Issues from Paper 1 and Paper 2 raised in the Principal Examiner’s Report

There are three versions of Paper 1 and three versions of Paper 2 to accommodate the candidates in the different time zones around the world. This document highlights those questions across all three versions of each paper that have been identified by the Principal Examiner as being found difficult by candidates.

Key messages• The use of precise terms in physics is important when stating definitions and explaining phenomena.

Candidates at this level should be encouraged to use the correct words or terms in formal definitions.Language that is often acceptable in everyday speech does not necessarily convey correct meanings.For example, in physics, there are clear differences between distance and displacement, speed and velocity.

• Candidates should be encouraged to start numerical questions with the defining equation and give their working at each stage. In this way partial credit can be awarded if the final answer is incorrect.

• Power-of-ten errors are a relatively common cause of lost credit. Candidates should be encouraged to take a moment to consider whether their answers are of a reasonable order of magnitude, as this check can often detect mistakes with powers-of-ten or arithmetic.

General commentsThe marks scored by candidates varied over almost the whole mark range. There were parts of some questions that were testing. Well-prepared candidates were able to show their understanding of the relevant concepts. Other parts of questions were straightforward and allowed weaker candidates to gain credit. Candidates were generally competent in dealing with calculations, although a common cause of incorrect answers was in power-of-ten errors. In descriptive questions many candidates would have benefited from having more thorough knowledge and understanding to give accurate and precise answers.

Definitions and explanations – the issue that these questions have in common is the generally imprecise language used by candidates to answer them.

9072/21 Q1aiDefine velocity. [1 mark]

Examiner report: Velocity should be defined in terms of rate of change of displacement. In many scripts, reference was made to ‘distance moved’. Where a ratio is involved, as in this case, candidates should be discouraged from using loose wording such as ‘over time’.

9072/21 Q1aiiDistinguish between speed and velocity. [2 marks]

Examiner report: Rather than refer to ‘magnitude’ and ‘magnitude and direction’, many candidates gave their answers by categorising the quantities as a scalar or a vector. Stating that velocity is ‘speed in a given direction’ does not distinguish clearly between the two.

9702/21 Q2aiDefine power.

Examiner report: Candidates should appreciate that power involves a transfer of energy. Thus, power should be defined in terms of energy transfer or work done, not just ‘energy’.

9702/21 Q2aiiUse your definition in (i) to show that power may also be expressed as the product of force and velocity.

Examiner report: In any derivation, it is important to explain the working, especially the symbols used. It was very common to find that this derivation was completed using symbols without explanation, or that nonstandard symbols were used rather than those listed in the syllabus.

9702/21 Q3aiA uniform plank AB of length 5.0 m and weight 200 N is placed across a stream, as shown in Fig. 3.1.

A man of weight 880 N stands a distance x from end A. The ground exerts a vertical force FA on the plank at end A and a vertical force FB on the plank at end B.As the man moves along the plank, the plank is always in equilibrium.Explain why the sum of the forces FA and FB is constant no matter where the man stands on the plank.

Examiner report: There were very few complete answers and many candidates merely stated that ‘upward forces equal downward forces’. Candidates should be advised to use more precise terminology. Stronger candidates discussed the direction and summation of the various forces.

9702/21 Q5aExplain what is meant by the following quantities for a wave on the surface of water:(i) displacement and amplitude(ii) frequency and time period. [4 marks]

Examiner report: Some candidates lost credit as a result of imprecise terminology. This is relatively common inquestions on waves. Candidates would benefit from learning the meanings of terms used to describe waves in more detail.

9702/21 Q5ciiState and explain whether the waves on the surface of the water shown in Fig. 5.1 are transverse or longitudinal.

[1 mark]

Examiner report: The great majority of answers included a reference to transverse waves. However, the reasoning was frequently inadequate. Candidates were expected to state that the direction of the vibrations is normal to the direction of travel of wave energy. Inappropriate statements such as ‘the particles move at right angles to the direction of the wave’ were common.

9207/21 Q6aDistinguish between electromotive force (e.m.f.) and potential difference (p.d.). [2 marks]

Examiner report: Many answers did not make reference to energy changes. Where energy changes were identified, few went on to state that both e.m.f. and p.d. are defined in terms of some form of energy change per unit charge.

9207/21 Q7aState what is meant byα-particle: β-particle:

γ-radiation: [2 marks]

Examiner report: In many answers, α-particles were stated as being helium ions or particles. It is more precise to refer to helium nuclei. Similarly, γ-radiation was often said to be ‘energy’, whereas a more precise answer is to say that it is electromagnetic radiation or photons.

9207/22 Q3aState Newton’s first law of motion. [1 mark]

Examiner report: Many candidates were imprecise in this definition. The majority omitted constant speed in a straight line or the resultant external force.

9207/22 Q3biA box slides down a slope, as shown in Fig. 3.1.

The angle of the slope to the horizontal is 20°. The box has a mass of 65 kg. The total resistive force R acting on the box is constant as it slides down the slope.State the names and directions of the other two forces acting on the box. [2 marks]

Examiner report: Many candidates gave imprecise directions for the two forces or no directions at all. A normal force upwards and a weight downwards were considered to be given with insufficient detail, as it is unclear whether “up” and “down” refer to the slope or to the vertical.

9702/22 Q7aA laser is placed in front of a double slit, as shown in Fig. 7.1.

The laser emits light of frequency 670 THz. Interference fringes are observed on the screen.Explain how the interference fringes are formed. [3 marks]

Examiner report: The answers often lacked the detail required at this level. ‘Rays’ or ‘light’ were terms often used instead of waves. Candidates should be encouraged to use correct terms when asked for an explanation. The need for the waves to be coherent after passing through the slits was seldom described by candidates. The description of ‘crest on crest’ for a maximum is not sufficient at this level. A description involving path or phase difference between the waves that reach the screen should be encouraged. There were a significant number of candidates who confused path and phase difference. Repeating the words in the question without any further explanation will not gain any credit and should be discouraged.

9702/23 Q3a and biExplain what is meant by work done. [1 mark]

Examiner report: Many candidates gave an answer that was not sufficiently precise at this level. The explanation should refer to a force moving a body in the direction of the force. Work done being described merely as the ‘product of force and distance’ does not gain credit (this could also describe a moment).

A boy on a board B slides down a slope, as shown in Fig. 3.1.The angle of the slope to the horizontal is 30°. The total resistive force F acting on B is constant.

The angle of the slope to the horizontal is 30°. The total resistive force F acting on B is constant.State a word equation that links the work done by the force F on B to the changes in potential and kinetic energy.

[1 mark]Examiner report: Very few candidates followed the instruction to give a word equation in terms of the quantities work done, change in kinetic energy and change in potential energy. Those that did were often not specific with the sign of the change for the kinetic and potential energies, and hence it was not clear whether the change was a gain or a loss.

9702/23 Q4aA spring hangs vertically from a point P, as shown in Fig. 4.1.

Explain how the apparatus in Fig. 4.1 may be used to determine the load on the spring at the elastic limit. [2 marks]

Examiner report: Very few candidates described a method of determining the load at the elastic limit. A majority described how the limit of proportionality or Hooke’s law could be determined. There were a small number who described adding loads in small amounts and removing the load at each stage to see if the spring returned to its original length.

9702/23 Q6aii, aiii and cA hollow tube is used to investigate stationary waves. The tube is closed at one end and open at the other end. A loudspeaker connected to a signal generator is placed near the open end of the tube, as shown in Fig. 6.1.

The tube has length L. The frequency of the signal generator is adjusted so that the loudspeaker produces a progressive wave of frequency 440 Hz. A stationary wave is formed in the tube. A representation of this stationary wave is shown in Fig. 6.1.Two points P and Q on the stationary wave are labelled.

Explain how the stationary wave is formed in the tube. [3 marks]

Examiner report: There were many answers that described the general formation of stationary waves without any reference to the situation in the question. Candidates should be encouraged to describe phenomena with specific reference to the question. Very few candidates gained full credit for this part.

State the direction of the oscillations of an air particle at point P. [1 mark]

Examiner report: A small minority of candidates recognised that the stationary wave in this question had been formed by a sound wave and that the vibration of particles was parallel to the length of the tube.

State the phase difference between points P and Q on the stationary wave. [1 mark]

Examiner report: The concept of phase difference within a loop of a stationary wave was poorly understood by the majority of candidates. Candidates would benefit from further study of the phase relationships in a stationary wave.

Base quantities and units and uncertainties – many questions in these sections were answered well, but the Principal examiner has identified a significant number as causing problems for candidates.

9702/11 Q1Which pair of units contains one derived unit and one SI base unit?

A ampere coulombB kilogram kelvinC metre secondD newton pascal

Examiner report: This question tested straightforward recall but many candidates did not choose the correct answer.

9702/11 Q2What is equivalent to 2000 microvolts?

A 2μJC-1 B 2 mV C 2 pV D 2000 mV

Examiner report: This type of routine question needs to be practised more carefully, not only for the multiple choice paper but also for the written papers. Candidates often do not notice impossibly incorrect answers because of incorrect units or powers of ten.

9702/13Which quantity can be measured in electronvolts (eV)?

A electric chargeB electric potentialC energyD power

Examiner report: Many candidates chose answer B, perhaps connecting “electro...” or “...volt” in the stem with the electrical answers.

9702/23 Q1aUnderline all the base quantities in the following list. [2 marks]

ampere charge current mass second temperature weight

Examiner report: A considerable number of answers included the base units instead of the base quantities. A significant number considered charge to be a base quantity. Mass was frequently omitted from the list of correct answers.

9702/11 Q5The Young modulus of the material of a wire is to be found. The Young modulus E is given by the equation below.

E = 4 Fl πd2x

The wire is extended by a known force and the following measurements are made.Which measurement has the largest effect on the uncertainty in the value of the calculated Young modulus?

Examiner report: Many candidates did not double the percentage uncertainty in the diameter.

9702/12 Q5A thermometer can be read to an accuracy of ± 0.5 °C. This thermometer is used to measure a temperature rise from 40 °C to 100 °C.What is the percentage uncertainty in the measurement of the temperature rise?

A 0.5 % B 0.8 % C 1.3 % D 1.7 %

Examiner report: Almost half of the candidates assumed that the uncertainty was 0.5 °C in the temperature difference of 60 °C, but this is only half of the uncertainty since this happens at both the low and the high temperatures. The correct answer is D.

9702/22 Q2bi and iiA coin is made in the shape of a thin cylinder, as shown in Fig. 2.1.

Fig. 2.2 shows the measurements made in order to determine the density ρ of the material used to make the coin.

Calculate the percentage uncertainty in ρ. [3 marks]

Examiner report: The majority of candidates found this part difficult. A significant proportion of candidates did not seem to understand the process for calculating a total percentage uncertainty. Common mistakes were to give the fractional uncertainty instead of the percentage uncertainty and not including all the quantities used to calculate the density. Many candidates did not use twice the percentage uncertainty for the diameter or used the same uncertainty in the radius as was given for the diameter.

State the value of ρ with its actual uncertainty. [1 mark]

Examiner report: Many candidates found this part difficult. The majority of answers gave two or more significant figures for the uncertainty or an answer for the density that did not have a value to the same decimal place as the uncertainty.

Projectile motion – although some aspects of these questions were answered well the independence of the horizontal and vertical components seems to have caused some problems.

9702/11 Q6A tennis ball is thrown horizontally in air from the top of a tall building.If the effect of air resistance is not negligible, what happens to the horizontal and vertical components of the ball’s velocity?

Examiner report: Many candidates assumed that the horizontal component of the velocity was constant. These candidates would have benefited from careful reading of the question, noting that “air resistance is not negligible”.

9702/11 Q7An object is thrown with velocity 5.2 m s–1 vertically upwards on the Moon. The acceleration due to gravity on the Moon is 1.62 ms-2 What is the time taken for the object to return to its starting point?

A 2.5 s B 3.2 s 4.5 s D 6.4 s

Examiner report: A common mistake was to choose B. Candidates choosing B correctly calculated the time for the object to reach its maximum height, but then did not double this to give the total time taken to return to its starting point.

9702/22 Q4bi2A ball of mass 400 g is thrown with an initial velocity of 30.0 m s–1 at an angle of 45.0° to the horizontal, as shown in Fig. 4.1.

Calculate the maximum height H of the ball. [2 marks]

Examiner report: A significant number of candidates showed a basic misunderstanding of this type of motion. The common errors were to use the initial velocity instead of the component in the vertical direction, to equate the loss in kinetic energy with the gain in potential energy or to use a positive acceleration.

9702/23 Q2biiiA ball leaves point P at the top of a cliff with a horizontal velocity of 15 m s–1, as shown in Fig. 2.1.

The height of the cliff is 25 m. The ball hits the ground at point Q.Air resistance is negligible.Calculate the magnitude of the displacement of the ball at point Q from point P. (candidates have already shown that it takes 2.3 seconds for the ball to fall to the ground) [4 marks]

Examiner report: Candidates who had practised this type of calculation generally scored high marks. A considerablenumber of candidates were unable to calculate the horizontal displacement of the ball. Many used an equation of constant acceleration. The idea that the displacements in the horizontal and vertical directions could be treated independently was missed by a large majority of candidates.

The significance of direction – there are many situations where the vector nature of a quantity is important and this often seems to be missed.

9702/11 Q15A weight W hangs from a trolley that runs along a rail. The trolley moves horizontally through a distance p and simultaneously raises the weight through a height q.

As a result, the weight moves through a distance r from X to Y. It starts and finishes at rest.How much work is done on the weight during this process?

A Wp B W(p + q) C Wq D Wr

Examiner report: It is likely that many candidates would have realised that Wq was the gain in potential energy, but many candidates chose Wr as the work done. These candidates should be encouraged to remember that the work done is the change in potential energy, or alternatively that work done = force × distance moved in the direction of the force.

9702/11 Q29The diagram shows two points P and Q which lie 90° apart on a circle of radius r.A positive point charge at the centre of the circle creates an electric field of magnitude E at both P and Q.

Which expression gives the work done in moving a unit positive charge from P to Q?

A 0 B E x r C E x (πr/2) D E x (πr)

Examiner report: Answer C was common, and this showed a similar misunderstanding to Question 15. As the charge moves from P to Q, the force is always at right-angles to the motion, and so no work is done, giving A as the correct answer.

9702/12 Q12A tiny oil droplet with mass 6.9 × 10–13 kg is at rest in an electric field of electric field strength 2.1 × 107 N C–1, as shown.

The weight of the droplet is exactly balanced by the electrical force on the droplet.What is the charge on the droplet?

A 3.3 × 10–20 CB –3.3 × 10–20 CC 3.2 × 10–19 CD –3.2 × 10–19 C

Examiner report: Most did the arithmetic of this question correctly, but many candidates did not take into account the fact that the arrows on an electric field diagram show the direction of the force on a positive charge. Here the charge must be negative.

9702/13 Q31An electron enters a region of space where there is a uniform electric field E as shown. Initially, the electron is moving parallel to, and in the direction of, the electric field.What is the subsequent path and change of speed of the electron?

Examiner report: The force on the electron is opposite to the direction of its motion. This causes deceleration but the path will be linear, so the answer is A.

9207/12 Q16An escalator is 60 m long and lifts passengers through a vertical height of 30 m, as shown.

To drive the escalator against the forces of friction when there are no passengers requires a power of 2.0 kW.The escalator is used by passengers of average mass 60 kg and the power to overcome friction remains constant.How much power is required to drive the escalator when it is carrying 20 passengers and is travelling at 0.75 m s–1?

A 4.4 kW B 6.4 kW C 8.8 kW D 10.8 kW

Examiner report: The gain in potential energy per second depends on the vertical distance moved per second. Two common mistakes were to use 60 m rather than 30 m and forgetting to add on the 2.0 kW.

9702/21 Q1cThe direction of motion of the car in (b) at time t = 2.0 s is shown in Fig. 1.2.

On Fig. 1.2, show with arrows the directions of the acceleration (label this arrow A) and the resultant force (label this arrow F). [1 mark]

Examiner report: It was common to see the two arrows pointing in opposite directions. Candidates should bereminded that resultant force and acceleration are always in the same direction.

The particle nature of matter – There were not many questions on this topic in the June 2014 papers, but those that there were seemed to have caused problems.

9702/11 Q18There is one temperature, about 0.01 °C, at which water, water vapour and ice can co-exist in equilibrium.Which statement about the properties of the molecules at this temperature is correct?

A Ice molecules are closer to one another than water molecules.B The mean kinetic energy of water molecules is greater than the mean kinetic energy of ice molecules.C Water vapour molecules are less massive than water molecules.D Water vapour molecules have the same mean speed as both ice and water molecules.

Examiner report: It is worth emphasising to candidates that the molecules in a solid and those in a liquid have similar spacing.Ice is slightly less dense than water at 0 °C (it floats on water), so its molecules are slightly further apart than molecules in water.

9702/13 Q19When the water in a pond freezes, it changes from a liquid to a solid. When this occurs, it changes volume and exchanges energy with the surroundings.Which row is correct?

Examiner report: Water expands when it freezes and becomes ice. This should have enabled candidates to rule out options A and B.

Manipulating equations and using graphs – this covers a wide range of topics, but seems to cause problems in whichever context it comes up.

9702/11 Q24The speed v of waves in deep water is given by the equation

v2 = g 2π

where λ is the wavelength of the waves and g is the acceleration of free fall.A student measures the wavelength λ and the frequency f of a number of these waves.Which graph should he plot to give a straight line through the origin?

A f2 against B f against 2

C f against 1/D f2 against 1/

Examiner report: The manipulation of this equation into f2 = g/2πλ caused problems. A large number of candidates gave the incorrect answer C.

9702/12 Q29Two oppositely-charged horizontal metal plates are placed in a vacuum. A positively-charged particle starts from rest and moves from one plate to the other plate, as shown.

Which graph shows how the kinetic energy EK of the particle varies with the distance x moved from the positive plate?

Examiner report: The field is uniform and so the electrical potential energy decreases uniformly with distance. The kinetic energy therefore increases uniformly too, which is answer D. This answer could also be obtained by reasoning that a constant force gives a constant acceleration, and then using v2 = 2ax shows that the kinetic energy is proportional to x. Many candidates may have assumed that the horizontal axis represents time, and these candidates would have benefited from more careful reading of the question.

9702/13 Q10A tennis ball is dropped onto a table and bounces back up. The table exerts a force F on the ball.Which graph best shows the variation with time t of the force F while the ball is in contact with the table?

Examiner report: Candidates found this question difficult. The table does not exert any force on the ball at the instant they make contact or at the instant they separate. The force must gradually increase and then decrease between these points, so only C could be correct.

9702/13 Q16An electric motor has an input power Pin, useful output power Pout and efficiency η.

How much power is lost by the motor?

Electric circuits and networks – this was one of the most heavily examined sections of the syllabus and although many questions were answered well, those with a twist seemed to cause problems.

9702/11 Q31A battery, with a constant internal resistance, is connected to a resistor of resistance 250 Ω, as shown.

The current in the resistor is 40 mA for a time of 60 s. During this time 6.0 J of energy is lost in the internal resistance.What are the energy supplied to the external resistor during the 60 s and the e.m.f. of the battery?

Examiner report: Candidates found this question difficult. The energy transferred in the external resistor is I2Rt which gives 24 J. The battery must therefore supply a total of 30 J of energy to the whole circuit. The charge transferred is It (= 2.4 C) and so the e.m.f. of the battery is (30 J / 2.4 C) = 12.5 V, answer D.

9702/13 Q34A cell has an electromotive force (e.m.f.) of 6 V and internal resistance R. An external resistor, also of resistance R, is connected across this cell, as shown.

Power P is dissipated by the external resistor.The cell is replaced by a different cell that has an e.m.f. of 6 V and negligible internal resistance.What is the new power that is dissipated in the external resistor?

A 0.5P B P C 2P D 4P

Examiner report: Replacing the cell with a different cell with negligible internal resistance gives half the total resistance, so double the current. The power dissipated in the external resistor is I2 R, so it must increase by a factor of 4, giving D.

9702/11 Q33In the circuit below, a voltmeter of resistance Rv and an ammeter of resistance RA are used to measure the resistance R of the fixed resistor.

Which condition is necessary for an accurate value to be obtained for R?

A R is much smaller than RV.B R is much smaller than RA.C R is much greater than RV.D R is much greater than RA.

Examiner report: The resistance of the ammeter has no effect because the voltmeter is connected to measure the potential difference across only the fixed resistor. The voltmeter must have a much greater resistance than R, otherwise there will be a significant current in the voltmeter and the value obtained from V/I will not be an accurate value for R.

9702/23 Q5aExplain why the terminal potential difference (p.d.) of a cell with internal resistance may beless than the electromotive force (e.m.f.) of the cell. [2 marks]

Examiner report: There were numerous answers that did not make reference to energy being used in the battery or ‘lost volts’ in the battery due to the internal resistance. There were very few answers that added that this only occurs when the battery supplies a current. The idea that the terminal potential difference is numerically equal to the electromotive force when the battery is not supplying a current was seldom mentioned.

9702/11 Q36What is the total resistance between points P and Q in this network of resistors?

A 8 Ω B 16 Ω C 24 Ω D 32 Ω

Examiner report: This question requires careful thought. The two ‘corners’ of the circuit each have resistance 32 Ω, so their combined resistance in parallel is 16 Ω. This is in parallel with the diagonal of resistance 16 Ω, so the total resistance must be 8 Ω, answer A.

9702/13 Q37What is the current in the 40 Ω resistor of the circuit shown?

A zero B 0.13 A C 0.25 A D 0.50 A

Examiner report: The answer to this question can be reasoned by symmetry. The junctions on either side of the 40 Ω resistor must be at the same potential, so there is no current in this resistor.

9702/12 Q32The diagram shows an electric pump for a garden fountain connected by an 18 m cable to a 230 V mains electrical supply.

The performance of the pump is acceptable if the potential difference (p.d.) across it is at least 218 V. The current through it is then 0.83 A.

What is the maximum resistance per metre of each of the two wires in the cable if the pump is to perform acceptably?

A 0.40 Ω m–1 B 0.80 Ω m–1 C 1.3 Ω m–1 D 1.4 Ω m–1

Examiner report: Candidates found this question to be one of the most difficult on the paper. If the pump performs acceptably, the potential difference across the pump is 218 V, so the potential difference across the cables is 12 V. The current is 0.83 A so the cables have a total resistance of 14.5 Ω. The total length of the cables is 36 m, giving a resistance per unit length of 0.40 Ω m–1, which is answer A.

9702/12 Q33Cell X has an e.m.f. of 2.0 V and an internal resistance of 2.0 Ω. Cell Y has an e.m.f. of 1.6 V and an internal resistance of 1.2 Ω. These two cells are connected to a resistor of resistance 0.8 Ω, as shown.

What is the current in cell X?

A 0.10 A B 0.50 A C 0.90 A D 1.0 A

Examiner report: This question can be answered using Kirchhoff’s second law: (2.0 V – 1.6 V) = (2.0 Ω + 0.8 Ω + 1.2 Ω)I.Candidates must take care with the signs of the e.m.f. terms.

9702/21 Q6biv and 6cA battery of e.m.f. 12 V and internal resistance 0.50 Ω is connected to two identical lamps, as shown in Fig. 6.1.

Each lamp has constant resistance. The power rating of each lamp is 48 W when connected across a p.d. of 12 V.Calculate the power dissipated in one lamp. [2 marks]

Examiner report: Candidates found this part difficult, even allowing for error-carried-forward from an incorrect answer in (iii). The value of the current was often taken as the current from the battery, rather than the current in an individual lamp.

A third identical lamp is placed in parallel with the battery in the circuit of Fig. 6.1. Describe and explain the effect on the terminal p.d. of the battery. [2 marks]

Examiner report: Candidates who had an understanding of this topic usually gave a good explanation. Some thought that an additional lamp would increase the resistance of the circuit. Of those who stated that the total resistance would decrease and that the current would increase, many then incorrectly stated that the terminal p.d. would increase. Candidates should be reminded that increased current in a battery with internal resistance results in a lower terminal p.d.

9701/12 Q35A power supply and a solar cell are compared using the potentiometer circuit shown.

The e.m.f. produced by the solar cell is measured on the potentiometer.The potentiometer wire PQ is 100.0 cm long and has a resistance of 5.00 Ω. The power supply has an e.m.f. of 2.000 V and the solar cell has an e.m.f. of 5.00 mV.Which resistance R must be used so that the galvanometer reads zero when PS = 40.0 cm?

A 395 Ω B 795 Ω C 995 Ω D 1055 Ω

Examiner report: The potential difference across PS is given by 2.000 V × (0.400 × 5.00 Ω / [R + 5.00 Ω]). Equate this to the e.m.f. of the solar cell and rearrange to obtain R, which is 795 Ω, answer B.

9702/13 Q35The diagram shows a low-voltage circuit for heating the water in a fish tank.

The heater has a resistance of 3.0 Ω. The power supply has an e.m.f. of 12 V and an internal resistance of 1.0 Ω.

At which rate is energy supplied to the heater?

A 27 W B 36 W C 48 W D 64 W

Examiner report: The current is obtained from the e.m.f. and the total resistance, and is 3.0 A. The power in the heater is given by I2R, which is 27 W, answer A.

9702/13 Q36The diagrams show the same cell, ammeter, potentiometer and fixed resistor connected in different ways.

The distance d between the sliding contact and a particular end of the potentiometer is varied.The current measured is then plotted against the distance d.For which two circuits will the graphs be identical?

A W and X B W and Y C X and Y D Y and Z

Examiner report: Candidates found this question difficult. W and X will give identical graphs. In each of these arrangements, the resistance of the length d of the potentiometer is connected into the circuit but the right-hand end does not contribute.

9702/22 Q6b, 6cii and 6dA battery is connected in series with resistors X and Y, as shown in Fig. 6.1.

The resistance of X is constant. The resistance of Y is 6.0 Ω. The battery has electromotive force (e.m.f.) 24 V and zero internal resistance. A variable resistor of resistance R is connected in parallel with X.The current І from the battery is changed by varying R from 5.0 Ω to 20 Ω. The variation with R of І is shown in Fig. 6.2.

Use Fig. 6.2 to state and explain the variation of the p.d. across resistor Y as R is increased. Numerical values are not required. [2 marks]

Examiner report: A significant number of candidates understood that, if the current decreases in the circuit, the potential difference across Y must decrease. Many candidates did not use the graph that showed the current decreasing with

increasing R. There were many misinterpretations of the circuit which often came as a result of not reading the circuit or text carefully. The effect of R increasing was used in the equation V = IR to suggest that the p.d. would increase. The fact that there was an effect on the current of increasing R and that the V required was not that across R was missed by many of these candidates. A significant number of candidates understood that, if the current decreases in the circuit, the potential difference across Y must decrease. Many candidates did not use the graph that showed the current decreasing with increasing R. There were many misinterpretations of the circuit which often came as a result of not reading the circuit or text carefully. The effect of R increasing was used in the equation V = IR to suggest that the p.d. would increase. The fact that there was an effect on the current of increasing R and that the V required was not that across R was missed by many of these candidates.

For R = 6.0 Ω, calculate the resistance of X (they have already shown that VAB = 9.6V) [3 marks]

Examiner report: The majority of candidates found this calculation difficult. The determination of the resistance using the parallel formula caused most candidates a problem. A significant number of candidates determined the correct resistance for the parallel section but then found it difficult to determine the resistance of X or gave this answer as the resistance of X. There were some good answers from the stronger candidates.

State and explain qualitatively how the power provided by the battery changes as the resistance R is increased. [1 mark]

Examiner report: There were very few correct answers. Many candidates did not comment on the e.m.f. of the battery remaining constant.

Nuclear reactions – although candidates seem to have been quite adept at using nuclear reactions, questions involving only words caused some problems.

9702/11 Q39U++ is a doubly-ionised uranium atom. The uranium atom has a nucleon number of 235 and a proton number of 92.In a simple model of the atom, how many particles are in this ionised atom?

A 235 B 325 C 327 D 329

Examiner report: The nucleus contains 235 nucleons. A neutral atom would have 92 electrons, but this atom is doubly-ionised and positive, so it must have 90 electrons. There are 325 particles in total, answer B.

9702/12 Q38A nucleus X decays into a nucleus Y by emitting an alpha particle followed by two beta particles.Which statement about this nuclear decay is correct?

A Beta particle decay occurs when a proton changes into a neutron.B Nucleus Y has the same nucleon number as nucleus X.C Nucleus Y is an isotope of nucleus X.D The total mass of the products is equal to the mass of the initial nucleus X.

Examiner report: Some candidates gave D. The mass cannot be conserved because the decay releases energy. The correct answer C can be obtained by carefully writing out the nuclear equation with correct proton numbers for the alpha and beta particles, and this shows no overall change in proton number.

General questions – the questions in this section do not fit a particular pattern, but have caused problems for candidates.

9702/11 Q21The graph is a load-extension graph for a wire undergoing elastic deformation.

How much work is done on the wire to increase the extension from 10 mm to 20 mm?

A 0.028 J B 0.184 J C 0.28 J D 0.37 J

Examiner report: Candidates found this question difficult. Some candidates forgot g and some seem to have ignored thegradient of the line (i.e. that the load is not constant). A direct way to reach the answer is to use (2.85 × 9.8 × 0.01) = 0.28.

9702/21 Q6A radio-controlled toy car travels along a straight line for a time of 15 s.The variation with time t of the velocity v of the car is shown below.

What is the average velocity of the toy car for the journey shown by the graph?

A –1.5 m s–1 B 0.0 m s–1 C 4.0 m s–1 D 4.5 m s–1

Examiner report: Many candidates chose C. These candidates had determined the average speed. The question asked for the average velocity, which is zero: the car starts and finishes at the same point. Candidates should be encouraged to read the question carefully. In this case, there is a significant difference between speed and velocity.

9702/12 Q15A small mass is placed at point P on the inside surface of a smooth hemisphere. It is then released from rest. When it reaches the lowest point T, its speed is 4.0 m s–1.The diagram (not to scale) shows the speed of the mass at other points Q, R and S as it slides down. Air resistance is negligible.

The mass loses potential energy E in falling from P to T.At which point has the mass lost potential energy E/4?

A QB RC SD none of these

Examiner report: This is more easily seen in terms of kinetic energy. When the mass is at R, it has half the final speed so has gained energy E/4. This is when it has lost potential energy E/4.

9702/13 Q18A ball drops onto a horizontal surface and bounces elastically.What happens to the kinetic energy of the ball during the very short time that it is in contact with the surface?

A Most of the kinetic energy is lost as heat and sound energy.B The kinetic energy decreases to zero and then returns to its original value.C The kinetic energy remains constant because it is an elastic collision.D The kinetic energy remains constant in magnitude but changes direction.

Examiner report: Kinetic energy is a scalar so does not have a direction, so D cannot be the correct answer. Some candidates chose C, but this cannot be correct because the ball stops (during which time all the energy becomes elastic potential energy).

9702/12 Q24The diagram shows a vertical cross-section through a water wave moving from left to right.At which point is the water moving upwards with maximum speed?

Examiner report: This question was found to be difficult. Candidates would benefit from drawing on the exam paper. If they did draw a wave, at most a tenth of a wavelength further on, they would have seen that at A and B the water is moving downwards. At D it is just slightly upwards but the answer is clearly C.

9702/13 Q12A stationary body explodes into two components of masses m and 2m.The components gain kinetic energies X and Y respectively.

What is the ratio X/Y?

A 1/4 B 1/2 C 2/1 D 4/1

Examiner report: This question needs care. The speed of m is twice the speed of 2m because momentum is conserved. The kinetic energy of m is therefore multiplied by 4 as a result of the v2 term and divided by 2 because it has half the mass. This gives answer C.

9702/13 Q14A uniform solid cuboid of concrete of dimensions 0.50 m × 1.20 m × 0.40 m and weight 4000 N rests on a flat surface with the 1.20 m edge vertical as shown in diagram 1.

What is the minimum energy required to roll the cuboid through 90° to the position shown in diagram 2 with the 0.50 m edge vertical?

A 200 J B 400 J C 1400 J D 2600 J

Examiner report: This question is unusual and candidates found it difficult. The energy required depends on working out the rise in potential energy when the cuboid’s bottom right edge is the only contact with the ground and the top left edge is vertically above it. This gives a 5, 12, 13 triangle and the centre of mass is then 0.65 m above the surface.

9207/13 Q22The Mariana Trench in the Pacific Ocean has a depth of about 10 km.Assuming that sea water is incompressible and has a density of about 1020 kg m–3, what would be the approximate pressure at that depth?

A 105 Pa B 106 Pa C 107 Pa D 108 Pa

Examiner report: Some candidates found this difficult. The formula for pressure is given in the question paper, and candidates should be reminded to use the formulae given rather than relying on memory.

9207/13 Q26What is the approximate range of frequencies of infra-red radiation?

A 1 × 103 Hz to 1 × 109 HzB 1 × 109 Hz to 1 × 1011 HzC 1 × 1011 Hz to 1 × 1014 HzD 1 × 1014 Hz to 1 × 1017 Hz

Examiner report: This was a somewhat different question but the syllabus does require candidates to know the orders of magnitude of the wavelengths of different components of the electromagnetic spectrum. Candidates should also be able to change these to frequencies.

9702/22 5biiTwo wires P and Q of the same material and same original length l0 are fixed so that they hang vertically, as shown in Fig. 5.1.

The diameter of P is d and the diameter of Q is 2d. The same force F is applied to the lower end of each wire.Show your working and determine the ratio strain in P/strain in Q (candidates have already found stress in P/stress in Q)

Examiner report: The majority of candidates found this part very difficult. Very few realised that, as the wires were made from the same material, the value of the Young modulus would be the same. Candidates should be encouraged to think of the Young modulus as a property of a material itself, rather than being a property of a sample with specific size or shape. A significant number of candidates gave an answer without the necessary working.

9702/21 Q2biiA lorry moves up a road that is inclined at 9.0° to the horizontal, as shown in Fig. 2.1.The lorry has mass 2500 kg and is travelling at a constant speed of 8.5 m s−1. The force due to air resistance is negligible.

State two reasons why the rate of change of potential energy of the lorry is equal to the power calculated in (i).

Examiner report: There were very few answers where reference was made to constant kinetic energy because the speed is constant. Some stated that the kinetic energy would be zero. Likewise, very few candidates stated that no work would be done against air resistance. In a significant number of scripts, the answer was given as ‘constant speed with no air resistance’. It should be emphasised to candidates that paraphrasing the question does not score any credit.

9702/11 Q11The diagrams show two ways of hanging the same picture.

In both cases, a string is attached to the same points on the picture and looped symmetrically over a nail in a wall. The forces shown are those that act on the nail.In diagram 1, the string loop is shorter than in diagram 2.Which information about the magnitude of the forces is correct?

A R1 = R2 T1 = T2

B R1 = R2 T1 > T2

C R1 > R2 T1 < T2

D R1 < R2 T1 = T2

Examiner report: The forces R1 and R2 are always equal to the weight of the picture, so only A and B have a correct statement involving these forces. Consideration of the angles shows that T1 > T2, or alternatively candidates could simply rule out A by reasoning that T1 and T2 cannot be equal.

9702/22 Q6A tractor of mass 1000 kg is connected by a tow-bar to a trailer of mass 1000 kg. The total resistance to motion has a constant value of 4000 N. One quarter of this resistance acts on the trailer.When the tractor and trailer are moving along horizontal ground at a constant speed of 6 m s–1, what is the force exerted on the tractor by the tow-bar?

A 0 N B 1000 N C 3000 N D 4000 N

Examiner report: This question involves equilibrium of an object moving with constant velocity. The resultant force on the trailer is zero, so the force exerted by the tractor on the trailer is 1000 N. This is equal and opposite to the force exerted on the tractor by the tow-bar.