. The elastic limit is where the wire, once unloaded, stays permanently longer. Between the two the wire still returns to its original length when the load is removed, even though Hooke's law no longer holds.
Revise this topicDeformation of Solids
A student wrote · Deformation of SolidsA student wrote: 'A spring is stretched by 0.080 m by a force of 24 N, so the energy stored is 24 x 0.080 = 1.9 J.'
Why it loses marksMultiplying the force by the extension works only if the force is the same all the way through the extension. Here the force grows steadily from zero to 24 N as the spring stretches, so the work done is the area of the triangle under the graph, which is half of that product.
The answer that scoresThe energy stored is the area under the force-extension graph: Ep = 1/2 Fx = 0.5 x 24 x 0.080 = 0.96 J. Check it with the spring constant: k = F / x = 24 / 0.080 = 300 N m-1, and Ep = 1/2 kx^2 = 0.5 x 300 x (0.080)^2 = 0.96 J.
Revise this topicDeformation of Solids
A student wrote · WavesA student wrote: 'The intensity of a wave is proportional to its amplitude, so a wave with twice the amplitude has twice the intensity.'
Why it loses marksIntensity is proportional to the square of the amplitude, not to the amplitude. Doubling the amplitude multiplies the intensity by 2 x 2 = 4, and the relationship is written as I is proportional to A^2.
The answer that scoresThe intensity of a progressive wave is proportional to the square of its amplitude. A wave of amplitude 2.0 mm carrying an intensity of 0.50 W m-2, when its amplitude is changed to 4.0 mm, has an amplitude twice as large and therefore an intensity of 4 x 0.50 = 2.0 W m-2. Going the other way, to double the intensity you must increase the amplitude by a factor of 2^0.5, which is 1.41.
Revise this topicWaves
A student wrote · WavesA student wrote: 'The trace on the oscilloscope has a peak-to-peak height of 6.0 divisions and the y-gain is 2.0 V per division, so the amplitude of the signal is 12 V.'
Why it loses marksMultiplying the height by the y-gain gives the peak-to-peak voltage, which is the distance from the bottom of a trough to the top of a crest. Amplitude is measured from the centre line to a crest, so it is half of that value.
The answer that scoresPeak-to-peak voltage = 6.0 x 2.0 = 12 V, and the amplitude is half of that: 12 / 2 = 6.0 V. Write the two steps out separately in the exam, because the halving is worth a mark of its own. The same trap applies to the time-base, where the divisions for one complete cycle, not for a half cycle, give the period.
Revise this topicWaves
A student wrote · WavesA student wrote: 'The siren emits 500 Hz and the ambulance approaches at 30 m s-1, so f_o = 500 x 340 / (340 + 30) = 459 Hz, which is higher than 500 Hz.'
Why it loses marksThe wrong sign was used, and the answer does not even agree with the student's own statement that a higher frequency is heard. An approaching source moves into its own wavefronts, so the wavefronts are crowded together and the observed wavelength is shorter, which means the denominator must be smaller than v and the observed frequency larger than f_s.
The answer that scoresFor an approaching source use the minus sign: f_o = f_s v / (v - v_s) = 500 x 340 / (340 - 30) = 500 x 340 / 310 = 548 Hz, which is correctly higher than the source frequency. After the ambulance passes, the plus sign applies: f_o = 500 x 340 / (340 + 30) = 500 x 340 / 370 = 459 Hz, which is lower than 500 Hz. Always check the sign of the answer against the direction of motion.
Revise this topicWaves
A student wrote · WavesA student wrote: 'Plane-polarised light of intensity 400 W m-2 meets a filter at 60 degrees to its plane of polarisation, so I = 400 x cos 60 = 200 W m-2, and a beam of sound could be polarised in the same way.'
Why it loses marksMalus's law uses cos squared, not cos, so the transmitted intensity is 400 x (0.50)^2 = 100 W m-2 and not 200 W m-2. The second part is wrong for a different reason: polarisation means restricting oscillations to one plane, and sound is longitudinal, with its oscillations along the direction of travel, so there is no plane to restrict them to.
The answer that scoresI = I0 cos^2 theta = 400 x cos^2 60 = 400 x (0.50)^2 = 400 x 0.25 = 100 W m-2. Only transverse waves can be polarised, which is why polarising filters work on light, on microwaves from a transmitter and on the transverse waves in a stretched string, but never on sound.
Revise this topicWaves
A student wrote · SuperpositionA student wrote: 'Two waves only superpose if they are exactly in phase; if there is a phase difference between them nothing happens.'
Why it loses marksSuperposition always happens when two waves meet, whatever their phase difference. What the phase difference decides is the value of the resultant, not whether the waves add together at all.
The answer that scoresSuperposition occurs whenever two waves arrive at the same point: the resultant displacement is the vector sum of the individual displacements. If the waves arrive in phase the displacements add and give a maximum amplitude; if they arrive in antiphase the displacements subtract and the resultant is small, or zero when the amplitudes are equal. Every other phase difference gives an intermediate resultant between those two limits.
Revise this topicSuperposition
A student wrote · SuperpositionA student wrote: 'If the separation of the two slits is doubled, the fringes on the screen will be twice as far apart.'
Why it loses marksThe fringe spacing is inversely proportional to the slit separation, not directly proportional. Doubling a halves x, so the fringes crowd together.
The answer that scoresFrom x = lambda D / a, the fringe spacing is proportional to 1 / a. Doubling the slit separation from 0.45 mm to 0.90 mm halves the spacing, from 3.4 mm to 1.7 mm. To make the fringes wider, move the slits closer together or move the screen further away, since x is proportional to D as well.
Revise this topicSuperposition
A student wrote · SuperpositionA student wrote: 'In a stationary wave on a string, energy is carried along the string from the vibrator to the fixed end, where the nodes absorb it.'
Why it loses marksA stationary wave transfers no net energy along the wave. Energy is not absorbed at the nodes, and the nodes are not points where the string has stopped.
The answer that scoresThe incident wave and the reflected wave carry equal energy in opposite directions, so there is no net transfer of energy along the string. Energy oscillates within the pattern, changing between the kinetic energy of the moving string and the potential energy stored in it as it stretches. The string vibrates everywhere except at the nodes, reaching its greatest displacement at the antinodes.
Revise this topicSuperposition
A student wrote · SuperpositionA student wrote: 'A grating has 3.00 x 10^5 lines per metre, so the slit spacing is d = 3.00 x 10^5 m.'
Why it loses marksThe slit spacing is the distance between adjacent lines, which is the reciprocal of the number of lines per metre, not the number itself. Substituting the number directly gives an answer wrong by a factor of 10^10.
The answer that scoresd = 1 / N = 1 / (3.00 x 10^5) = 3.33 x 10-6 m. This value is then used in d sin theta = n lambda. If the grating were quoted in lines per millimetre, N would first be converted to lines per metre: 600 lines per mm is 6.00 x 10^5 lines per metre, so d = 1.67 x 10-6 m.
Revise this topicSuperposition
A student wrote · ElectricityA student wrote: 'The current from the battery is used up as it goes round the circuit, so the current after the lamp is smaller than the current before it.'
Why it loses marksCharge is conserved. The current is the rate of flow of charge, and in a series circuit the same current passes through every component, because whatever charge leaves one terminal of the source must return to the other terminal at the same rate. What falls as the charge goes round the circuit is the energy of the carriers, not the number of them.
The answer that scoresThe current is the same on both sides of the lamp, because the same number of coulombs passes each point each second. Energy is what is transferred: the p.d. across the lamp is the energy transferred per unit charge, so each coulomb gives up W = QV of energy to the lamp. Adding a second identical lamp in series does not use up the current part by part; it increases the total resistance, so the current through both lamps is smaller than the current through one.
Revise this topicElectricity
A student wrote · ElectricityA student wrote: 'The electrons in a wire travel at about 3.00 x 10^8 m s-1, which is why a lamp lights as soon as the switch is closed.'
Why it loses marksThe mean drift velocity of the free electrons in a metal is only of the order of 10-4 m s-1, so the electrons arriving at the lamp are not the ones that left the battery. The lamp lights immediately because the electric field is set up throughout the circuit at close to the speed of light, so every free electron in the circuit, including those already inside the filament, begins to drift almost at once. The random speed of the electrons is of the order of 10^5 m s-1, but random motion alone transfers no net charge.
The answer that scoresFor a copper wire of cross-sectional area A = 2.0 x 10-6 m2 carrying I = 2.72 A, with n = 8.5 x 10^28 m-3 and q = 1.60 x 10-19 C: v = I / (Anq) = 2.72 / (2.0 x 10-6 x 8.5 x 10^28 x 1.60 x 10-19) = 2.72 / (2.72 x 10^4) = 1.0 x 10-4 m s-1. The electrons creep along at a tiny fraction of a millimetre per second, while the signal travels with the electric field at about 3.00 x 10^8 m s-1.
Revise this topicElectricity
A student wrote · ElectricityA student wrote: 'A filament lamp obeys Ohm's law, because R = V / I can be worked out for it at every point on its characteristic.'
Why it loses marksR = V / I is the definition of resistance and it can be calculated for any component, so being able to work it out proves nothing. Ohm's law is the statement that the current is proportional to the p.d., which is only true when the temperature stays constant. In a filament lamp the temperature rises as the current rises, so I is not proportional to V and the lamp is not ohmic.
The answer that scoresCalculate R = V / I at several points on the lamp's characteristic and you will find it increasing, from a small value when the filament is cold to a much larger one when it is white hot, so the graph is a curve through the origin and not a straight line. Ohm's law is obeyed at constant temperature, which is why constantan or nichrome resistors are used when a straight-line characteristic is needed. The temperature dependence of the resistance is exactly what the lamp's characteristic demonstrates.
Revise this topicElectricity
A student wrote · ElectricityA student wrote: 'The wire has a cross-sectional area of 0.50 mm2, that is 0.50 x 10-3 m2, so with rho = 1.7 x 10-8 ohm m and L = 1.0 m the resistance is R = (1.7 x 10-8 x 1.0) / (0.50 x 10-3) = 3.4 x 10-5 ohm.'
Why it loses marksThe conversion of mm2 to m2 is wrong. Because 1 mm = 1 x 10-3 m, one square millimetre is (1 x 10-3 m)2 = 1.0 x 10-6 m2, so the conversion factor has to be squared. Using 0.50 x 10-3 m2 in place of 5.0 x 10-7 m2 makes the area a thousand times too large and the resistance a thousand times too small.
The answer that scores1 mm2 = (1 x 10-3 m)2 = 1.0 x 10-6 m2, so 0.50 mm2 = 0.50 x 1.0 x 10-6 m2 = 5.0 x 10-7 m2. Then R = rho L / A = (1.7 x 10-8 x 1.0) / (5.0 x 10-7) = 3.4 x 10-2 ohm, that is 0.034 ohm. Always square the length conversion when an area is given in mm2, and check the size of the answer: a copper connecting wire a metre long has a resistance of a few hundredths of an ohm, not a few ten-thousandths.
Revise this topicElectricity
A student wrote · D.C. CircuitsA student wrote: 'A voltmeter connected across the terminals of a cell always reads the e.m.f. of the cell, so a reading of 1.50 V means the e.m.f. is 1.50 V even when the cell is lighting a lamp.'
Why it loses marksThe terminal p.d. equals the e.m.f. only when no current is being drawn. As soon as the cell delivers current there are lost volts Ir inside the cell, and the terminal p.d. is lower than the e.m.f. Reading the e.m.f. from a working circuit is the standard version of this error.
The answer that scoresUse the terminal p.d. formula V = E - Ir. For a cell with E = 1.50 V and r = 0.40 ohm delivering 0.50 A, the lost volts are Ir = 0.50 x 0.40 = 0.20 V, so the terminal p.d. is 1.50 - 0.20 = 1.30 V. The cell's e.m.f. is 1.50 V; the voltmeter reads 1.30 V while that current flows.
Revise this topicD.C. Circuits
A student wrote · D.C. CircuitsA student wrote: 'Two 6.0 ohm resistors connected in parallel have a combined resistance of 12 ohm, because resistances add together.'
Why it loses marksThe series rule has been applied to a parallel arrangement. In parallel the two resistors are across the same two junctions, so the same p.d. drives current through two paths at once and the total current from the source rises, which makes the combined resistance fall rather than rise.
The answer that scoresFor resistors in parallel, 1 / R = 1 / R1 + 1 / R2, so 1 / R = 1 / 6.0 + 1 / 6.0 = 2 / 6.0 = 1 / 3.0, giving R = 3.0 ohm. For n equal resistors in parallel, R is the resistance of one divided by n, so two 6.0 ohm resistors in parallel give 6.0 / 2 = 3.0 ohm, which is smaller than either one of them. In series they would give 12 ohm.
Revise this topicD.C. Circuits
A student wrote · D.C. CircuitsA student wrote: 'At a junction the current flowing in is larger than the current flowing out, because the components in the branches use up some of the current as they transfer energy.'
Why it loses marksA component transfers energy from the charge passing through it, but it does not consume the charge itself. Charge is conserved, so in a steady-state circuit no charge can accumulate at a junction and the rate at which it arrives must equal the rate at which it leaves.
The answer that scoresState Kirchhoff's first law: the sum of the currents entering a junction equals the sum of the currents leaving it. At a junction where a current of 3.0 A arrives and the branches carry 1.0 A and 2.0 A, 1.0 + 2.0 = 3.0 A, so the charges balance exactly. The components change how the energy is carried away, not how much charge flows.
Revise this topicD.C. Circuits
A student wrote · D.C. CircuitsA student wrote: 'In a potential divider the p.d. across the larger resistance is smaller, and when an NTC thermistor in the divider is warmed its resistance increases so the p.d. across it increases.'
Why it loses marksBoth halves are wrong. In a series divider the same current flows through each resistor and V = IR, so the larger resistance takes the larger share of the supply p.d. And an NTC thermistor's resistance falls as its temperature rises, not rises.
The answer that scoresFor two resistors in series across a supply, V_out / V_in = R2 / (R1 + R2), so the biggest resistance takes the biggest share of the p.d. When an NTC thermistor is warmed its resistance falls, so the current in the divider rises; the p.d. across the thermistor therefore falls, the p.d. across the fixed resistor rises, and the two still add up to the supply p.d. by Kirchhoff's second law.
Revise this topicD.C. Circuits
A student wrote · Particle PhysicsA student wrote: 'Most of the alpha particles were deflected through large angles, which shows that the nucleus is large and fills most of the volume of the atom.'
Why it loses marksThe conclusion is exactly the opposite of the observation. If the nucleus filled most of the atom, nearly every alpha particle would come close to one and be deflected, and the results would show large deflections everywhere. What was actually seen was that almost all the alpha particles passed straight through, and only about 1 in 8000 was deflected through more than 90 degrees.
The answer that scoresAlmost all the alpha particles passed straight through the foil, which shows that the atom is mostly empty space. Only about 1 in 8000 was deflected through more than 90 degrees, and those must have approached a concentrated region of positive charge, so the nucleus carries a large positive charge. Because the large deflections are so rare, that nucleus must be very small compared with the atom, about 10000 times smaller in diameter, and it must contain nearly all of the atom's mass.
Revise this topicParticle Physics
A student wrote · Particle PhysicsA student wrote: 'In beta-minus decay a neutron changes into a proton and an electron, and a neutrino is emitted so that energy is conserved.'
Why it loses marksThe right particle is named in the wrong place. Beta-minus decay emits an electron together with an electron antineutrino; it is beta-plus decay, which emits a positron, that produces an electron neutrino. The (anti)neutrino is needed because the energy released is shared between it and the beta particle, which is why beta particles have a continuous range of energies.
The answer that scoresIn beta-minus decay a neutron changes into a proton and an electron, and an electron antineutrino is emitted as well: n -> p + beta-minus particle + electron antineutrino. In beta-plus decay a proton changes into a neutron and a positron, and an electron neutrino is emitted: p -> n + positron + electron neutrino. The (anti)neutrino carries away some of the energy and momentum, which is why the beta-particle energies are continuous rather than discrete.
Revise this topicParticle Physics
A student wrote · Particle PhysicsA student wrote: 'A proton is made of one up quark and two down quarks (udd) and a neutron is made of two up quarks and one down quark (uud).'
Why it loses marksThe quark compositions have been swapped, and the charges prove it. udd gives (+2/3) + (-1/3) + (-1/3) = 0, which is the charge of a neutron, not of a proton. uud gives (+2/3) + (+2/3) + (-1/3) = +1e, which is the charge of a proton. Always check the total charge against the particle you are describing.
The answer that scoresA proton is up, up, down (uud), charge (+2/3) + (+2/3) + (-1/3) = +1e. A neutron is up, down, down (udd), charge (+2/3) + (-1/3) + (-1/3) = 0. Both are baryons, made of three quarks; the extra down quark in the neutron is what makes it neutral and slightly heavier than the proton.
Revise this topicParticle Physics
A student wrote · Particle PhysicsA student wrote: 'An antiparticle has the same charge as its particle but the opposite mass, so a positron is heavier than an electron and a proton is heavier than an antiproton.'
Why it loses marksAn antiparticle has the same mass and the opposite charge. The masses are identical to the last measurement, and experiments on matter and antimatter have confirmed this to a very high precision. Reversing the mass is not only wrong physics, it also contradicts the annihilation of a positron with an electron, in which the total mass of the two equal particles is converted into two photons, each of energy equal to the rest energy of one electron.
The answer that scoresAn antiparticle has the same mass as the corresponding particle and a charge of the opposite sign. The positron is the antiparticle of the electron: it has a mass of 9.11 x 10-31 kg, the same as an electron, and a charge of +1e instead of -1e. When a positron meets an electron the two annihilate, and the two gamma-ray photons produced each carry away 0.511 MeV, the rest energy of one electron.
Revise this topicParticle Physics
A Level Content
A student wrote · Motion in a circleA student wrote: 'When a cyclist goes round a bend there is an outward force pushing the bicycle away from the centre of the bend, and that outward force balances the centripetal force.'
Why it loses marksThere is no outward force on the cyclist. In the frame of the ground the only horizontal force is friction between the tyres and the road, and it acts towards the centre of the bend. Introducing a balancing outward force leaves no resultant force, so it cannot explain why the cyclist accelerates towards the centre at all.
The answer that scoresThe centripetal force is the friction between the tyres and the road, directed towards the centre of the bend. It is not balanced by anything: it is the resultant force, and it produces the centripetal acceleration a = v^2 / r towards the centre. The cyclist feels pushed outwards only because the bicycle is turning while the rider tends to carry on in a straight line, which is Newton's first law.
Revise this topicMotion in a circle
A student wrote · Motion in a circleA student wrote: 'The centripetal acceleration of a rider on a carousel is zero, because the rider moves at a constant speed.'
Why it loses marksConstant speed does not mean zero acceleration. The velocity of the rider is changing direction all the time, so the velocity vector is changing and there is an acceleration. The magnitude of the velocity is constant, but a vector needs both magnitude and direction.
The answer that scoresFor a rider 6.5 m from the axis of a carousel turning with omega = 0.524 rad s-1, the centripetal acceleration is a = r omega^2 = 6.5 x (0.524)^2 = 1.8 m s-2, directed towards the axis of rotation. It is this acceleration, and the force that causes it, that keeps the rider on the circular path instead of moving off along the tangent.
Revise this topicMotion in a circle
A student wrote · Motion in a circleA student wrote: 'A satellite in a circular orbit is weightless, so the gravitational force on it must be zero.'
Why it loses marksThe gravitational force on a satellite is far from zero; it is the centripetal force that keeps the satellite in its orbit. Weightlessness inside the satellite is a sensation, caused by the satellite and everything in it accelerating towards the Earth at the same rate, so nothing pushes on anything else.
The answer that scoresThe gravitational force acts towards the centre of the Earth and provides the centripetal force, F = mv^2 / r. For the International Space Station, at a radius of about 6.8 x 10^6 m, the gravitational field strength is about 8.7 N kg-1, so a 70 kg astronaut still has a weight of about 610 N. The astronaut feels weightless because both are in free fall together.
Revise this topicMotion in a circle
A student wrote · Motion in a circleA student wrote: 'A wheel of radius 0.30 m rolls along the ground at 4.0 m s-1, so its angular speed is 4.0 x 0.30 = 1.2 rad s-1.'
Why it loses marksThe equation v = r omega has to be rearranged before the numbers go in. The student multiplied instead of dividing, and a quick check shows why that is wrong: 1.2 rad s-1 would give a speed of only 0.36 m s-1 for a point on the rim, not 4.0 m s-1.
The answer that scoresFrom v = r omega, omega = v / r = 4.0 / 0.30 = 13 rad s-1. Always rearrange the equation with symbols first and then substitute, and check the answer by putting it back into v = r omega: 0.30 x 13 = 3.9 m s-1, which agrees with 4.0 m s-1 to the significant figures of the data.
Revise this topicMotion in a circle
A student wrote · Gravitational fieldsA student wrote: 'g = G M / r2, so at the Earth's surface g = (6.67 x 10-11 x 5.97 x 10^24) / (6.37 x 10^6) = 6.3 x 10^7 N kg-1.'
Why it loses marksThe radius was not squared. The equation contains r2, not r, so substituting the radius itself makes the answer about 6 x 10^6 times too large. A field strength of 6.3 x 10^7 N kg-1 would mean every kilogram on the Earth's surface weighing 63 million newtons, which should have alerted the student immediately.
The answer that scoresg = (6.67 x 10-11 x 5.97 x 10^24) / (6.37 x 10^6)2 = (3.98 x 10^14) / (4.06 x 10^13) = 9.81 N kg-1. This is the standard value of the field strength at the Earth's surface, and the value of g = 9.81 m s-2 used in free-fall calculations is the same number in different units.
Revise this topicGravitational fields
A student wrote · Gravitational fieldsA student wrote: 'The gravitational force on a satellite is G M m / r2 and the centripetal force is m v2 / r. Doubling the mass of the satellite in the same orbit therefore doubles the force needed, so the orbital speed must double.'
Why it loses marksThe two expressions were not equated. Setting them equal gives G M m / r2 = m v2 / r, and the satellite's mass cancels from both sides. The orbital speed is v = sqrt(G M / r), which contains no reference to the satellite at all, so changing the mass of the satellite changes the force needed and the force available by the same factor and leaves the orbit unchanged.
The answer that scoresG M m / r2 = m v2 / r gives v = sqrt(G M / r), which depends only on the mass of the central body and the radius of the orbit. Doubling the mass of the satellite doubles the gravitational force on it, but it also doubles the centripetal force needed for the same path, so the speed and the period are unchanged. This is why a heavy satellite and a light one can share the same orbit.
Revise this topicGravitational fields
A student wrote · Gravitational fieldsA student wrote: 'The gravitational potential at the Earth's surface is +6.25 x 10^7 J kg-1, because the potential energy of a mass near the Earth is positive.'
Why it loses marksThe sign is wrong, and the confusion comes from mixing up potential with the energy of an object lifted a few metres. Gravitational potential is defined from zero at infinity, and because gravity is attractive the work done in bringing a test mass in from infinity is negative, so every gravitational potential is negative. Gravitational potential energy EP = m x phi has the same sign as the potential.
The answer that scoresphi = -G M / r = -(6.67 x 10-11 x 5.97 x 10^24) / (6.37 x 10^6) = -6.25 x 10^7 J kg-1. The potential is negative at every finite distance and increases towards zero as r increases. A mass lifted from the surface gains energy because the potential becomes less negative, and the change in potential energy is positive even though both values of the potential are negative.
Revise this topicGravitational fields
A student wrote · Gravitational fieldsA student wrote: 'The Moon stays in orbit around the Earth because there is no gravity in space and the Earth's rotation carries the Moon around with it.'
Why it loses marksBoth claims are wrong. There is no place where gravity suddenly stops, and the inverse square law means the Earth's field is still appreciable at the distance of the Moon. A satellite is not carried round by the Earth's rotation; it is falling freely in the Earth's field, and it is the gravitational force towards the Earth's centre that provides the centripetal force for its circular path.
The answer that scoresAt the Moon's distance of 3.84 x 10^8 m the Earth's field strength is g = G M / r2 = (6.67 x 10-11 x 5.97 x 10^24) / (3.84 x 10^8)2 = 2.70 x 10-3 N kg-1. That field supplies the centripetal acceleration a = v2 / r, so v = sqrt(g r) = sqrt(2.70 x 10-3 x 3.84 x 10^8) = 1.02 x 10^3 m s-1, which is exactly the Moon's orbital speed and gives a period of 27.4 days.
Revise this topicGravitational fields
A student wrote · TemperatureA student wrote: 'The temperature of the water rises from 20 degrees C to 50 degrees C, so the temperature rise is 30 + 273.15 = 303.15 K, and Q = m c delta theta gives a much bigger answer.'
Why it loses marksThe kelvin conversion applies to an absolute temperature, not to a temperature difference. The kelvin and the degree Celsius are the same size, so a rise of 30 degrees C is a rise of 30 K. Adding 273.15 to every temperature in sight is one of the commonest errors on this topic and it changes the answer by a factor of about ten.
The answer that scoresUse the temperature change only: delta theta = 50 - 20 = 30 degrees C = 30 K. Then Q = m c delta theta with delta theta = 30 K. For 2.0 kg of water that is 2.0 x 4200 x 30 = 252 000 J, or 2.5 x 10^5 J, not a value ten times larger.
Revise this topicTemperature
A student wrote · TemperatureA student wrote: 'The energy needed to boil away 0.50 kg of water already at 100 degrees C is Q = m c delta theta = 0.50 x 4200 x 100 = 210 000 J.'
Why it loses marksThe water is already at 100 degrees C, so its temperature does not change at all while it boils: there is no temperature rise to put into the specific heat capacity equation. Energy is still being supplied, but it goes into the potential energy of the molecules as they are pulled apart, not into their kinetic energy.
The answer that scoresBoiling is a change of state, so use the latent heat: Q = m L = 0.50 x 2.26 x 10^6 = 1 130 000 J, or 1.13 x 10^6 J. That is more than five times the answer the student gave, which is the point - converting water at its boiling point into steam is far more expensive than heating it up to the boiling point.
Revise this topicTemperature
A student wrote · TemperatureA student wrote: 'Absolute zero is 0 degrees C, because that is where the Celsius scale starts.'
Why it loses marks0 degrees C is merely the ice point, the temperature of melting ice, and it is a convenient but arbitrary place to start the Celsius scale. Absolute zero is the genuinely lowest temperature possible, and it lies far below the ice point: it is -273.15 degrees C, which is 0 K. Nothing can be colder than absolute zero, so the kelvin scale has no negative values.
The answer that scoresAbsolute zero is the lowest possible temperature, 0 K or -273.15 degrees C, at which the molecules of a substance have the least random kinetic energy they can have. The Celsius scale is offset from the kelvin scale by 273.15, so T / K = theta / degrees C + 273.15. The ice point, 0 degrees C, is 273.15 K, not absolute zero.
Revise this topicTemperature
A student wrote · TemperatureA student wrote: 'The specific heat capacity of water is 4200 J K-1, so 1 kg of water needs 4200 J to become 1 degree C hotter.'
Why it loses marksThe unit is wrong. Specific heat capacity is defined per unit mass, so the unit must contain the kilogram: J kg-1 K-1. J K-1 is the unit of heat capacity, the energy needed to raise the temperature of the whole object by 1 K, which depends on how much of the substance there is and is not a property of the substance itself.
The answer that scoresThe specific heat capacity of water is 4200 J kg-1 K-1, which means 4200 J raises the temperature of 1 kg of water by 1 K, or 8400 J raises 1 kg by 2 K. The rest of the student's sentence is correct; only the unit needs fixing, but the unit carries a mark here because it shows what the number means.
Revise this topicTemperature
A student wrote · Ideal GasesA student wrote: 'A gas at 20 degrees C is twice as hot as a gas at 10 degrees C, so its molecules have twice the average kinetic energy.'
Why it loses marksThe kinetic energy is proportional to the thermodynamic temperature, and degrees C is not a ratio scale. In kelvin the two temperatures are 293 K and 283 K, and the ratio of the energies is 293 / 283 = 1.04, not 2.
The answer that scoresConvert to kelvin first: T / K = theta / degrees C + 273.15, so the temperatures are 293 K and 283 K. Since the average translational kinetic energy of a molecule is (3/2)kT, the ratio of the energies is 293 / 283 = 1.04, a change of only about four per cent.
Revise this topicIdeal Gases
A student wrote · Ideal GasesA student wrote: 'The average kinetic energy of one mole of an ideal gas at 300 K is (3/2)kT = 6.21 x 10-21 J.'
Why it loses marksThe expression (3/2)kT is the average translational kinetic energy of a single molecule. A mole contains 6.02 x 10^23 molecules, so the value for a mole is enormously larger.
The answer that scoresFor one molecule, (3/2)kT = 1.5 x 1.38 x 10-23 x 300 = 6.21 x 10-21 J. For one mole, multiply by the Avogadro constant, which gives (3/2)RT = 1.5 x 8.31 x 300 = 3.74 x 10^3 J.
Revise this topicIdeal Gases
A student wrote · Ideal GasesA student wrote: 'In the kinetic theory derivation the pressure on the wall is 2m cx, because the molecule's momentum changes by 2m cx at each collision.'
Why it loses marks2m cx is a change in momentum, not a force, and the force depends on how often the molecule hits the wall. The mark is for using force = rate of change of momentum with the correct time between collisions.
The answer that scoresThe change in momentum per collision is 2m cx and the time between collisions with the same face is 2L / cx, so the average force is F = 2m cx / (2L / cx) = m cx^2 / L. For N molecules the total force is N m <cx^2> / L, so p = F / L^2 = N m <cx^2> / V, and using <cx^2> = (1/3)<c^2> gives pV = (1/3) N m <c^2>.
Revise this topicIdeal Gases
A student wrote · Ideal GasesA student wrote: 'A fixed mass of gas is compressed at constant temperature from 200 cm3 to 100 cm3, so the pressure falls to half of its original value.'
Why it loses marksBoyle's law says that pV is constant at constant temperature, so p and V are inversely proportional. Halving the volume doubles the pressure, not halves it.
The answer that scoresUsing p1V1 = p2V2, a gas at 1.0 x 10^5 Pa in 200 cm3 compressed to 100 cm3 has a pressure p2 = (1.0 x 10^5 x 200) / 100 = 2.0 x 10^5 Pa. A graph of p against 1/V for such a gas is a straight line through the origin, which shows that p is proportional to 1 / V.
Revise this topicIdeal Gases
A student wrote · ThermodynamicsA student wrote: 'A gas expands at constant pressure and does 450 J of work on the piston, so W = +450 J and Delta U = q + 450.'
Why it loses marksCambridge writes the first law with W as the work done on the system. When the gas itself does the work, energy leaves the gas, so W must be negative. Substituting +450 J makes the internal energy come out 900 J larger than it really is.
The answer that scoresThe work done by the gas is p Delta V = 450 J, so the work done on the gas is W = -450 J. If 600 J is transferred to the gas by heating at the same time, Delta U = q + W = 600 - 450 = +150 J, so the internal energy increases by 150 J.
Revise this topicThermodynamics
A student wrote · ThermodynamicsA student wrote: 'A gas expands by 20 cm3 at a constant pressure of 1.0 x 10^5 Pa, so the work done by the gas is 1.0 x 10^5 x 20 = 2.0 x 10^6 J.'
Why it loses marksThe volume is still in cm3. In SI units the volume change must be in m3: 20 cm3 = 20 x 10^-6 m3 = 2.0 x 10^-5 m3. Leaving the volume in cm3 multiplies the answer by a million, and the mark for the correct unit is lost as well.
The answer that scoresDelta V = 20 x 10^-6 = 2.0 x 10^-5 m3, so the work done by the gas is p Delta V = 1.0 x 10^5 x 2.0 x 10^-5 = 2.0 J, and the work done on the gas is -2.0 J.
Revise this topicThermodynamics
A student wrote · ThermodynamicsA student wrote: 'The internal energy of a substance can only change if its temperature changes.'
Why it loses marksInternal energy is the sum of the random kinetic AND potential energies of the molecules. During melting or boiling the temperature is constant but the internal energy rises, because the energy transferred by heating goes into the potential energy of the molecules. Work done on a gas raises its internal energy as well, and the first law covers both routes.
The answer that scoresInternal energy is the sum of the random kinetic and potential energies associated with the molecules. While ice melts at 0 degrees C the molecules gain potential energy as the forces between them are pulled apart, so the internal energy increases with no change of temperature, and a gas can have its internal energy raised by work done on it with q = 0.
Revise this topicThermodynamics
A student wrote · ThermodynamicsA student wrote: 'In one complete cycle of an engine Delta U = q + W cannot be zero, because the gas transfers energy to its surroundings by heating as it cools.'
Why it loses marksInternal energy is determined by the state of the system. After a complete cycle the gas is back in exactly the same state, with the same pressure, volume and temperature, so its internal energy is back to its starting value and Delta U must be zero. The energy given out by heating is balanced by the energy taken in by heating from the fuel and the net work done on the gas.
The answer that scoresDelta U = 0 over a complete cycle because internal energy is determined by the state of the system. Over the cycle, the net energy transferred to the gas by heating plus the net work done on the gas is zero: the net work done by the gas on its surroundings equals the net energy supplied to it by heating from the fuel.
Revise this topicThermodynamics
A student wrote · OscillationsA student wrote: 'At the point of maximum displacement the velocity and the acceleration of the mass are both zero, because the mass is momentarily at rest.'
Why it loses marksOnly the velocity is zero at the extremes. The acceleration there is as large as it ever gets, because a = -omega^2 x and the displacement magnitude is at its maximum.
The answer that scoresAt maximum displacement x = x0, so the speed is zero but the acceleration has magnitude omega^2 x0 and is directed towards the equilibrium position. The two quantities are out of step: the acceleration is zero where the speed is greatest, which is at the equilibrium position, and the speed is zero where the acceleration is greatest, which is at the extremes.
Revise this topicOscillations
A student wrote · OscillationsA student wrote: 'An oscillator has a frequency of 4.0 Hz, so its acceleration is a = -(4.0)^2 x.'
Why it loses marksThe equation a = -omega^2 x needs the angular frequency, not the frequency. The two differ by a factor of 2 pi, so omega is always greater than f and the acceleration comes out far too small if f is used.
The answer that scoresConvert first: omega = 2 pi f = 2 x 3.14 x 4.0 = 25 rad s-1. Then a = -omega^2 x = -(25)^2 x = -625 x, with x in metres and a in m s-2. The maximum speed comes out of the same constant, v0 = omega x0 = 25 x0, and the total energy is E = 1/2 m omega^2 x0^2 = 1/2 m x 625 x x0^2.
Revise this topicOscillations
A student wrote · OscillationsA student wrote: 'Doubling the amplitude of an oscillation doubles the total energy, because the energy is proportional to the amplitude.'
Why it loses marksThe total energy is proportional to the amplitude squared, E = 1/2 m omega^2 x0^2, not to the amplitude. Doubling the amplitude multiplies the total energy by four, because 2 squared is 4.
The answer that scoresE = 1/2 m omega^2 x0^2, so doubling x0 multiplies the energy by 2^2 = 4. What does double is the maximum speed, v0 = omega x0, because the energy depends on the square of the speed as well as on the square of the amplitude: E = 1/2 m v0^2 = 1/2 m (omega x0)^2, which is the same expression.
Revise this topicOscillations
A student wrote · OscillationsA student wrote: 'A heavily damped system returns to its equilibrium position more quickly than a critically damped one, because the damping force is larger.'
Why it loses marksThe damping force opposes the motion, so a very large one slows the return down. It is critical damping, not heavy damping, that gives the fastest return to equilibrium without overshoot.
The answer that scoresCritical damping returns the system to equilibrium in the shortest time without it crossing the equilibrium position, and it is used for door closers, the needle of a moving-coil galvanometer and car suspension. Heavy damping also avoids overshoot, but the very large resistive force prevents the system from building up speed, so the return to equilibrium is much slower. Light damping is the only one of the three that still oscillates, with the amplitude decaying gradually.
Revise this topicOscillations
A student wrote · Electric FieldsA student wrote: 'A charge of +2.0 x 10-6 C placed in a field of strength 5.0 x 10^3 V m-1 experiences a force of 5.0 x 10^3 N.'
Why it loses marksThe student has used the field strength as if it were the force. Field strength is the force on each unit of charge, so the force depends on how much charge is placed in the field.
The answer that scoresUse F = qE. F = 2.0 x 10-6 x 5.0 x 10^3 = 1.0 x 10-2 N, in the direction of the field because the charge is positive. The field strength itself is 5.0 x 10^3 N C-1 regardless of the charge placed in it.
Revise this topicElectric Fields
A student wrote · Electric FieldsA student drew the field between two oppositely charged parallel plates as a set of curved lines that fan out and get further apart as they cross the gap, saying that the field must be getting weaker.
Why it loses marksBetween the plates of a parallel-plate arrangement the field is uniform, so the lines must be straight, parallel and equally spaced, meeting both plates at right angles. Only near the open edges does the field fringe outwards, and that region is not the part being asked about.
The answer that scoresBetween the plates draw straight parallel lines, equally spaced, running from the positive plate to the negative plate and perpendicular to both. Equal spacing means equal strength everywhere, so the field is uniform. Fringing should appear only at the edges, and no two lines may ever cross.
Revise this topicElectric Fields
A student wrote · Electric FieldsA student wrote: 'Two point charges are 0.10 m apart and repel each other with a force of 0.90 N. If they are moved 0.20 m apart the force becomes 0.45 N, because the distance has doubled.'
Why it loses marksCoulomb's law is an inverse square law: the force is proportional to 1 / r2, not to 1 / r. Doubling the separation divides the force by 2^2 = 4, so it falls to one quarter of its original value, not one half.
The answer that scoresF = Q1Q2 / (4 pi epsilon0 r2). Doubling r multiplies F by 1 / 2^2 = 1 / 4, so the force becomes 0.90 / 4 = 0.22 N. The same inverse square reasoning applies to the field of a point charge, E = Q / (4 pi epsilon0 r2).
Revise this topicElectric Fields
A student wrote · Electric FieldsA student wrote: 'An electron released from rest between two charged plates moves towards the negative plate, because the electric field points from the positive plate to the negative plate.'
Why it loses marksThe direction of travel is set by the force, not by the field. The force is F = qE, and the charge on an electron is negative, so the force on it points the opposite way to the field.
The answer that scoresThe field points from the positive plate to the negative plate, so the force on an electron, which has a charge of -1.60 x 10-19 C, points from the negative plate towards the positive plate. The electron accelerates towards the positive plate with a = eE / m, gaining kinetic energy from the work done on it, qV, as it crosses the potential difference.
Revise this topicElectric Fields
A student wrote · CapacitanceA student wrote: 'A 470 microfarad capacitor charged to 6.0 V stores W = C V^2 = 470 x 10-6 x 6.0^2 = 0.017 J.'
Why it loses marksThe energy stored is one half of the charge multiplied by the potential difference, W = 1/2 Q V = 1/2 C V^2. The potential difference rises from zero to V while the capacitor charges, so only the average potential difference, V / 2, does work on the charge. Leaving out the one half doubles the answer.
The answer that scoresW = 1/2 C V^2 = 0.5 x 470 x 10-6 x 6.0^2 = 8.5 x 10-3 J. The charge stored is Q = C V = 470 x 10-6 x 6.0 = 2.8 x 10-3 C, and W = 1/2 Q V = 0.5 x 2.8 x 10-3 x 6.0 = 8.4 x 10-3 J, which agrees to two significant figures. Using two routes to the same energy is a quick check that the factor of one half is in the right place.
Revise this topicCapacitance
A student wrote · CapacitanceA student wrote: 'In series the combined capacitance is the sum of the capacitances, so 6.0 microfarad and 3.0 microfarad in series give 9.0 microfarad.'
Why it loses marksCapacitors in series store the same charge, not the same potential difference, so it is the reciprocals of the capacitances that add: 1 / C = 1 / C1 + 1 / C2. Adding the capacitances directly is the rule for a parallel connection, and in series it gives an answer larger than either capacitor, which cannot be right.
The answer that scores1 / C = 1 / 6.0 + 1 / 3.0 = 0.50 per microfarad, so C = 1 / 0.50 = 2.0 microfarad. A series combination always has a smaller capacitance than the smallest capacitor in it, because the same charge is pushed across two potential differences that add up to the supply voltage. In parallel the capacitances really do add: 6.0 + 3.0 = 9.0 microfarad.
Revise this topicCapacitance
A student wrote · CapacitanceA student wrote: 'A capacitor that stores 4.8 x 10-4 C when the potential difference across it is 12 V has a capacitance of C = Q / V = 4.8 x 10-4 / 12 = 4.0 x 10-5 microfarad.'
Why it loses marksDividing a charge in coulomb by a potential difference in volts gives a capacitance in farad, not in microfarad, so the numerical value 4.0 x 10-5 is a value in farad. Changing it to microfarad means multiplying by 10^6, and the student has simply tagged the wrong unit onto the number.
The answer that scoresC = 4.0 x 10-5 F = 4.0 x 10-5 x 10^6 microfarad = 40 microfarad. A capacitance of 4.0 x 10-5 microfarad would be 4.0 x 10-11 F, which is 40 picofarad, a million times too small. Always decide the unit before you convert the number, and write the unit with the final answer.
Revise this topicCapacitance
A student wrote · CapacitanceA student wrote: 'A capacitor is completely discharged after one time constant, because the time constant is the time taken for the capacitor to lose its charge.'
Why it loses marksThe time constant tau = RC is the time for the charge, potential difference or current to fall to 1 / e, about 0.37, of its initial value, not the time to reach zero. An exponential curve approaches zero without ever quite reaching it, so there is no time at which the capacitor is completely discharged.
The answer that scoresAfter one time constant the potential difference has fallen to 0.37 of its initial value, after two to 0.14, after three to 0.050 and after five time constants to about 0.0067 of it, less than one per cent, which is treated as fully discharged. A 47 microfarad capacitor discharging through 200 kohm has tau = 9.4 s, so the 12 V across it falls to 4.4 V after 9.4 s and to 0.60 V after 28.2 s.
Revise this topicCapacitance
A student wrote · Magnetic FieldsA student wrote: 'The force on a current-carrying wire in a magnetic field is F = BIL, so the force is the same whatever angle the wire makes with the field.'
Why it loses marksThe equation F = BIL applies only when the wire is at right angles to the field. In general the force is F = BIL sin theta, and the sine factor changes the answer whenever the wire is not perpendicular to the field.
The answer that scoresThe force is F = BIL sin theta, where theta is the angle between the wire and the direction of the field. The force is greatest when the wire is at right angles to the field, where sin theta is 1 and F = BIL, and it falls to zero when the wire lies along the field, where sin theta is 0. The force is always at right angles to both the wire and the field.
Revise this topicMagnetic Fields
A student wrote · Magnetic FieldsA student wrote: 'A coil of cross-sectional area 0.50 m2 sits in a uniform magnetic field of flux density 0.30 T, so the magnetic flux through the coil is B x A = 0.30 x 0.50 = 0.15 Wb.'
Why it loses marksThe student used the whole area, which would be correct only if the plane of the coil were at right angles to the field. The field makes an angle of 30 degrees with the plane of the coil, so the effective area at right angles to the field is smaller than the full area, and 0.15 Wb is too large.
The answer that scoresflux = B x A x cos theta, where theta is the angle between the field and the normal to the plane of the coil. The field lies at 30 degrees to the plane, so theta is 60 degrees and flux = 0.30 x 0.50 x cos 60 degrees = 0.15 x 0.50 = 0.075 Wb. Always use the area perpendicular to the field, or the component of the flux density perpendicular to the coil.
Revise this topicMagnetic Fields
A student wrote · Magnetic FieldsA student wrote: 'When a magnet is pushed into a coil, the induced current flows in the direction that opposes the magnetic flux of the magnet.'
Why it loses marksLenz's law says that the induced e.m.f. opposes the change in flux linkage, not the flux itself. A statement about opposing the flux does not explain why a magnet held still inside a coil produces no current at all.
The answer that scoresLenz's law states that the induced e.m.f. is in such a direction as to oppose the change producing it. Pushing the magnet in increases the flux linkage, so the induced current is in the direction whose own field reduces that increase; pulling the magnet out decreases the flux linkage, so the induced current reverses. Holding the magnet still produces no e.m.f., because the flux linkage is not changing. The law follows from the conservation of energy, since work must be done against the opposing force.
Revise this topicMagnetic Fields
A student wrote · Magnetic FieldsA student wrote: 'A charged particle moving in a uniform magnetic field gains kinetic energy, because the magnetic field does work on the charge.'
Why it loses marksThe magnetic force F = BQv sin theta is always at right angles to the velocity of the charge. A force at right angles to the motion does no work on the particle, so the kinetic energy cannot change.
The answer that scoresThe force on the charge is perpendicular to its velocity at every instant, so no work is done on the particle and its speed, and therefore its kinetic energy, stays the same. The force acts as a centripetal force, so only the direction of motion changes and the particle travels in a circle of radius r = mv/(BQ). What a magnetic field cannot do is change the speed of a charged particle.
Revise this topicMagnetic Fields
A student wrote · Alternating CurrentsA student wrote: 'The mains voltage is 230 V, so the potential difference across the insulation reaches 230 V at the peak of the cycle.'
Why it loses marks230 V is the root-mean-square value, the value of direct voltage that would deliver energy to a resistor at the same average rate. The waveform actually rises to a peak that is sqrt(2) times larger, and it is the peak that the insulation has to withstand without breaking down.
The answer that scoresThe peak value is V0 = V r.m.s. x sqrt(2) = 230 x 1.41 = 325 V, and the peak-to-peak value is 650 V. Insulation, breakdown and peak electric field questions use the peak value; mean power and energy questions use the r.m.s. value, so mean power = V r.m.s.^2 / R.
Revise this topicAlternating Currents
A student wrote · Alternating CurrentsA student wrote: 'A sinusoidal current of peak value 3.0 A flows in a 40 ohm resistor, so the power dissipated is 3.0^2 x 40 = 360 W.'
Why it loses marksThat is the maximum power, reached only at the two instants in each cycle when the current is at its peak value. What heats the resistor is the average rate of energy transfer over the whole cycle, and the current is below its peak for most of the time.
The answer that scoresThe mean power is half the maximum power: (1/2) x 3.0^2 x 40 = 180 W. Equivalently, the r.m.s. current is 3.0 / sqrt(2) = 2.1 A and I r.m.s.^2 R = 2.1^2 x 40 = 180 W. Both routes agree, and either is accepted.
Revise this topicAlternating Currents
A student wrote · Alternating CurrentsA student wrote: 'A step-down transformer can be used to reduce the steady 12 V output of a battery to 6 V for a small lamp.'
Why it loses marksA transformer works by mutual induction, so it needs a changing magnetic flux linking the secondary coil. The steady direct current from a battery produces a steady flux, so there is no changing flux linkage and no induced e.m.f. in the secondary at all, however many turns it has.
The answer that scoresA transformer only works on an alternating supply. To change the size of a steady direct voltage you must first convert it to alternating current with an inverter, which switches the direct current on and off electronically, then transform it and rectify the result back to direct current. That is exactly what a laptop power supply or a phone charger does, using a high-frequency transformer to keep the component small.
Revise this topicAlternating Currents
A student wrote · Alternating CurrentsA student wrote: 'Half-wave and full-wave rectification of the same sinusoidal supply give the same mean power in the load, because the peak value of the output is the same in both cases.'
Why it loses marksThe peak value is indeed the same, but a half-wave rectifier delivers energy to the load only during half of each cycle, so one entire half cycle contributes nothing at all. Averaging the square of the output over a whole cycle therefore halves the answer, and the heat produced is halved with it.
The answer that scoresFor full-wave rectification the mean of sin^2(omega t) over a cycle is 1/2, so the mean power is (1/2) I0^2 R. For half-wave rectification the load is connected for only half the time, so the mean square is a quarter of the peak square and the mean power is (1/4) I0^2 R, exactly half of the full-wave value. Full-wave rectification therefore delivers twice the power from the same supply and gives a much smoother output, with twice as many pulses per cycle.
Revise this topicAlternating Currents
A student wrote · Quantum PhysicsA student wrote: 'Increasing the intensity of the light increases the maximum kinetic energy of the photoelectrons, because more energy arrives each second.'
Why it loses marksThe energy carried away by one photoelectron comes from one photon, and the energy of a photon depends only on the frequency of the light. Intensity changes the number of photons arriving each second, so it changes the number of electrons released and the photoelectric current, not the energy each one leaves with.
The answer that scoresThe maximum kinetic energy depends only on the frequency and on the metal: KE max = hf - work function energy. Doubling the intensity doubles the number of photons per second, so twice as many electrons are released and the current doubles, but each electron still receives only one photon's worth of energy hf, so KE max and the stopping potential are unchanged.
Revise this topicQuantum Physics
A student wrote · Quantum PhysicsA student calculated the energy of a photon of wavelength 500 nm as E = hc / lambda = (6.63 x 10-34 x 3.00 x 10^8) / 500 = 4.0 x 10-28 J.
Why it loses marksThe wavelength was left in nanometres. Every quantity put into an equation must be in SI units, so the wavelength has to be converted into metres before it is substituted, and 500 nm is 5.00 x 10-7 m, not 500 m.
The answer that scores500 nm = 500 x 10-9 m = 5.00 x 10-7 m, so E = (6.63 x 10-34 x 3.00 x 10^8) / (5.00 x 10-7) = 3.98 x 10-19 J. That is about 2.5 eV, the sort of energy a visible photon should carry, whereas 4.0 x 10-28 J is far too small to be physically sensible.
Revise this topicQuantum Physics
A student wrote · Quantum PhysicsA student wrote: 'If the light is intense enough, photoelectrons will be emitted from a metal even when the frequency is below the threshold frequency.'
Why it loses marksBelow the threshold frequency every photon arriving has an energy hf that is smaller than the work function energy of the metal. One photon is absorbed by one electron and a photon cannot be split or shared with others, so no electron can ever collect enough energy to escape, however many photons arrive or however long the metal is illuminated.
The answer that scoresNo emission occurs at any intensity below the threshold frequency. Emission begins only when f is greater than f0, which is the same condition as lambda being shorter than lambda0. Above the threshold, increasing the intensity increases the number of photoelectrons released each second and so the current, but it leaves the maximum kinetic energy unchanged.
Revise this topicQuantum Physics
A student wrote · Quantum PhysicsA student wrote: 'When a hydrogen electron falls from the n = 3 level to the n = 2 level, the atom emits a photon of energy 3.40 + 1.51 = 4.91 eV.'
Why it loses marksThe photon energy is the difference between the two levels, not their sum. The levels are quoted as negative energies measured from the ionisation limit at 0, and the energy released is E(higher level) - E(lower level) = -1.51 - (-3.40) = 1.89 eV.
The answer that scoresThe energy released is 1.89 eV, which is 1.89 x 1.60 x 10-19 = 3.02 x 10-19 J. The wavelength is lambda = hc / energy = (6.63 x 10-34 x 3.00 x 10^8) / (3.02 x 10-19) = 6.58 x 10-7 m, a red line of the Balmer series, and it lies in the visible region because 658 nm is inside the range of visible wavelengths.
Revise this topicQuantum Physics
A student wrote · Nuclear PhysicsA student wrote: 'The mass defect of a nucleus is the mass that is destroyed when the nucleus forms, so a helium nucleus is lighter than its protons and neutrons because some of them have disappeared.'
Why it loses marksNo nucleons disappear. A helium nucleus contains exactly two protons and two neutrons, and if you count them you get the same six quarks-plus energy bookkeeping as before. The mass defect is a difference between two masses, not a loss of particles, and it is accounted for by the binding energy: the nucleus is in a lower energy state, and a lower energy state means a lower mass.
The answer that scoresThe mass defect is the difference between the total mass of the separate protons and neutrons and the mass of the nucleus: delta m = (2 x 1.00728 + 2 x 1.00867) - 4.00151 = 4.03190 - 4.00151 = 0.03039 u. The nucleons are all still there. The missing mass appears as the binding energy, E = c2 x delta m = 9.00 x 10^16 x 0.03039 x 1.66 x 10-27 = 4.54 x 10-12 J, which is 28.4 MeV. That is the energy you would have to supply to pull the helium nucleus apart again.
Revise this topicNuclear Physics
A student wrote · Nuclear PhysicsA student wrote: 'The half-life of iodine-131 is 8.0 days, so its decay constant is lambda = 0.693 x 8.0 = 5.5 per day.'
Why it loses marksThe relation is lambda = 0.693 / t(1/2), so the half-life is divided into 0.693, not multiplied by it. The student's answer is also physically wrong: a long half-life means a stable isotope and a small decay constant, so a large half-life must give a small lambda, not a large one. The unit is wrong too, because the activity in a calculation will be in Bq and needs a decay constant in s-1.
The answer that scoresConvert the half-life to seconds: 8.0 x 24 x 60 x 60 = 6.9 x 10^5 s. Then lambda = 0.693 / 6.9 x 10^5 = 1.0 x 10-6 s-1. The decay constant is a probability per second for a single nucleus, so it must be much smaller than 1, and it must carry the unit s-1.
Revise this topicNuclear Physics
A student wrote · Nuclear PhysicsA student wrote: 'After four half-lives a source has decayed four times, so its activity is zero and nothing radioactive is left.'
Why it loses marksEach half-life halves what remains, it does not subtract a fixed amount. After four half-lives a sixteenth of the nuclei are still undecayed, and an exponential curve never reaches the axis, so there is always some radioactive material left. Saying the activity is zero after four half-lives would also break the exponential law, which has no time at which x becomes exactly zero.
The answer that scoresThe fraction remaining after n half-lives is (1/2) raised to the power n, so after four half-lives 1/16 of the nuclei remain and the activity has fallen to 1/16 of its original value. An activity of 8.0 x 10^9 Bq would fall to 5.0 x 10^8 Bq. The activity keeps approaching zero but never reaches it, which is why radioactive waste remains hazardous for many half-lives.
Revise this topicNuclear Physics
A student wrote · Nuclear PhysicsA student plotted the count rate from a source against time, read 700 counts per minute at t = 0 and 175 counts per minute after 12 minutes, and wrote: 'The count rate halves twice, from 700 to 350 to 175, so the half-life is 12 / 2 = 6.0 minutes.'
Why it loses marksThe readings still contain background radiation, which does not decay. Halving the raw readings counts the background in at the start and then forces it to halve as well, which it does not do, so the number of halvings is underestimated and the half-life comes out too long. The background must be measured with the source removed and subtracted from every reading before the curve is used.
The answer that scoresWith a background count rate of 100 counts per minute, the corrected readings are 700 - 100 = 600 and 175 - 100 = 75 counts per minute. The halvings are 600 -> 300 -> 150 -> 75, which is three half-lives in 12 minutes, so the half-life is 12 / 3 = 4.0 minutes, not 6.0 minutes. The curve should also have been plotted to check that it falls smoothly and flattens at the background level rather than at zero.
Revise this topicNuclear Physics
A student wrote · Medical physicsA student wrote: 'A boundary is 3.0 cm below the surface of the tissue and the speed of ultrasound is 1540 m s-1, so the echo is received 0.030 / 1540 = 19 microseconds after the pulse is transmitted.'
Why it loses marksThe pulse does not stop at the boundary. It travels down to the boundary and back again, so the path length is twice the depth. Using the depth as the path length halves the calculated time.
The answer that scoresThe path length is 2 x 0.030 = 0.060 m, so the time is 0.060 / 1540 = 3.9 x 10-5 s = 39 microseconds. Going the other way, a measured time of 39 microseconds corresponds to a depth of 1540 x 39 x 10-6 / 2 = 0.030 m = 3.0 cm, which agrees. Always write depth = c t / 2 and substitute the time, never the depth.
Revise this topicMedical physics
A student wrote · Medical physicsA student wrote: 'In PET scanning the positrons emitted by the tracer travel out of the body and are detected by the ring of detectors, so the position of the tracer can be worked out.'
Why it loses marksA positron is a charged particle, so it interacts strongly with the tissue it passes through and has a range of only about a millimetre. It annihilates long before it reaches the surface, and it is the two gamma-ray photons produced in the annihilation that escape and are detected.
The answer that scoresThe positron travels about 1 mm through the tissue and then annihilates with an electron. The annihilation converts the two rest masses into two gamma-ray photons of about 5.1 x 10^5 eV each, travelling in opposite directions. Those photons are penetrating enough to leave the body, and a coincidence recorded by two detectors on opposite sides of the patient shows that an annihilation took place somewhere on the line between them.
Revise this topicMedical physics
A student wrote · Medical physicsA student wrote: 'An X-ray beam loses half of its intensity in the first 2 cm of bone, so after 4 cm of bone the intensity is zero because all of it has been absorbed.'
Why it loses marksAttenuation is exponential, not a fixed loss per centimetre. Each successive layer of material transmits the same fraction of what reaches it, so the intensity falls by an ever smaller amount and never reaches zero.
The answer that scoresUse I = I0 e^(-mu x). With mu = 60 m-1, 2.0 cm gives I / I0 = e^(-1.2) = 0.30, so 30% is transmitted, and 4.0 cm gives I / I0 = e^(-60 x 0.040) = e^(-2.4) = 0.091, so 9.1% is transmitted. The intensity is never zero, because e^(-mu x) only approaches zero as x becomes very large.
Revise this topicMedical physics
A student wrote · Medical physicsA student wrote: 'A higher ultrasound frequency gives a clearer image because it is not attenuated as much, so more of it reaches the deeper tissues.'
Why it loses marksThe attenuation coefficient increases with frequency, so a higher-frequency beam loses intensity faster, not more slowly. Its advantage is a shorter wavelength, which gives better resolution and lets two nearby boundaries be seen as separate echoes.
The answer that scoresLarger mu means a smaller I for the same value of x, so higher-frequency ultrasound penetrates less deeply. The benefit of the higher frequency is resolution: two boundaries closer together than about one wavelength merge into a single echo. Choosing a frequency is therefore a compromise between resolution and depth - about 2 MHz to 5 MHz for an abdomen, but 10 MHz to 20 MHz for an eye.
Revise this topicMedical physics
A student wrote · Astronomy and CosmologyA student wrote: 'The H-beta line of hydrogen has a laboratory wavelength of 486.1 nm and is observed at 489.0 nm in a galaxy spectrum, so the recession speed is v = c x lambda / delta_lambda = 3.00 x 10^8 x 486.1 / 2.9 = 5.0 x 10^10 m s-1.'
Why it loses marksThe ratio has been turned upside down. The redshift is the change in wavelength divided by the original laboratory wavelength, delta_lambda / lambda, and the answer of 5.0 x 10^10 m s-1 is about 170 times the speed of light, which is impossible and should have been spotted at once.
The answer that scoresdelta_lambda = 489.0 - 486.1 = 2.9 nm, so v = c delta_lambda / lambda = 3.00 x 10^8 x (2.9 / 486.1) = 1.8 x 10^6 m s-1. The redshift fraction is 2.9 / 486.1 = 5.97 x 10^-3, well under one per cent, which is typical for a galaxy a few tens of Mpc away.
Revise this topicAstronomy and Cosmology
A student wrote · Astronomy and CosmologyA student wrote: 'The redshift of the distant galaxies shows that they are rushing through space away from the Milky Way, so the Milky Way must be at the centre of the Universe.'
Why it loses marksThe redshift of a distant galaxy is not the Doppler effect of a galaxy flying through space. Space itself is expanding, and the wavelengths of the light are stretched while the light is travelling. The second half of the statement is also wrong: the pattern of recession speeds proportional to distance looks the same from every galaxy, so there is no centre, and the Milky Way is in no way special.
The answer that scoresThe lines are at longer wavelengths than their laboratory values because the space the light travels through expands while the light is in transit, stretching the wavelength. Measured from any galaxy, every distant galaxy is receding with a speed proportional to its distance, v = H0 d. That pattern has no centre and no edge, and it is the expansion of space itself rather than the motion of the Milky Way that produces the redshift.
Revise this topicAstronomy and Cosmology
A student wrote · Astronomy and CosmologyA student wrote: 'The Hubble constant is 68 km s-1 Mpc-1, so the age of the Universe is 1 / 68 = 0.015 years.'
Why it loses marksThe Hubble constant in the form km s-1 Mpc-1 is not in s-1, so its reciprocal is not a time in years and does not even have the right units. The distance unit Mpc must be converted to metres and the speed unit km s-1 to m s-1 before the reciprocal means anything.
The answer that scoresH0 = 68 km s-1 Mpc-1 = 6.8 x 10^4 m s-1 / 3.1 x 10^22 m = 2.2 x 10^-18 s-1. Then t = 1 / H0 = 4.5 x 10^17 s, and dividing by 3.15 x 10^7 s per year gives 1.4 x 10^10 years, about 14 billion years. The unit check matters: 1 / H0 in s-1 gives seconds, and seconds only become years after you divide by the number of seconds in a year.
Revise this topicAstronomy and Cosmology
A student wrote · Astronomy and CosmologyA student wrote: 'Sirius looks brighter than Betelgeuse in the night sky, so Sirius must have the greater luminosity.'
Why it loses marksApparent brightness is the flux intensity received at the Earth, which depends on the distance as well as on the luminosity. A nearby star of modest luminosity can easily outshine a distant supergiant, so flux intensity on its own says nothing about luminosity.
The answer that scoresLuminosity and flux intensity are linked by F = L / (4 pi d^2), so you cannot compare luminosities without the distances. Sirius is the brightest star in the night sky largely because it is only 8.6 ly away, with a luminosity of about 25 times the Sun's, whereas Betelgeuse is far more luminous, radiating tens of thousands of times as much power as the Sun, but it is about 550 ly away and so looks much fainter. Only for two stars at the same distance does the brighter one have the greater luminosity.
Revise this topicAstronomy and Cosmology