State what is meant by gravitational field strength.

Show that V = –g(R + h).

Draw a graph, on the axes, to show the variation of the gravitational potential V of the planet with height h above the surface of the planet.

A planet has a radius of 3.1 × 10⁶ m. At a point P a distance 2.4 × 10⁷ m above the surface of the planet the gravitational field strength is 2.2 N kg^–1. Calculate the gravitational potential at point P, include an appropriate unit for your answer.

The diagram shows the path of an asteroid as it moves past the planet.

When the asteroid was far away from the planet it had negligible speed. Estimate the speed of the asteroid at point P as defined in (b).

The mass of the asteroid is 6.2 × 10¹² kg. Calculate the gravitational force experienced by the planet when the asteroid is at point P.

the «gravitational» force per unit mass exerted on a point/small/test mass

[1 mark]

at height h potential is V = –$\frac{GM}{(R+h)}$

field is g = $\frac{GM}{{(R+h)}^2}$

«dividing gives answer»

Do not allow an answer that starts with g = –$\frac{\mathrm{\Delta }V}{\mathrm{\Delta }r}$ and then cancels the deltas and substitutes R + h

[2 marks]

correct shape and sign

non-zero negative vertical intercept

[2 marks]

V = «–2.2 × (3.1 × 10⁶ + 2.4 × 10⁷) =» «–» 6.0 × 10⁷ J kg^–1

Unit is essential

Allow eg MJ kg^–1 if power of 10 is correct

Allow other correct SI units eg m²s^–2, N m kg^–1

[1 mark]

total energy at P = 0 / KE gained = GPE lost

«$\frac{1}{2}$mv² + mV = 0 ⇒» v = $\sqrt{-2V}$

v = «$\sqrt{2\times 6.0\times {10}^7}$ =» 1.1 × 10⁴ «ms^–1»

Award [3] for a bald correct answer

Ignore negative sign errors in the workings

Allow ECF from 6(b)

[3 marks]

ALTERNATIVE 1

force on asteroid is «6.2 × 10¹² × 2.2 =» 1.4 × 10¹³ «N»

«by Newton’s third law» this is also the force on the planet

ALTERNATIVE 2

mass of planet = 2.4 x 10²⁵ «kg» «from V = –$\frac{GM}{(R+h)}$»

force on planet «$\frac{GMm}{{(R+h)}^2}$» = 1.4 × 10¹³ «N»

MP2 must be explicit

[2 marks]

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]

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 what is meant by the gravitational potential at the surface of a planet.

An unpowered projectile is fired vertically upwards into deep space from the surface of planet Venus. Assume that the gravitational effects of the Sun and the other planets are negligible.

The following data are available.

(i) Determine the initial kinetic energy of the projectile.

(ii) Describe the subsequent motion of the projectile until it is effectively beyond the gravitational field of Venus.

the «gravitational» work done «by an external agent» per/on unit mass/kg

Allow definition in terms of reverse process of moving mass to infinity eg “work done on external agent by…”.
Allow “energy” as equivalent to “work done”

in moving a «small» mass from infinity to the «surface of» planet / to a point

N.B.: on SL paper Q5(a)(i) and (ii) is about “gravitational field”.

i
escape speed
Care with ECF from MP1.

v = «$\sqrt{\left(\frac{2\text{GM}}{R}\right)}=$»

$\sqrt{\left(\frac{2\times 6.67\times {10}^{-11}\times 4.87\times {10}^{24}}{6.05\times {10}^6}\right)}$ or 1.04×10⁴«m s^–1»

or «1.1 × 1.04 × 10⁴ m s^-1»= 1.14 × 10⁴ «m s^–1»

KE = «0.5 × 3500 × (1.1 × 1.04 × 10⁴ m s^–1)² =» 2.27×10¹¹ «J»

Award [1 max] for omission of 1.1 – leads to 1.88×10¹¹ m s^-1.
Award [2] for a bald correct answer.

ii
Velocity/speed decreases / projectile slows down «at decreasing rate»

«magnitude of» deceleration decreases «at decreasing rate»
Mention of deceleration scores MP1 automatically.

velocity becomes constant/non-zero
OR
deceleration tends to zero

Accept “negative acceleration” for “deceleration”.

Must see “velocity” not “speed” for MP3.

Outline why the gravitational potential is negative.

The gravitational potential due to the Sun at a distance r from its centre is V_S. Show that

rV_S = constant.

Calculate the gravitational potential energy of the Earth in its orbit around the Sun. Give your answer to an appropriate number of significant figures.

Calculate the total energy of the Earth in its orbit.

An asteroid strikes the Earth and causes the orbital speed of the Earth to suddenly decrease. Suggest the ways in which the orbit of the Earth will change.

Outline, in terms of the force acting on it, why the Earth remains in a circular orbit around the Sun.

potential is defined to be zero at infinity

so a positive amount of work needs to be supplied for a mass to reach infinity

V_S = $-\frac{GM}{r}$ so r x V_S «= –GM» = constant because G and M are constants

GM = 1.33 x 10²⁰ «J m kg^–1»

GPE at Earth orbit «= –$\frac{1.33\times {10}^{20}\times 6.0\times {10}^{24}}{1.5\times {10}^{11}}$» = «–» 5.3 x 10³³ «J»

Award [1 max] unless answer is to 2 sf.

Ignore addition of Sun radius to radius of Earth orbit.

ALTERNATIVE 1
work leading to statement that kinetic energy $\frac{GMm}{2r}$, AND kinetic energy evaluated to be «+» 2.7 x 10³³ «J»

energy «= PE + KE = answer to (b)(ii) + 2.7 x 10³³» = «–» 2.7 x 10³³ «J»

ALTERNATIVE 2
statement that kinetic energy is $=-\frac{1}{2}$ gravitational potential energy in orbit

so energy «$=\frac{\text{answer to (b)(ii)}}{2}$» = «–» 2.7 x 10³³ «J»

Various approaches possible.

«KE will initially decrease so» total energy decreases
OR
«KE will initially decrease so» total energy becomes more negative

Earth moves closer to Sun

new orbit with greater speed «but lower total energy»

changes ellipticity of orbit

centripetal force is required

and is provided by gravitational force between Earth and Sun

Award [1 max] for statement that there is a “centripetal force of gravity” without further qualification.

Satellite X orbits 6600 km from the centre of the Earth.

Mass of the Earth = 6.0 x 10²⁴ kg

Show that the orbital speed of satellite X is about 8 km s^–1.

the orbital times for X and Y are different.

satellite Y requires a propulsion system.

The cable between the satellites cuts the magnetic field lines of the Earth at right angles.

Explain why satellite X becomes positively charged.

Satellite X must release ions into the space between the satellites. Explain why the current in the cable will become zero unless there is a method for transferring charge from X to Y.

The magnetic field strength of the Earth is 31 μT at the orbital radius of the satellites. The cable is 15 km in length. Calculate the emf induced in the cable.

Estimate the value of k in the following expression.

T = $2\pi \sqrt{\frac{m}{k}}$

Give an appropriate unit for your answer. Ignore the mass of the cable and any oscillation of satellite X.

Describe the energy changes in the satellite Y-cable system during one cycle of the oscillation.

«$v=\sqrt{\frac{G{M}_E}{r}}$» = $\sqrt{\frac{6.67\times {10}^{-11}\times 6.0\times {10}^{24}}{6600\times {10}^3}}$

7800 «m s^–1»

Full substitution required

Must see 2+ significant figures.

Y has smaller orbit/orbital speed is greater so time period is less

Allow answer from appropriate equation

Allow converse argument for X

to stop Y from getting ahead

to remain stationary with respect to X

otherwise will add tension to cable/damage satellite/pull X out of its orbit

cable is a conductor and contains electrons

electrons/charges experience a force when moving in a magnetic field

use of a suitable hand rule to show that satellite Y becomes negative «so X becomes positive»

Alternative 2

cable is a conductor

so current will flow by induction flow when it moves through a B field

use of a suitable hand rule to show current to right so «X becomes positive»

Marks should be awarded from either one alternative or the other.

Do not allow discussion of positive charges moving towards X

electrons would build up at satellite Y/positive charge at X

preventing further charge flow

by electrostatic repulsion

unless a complete circuit exists

«ε = Blv =» 31 x 10^–6 x 7990 x 15000

3600 «V»

Allow 3700 «V» from v = 8000 m s^–1.

use of k = «$\frac{4{\pi }^{2}m}{T^2}=$» $\frac{4\times {\pi }^2\times 350}{{5.2}^2}$

510

N m^–1 or kg s^–2

Allow MP1 and MP2 for a bald correct answer

Allow 500

Allow N/m etc.

E_p in the cable/system transfers to E_k of Y

and back again twice in each cycle

Exclusive use of gravitational potential energy negates MP1

Calculate, for the surface of $\text{Io}$, the gravitational field strength g_Io due to the mass of $\text{Io}$. State an appropriate unit for your answer.

Show that the $\frac{\mathrm{gravitational} \mathrm{potential} \mathrm{due} \mathrm{to} \mathrm{Jupiter} \mathrm{at} \mathrm{the} \mathrm{orbit} \mathrm{of} \mathrm{Io}}{ \mathrm{gravitational} \mathrm{potential} \mathrm{due} \mathrm{to} \mathrm{Io} \mathrm{at} \mathrm{the} \mathrm{surface} \mathrm{of} \mathrm{Io}}$ is about 80.

Outline, using (b)(i), why it is not correct to use the equation $\sqrt{\frac{2G\times \text{mass of Io}}{\text{radius of Io}}}$ to calculate the speed required for the spacecraft to reach infinity from the surface of $\text{Io}$.

An engineer needs to move a space probe of mass 3600 kg from Ganymede to Callisto. Calculate the energy required to move the probe from the orbital radius of Ganymede to the orbital radius of Callisto. Ignore the mass of the moons in your calculation. 

$«\frac{GM}{r^2}=\frac{6.67\times {10}^{-11}\times 8.9\times {10}^{22}}{{\left(1.8\times {10}^6\right)}^2}=»1.8$ ✓

N kg⁻¹  OR  m s⁻²  ✓

$\frac{1.9\times {10}^{27}}{4.9\times {10}^8}$  AND  $\frac{8.9\times {10}^{22}}{1.8\times {10}^6}$ seen ✓

$«\frac{1.9\times {10}^{27}\times 1.8\times {10}^6}{4.9\times {10}^8\times 8.9\times {10}^{22}}=»78$  ✓

For MP1, potentials can be seen individually or as a ratio.

«this is the escape speed for $\text{Io}$ alone but» gravitational potential / field of Jupiter must be taken into account  ✓

OWTTE

$-G{M}_{\text{Jupiter}}\left(\frac{1}{1.88\times {10}^9}-\frac{1}{1.06\times {10}^9}\right)=«5.21\times {10}^7 {\text{J\hspace{0.05em}kg}}^{-1}»$  ✓

« multiplies by 3600 kg to get » 1.9 × 10¹¹«J» ✓

Award [2] marks if factor of ½ used, taking into account orbital kinetic energies, leading to a final answer of 9.4 x 10¹⁰«J».

Allow ECF from MP1

Award [2] marks for a bald correct answer.

Explain why the electric potential decreases from A to B.

Draw, on the axes, the variation of electric potential $V$ with distance $r$ from the centre of the sphere.

Calculate the electric potential difference between points A and B.

Determine the charge $Q$ of the sphere.

The concept of potential is also used in the context of gravitational fields. Suggest why scientists developed a common terminology to describe different types of fields.

ALTERNATIVE 1
work done on moving a positive test charge in any outward direction is negative ✓
potential difference is proportional to this work «so $V$ decreases from A to B» ✓

ALTERNATIVE 2
potential gradient is directed opposite to the field so inwards ✓
the gradient indicates the direction of increase of $V$ «hence $V$ increases towards the centre/decreases from A to B» ✓

ALTERNATIVE 3
$V=\frac{kQ}{R}$ so as $r$ increases $V$ decreases ✓
$V$ is positive as $Q$ is positive ✓

ALTERNATIVE 4
the work done per unit charge in bringing a positive charge from infinity ✓
to point B is less than point A ✓

curve decreasing asymptotically for $r>R$ ✓

non $-$ zero constant between $0$ and $R$ ✓

$«\frac{W}{q}=\frac{1.7\times {10}^{-16}}{1.60\times {10}^{-19}}=»1.1\times {10}^3 «\mathrm{V}»$ ✓

$8.99\times {10}^9\times Q\times \left(\frac{1}{5.0\times {10}^{-2}}-\frac{1}{1.0\times {10}^{-1}}\right)=1.1\times {10}^3$ ✓

$Q=1.2\times {10}^{-8} «\mathrm{C}»$ ✓

to highlight similarities between «different» fields ✓

The majority who answered in terms of potential gained one mark. Often the answers were in terms of work done rather than work done per unit charge or missed the fact that the potential is positive.

This was well answered.

Most didn't realise that the key to the answer is the definition of potential or potential difference and tried to answer using one of the formulae in the data booklet, but incorrectly.

Even though many were able to choose the appropriate formula from the data booklet they were often hampered in their use of the formula by incorrect techniques when using fractions.

This was generally well answered with only a small number of answers suggesting greater international cooperation.

Outline how this diagram shows that the gravitational field strength of planet X decreases with distance from the surface.

The diagram shows part of the surface of planet X. The gravitational potential at the surface of planet X is –3V and the gravitational potential at point Y is –V.

Sketch on the grid the equipotential surface corresponding to a gravitational potential of –2V.

A meteorite, very far from planet X begins to fall to the surface with a negligibly small initial speed. The mass of planet X is 3.1 × 10²¹ kg and its radius is 1.2 × 10⁶ m. The planet has no atmosphere. Calculate the speed at which the meteorite will hit the surface.

At the instant of impact the meteorite which is made of ice has a temperature of 0 °C. Assume that all the kinetic energy at impact gets transferred into internal energy in the meteorite. Calculate the percentage of the meteorite’s mass that melts. The specific latent heat of fusion of ice is 3.3 × 10⁵ J kg^–1.

the field lines/arrows are further apart at greater distances from the surface

circle centred on Planet X
three units from Planet X centre

loss in gravitational potential = $\frac{6.67\times {10}^{-11}\times 3.1\times {10}^{21}}{1.2\times {10}^6}$

«= 1.72 × 10⁵ JKg⁻¹»

equate to $\frac{1}{2}$v²

v = 590 «m s⁻¹»

Allow ECF from MP1.

available energy to melt one kg 1.72 × 10⁵ «J»

fraction that melts is $\frac{1.72\times {10}^5}{3.3\times {10}^5}$ = 0.52 OR 52%

Allow ECF from MP1.

Allow 53% from use of 590 ms^-1.

Show that the intensity of the solar radiation at the location of Titan is 16 W m⁻².

Titan has an atmosphere of nitrogen. The albedo of the atmosphere is 0.22. The surface of Titan may be assumed to be a black body. Explain why the average intensity of solar radiation absorbed by the whole surface of Titan is 3.1 W m⁻².

Show that the equilibrium surface temperature of Titan is about 90 K.

The mass of Titan is 0.025 times the mass of the Earth and its radius is 0.404 times the radius of the Earth. The escape speed from Earth is 11.2 km s⁻¹. Show that the escape speed from Titan is 2.8 km s⁻¹.

The orbital radius of Titan around Saturn is $R$ and the period of revolution is $T$.

Show that $T^2=\frac{4{\mathrm{\pi }}^2{R}^{ 3}}{GM}$ where $M$ is the mass of Saturn.

The orbital radius of Titan around Saturn is 1.2 × 10⁹m and the orbital period is 15.9 days. Estimate the mass of Saturn.

Show that the mass of a nitrogen molecule is 4.7 × 10⁻²⁶ kg.

Estimate the root mean square speed of nitrogen molecules in the Titan atmosphere. Assume an atmosphere temperature of 90 K.

Discuss, by reference to the answer in (b), whether it is likely that Titan will lose its atmosphere of nitrogen.

incident intensity $\frac{1360}{9.3^2}$ OR $15.7\approx 16$ «W m⁻²» ✓

Allow the use of 1400 for the solar constant.

exposed surface is ¼ of the total surface ✓

absorbed intensity = (1−0.22) × incident intensity ✓

0.78 × 0.25 × 15.7  OR  3.07 «W m⁻²» ✓

Allow 3.06 from rounding and 3.12 if they use 16 W m⁻².

σT ⁴ = 3.07

OR

T = 86 «K» ✓

$v=«\sqrt{\frac{2GM}{R}}=»\sqrt{\frac{0.025}{0.404}}\times 11.2$

OR

$2.79$ «km s⁻¹» ✓

correct equating of gravitational force / acceleration to centripetal force / acceleration ✓

correct rearrangement to reach the expression given ✓

Allow use of $\sqrt{\frac{GM}{R}}=\frac{2\mathrm{\pi }R}{T}$ for MP1.

$T=15.9\times 24\times 3600$ «s» ✓

$M=\frac{4{\mathrm{\pi }}^2{\left(1.2\times {10}^9\right)}^3}{6.67\times {10}^{-11}\times {\left(15.9\times 24\times 3600\right)}^2}=5.4\times {10}^{26}$«kg» ✓

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

$m=\frac{28\times {10}^{-3}}{6.02\times {10}^{23}}$

OR

$4.65\times {10}^{-26}$ «kg» ✓

$«\frac{1}{2}m{v}^2=\frac{3}{2}kT\Rightarrow »v=\sqrt{\frac{3kT}{m}}$ ✓

$v=«\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 90}{4.651\times {10}^{-26}}}=»283\approx 300$ «ms⁻¹» ✓

Award [2] marks for a bald correct answer.

Allow 282 from a rounded mass.

no, molecular speeds much less than escape speed ✓

Allow ECF from incorrect (d)(ii).

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

Show that the charge on the surface of the sphere is +18 μC.

Describe, in terms of electron flow, how the smaller sphere becomes charged.

Predict the charge on each sphere.

$Q=«\frac{VR}{k}=»\frac{3.4\times {10}^5\times 0.48}{8.99\times {10}^9}$

OR

$Q=18.2$ «μC» ✓

electrons leave the small sphere «making it positively charged» ✓

$k\frac{q_1}{48}=k\frac{q_2}{24}\Rightarrow q_1=2q_2$ ✓

$q_1+q_2=18$ ✓

so $q_1=12$ «μC», $q_2=6.0$ «μC» ✓

Award [3] marks for a bald correct answer.

Show that the total energy of the planet is given by the equation shown.

$E=\frac{1}{2}mV$

Suppose the star could contract to half its original radius without any loss of mass. Discuss the effect, if any, this has on the total energy of the planet.

The diagram shows some of the electric field lines for two fixed, charged particles X and Y.

The magnitude of the charge on X is $Q$ and that on Y is $q$. The distance between X and Y is 0.600 m. The distance between P and Y is 0.820 m.

At P the electric field is zero. Determine, to one significant figure, the ratio $\frac{Q}{q}$.

$E=\frac{1}{2}m\frac{GM}{r}-\frac{GMm}{r}=-\frac{1}{2}\frac{GMm}{r}$  ✔

comparison with $V=-\frac{GM}{r}$   ✔

«to give answer»

ALTERNATIVE 1

«at the position of the planet» the potential depends only on the mass of the star /does not depend on the radius of the star ✔

the potential will not change and so the energy will not change ✔

ALTERNATIVE 2

r / distance between the centres of the objects / orbital radius remains unchanged ✔

since $E_{Total}=-\frac{1}{2}\frac{GMm}{r}$, energy will not change ✔

$\frac{kQ}{{(0.600+0.820)}^2}=\frac{kq}{{0.820}^2}$  ✔

$\frac{Q}{q}=«\frac{{(0.600+0.820)}^2}{{0.820}^2}=2.9988\approx »3$  ✔

This was generally well answered but with candidates sometimes getting in to trouble over negative signs but otherwise producing well-presented answers.

A large number of candidates thought that the total energy of the planet would change, mostly double.

The majority of candidates had an idea of the basic technique here but it was surprisingly common to see the squared missing from the expression for field strengths.

State how the density of a nucleus varies with the number of nucleons in the nucleus.

Show that the nuclear radius of phosphorus-31 (${}_{15}^{31}\text{P}$) is about 4 fm.

State the maximum distance between the centres of the nuclei for which the production of ${}_{15}^{32}\text{P}$ is likely to occur.

Determine, in J, the minimum initial kinetic energy that the deuterium nucleus must have in order to produce ${}_{15}^{32}\text{P}$. Assume that the phosphorus nucleus is stationary throughout the interaction and that only electrostatic forces act.

${}_{15}^{32}\text{P}$ undergoes beta-minus (β^–) decay. Explain why the energy gained by the emitted beta particles in this decay is not the same for every beta particle.

State what is meant by decay constant.

In a fresh pure sample of ${}_{15}^{32}\text{P}$ the activity of the sample is 24 Bq. After one week the activity has become 17 Bq. Calculate, in s^–1, the decay constant of ${}_{15}^{32}\text{P}$.

it is constant ✔

R = $\text{1}\text{.20}\times {10}^{-15}\times {31}^{\frac{1}{3}}=3.8\times {10}^{-15}$ «m» ✔

Must see working and answer to at least 2SF

separation for interaction = 5.3 or 5.5 «fm» ✔

energy required = $\frac{15e^2}{4\pi {\epsilon }_0\times 5.3\times {10}^{-15}}$ ✔

= 6.5 / 6.6 ×10⁻¹³ OR 6.3 ×10⁻¹³«J» ✔

Allow ecf from (b)(i)

«electron» antineutrino also emitted ✔

energy split between electron and «anti»neutrino ✔

probability of decay of a nucleus ✔

OR

the fraction of the number of nuclei that decay

in one/the next second

OR

per unit time ✔

1 week = 6.05 × 10⁵ «s»

17 = $24{\text{e}}^{-\lambda \times 6.1\times {10}^5}$ ✔

5.7 × 10⁻⁷ «s^–1» ✔

Award [2 max] if answer is not in seconds

If answer not in seconds and no unit quoted award [1 max] for correct substitution into equation (MP2)

Show that the speed $v$ of the electron with mass $m$, is given by $v=\sqrt{\frac{2k{e}^2}{mr}}$.

Hence, deduce that the total energy of the electron is given by $E_{\mathrm{TOT}}=-\frac{k{e}^2}{r}$.

In this model the electron loses energy by emitting electromagnetic waves. Describe the predicted effect of this emission on the orbital radius of the electron.

Show that the de Broglie wavelength $\lambda$ of the electron in the $n=3$ state is  $\lambda =5.1\times {10}^{-10}$ m.

The formula for the de Broglie wavelength of a particle is $\lambda =\frac{h}{mv}$.

Estimate for $n=3$, the ratio $\frac{\mathrm{circumference} \mathrm{of} \mathrm{orbit}}{\mathrm{de} \mathrm{Broglie} \mathrm{wavelength} \mathrm{of} \mathrm{electron}}$.

State your answer to one significant figure.

The description of the electron is different in the Schrodinger theory than in the Bohr model. Compare and contrast the description of the electron according to the Bohr model and to the Schrodinger theory.

equating centripetal to electrical force $\frac{2k{e}^2}{r^2}=\frac{m{v}^2}{r}$ to get result ✔

uses (a)(i) to state $E_{\mathrm{k}}=\frac{k{e}^2}{r}$ OR states $E_{\mathrm{p}}=-\frac{2k{e}^2}{r}$ ✔

adds « $E_{\mathrm{TOT}}=E_{\mathrm{k}}+E_{\mathrm{p}}=\frac{k{e}^2}{r}-\frac{2k{e}^2}{r}$» to get the result ✔

the total energy decreases
OR
by reference to $E_{\mathrm{TOT}}=-\frac{k{e}^2}{r}$ ✔

the radius must also decrease ✔

NOTE: Award [0] for an answer concluding that radius increases

with $n=3, v=«\sqrt{\frac{2\times 8.99\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{9.11\times {10}^{-31}\times 9\times 2.7\times {10}^{-11}}}=» 1.44\times {10}^6 «\mathrm{m} {\mathrm{s}}^{-1}»$✔

$\lambda =\frac{6.63\times {10}^{-34}}{9.11\times {10}^{-31}\times 1.44\times {10}^6}$  OR  $\lambda =5.05\times {10}^{-10} «\mathrm{m}»$✔

$\frac{{\displaystyle 2\pi r}}{{\displaystyle \lambda }}=«\frac{{\displaystyle 2\pi \times 9\times 2.7\times {10}^{-11}}}{{\displaystyle 5.1\times {10}^{-10}}}=2.99»\cong 3$ ✔

NOTE: Allow ECF from (b)(i)

reference to fixed orbits/specific radii OR quantized angular momentum in Bohr model ✔

electron described by a wavefunction/as a wave in Schrödinger model OR as particle in Bohr model ✔

reference to «same» energy levels in both models ✔

reference to «relationship between wavefunction and» probability «of finding an electron in a point» in Schrödinger model ✔

Calculate the total length of aluminium foil that the student will require.

The plastic film begins to conduct when the electric field strength in it exceeds 1.5 MN C^–1. Calculate the maximum charge that can be stored on the capacitor.

The resistor R in the circuit has a resistance of 1.2 kΩ. Calculate the time taken for the charge on the capacitor to fall to 50 % of its fully charged value.

The ammeter is replaced by a coil. Explain why there will be an induced emf in the coil while the capacitor is discharging.

Suggest one change to the discharge circuit, apart from changes to the coil, that will increase the maximum induced emf in the coil.

length = $\frac{d\times C}{\text{width}\times \epsilon }$ ✔

= 0.33 «m» ✔

so 0.66/0.67 «m» «as two lengths required» ✔

1.5 × 10⁶ × 55 × 10^-6 = 83 «V» ✔

q «= CV»= 5.6 × 10^-6 «C»✔

$0.5={\text{e}}^{-\frac{t}{RC}}={\text{e}}^{-\frac{t}{1200\times 6.8\times {10}^{-8}}}$

t = «−» 1200 × 6.8 × 10⁻⁸ × ln0.5 ✔

5.7 × 10⁻⁵ «s» ✔

OR

use of t $\frac{1}{2}$ = RC × ln2 ✔

1200 × 6.8 × 10⁻⁸ × 0.693 ✔

5.7 × 10⁻⁵ «s» ✔

mention of Faraday’s law ✔

indicating that changing current in discharge circuit leads to change in flux in coil/change in magnetic field «and induced emf» ✔

decrease/reduce ✔

resistance (R) OR capacitance (C) ✔

Many candidates were able to use the proper equation to calculate the length of one piece of aluminum foil for the first two marks, but very few doubled the length for the final mark.

This question was challenging for many candidates. While some candidates were able to use proper equations for capacitors to determine the charge some of the candidates attempted to use electrostatic equations for the electric field around a point charge to solve this problem.

This question was also challenging for many candidates, with not an insignificant number leaving it blank. The candidates who did attempt it generally set up a correct equation, but ran into some simple calculation and power of ten errors. Some candidates attempted to solve the equation using basic circuit equations, which did not receive any marks.

This is an explain question, so there was an expectation for a fairly detailed response. Many candidates missed the fact that the discharging capacitor is causing the current in the coil to change in time, and that this is what is inducing the emf in the coil. Many simply stated that the current created a magnetic field with not complete explanation of induction.

Candidates who recognized that something about the discharge circuit (not the charging circuit) needed to be changed generally suggested that something had to change with the resistance or capacitance. It should be noted that even though this was the last question on the exam, it was attempted at a higher rate than many of the other questions on the exam.

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.

Explain why a centripetal force is needed for the planet to be in a circular orbit.

Calculate the value of the centripetal force.

Show that the gravitational potential due to the planet and the star at the surface of the planet is about −5 × 10⁹J kg⁻¹.

Estimate the escape speed of the spacecraft from the planet–star system.

«circular motion» involves a changing velocity ✓

«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓

there must be a force/acceleration towards centre/star ✓

without a centripetal force the planet will move in a straight line ✓

$F=\frac{(6.67\times {10}^{-11})(8\times {10}^{24})(3.2\times {10}^{30})}{(4.4\times {10}^{10}{)}^2}=8.8\times {10}^{23}$ «N» ✓

V_planet = «−»$\frac{(6.67\times {10}^{-11})(8\times {10}^{24})}{9.1\times {10}^6}=$«−» 5.9 × 10⁷ «J kg⁻¹» ✓

V_star = «−»$\frac{(6.67\times {10}^{-11})(3.2\times {10}^{30})}{4.4\times {10}^{10}}=$«−» 4.9 × 10⁹ «J kg⁻¹» ✓

V_planet + V_star = «−» 4.9 «09» × 10⁹ «J kg⁻¹» ✓

Must see substitutions and not just equations.

use of v_esc = $\sqrt{2V}$ ✓

v = 9.91 × 10⁴ «m s⁻¹» ✓

Calculate the electric potential at O.

Sketch, on the axes, the variation of the electric potential V with distance between X and Y.

Identify the direction of the resultant force acting on Z as it oscillates.

Deduce whether the motion of Z is simple harmonic.

use of $\frac{kQ}{r}$ ✓

$\frac{\left(8.99\times {10}^9\right)\left(4\times {10}^{-6}\right)}{0.15}$ OR 240 «kV» for one charge calculated ✓

480 «kV» for both ✓

MP1 can be seen or implied from calculation.

Allow ECF from MP2 for MP3.

symmetric curve around 0 with potential always positive, “bowl shape up” and curve not touching the horizontal axis. ✓

clear asymptotes at X and Y ✓

force is towards O ✓

always ✓

ALTERNATIVE 1

motion is not SHM ✓

«because SHM requires force proportional to r and» this force depends on $\frac{1}{r^2}$ ✓

ALTERNATIVE 2

motion is not SHM ✓

energy-distance «graph must be parabolic for SHM and this» graph is not parabolic ✓

This question was generally well approached. Two common errors were either starting with the wrong equation (electric potential energy or Coulomb's law) or subtracting the potentials rather than adding them.

Very few candidates drew a graph that was awarded two marks. Many had a generally correct shape, but common errors were drawing the graph touching the x-axis at O and drawing a general parabola with no clear asymptotes.

Many candidates were able to identify the direction of the force on the particle at position Z, but a common error was to miss that the question was about the direction as the particle was oscillating. Examiners were looking for a clear understanding that the force was always directed toward the equilibrium position, and not just at the moment shown in the diagram.

This was a challenging question for candidates. Most simply assumed that because the charge was oscillating that this meant the motion was simple harmonic. Some did recognize that it was not, and most of those candidates correctly identified that the relationship between force and displacement was an inverse square.