Physics 9702 / 43 — October/November 2024

DifficultyEasy

Worked solution

Answer

For two point masses $m_1$ and $m_2$ separated by distance $r$ , the gravitational force is attractive and acts along the line joining them, with magnitude

F = \frac{G m_1 m_2}{r^2}.

Final answer

Attractive force between two point masses: F = G m1 m2 / r^2 along the line joining them.

Detailed explanation

Background Concept

Newton’s law of gravitation describes the force that any two masses exert on each other. It is an inverse-square law: doubling the separation reduces the force by a factor of $4$ . The force depends on:

the product of the masses $m_1 m_2$ (more mass gives a larger force),
the square of the separation $r^2$ ,
the universal gravitational constant $G$ .

The force is always attractive and acts along the straight line joining the two masses.

Understanding the Question

You are asked to state the law. For full credit you need both:

the relationship/equation, and
a clear statement about direction/nature (along the line joining the masses and attractive).

Approach

Write the standard formula for gravitational force between two point masses and add the key descriptive statements (attractive, along the line joining centres).

Step-by-Step Reasoning

Consider masses $m_1$ and $m_2$ separated by distance $r$ .
The magnitude of the force is proportional to $m_1 m_2$ and inversely proportional to $r^2$ .
Introduce the proportionality constant $G$ :

F = \frac{G m_1 m_2}{r^2}.

State that the force acts along the line joining the masses and is attractive.

Key Takeaways

Gravitational force is an inverse-square law.
The force is always attractive and directed along the line joining the masses.

Common Mistakes

Writing $F \propto 1/r$ instead of $1/r^2$ .
Omitting that the force is attractive.
Not stating it acts along the line joining the masses (directional information).

Things to Be Careful About

The $r$ in the formula is the separation between the centres of mass (for spheres, centre-to-centre distance).
Use correct symbols and avoid mixing up $g$ (field strength) with $G$ (gravitational constant).

Techniques used

state Newton's inverse-square force lawspecify direction and nature (attractive) of the force

(b)

A planet may be considered as a uniform sphere.

A satellite is in circular orbit of period $T$ around the planet at a height $h$ above the surface. The height of the orbit can be adjusted by use of the satellite’s rocket engines.

Fig. 1.1 shows the variation with $h$ of $T^{\frac{2}{3}}$ .

Fig. 1.1

10M

(i)

By reference to forces, explain why the orbit of the satellite is circular.

DifficultyMedium-Easy

Worked solution

Answer

The satellite experiences gravitational attraction towards the planet’s centre. This inward force provides the centripetal force, so the acceleration is towards the centre and changes only the direction of the velocity, keeping the satellite in a circular path.

Final answer

Gravity acts towards the centre and provides the centripetal force/acceleration, so the velocity direction changes continuously, giving a circular orbit.

Detailed explanation

Background Concept

For an object to move in a circle at constant speed, its velocity must continuously change direction. This requires an acceleration towards the centre of the circle, called the centripetal acceleration:

a_c = \frac{v^2}{r} = r\omega^2.

By Newton’s second law, an acceleration implies a resultant force. The required inward force is the centripetal force:

F_c = m a_c.

In a satellite orbit, the gravitational force is directed towards the planet’s centre, so it can act as the centripetal force.

Understanding the Question

The question asks you to explain using forces why the satellite’s orbit is circular. You must mention:

the direction of the gravitational force (towards the centre), and
that it provides the centripetal force/acceleration needed for circular motion.

Approach

Describe the force on the satellite (gravity towards the planet’s centre) and connect it to the requirement for circular motion (centripetal force towards the centre).

Step-by-Step Reasoning

diagram-b-i-1

The only significant force on the satellite (ignoring thrust during adjustment) is the gravitational attraction from the planet.
This force always points towards the planet’s centre.
A force always towards the centre gives an acceleration towards the centre.
An inward (centripetal) acceleration changes the direction of the velocity vector continuously, producing circular motion.

Key Takeaways

Circular motion requires a centripetal acceleration towards the centre.
The centripetal force is simply the resultant force towards the centre.
For satellites, gravity supplies this centripetal force.

Common Mistakes

Saying “there is no force” because the speed is constant (direction is still changing, so acceleration exists).
Claiming gravity acts tangentially (it acts radially inward).
Mixing up “centrifugal force” (not used here in an inertial frame) with centripetal force.

Things to Be Careful About

The orbit is circular only if the speed corresponds exactly to the radius so that gravity equals the required centripetal force.
If thrust is being used, there can be additional forces, but the explanation still centres on the inward resultant producing centripetal acceleration.

Techniques used

identify the centripetal force as the resultant toward the centrerelate gravitational attraction to centripetal accelerationlink inward force to continuous change of velocity direction

(ii)

Use Newton’s law of gravitation to show that $h$ and $T$ are related by

(h + B)^3 = \frac{GA}{4\pi^2} T^2

where $G$ is the gravitational constant and $A$ and $B$ are constants that depend on the properties of the planet.

DifficultyMedium

Worked solution

Working

Let orbital radius $r = h + B$ .

Gravitational force provides centripetal force:

\frac{GA m}{r^2} = \frac{mv^2}{r}

so

\frac{GA}{r^2} = \frac{v^2}{r}.

With $v = \frac{2\pi r}{T}$ ,

\frac{GA}{r^2} = \frac{1}{r}\left(\frac{2\pi r}{T}\right)^2 = \frac{4\pi^2 r}{T^2}.

Hence

r^3 = \frac{GA}{4\pi^2}T^2.

Since $r=h+B$ ,

(h+B)^3 = \frac{GA}{4\pi^2}T^2.

Answer

Shown.

Final answer

(h + B)^3 = (G A / (4 pi^2)) T^2

Detailed explanation

Background Concept

A satellite in a circular orbit has a centripetal acceleration towards the centre:

a_c = \frac{v^2}{r}.

The required centripetal force is:

F_c = m\frac{v^2}{r}.

For an orbit around a planet, the gravitational force on a satellite of mass $m$ due to a planet of mass $M$ at separation $r$ is:

F_g = \frac{GMm}{r^2}.

In a circular orbit, gravity is the centripetal force, so $F_g = F_c$ . Also, the orbital speed relates to period $T$ by

v = \frac{2\pi r}{T}.

Understanding the Question

You are asked to use Newton’s gravitation law to derive a relationship between the orbital height $h$ and the period $T$ . The constants $A$ and $B$ “depend on the properties of the planet”, so they represent quantities like the planet’s mass and radius (not properties of the satellite).

The target result is a Kepler-type relation:

(h+B)^3 \propto T^2.

Approach

Express the orbital radius as $r = h + B$ .
Set gravitational force equal to centripetal force.
Use $v = 2\pi r/T$ to eliminate $v$ .
Rearrange to obtain $r^3 = (GA/4\pi^2)T^2$ , then replace $r$ with $(h+B)$ .

Step-by-Step Reasoning

Let the planet have mass $A$ (a constant) and satellite have mass $m$ . Let the orbital radius be

r = h + B.

Gravitational force magnitude on the satellite is

F_g = \frac{GA m}{r^2}.

For uniform circular motion, centripetal force needed is

F_c = \frac{mv^2}{r}.

Set $F_g = F_c$ :

\frac{GA m}{r^2} = \frac{mv^2}{r}.

Cancel $m$ :

\frac{GA}{r^2} = \frac{v^2}{r}.

Substitute orbital speed $v = 2\pi r/T$ :

\frac{GA}{r^2} = \frac{1}{r}\left(\frac{2\pi r}{T}\right)^2 = \frac{4\pi^2 r}{T^2}.

Rearrange:

GA T^2 = 4\pi^2 r^3

so

r^3 = \frac{GA}{4\pi^2}T^2.

Replace $r$ with $h+B$ :

(h+B)^3 = \frac{GA}{4\pi^2}T^2.

This matches the required expression.

Key Takeaways

Circular orbit condition: gravity provides the centripetal force.
Combining $F_g = GMm/r^2$ with $v = 2\pi r/T$ leads to $r^3 \propto T^2$ .
The satellite’s mass cancels: the period depends on the planet and the orbital radius only.

Common Mistakes

Using $v = 2\pi/T$ (missing the factor of $r$ ).
Equating $GMm/r^2$ to $mv^2$ (missing division by $r$ in centripetal force).
Forgetting that $r$ is measured from the planet’s centre, not the surface.

Things to Be Careful About

Keep $r$ as the centre-to-centre distance throughout; only at the end replace $r$ by $(h+B)$ .
Algebra: when substituting $v = 2\pi r/T$ , ensure $(2\pi r/T)^2$ gives a factor $r^2$ .

Techniques used

equate gravitational force to centripetal forcesubstitute v = 2 pi r / T into centripetal force expressionrearrange algebra to obtain the required orbital period relation

(iii)

Use the gradient and intercept of the line in Fig. 1.1 to determine values for $A$ and $B$ . Give units with your answers.

$A$ = ______ unit ______
$B$ = ______ unit ______

5M Similar

DifficultyMedium-Hard

Worked solution

Working

From

(h+B)^3 = \frac{GA}{4\pi^2}T^2

T^2 = \frac{4\pi^2}{GA}(h+B)^3

Take power $\frac{1}{3}$ :

T^{2/3} = \left(\frac{4\pi^2}{GA}\right)^{1/3}(h+B).

So, with $y=T^{2/3}$ and $x=h$ :

y = kx + kB, \qquad k = \left(\frac{4\pi^2}{GA}\right)^{1/3}.

From the graph, intercept $\approx 360\ \text{s}^{2/3}$ .
Using points $(0,360)$ and $(12,1300)$ on the plotted axes,

\text{gradient on axes} = \frac{1300-360}{12-0} = 78.3\ \frac{\text{s}^{2/3}}{10^6\ \text{m}}.

Hence

k = \frac{78.3}{10^6} = 7.83\times 10^{-5}\ \text{s}^{2/3}\ \text{m}^{-1}.

Then

B = \frac{\text{intercept}}{k} = \frac{360}{7.83\times 10^{-5}} = 4.6\times 10^6\ \text{m}.

Also

k^3 = \frac{4\pi^2}{GA} \Rightarrow A = \frac{4\pi^2}{Gk^3}.

With $G = 6.67\times 10^{-11}$ ,

A = \frac{4\pi^2}{(6.67\times 10^{-11})(7.83\times 10^{-5})^3} \approx 1.2\times 10^{24}\ \text{kg}.

Answer

$A \approx 1.2\times 10^{24}\ \text{kg}$

$B \approx 4.6\times 10^{6}\ \text{m}$

Final answer

A ≈ 1.2 × 10^24 kg, B ≈ 4.6 × 10^6 m

Detailed explanation

Background Concept

For circular orbits,

(h+B)^3 = \frac{GA}{4\pi^2}T^2

is the same structure as Kepler’s third law in Newtonian form:

r^3 = \frac{GM}{4\pi^2}T^2,

where $r$ is the orbital radius from the centre. If we let $r = h + B$ , then $B$ represents the planet’s radius and $A$ represents the planet’s mass.

To use a straight-line graph, we rearrange into the form $y = mx + c$ . A very useful manipulation is taking the cube root of both sides so that $T^{2/3}$ becomes proportional to $(h+B)$ .

Understanding the Question

You are given a graph of $T^{2/3}$ (vertical axis) against $h$ (horizontal axis), where the horizontal axis is scaled as $h/10^6\ \text{m}$ . The line is straight, so we can read:

the gradient, and
the intercept at $h=0$ .

You must then use these to find numerical values (with units) for the constants $A$ and $B$ .

Approach

Convert the derived relationship into a linear form:

T^{2/3} = k(h+B).

Match to $y = kx + kB$ , so:

gradient $= k$ ,
intercept $= kB$ .

Use the graph to find gradient and intercept (taking care with the $10^6\ \text{m}$ x-axis scaling).
Calculate:

$B = \text{intercept}/k$ ,
$A$ from $k = \left(\frac{4\pi^2}{GA}\right)^{1/3}$ .

Step-by-Step Reasoning

Start from the given result:

(h+B)^3 = \frac{GA}{4\pi^2}T^2.

Rearrange for $T^2$ :

T^2 = \frac{4\pi^2}{GA}(h+B)^3.

Take the cube root of both sides. The key point is that

(T^2)^{1/3} = T^{2/3}.

So:

T^{2/3} = \left(\frac{4\pi^2}{GA}\right)^{1/3}(h+B).

Define

k = \left(\frac{4\pi^2}{GA}\right)^{1/3}.

Then

T^{2/3} = k h + kB.

This is a straight line with gradient $k$ and y-intercept $kB$ .

Read two points from the straight line on the graph. Using the described points approximately:

intercept at $h=0$ is $\approx 360\ \text{s}^{2/3}$ ,
at $h/10^6\ \text{m} = 12$ , $T^{2/3} \approx 1300\ \text{s}^{2/3}$ .

Gradient using plotted-axis units:

\text{gradient (per x-axis unit)} = \frac{1300-360}{12} = 78.3\ \frac{\text{s}^{2/3}}{(10^6\ \text{m})}.

Convert to per metre:

k = 78.3\times 10^{-6} = 7.83\times 10^{-5}\ \text{s}^{2/3}\ \text{m}^{-1}.

Use intercept $= kB$ to find $B$ :

B = \frac{360}{7.83\times 10^{-5}} \approx 4.6\times 10^6\ \text{m}.

So $B$ is a length (consistent with being a planetary radius).

Use the definition of $k$ to find $A$ .
Cube both sides:

k^3 = \frac{4\pi^2}{GA} \Rightarrow A = \frac{4\pi^2}{Gk^3}.

Insert $G = 6.67\times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ and $k = 7.83\times 10^{-5}\ \text{s}^{2/3}\ \text{m}^{-1}$ :

A \approx 1.2\times 10^{24}\ \text{kg}.

This has units of mass, consistent with being the planet’s mass.

Key Takeaways

Linearising is often done by taking powers/logs until you get $y = mx + c$ .
From $T^{2/3} = k(h+B)$ : gradient gives $k$ and intercept gives $kB$ .
Always account for axis scaling factors (here $h/10^6\ \text{m}$ ).

Common Mistakes

Forgetting the x-axis is $h/10^6\ \text{m}$ , leading to a value of $k$ too large by a factor of $10^6$ .
Using $T^{1/3}$ instead of $T^{2/3}$ when taking the cube root.
Taking $B$ equal to the intercept directly (it is intercept divided by gradient).
Missing units for $A$ and $B$ .

Things to Be Careful About

Read points from the best-fit line, not individual scattered points.
Use a large triangle on the graph for the gradient to reduce percentage reading error.
Check unit consistency: since $T^{2/3} = k(h+B)$ , $k$ must have units $\text{s}^{2/3}\ \text{m}^{-1}$ , and $B$ must be in $\text{m}$ ; then $A$ comes out in $\text{kg}$ .

Techniques used

linearise the orbital period relation into y = mx + c formextract gradient and intercept from a straight-line graphsolve for constants using gradient-intercept relationshipscheck and assign correct SI units

Q2MediumTemperature

(a)

Define specific heat capacity.

DifficultyEasy

Worked solution

Answer

Specific heat capacity is the thermal energy required to raise the temperature of $1.0\ \text{kg}$ of a substance by $1.0\ \text{K}$ (or $1.0\ ^\circ\text{C}$ ), with no change of state.

Final answer

Thermal energy required to raise the temperature of 1.0 kg of a substance by 1.0 K (no change of state).

Detailed explanation

Background Concept

The specific heat capacity $c$ of a material tells you how much energy is needed to change its temperature. When a mass $m$ changes temperature by $\Delta T$ , the thermal energy transferred (heating/cooling) is

Q = mc\Delta T

$Q$ is energy transferred by heating (or removed by cooling), in joules (J).
$m$ is mass in kg.
$\Delta T$ is the temperature change in K (numerically the same as in $^\circ\text{C}$ ).
$c$ is in $\text{J kg}^{-1}\ \text{K}^{-1}$ .

The definition always assumes no change of state; otherwise energy can go into latent heat instead of temperature change.

Understanding the Question

You are asked to define specific heat capacity. So you must give a clear, standard statement mentioning:

energy required,
per unit mass ( $1\ \text{kg}$ ),
per unit temperature rise ( $1\ \text{K}$ ),
and that there is no change of state.

Approach

Use the standard textbook definition linked to $Q = mc\Delta T$ , and express it in words with correct units.

Step-by-Step Reasoning

From $Q = mc\Delta T$ , rearrange conceptually:

c = \frac{Q}{m\Delta T}

So $c$ is “energy per kilogram per kelvin”. To make this a definition, specify $m = 1\ \text{kg}$ and $\Delta T = 1\ \text{K}$ and state the “no change of state” condition.

Key Takeaways

Specific heat capacity is energy needed per kg per K temperature rise.
Units: $\text{J kg}^{-1}\ \text{K}^{-1}$ .
Must exclude change of state.

Common Mistakes

Missing “per unit mass” or “per unit temperature rise”.
Forgetting to state “no change of state”.
Confusing with specific latent heat (energy for change of state at constant temperature).

Things to Be Careful About

Use “raise the temperature” (or “change temperature by”) and specify $1\ \text{kg}$ and $1\ \text{K}$ .
$1\ \text{K}$ temperature change equals $1\ ^\circ\text{C}$ temperature change, but the unit in $c$ must be $\text{K}^{-1}$ (or $^\circ\text{C}^{-1}$ sometimes accepted).

Techniques used

state a precise definition of a thermodynamic quantityrelate energy transfer to mass and temperature change

(b)

Two solid blocks X and Y are made from different metals. The blocks have different initial temperatures. Block Y is initially at room temperature.

The blocks are placed in direct thermal contact with each other at time $t = 0$ . Fig. 2.1 shows the variation with $t$ of the temperatures of the two blocks.

Fig. 2.1

6M

(i)

State three conclusions that may be drawn from Fig. 2.1. The conclusions may be qualitative or quantitative.

1 ______
2 ______
3 ______

DifficultyMedium-Easy

Worked solution

Answer

Block X is initially hotter than block Y (about $85\ ^\circ\text{C}$ for X and $25\ ^\circ\text{C}$ for Y at $t=0$ ).
Heat is transferred from X to Y, so X cools and Y warms.
Both blocks reach the same final temperature of about $40\ ^\circ\text{C}$ (thermal equilibrium) after about $2.5\ \text{min}$ .

Final answer

Example conclusions: X starts at ~85°C and Y at ~25°C; heat flows from X to Y so X cools and Y warms; both reach equilibrium at ~40°C after ~2.5 min.

Detailed explanation

Background Concept

When two objects at different temperatures are placed in thermal contact, thermal energy transfers from the hotter object to the cooler one. The transfer continues until thermal equilibrium is reached: both objects have the same temperature and there is no net heat flow.

Temperature–time graphs often show that the rate of temperature change is largest at the start (big temperature difference) and then reduces as the temperatures get closer.

Understanding the Question

The graph shows temperature of block X and block Y against time after they are put in contact at $t=0$ .

You must state three conclusions you can justify from the curves. These can be:

qualitative (e.g. “heat flows from X to Y”), or
quantitative (e.g. initial and final temperatures, time to reach equilibrium).

Approach

Read key features from the graph:

initial temperatures at $t=0$ ,
direction of change for each block,
final common temperature (plateau) and approximately when it is reached.

Convert these into clear conclusions about heat transfer and equilibrium.

Step-by-Step Reasoning

At $t=0$ , the curve for X is high (about $85\ ^\circ\text{C}$ ) while Y starts lower (about $25\ ^\circ\text{C}$ ). So X is initially hotter.
After contact, X’s temperature decreases and Y’s temperature increases. That indicates energy is leaving X and entering Y: net heat flow is from hot to cold.
Both curves approach and then level off at the same temperature (about $40\ ^\circ\text{C}$ ). A shared constant temperature means equilibrium.
The curves become essentially flat at about $2.5\ \text{min}$ , so equilibrium is reached around then.

Any three of these well-stated points score credit.

Key Takeaways

Hotter to cooler: direction of heat transfer can be inferred from the changing temperatures.
Same final temperature: indicates thermal equilibrium.
Time to plateau can be estimated from the graph.

Common Mistakes

Claiming the temperatures become equal immediately at the crossing point, instead of recognising they gradually approach a common value.
Giving conclusions not supported by the graph (e.g. exact values with unrealistic precision).
Forgetting to mention equilibrium / “same final temperature” as a key conclusion.

Things to Be Careful About

Quote values as approximate when reading from a graph.
“Room temperature” for Y at $t=0$ is consistent with about $25\ ^\circ\text{C}$ as shown.
The plateau temperature is common to both blocks, not separate final temperatures.

Techniques used

interpret a temperature–time graph qualitativelyread approximate numerical values from a graphinfer thermal equilibrium from a constant final temperature

(ii)

The ratio $\frac{\text{mass of block Y}}{\text{mass of block X}}$ is equal to 1.3.

The metal in block Y has a specific heat capacity of $901\ \text{J kg}^{-1}\ \text{K}^{-1}$ .

Determine the specific heat capacity of the metal in block X.

specific heat capacity = ______ $\text{J kg}^{-1}\ \text{K}^{-1}$

DifficultyMedium

Worked solution

Working

From Fig. 2.1: $T_X$ changes from $85\ ^\circ\text{C}$ to $40\ ^\circ\text{C}$ and $T_Y$ changes from $25\ ^\circ\text{C}$ to $40\ ^\circ\text{C}$ .

Assuming no heat loss to surroundings,

m_X c_X (85-40) = m_Y c_Y (40-25)

Given $\dfrac{m_Y}{m_X}=1.3$ and $c_Y = 901\ \text{J kg}^{-1}\ \text{K}^{-1}$ :

c_X = \frac{m_Y}{m_X} c_Y \frac{15}{45} = 1.3\times 901 \times \frac{1}{3} = 3.90\times 10^2\ \text{J kg}^{-1}\ \text{K}^{-1}

Answer

$3.90\times 10^2\ \text{J kg}^{-1}\ \text{K}^{-1}$

Final answer

3.90 × 10^2 J kg^-1 K^-1

Detailed explanation

Background Concept

When two objects exchange heat and are thermally isolated from the surroundings, the thermal energy lost by the hotter object equals the thermal energy gained by the cooler object:

Q_{\text{lost}} = Q_{\text{gained}}

For a temperature change (no change of state),

Q = mc\Delta T

where $\Delta T$ is the magnitude of the temperature change. In a heat-exchange problem:

the hotter object cools, so it loses energy,
the cooler object warms, so it gains energy.

Understanding the Question

From the graph:

Block X starts at about $85\ ^\circ\text{C}$ and ends at $40\ ^\circ\text{C}$ .
Block Y starts at $25\ ^\circ\text{C}$ (room temperature) and ends at $40\ ^\circ\text{C}$ .

You are also given:

$\dfrac{m_Y}{m_X} = 1.3$ ,
$c_Y = 901\ \text{J kg}^{-1}\ \text{K}^{-1}$ .

You must find $c_X$ .

Approach

Read $\Delta T$ for each block from the graph.
Write energy balance (assuming negligible heat loss): $m_X c_X \Delta T_X = m_Y c_Y \Delta T_Y$
Rearrange to solve for $c_X$ and substitute the mass ratio.

Step-by-Step Reasoning

1) Temperature changes from the graph

For X: $85\ ^\circ\text{C} \to 40\ ^\circ\text{C}$ so

\Delta T_X = 85 - 40 = 45\ \text{K}

For Y: $25\ ^\circ\text{C} \to 40\ ^\circ\text{C}$ so

\Delta T_Y = 40 - 25 = 15\ \text{K}

(Using K is fine because temperature differences in K and $^\circ\text{C}$ are numerically the same.)

2) Apply conservation of energy
Assuming all heat lost by X is gained by Y:

m_X c_X (45) = m_Y c_Y (15)

3) Solve for $c_X$

c_X = \frac{m_Y}{m_X} c_Y \frac{15}{45}

Substitute $\dfrac{m_Y}{m_X} = 1.3$ and $c_Y = 901$ :

c_X = 1.3\times 901\times \frac{15}{45}

Since $\dfrac{15}{45} = \dfrac{1}{3}$ :

c_X = 1.3\times 901\times \frac{1}{3} \approx 3.90\times 10^2\ \text{J kg}^{-1}\ \text{K}^{-1}

Key Takeaways

Use $Q=mc\Delta T$ for each object.
In an isolated exchange: $m_X c_X \Delta T_X = m_Y c_Y \Delta T_Y$ .
Temperature changes must be read from the graph and used consistently.

Common Mistakes

Using the wrong temperature changes (e.g. mixing up $85-25$ instead of using the final equilibrium temperature).
Forgetting the mass ratio and treating $m_X=m_Y$ .
Writing $m_X c_X \Delta T_X + m_Y c_Y \Delta T_Y = 0$ but then mishandling signs; using magnitudes avoids this.
Giving the unit as $\text{J kg}^{-1}\ ^\circ\text{C}^{-1}$ is usually acceptable, but omitting units entirely loses marks.

Things to Be Careful About

The final temperature must be the common equilibrium value ( $40\ ^\circ\text{C}$ ), not the time when curves are close.
Use a sensible number of significant figures (typically 2–3 s.f. based on graph readings and given data).
The method assumes negligible heat loss to the surroundings; this is standard unless the question suggests otherwise.

Techniques used

apply conservation of energy in thermal interactionsuse the equation Q = mcΔT for each bodyrearrange an algebraic equation to solve for an unknown constantsubstitute a mass ratio into an energy-balance equation

Q3MediumIdeal Gases

(a)

2M

(i)

State what is meant by the Avogadro constant.

(ii)

State the relationship between the Avogadro constant $N_A$ , the molar gas constant $R$ and the Boltzmann constant $k$ .

(b)

Two samples X and Y of ideal gases are both at thermodynamic temperature $T$ .

Sample X has volume $V$ and consists of $N$ molecules, each of mass $m$ .
Sample Y has volume $2V$ and consists of $2N$ molecules, each of mass $2m$ .

6M

(i)

Complete Table 3.1 by giving expressions, in terms of some or all of $N$ , $m$ , $T$ , $V$ and the constants in (a)(ii), for the quantities indicated.

Table 3.1

	sample X	sample Y
pressure
amount of substance
mean-square speed of molecules
internal energy

4M Similar

DifficultyMedium

Worked solution

Working

For an ideal gas,

pV = NkT

and

\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT \Rightarrow \langle c^2 \rangle = \frac{3kT}{m}

Total internal energy (ideal gas):

U = \frac{3}{2}NkT

Answer

quantity	sample X	sample Y
pressure	$p_X = \dfrac{NkT}{V}$	$p_Y = \dfrac{(2N)kT}{2V} = \dfrac{NkT}{V}$
amount of substance	$n_X = \dfrac{N}{N_A}$	$n_Y = \dfrac{2N}{N_A}$
mean-square speed	$\langle c^2 \rangle_X = \dfrac{3kT}{m}$	$\langle c^2 \rangle_Y = \dfrac{3kT}{2m}$
internal energy	$U_X = \dfrac{3}{2}NkT$	$U_Y = \dfrac{3}{2}(2N)kT = 3NkT$

Final answer

See table in working (pX = pY = NkT/V; nX = N/NA, nY = 2N/NA; <c^2>X = 3kT/m, <c^2>Y = 3kT/(2m); UX = (3/2)NkT, UY = 3NkT).

Detailed explanation

Background Concept

For ideal gases, the macroscopic variables are related by the ideal gas equation.

Using molecules:

pV = NkT

Using moles:

pV = nRT

with

N = nN_A \quad \text{and} \quad k = \frac{R}{N_A}

Kinetic theory links temperature to molecular motion. For an ideal gas (translational motion), the mean translational kinetic energy per molecule is:

\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT

So:

\langle c^2 \rangle = \frac{3kT}{m}

Internal energy $U$ of an ideal gas depends only on temperature. For a monatomic ideal gas (or considering translational degrees of freedom only):

U = N \times \left(\frac{3}{2}kT\right) = \frac{3}{2}NkT

Understanding the Question

You have two ideal-gas samples at the same temperature $T$ :

Sample X: volume $V$ , number of molecules $N$ , mass per molecule $m$ .
Sample Y: volume $2V$ , number of molecules $2N$ , mass per molecule $2m$ .

You must fill a table with expressions for:

pressure,
amount of substance,
mean-square speed $\langle c^2 \rangle$ ,
internal energy.

All answers must be in terms of $N, m, T, V$ and constants ( $N_A, R, k$ ).

Approach

Handle each row independently:

Use $pV = NkT$ for pressure.
Use $n = N/N_A$ for amount of substance.
Use $\langle c^2 \rangle = 3kT/m$ and substitute the molecular mass for each sample.
Use $U = \frac{3}{2}NkT$ and substitute the number of molecules for each sample.

A key skill here is tracking how changes in $N$ , $V$ , and $m$ affect each derived quantity.

Step-by-Step Reasoning

Pressure

Sample X:

p_X V = NkT \Rightarrow p_X = \frac{NkT}{V}

Sample Y has $N \to 2N$ and $V \to 2V$ (temperature unchanged):

p_Y (2V) = (2N)kT \Rightarrow p_Y = \frac{(2N)kT}{2V} = \frac{NkT}{V}

So the pressures are the same.

Amount of substance

Use $N = nN_A$ .

Sample X:

n_X = \frac{N}{N_A}

Sample Y:

n_Y = \frac{2N}{N_A}

Mean-square speed

From kinetic theory:

\langle c^2 \rangle = \frac{3kT}{m_{\text{molecule}}}

Sample X has molecule mass $m$ :

\langle c^2 \rangle_X = \frac{3kT}{m}

Sample Y has molecule mass $2m$ :

\langle c^2 \rangle_Y = \frac{3kT}{2m}

So heavier molecules move more slowly (lower mean-square speed) at the same temperature.

Internal energy

Total internal energy (for translational motion):

U = \frac{3}{2}NkT

Sample X:

U_X = \frac{3}{2}NkT

Sample Y has $2N$ molecules:

U_Y = \frac{3}{2}(2N)kT = 3NkT

Even though each molecule in Y is heavier, at the same temperature each molecule still has average translational kinetic energy $\frac{3}{2}kT$ , so doubling the number of molecules doubles the total internal energy.

Key Takeaways

$p$ depends on number density $N/V$ and $T$ via $p = NkT/V$ .
$n = N/N_A$ converts molecules to moles.
At fixed $T$ , mean-square speed scales as $1/m$ .
For an ideal gas, internal energy depends only on $T$ and number of molecules (for the model used): $U \propto NT$ .

Common Mistakes

Using $pV = nRT$ but forgetting to convert $n$ correctly from $N$ .
Writing $\langle c^2 \rangle = 3kT/m$ and then forgetting to replace $m$ with $2m$ for sample Y.
Confusing mean-square speed $\langle c^2 \rangle$ with r.m.s. speed $c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$ .
Assuming internal energy depends on molecular mass; for an ideal gas at fixed $T$ , average kinetic energy per molecule depends only on $T$ .

Things to Be Careful About

Keep clear: in this question $m$ is the mass of one molecule in sample X, not the total mass.
Check cancellations carefully: for pressure in sample Y, both $N$ and $V$ double so $N/V$ is unchanged.
If using $R$ instead of $k$ , ensure you also use $n$ and not $N$ (or explicitly convert using $k = R/N_A$ ).

Techniques used

apply the ideal gas equation in molecular and molar formsconvert between amount of substance and number of molecules using the Avogadro constantrelate mean-square speed to temperature using kinetic theorydetermine internal energy from average molecular kinetic energy

(ii)

The temperature of sample X is now varied.

On Fig. 3.1, sketch the variation with thermodynamic temperature of the root-mean square (r.m.s.) speed of the molecules of the gas.

Fig. 3.1

DifficultyMedium-Easy

Worked solution

Working

For sample X,

\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT \Rightarrow c_{\text{rms}} = \sqrt{\langle c^2 \rangle} = \sqrt{\frac{3kT}{m}}

So $c_{\text{rms}} \propto \sqrt{T}$ .

Answer

diagram-bii-1

Final answer

A curve through the origin with c_rms proportional to sqrt(T) (increasing with decreasing gradient).

Detailed explanation

Background Concept

Kinetic theory relates temperature to the average translational kinetic energy of molecules:

\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT

The r.m.s. speed is defined by:

c_{\text{rms}} = \sqrt{\langle c^2 \rangle}

Combining gives:

c_{\text{rms}} = \sqrt{\frac{3kT}{m}}

So r.m.s. speed increases with temperature, but as a square-root, not a straight line.

Understanding the Question

Only sample X is considered and its temperature $T$ is varied. You must sketch $c_{\text{rms}}$ (y-axis) against thermodynamic temperature $T$ (x-axis) on axes that start at the origin.

The key features the examiner is looking for are:

correct dependence: $c_{\text{rms}} \propto \sqrt{T}$ ,
correct shape: increases but with decreasing gradient (concave down),
passes through the origin.

Approach

Write the expression for $c_{\text{rms}}$ in terms of $T$ .
Identify the functional form ( $\sqrt{T}$ ).
Sketch a curve that starts at $(0,0)$ and increases, flattening gradually.

Step-by-Step Reasoning

From kinetic theory:

\langle c^2 \rangle = \frac{3kT}{m}

Therefore:

c_{\text{rms}} = \sqrt{\langle c^2 \rangle} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3k}{m}}\,\sqrt{T}

Since $\sqrt{\frac{3k}{m}}$ is a constant for sample X, we have:

c_{\text{rms}} \propto \sqrt{T}

Graph features:

At $T=0$ , $c_{\text{rms}}=0$ , so it goes through the origin.
As $T$ increases, $c_{\text{rms}}$ increases.
Because it is a square-root curve, the slope $\mathrm{d}c_{\text{rms}}/\mathrm{d}T$ decreases with $T$ (the curve becomes less steep).

diagram-bii-1

Key Takeaways

For a fixed gas (fixed $m$ ), $c_{\text{rms}} \propto \sqrt{T}$ .
Square-root graphs rise quickly at first, then gradually flatten.

Common Mistakes

Drawing a straight line (implies $c_{\text{rms}} \propto T$ , incorrect).
Drawing a curve that is concave up (gradient increasing), opposite to $\sqrt{T}$ .
Not passing through the origin even though the axes start at $(0,0)$ .

Things to Be Careful About

Use thermodynamic temperature in kelvin, so $T=0$ is a meaningful intercept.
The question asks for r.m.s. speed, not mean-square speed; the graph should follow $\sqrt{T}$ not $T$ .

Techniques used

use kinetic theory to relate r.m.s. speed to temperaturetranslate a proportionality into the correct graph shapesketch a curve with correct intercept and changing gradient

Q4MediumOscillations

(a)

State what is meant by simple harmonic motion.

DifficultyEasy

Worked solution

Answer

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and is directed towards the equilibrium position:

a \propto -x \quad (a = -\omega^2 x)

Final answer

Acceleration is proportional to displacement from equilibrium and opposite in direction (a = -ω^2 x).

Detailed explanation

Background Concept

In oscillations, simple harmonic motion (SHM) is a special type of periodic motion where the restoring effect gets stronger the further you are from equilibrium.

The defining mathematical condition is:

a = -\omega^2 x

where:

$x$ is the displacement from the equilibrium position,
$a$ is the acceleration,
$\omega$ is the angular frequency (a constant for that oscillation),
the negative sign means the acceleration is always towards equilibrium (opposite to $x$ ).

Understanding the Question

The question asks for the meaning of SHM, so you must state the relationship between acceleration and displacement, including the direction.

Approach

Give the definition in words (proportional to displacement and directed towards equilibrium) and/or give the standard SHM equation $a=-\omega^2 x$ .

Step-by-Step Reasoning

“Proportional to displacement” means if $x$ doubles, $a$ doubles.
“Directed towards equilibrium” means the acceleration always acts to reduce the displacement, which is why there is a negative sign:

a \propto -x

This is commonly written as:

a = -\omega^2 x

Key Takeaways

SHM is defined by $a$ being proportional to $x$ .
The acceleration must be opposite to the displacement (restoring).

Common Mistakes

Saying “force proportional to displacement” without mentioning it is towards equilibrium (missing the negative sign idea).
Confusing proportionality with “equal to displacement”.
Stating velocity is proportional to displacement (not the definition).

Things to Be Careful About

$x$ must be measured from the equilibrium position, not from an end point.
Direction matters: you need the idea of “opposite direction” (or the minus sign).

Techniques used

state the defining condition for simple harmonic motionexpress the SHM condition using a proportionality and a sign to indicate direction

(b)

A block is suspended from a spring, as shown in Fig. 4.1.

Fig. 4.1

The block is pulled down and released at time $t = 0$ . It then oscillates vertically with simple harmonic motion.

Fig. 4.2 shows the variation of the velocity $v$ of the block with height $h$ of the base of the block above the floor.

Fig. 4.2

9M

(i)

Determine the amplitude, in cm, of the oscillations.

amplitude = ______ $\text{cm}$

DifficultyEasy

Worked solution

Working

From the graph, $h_{\max} = 9.5\ \text{cm}$ and $h_{\min} = 3.5\ \text{cm}$ .

A = \frac{h_{\max}-h_{\min}}{2} = \frac{9.5-3.5}{2} = 3.0\ \text{cm}

Answer

$3.0\ \text{cm}$

Final answer

3.0 cm

Detailed explanation

Background Concept

For SHM, the amplitude $A$ is the maximum displacement from the equilibrium position.

If you can read the maximum and minimum values of the displacement coordinate (here the height $h$ ), then:

A = \frac{h_{\max}-h_{\min}}{2}

because $h_{\max}-h_{\min}$ is the total peak-to-peak range.

Understanding the Question

The graph given is $v$ against $h$ . The turning points of the motion occur where $v=0$ , which correspond to the extreme values of $h$ .

So you must read the extreme $h$ values from the ellipse and take half their difference.

Approach

Identify $h_{\max}$ and $h_{\min}$ (where $v=0$ on the ellipse).
Calculate $A=(h_{\max}-h_{\min})/2$ .

Step-by-Step Reasoning

From the ellipse on the $v$ – $h$ graph:

At the right-most intercept with $v=0$ , $h_{\max}=9.5\ \text{cm}$ .
At the left-most intercept with $v=0$ , $h_{\min}=3.5\ \text{cm}$ .

Peak-to-peak range:

h_{\max}-h_{\min} = 9.5-3.5 = 6.0\ \text{cm}

Amplitude is half of this:

A = \frac{6.0}{2} = 3.0\ \text{cm}

Key Takeaways

Turning points have $v=0$ .
Amplitude is half of the maximum-to-minimum displacement range.

Common Mistakes

Giving $6.0\ \text{cm}$ as the amplitude (that is peak-to-peak, not amplitude).
Using the centre value ( $6.5\ \text{cm}$ ) as the amplitude (that is the equilibrium height).

Things to Be Careful About

The amplitude is measured from equilibrium, not from the floor.
Read values carefully from axis labels and units (cm here).

Techniques used

identify maximum and minimum displacement from a graphuse half the peak-to-peak value to find amplitude

(ii)

Show that the angular frequency of the oscillations is $3.2\ \text{rad s}^{-1}$ .

DifficultyMedium-Easy

Worked solution

Working

For SHM,

v_{\max} = \omega A

From the graph, $v_{\max} = 9.6\ \text{cm s}^{-1}$ and from (i) $A = 3.0\ \text{cm}$ .

\omega = \frac{v_{\max}}{A} = \frac{9.6}{3.0} = 3.2\ \text{rad s}^{-1}

Answer

$\omega = 3.2\ \text{rad s}^{-1}$

Final answer

3.2 rad s^-1

Detailed explanation

Background Concept

In SHM, the speed is maximum at the equilibrium position and zero at the turning points. A key result is:

v_{\max} = \omega A

where:

$A$ is the amplitude,
$\omega$ is the angular frequency,
$v_{\max}$ is the maximum speed.

This comes from differentiating $x = A\sin(\omega t)$ , giving $v = A\omega\cos(\omega t)$ , so the maximum value is $A\omega$ .

Understanding the Question

The $v$ – $h$ graph is an ellipse for SHM. The top and bottom of the ellipse give the maximum speed (positive and negative). The width gives the amplitude in $h$ .

You are asked to show $\omega = 3.2\ \text{rad s}^{-1}$ , so you must use a relationship and substitute the read-off values.

Approach

Read $v_{\max}$ from the graph.
Use the amplitude $A$ from part (i).
Apply $v_{\max}=\omega A$ and rearrange for $\omega$ .

Step-by-Step Reasoning

From the graph:

The largest magnitude of velocity is $9.6\ \text{cm s}^{-1}$ , so $v_{\max}=9.6\ \text{cm s}^{-1}$ .
The amplitude from part (i) is $A=3.0\ \text{cm}$ .

Use the SHM maximum speed relation:

v_{\max} = \omega A

Rearrange:

\omega = \frac{v_{\max}}{A}

Substitute:

\omega = \frac{9.6}{3.0} = 3.2\ \text{s}^{-1}

Angular frequency is conventionally written with unit $\text{rad s}^{-1}$ (radian is dimensionless but stated).

Key Takeaways

In SHM, $v_{\max}$ occurs at equilibrium and equals $\omega A$ .
The ellipse in a $v$ – $x$ plot encodes $A$ horizontally and $v_{\max}$ vertically.

Common Mistakes

Using peak-to-peak displacement $6.0\ \text{cm}$ instead of amplitude $3.0\ \text{cm}$ .
Mixing units (e.g. converting only one of cm to m). It’s fine to keep both in cm as long as you are consistent.
Forgetting the unit for $\omega$ .

Things to Be Careful About

Use magnitudes: $v_{\max}$ is a positive number even though the graph shows $\pm 9.6$ .
If you convert to SI, convert both: $9.6\ \text{cm s}^{-1}=0.096\ \text{m s}^{-1}$ and $3.0\ \text{cm}=0.030\ \text{m}$ gives the same $\omega$ .

Techniques used

use the SHM relation between maximum speed, angular frequency and amplituderead the maximum speed from a velocity–displacement graphsubstitute into an SHM formula and evaluate

(iii)

Calculate the period $T$ of the oscillations.

$T$ = ______ $\text{s}$

(iv)

On Fig. 4.3, sketch the variation of $h$ with time $t$ from $t = 0$ to $t = 6.0\ \text{s}$ .

Fig. 4.3

4M Similar

DifficultyMedium

Worked solution

diagram-biv

Answer

Sketch a sinusoidal $h$ – $t$ graph with mean (equilibrium) height $6.5\ \text{cm}$ and amplitude $3.0\ \text{cm}$ , so $h$ varies between $3.5\ \text{cm}$ and $9.5\ \text{cm}$ .

At $t=0$ , the block is released from the lowest point: $h=3.5\ \text{cm}$ with zero gradient.

Use $T = 2.0\ \text{s}$ (from (iii)) to mark turning points every $T/2 \approx 1.0\ \text{s}$ and repeat the sinusoid up to $t=6.0\ \text{s}$ (about 3 cycles), ending close to the minimum again.

Final answer

Sinusoidal h–t curve about 6.5 cm with amplitude 3.0 cm, starting at h = 3.5 cm at t = 0 and period ≈ 2.0 s (3 cycles to 6 s).

Detailed explanation

Background Concept

SHM displacement varies sinusoidally with time:

x = A\sin(\omega t) \quad \text{or} \quad x = A\cos(\omega t)

The choice of sine/cosine depends on the starting position at $t=0$ .

Key facts:

The equilibrium position is the mid-point of the motion.
Turning points occur at maximum/minimum displacement, where velocity is zero.
One complete cycle takes time $T$ , and adjacent turning points are separated by $T/2$ .

Understanding the Question

You are given a $v$ – $h$ ellipse. From it you can read:

equilibrium height (centre of ellipse): $h_0 = 6.5\ \text{cm}$ ,
amplitude: $A = 3.0\ \text{cm}$ , so $h_{\min}=3.5\ \text{cm}$ and $h_{\max}=9.5\ \text{cm}$ ,
period: $T \approx 2.0\ \text{s}$ .

The block is “pulled down and released at $t=0$ ” meaning:

it starts at the lowest height (a turning point),
so initial velocity is zero and the $h$ – $t$ graph has zero gradient at $t=0$ .

You must sketch $h$ against $t$ from $0$ to $6.0\ \text{s}$ with the correct midline, amplitude and timing.

Approach

Draw the midline at $h=6.5\ \text{cm}$ .
Draw the maximum and minimum at $9.5\ \text{cm}$ and $3.5\ \text{cm}$ .
Because it is released from the lowest point, start the curve at $h=3.5\ \text{cm}$ with a flat tangent.
Use $T\approx 2.0\ \text{s}$ to space the oscillations: peaks every $T$ and adjacent turning points every $T/2$ .

Step-by-Step Reasoning

From the $v$ – $h$ ellipse:

Centre at $h=6.5\ \text{cm}$ gives equilibrium height $h_0=6.5\ \text{cm}$ .
Left and right intercepts give $h_{\min}=3.5\ \text{cm}$ and $h_{\max}=9.5\ \text{cm}$ .

The initial condition “pulled down and released” implies:

at $t=0$ the block is at $h_{\min}=3.5\ \text{cm}$ ,
and $v=0$ at that instant, so slope $\mathrm{d}h/\mathrm{d}t = 0$ .

A suitable model is:

h = h_0 - A\cos(\omega t)

because at $t=0$ , $\cos(0)=1$ so $h=h_0-A=3.5\ \text{cm}$ .

Now place key times using $T\approx 2.0\ \text{s}$ :

At $t=T/4 \approx 0.50\ \text{s}$ , $h=h_0$ (crossing equilibrium, moving upward).
At $t=T/2 \approx 1.0\ \text{s}$ , $h=h_{\max}=9.5\ \text{cm}$ (next turning point).
At $t=3T/4 \approx 1.5\ \text{s}$ , back through $h_0$ (moving downward).
At $t=T \approx 2.0\ \text{s}$ , back to $h_{\min}=3.5\ \text{cm}$ .

Repeat this pattern up to $t=6.0\ \text{s}$ (about 3 full periods), so the curve completes about three full sinusoidal cycles and is near the minimum again at $t=6\ \text{s}$ .

diagram-biv

Key Takeaways

Equilibrium on a $v$ – $x$ ellipse is the centre of the ellipse.
Turning points correspond to $v=0$ .
“Released from rest” means the $x$ – $t$ graph starts at an extreme with zero gradient.
Period sets the horizontal spacing of the oscillation.

Common Mistakes

Starting at the equilibrium position instead of the lowest point.
Drawing a triangle wave or non-sinusoidal curve.
Using the wrong midline (e.g. centred at $h=0$ instead of $h=6.5\ \text{cm}$ ).
Getting the period wrong (e.g. using $T=1/\omega$ ).

Things to Be Careful About

Label values on the y-axis consistently: the curve must reach about $3.5\ \text{cm}$ and $9.5\ \text{cm}$ .
Make sure the curve has zero slope at turning points.
Over $0$ to $6\ \text{s}$ , there should be about three complete cycles if $T\approx 2\ \text{s}$ .

Techniques used

identify the equilibrium position and turning points from a velocity–displacement graphdetermine initial phase from the described initial conditionuse period to place key points on a displacement–time sketchsketch a sinusoidal curve with correct amplitude and mean value

Q5MediumElectric Fields

(a)

State the relationship between electric field and electric potential.

DifficultyEasy

Worked solution

Answer

Electric field is the (negative) potential gradient:

\vec{E} = -\nabla V

Along a straight line,

E = -\frac{\mathrm{d}V}{\mathrm{d}x}

Final answer

Electric field is the negative potential gradient: \vec{E} = -\nabla V (so along a line E = -dV/dx).

Detailed explanation

Background Concept

Electric potential $V$ at a point is the work done per unit positive test charge in bringing it from infinity to that point. Electric field $\vec{E}$ is related to how rapidly the potential changes with position.

For any direction, the component of electric field in that direction equals the negative rate of change of potential in that direction. In vector form this is

\vec{E} = -\nabla V

If we only move along one straight line (say the $x$ -axis), this becomes

E_x = -\frac{\mathrm{d}V}{\mathrm{d}x}

The negative sign means the field points from high potential to low potential.

Understanding the Question

You are asked to state the relationship between electric field and electric potential. Since it says “state”, no numerical working is needed; just the correct equation and sign.

Approach

Write the standard “field = negative potential gradient” relationship either in full vector form or as a 1D derivative along a line.

Step-by-Step Reasoning

Recognise that the electric field is linked to the spatial rate of change of potential.
State the relationship:

\vec{E} = -\nabla V

Optionally, give the 1D form (often used in A-level problems):

E = -\frac{\mathrm{d}V}{\mathrm{d}x}

Key Takeaways

Electric field is the negative gradient of potential.
Field direction is towards decreasing potential.

Common Mistakes

Missing the minus sign (this loses the direction information).
Writing $E = Vd$ or $E = V/d$ without stating it is for a uniform field only.

Things to Be Careful About

$E = V/d$ is only valid for a uniform field between parallel plates; the general relationship is the gradient form.
Use vector form $\vec{E} = -\nabla V$ if the field is not restricted to one line.

Techniques used

use the definition of electric field as potential gradientapply the sign convention that the field points in the direction of decreasing potential

(b)

Two charged isolated insulating spheres X and Y are near to each other, as shown in Fig. 5.1.

Fig. 5.1

$P$ is a point on the line joining the centres of the spheres.

Explain why it is not possible for the total electric potential and the resultant electric field to simultaneously be zero at point $P$ .

DifficultyMedium

Worked solution

Answer

Electric potential adds as a scalar. For $V$ to be zero at $P$ due to two charges, the charges must be of opposite sign (their potentials must cancel).

Electric field adds as a vector. For opposite signs, the fields at a point between the charges are in the same direction (from $+$ to $-$ ), so they cannot cancel and hence resultant $E \neq 0$ .

Conversely, if resultant $E = 0$ at $P$ (between the spheres), the charges must be the same sign so that their fields oppose there; then the potentials have the same sign and cannot sum to zero. Therefore $V$ and $E$ cannot both be zero at $P$ .

Final answer

Zero potential at P requires opposite-sign charges, but then the fields at a point between them are in the same direction and cannot cancel; if the field were zero between them the charges would have to be like, making the potentials the same sign and not able to sum to zero.

Detailed explanation

Background Concept

Electric potential $V$ is a scalar quantity, so the total potential at a point is the algebraic sum of the individual potentials:

V_{\text{total}} = V_1 + V_2

Electric field $\vec{E}$ is a vector, so the resultant field is the vector sum:

\vec{E}_{\text{total}} = \vec{E}_1 + \vec{E}_2

For point charges, the potential is

V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

and the field is

E = \frac{1}{4\pi\varepsilon_0}\frac{|Q|}{r^2}

with direction: away from positive charge and towards negative charge.

Understanding the Question

Two isolated insulating spheres (so charges stay on each sphere) are near each other. Point $P$ lies on the line between their centres.

You must explain why it is impossible for both:

total electric potential at $P$ to be zero, and
resultant electric field at $P$ to be zero,

at the same time.

The key is that potentials cancel by sign (scalar), but fields cancel by direction (vector).

Approach

Consider the two possible ways to get one quantity to be zero, and show that each implies conditions that prevent the other from being zero:

If $V_{\text{total}}=0$ between two charges, what must the charge signs be?
If $\vec{E}_{\text{total}}=0$ between two charges, what must the charge signs be?
Then compare these requirements.

Step-by-Step Reasoning

diagram-b-1

Condition for $V_{\text{total}}=0$ at $P$
- For two charges, the potentials at $P$ are proportional to $Q/r$ and carry the sign of the charge.
- To add to zero, one contribution must be positive and the other negative.
- Therefore the two charges must be opposite sign.
What happens to the electric fields between opposite-sign charges?
- Take the common arrangement: left charge positive, right charge negative.
- At a point between them:
  - the field due to the positive charge points away from it (to the right),
  - the field due to the negative charge points towards it (also to the right).
- So both field contributions point the same direction, meaning they add.
- Therefore the resultant field cannot be zero: $\vec{E}_{\text{total}} \neq 0$ .
Conversely, condition for $\vec{E}_{\text{total}}=0$ at $P$ (between the charges)
- For the field vectors to cancel between the charges, they must be in opposite directions.
- This happens only if both charges are like charges (both positive or both negative), so that one field at $P$ points left and the other points right.
- But then the potentials at $P$ have the same sign and therefore cannot sum to zero.
Therefore, the requirements for $V_{\text{total}}=0$ and $\vec{E}_{\text{total}}=0$ are mutually incompatible at a point between the spheres.

Key Takeaways

Potential is a scalar: cancellation requires opposite signs.
Electric field is a vector: cancellation requires equal magnitudes in opposite directions.
Between two charges, the direction pattern of $\vec{E}$ depends on whether charges are like or unlike.

Common Mistakes

Assuming “if potential is zero then field is zero”; zero potential just means the scalar sum is zero, not that the slope (field) is zero.
Forgetting field directions and treating field as a scalar.
Thinking fields between unlike charges can oppose (they do not; they point from $+$ to $-$ ).

Things to Be Careful About

The question specifies $P$ is on the line joining the centres and (from the diagram) between the spheres; the direction arguments above are for that region.
Outside the region between charges, field directions change; always decide directions using “away from +, towards −” at the point of interest.

Techniques used

apply the superposition principle for electric potential (scalar addition)apply the superposition principle for electric field (vector addition)infer the signs of charges from a zero-potential conditioncompare field directions for like and unlike charges along the line joining them

(c)

The magnitudes of the charges on spheres X and Y in Fig. 5.1 are $Q$ and $2Q$ respectively. The spheres may be considered as point charges at their centres.

Point $P$ is a distance $x$ from the centre of sphere X.

The electric potential at point $P$ is zero.

5M

(i)

Show that the distance $y$ of point $P$ from the centre of sphere Y is equal to $2x$ .

DifficultyMedium-Easy

Worked solution

Working

For $V_P = 0$ , the charges must be opposite sign, so

\frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{x}-\frac{2Q}{y}\right)=0

\frac{Q}{x}=\frac{2Q}{y}\ \Rightarrow\ y=2x

Answer

$y = 2x$

Final answer

y = 2x

Detailed explanation

Background Concept

The electric potential at a point due to a point charge $Q$ a distance $r$ away is

V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

Potential is a scalar, so for multiple charges we add the potentials (with sign):

V_{\text{total}} = \sum V_i

A total potential of zero means the positive and negative contributions cancel.

Understanding the Question

You are told the magnitudes of the charges are $Q$ on X and $2Q$ on Y, and that point $P$ is at distance $x$ from X and distance $y$ from Y. The total potential at $P$ is zero.

You must show that this condition forces a particular distance ratio, namely $y = 2x$ .

Approach

Write the expression for $V_P$ as the sum of the potentials from X and Y.
Use the fact that $V_P=0$ , so the two contributions must be equal in magnitude and opposite in sign.
Solve the resulting equation for $y$ in terms of $x$ .

Step-by-Step Reasoning

Potential at $P$ due to X (magnitude $Q$ , distance $x$ ):

V_X = \frac{1}{4\pi\varepsilon_0}\frac{Q}{x}

Potential at $P$ due to Y (magnitude $2Q$ , distance $y$ ):

V_Y = \frac{1}{4\pi\varepsilon_0}\frac{\pm 2Q}{y}

For the total to be zero, the charges must be opposite sign, so take $V_Y$ as negative relative to $V_X$ :

V_P = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{x}-\frac{2Q}{y}\right)=0

Multiply through by $4\pi\varepsilon_0$ (non-zero) and simplify:

\frac{Q}{x} = \frac{2Q}{y}

Cancel $Q$ :

\frac{1}{x} = \frac{2}{y}

Rearrange:

y = 2x

Key Takeaways

Use superposition: total potential is the algebraic sum of individual potentials.
Zero total potential implies opposite signs and equal magnitudes of the potential contributions.
For point charges, potential depends on $1/r$ (not $1/r^2$ ).

Common Mistakes

Using $1/r^2$ instead of $1/r$ for potential.
Forgetting that to get $V=0$ the charges must be opposite sign.
Dropping the common factor $\frac{1}{4\pi\varepsilon_0}$ incorrectly (it cancels safely).

Things to Be Careful About

The question says magnitudes are $Q$ and $2Q$ ; the sign must be inferred from $V_P=0$ .
Keep $x$ and $y$ clearly associated with the correct sphere.

Techniques used

write electric potential due to point chargesapply the superposition principle for potentialset the algebraic sum of potentials to zero and solve for distance ratio

(ii)

State an expression, in terms of $Q$ , $x$ and the permittivity of free space $\varepsilon_0$ , for the electric field strength $E_X$ at $P$ due to sphere X.

$E_X$ = ______

(iii)

Determine an expression, in terms of $Q$ , $x$ and $\varepsilon_0$ , for the resultant electric field strength $E$ at point $P$ due to the two spheres.

$E$ = ______

DifficultyMedium-Easy

Worked solution

Working

From (i), $y = 2x$ .

E_X = \frac{1}{4\pi\varepsilon_0}\frac{Q}{x^2}

E_Y = \frac{1}{4\pi\varepsilon_0}\frac{2Q}{y^2} = \frac{1}{4\pi\varepsilon_0}\frac{2Q}{(2x)^2} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{2x^2}

Since $V_P=0$ the charges are opposite sign, so at a point between them the fields are in the same direction and add:

E = E_X + E_Y = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{x^2}+\frac{Q}{2x^2}\right) = \frac{1}{4\pi\varepsilon_0}\frac{3Q}{2x^2}

Answer

E = \frac{1}{4\pi\varepsilon_0}\frac{3Q}{2x^2}

Final answer

E = (1/(4π ε0)) · (3Q/(2x^2))

Detailed explanation

Background Concept

Electric field from a point charge obeys an inverse-square law:

E = \frac{1}{4\pi\varepsilon_0}\frac{|Q|}{r^2}

The resultant electric field is found by vector addition (superposition). On a single straight line, the vectors are collinear, so you either add magnitudes (same direction) or subtract magnitudes (opposite directions).

Also, from earlier parts: zero total potential at a point requires opposite-sign charges, because potential is scalar.

Understanding the Question

You know sphere X has charge magnitude $Q$ and sphere Y has charge magnitude $2Q$ . Point $P$ is at distance $x$ from X, and from part (i) you showed the distance from Y is $y=2x$ .

You must find the resultant electric field strength at $P$ due to both spheres.

Approach

Compute $E_X$ at distance $x$ .
Compute $E_Y$ at distance $y=2x$ .
Decide whether to add or subtract based on directions. Since $V_P=0$ , the charges are opposite sign, so at a point between them the fields point the same way and must be added.

Step-by-Step Reasoning

Field due to X:

E_X = \frac{1}{4\pi\varepsilon_0}\frac{Q}{x^2}

Field due to Y at distance $y$ :

E_Y = \frac{1}{4\pi\varepsilon_0}\frac{2Q}{y^2}

Using $y = 2x$ :

E_Y = \frac{1}{4\pi\varepsilon_0}\frac{2Q}{(2x)^2} = \frac{1}{4\pi\varepsilon_0}\frac{2Q}{4x^2} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{2x^2}

Direction decision:

Opposite-sign charges produce electric fields at points between them that both point from the positive charge towards the negative charge.
Hence the field contributions at $P$ are in the same direction and add.

So the resultant magnitude is

E = E_X + E_Y = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{x^2}+\frac{Q}{2x^2}\right) = \frac{1}{4\pi\varepsilon_0}\frac{3Q}{2x^2}

(If direction were requested: it is towards the negatively charged sphere.)

Key Takeaways

Use $E \propto 1/r^2$ for fields and $V \propto 1/r$ for potentials.
Use the potential condition to infer charge signs, which then fixes whether field vectors add or subtract.
When points lie on the line joining two charges, vector addition becomes 1D (sign/direction matters).

Common Mistakes

Using $y=2x$ but forgetting to square it in $E_Y \propto 1/y^2$ .
Subtracting fields when the charges must be opposite sign (between opposite charges, the fields add).
Treating potential cancellation as implying field cancellation.

Things to Be Careful About

Keep track of whether $Q$ is being used as a magnitude or includes sign. The final expression for field strength (magnitude) should be positive.
Ensure the algebra inside the brackets is combined correctly:

\frac{1}{x^2}+\frac{1}{2x^2} = \frac{3}{2x^2}

Techniques used

use the result y = 2x to substitute distancescalculate each field contribution using the inverse-square lawadd electric field contributions as collinear vectors

Q6MediumAlternating CurrentsCapacitance

(a)

3M

(i)

State what is meant by rectification of an alternating voltage.

(ii)

State the difference between half-wave rectification and full-wave rectification.

DifficultyMedium-Easy

Worked solution

Answer

Half-wave rectification uses only one half-cycle of the a.c. input (the other half-cycle is blocked), so the output has gaps.
Full-wave rectification uses both half-cycles, with one half-cycle reversed so the output is always the same polarity.

Final answer

Half-wave uses one half-cycle only; full-wave uses both half-cycles with one reversed so output is same polarity throughout.

Detailed explanation

Background Concept

A rectified signal is one that stays the same polarity. The difference between half-wave and full-wave rectification is how much of the input waveform is used:

In half-wave rectification, the diode conducts on only one polarity, so only one half of each cycle appears at the output.
In full-wave rectification, the circuit arranges for current through the load to flow in the same direction during both halves of the input cycle (commonly via a bridge rectifier).

Understanding the Question

You need to state a clear difference. Since it is 2 marks, two distinct points are expected.

Approach

Describe (1) which half-cycles are used and (2) what that means for the output polarity / gaps / ripple frequency.

Step-by-Step Reasoning

Half-wave: diode forward biases on (say) the positive half-cycle, so current flows; on the negative half-cycle the diode is reverse-biased, so current is (approximately) zero. Output exists only for half the time.
Full-wave: circuitry flips the negative half-cycle so it becomes positive across the load; therefore every half-cycle contributes to the output, and the ripple repeats twice as often.

Key Takeaways

Half-wave: uses one half-cycle; lower average output; ripple frequency equals input frequency.
Full-wave: uses both half-cycles; higher average output; ripple frequency is doubled.

Common Mistakes

Saying “full-wave gives a bigger voltage”: it generally gives a larger mean output, but the key distinguishing idea is using both halves.
Confusing rectification with smoothing (capacitor action).

Things to Be Careful About

Use wording like “one half-cycle blocked” and “both half-cycles used (one reversed)”. That maps directly to typical mark points.

Techniques used

compare output waveforms for different rectifier typesstate how many half-cycles contribute to the output

(b)

3M

(i)

Complete Fig. 6.1 to show a circuit that produces half-wave rectification of an alternating input voltage $V_{IN}$ to produce output voltage $V_{OUT}$ across the resistor R.

Fig. 6.1

DifficultyMedium-Easy

Worked solution

diagram-6b-i

Answer

A single diode is connected in series between one input terminal and the top of the parallel (C)-(R) branch, oriented so it conducts when (V_{IN}) makes the top of the (C)-(R) branch positive. The other input terminal is connected directly to the bottom of the (C)-(R) branch.

Final answer

Series diode feeding the parallel C–R load, oriented to conduct on the positive half-cycle; return directly to the other input terminal.

Detailed explanation

Background Concept

A diode conducts significantly only when forward-biased, so it can be used to block one half-cycle of an a.c. input. For half-wave rectification, a single diode in series with the load is sufficient.

A capacitor placed across the load acts as a reservoir: it charges when the diode conducts and discharges through the load when the diode is off, reducing ripple.

Understanding the Question

Fig. 6.1 already contains (C) and (R) in parallel and shows where (V_{OUT}) is measured (across (R)). The missing part is the rectifying component(s) and wiring from the a.c. input terminals to the load.

You must produce half-wave rectification, so only one diode is required.

Approach

Put a diode in series with the load so that current can flow for only one polarity of (V_{IN}).
Connect the other side of the a.c. supply to the return (bottom node) of the load.
Ensure the diode orientation makes the output positive during the conducting half-cycle (consistent with the later graphs showing (V_{OUT}) reaching (+12\ \text{V})).

Step-by-Step Reasoning

diagram-6b-i

Choose the top node of the parallel (C) and (R) as the output positive node.
Insert a diode between the appropriate input terminal and this top node.
Orient the diode so that, when the input is positive, it is forward-biased and charges the capacitor / supplies the resistor.
Connect the other input terminal directly to the bottom node of the parallel (C) and (R).

Key Takeaways

Half-wave rectifier: one series diode.
Smoothing capacitor: placed in parallel with the load resistor.

Common Mistakes

Putting the diode in parallel with (R) (this does not rectify the supply to the load correctly).
Reversing the diode direction so the output would be negative when it conducts (inconsistent with the given (V_{OUT}) graph).
Leaving the circuit incomplete by not providing a return connection to the second input terminal.

Things to Be Careful About

The output (V_{OUT}) is across (R), so the diode must feed the node connected to the top of (R).
A single diode gives half-wave rectification; a bridge arrangement is for full-wave rectification.

Techniques used

select a diode arrangement to block one half-cycleplace a smoothing capacitor in parallel with the loadidentify where to measure the output voltage

(ii)

State the purpose of the capacitor C in the circuit of Fig. 6.1.

DifficultyEasy

Worked solution

Answer

The capacitor charges when the diode conducts and then discharges through (R) when the diode is off, so it smooths (V_{OUT}) (reduces the ripple).

Final answer

To smooth the rectified output by charging on peaks and discharging through R between peaks.

Detailed explanation

Background Concept

In a rectifier circuit, the output can be a pulsating d.c. rather than a steady voltage. A capacitor connected across the load can reduce these fluctuations:

When the supply (through the diode) provides a high voltage, the capacitor charges up towards that peak voltage.
When the diode stops conducting (input reverses or falls), the capacitor discharges through the load resistor, providing current and maintaining the output voltage.

This is called smoothing (reducing ripple).

Understanding the Question

You are asked for the purpose of (C) in Fig. 6.1, which contains (C) in parallel with (R). Since (V_{OUT}) is measured across (R), anything across (R) affects the output waveform.

Approach

State the key idea: charge on conduction, discharge between conduction intervals, giving a more constant output.

Step-by-Step Reasoning

During the conducting half-cycle, the diode is forward-biased and the capacitor charges quickly to near the maximum output.
During the non-conducting half-cycle, the diode is reverse-biased so the supply is effectively disconnected; the capacitor then discharges through (R), keeping (V_{OUT}) from dropping to zero.

Key Takeaways

Capacitor across load = smoothing capacitor.
It reduces ripple by acting as a temporary energy store.

Common Mistakes

Saying the capacitor “rectifies” the voltage: rectification is done by the diode.
Saying it “increases the peak voltage”: it primarily affects how quickly the voltage falls between peaks.

Things to Be Careful About

Mention both charging and discharging for full credit.
Use the term “smooth” or “reduce ripple” to align with standard mark schemes.

Techniques used

explain capacitor charging and discharging in a rectifier circuitlink capacitor action to reduced ripple in the output

(c)

The input voltage $V_{IN}$ in Fig. 6.1 is a square wave. Fig. 6.2 shows the variation of $V_{IN}$ with time $t$ .

Fig. 6.2

Fig. 6.3 shows the variation of $V_{OUT}$ with $t$ .

Fig. 6.3

The maximum energy stored in the capacitor is $0.041\ \text{J}$ .

5M

(i)

Show that the capacitance of C is $570\ \mu\text{F}$ .

DifficultyMedium-Easy

Worked solution

Working

Maximum (V_{OUT}) (and capacitor p.d.) is (12\ \text{V}).

E = \frac{1}{2}CV^2

C = \frac{2E}{V^2} = \frac{2(0.041)}{12^2} = \frac{0.082}{144} = 5.7 \times 10^{-4}\ \text{F}

C = 5.7 \times 10^{-4}\ \text{F} = 5.7 \times 10^{2}\ \mu\text{F} \approx 570\ \mu\text{F}

Answer

(C \approx 570\ \mu\text{F}).

Final answer

570 \muF

Detailed explanation

Background Concept

A capacitor stores electrical energy in its electric field when there is a potential difference across it. The energy stored is

E = \frac{1}{2}CV^2

where:

(E) is energy in joules (J)
(C) is capacitance in farads (F)
(V) is the potential difference across the capacitor in volts (V)

This formula applies at any instant for a capacitor at voltage (V).

Understanding the Question

You are told the maximum energy stored in the capacitor is (0.041\ \text{J}). From the graph, the maximum output voltage is (12\ \text{V}). Since (C) is in parallel with (R), the capacitor voltage equals (V_{OUT}), so the maximum capacitor voltage is also (12\ \text{V}).

You must show the capacitance is (570\ \mu\text{F}).

Approach

Use (E = \tfrac{1}{2}CV^2), rearrange for (C), substitute (E) and (V), then convert F to (\mu\text{F}).

Step-by-Step Reasoning

Start with

E = \frac{1}{2}CV^2

Rearrange:

C = \frac{2E}{V^2}

Substitute (E = 0.041\ \text{J}) and (V = 12\ \text{V}):

C = \frac{2(0.041)}{12^2} = \frac{0.082}{144} = 5.694\times 10^{-4}\ \text{F}

Convert to microfarads using (1\ \mu\text{F} = 10^{-6}\ \text{F}):

5.694 \times 10^{-4}\ \text{F} = 5.694 \times 10^{2}\ \mu\text{F} \approx 570\ \mu\text{F}.

Key Takeaways

Read the relevant voltage from the graph (maximum capacitor p.d.).
Use (E = \tfrac{1}{2}CV^2) and rearrange carefully.
Be fluent converting between F and (\mu\text{F}).

Common Mistakes

Using (V = 24\ \text{V}) because the input goes from (+12) to (-12): the capacitor voltage is not 24 V; it is the p.d. across it at maximum, which is 12 V.
Forgetting the factor (\tfrac{1}{2}).
Converting to (\mu\text{F}) incorrectly (wrong power of ten).

Things to Be Careful About

The capacitor is across the output, not directly across the input source, so use (V_{OUT}) maximum.
Quote the final answer to an appropriate number of significant figures (here (570\ \mu\text{F}) matches the target value).

Techniques used

apply the capacitor energy equationrearrange to solve for capacitanceconvert between farads and microfarads

(ii)

Determine the resistance of R.

resistance = ______ $\Omega$

DifficultyMedium

Worked solution

Working

During the diode-off interval, the capacitor discharges through (R) from (12\ \text{V}) to (8\ \text{V}) in (0.010\ \text{s}).

V = V_0 e^{-t/RC}

8 = 12 e^{-0.010/RC}

e^{-0.010/RC} = \frac{2}{3} \quad\Rightarrow\quad RC = \frac{0.010}{\ln(3/2)} = 2.47 \times 10^{-2}\ \text{s}

With (C = 5.7 \times 10^{-4}\ \text{F}),

R = \frac{RC}{C} = \frac{2.47 \times 10^{-2}}{5.7 \times 10^{-4}} = 4.3 \times 10^{1}\ \Omega

Answer

(R \approx 43\ \Omega).

Final answer

43 \Omega

Detailed explanation

Background Concept

When a capacitor discharges through a resistor, the voltage across the capacitor (and the resistor, if they are in parallel) falls exponentially:

V = V_0 e^{-t/RC}

where:

(V_0) is the initial voltage at the start of the discharge
(V) is the voltage after time (t)
(R) is the resistance, (C) is the capacitance
(RC) is the time constant (\tau)

In a rectifier with smoothing capacitor, the capacitor charges when the diode conducts and discharges through the load when the diode is reverse-biased, producing the “droop” (ripple) between peaks.

Understanding the Question

From Fig. 6.2, the input is +12 V for 0.01 s and then −12 V for 0.01 s (period 0.02 s). With a half-wave rectifier, the diode conducts during the +12 V part, charging the capacitor to 12 V.

From Fig. 6.3, during the next 0.01 s (when the diode is off), the output (and capacitor voltage) falls from 12 V to 8 V.

You are asked to find (R). You already found (C \approx 570\ \mu\text{F}) in part (c)(i).

Approach

Identify the discharge interval: (t = 0.010\ \text{s}).
Use the exponential discharge equation with (V_0 = 12\ \text{V}) and (V = 8\ \text{V}) to find (RC).
Divide by (C) to find (R).

Step-by-Step Reasoning

Use the discharge equation:

V = V_0 e^{-t/RC}

Substitute the values from the graph:

Initial at start of discharge: (V_0 = 12\ \text{V})
After (t = 0.010\ \text{s}): (V = 8\ \text{V})

So:

8 = 12 e^{-0.010/RC}

Rearrange to isolate the exponential:

e^{-0.010/RC} = \frac{8}{12} = \frac{2}{3}

Take natural logs:

-\frac{0.010}{RC} = \ln\left(\frac{2}{3}\right) = -\ln\left(\frac{3}{2}\right)

So:

RC = \frac{0.010}{\ln(3/2)} \approx 2.47 \times 10^{-2}\ \text{s}

Now use (C = 570\ \mu\text{F} = 5.70 \times 10^{-4}\ \text{F}):

R = \frac{RC}{C} = \frac{2.47 \times 10^{-2}}{5.70 \times 10^{-4}} \approx 4.33 \times 10^{1}\ \Omega \approx 43\ \Omega

Note: the plotted fall looks linear, but the physical discharge of a capacitor through a resistor is exponential; over a small time interval and modest drop it can look approximately straight on a simple sketch.

Key Takeaways

In smoothing circuits, the “droop” happens when the diode is off and the capacitor discharges through (R).
Use (V = V_0 e^{-t/RC}) to link the voltage ratio to the time constant.
Once (RC) is found, (R) follows from (R = (RC)/C).

Common Mistakes

Using (t = 0.020\ \text{s}) instead of (0.010\ \text{s}): discharge happens only during the half-cycle when the diode does not conduct.
Using a linear discharge formula (there isn’t one for an (RC) circuit with a resistor load).
Forgetting to convert (\mu\text{F}) to F before calculating (R).

Things to Be Careful About

Take (V_0) and (V) from the correct points on the (V_{OUT}) graph.
Use natural log (\ln), not (\lg).
Keep units consistent: (t) in seconds, (C) in farads, giving (R) in ohms.

Techniques used

use the exponential discharge equation for a capacitorextract a time interval from a waveform graphsolve for the time constant and then the resistance

Q7MediumMagnetic Fields

(a)

Define magnetic flux density.

DifficultyEasy

Worked solution

Answer

Magnetic flux density $B$ is the force per unit length per unit current on a straight conductor placed perpendicular to the field:

B = \frac{F}{IL} \quad (\text{for } \theta = 90^\circ)

Final answer

Magnetic flux density is the force per unit length per unit current on a straight conductor at right angles to the field (B = F/IL for θ = 90°).

Detailed explanation

Background Concept

Magnetic flux density $B$ measures how strong a magnetic field is. Its practical (operational) definition comes from the force a magnetic field exerts on a current-carrying conductor:

F = BIL\sin\theta

where $F$ is the magnetic force on a straight conductor of length $L$ carrying current $I$ , and $\theta$ is the angle between the conductor and the magnetic field.

Understanding the Question

You are asked to define magnetic flux density. For full credit you must give the standard A-Level definition linked to the force on a current-carrying conductor, and include the condition that the conductor is perpendicular to the field.

Approach

Start from $F = BIL\sin\theta$ . Rearrange to make $B$ the subject, and state the definition for the special case $\theta = 90^\circ$ so that $\sin\theta = 1$ .

Step-by-Step Reasoning

From

F = BIL\sin\theta

When the wire is perpendicular to the field, $\theta = 90^\circ$ and $\sin\theta = 1$ , so

F = BIL

Rearranging gives

B = \frac{F}{IL}

So $B$ is the force per unit current per unit length on a straight wire at right angles to the field.

Key Takeaways

The exam definition of $B$ is tied to the force on a current-carrying conductor.
Always include the condition perpendicular to the field.

Common Mistakes

Omitting “per unit length” or “per unit current”.
Forgetting to state the wire is at $90^\circ$ to the field (or implying a general angle).
Giving a different formula (e.g. for a moving charge) instead of the conductor definition.

Things to Be Careful About

The definition is for a uniform field and a straight conductor.
Use clear wording: “force per unit length per unit current” is what the mark scheme typically looks for.

Techniques used

state the operational definition of magnetic flux density using force on a current-carrying conductorspecify the condition of perpendicular orientation for the definition

(b)

A long, straight wire carries a current into the page, as shown in Fig. 7.1.

Fig. 7.1

On Fig. 7.1, draw four field lines to represent the magnetic field around the wire due to the current in it.

(c)

Two identical wires X and Y are placed parallel to each other. The wires both carry current into the page, as shown in Fig. 7.2.

Fig. 7.2

6M

(i)

Explain why the two wires exert a magnetic force on each other.

DifficultyMedium-Easy

Worked solution

Answer

Current in X produces a magnetic field at Y. Since Y carries a current, it experiences a force in this field ( $F = BIL$ ). Similarly, current in Y produces a field at X so X also experiences a force.

Final answer

Each current produces a magnetic field that acts on the other current-carrying wire, so each wire experiences a force (F = BIL).

Detailed explanation

Background Concept

There are two key facts:

A current in a wire produces a magnetic field around it.
A wire carrying current in an external magnetic field experiences a force:

F = BIL\sin\theta

For long parallel wires, the magnetic field from one wire at the position of the other is perpendicular to the other wire, so $\theta = 90^\circ$ and $\sin\theta = 1$ .

Understanding the Question

Two parallel wires X and Y both carry current into the page. The question asks why they exert forces on each other, not the direction yet. You must mention both “field produced” and “force on current in a field”.

Approach

Explain the interaction in two steps:

Wire X creates a magnetic field where wire Y is.
Wire Y (carrying current) experiences a force due to that magnetic field.
Then state the same argument the other way round (Y’s field acts on X).

Step-by-Step Reasoning

Because X carries current, it produces a magnetic field in the space around it.
At the location of wire Y, there is therefore a magnetic flux density $B$ due to wire X.
Wire Y carries current, so in magnetic field $B$ it experiences a force given by $F = BIL$ (since the wire is perpendicular to the field direction there).
Similarly, Y produces a magnetic field at X, so X experiences a force as well.

Key Takeaways

Mutual forces between parallel currents come from: current creates field + field exerts force on current.

Common Mistakes

Saying only “they attract” without explaining the mechanism.
Referring to electric forces instead of magnetic forces.
Forgetting that the field from one wire exists at the position of the other wire.

Things to Be Careful About

Use “magnetic field” and “current-carrying conductor experiences a force” language explicitly; that is what the marks usually target.

Techniques used

state that a current produces a magnetic field around a conductorapply the force on a current-carrying conductor in an external magnetic fieldlink the two interactions to explain mutual forces between parallel wires

(ii)

On Fig. 7.2, draw an arrow to show the direction of the magnetic force exerted on wire X. Label your arrow F.

(iii)

The current in X is double the current in Y.

State how the magnetic force exerted on wire Y compares with the magnetic force exerted on wire X.

DifficultyMedium

Worked solution

Answer

The forces are equal in magnitude (and opposite in direction).

Since $B_X \propto I_X$ and $B_Y \propto I_Y$ ,

F_X = B_Y I_X L \propto I_Y I_X

F_Y = B_X I_Y L \propto I_X I_Y

so $F_X = F_Y$ .

Final answer

Equal magnitude (opposite directions).

Detailed explanation

Background Concept

For long parallel wires, the magnetic field due to one wire at the position of the other is proportional to the current in the source wire:

B \propto I

The force on a length $L$ of the other wire carrying current $I$ is

F = BIL

So the mutual force depends on the product of the two currents.

Understanding the Question

You are told $I_X$ is double $I_Y$ . It is tempting to think “bigger current means bigger force”, but each force depends on both currents: the field is produced by one current and the force acts on the other current.

Approach

Write expressions for:

Force on X due to the magnetic field created by Y.
Force on Y due to the magnetic field created by X.
Use proportionality to compare the magnitudes.

Step-by-Step Reasoning

Let $B_Y$ be the magnetic flux density at X due to current in wire Y. Then

B_Y \propto I_Y

Force on X is

F_X = B_Y I_X L \propto I_Y I_X

Similarly, let $B_X$ be the magnetic flux density at Y due to current in wire X. Then

B_X \propto I_X

Force on Y is

F_Y = B_X I_Y L \propto I_X I_Y

These are the same product, so the magnitudes are equal:

F_X = F_Y

The directions are opposite because the forces act towards each other (action–reaction pair along the line joining the wires).

Key Takeaways

For parallel wires, $F \propto I_X I_Y$ .
Even if the currents differ, the mutual forces are equal in magnitude and opposite in direction.

Common Mistakes

Saying the wire with larger current experiences a larger force.
Forgetting to state “opposite direction” when comparing forces.

Things to Be Careful About

The equal-and-opposite conclusion is consistent with Newton’s third law, but you should still show (or state) that each force depends on both currents, not just one.

Techniques used

use proportionality of magnetic field around a wire with currentapply F = BIL to each wire in the other wire’s fieldapply Newton's third law to compare mutual forces

(iv)

The direction of the current in both wires is now reversed.

State, with a reason, the effect of this change on the direction of the force on wire X.

Q8MediumQuantum Physics

A polished sheet of magnesium in a vacuum emits electrons when it is illuminated by ultraviolet radiation.

(a)

State the name of this phenomenon.

(b)

For emission of electrons to occur, the frequency of the ultraviolet radiation must be at least $8.8 \times 10^{14}\ \text{Hz}$ .

5M

(i)

Calculate the work function energy of magnesium.

work function energy = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

At threshold, $\phi = hf_0$ .

\phi = (6.63 \times 10^{-34})(8.8 \times 10^{14}) = 5.83 \times 10^{-19}\ \text{J}

Answer

$\phi \approx 5.8 \times 10^{-19}\ \text{J}$

Final answer

5.8 × 10^-19 J

Detailed explanation

Background Concept

For the photoelectric effect,

hf = \phi + E_{\text{max}}

$hf$ is the photon energy.
$\phi$ is the work function (minimum energy needed to liberate an electron from the metal surface).
$E_{\text{max}}$ is the maximum kinetic energy of emitted electrons.

At the threshold frequency $f_0$ , electrons are just emitted with zero maximum kinetic energy, so $E_{\text{max}}=0$ and hence

\phi = hf_0.

Understanding the Question

You are told that for emission to occur, the frequency must be at least $8.8 \times 10^{14}\ \text{Hz}$ . That value is the threshold frequency $f_0$ . The task is to calculate the work function energy of magnesium in joules.

Approach

Use the threshold condition $\phi = hf_0$ and substitute the given $f_0$ with Planck’s constant $h$ .

Step-by-Step Reasoning

Identify threshold frequency: $f_0 = 8.8 \times 10^{14}\ \text{Hz}.$
Apply threshold relation: $\phi = hf_0.$
Substitute $h = 6.63 \times 10^{-34}\ \text{J s}$ : $\phi = (6.63 \times 10^{-34})(8.8 \times 10^{14})$ Combine powers of ten: $10^{-34} \times 10^{14} = 10^{-20}$ , and multiply the numbers: $6.63 \times 8.8 = 58.3$ so $\phi = 58.3 \times 10^{-20} = 5.83 \times 10^{-19}\ \text{J}.$

Key Takeaways

At threshold: $E_{\text{max}}=0$ so $\phi=hf_0$ .
Always include the unit (joules here).

Common Mistakes

Using $f=8.8\times 10^{14}$ as if it were the incident frequency in $hf=\phi+E_{\text{max}}$ with a non-zero $E_{\text{max}}$ .
Forgetting that $h$ is in $\text{J s}$ so the product $hf$ is in joules.

Things to Be Careful About

Use the given minimum frequency as $f_0$ (threshold), not as just “a frequency”.
Significant figures: quoting $\phi$ to 2–3 s.f. is appropriate given $f_0$ is 2 s.f.

Techniques used

use the threshold condition for the photoelectric effectapply the photon energy equationsubstitute values and express the result in standard form

(ii)

For ultraviolet radiation with a frequency of $11 \times 10^{14}\ \text{Hz}$ , calculate the maximum speed of the emitted electrons.

maximum speed = ______ $\text{m s}^{-1}$

DifficultyMedium

Worked solution

Working

E_{\text{max}} = hf - \phi

E_{\text{max}} = (6.63 \times 10^{-34})(11 \times 10^{14}) - (5.83 \times 10^{-19}) = 1.46 \times 10^{-19}\ \text{J}

E_{\text{max}} = \frac{1}{2}m_e v_{\text{max}}^2 \Rightarrow v_{\text{max}} = \sqrt{\frac{2E_{\text{max}}}{m_e}}

v_{\text{max}} = \sqrt{\frac{2(1.46 \times 10^{-19})}{9.11 \times 10^{-31}}} = 5.66 \times 10^{5}\ \text{m s}^{-1}

Answer

$v_{\text{max}} \approx 5.7 \times 10^{5}\ \text{m s}^{-1}$

Final answer

5.7 × 10^5 m s^-1

Detailed explanation

Background Concept

Einstein’s photoelectric equation links the photon energy to the maximum kinetic energy of the emitted electrons:

hf = \phi + E_{\text{max}}

Rearranging,

E_{\text{max}} = hf - \phi.

The maximum kinetic energy is also related to the maximum electron speed:

E_{\text{max}} = \frac{1}{2}m_e v_{\text{max}}^2.

Understanding the Question

The incident ultraviolet frequency is now $f = 11 \times 10^{14}\ \text{Hz}$ . Using the work function found in (i), you must calculate the maximum speed of emitted electrons.

Approach

Find photon energy $hf$ .
Subtract the work function $\phi$ to get $E_{\text{max}}$ .
Convert kinetic energy into speed using $E_k = \tfrac{1}{2}mv^2$ .

Step-by-Step Reasoning

Photon energy:

hf = (6.63 \times 10^{-34})(11 \times 10^{14}) = 7.29 \times 10^{-19}\ \text{J}.

Use $\phi = 5.83 \times 10^{-19}\ \text{J}$ from part (i):

E_{\text{max}} = 7.29 \times 10^{-19} - 5.83 \times 10^{-19} = 1.46 \times 10^{-19}\ \text{J}.

Convert to speed using electron mass $m_e = 9.11 \times 10^{-31}\ \text{kg}$ :

v_{\text{max}} = \sqrt{\frac{2E_{\text{max}}}{m_e}} = \sqrt{\frac{2(1.46 \times 10^{-19})}{9.11 \times 10^{-31}}}.

Calculate inside the square root:

\frac{2(1.46 \times 10^{-19})}{9.11 \times 10^{-31}} \approx 3.20 \times 10^{11}

so

v_{\text{max}} \approx \sqrt{3.20 \times 10^{11}} \approx 5.66 \times 10^{5}\ \text{m s}^{-1}.

Key Takeaways

Higher frequency means larger $hf$ , so (above threshold) electrons can leave with non-zero kinetic energy.
To get a speed from energy, use $E_k = \tfrac{1}{2}mv^2$ .

Common Mistakes

Using $E_{\text{max}} = hf$ (forgetting to subtract $\phi$ ).
Using the proton mass instead of the electron mass.
Forgetting the square root when solving for $v$ .

Things to Be Careful About

Keep track of powers of ten carefully when subtracting energies in standard form.
$E_{\text{max}}$ must be positive; if $f < f_0$ then emission does not occur (but here $11 \times 10^{14} > 8.8 \times 10^{14}$ so it does).

Techniques used

apply the photoelectric equationcalculate maximum kinetic energy from photon energy and work functionuse kinetic energy to determine particle speed

(c)

The frequency $f$ of the ultraviolet radiation incident on the magnesium sheet is varied between $8.0 \times 10^{14}\ \text{Hz}$ and $11 \times 10^{14}\ \text{Hz}$ .

On Fig. 8.1, sketch the variation with $f$ of the maximum kinetic energy $E_{MAX}$ of the emitted electrons. Use the space below for any working that you need.

Fig. 8.1

DifficultyMedium

Worked solution

Working

E_{\text{max}} = hf - \phi = h(f - f_0)

so $E_{\text{max}}=0$ for $f \le f_0$ with $f_0 = 8.8 \times 10^{14}\ \text{Hz}$ .

At $f = 11 \times 10^{14}\ \text{Hz}$ ,

E_{\text{max}} = 1.46 \times 10^{-19}\ \text{J}

Answer

graph-8c

Final answer

See sketch

Detailed explanation

Background Concept

The photoelectric equation can be written as

E_{\text{max}} = hf - \phi.

Since $\phi = hf_0$ , this becomes

E_{\text{max}} = h(f - f_0).

This tells you:

For $f < f_0$ , $E_{\text{max}}$ would be negative, which is unphysical, so no electrons are emitted (so effectively $E_{\text{max}}=0$ and emission does not occur).
For $f \ge f_0$ , $E_{\text{max}}$ increases linearly with $f$ .
The gradient of the straight line is $h$ .
The x-intercept is $f_0$ .

Understanding the Question

You vary the incident frequency between $8.0 \times 10^{14}\ \text{Hz}$ and $11 \times 10^{14}\ \text{Hz}$ , and must sketch how the maximum kinetic energy $E_{\text{max}}$ depends on $f$ .

The axes on Fig. 8.1 are scaled as:

x-axis: $f / 10^{14}\ \text{Hz}$ from 8.0 to 11.0
y-axis: $E_{\text{max}} / 10^{-19}\ \text{J}$ from 0 to 2.0

Threshold frequency is $f_0 = 8.8 \times 10^{14}\ \text{Hz}$ (so at $f/10^{14} = 8.8$ , $E_{\text{max}}=0$ ).

Approach

Mark the threshold point $(f_0, 0)$ .
Below $f_0$ , draw the graph along the x-axis (no emission / zero maximum KE shown).
Above $f_0$ , draw a straight line with gradient $h$ .
Use a second point to set the line position, e.g. at $f = 11 \times 10^{14}\ \text{Hz}$ where from part (b)(ii) we have $E_{\text{max}} = 1.46 \times 10^{-19}\ \text{J}$ .

Step-by-Step Reasoning

Threshold frequency:

f_0 = 8.8 \times 10^{14}\ \text{Hz} \Rightarrow \left(\frac{f}{10^{14}\ \text{Hz}}\right) = 8.8.

So the straight-line part must meet the x-axis at $x=8.8$ .

For $f$ from $8.0$ to $8.8 \times 10^{14}\ \text{Hz}$ , there is no emission, so $E_{\text{max}}$ is shown as $0$ (line along the x-axis).
For $f > 8.8 \times 10^{14}\ \text{Hz}$ , use

E_{\text{max}} = h(f - f_0)

which is a straight line.

Use the point at $f = 11 \times 10^{14}\ \text{Hz}$ :

E_{\text{max}} = 1.46 \times 10^{-19}\ \text{J}

So on the given y-scale, this is $y = 1.46$ at $x=11.0$ .

Draw a straight line from $(8.8, 0)$ to $(11.0, 1.46)$ .

graph-8c

Key Takeaways

The $E_{\text{max}}$ vs $f$ graph is linear above threshold with gradient $h$ .
The x-intercept gives the threshold frequency $f_0$ .
Below $f_0$ , no electrons are emitted (graph remains on the axis).

Common Mistakes

Drawing the straight line through the origin instead of crossing the x-axis at $f_0$ .
Extending the straight line into the region $f < f_0$ with negative $E_{\text{max}}$ .
Plotting a curve instead of a straight line.

Things to Be Careful About

Use the axes scales correctly: $f$ is labelled in units of $10^{14}\ \text{Hz}$ and $E_{\text{max}}$ in units of $10^{-19}\ \text{J}$ .
The line should pass through the correct intercept at $8.8$ and be consistent with the calculated value at $11.0$ (about $1.46$ on the y-axis).

Techniques used

use the photoelectric equation to relate kinetic energy to frequencyinterpret threshold frequency as the x-interceptsketch a piecewise linear graph with correct gradient and intercept

Q9MediumNuclear PhysicsMedical Physics

Fluorine-18 ( $^{18}_{9}\text{F}$ ) decays by beta-plus ( $\beta^+$ ) emission with a half-life of 110 minutes.

(a)

5M

(i)

State the name of the beta-plus particle.

(ii)

Show that the decay constant of fluorine-18 is $1.05 \times 10^{-4}\ \text{s}^{-1}$ .

DifficultyMedium-Easy

Worked solution

Working

T_{1/2} = 110\ \text{min} = 110 \times 60 = 6600\ \text{s}

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{6600} = 1.05 \times 10^{-4}\ \text{s}^{-1}

Answer

(\lambda = 1.05 \times 10^{-4}\ \text{s}^{-1})

Final answer

1.05 \times 10^{-4} s^{-1}

Detailed explanation

Background Concept

Radioactive decay is exponential. The number of undecayed nuclei (N) decreases with time according to

N = N_0 e^{-\lambda t}

where (\lambda) is the decay constant (probability per unit time that a nucleus decays).

Half-life (T_{1/2}) is the time taken for (N) (or activity (A)) to fall to half its initial value. The link between half-life and decay constant is

\lambda = \frac{\ln 2}{T_{1/2}}.

Understanding the Question

You are given (T_{1/2} = 110\ \text{min}) and asked to show (\lambda) is (1.05 \times 10^{-4}\ \text{s}^{-1}). The key detail is that (\lambda) is in (\text{s}^{-1}), so the half-life must be converted to seconds.

Approach

Convert minutes to seconds.
Substitute into (\lambda = \ln 2 / T_{1/2}).
Write the result in standard form with unit (\text{s}^{-1}).

Step-by-Step Reasoning

Convert the half-life:

T_{1/2} = 110\ \text{min} = 110 \times 60 = 6600\ \text{s}.

Now use the half-life relation:

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{6600}.

Calculate:

\lambda = 1.05 \times 10^{-4}\ \text{s}^{-1}.

This matches the required value.

Key Takeaways

Always convert (T_{1/2}) into seconds if (\lambda) is required in (\text{s}^{-1}).
Use (\lambda = \ln 2 / T_{1/2}) for quick half-life (\leftrightarrow) decay constant conversions.

Common Mistakes

Forgetting to convert minutes to seconds, giving (\lambda) off by a factor of 60.
Using (\log_{10}) instead of (\ln) (must be natural log).

Things to Be Careful About

Quote (\lambda) with the unit (\text{s}^{-1}).
Keep enough significant figures (here 3 s.f. matches the given shown value).

Techniques used

convert half-life into SI unitsuse $\lambda = \ln 2 / T_{1/2}$evaluate an expression in standard form

(iii)

Determine the activity of $2.1 \times 10^{-12}\ \text{kg}$ of fluorine-18.

activity = ______ $\text{Bq}$

DifficultyMedium

Worked solution

Working

M = 18\ \text{g mol}^{-1} = 0.018\ \text{kg mol}^{-1}

n = \frac{m}{M} = \frac{2.1 \times 10^{-12}}{0.018} = 1.17 \times 10^{-10}\ \text{mol}

N = n N_A = (1.17 \times 10^{-10})(6.02 \times 10^{23}) = 7.02 \times 10^{13}

A = \lambda N = (1.05 \times 10^{-4})(7.02 \times 10^{13}) = 7.37 \times 10^{9}\ \text{Bq}

Answer

(A \approx 7.4 \times 10^{9}\ \text{Bq})

Final answer

7.4 \times 10^9 Bq

Detailed explanation

Background Concept

Activity (A) is the number of decays per second:

A = \lambda N

where:

(\lambda) is the decay constant in (\text{s}^{-1})
(N) is the number of undecayed nuclei present.

To find (N) from a mass (m) of an isotope, use moles:

n = \frac{m}{M}, \quad N = n N_A

where (M) is the molar mass and (N_A) is the Avogadro constant.

Understanding the Question

You are given a mass of fluorine-18, (2.1 \times 10^{-12}\ \text{kg}), and you have already found (\lambda = 1.05 \times 10^{-4}\ \text{s}^{-1}). You must calculate the activity in Bq.

So the task is: convert the tiny mass into number of nuclei (N), then multiply by (\lambda).

Approach

Use (M = 18\ \text{g mol}^{-1}) for fluorine-18 (numerically equal to the nucleon number).
Convert (M) to (\text{kg mol}^{-1}) to match the given mass.
Find moles (n = m/M).
Find number of nuclei (N = n N_A).
Find activity (A = \lambda N).

Step-by-Step Reasoning

1) Molar mass
Fluorine-18 has molar mass approximately (18\ \text{g mol}^{-1}):

M = 18\ \text{g mol}^{-1} = 0.018\ \text{kg mol}^{-1}.

2) Convert mass to moles

n = \frac{m}{M} = \frac{2.1 \times 10^{-12}\ \text{kg}}{0.018\ \text{kg mol}^{-1}} = 1.17 \times 10^{-10}\ \text{mol}.

3) Convert moles to number of nuclei

N = n N_A = (1.17 \times 10^{-10})(6.02 \times 10^{23}) = 7.02 \times 10^{13}.

(Units check: mol (\times) nuclei per mol (\rightarrow) nuclei.)

4) Calculate activity

A = \lambda N = (1.05 \times 10^{-4})(7.02 \times 10^{13}) = 7.37 \times 10^{9}\ \text{s}^{-1}.

Since (1\ \text{Bq} = 1\ \text{s}^{-1}),

A \approx 7.4 \times 10^{9}\ \text{Bq}.

Key Takeaways

Activity depends on both how many nuclei you have and how quickly each nucleus tends to decay.
Convert mass (\rightarrow) moles (\rightarrow) number of nuclei (\rightarrow) activity.

Common Mistakes

Using (M = 18\ \text{kg mol}^{-1}) instead of (0.018\ \text{kg mol}^{-1}) (factor of 1000 error).
Forgetting that (\text{Bq} = \text{s}^{-1}).
Using the proton number (9) as the molar mass instead of 18.

Things to Be Careful About

Keep powers of ten under control; it’s easy to slip by (10^{\pm 1}) in standard form.
Use a sensible number of significant figures (here 2 s.f. is reasonable because the mass is 2 s.f.).

Techniques used

convert mass to amount of substance using molar massuse Avogadro's constant to find number of nucleiapply $A = \lambda N$

(b)

A small sample of fluorine-18 injected into the body acts as a tracer for use in medical imaging.

5M

(i)

Describe how the interaction of a $\beta^+$ particle with an electron in the body enables the formation of an image.

DifficultyMedium

Worked solution

Answer

The emitted (\beta^+) particle (positron) annihilates with an electron, producing two (\gamma)-ray photons.

The two photons are emitted in opposite directions.

A ring of detectors detects the two (\gamma) photons simultaneously (coincidence), locating the line along which the annihilation occurred; many events build up an image of tracer distribution.

Final answer

Positron annihilates with electron producing two opposite (\gamma) photons; coincidence detection locates events and many detections form the image.

Detailed explanation

Background Concept

In positron emission tomography (PET), a radioactive tracer emits positrons ((\beta^+)). A positron is the antiparticle of the electron. When a positron meets an electron, they annihilate:

e^{+} + e^{-} \rightarrow \gamma + \gamma

The annihilation converts the particles’ mass (and any kinetic energy) into two (\gamma)-ray photons. Because momentum must be conserved and the initial momentum is usually small, the two photons travel in nearly opposite directions.

PET scanners use rings of (\gamma) detectors. When two detectors opposite each other detect photons at the same time (a coincidence event), the system assumes the annihilation occurred somewhere along the straight line joining those detectors (line of response). By collecting many such lines, the tracer concentration can be reconstructed into an image.

Understanding the Question

The question asks how interaction of the (\beta^+) particle with an electron in the body allows an image to be formed. So you must describe:

annihilation, 2) two (\gamma) photons emitted oppositely, 3) detection and reconstruction into an image.

Approach

Explain the physical chain: (\beta^+) emission (\rightarrow) positron travels a short distance (\rightarrow) annihilation with electron (\rightarrow) two (\gamma) photons (\rightarrow) coincidence detection (\rightarrow) image of tracer distribution.

Step-by-Step Reasoning

Fluorine-18 decays by (\beta^+) emission, producing a positron.
The positron quickly loses energy in tissue and then meets an electron.
Positron and electron annihilate, producing two (\gamma) photons.
The photons are emitted in opposite directions (approximately (180^{\circ}) apart) due to momentum conservation.
A PET scanner detects these two photons in coincidence using detectors around the body.
Each coincidence detection defines a line through the body where the annihilation happened; combining a very large number of such events reconstructs a map (image) of where the tracer is concentrated.

diagram-9b1

Key Takeaways

PET imaging relies on positron-electron annihilation producing two back-to-back (\gamma) photons.
Coincidence detection provides positional information; many events produce the final image.

Common Mistakes

Saying the (\beta^+) particle is detected directly outside the body (positrons are quickly absorbed and do not escape).
Mentioning only “gamma rays are produced” without stating that two photons are emitted in opposite directions.
Not linking detection to how position information is obtained (coincidence/line of response).

Things to Be Careful About

Use the word “annihilation” (specific process) rather than just “collision”.
State the key geometrical idea: opposite directions + coincidence gives a line in the body.
Keep the explanation focused on image formation, not just decay.

Techniques used

describe particle-antiparticle annihilationuse conservation of momentum to state back-to-back photon emissionrelate coincidence detection to image reconstruction

(ii)

Suggest why 110 minutes is a suitable half-life for a nuclide used as a tracer in medical diagnosis.

DifficultyMedium-Easy

Worked solution

Answer

Long enough for the tracer to be injected and to reach the required organs and for imaging to be carried out.

Short enough that the activity falls quickly afterwards, so the patient receives a smaller radiation dose / is not radioactive for long.

Final answer

Long enough for imaging; short enough to minimise dose and time radioactive.

Detailed explanation

Background Concept

For a medical tracer, the half-life must be chosen to satisfy two competing requirements:

Measurement requirement: activity must remain high enough during the scan to detect sufficient (\gamma) photons and form a clear image.
Safety requirement: the nuclide should not remain in the patient for too long; shorter half-life reduces total radiation dose.

The half-life sets the timescale over which activity decreases: after one half-life, activity halves; after two, it is one quarter, etc.

Understanding the Question

You are asked why (110\ \text{min}) is suitable for fluorine-18 as a diagnostic tracer. You need to suggest reasons connected to practical scanning time and patient dose.

Approach

Give two distinct points:

why it is not too short (enough time to administer and image),
why it is not too long (decays away reasonably quickly, reducing dose).

Step-by-Step Reasoning

Long enough: 110 minutes gives time for preparation, injection, transport in the blood, uptake in the target tissue, and the PET scan itself before activity drops too much.
Short enough: after a few hours, several half-lives have passed so the activity is much lower; this reduces the total radiation dose and the time the patient remains significantly radioactive.

Key Takeaways

Suitable tracer half-life is a compromise between detectable count rate and minimising radiation exposure.

Common Mistakes

Saying only “short half-life is safer” without also stating it must be long enough to measure.
Claiming “long half-life gives better images” (it can, but increases patient dose; you must balance both).

Things to Be Careful About

Make the two points clearly different (practical imaging time vs radiation dose/time in body).
Avoid vague statements like “it is convenient”; explain what it is convenient for.

Techniques used

relate half-life to detectable activity over timebalance image acquisition needs against radiation dosestate practical constraints on tracer use

Q10Medium-EasyAstronomy and Cosmology

(a)

Explain how redshift leads to the idea that the Universe is expanding.

DifficultyMedium-Easy

Worked solution

Answer

Redshift means spectral lines are observed at longer wavelength than when emitted.
This is a Doppler effect, so the source galaxy is moving away from the Earth.
Since most distant galaxies show redshift (and larger redshift for greater distance), galaxies are receding from each other, so the Universe is expanding.

Final answer

Redshift (longer observed wavelength) implies galaxies are receding (Doppler effect); widespread/greater redshift for distant galaxies implies galaxies are moving apart, so the Universe is expanding.

Detailed explanation

Background Concept

Light from a source has characteristic spectral lines at well-defined wavelengths. If the source and observer move relative to each other along the line of sight, the observed wavelength changes due to the Doppler effect.

Define redshift parameter

z = \frac{\Delta \lambda}{\lambda_0} = \frac{\lambda_{\text{obs}}-\lambda_{\text{emit}}}{\lambda_{\text{emit}}}

For light, if $\lambda_{\text{obs}} > \lambda_{\text{emit}}$ , then $z>0$ and the source is receding.

Understanding the Question

You are asked to explain (not calculate) how the observation that spectral lines from galaxies are shifted towards the red (longer wavelengths) leads to the idea that the Universe is expanding.

Approach

State what redshift means in terms of wavelength.
Interpret it using Doppler effect: redshift corresponds to recession.
Generalise to many galaxies: most are redshifted, and redshift increases with distance, so galaxies are moving away from one another, implying expansion.

Step-by-Step Reasoning

When a known spectral line is found at a longer wavelength than its laboratory/emitted value, the spectrum is said to be redshifted.
The Doppler effect for waves means a receding source produces longer observed wavelength (and lower observed frequency).
Therefore, redshift is evidence that the galaxy is moving away from Earth.
Observations show this is true for very many galaxies in all directions, and the more distant a galaxy is, the bigger its redshift (hence bigger recession speed).
If objects in every direction are receding and the recession speed increases with distance, the natural interpretation is that the scale of the Universe is increasing: space is expanding and galaxies are being carried apart.

Key Takeaways

Redshift $\Rightarrow$ longer wavelength than emitted.
Longer wavelength for light is interpreted as Doppler recession.
Widespread recession (especially increasing with distance) supports an expanding Universe.

Common Mistakes

Saying redshift is due to absorption or scattering rather than Doppler motion.
Forgetting to mention that redshift corresponds to the galaxy moving away.
Claiming “Earth is at the centre” rather than recognising expansion occurs uniformly (redshift seen in all directions).

Things to Be Careful About

The argument is qualitative here: you do not need formulas, but you must explicitly connect redshift to recession and then to expansion.
Avoid mixing up frequency shift and wavelength shift: for redshift, wavelength increases and frequency decreases.

Techniques used

interpret a spectral line shift as a Doppler effectlink recessional motion of galaxies to expansion of spaceuse the trend of redshift with distance to justify expansion

(b)

Stars in a distant galaxy emit radiation. The total luminosity of the stars in the galaxy is $1.90 \times 10^{36}\ \text{W}$ .

The emission spectrum of the radiation contains a line X at a wavelength of $658\ \text{nm}$ .

Radiation from the galaxy is observed on the Earth. The observed radiation has a radiant flux intensity of $8.42 \times 10^{-16}\ \text{W m}^{-2}$ . In the observed emission spectrum, line X is at a wavelength of $726\ \text{nm}$ .

Determine:

4M

(i)

the distance $d$ of the galaxy from the Earth

$d$ = ______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Working

Use

F = \frac{L}{4\pi d^2}

so

d = \sqrt{\frac{L}{4\pi F}}

d = \sqrt{\frac{1.90\times 10^{36}}{4\pi\left(8.42\times 10^{-16}\right)}} = 1.34\times 10^{25}\ \text{m}

Answer

$1.34\times 10^{25}\ \text{m}$

Final answer

1.34 × 10^25 m

Detailed explanation

Background Concept

Luminosity $L$ is the total power emitted by a source (in $\text{W}$ ). Radiant flux intensity $F$ (often called flux) is the power received per unit area at distance $d$ from the source (in $\text{W m}^{-2}$ ).

For isotropic emission (radiation spreading out equally in all directions), the power passes through an imaginary sphere of radius $d$ centred on the source. The surface area is $4\pi d^2$ , so the flux is

F = \frac{L}{4\pi d^2}.

Understanding the Question

You are given $L = 1.90\times 10^{36}\ \text{W}$ for the galaxy and the flux measured at Earth $F = 8.42\times 10^{-16}\ \text{W m}^{-2}$ . You must find the distance $d$ from Earth to the galaxy.

Approach

Use the inverse-square relation

F = \frac{L}{4\pi d^2}

and rearrange for $d$ :

d = \sqrt{\frac{L}{4\pi F}}.

Then substitute the given numbers and keep the final answer in standard form with unit $\text{m}$ .

Step-by-Step Reasoning

Start from

F = \frac{L}{4\pi d^2}.

Rearrange:

4\pi d^2 = \frac{L}{F} \quad \Rightarrow \quad d^2 = \frac{L}{4\pi F} \quad \Rightarrow \quad d = \sqrt{\frac{L}{4\pi F}}.

Substitute values:

d = \sqrt{\frac{1.90\times 10^{36}}{4\pi\left(8.42\times 10^{-16}\right)}}.

Calculate the denominator:

4\pi F \approx 12.57 \times 8.42\times 10^{-16} \approx 1.06\times 10^{-14}.

Then

\frac{L}{4\pi F} \approx \frac{1.90\times 10^{36}}{1.06\times 10^{-14}} \approx 1.80\times 10^{50}.

Finally,

d = \sqrt{1.80\times 10^{50}} \approx 1.34\times 10^{25}\ \text{m}.

Key Takeaways

Use the inverse-square law $F = L/(4\pi d^2)$ for radiation spreading out spherically.
Distance comes from a square root after rearranging.
Track units: $\text{W}/(\text{W m}^{-2}) = \text{m}^2$ , so the square root gives metres.

Common Mistakes

Forgetting the $4\pi$ factor (using $F=L/d^2$ ).
Rearranging incorrectly (e.g. $d = L/(4\pi F)$ without the square root).
Losing powers of ten when taking the square root (remember $\sqrt{10^{50}} = 10^{25}$ ).

Things to Be Careful About

The formula assumes the luminosity is emitted equally in all directions (isotropic).
Give the answer to appropriate significant figures (here 3 s.f. is consistent with the data).
Ensure you use $\text{W m}^{-2}$ for flux, not total power received.

Techniques used

apply the inverse-square law for flux from a luminous sourcerearrange an equation to solve for distanceuse standard form and square roots correctly

(ii)

the speed $v$ of the galaxy relative to the Earth.

$v$ = ______ $\text{m s}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

Redshift

z = \frac{\lambda_{\text{obs}}-\lambda_{\text{emit}}}{\lambda_{\text{emit}}} = \frac{726-658}{658} = 0.103

For small $z$ ,

v = zc = 0.103\left(3.00\times 10^8\right) = 3.10\times 10^7\ \text{m s}^{-1}

Answer

$3.10\times 10^7\ \text{m s}^{-1}$

Final answer

3.10 × 10^7 m s^-1

Detailed explanation

Background Concept

A shift in the wavelength of a spectral line indicates relative motion along the line of sight (Doppler effect). The redshift $z$ is defined as

z = \frac{\lambda_{\text{obs}}-\lambda_{\text{emit}}}{\lambda_{\text{emit}}}.

For galaxies where $z$ is not too large, A Level Physics typically uses the approximation

\frac{v}{c} \approx z \quad \Rightarrow \quad v \approx zc,

where $c = 3.00\times 10^8\ \text{m s}^{-1}$ .

Understanding the Question

Line X is emitted at $658\ \text{nm}$ but observed at $726\ \text{nm}$ . You need the galaxy’s speed relative to Earth (recession speed), so you should compute the redshift and then convert it into $v$ .

Approach

Calculate $z$ using the wavelength shift.
Use $v \approx zc$ (since $z\approx 0.10$ is small enough for this syllabus approximation).

Step-by-Step Reasoning

Compute the wavelength change:

\Delta \lambda = 726\ \text{nm} - 658\ \text{nm} = 68\ \text{nm}.

Now the redshift:

z = \frac{68}{658} = 0.103.

Convert to speed:

v = zc = 0.103\times 3.00\times 10^8 = 3.10\times 10^7\ \text{m s}^{-1}.

This is a recession speed because $\lambda_{\text{obs}} > \lambda_{\text{emit}}$ .

Key Takeaways

Redshift uses the fractional change in wavelength.
For small redshift, use $v \approx zc$ .
A longer observed wavelength means the source is moving away.

Common Mistakes

Using $z = \lambda_{\text{emit}}/\lambda_{\text{obs}}$ (wrong definition).
Using $\Delta\lambda/\lambda_{\text{obs}}$ instead of $\Delta\lambda/\lambda_{\text{emit}}$ .
Forgetting that $c$ is in $\text{m s}^{-1}$ (though the nm units cancel in $z$ ).

Things to Be Careful About

You do not need to convert nm to m because $z$ is a ratio, but be consistent.
If a question indicates large $z$ , a relativistic Doppler formula may be required; here the small- $z$ approximation is intended.

Techniques used

calculate redshift from observed and emitted wavelengthsuse the low-redshift approximation v = zchandle unit conversions and standard form

(c)

Observations of many galaxies, such as the one in (b), lead to many pairs of values of $d$ and $v$ . Plotting these values reveals a trend.

3M

(i)

On Fig. 10.1, sketch the variation of $v$ with $d$ .

Fig. 10.1

(ii)

State the name of the quantity represented by the gradient of the line in Fig. 10.1.

gradient = ______