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 centre of the ball, still carrying a charge of $1.2\times {10}^{-6} \text{C}$, is now placed $0.40 \text{m}$ from a point charge Q. The charge on the ball acts as a point charge at the centre of the ball.

P is the point on the line joining the charges where the electric field strength is zero.
The distance PQ is $0.22 \text{m}$.

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

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 the 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.

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

$«+»1.8\times {10}^{-6} «\text{C}»$ ✓

2sf ✓

Do not award MP2 if charge is negative

Any answer given to 2 sig figs scores MP3

The work done to move a particle of charge 0.25 μC from one point in an electric field to another is 4.5 μJ. Calculate the magnitude of the potential difference between the two points.

Determine the force on Q at the instant it is released.

Describe the motion of Q after release.

On the diagram draw an arrow to show the direction of the magnetic field at Q due to wire X alone.

Determine the magnitude and direction of the resultant magnetic field at Q.

«$V=\frac{4.5}{0.25}=$» 18 «V» ✓

$F=\frac{8.99\times {10}^9\times 68\times {10}^{-6}\times 0.25\times {10}^{-6}}{0.{48}^2}$ ✓

$F=0.66$ «N» ✓

Award [2] marks for a bald correct answer.

Allow symbolic k in substitutions for MP1.

Do not allow ECF from incorrect or not squared distance.

Q moves to the right/away from P «along a straight line»

OR

Q is repelled from P ✓

with increasing speed/Q accelerates ✓

acceleration decreases ✓

arrow of any length as shown ✓

«using components or Pythagoras to get» B = 21 «mT» ✓

directed «horizontally» to the right ✓

If no unit seen, assume mT.

(i) State how the resistance of T varies with the current going through T.

(ii) Deduce, without a numerical calculation, whether R or T has the greater resistance at I=0.40 A.

Components R and T are placed in a circuit. Both meters are ideal.

Slider Z of the potentiometer is moved from Y to X.

(i) State what happens to the magnitude of the current in the ammeter.

(ii) Estimate, with an explanation, the voltmeter reading when the ammeter reads 0.20 A.

i

R_T decreases with increasing I

OR

R_T and I are negatively correlated

Must see reference to direction of change of current in first alternative.
Do not allow “inverse proportionality”.
May be worth noting any marks on graph relating to 7bii

ii

at 0.4 A: V_R > V_T or V_R= 5.6 V and V_T = 5.3 V

Award [0] for a bald correct answer without deduction or with incorrect reasoning.

Ignore any references to graph gradients.

so R_R >R_T  because V = IR / V∝ R «and I same for both»

Both elements must be present for MP2 to be awarded.

i

decreases
OR
becomes zero at X

ii

realization that V is the same for R and T
OR
identifies that currents are 0.14 A and 0.06 A

Award [0] if pds 2.8 V and 3.7 V or 1.4 V and 2.6V are used in any way. Otherwise award [1 max] for a bald correct answer. Explanation expected.

2 V = 2 V OR 2.0 V

Each rod is to have a resistance no greater than 0.10 Ω. Calculate, in m, the minimum radius of each rod. Give your answer to an appropriate number of significant figures.

Calculate the maximum number of lamps that can be connected between the rods. Neglect the resistance of the rods.

One advantage of this system is that if one lamp fails then the other lamps in the circuit remain lit. Outline one other electrical advantage of this system compared to one in which the lamps are connected in series.

ALTERNATIVE 1:

$r=\sqrt{\frac{\rho l}{\pi \text{R}}}$ OR $\sqrt{\frac{7.2\times {10}^{-7}\times 12.5}{\pi \times 0.1}}$ ✔

r = 5.352 × 10⁻³ ✔

5.4 × 10⁻³ «m» ✔

ALTERNATIVE 2:

$A=\frac{7.2\times {10}^{-7}\times 12.5}{0.1}$ ✔

r = 5.352 × 10⁻³ ✔

5.4 × 10⁻³ «m» ✔

current in lamp = $\frac{5}{24}$ «= 0.21» «A»

OR

n = 24 × $\frac{8}{5}$ ✔

so «38.4 and therefore» 38 lamps ✔

when adding more lamps in parallel the brightness stays the same ✔

when adding more lamps in parallel the pd across each remains the same/at the operating value/24 V ✔

when adding more lamps in parallel the current through each remains the same ✔

lamps can be controlled independently ✔

the pd across each bulb is larger in parallel ✔

the current in each bulb is greater in parallel ✔

lamps will be brighter in parallel than in series ✔

In parallel the pd across the lamps will be the operating value/24 V ✔

Accept converse arguments for adding lamps in series:

when adding more lamps in series the brightness decreases

when adding more lamps in series the pd decreases

when adding more lamps in series the current decreases

lamps can’t be controlled independently

the pd across each bulb is smaller in series

the current in each bulb is smaller in series

in series the pd across the lamps will less than the operating value/24 V

Do not accept statements that only compare the overall resistance of the combination of bulbs.

State the quark structures of a meson and a baryon.

Explain which interaction is responsible for this decay.

Draw arrow heads on the lines representing $\overline{u}$ and d in the ${\pi }^{-}$.

Identify the exchange particle in this decay.

Outline one benefit of international cooperation in the construction or use of high-energy particle accelerators.

Meson: quark-antiquark pair
Baryon: 3 quarks

Alternative 1

strange quark changes «flavour» to an up quark

changes in quarks/strangeness happen only by the weak interaction

Alternative 2

Strangeness is not conserved in this decay «because the strange quark changes to an up quark»

Strangeness is not conserved during the weak interaction

Do not allow a bald answer of weak interaction.

arrows drawn in the direction shown

Both needed for [1] mark.

W ⁻

Do not allow W or W⁺.

it lowers the cost to individual nations, as the costs are shared

international co-operation leads to international understanding OR historical example of co-operation OR co-operation always allows science to proceed

large quantities of data are produced that are more than one institution/research group can handle co-operation allows effective analysis

Any one.

Calculate the current in the copper cable.

Calculate the resistance of the cable.

Explain, in terms of electrons, what happens to the resistance of the cable as the temperature of the cable increases.

The heater changes the temperature of the water by 35 K. The specific heat capacity of water is 4200 J kg^–1 K^–1.

Determine the rate at which water flows through the shower. State an appropriate unit for your answer.

I «=$\frac{8.5\times {10}^3}{240}$» =35«A»

R = $\frac{1.7\times {10}^{-8}\times 10}{6.0\times {10}^{-6}}$

= 0.028 «Ω»

Allow missed powers of 10 for MP1.

«as temperature increases» there is greater vibration of the metal atoms/lattice/lattice ions

OR

increased collisions of electrons

drift velocity decreases «so current decreases»

«as V constant so» R increases

Award [0] for suggestions that the speed of electrons increases so resistance decreases.

recognition that power = flow rate × cΔT

flow rate «$=\frac{\text{power}}{c\mathrm{\Delta }T}$» $=\frac{8.5\times {10}^3}{4200\times 35}$

= 0.058 «kg s^–1»

kg s⁻¹ / g s⁻¹ / l s⁻¹ / ml s⁻¹ / m³ s⁻¹

Allow MP4 if a bald flow rate unit is stated. Do not allow imperial units.

Outline why component X is considered non-ohmic.

Determine the resistance of the variable resistor.

Calculate the power dissipated in the circuit.

State the range of current that the ammeter can measure as the slider S of the potential divider is moved from Q to P.

Describe, by reference to your answer for (c)(i), the advantage of the potential divider arrangement over the arrangement in (b).

current is not «directly» proportional to the potential difference
OR
resistance of X is not constant
OR
resistance of X changes «with current/voltage» ✓

ALTERNATIVE 1

voltage across X$=2.3 «\mathrm{V}»$ ✓

voltage across R$«=4.0-2.3»=1.7 «\mathrm{V}»$ ✓

resistance of variable resistor $«=\frac{1.7}{0.020}»=85 «\mathrm{\Omega }»$ ✓

ALTERNATIVE 2

overall resistance $«=\frac{4.0}{0.020}»=200 «\mathrm{\Omega }»$ ✓

resistance of X $«=\frac{2.3}{0.020}»=115 «\mathrm{\Omega }»$ ✓

resistance of variable resistor $«=200-115»=85 «\mathrm{\Omega }»$ ✓

power $«=4.0\times 0.020»=0.080 «\mathrm{W}»$ ✓

from $0$ to $60 \mathrm{mA}$ ✓

allows zero current through component X / potential divider arrangement ✓

provides greater range «of current through component X» ✓

The resistance of the carbon film is 82 Ω. The resistivity of carbon is 4.1 x 10^–5 Ω m. Calculate the length l of the film.

The film must dissipate a power less than 1500 W from each square metre of its surface to avoid damage. Calculate the maximum allowable current for the resistor.

State why knowledge of quantities such as resistivity is useful to scientists.

The current direction is now changed so that charge flows vertically through the film.

Deduce, without calculation, the change in the resistance.

Draw a circuit diagram to show how you could measure the resistance of the carbon-film resistor using a potential divider arrangement to limit the potential difference across the resistor.

«l = $\frac{RA}{\rho }=\frac{82\times 8\times {10}^{-3}\times 2\times {10}^{-6}}{4.1\times {10}^{-5}}$»

0.032 «m»

power = 1500 × 8 × 10^–3 × 0.032 «= 0.384»

«current ≤ $\sqrt{\frac{\text{power}}{\text{resistance}}}=\sqrt{\frac{0.384}{82}}$»

0.068 «A»

Be aware of ECF from (a)(i)

Award [1] for 4.3 «A» where candidate has not calculated area

quantities such as resistivity depend on the material

OR

they allow the selection of the correct material

OR

they allow scientists to compare properties of materials

as area is larger and length is smaller

resistance is «very much» smaller

Award [1 max] for answers that involve a calculation

complete functional circuit with ammeter in series with resistor and voltmeter across it

potential divider arrangement correct

eg:

The molar mass of water is 18 g mol⁻¹. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas.

State one assumption of the kinetic model of an ideal gas.

Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer.

Explain why the temperature of water remains at 100 °C during this time.

The heater is removed and a mass of 0.30 kg of pasta at −10 °C is added to the boiling water.

Determine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible.

Specific heat capacity of pasta = 1.8 kJ kg⁻¹ K⁻¹
Specific heat capacity of water = 4.2 kJ kg⁻¹ K⁻¹

Show that each resistor has a resistance of about 30 Ω.

Calculate the power transferred by the heater when both switches are closed.

E_k = « $\frac{3}{2}(1.38\times {10}^{-23})\left(373\right)$» = $7.7\times {10}^{-21}$ «J» ✓

v = «$\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 6.02\times {10}^{23}\times 373}{0.018}}$» = 720 «m s⁻¹» ✓

particles can be considered points «without dimensions» ✓

no intermolecular forces/no forces between particles «except during collisions»✓

the volume of a particle is negligible compared to volume of gas ✓

collisions between particles are elastic ✓

time between particle collisions are greater than time of collision ✓

no intermolecular PE/no PE between particles  ✓

Accept reference to atoms/molecules for “particle”

«mL = P t» so «$L=\frac{1600\times 200}{0.14}$» = 2.3 x 10⁶ «J kg^-1» ✓

J kg⁻¹ ✓

«all» of the energy added is used to increase the «intermolecular» potential energy of the particles/break «intermolecular» bonds/OWTTE ✓

Accept reference to atoms/molecules for “particle” 

use of mcΔT ✓

0.86 × 4200 × (100 – T) = 0.3 × 1800 × (T +10) ✓

T_eq = 85.69«°C» ≅ 86«°C» ✓

Accept T_eq in Kelvin (359 K).

$P=\frac{v^2}{R} \mathrm{so} \frac{{220}^2}{1600} \mathrm{so} R=30.25$ «Ω» ✓

Must see either the substituted values OR a value for R to at least three s.f.

use of parallel resistors addition so R_eq = 15 «Ω» ✓

P = 3200 «W» ✓

State the direction of the magnetic field.

Calculate, in N, the magnitude of the magnetic force acting on the electron.

Explain why the electron moves at constant speed.

Explain why the electron moves on a circular path.

out of the page plane / ⊙

Do not accept just “up” or “outwards”.

[1 mark]

1.60 × 10^–19 × 6.8 × 10⁵ × 8.5 = 9.2 × 10^–13 «N»

[1 mark]

the magnetic force does not do work on the electron hence does not change the electron’s kinetic energy

OR

the magnetic force/acceleration is at right angles to velocity

[1 mark]

the velocity of the electron is at right angles to the magnetic field

(therefore) there is a centripetal acceleration / force acting on the charge

OWTTE

[2 marks]

Calculate the resistance of the conductor.

Calculate the drift speed v of the electrons in the conductor in cm s^–1. State your answer to an appropriate number of significant figures.

1.7 × 10^–8 × $\frac{0.10}{{(0.02\times {10}^{-2})}^2}$

0.043 «Ω»

[2 marks]

v «= $\frac{I}{neA}$» = $\frac{2}{8.5\times {10}^{22}\times 1.60\times {10}^{-19}\times {0.02}^2}$

0.368 «cms^–1»

0.37 «cms^–1»

Award [2 max] if answer is not expressed to 2 sf.

[3 marks]

State what is meant by the emf of a cell.

Show that the resistance of the wire AC is 28 Ω.

Determine E.

the work done per unit charge

in moving charge from one terminal of a cell to the other / all the way round the circuit

Award [1] for “energy per unit charge provided by the cell”/“power per unit current”

Award [1] for “potential difference across the terminals of the cell when no current is flowing” 

Do not accept “potential difference across terminals of cell”

[2 marks]

the resistance is proportional to length / see 0.35 AND 1«.00»

so it equals 0.35 × 80

«= 28 Ω»

[2 marks]

current leaving 12 V cell is $\frac{12}{80}$ = 0.15 «A»

OR

E = $\frac{12}{80}$ × 28

E = «0.15 × 28 =» 4.2 «V»

Award [2] for a bald correct answer

Allow a 1sf answer of 4 if it comes from a calculation.

Do not allow a bald answer of 4 «V»

Allow ECF from incorrect current

[2 marks]

Label with arrows on the diagram the magnetic force F on the proton. 

Label with arrows on the velocity vector v of the proton.

The speed of the proton is 2.16 × 10⁶ m s^-1 and the magnetic field strength is 0.042 T. For this proton, determine, in m, the radius of the circular path. Give your answer to an appropriate number of significant figures.

F towards centre ✔

v tangent to circle and in the direction shown in the diagram ✔

«$qvB=\frac{m{v}^2}{R}\Rightarrow »R=\frac{mv}{qB}/\frac{1.673\times {10}^{-27}\times 2.16\times {10}^6}{1.60\times {10}^{-19}\times 0.042}$  ✔

R = 0.538 «m»✔

R = 0.54 «m» ✔

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

This was generally well answered although usually to 3 sf. Common mistakes were to substitute 0.042 for F and 1 for q. Also some candidates tried to answer in terms of electric fields.

Explain why the path of the proton is a circle.

Show that the radius of the path is about 6 cm.

Calculate the time for one complete revolution.

Explain why the kinetic energy of the proton is constant.

magnetic force is to the left «at the instant shown»
OR
explains a rule to determine the direction of the magnetic force ✔

force is perpendicular to velocity/«direction of» motion
OR
force is constant in magnitude ✔

force is centripetal/towards the centre ✔

NOTE: Accept reference to acceleration instead of force

$qvB=\frac{m{v}^2}{R}$✔

$R=\frac{1.67\times {10}^{-27}\times 2.0\times {10}^6}{1.6\times {10}^{-19}\times 0.35}$ OR 0.060 « m »

NOTE: Award MP2 for full replacement or correct answer to at least 2 significant figures

$T=\frac{2\mathrm{\pi }R}{v}$✔

$T=«\frac{2\mathrm{\pi }\times 0.06}{2.0\times {10}^6}=1.9\times {10}^{-7}$ « s » ✔

NOTE: Award [2] for bald correct answer

ALTERNATIVE 1
work done by force is change in kinetic energy ✔
work done is zero/force perpendicular to velocity ✔

NOTE: Award [2] for a reference to work done is zero hence E_k remains constant

ALTERNATIVE 2
proton moves at constant speed ✔
kinetic energy depends on speed ✔

NOTE: Accept mention of speed or velocity indistinctly in MP2

Show that the electric field strength due to the point charge at the position of the electron is 3.4 × 10⁸ N C^–1.

Calculate the magnitude of the initial acceleration of the electron.

Describe the subsequent motion of the electron.

$E=\frac{k\times q}{r^2}$ ✔

$E=\frac{8.99\times {10}^9\times 6.0\times {10}^{-3}}{0.4^2}$  OR  $E=3.37\times {10}^8 «\mathrm{N} {\mathrm{C}}^{-1}»$ ✔

NOTE: Ignore any negative sign.

$F=q\times E$ OR  $F=1.6\times {10}^{-19}\times 3.4\times {10}^8=5.4\times {10}^{-11} «\mathrm{N}»$ ✔

$a=«\frac{5.4\times {10}^{-11}}{9.1\times {10}^{-31}}=» 5.9\times {10}^{19}« \mathrm{m} {\mathrm{s}}^{-2}»$✔

NOTE: Ignore any negative sign.
Award [1] for a calculation leading to $a=« \mathrm{m} {\mathrm{s}}^{-2}»$
Award [2] for bald correct answer

the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔

NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force as attractive so concludes that the acceleration will increase

The copper wires and insulator are both exposed to an electric field. Discuss, with reference to charge carriers, why there is a significant electric current only in the copper wires.

Calculate the radius of each wire.

There is a current of 730 A in the cable. Show that the power loss in 1 m of the cable is about 30 W.

When the current is switched on in the cable the initial rate of rise of temperature of the cable is 35 mK s^–1. The specific heat capacity of copper is 390 J kg^–1 K^–1. Determine the mass of a length of one metre of the cable.

when an electric field is applied to any material «using a cell etc» it acts to accelerate any free electrons

electrons are the charge carriers «in copper»

Accept “free/valence/delocalised electrons”.

metals/copper have many free electrons whereas insulators have few/no free electrons/charge carriers

area = $\frac{1.7\times {10}^{-3}\times 35\times {10}^3}{64}$ «= 9.3 x 10^–6 m²»

«resistance of cable = 2Ω»

power dissipated in cable = 730² x 2 «= 1.07 MW»

power loss per meter $=\frac{1.07\times {10}^{-6}}{35\times {10}^3}$ or 30.6 «W m^–1»

Allow [2] for a solution where the resistance per unit metre is calculated using resistivity and answer to (b)(i) (resistance per unit length of cable =5.7 x 10^–5 m)

30 = m x 390 x 3.5 x 10^–2

2.2 k«g»

Correct answer only.

Calculate the potential difference across P. 

The voltmeter reads zero. Determine the resistance of S.

Deduce the resistance of this new cylinder when it has been reshaped.

Outline, without calculation, the change in the total power dissipated in Q and the new cylinder after it has been reshaped.

ALTERNATIVE 1

attempt to use potential divider equation or similar method ✓
«$6.0\times \frac{40}{\left(40+60\right)}$»= 2.4 «V» ✓

ALTERNATIVE 2

«current = $\frac{6}{60+40}$» = 0.06 «A» ✓

40 x 0.06 = 2.4 «V» ✓

ALTERNATIVE 1

Pd across Q = 2.4 V so I = 0.15 « A » ✓

and pd across S is 6.0 – 2.4 = 3.6 « V » ✓

$R=\frac{V}{I}=\frac{3.6}{0.15}=24$ «Ω» ✓

ALTERNATIVE 2

pd at PR junction = pd at QS junction ✓

so $\frac{S}{Q}=\frac{R}{P}$  OR  $S=\frac{16\times 60}{40}$✓

24 «Ω» ✓

ALTERNATIVE 3

$I=\frac{40\times 0.06}{16}=0.15$ «A» ✓

$6.0=\left(16+\text{S}\right)\left(0.15\right)$  OR  $2.4=\left(6\right)\left(\frac{16}{S+16}\right)$ ✓

R = 24 «Ω» ✓

Allow ECF for MP3 from incorrect MP1 or MP2.

recognition that $4L$ leads to $4R$ / «$\frac{\rho 4L}{A}$» = $4R$ ✓

«because the volume of S is constant new area is» $\frac{A}{4}$✓

16 x 24 = 384 «Ω» ✓

«total» power has decreased ✓

Because current in the branch has decreased «and P=I²R »

OR

Because resistance has increased in branch «and $P=\frac{V^2}{R}$» ✓

Allow opposite argument as ECF from (c)(i) (if candidate deduces a lower resistance).

Allow “power doesn’t change” if candidate has no change of resistance from (b) to (c)(i).

This was generally well done at the higher level. Some SL candidates struggled to calculate the correct current but earned the second marking point through ECF.

Many candidates struggled with this question. A very common mistake was to assume the current was the same in each branch, leading to a resistance of 84 Ohms. The placement of the voltmeter may have caused some confusion for candidates, and they may not have understood what the zero reading was indicating. It is important that candidates understand what voltmeters are actually reading and are familiar with different placements in circuits.

Most candidates recognized that increasing the length of the conductive putty would increase the resistance by a factor of four, but very few considered that if the volume of the putty remained constant that the cross-sectional surface area would decrease as well.

This was an item that caused a bit of confusion for candidates. The prompt asks for a comparison of the power in the branch before and after changing the length of the conductive putty. Many candidates correctly identified that the power in Q would decrease, but either did not discuss the power in the whole branch or were not clear that the power in the putty would decrease as well.

Identify the laws of conservation that are represented by Kirchhoff’s circuit laws.

State the emf of the cell.

Deduce the internal resistance of the cell.

The voltmeter is used in another circuit that contains two secondary cells.

Cell A has an emf of 10 V and an internal resistance of 1.0 Ω. Cell B has an emf of 4.0 V and an internal resistance of 2.0 Ω.

Calculate the reading on the voltmeter.

Outline why electricity is a secondary energy source.

Some fuel sources are renewable. Outline what is meant by renewable.

A fully charged cell of emf 6.0 V delivers a constant current of 5.0 A for a time of 0.25 hour until it is completely discharged.

The cell is then re-charged by a rectangular solar panel of dimensions 0.40 m × 0.15 m at a place where the maximum intensity of sunlight is 380 W m⁻².

The overall efficiency of the re-charging process is 18 %.

Calculate the minimum time required to re-charge the cell fully.

Outline why research into solar cell technology is important to society.

« conservation of » charge ✓

« conservation of » energy ✓

Allow [1] max if they explicitly refer to Kirchhoff’ laws linking them to the conservation laws incorrectly.

12 V ✓

I = 2.0 A OR 12 = I (r +4) OR 4 = Ir OR 8 = 4I ✓

«Correct working to get » r = 2.0 «Ω» ✓

Allow ECF from (b)(i)

Loop equation showing EITHER correct voltages, i.e., 10 – 4 on one side or both emfs positive on different sides of the equation OR correct resistances, i.e. I (1 + 2) ✓

10−4 = I (1 + 2) OR I = 2.0 «A» seen✓

V = 8.0 «V» ✓

Allow any valid method

is generated from primary/other sources ✓

«a fuel » that can be replenished/replaced within a reasonable time span

OR

«a fuel» that can be replaced faster than the rate at which it is consumed

OR

renewables are limitless/never run out

OR

«a fuel» produced from renewable sources

OR

gives an example of a renewable (biofuel, hydrogen, wood, wind, solar, tidal, hydro etc..) ✓

OWTTE

ALTERNATIVE 1

«energy output of the panel =» VIt OR 6 x 5 x 0.25 x 3600 OR 27000 «J» ✓

«available power =» 380 x 0.4 x 0.15 x 0.18 OR 4.1 «W» ✓

$t=$ «$\frac{27000}{4.1}$=» 6600 «s» ✓

ALTERNATIVE 2

«energy needed from Sun =» $\frac{Vlt}{eff}$ OR $\frac{6\times 5\times 0.25\times 3600}{0.18}$ OR 150000 «J» ✓

« incident power=» 380 x 0.4 x 0.15 OR 22.8 «W» ✓

$t=$ «$\frac{150000}{22.8}$=» 6600 «s» ✓

Allow ECF for MP3

Accept final answer in minutes (110) or hours (1.8).

coherent reason ✓

e.g., to improve efficiency, is non-polluting, is renewable, does not produce greenhouse gases, reduce use of fossil fuels

Do not allow economic reasons

a) Most just stated Kirchhoff's laws rather than the underlying laws of conservation of energy and charge, basically describing the equations from the data booklet. When it came to guesses, energy and momentum were often the two, although even a baryon and lepton number conservation was found. It cannot be emphasised enough the importance of correctly identifying the command verb used to introduce the question. In this case, identify, with the specific reference to conservation laws, seem to have been explicit tips not picked up by some candidates.

bi) This was probably the easiest question on the paper and almost everybody got it right. 12V. Some calculations were seen, though, that contradict the command verb used. State a value somehow implies that the value is right in front to be read or interpreted suitably.

bii) In the end a lot of the answers here were correct. Some obtained 2 ohms and were able to provide an explanation that worked. A very few negative answers were found, suggesting that some candidates work mechanically without properly reflecting in the nature of the value obtained.

ci) A lot of candidates figured out they had to do some sort of loop here but most had large currents in the voltmeter. Currents of 2 A and 10 A simultaneously were common. Some very good and concise work was seen though, leading to correct steps to show a reading of 8V.

cii) This question was cancelled due to an internal reference error. The paper total was adjusted in grade award. This is corrected for publication and future teaching use.

di) The vast majority of candidates could explain why electricity was a secondary energy source.

dii) An ideal answer was that renewable fuels can be replenished faster than they are consumed. However, many imaginative alternatives were accepted.

ei) This question was often very difficult to mark. Working was often scattered all over the answer box. Full marks were not that common, most candidates achieved partial marks. The commonest problem was determining the energy required to charge the battery. It was also common to see a final calculation involving a power divided by a power to calculate the time.

eii) Almost everybody could give a valid reason why research into solar cells was important. Most answers stated that solar is renewable. There were very few that didn't get a mark due to discussing economic reasons.

Determine the initial acceleration of the spacecraft.

Estimate the maximum speed of the spacecraft.

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.

Outline what is meant by the gravitational field strength at a point.

Newton’s law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting a spherical planet of uniform density.

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» ✔

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» ✔

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. ✔

force per unit mass ✔

acting on a small/test/point mass «placed at the point in the field» ✔

satellite has a much smaller mass/diameter/size than the planet «so approximates to a point mass» ✔ 

The switch S is initially open. Calculate the total power dissipated in the circuit.

The switch is now closed. State, without calculation, why the current in the cell will increase. 

The switch is now closed. Deduce the ratio $\frac{\text{power dissipated in Y with S open}}{\text{power dissipated in Y with S closed}}$.

total resistance of circuit is 8.0 «Ω» ✔

P = $\frac{{12}^2}{8.0}$ =18 «W» ✔

«a resistor is now connected in parallel» reducing the total resistance

OR

current through YZ unchanged and additional current flows through X ✔

evidence in calculation or statement that pd across Y/current in Y is the same as before ✔

so ratio is 1 ✔

Most candidates scored both marks. ECF was awarded for those who didn’t calculate the new resistance correctly. Candidates showing clearly that they were attempting to calculate the new total resistance helped examiners to award ECF marks.

Most recognised that this decreased the total resistance of the circuit. Answers scoring via the second alternative were rare as the statements were often far too vague.

Very few gained any credit for this at both levels. Most performed complicated calculations involving the total circuit and using 12V – they had not realised that the question refers to Y only.

Explain why the output potential difference to the external circuit and the output emf of the photovoltaic cell are different.

Calculate the internal resistance of the photovoltaic cell for the maximum intensity condition using the model for the cell.

The maximum intensity of sunlight incident on the photovoltaic cell at the place on the Earth’s surface is 680 W m⁻².

A measure of the efficiency of a photovoltaic cell is the ratio

$\frac{\text{energy available every second to the external circuit}}{\text{energy arriving every second at the photovoltaic cell surface}}.$

Determine the efficiency of this photovoltaic cell when the intensity incident upon it is at a maximum.

State two reasons why future energy demands will be increasingly reliant on sources such as photovoltaic cells.

there is a potential difference across the internal resistance
OR
there is energy/power dissipated in the internal resistance ✓

when there is current «in the cell»/as charge flows «through the cell» ✓

Allow full credit for answer based on $V=\epsilon -Ir$

ALTERNATIVE 1
pd dropped across cell $=6.5 «\text{V}»$✓

internal resistance $\mathrm{=}\frac{6.5}{0.9}$ ✓

$7.2 «\text{Ω}»$✓

ALTERNATIVE 2

$\epsilon =I(R+r)$ so $\epsilon =V+Ir$ ✓

$21.0=14.5+0.9\times r$ ✓

$7.2 «\text{Ω}»$ ✓

Alternative solutions are possible

Award [3] marks for a bald correct answer

power arriving at cell = 680 x 0.35 x 0.45 = «107 W» ✓

power in external circuit = 14.5 x 0.9 = «13.1 W» ✓

efficiency = 0.12 OR 12 % ✓

Award [3] marks for a bald correct answer

Allow ECF for MP3

«energy from Sun/photovoltaic cells» is renewable
OR
non-renewable are running out ✓

non-polluting/clean ✓

no greenhouse gases
OR
does not contribute to global warming/climate change ✓

OWTTE

Do not allow economic aspects (e.g. free energy)

Show that the time taken for the battery to discharge is about 3 × 10³ s.

Deduce that the average power output of the battery is about 240 W.

Friction and air resistance act on the bicycle and the girl when they move. Assume that all the energy is transferred from the battery to the electric motor. Determine the total average resistive force that acts on the bicycle and the girl.

Calculate the component of weight for the bicycle and girl acting down the slope.

The battery continues to give an output power of 240 W. Assume that the resistive forces are the same as in (a)(iii).

Calculate the maximum speed of the bicycle and the girl up the slope.

On another journey up the slope, the girl carries an additional mass. Explain whether carrying this mass will change the maximum distance that the bicycle can travel along the slope.

Determine the internal resistance of the battery.

Calculate the emf of one cell.

Calculate the internal resistance of one cell.

time taken $\frac{2.0\times {10}^4}{7}$«= 2860 s» = 2900«s» ✔

Must see at least two s.f.

use of E = qV OR energy = 4.3 × 10³ × 16 «= 6.88 × 10⁵ J» ✔

power = 241 «W» ✔

Accept 229 W − 241 W depending on the exact value of t used from ai.

Must see at least three s.f.

use of power = force × speed OR force × distance = power × time ✔

«34N» ✔

Award [2] for a bald correct answer.

Accept 34 N – 36 N.

66 g sin(3°) = 34 «N» ✔

total force 34 + 34 = 68 «N» ✔
3.5 «ms^-1»✔

If you suspect that the incorrect reference in this question caused confusion for a particular candidate, please refer the response to the PE.

Look for ECF from aiii and bi.

Accept 3.4 − 3.5 «ms^-1».

Award [0] for solutions involving use of KE.

Award [0] for v = 7 ms^-1.

Award [2] for a bald correct answer.

«maximum» distance will decrease OWTTE ✔

because opposing/resistive force has increased
OR
because more energy is transferred to GPE
OR
because velocity has decreased
OR
increased mass means more work required «to move up the hill» ✔

V dropped across battery OR R_circuit = 1.85 Ω ✔

so internal resistance = $\frac{4.0}{6.5}$ = 0.62«Ω» ✔

For MP1 allow use of internal resistance equations that leads to 16V − 12V (=4V).

Award [2] for a bald correct answer.

$\frac{16}{5}$ = 3.2 «V» ✔

ALTERNATIVE 1:

2.5r = 0.62 ✔

r = 0.25 «Ω» ✔

ALTERNATIVE 2:

$\frac{0.62}{5}$ = 0.124 «Ω» ✔

r = 2(0.124)= 0.248 «Ω» ✔

Allow ECF from (d) and/or e(i).

This question was generally well answered. Candidates should be reminded on questions where a given value is being calculated that they should include an unrounded answer. This whole question set was a blend of electricity and mechanics concepts, and it was clear that some candidates struggled with applying the correct concepts in the various sub-questions.

Many candidates struggled with this question. They either simply calculated the weight, used the cosine rather than the sine function, or failed to multiply by the acceleration due to gravity. Candidates need to be able to apply free-body diagram skills in a variety of “real world” situations.

This question was well answered in general, with the vast majority of candidates specifying that the maximum distance would decrease. This is an “explain” command term, so the examiners were looking for a detailed reason why the distance would decrease for the second marking point. Unfortunately, some candidates simply wrote that because the mass increased so did the weight without making it clear why this would change the maximum distance.