Determine the coefficient of dynamic friction between the stone and the ice during the last 14.0 s of the stone’s motion.

The diagram shows the stone during its motion after release.

Label the diagram to show the forces acting on the stone. Your answer should include the name, the direction and point of application of each force.

ALTERNATIVE 1

«deceleration» $=\frac{3.41}{14.0}$ «$=0.243\text{m}{\text{s}}^{-2}$»

F = 0.243 × m

$\mu =\frac{0.243\times m}{m\times 9.81}=0.025$

ALTERNATIVE 2

distance travelled after release = 23.85 «m»
KE lost = 5.81m «J»

${\mu }_{\text{d}}=\frac{\text{KE lost}}{mg\times \text{distance}}=\frac{5.81m}{23.85mg}=0.025$

Award [3] for a bald correct answer.

Ignore sign in acceleration.

Allow ECF from (a) (note that $\mu =0.0073$ x candidate answer to (a) ).

Ignore any units in answer.

Condone omission of m in solution.

Allow g = 10 N kg^–1 (gives 0.024).

normal force, upwards, ignore point of application

Force must be labeled for its mark to be awarded. Blob at poa not required.
Allow OWTTE for normal force.  Allow N, R, reaction.
The vertical forces must lie within the middle third of the stone

weight/weight force/force of gravity, downwards, ignore point of application

Allow mg, W but not “gravity”.

Penalise gross deviations from vertical/horizontal once only

friction/resistive force, to left, at bottom of stone, point of application must be on the interface between ice and stone

Allow F, μR. Only allow arrows/lines that lie on the interface. Take the tail of the arrow as the definitive point of application and expect line to be drawn horizontal.

Award [2 max] if any force arrow does not touch the stone

Do not award MP3 if a “driving force” is shown acting to the right. This need not be labelled to disqualify the mark. Treat arrows labelled “air resistance” as neutral.

N.B: Diagram in MS is drawn with the vertical forces not direction of travel collinear for clarity

Outline why a force acts on the airboat due to the fan blade.

Show that a mass of about 240 kg of air moves through the fan every second.

Show that the tension in the rope is about 5 kN.

Explain why the airboat has a maximum speed under these conditions.

Estimate the distance the airboat travels to reach its maximum speed.

Deduce the mass of the airboat.

ALTERNATIVE 1

there is a force «by the fan» on the air / air is accelerated «to the rear» ✓

by Newton 3 ✓

there is an «equal and» opposite force on the boat ✓

ALTERNATIVE 2

air gains momentum «backward» ✓

by conservation of momentum / force is rate of change in momentum ✓

boat gains momentum in the opposite direction ✓

Accept a reference to Newton’s third law, e.g. N’3, or any correct statement of it for MP2 in ALT 1.

Allow any reasonable choice of object where the force of the air is acting on, e.g., fan or blades.

$\pi R^2$ OR «mass of air through system per unit time =» $Av\rho$ seen ✓

244 «kg s⁻¹» ✓

Accept use of Energy of air per second = 0.5 ρΑv³ = 0.5 mv² for MP1.

«force = Momentum change per sec = $A{v}^2\rho$ = » 244 x 20 OR 4.9 «kN» ✓

Allow use of 240

resistive forces increase with speed  OR  resistive forces/drag equal forward thrust ✓

acceleration/net force becomes zero/speed remains constant ✓

recognition that area under the graph is distance covered ✓

«Distance =» 480 - 560 «m» ✓

Accept graphical evidence or calculation of correct geometric areas for MP1.

MP2 is numerical value within range.

calculation of acceleration as gradient at t = 0 «= 1 m s^-2» ✓

use of F=ma OR $\frac{4900}{1}$seen ✓

4900 «kg» ✓

MP1 can be shown on the graph.

Allow an acceleration in the range 1 – 1.1 for MP2 and consistent answer for MP3

Allow ECF from MP1.

Allow use of average acceleration = $\frac{18}{40}$

or assumption of constant force to obtain 11000 «kg» for [2]

Allow use of 4800 or 5000 for MP2

The majority succeeded in making use of Newton's third law to explain the force on the boat. The question was quite well answered but sequencing of answers was not always ideal. There were some confusions about the air hitting the bank and bouncing off to hit the boat. A small number thought that the wind blowing the fan caused the force on the boat.

bi) This was generally well answered with candidates either starting from the wind turbine formula given in the data booklet or with the mass of the air being found using $\rho Av$.

1bii) Well answered by most candidates. Some creative work to end up with 240 was found in scripts.

1ci) Many candidates gained credit here for recognising that the resistive force eventually equalled the drag force and most were able to go on to link this to e.g. zero acceleration. Some had not read the question properly and assumed that the rope was still tied. There was one group of answers that stated something along the lines of "as there is no rope there is nothing to stop the boat so it can go at max speed.

1cii) A slight majority did not realise that they had to find the area under the velocity-time graph, trying equations of motion for non-linear acceleration. Those that attempted to calculate the area under the graph always succeeded in answering within the range.

1ciii) Use of the average gradient was common here for the acceleration. However, there also were answers that attempted to calculate the mass via a kinetic energy calculation that made all sorts of incorrect assumptions. Use of average acceleration taken from the gradient of the secant was also common.

Calculate, in m, the length of the thread. State your answer to an appropriate number of significant figures.

Label on the graph with the letter X a point where the speed of the pendulum is half that of its initial speed.

The mass of the pendulum bob is 75 g. Show that the maximum speed of the bob is about 0.7 m s^–1.

Calculate the speed of the combined masses immediately after the collision.

Show that the collision is inelastic.

Sketch, on the axes, a graph to show the variation of gravitational potential energy with time for the bob and the object after the collision. The data from the graph used in (a) is shown as a dashed line for reference.

The speed after the collision of the bob and the object was measured using a sensor. This sensor emits a sound of frequency f and this sound is reflected from the moving bob. The sound is then detected by the sensor as frequency f′.

Explain why f and f′ are different.

identifies T as 2.25 s ✔

L = 1.26 m ✔

1.3 / 1.26 «m» ✔

Accept any number of s.f. for MP2.

Accept any answer with 2 or 3 s.f. for MP3.

X labels any point on the curve where E_K   $\frac{1}{4}$ of maximum/5 mJ ✔

$\frac{1}{2}$ mv² = 20 × 10⁻³ seen OR $\frac{1}{2}$ × 7.5 × 10^-2 × v² = 20 × 10^-3 ✔

0.73 «m s⁻¹» ✔

Must see at least 2 s.f. for MP2.

0.40 «m s^-1» ✔

initial energy 24 mJ and final energy 12 mJ ✔

energy is lost/unequal /change in energy is 12 mJ ✔

inelastic collisions occur when energy is lost ✔

graph with same period but inverted ✔

amplitude one half of the original/two boxes throughout (by eye) ✔

mention of Doppler effect ✔

there is a change in the wavelength of the reflected wave ✔

because the wave speed is constant, there is a change in frequency ✔

This question was well approached by candidates. The noteworthy mistakes were not reading the correct period of the pendulum from the graph, and some simple calculation and mathematical errors. This question also had one mark for writing an answer with the correct number of significant digits. Candidates should be aware to look for significant digit question on the exam and can write any number with correct number of significant digits for the mark.

This question was well answered. This is a “show that” question so candidates needed to clearly show the correct calculation and write an answer with at least one significant digit more than the given answer. Many candidates failed to appreciate that the energy was given in mJ and the mass was in grams, and that these values needed to be converted before substitution.

Candidates fell into some broad categories on this question. This was a “show that” question, so there was an expectation of a mathematical argument. Many were able to successfully show that the initial and final kinetic energies were different and connect this to the concept of inelastic collisions. Some candidates tried to connect conservation of momentum unsuccessfully, and some simply wrote an extended response about the nature of inelastic collisions and noted that the bobs stuck together without any calculations. This approach was awarded zero marks.

Many candidates drew graphs that received one mark for either recognizing the phase difference between the gravitational potential energy and the kinetic energy, or for recognizing that the total energy was half the original energy. Few candidates had both features for both marks.

This question was essentially about the Doppler effect, and therefore candidates were expected to give a good explanation for why there is a frequency difference. As with all explain questions, the candidates were required to go beyond the given information. Very few candidates earned marks beyond just recognizing that this was an example of the Doppler effect. Some did discuss the change in wavelength caused by the relative motion of the bob, although some candidates chose very vague descriptions like “the waves are all squished up” rather than using proper physics terms. Some candidates simply wrote and explained the equation from the data booklet, which did not receive marks. It should be noted that this was a three mark question, and yet some candidates attempted to answer it with a single sentence.

The charge per unit area on the surface of the wall is σ. It can be shown that the electric field strength E due to the charge on the wall is given by the equation

$E=\frac{\sigma }{2{\epsilon }_0}$.

Demonstrate that the units of the quantities in this equation are consistent.

The thread makes an angle of 30° with the vertical wall. The ball has a mass of 0.025 kg.

Determine the horizontal force that acts on the ball.

The charge on the ball is 1.2 × 10⁻⁶C. Determine σ.

The thread breaks. Explain the initial subsequent motion of the ball.

Calculate the charge on Q. State your answer to an appropriate number of significant figures.

Outline, without calculation, whether or not the electric potential at P is zero.

identifies units of $\sigma$ as ${\text{C\hspace{0.05em}m}}^{-2}$ ✓

$\frac{C}{m^2}\times \frac{N{m}^2}{C^2}$ seen and reduced to ${\text{N\hspace{0.05em}C}}^{-1}$ ✓

Accept any analysis (eg dimensional) that yields answer correctly

horizontal force $F$ on ball $=T \sin 30$ ✓

$T=\frac{mg}{\cos 30}$ ✓

$F «=mg \tan 30 = 0.025\times 9.8 \times \tan 30» = 0.14 «\text{N}»$ ✓

Allow g = 10 N kg⁻¹

Award [3] marks for a bald correct answer.

Award [1max] for an answer of zero, interpreting that the horizontal force refers to the horizontal component of the net force.

$E=\frac{0.14}{1.2\times {10}^{-6}}«=1.2\times {10}^5»$ ✓

$\sigma =«\frac{2\times 8.85\times {10}^{-12}\times 0.14}{1.2\times {10}^{-6}}»=2.1\times {10}^{-6} «{\text{C\hspace{0.05em}m}}^{-2}»$ ✓

Allow ECF from the calculated F in (b)(i)

Award [2] for a bald correct answer.

horizontal/repulsive force and vertical force/pull of gravity act on the ball ✓

so ball has constant acceleration/constant net force ✓

motion is in a straight line ✓

at 30° to vertical away from wall/along original line of thread ✓

$\frac{Q}{0.{22}^2}=\frac{1.2\times {10}^{-6}}{0.{18}^2}$ ✓

$«+»1.8\times {10}^{-6} «\text{C}»$✓

2sf ✓

Do not award MP2 if charge is negative

Any answer given to 2 sig figs scores MP3

work must be done to move a «positive» charge from infinity to P «as both charges are positive»
OR
reference to both potentials positive and added
OR
identifies field as gradient of potential and with zero value ✓

therefore, point P is at a positive / non-zero potential ✓

Award [0] for bald answer that P has non-zero potential

Outline what is meant by electric field strength.

An electron is placed at X and released from rest. Draw, on the diagram, the direction of the force acting on the electron due to the field.

The electron is replaced by a proton which is also released from rest at X. Compare, without calculation, the motion of the electron with the motion of the proton after release. You may assume that no frictional forces act on the electron or the proton.

force per unit charge

acting on a small/test positive charge

horizontally to the left

Arrow does not need to touch X

proton moves to the right/they move in opposite directions

force on each is initially the same

proton accelerates less than electron initially «because mass is greater»

field is stronger on right than left «as lines closer»

proton acceleration increases «as it is moving into stronger field»

OR

electron acceleration decreases «as it is moving into weaker field»

Allow ECF from (b)

Accept converse argument for electron

At position B the rope starts to extend. Calculate the speed of the block at position B.

Determine the magnitude of the average resultant force acting on the block between B and C.

Sketch on the diagram the average resultant force acting on the block between B and C. The arrow on the diagram represents the weight of the block.

Calculate the magnitude of the average force exerted by the rope on the block between B and C.

between A and B.

between B and C.

The length reached by the rope at C is 77.4 m. Suggest how energy considerations could be used to determine the elastic constant of the rope.

Calculate the time taken for the block to return to the equilibrium position for the first time. 

Calculate the speed of the block as it passes the equilibrium position. 

use of conservation of energy

OR

v² = u² + 2as

v = «$\sqrt{2\times 60.0\times 9.81}$» = 34.3 «ms^–1»

[2 marks]

use of impulse F_ave × Δt = Δp

OR

use of F = ma with average acceleration

OR

F = $\frac{80.0\times 34.3}{0.759}$

3620«N»

Allow ECF from (a).

[2 marks]

upwards

clearly longer than weight

For second marking point allow ECF from (b)(i) providing line is upwards.

[2 marks]

3620 + 80.0 × 9.81

4400 «N»

Allow ECF from (b)(i).

[2 marks]

(loss in) gravitational potential energy (of block) into kinetic energy (of block)

Must see names of energy (gravitational potential energy and kinetic energy) – Allow for reasonable variations of terminology (eg energy of motion for KE).

[1 mark]

(loss in) gravitational potential and kinetic energy of block into elastic potential energy of rope

See note for 1(c)(i) for naming convention.

Must see either the block or the rope (or both) mentioned in connection with the appropriate energies.

[1 mark]

k can be determined using EPE = $\frac{1}{2}$kx²

correct statement or equation showing

GPE at A = EPE at C

OR

(GPE + KE) at B = EPE at C

Candidate must clearly indicate the energy associated with either position A or B for MP2.

[2 marks]

T = 2π$\sqrt{\frac{80.0}{400}}$ = 2.81 «s»

time = $\frac{T}{4}$ = 0.702 «s»

Award [0] for kinematic solutions that assume a constant acceleration.

[2 marks]

ALTERNATIVE 1

ω = $\frac{2\pi }{2.81}$ = 2.24 «rad s^–1»

v = 2.24 × 3.50 = 7.84 «ms^–1»

ALTERNATIVE 2

$\frac{1}{2}$kx² = $\frac{1}{2}$mv² OR $\frac{1}{2}$400 × 3.5² = $\frac{1}{2}$80v²

v = 7.84 «ms^–1»

Award [0] for kinematic solutions that assume a constant acceleration.

Allow ECF for T from (e)(i).

[2 marks]

State the direction of the resultant force on the ball.

On the diagram, construct an arrow of the correct length to represent the weight of the ball.

Show that the magnitude of the net force F on the ball is given by the following equation.

                                          $$F=\frac{mg}{\tan \theta }$$

The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.

Outline whether this ball can move on a horizontal circular path of radius equal to the radius of the bowl.

Outline why the ball will perform simple harmonic oscillations about the equilibrium position.

Show that the period of oscillation of the ball is about 6 s.

The amplitude of oscillation is 0.12 m. On the axes, draw a graph to show the variation with time t of the velocity v of the ball during one period.

A second identical ball is placed at the bottom of the bowl and the first ball is displaced so that its height from the horizontal is equal to 8.0 m.

The first ball is released and eventually strikes the second ball. The two balls remain in contact. Determine, in m, the maximum height reached by the two balls.

towards the centre «of the circle» / horizontally to the right

Do not accept towards the centre of the bowl

[1 mark]

downward vertical arrow of any length

arrow of correct length

Judge the length of the vertical arrow by eye. The construction lines are not required. A label is not required

eg: 

[2 marks]

ALTERNATIVE 1

F = N cos θ

mg = N sin θ

dividing/substituting to get result

ALTERNATIVE 2

right angle triangle drawn with F, N and W/mg labelled

angle correctly labelled and arrows on forces in correct directions

correct use of trigonometry leading to the required relationship

tan θ = $\frac{\text{O}}{A}=\frac{mg}{F}$

[3 marks]

$\frac{mg}{\tan \theta }$ = m$\frac{v^2}{r}$

r = R cos θ

v = $\sqrt{\frac{gR{\cos }^2\theta }{\sin \theta }}/\sqrt{\frac{gR\cos \theta }{\tan \theta }}/\sqrt{\frac{9.81\times 8.0\cos 22}{\tan 22}}$

v = 13.4/13 «ms ^–1»

Award [4] for a bald correct answer 

Award [3] for an answer of 13.9/14 «ms ^–1». MP2 omitted

[4 marks]

there is no force to balance the weight/N is horizontal

so no / it is not possible

Must see correct justification to award MP2

[2 marks]

the «restoring» force/acceleration is proportional to displacement

Direction is not required

[1 mark]

ω = «$\sqrt{\frac{g}{R}}$» = $\sqrt{\frac{9.81}{8.0}}$ «= 1.107 s^–1»

T = «$\frac{2\pi }{\omega }$ = $\frac{2\pi }{1.107}$ =» 5.7 «s»

Allow use of or g = 9.8 or 10

Award [0] for a substitution into T = 2π$\sqrt{\frac{I}{g}}$

[2 marks]

sine graph

correct amplitude «0.13 m s^–1»

correct period and only 1 period shown

Accept ± sine for shape of the graph. Accept 5.7 s or 6.0 s for the correct period.

Amplitude should be correct to ±$\frac{1}{2}$ square for MP2

eg: v /m s^–1   

[3 marks]

speed before collision v = «$\sqrt{2gR}$ =» 12.5 «ms^–1»

«from conservation of momentum» common speed after collision is $\frac{1}{2}$ initial speed «v_c = $\frac{12.5}{2}$ = 6.25 ms^–1»

h = «$\frac{{v_c}^2}{2g}=\frac{{6.25}^2}{2\times 9.81}$» 2.0 «m»

Allow 12.5 from incorrect use of kinematics equations

Award [3] for a bald correct answer

Award [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.

Allow ECF from MP1

Allow ECF from MP2

[3 marks]

State the value of the resultant force on the aircraft when hovering.

Outline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved.

Determine $v$. State your answer to an appropriate number of significant figures.

Calculate the power transferred to the air by the aircraft.

The package and string are now released and fall to the ground. The lift force on the aircraft remains unchanged. Calculate the initial acceleration of the aircraft.

zero ✓

Blades exert a downward force on the air ✓

air exerts an equal and opposite force on the blades «by Newton’s third law»
OR
air exerts a reaction force on the blades «by Newton’s third law» ✓

Downward direction required for MP1.

«lift force/change of momentum in one second» $=1.7v$ ✓

$1.7v=\left(0.95+0.45\right)\times 9.81$ ✓

$v=8.1 «{\mathrm{ms}}^{-1}»$ AND answer expressed to $2$ sf only ✓

Allow $\mathit{8}\mathit{.}\mathit{2}$ from $g\mathit{=}\mathit{10}\mathit{ }m{s}^{\mathit{-}\mathit{2}}$.

ALTERNATIVE 1

power $«=\mathrm{rate} \mathrm{of} \mathrm{energy} \mathrm{transfer} \mathrm{to} \mathrm{the} \mathrm{air}=\frac{1}{2}\frac{∆m}{∆t}{v}^2»=\frac{1}{2}\times 1.7\times 8.1^2$ ✓

$=56 «\mathrm{W}»$ ✓

ALTERNATIVE 2

Power $«=\mathrm{Force} \times v \mathrm{ave}»=\left(0.95+0.45\right)\times 9.81\times \frac{8.1}{2}$ ✓ 

$=56 «\mathrm{W}»$ ✓

vertical force = lift force – weight OR $=0.45\times 9.81$ OR $=4.4 «\mathrm{N}»$✓

acceleration $=\frac{0.45\times 9.81}{0.95}=4.6 «{\mathrm{ms}}^{-2}»$✓

This was generally answered well with the most common incorrect answer being the weight of the aircraft and package. The question uses the command term 'state' which indicates that the answer requires no working.

The question required candidates to apply Newton's third law to a specific situation. Candidates who had learned the 'action and reaction' version of Newton's third law generally did less well than those who had learned a version describing 'object A exerting a force on object B' etc. Some answers lacked detail of what was exerting the force and in which direction.

This was answered well with many getting full marks. A small number gave the wrong number of significant figures and some attempted to answer using kinematics equations or kinetic energy.

HL only. It was common to see answers that neglected to average the velocity and consequently arrived at an answer twice the size of the correct one. This was awarded 1 of the 2 marks.

Well done by a good number of candidates. Many earned a mark by simply using the correct mass to find an acceleration even though the force was incorrect.

Outline the origin of the force that acts on Phobos.

Outline why this force does no work on Phobos.

The orbital period T of a moon orbiting a planet of mass M is given by

$$\frac{R^3}{T^2}=kM$$

where R is the average distance between the centre of the planet and the centre of the moon.

Show that $k=\frac{G}{4{\pi }^2}$

The following data for the Mars–Phobos system and the Earth–Moon system are available:

Mass of Earth = 5.97 × 10²⁴ kg

The Earth–Moon distance is 41 times the Mars–Phobos distance.

The orbital period of the Moon is 86 times the orbital period of Phobos.

Calculate, in kg, the mass of Mars.

The graph shows the variation of the gravitational potential between the Earth and Moon with distance from the centre of the Earth. The distance from the Earth is expressed as a fraction of the total distance between the centre of the Earth and the centre of the Moon.

Determine, using the graph, the mass of the Moon.

gravitational attraction/force/field «of the planet/Mars» ✔

Do not accept “gravity”.

the force/field and the velocity/displacement are at 90° to each other OR

there is no change in GPE of the moon/Phobos ✔

ALTERNATE 1

«using fundamental equations»

use of Universal gravitational force/acceleration/orbital velocity equations ✔

equating to centripetal force or acceleration. ✔

rearranges to get $k=\frac{G}{4{\pi }^2}$  ✔

ALTERNATE 2

«starting with $\frac{R^3}{T^2}=kM$»

substitution of proper equation for T from orbital motion equations ✔

substitution of proper equation for M OR R from orbital motion equations ✔

rearranges to get $k=\frac{G}{4{\pi }^2}$  ✔

$m_{\text{Mars}}={\left(\frac{R_{\text{Mars}}}{R_{\text{Earth}}}\right)}^3{\left(\frac{T_{\text{Earth}}}{T_{Mars}}\right)}^2{m}_{\text{Earth}}$ or other consistent re-arrangement ✔

6.4 × 10²³ «kg» ✔

read off separation at maximum potential 0.9 ✔

equating of gravitational field strength of earth and moon at that location OR ✔

7.4 × 10²² «kg» ✔

Allow ECF from MP1

This was generally well answered, although some candidates simply used the vague term “gravity” rather than specifying that it is a gravitational force or a gravitational field. Candidates need to be reminded about using proper physics terms and not more general, “every day” terms on the exam.

Some candidates connected the idea that the gravitational force is perpendicular to the velocity (and hence the displacement) for the mark. It was also allowed to discuss that there is no change in gravitational potential energy, so therefore no work was being done. It was not acceptable to simply state that the net displacement over one full orbit is zero. Unfortunately, some candidates suggested that there is no net force on the moon so there is no work done, or that the moon is so much smaller so no work could be done on it.

This was another “show that” derivation. Many candidates attempted to work with universal gravitation equations, either from memory or the data booklet, to perform this derivation. The variety of correct solution paths was quite impressive, and many candidates who attempted this question were able to receive some marks. Candidates should be reminded on “show that” questions that it is never allowed to work backwards from the given answer. Some candidates also made up equations (such as T = 2𝝿r) to force the derivation to work out.

This question was challenging for candidates. The candidates who started down the correct path of using the given derived value from 5bi often simply forgot that the multiplication factors had to be squared and cubed as well as the variables.

This question was left blank by many candidates, and very few who attempted it were able to successfully recognize that the gravitational fields of the Earth and Moon balance at 0.9r and then use the proper equation to calculate the mass of the Moon.

Define impulse.

Show that the kinetic energy of the egg just before impact is about 0.6 J.

The egg comes to rest in a time of 55 ms. Determine the magnitude of the average decelerating force that the ground exerts on the egg.

Explain why the egg is likely to break when dropped onto concrete from the same height.

force × time

OR

change in momentum ✔

E_k = mgh =  0.058 × 9.81 ×1.1 = 0.63 J ✔

Allow use of g = 10 m s⁻² (which gives 0.64 «J»)

Substitution and at least 2 SF must be shown

ALTERNATIVE 1:

initial momentum = mv = $\sqrt{2\times 0.058\times 0.63}$ «= 0.27 kg m s⁻¹»

OR

mv = $0.058\times \sqrt{2\times 9.81\times 1.1}$ «= 0.27 kg m s⁻¹» ✔

force = «$\frac{\text{change in momentum}}{\text{time}}$ =» $\frac{0.27}{0.055}$ ✔

4.9 «N» ✔

F − mg = 4.9 so F = 5.5 «N» ✔

ALTERNATIVE 2:

«E_k = $\frac{1}{2}$mv² = 0.63 J» v = 4.7 m s⁻¹ ✔

acceleration = «$\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$ =» $\frac{4.7}{55\times {10}^{-3}}$ = «85 m s⁻²» ✔

4.9 «N» ✔

F − mg = 4.9 so F= 5.5 «N» ✔

Accept negative acceleration and force.

ALTERNATIVE 1:

concrete reduces the stopping time/distance ✔

impulse/change in momentum same so force greater

OR

work done same so force greater ✔

ALTERNATIVE 2:

concrete reduces the stopping time ✔

deceleration is greater so force is greater ✔

Allow reverse argument for grass.

State what is meant by the binding energy of a nucleus.

Draw, on the axes, a graph to show the variation with nucleon number $A$ of the binding energy per nucleon, $\frac{\text{BE}}{A}$. Numbers are not required on the vertical axis.

Identify, with a cross, on the graph in (a)(ii), the region of greatest stability.

Some unstable nuclei have many more neutrons than protons. Suggest the likely decay for these nuclei.

Show that the energy released in this decay is about 6 MeV.

The plutonium nucleus is at rest when it decays.

Calculate the ratio $\frac{\text{kinetic energy of alpha particle}}{\text{kinetic energy of uranium}}$.

Estimate the power, in kW, that is available from the plutonium at launch.

The spacecraft will take 7.2 years (2.3 × 10⁸ s) to reach a planet in the solar system. Estimate the power available to the spacecraft when it gets to the planet.

 State and explain what happens to the kinetic energy of an emitted photoelectron.

 State and explain what happens to the rate at which charge leaves the metallic surface.

the energy needed to «completely» separate the nucleons of a nucleus

OR

the energy released when a nucleus is assembled from its constituent nucleons ✓

Accept reference to protons and neutrons.

curve rising to a maximum between 50 and 100 ✓

curve continued and decreasing ✓

Ignore starting point.

Ignore maximum at alpha particle.

At a point on the peak of their graph ✓

beta minus «decay» ✓

correct mass numbers for uranium (234) and alpha (4) ✓

$234\times 7.600+4\times 7.074-238\times 7.568$ «MeV» ✓

energy released 5.51 «MeV» ✓

Ignore any negative sign.

«$\frac{K{E}_{\alpha }}{K{E}_U}=$»$\frac{\frac{p^2}{2m_{\alpha }}}{\frac{p^2}{2m_U}}$  OR  $\frac{m_U}{m_{\alpha }}$ ✓

«$\frac{234}{4}=$» $58.5$ ✓

Award [2] marks for a bald correct answer.

Accept $\frac{117}{2}$ for MP2.

number of nuclei present $=\frac{33\times {10}^3}{238}\times 6.02\times {10}^{23}«=8.347\times {10}^{25}»$ ✓

initial activity is $\lambda N_0=2.5\times {10}^{-10}\times 8.347\times {10}^{25}«=2.08\times {10}^{16} \text{Bq}»$ ✓

power is $2.08\times {10}^{16}\times 5.51\times {10}^6\times 1.6\times {10}^{-19}\approx 18$ «kW» ✓

Allow a final answer of 20 kW if 6 MeV used.

Allow ECF from MP1 and MP2.

available power after time t is $P_0{e}^{-\lambda t}$ ✓

$18e^{-2.50\times {10}^{-10}\times 2.3\times {10}^8}=17.0$ «kW» ✓

MP1 may be implicit.

Allow ECF from (c)(i).

Allow 17.4 kW from unrounded power from (c)(i).

Allow 18.8 kW from 6 MeV.

stays the same ✓

as energy depends on the frequency of light ✓

Allow reference to wavelength for MP2.

Award MP2 only to answers stating that KE decreases due to Doppler effect.

decreases ✓

as number of photons incident decreases ✓

Determine the initial acceleration of the spacecraft.

(i) Estimate the maximum speed of the spacecraft.

(ii) Outline why the answer to (i) is an estimate.

Outline why scientists sometimes use estimates in making calculations.

Outline why the ions are likely to spread out.

Explain what effect, if any, this spreading of the ions has on the acceleration of the spacecraft.

change in momentum each second = 6.6 × 10⁻⁶ × 5.2 × 10⁴ «= 3.4 × 10⁻¹kg m s⁻¹» ✔

acceleration = «$\frac{3.4\times {10}^{-1}}{740}$ =» 4.6 × 10⁻⁴ «m s⁻²» ✔

(i) ALTERNATIVE 1:

(considering the acceleration of the spacecraft)

time for acceleration = $\frac{30}{6.6\times {10}^{-6}}$ = «4.6 × 10⁶» «s» ✔

max speed = «answer to (a) × 4.6 × 10⁶ =» 2.1 × 10³ «m s⁻¹» ✔

ALTERNATIVE 2:

(considering the conservation of momentum)

(momentum of 30 kg of fuel ions = change of momentum of spacecraft)

30 × 5.2 × 10⁴ = 710 × max speed ✔

max speed = 2.2 × 10³ «m s⁻¹» ✔

(ii) as fuel is consumed total mass changes/decreases so acceleration changes/increases
OR
external forces (such as gravitational) can act on the spacecraft so acceleration isn’t constant ✔

problem may be too complicated for exact treatment ✔

to make equations/calculations simpler ✔

when precision of the calculations is not important ✔

some quantities in the problem may not be known exactly ✔

ions have same (sign of) charge ✔

ions repel each other ✔

the forces between the ions do not affect the force on the spacecraft. ✔

there is no effect on the acceleration of the spacecraft. ✔

Write down the maximum magnitude of the rate of change of flux linked with the coil.

State the fundamental SI unit for your answer to (a)(i).

Explain why the graph becomes negative.

Part of the graph is above the t-axis and part is below. Outline why the areas between the t-axis and the curve for these two parts are likely to be the same.

Predict the changes to the graph when the magnet is dropped from a lower height above the coil.

«−» 5.0 «mV»  OR  5.0 × 10⁻³ «V» ✓

Accept 5.1

kg m²A⁻¹s⁻³ ✓

ALTERNATIVE 1

Flux linkage is represented by magnetic field lines through the coil ✓

when magnet has passed through the coil / is moving away ✓

flux «linkage» is decreasing ✓

suitable comment that it is the opposite when above ✓

when the magnet goes through the midpoint the induced emf is zero ✓

ALTERNATIVE 2

reference to / states Lenz’s law ✓

when magnet has passed through the coil / is moving away ✓

«coil attracts outgoing S pole so» induced field is downwards ✓

before «coil repels incoming N pole so» induced field is upwards
OR
induced field has reversed ✓

when the magnet goes through the midpoint the induced emf is zero ✓

OWTTE

area represents the total change in flux «linkage» ✓

the change in flux is the same going in and out ✓

«when magnet is approaching» flux increases to a maximum ✓

«when magnet is receding» flux decreases to zero ✓

«so areas must be the same»

magnet moves slower ✓

overall time «for interaction» will be longer ✓

peaks will be smaller ✓

areas will be the same as before ✓

Allow a graphical interpretation for MP2 as “graph more spread out”

ai
Well answered, with wrong answers stating 8 for the difference or 3 without realising that the sign does not matter.

aii
Very few candidates managed to get the correct fundamental SI unit for V. All kinds of errors were observed, from power errors to the use of C as a fundamental unit instead of A.

bi) Most scored best by marking using an alternative method introduced to the markscheme in standardisation. There were some confused and vague comments. Clear, concise answers were rare.

bii) It was common to see conservation of energy invoked here with suggestions that energy was the area under the graph. Many candidates described the shapes to explain why the areas were the same rather than talking about the physics e.g. one peak is short and fat and the other is tall and thin so they balance out.

c) A surprising number didn't pick up on the fact that the magnet would be moving slower. As a result, they discussed everything happening sooner, i.e. the interaction with the magnet and the coil, and that led onto things happening quicker so peaks being bigger.

Show that the energy of photons from the UV lamp is about 10 eV.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

Suggest, with reference to conservation of energy, how the variable voltage source can be used to stop all emitted electrons from reaching the collecting plate.

The variable voltage can be adjusted so that no electrons reach the collecting plate. Write down the minimum value of the voltage for which no electrons reach the collecting plate.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV. Calculate the speed of the electron at the collecting plate.

E₁ = –13.6 «eV» E₂ = – $\frac{13.6}{4}$ = –3.4 «eV»

energy of photon is difference E₂ – E₁ = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]

10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10^–19 = 7.8 × 10^–19 «J»

Allow 5.1 if 10.2 is used to give 8.2×10⁻¹⁹ «J».

EPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference of electrons between surface and plate.

[2 marks]

4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

two equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]

kinetic energy at collecting plate = 0.9 «eV»

speed = «$\sqrt{\frac{2\times 0.9\times 1.6\times {10}^{-19}}{9.11\times {10}^{-31}}}$» = 5.6 × 10⁵ «ms^–1»

Allow ECF from MP1

[2 marks]

Calculate the wavelength of the wave.

Determine, for particle P, the magnitude and direction of the acceleration at t = 2.0 m s.

State the phase difference between the two waves.

Identify a time at which the displacement of P is zero.

Estimate the amplitude of the resultant wave.

Calculate the length of the tube.

A particle in the tube has its equilibrium position at the open end of the tube.
State and explain the direction of the velocity of this particle at time $t=\frac{T}{8}$.

Draw on the diagram the standing wave at time $t=\frac{T}{4}$.

$T=4\times {10}^{-3}$«s» or $f=250$«Hz» ✓

$\lambda =340\times 4.0\times {10}^{-3}=1.36\approx 1.4$«m» ✓

Allow ECF from MP1.
Award [2] for a bald correct answer.

$\varpi =«\frac{2\mathrm{\pi }}{T}=»\frac{2\mathrm{\pi }}{4\times {10}^{-3}}$  OR  $1.57\times {10}^3$ «s⁻¹» ✓

$\text{a}=«{\varpi }^2{x}_0={\left(1.57\times {10}^3\right)}^2\times 6\times {10}^{-6}=14.8\approx »15$ «ms⁻²» ✓

«opposite to displacement so» to the right ✓

«±» $\frac{\mathrm{\pi }}{2}/90°$  OR  $\frac{3\mathrm{\pi }}{2}/270°$ ✓

1.5 «ms» ✓

8.0 OR 8.5 «μm» ✓

From the graph on the paper, value is 8.0. From the calculated correct trig functions, value is 8.49.

L = «$\frac{3}{4}\lambda =$» 0.90 «m» ✓

to the right ✓

displacement is getting less negative

OR

change of displacement is positive ✓

horizontal line drawn at the equilibrium position ✓

Calculate the average force exerted by the racquet on the ball.

Calculate the average power delivered to the ball during the impact.

Calculate the time it takes the tennis ball to reach the net.

Show that the tennis ball passes over the net.

Determine the speed of the tennis ball as it strikes the ground.

A student models the bounce of the tennis ball to predict the angle θ at which the ball leaves a surface of clay and a surface of grass.

The model assumes

• during contact with the surface the ball slides.
• the sliding time is the same for both surfaces.
• the sliding frictional force is greater for clay than grass.
• the normal reaction force is the same for both surfaces.

Predict for the student’s model, without calculation, whether θ is greater for a clay surface or for a grass surface.

$F=\frac{\mathrm{\Delta }mv}{\mathrm{\Delta }t}/m\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}/\frac{0.058\times 64.0}{25\times {10}^{-3}}$  ✔

$F$ = 148 «N»≈150«N»  ✔

ALTERNATIVE 1

$P=\frac{\frac{1}{2}m{v}^2}{t}/\frac{\frac{1}{2}\times 0.058\times {64.0}^2}{25\times {10}^{-3}}$  ✔

$P=4700/4800$«$\text{W}$»  ✔

ALTERNATIVE 2

$P=\text{average}Fv\text{ / 148}\times \frac{64.0}{2}$  ✔

$P=4700/4800$«$\text{W}$»  ✔

horizontal component of velocity is $64.0\times \cos 7^{\circ }=63.52$«$\text{m }{\text{s}}^{-1}$»   ✔ 

$t=$«$\frac{11.9}{63.52}$»$\text{0}\text{.187/0}\text{.19}$«$\text{s}$»  ✔

ALTERNATIVE 1

u_y=64sin7/7.80«ms^–1»  ✔

decrease in height = 7.80 × 0.187 + $\frac{1}{2}$ × 9.81 × 0.187² / 1.63«m»  ✔

final height = «2.80 – 1.63» = 1.1/1.2«m»  ✔

«higher than net so goes over»

ALTERNATIVE 2

vertical distance to fall to net «=2.80 – 0.91» = 1.89«m»  ✔

time to fall this distance found using «$1.89=7.8t+\frac{1}{2}\times 9.81\times t^2$»

$t$ = 0.21«s»  ✔

0.21«s» > 0.187«s»   ✔

«reaches the net before it has fallen far enough so goes over»

ALTERNATIVE 1

Initial KE + PE = final KE /

$\frac{1}{2}\times 0.058\times {64}^2+0.058\times 9.81\times 2.80=\frac{1}{2}\times 0.058\times v^2$  ✔

$v=64.4$«$\text{m }{\text{s}}^{-1}$»  ✔

ALTERNATIVE 2

$v_v=$«$\sqrt{{7.8}^2+2\times 9.81\times 2.8}$» = 10.8«$\text{m }{\text{s}}^{-1}$»  ✔

«$v=\sqrt{{63.5}^2+{10.8}^2}$»

$v=64.4$«$\text{m }{\text{s}}^{-1}$»  ✔

so horizontal velocity component at lift off for clay is smaller ✔

normal force is the same so vertical component of velocity is the same ✔

so bounce angle on clay is greater ✔

At both HL and SL many candidates scored both marks for correctly answering this. A straightforward start to the paper. For those not gaining both marks it was possible to gain some credit for calculating either the change in momentum or the acceleration. At SL some used 64 ms-1 as a value for a and continued to use this value over the next few parts to the question.

This was well answered although a significant number of candidates approached it using P = Fv but forgot to divide v by 2 to calculated the average velocity. This scored one mark out of 2.

This question scored well at HL but less so at SL. One common mistake was to calculate the direct distance to the top of the net and assume that the ball travelled that distance with constant speed. At SL particularly, another was to consider the motion only when the ball is in contact with the racquet.

There were a number of approaches students could take to answer this and examiners saw examples of them all. One approach taken was to calculate the time taken to fall the distance to the top of the net and to compare this with the time calculated in bi) for the ball to reach the net. This approach, which is shown in the mark scheme, required solving a quadratic in t which is beyond the mathematical requirements of the syllabus. This mathematical technique was only required if using this approach and not required if, for example, calculating heights.

A common mistake was to forget that the ball has a vertical acceleration. Examiners were able to award credit/ECF for correct parts of an otherwise flawed method.

This proved difficult for candidates at both HL and SL. Many managed to calculate the final vertical component of the velocity of the ball.

As the command term in this question is ‘predict’ a bald answer of clay was acceptable for one mark. This was a testing question that candidates found demanding but there were some very well-reasoned answers. The most common incorrect answer involved suggesting that the greater frictional force on the clay court left the ball with less kinetic energy and so a smaller angle. At SL many gained the answer that the angle on clay would be greater with the argument that frictional force is greater and so the distance the ball slides is less.

Outline what is meant by electric potential at a point.

The electric potential at a point a distance 2.8 m from the centre of the sphere is 7.71 kV. Determine the radius of the sphere.

Comment on the angle at which the object meets equipotential surfaces around the sphere.

Show that the kinetic energy of the object is about 0.7 mJ.

Determine whether the object will reach the surface of the sphere.

the work done per unit charge ✓

In bringing a small/point/positive/test «charge» from infinity to the point ✓

Allow use of energy per unit charge for MP1

use of Vr = constant ✓

0.40 m ✓

Allow [1] max if r + 2.8 used to get 0.47 m.

Allow [2] marks if they calculate Q at one potential and use it to get the distance at the other potential.

90° / perpendicular ✓

$\frac{1}{2}\times 0.14\times {10}^{-3}\times 3.1^2$  OR  0.67 «mJ» seen ✓

«p.d. between point and sphere surface = » (53.9 kV – 7.71) «kV»  OR  46.2 «kV» seen ✓

«energy required =» VQ « = 46 200 × 2.4 × 10^-8» = 1.11 mJ ✓

this is greater than kinetic energy so will not reach sphere ✓

MP3 is for a conclusion consistent with the calculations shown.

Allow ECF from MP1

a) Well answered.

b) Generally, well answered, but there were quite a few using r + 2.8.

ci) Very few had problems to recognize the perpendicular angle

cii) Good simple calculation

ciii) Many had a good go at this, but a significant number tried to answer it based on forces.

Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs.

The slits are separated by 1.5 mm and the laser light has a wavelength of 6.3 x 10^–7 m. The slits are 5.0 m from the train track. Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum.

Estimate the speed of the train.

Determine the width of one of the slits.

Suggest the variation in the output voltage from the light sensor that will be observed as the train moves beyond the first diffraction minimum.

In another experiment the student replaces the light sensor with a sound sensor. The train travels away from a loudspeaker that is emitting sound waves of constant amplitude and frequency towards a reflecting barrier.

The graph shows the variation with time of the output voltage from the sounds sensor.

Explain how this effect arises.

«light» superposes/interferes

pattern consists of «intensity» maxima and minima
OR
consisting of constructive and destructive «interference»

voltage peaks correspond to interference maxima

«$s=\frac{\lambda D}{d}=\frac{6.3\times {10}^{-7}\times 5.0}{1.5\times {10}^{-3}}=$» 2.1 x 10^–3 «m» 

If no unit assume m.
Correct answer only.

correct read-off from graph of 25 m s

v = «$\frac{x}{t}=\frac{2.1\times {10}^{-3}}{25\times {10}^{-3}}=$» 8.4 x 10^–2 «m s^–1»

Allow ECF from (b)(i)

angular width of diffraction minimum = $\frac{0.13}{5.0}$ «= 0.026 rad»

slit width = «$\frac{\lambda }{d}=\frac{6.3\times {10}^{-7}}{0.026}=$» 2.4 x 10^–5 «m»

Award [1 max] for solution using 1.22 factor.

«beyond the first diffraction minimum» average voltage is smaller

«voltage minimum» spacing is «approximately» same
OR
rate of variation of voltage is unchanged

OWTTE

«reflection at barrier» leads to two waves travelling in opposite directions 

mention of formation of standing wave

maximum corresponds to antinode/maximum displacement «of air molecules»
OR
complete cancellation at node position