9702/41 Oct/Nov 2024 — Physics Solved Past Paper

Q1MediumGravitational FieldsMotion in a Circle

(a)

State Newton’s law of gravitation.

DifficultyEasy

Worked solution

Answer

For two point (or spherical) masses $m_1$ and $m_2$ separated by distance $r$ , the gravitational force has magnitude

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

and acts along the line joining their centres, attracting each mass towards the other.

Final answer

F = G m1 m2 / r^2, attractive along the line joining the centres.

Detailed explanation

Background Concept

Newton's law of gravitation describes the gravitational force between two masses. For masses $m_1$ and $m_2$ with separation $r$ , the magnitude of the force is proportional to the product of the masses and inversely proportional to the square of the separation. The constant of proportionality is the gravitational constant $G$ .

Understanding the Question

You are asked to state the law, so you need the formula for the magnitude and a statement about the direction/nature of the force (that it is attractive and acts along the line joining the masses).

Approach

Write the inverse-square formula and then add the direction: along the line joining the centres, towards each other.

Step-by-Step Reasoning

The magnitude is

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

The force is attractive, so each mass experiences a force directed towards the other mass, along the line joining their centres.

Key Takeaways

Gravitational force follows an inverse-square dependence on separation.
The force acts along the line joining the centres and is always attractive.

Common Mistakes

Omitting the $r^2$ (writing $F \propto 1/r$ ).
Forgetting to state the direction (line of centres) or that the force is attractive.
Mixing up the symbols (e.g. writing $g$ instead of $G$ ).

Things to Be Careful About

$r$ is the separation between the centres of the masses.
The law is exact for point masses; for spherically symmetric masses it applies using centre-to-centre separation.

Techniques used

state Newton's inverse-square law of gravitationspecify the direction of the gravitational 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.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The satellite experiences a gravitational force towards the centre of the planet. This force provides the centripetal (resultant) force, so the satellite has centripetal acceleration towards the centre and hence follows a circular path.

Final answer

Gravity acts towards the centre and provides the centripetal force, giving centripetal acceleration so the orbit is circular.

Detailed explanation

Background Concept

For motion in a circle at constant speed, the velocity continually changes direction. A change in velocity means there must be an acceleration. In uniform circular motion this acceleration is always directed towards the centre of the circle and is called the centripetal acceleration. Therefore the resultant force must also be towards the centre:

F_\text{centripetal} = \frac{mv^2}{r} = mr\omega^2.

Understanding the Question

The question asks you to explain using forces why the satellite's orbit is circular. The key idea is that the gravitational force from the planet is always directed towards the planet's centre, so it acts as the centripetal force needed for circular motion.

Approach

Identify the force acting on the satellite (gravity).
State its direction (towards the planet's centre).
Link that to centripetal force/acceleration, concluding the motion is circular.

Step-by-Step Reasoning

orbit-fbd

The planet exerts a gravitational force on the satellite directed radially inward (towards the planet's centre).
For a circular orbit, the required resultant force must be towards the centre at all times (centripetal force).
Since gravity provides this inward resultant force, the satellite has centripetal acceleration and its path is circular (with its velocity tangential to the orbit at each point).

Key Takeaways

Circular motion requires a centripetal acceleration and therefore a centripetal (inward) resultant force.
In an orbit, gravity provides that inward force.

Common Mistakes

Saying "there is no force" because the satellite is in space.
Claiming the force is tangential (it is radial, towards the centre).
Confusing "centripetal force" as a new force rather than the resultant of real forces (here, gravity).

Things to Be Careful About

The satellite can have constant speed but still accelerate because its velocity direction changes.
The explanation must mention force direction towards the centre, not just "it goes round".

Techniques used

identify the centripetal force as the resultant force towards the centrelink gravitational attraction to centripetal acceleration in orbit

(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.

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Let the planet have mass $A$ and radius $B$ . Orbital radius

r = h + B.

Gravitational force on satellite of mass $m$ :

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

For a circular orbit,

F_c = \frac{mv^2}{r}, \qquad v = \frac{2\pi r}{T}.

So

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

Cancel $m$ and rearrange:

GA = \frac{4\pi^2 r^3}{T^2} \;\Rightarrow\; (r)^3 = \frac{GA}{4\pi^2}T^2.

With $r = h+B$ ,

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

Answer

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

Final answer

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

Detailed explanation

Background Concept

A satellite in a circular orbit requires a centripetal force

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

directed towards the centre of the circle. For an orbit, the only significant force is gravity from the planet:

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

In a stable circular orbit, gravity provides exactly the centripetal force, so $F_g = F_c$ . Also, for one orbit of circumference $2\pi r$ completed in period $T$ ,

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

Understanding the Question

You are told the planet is a uniform sphere, so it behaves like a point mass at its centre for external gravitational forces. The satellite is at height $h$ above the surface, so its distance from the centre is the planet radius plus $h$ . The question asks you to derive a relation between $h$ and $T$ in the given form, identifying constants $A$ and $B$ that depend only on the planet.

Approach

Write the orbital radius as $r = h + B$ where $B$ is the planet radius.
Write gravitational force in terms of planet mass $A$ .
Write centripetal force for circular motion and express $v$ in terms of $r$ and $T$ .
Equate forces and rearrange to isolate $(h+B)^3$ in terms of $T^2$ .

Step-by-Step Reasoning

Distance from the planet centre:

r = h + B.

Gravitational force magnitude on satellite of mass $m$ :

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

Centripetal force needed for a circular path of radius $r$ :

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

The orbital speed in terms of period:

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

Substitute $v$ into $F_c$ :

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

Set $F_g = F_c$ :

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

Cancel $m$ and rearrange:

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

Finally replace $r$ with $h+B$ to match the required form.

Key Takeaways

For circular orbits: gravitational force provides the centripetal force.
Combining $F_g \propto 1/r^2$ with $F_c \propto v^2/r$ and $v = 2\pi r/T$ leads to $r^3 \propto T^2$ .

Common Mistakes

Using $r = h$ instead of $r = h + B$ .
Forgetting to substitute $v = 2\pi r/T$ (or mixing up $2\pi$ factors).
Algebra slip: not getting $r^3$ proportional to $T^2$ .

Things to Be Careful About

The $r$ in the gravitational equation is centre-to-centre distance.
Cancel the satellite mass $m$ correctly; orbital period does not depend on the satellite mass.

Techniques used

equate gravitational force to centripetal forcesubstitute v = 2 pi r / Trearrange to obtain the r-cubed proportionality with T-squared

(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

Worked solution

Working

From

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

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

where

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

From Fig. 1.1, using points $(h/10^6,\,T^{2/3})=(0,360)$ and $(10,1160)$ :

m = \frac{1160-360}{(10-0)\times 10^6} = 8.0\times 10^{-5}\ \text{s}^{2/3}\ \text{m}^{-1}

Intercept $c = 360\ \text{s}^{2/3}$ , so

B = \frac{c}{m} = \frac{360}{8.0\times 10^{-5}} = 4.5\times 10^6\ \text{m}.

Also

m^3 = \frac{4\pi^2}{GA} \;\Rightarrow\; A = \frac{4\pi^2}{Gm^3}.

m^3 = (8.0\times 10^{-5})^3 = 5.12\times 10^{-13}\ \text{s}^2\ \text{m}^{-3}

A = \frac{4\pi^2}{(6.67\times 10^{-11})(5.12\times 10^{-13})} = 1.16\times 10^{24}\ \text{kg}.

Answer

$A = 1.16 \times 10^{24}\ \text{kg}$

$B = 4.5 \times 10^6\ \text{m}$

Final answer

A = 1.16 × 10^24 kg, B = 4.5 × 10^6 m

Detailed explanation

Background Concept

When a relation can be rearranged into the straight-line form

y = mx + c,

a plot of $y$ against $x$ has gradient $m$ and intercept $c$ . Here, the orbital derivation gives

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

To match the plotted graph of $T^{2/3}$ against $h$ , we take the cube root of both sides:

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

So a straight-line graph is expected.

Understanding the Question

You are given a graph of $T^{2/3}$ (vertical axis) against $h$ (horizontal axis, but scaled as $h/10^6\ \text{m}$ ). You must use the gradient and intercept of the straight line to find the constants $A$ and $B$ in the equation. From the physics, $A$ corresponds to the planet mass and $B$ to the planet radius.

Approach

Linearise the given relationship to show it has the form $T^{2/3} = mh + c$ .
Identify $m$ $m$ and $c$ $c$ in terms of $A$ $A$ and $B$ $B$ :
- $m = \left(\frac{4\pi^2}{GA}\right)^{1/3}$
- $c = mB$
Read two well-separated points to calculate the gradient and read the intercept.
Convert the gradient into SI units because the $h$ -axis uses $h/10^6\ \text{m}$ .
Solve for $B = c/m$ and then for $A$ from the expression for $m$ .

Step-by-Step Reasoning

Start from

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

Rearrange for $T^{2/3}$ :

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

Define

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

so

T^{2/3} = m(h+B) = mh + mB.

This matches $y = mx + c$ with $y = T^{2/3}$ , $x = h$ , and $c = mB$ .

Read gradient from the graph.
Using the given approximate points $(0,360)$ and $(10,1160)$ on the plotted axes:

Change in $T^{2/3}$ :

\Delta y = 1160 - 360 = 800\ \text{s}^{2/3}.

Change in the plotted $x$ coordinate is $10$ , but each $1$ on the axis represents $10^6\ \text{m}$ . So

\Delta h = 10\times 10^6\ \text{m}.

Therefore the gradient in SI units is

m = \frac{\Delta y}{\Delta h} = \frac{800}{10\times 10^6} = 8.0\times 10^{-5}\ \text{s}^{2/3}\ \text{m}^{-1}.

Find $B$ from the intercept.
The intercept at $h=0$ is $c = 360\ \text{s}^{2/3}$ . Since $c = mB$ ,

B = \frac{c}{m} = \frac{360}{8.0\times 10^{-5}} = 4.5\times 10^6\ \text{m}.

Find $A$ from the gradient expression.
We have

m^3 = \frac{4\pi^2}{GA} \quad\Rightarrow\quad A = \frac{4\pi^2}{Gm^3}.

Compute

m^3 = (8.0\times 10^{-5})^3 = 5.12\times 10^{-13}\ \text{s}^2\ \text{m}^{-3}.

Then

A = \frac{4\pi^2}{(6.67\times 10^{-11})(5.12\times 10^{-13})} = 1.16\times 10^{24}\ \text{kg}.

(The unit comes out as kg because $G$ has units $\text{m}^3\ \text{kg}^{-1}\ \text{s}^{-2}$ .)

Key Takeaways

Taking powers of an equation can linearise it; here $T^{2/3}$ makes a straight line with $h$ .
The gradient gives a combination of constants; the intercept then separates out $B$ .
Always correct for axis scaling when converting gradients into SI units.

Common Mistakes

Using the plotted x-values as metres (forgetting the $10^6$ scale).
Taking the intercept as $B$ directly (it is $mB$ , not $B$ ).
Rearranging wrongly (e.g. using $m = (GA/4\pi^2)^{1/3}$ instead of its reciprocal).
Missing or incorrect units for $A$ and $B$ .

Things to Be Careful About

Choose two points far apart on the best-fit line to reduce gradient uncertainty.
Keep track of units: gradient here must be in $\text{s}^{2/3}\ \text{m}^{-1}$ .
Use an appropriate number of significant figures consistent with the graph reading precision.

Techniques used

linearise the orbit equation to match y = mx + cextract gradient and intercept from a straight-line graphaccount for axis scaling when converting gradient to SI unitsrearrange algebraically to solve for constants and units

Q2Medium-EasyTemperature

(a)

Define specific heat capacity.

2M Similar

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

Specific heat capacity $c$ measures how much energy is needed to change a substance’s temperature.

The relationship is

\Delta E = mc\Delta T

where $\Delta E$ is thermal energy transferred, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the temperature change.

Understanding the Question

You are asked to define specific heat capacity, so you must give a clear verbal statement including:

the mass (per kilogram),
the temperature rise (per kelvin),
that it is energy required (thermal energy supplied),
and that there is no change of state.

Approach

Write the standard textbook definition in words. You can also connect it to $\Delta E = mc\Delta T$ to ensure you include all required components.

Step-by-Step Reasoning

From

\Delta E = mc\Delta T

if $m = 1\ \text{kg}$ and $\Delta T = 1\ \text{K}$ , then $c$ equals the energy required per kilogram per kelvin. Mentioning “no change of state” distinguishes heating within a phase from energy used in melting/boiling (latent heat).

Key Takeaways

Specific heat capacity is energy per unit mass per unit temperature rise.
Unit is $\text{J kg}^{-1}\ \text{K}^{-1}$ .

Common Mistakes

Missing “per kilogram” or “per kelvin”.
Saying “heat” without clarifying it means energy transferred.
Confusing specific heat capacity with specific latent heat (which is energy per kg for a change of state at constant temperature).

Things to Be Careful About

Use $1\ \text{K}$ (or equivalently $1\ ^\circ\text{C}$ rise) but do not imply the temperature itself is $1\ ^\circ\text{C}$ .
Include “no change of state” for full credit.

Techniques used

state the definition of specific heat capacityspecify mass, temperature change and no change of state conditions

(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 ______

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

At $t=0$ , block $X$ is at about $85\ ^\circ\text{C}$ and block $Y$ is at about $25\ ^\circ\text{C}$ .
The blocks reach the same final temperature (thermal equilibrium) of about $40\ ^\circ\text{C}$ .
Thermal equilibrium is reached at about $t \approx 2.5\ \text{min}$ (the two curves meet), and the rate of temperature change decreases with time (curves become less steep).

Final answer

Example conclusions: X starts ~85 °C, Y starts ~25 °C; equilibrium temperature ~40 °C; equilibrium reached at ~2.5 min (rate of change decreases with time).

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 object.

The transfer continues until both objects reach the same temperature.
This common temperature is called the thermal equilibrium temperature.
The rate of heat transfer (and so the rate of temperature change) is typically greatest at the start because the temperature difference is largest, and then decreases as the temperature difference becomes smaller.

Understanding the Question

You are given a temperature–time graph for two metal blocks $X$ and $Y$ placed in direct contact at $t=0$ .

You must state three conclusions that can be read from the graph. These may be:

numerical readings (initial temperatures, equilibrium temperature, time to reach equilibrium), and/or
qualitative statements (which direction heat flows, rate changes over time, whether equilibrium is reached).

Approach

Read key features from the graph:

Read both initial temperatures at $t=0$ .
Find the final common temperature (where both curves level off / meet).
Estimate the time at which both temperatures become equal (curves intersect), and/or describe how the slopes change.

Step-by-Step Reasoning

At $t=0$ , the curve labelled $X$ begins at about $85\ ^\circ\text{C}$ , and the curve labelled $Y$ begins at about $25\ ^\circ\text{C}$ .
Both curves move towards the same temperature, and they approach/meet at about $40\ ^\circ\text{C}$ . This is the equilibrium temperature.
The curves intersect at roughly $t \approx 2.5\ \text{min}$ , so that is when both blocks have the same temperature.
The curves are steep at small $t$ and become less steep later, showing the rate of temperature change decreases with time as the temperature difference reduces.

(Any three distinct, correct statements like these would score.)

Key Takeaways

The equilibrium temperature is the common final temperature.
The intersection time indicates when both objects have the same temperature.
A decreasing slope magnitude indicates decreasing rate of energy transfer.

Common Mistakes

Stating “equilibrium at $40\ ^\circ\text{C}$ ” but not identifying it as both blocks having the same temperature.
Confusing the time when curves are close with the time they are equal (need the intersection).
Giving three statements that are not distinct (e.g. repeating the same point in different words).

Things to Be Careful About

Temperature differences can be read in $^\circ\text{C}$ directly from the graph; for differences, $1\ \text{K} = 1\ ^\circ\text{C}$ .
Quote approximate values consistent with the graph scale (do not overstate precision).

Techniques used

extract initial temperatures from a temperature–time graphidentify the equilibrium temperature from where curves meetdescribe how rate of temperature change varies with time

(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 = ______ $\text{J kg}^{-1}\ \text{K}^{-1}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

At equilibrium, heat lost by $X$ = heat gained by $Y$ :

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}

c_X = 1.3 \times 901 \times \frac{15}{45} = 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 bodies exchange thermal energy in an isolated situation (no significant energy loss to surroundings), energy conservation gives:

\text{thermal energy lost by hot object} = \text{thermal energy gained by cold object}

For each object,

\Delta E = mc\Delta T

where $\Delta T$ is the magnitude of the temperature change. Temperature differences in $^\circ\text{C}$ are equal to temperature differences in K.

Understanding the Question

From the graph:

Block $X$ cools from about $85\ ^\circ\text{C}$ to the equilibrium temperature $40\ ^\circ\text{C}$ .
Block $Y$ warms from $25\ ^\circ\text{C}$ (room temperature) to $40\ ^\circ\text{C}$ .

You are told:

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

and you must find $c_X$ .

Approach

Write energy balance: $m_X c_X \Delta T_X = m_Y c_Y \Delta T_Y$ .
Read $\Delta T_X$ and $\Delta T_Y$ from the graph.
Use the given mass ratio to eliminate $m_X$ and $m_Y$ .
Solve for $c_X$ .

Step-by-Step Reasoning

diagram-bii-1

Temperature changes:

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

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

Energy conservation (assuming negligible losses):

m_X c_X (45) = m_Y c_Y (15)

Rearrange 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{1}{3}

c_X \approx 390\ \text{J kg}^{-1}\ \text{K}^{-1}

This is reasonable: many metals have $c$ values of a few hundred $\text{J kg}^{-1}\ \text{K}^{-1}$ .

Key Takeaways

For two objects exchanging heat, use $m_1 c_1 \Delta T_1 = m_2 c_2 \Delta T_2$ .
Read $\Delta T$ values from the graph using the equilibrium temperature.
Use mass ratios to simplify without needing the actual masses.

Common Mistakes

Swapping the temperature changes (using $85-25$ instead of changes to equilibrium).
Forgetting to use the mass ratio, or inverting it (using $m_X/m_Y$ instead of $m_Y/m_X$ ).
Using final temperature as $25\ ^\circ\text{C}$ or $85\ ^\circ\text{C}$ instead of the equilibrium value $40\ ^\circ\text{C}$ .
Adding the energies instead of equating loss to gain.

Things to Be Careful About

Use magnitudes of temperature change; do not introduce negative signs if you are using “lost = gained”.
Ensure consistent units: $c$ in $\text{J kg}^{-1}\ \text{K}^{-1}$ .
Do not over-round mid-calculation; round to 2–3 significant figures at the end.

Techniques used

apply conservation of energy to heat exchangeuse $\Delta E = mc\Delta T$ for each blockuse a mass ratio to eliminate unknown masses

Q3MediumIdeal GasesThermodynamics

(a)

2M

(i)

State what is meant by the Avogadro constant.

1M Similar

(ii)

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

1M Similar

(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

Pressure:

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

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

Amount of substance:

n_X = \frac{N}{N_A},\quad n_Y = \frac{2N}{N_A}

Mean-square speed:

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

Internal energy:

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

Answer

$\displaystyle p_X = \frac{NkT}{V},\ p_Y = \frac{NkT}{V}$

$\displaystyle n_X = \frac{N}{N_A},\ n_Y = \frac{2N}{N_A}$

$\displaystyle \langle c^2 \rangle_X = \frac{3kT}{m},\ \langle c^2 \rangle_Y = \frac{3kT}{2m}$

$\displaystyle U_X = \frac{3}{2}NkT,\ U_Y = 3NkT$

Final answer

pX = NkT/V, 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 an ideal gas, several key relationships connect macroscopic variables to microscopic ones:

Ideal gas equation (molecular form):

pV = NkT

Amount of substance $n$ and Avogadro constant:

N = nN_A \Rightarrow n = \frac{N}{N_A}

Kinetic theory link between pressure and mean-square speed:

pV = \frac{1}{3}Nm\langle c^2 \rangle

Combining this with $pV = NkT$ gives

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

Internal energy of an ideal gas (at this level, taken as translational kinetic energy only): average kinetic energy per molecule is $\frac{3}{2}kT$ , so

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

Understanding the Question

Both samples are at the same temperature $T$ .

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

You must fill a table with expressions for pressure, amount of substance, mean-square speed, and internal energy, for each sample.

Approach

Use the most direct formula for each row:

pressure: use $pV = NkT$ .
amount of substance: use $n = N/N_A$ .
mean-square speed: use $\langle c^2 \rangle = 3kT/m$ and substitute the appropriate molecular mass.
internal energy: use $U = \frac{3}{2}NkT$ with the appropriate number of molecules.

Step-by-Step Reasoning

Pressure

For X:

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

For Y, note that both $N$ and $V$ have doubled:

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

So both have the same pressure.

Amount of substance

Use $n = N/N_A$ .

For X:

n_X = \frac{N}{N_A}

For Y, there are $2N$ molecules:

n_Y = \frac{2N}{N_A}

Mean-square speed

From kinetic theory:

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

For X, molecular mass is $m$ :

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

For Y, molecular mass is $2m$ :

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

So at the same temperature, heavier molecules have a smaller mean-square speed.

Internal energy

Average kinetic energy per molecule is $\frac{3}{2}kT$ , so

U = \frac{3}{2}NkT

For X:

U_X = \frac{3}{2}NkT

For Y, number of molecules is $2N$ :

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

Note internal energy depends on the number of molecules and temperature, not on the volume.

Key Takeaways

Using $pV = NkT$ , if both $N$ and $V$ scale by the same factor at fixed $T$ , pressure is unchanged.
$n = N/N_A$ converts particle number to moles.
$\langle c^2 \rangle \propto T/m$ : at the same $T$ , heavier molecules move more slowly on average.
At this level, ideal-gas internal energy $U = \frac{3}{2}NkT$ depends only on $N$ and $T$ .

Common Mistakes

Using $pV = nRT$ but then forgetting to convert $N$ to $n$ (or mixing $N$ and $n$ ).
Writing $\langle c^2 \rangle = \frac{3RT}{m}$ (incorrect constant; should be $k$ with molecular mass per molecule, or $R$ with molar mass).
Thinking doubling volume at fixed temperature must halve pressure, without noticing that $N$ has also doubled.

Things to Be Careful About

Distinguish between molecular mass (per molecule, $m$ ) and molar mass (per mole).
Mean-square speed is $\langle c^2 \rangle$ (proportional to $T$ ), whereas r.m.s. speed is $\sqrt{\langle c^2 \rangle}$ (proportional to $\sqrt{T}$ ).

Techniques used

apply the ideal gas equation in microscopic formuse N = n N_A to relate amount of substance to number of moleculesuse kinetic theory to relate mean-square speed to temperature and molecular massuse internal energy of an ideal gas as proportional to N and T

(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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For sample X,

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

Answer

diagram-1

A curve through the origin that increases with $T$ with decreasing gradient (square-root shape).

Final answer

c_rms ∝ √T; curve through origin, increasing with decreasing gradient.

Detailed explanation

Background Concept

The r.m.s. speed is defined by

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

For an ideal gas, kinetic theory gives

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

so

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

This shows $c_{\text{rms}} \propto \sqrt{T}$ for a fixed gas (fixed molecular mass $m$ ).

Understanding the Question

Only sample X is relevant now, and its temperature $T$ is varied. You must sketch how $c_{\text{rms}}$ changes with thermodynamic temperature on axes starting at the origin.

Approach

Use the proportionality $c_{\text{rms}} \propto \sqrt{T}$ .
Translate that into a graph shape: it must pass through $(0,0)$ and rise, but not as a straight line.

Step-by-Step Reasoning

From

c_{\text{rms}} = \sqrt{\frac{3k}{m}}\,\sqrt{T}

When $T = 0\ \text{K}$ , $c_{\text{rms}} = 0$ in this idealised model, so the curve passes through the origin.
As $T$ increases, $c_{\text{rms}}$ increases.
Because it is a square-root relationship, the gradient decreases as $T$ gets larger (the curve is concave down).

diagram-1

Key Takeaways

$c_{\text{rms}} \propto \sqrt{T}$ for a given gas.
Square-root graphs rise quickly at first and then flatten (decreasing gradient).

Common Mistakes

Drawing a straight line (that would imply $c_{\text{rms}} \propto T$ ).
Drawing a curve that does not pass through the origin on a Kelvin-temperature axis.
Drawing a curve that gets steeper with $T$ (wrong curvature).

Things to Be Careful About

The temperature is thermodynamic (Kelvin) temperature, so the origin represents $0\ \text{K}$ .
The y-axis is speed, so it cannot be negative; only the first quadrant is relevant.

Techniques used

use the r.m.s. speed–temperature relationshipsketch a non-linear graph with correct qualitative featuresidentify intercept behaviour at T = 0 K

Q4MediumOscillations

(a)

State what is meant by simple harmonic motion.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

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

a \propto -x \qquad (\text{or } a = -\omega^2 x)

Final answer

Acceleration is proportional to displacement from equilibrium and directed towards equilibrium (a ∝ −x, or a = −ω²x).

Detailed explanation

Background Concept

In simple harmonic motion (SHM), the key physical idea is a restoring effect: whenever the object is displaced from its equilibrium position, something causes it to accelerate back towards equilibrium.

Mathematically, SHM is defined by the relationship between acceleration $a$ and displacement $x$ from equilibrium:

a = -\omega^2 x

where:

$x$ is the displacement from equilibrium (can be positive or negative),
$a$ is the acceleration,
$\omega$ is the angular frequency (a constant for that system).

The minus sign means the acceleration is always opposite in direction to the displacement (i.e. towards equilibrium).

Understanding the Question

The question asks for the meaning of SHM, so you must give the defining condition, not examples (spring, pendulum, etc.). It is a 2-mark definition, so you typically need:

proportionality to displacement, and
direction towards equilibrium (negative sign).

Approach

State the definition in words, and (optionally) give the equivalent mathematical form $a \propto -x$ or $a = -\omega^2 x$ .

Step-by-Step Reasoning

If an object is displaced by $x$ from equilibrium, then in SHM the acceleration magnitude increases in direct proportion to $|x|$ .
The acceleration must act towards equilibrium, so its direction is opposite to the displacement, giving $a \propto -x$ .
Writing the constant of proportionality as $\omega^2$ gives the standard form $a = -\omega^2 x$ .

Key Takeaways

SHM is defined by $a \propto -x$ .
You must include both proportionality and the restoring direction (negative sign / “towards equilibrium”).

Common Mistakes

Saying “motion is sinusoidal” without mentioning acceleration: sinusoidal motion is a consequence, not the definition.
Missing the direction (“towards equilibrium”) or the negative sign.
Saying velocity is proportional to displacement (false for SHM).

Things to Be Careful About

Displacement must be measured from equilibrium, not from an end point.
Use the word acceleration, not force, unless you explicitly connect it (since $F \propto -x$ implies $a \propto -x$ only if mass is constant).

Techniques used

state the defining condition for SHMexpress proportionality with a negative sign to indicate direction towards equilibrium

(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 = ______ $\text{cm}$

1M Similar

DifficultyEasy

Worked solution

Working

From the $v$ – $h$ graph, the extreme heights are $h_{\max} = 9.5\ \text{cm}$ 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 know the maximum and minimum values of the position coordinate (here height $h$ ), then:

the equilibrium height is the midpoint,
the amplitude is half the total range:

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

Understanding the Question

The diagram/graph shows $v$ plotted against $h$ . Even though it is not a direct $h$ – $t$ graph, the ellipse still tells you the turning points: where the speed is zero ( $v=0$ ), the block is at maximum or minimum height.

You need the amplitude in cm.

Approach

Read off the two heights at which $v=0$ on the ellipse (leftmost/rightmost intercepts with the $h$ -axis), then take half the difference.

Step-by-Step Reasoning

Turning points occur at $v=0$ .
From the graph, the ellipse meets the $h$ -axis at about $3.5\ \text{cm}$ and $9.5\ \text{cm}$ .
The equilibrium height is the midpoint:

\frac{3.5 + 9.5}{2} = 6.5\ \text{cm}

The amplitude is the displacement from equilibrium to an extreme:

A = 9.5 - 6.5 = 3.0\ \text{cm}

(or equivalently $A = (9.5-3.5)/2$ ).

Key Takeaways

Turning points: $v=0$ .
Amplitude: half the peak-to-peak position range.

Common Mistakes

Taking amplitude as $9.5\ \text{cm}$ or $3.5\ \text{cm}$ (these are absolute heights, not displacement from equilibrium).
Using the velocity maximum ( $9.6\ \text{cm s}^{-1}$ ) for amplitude.

Things to Be Careful About

Read the axis values carefully; amplitude is a difference of heights, not the centre value.
Keep units consistent (here cm is fine).

Techniques used

read maximum and minimum values from a graphcalculate amplitude as half the peak-to-peak displacement

(ii)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

For SHM,

v_{\max} = \omega A

So

\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 displacement can be written as

x = A\sin(\omega t)

Differentiating gives the velocity:

v = \frac{\mathrm{d}x}{\mathrm{d}t} = A\omega\cos(\omega t)

So the maximum speed occurs when $\cos(\omega t)=\pm 1$ , giving the key result:

v_{\max} = \omega A

This is true regardless of whether the motion is horizontal or vertical, as long as it is SHM.

Understanding the Question

You are given a $v$ – $h$ plot (often called a phase-space plot). The ellipse shows that velocity is greatest as the block passes through equilibrium height, and zero at the turning points.

You must “show that” $\omega = 3.2\ \text{rad s}^{-1}$ , so you need to show the calculation leading to that value.

Approach

Find the amplitude $A$ from the extreme $h$ values.
Read the maximum speed $v_{\max}$ from the top/bottom of the ellipse.
Use $v_{\max} = \omega A$ and rearrange.

Step-by-Step Reasoning

From part (i), $A = 3.0\ \text{cm}$ .
From the graph, the maximum speed is about $9.6\ \text{cm s}^{-1}$ .
Apply

v_{\max} = \omega A

Rearrange:

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

Substitute values:

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

The unit $\text{s}^{-1}$ is equivalent to $\text{rad s}^{-1}$ for angular frequency (radian is dimensionless).

Key Takeaways

Use $v_{\max} = \omega A$ to link speed and amplitude.
A $v$ – $x$ (or $v$ – $h$ ) ellipse contains both $A$ and $v_{\max}$ .

Common Mistakes

Using $v = \omega x$ (not true; it’s $v$ maximum that equals $\omega A$ ).
Mixing up equilibrium height (centre of ellipse) with amplitude.
Converting cm to m unnecessarily and then making a powers-of-ten mistake.

Things to Be Careful About

Ensure you use consistent length units for $v$ and $A$ (both in cm here, so they cancel correctly).
Quote $\omega$ with unit $\text{rad s}^{-1}$ (or $\text{s}^{-1}$ ).

Techniques used

use the SHM relation v_max = omega Aread maximum speed and amplitude from a graphrearrange to determine angular frequency

(iii)

Calculate the period $T$ of the oscillations.

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

2M Similar

(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-4-b-iv

Answer

Sketch a sinusoidal graph of $h$ against $t$ with:

midline at $h = 6.5\ \text{cm}$ ,
amplitude $3.0\ \text{cm}$ (so $h_{\min}=3.5\ \text{cm}$ and $h_{\max}=9.5\ \text{cm}$ ),
period $T \approx 2.0\ \text{s}$ ,
at $t=0$ , $h=3.5\ \text{cm}$ (released from lowest point) with zero gradient, then rising.

(So maxima at $t \approx 1.0, 3.0, 5.0\ \text{s}$ and minima at $t \approx 0, 2.0, 4.0, 6.0\ \text{s}$ .)

Final answer

Sinusoidal h–t graph about 6.5 cm with amplitude 3.0 cm, period ≈ 2.0 s, starting at h = 3.5 cm at t = 0 and rising.

Detailed explanation

Background Concept

For SHM, displacement (or here height) varies sinusoidally with time. If we measure displacement from equilibrium as $x$ , then a standard form is:

x = A\sin(\omega t + \phi)

The constants mean:

$A$ is amplitude,
$\omega$ sets the period $T = 2\pi/\omega$ ,
$\phi$ sets the phase (where in the cycle you start).

At a turning point (top or bottom), the velocity is zero, so the $h$ – $t$ graph has zero gradient there.

Understanding the Question

You must sketch $h$ against $t$ from $0$ to $6.0\ \text{s}$ .

From the given description and the $v$ – $h$ graph:

equilibrium (centre) height is $6.5\ \text{cm}$ ,
amplitude is $3.0\ \text{cm}$ , so $h$ ranges from $3.5$ to $9.5\ \text{cm}$ ,
period is about $2.0\ \text{s}$ ,
at $t=0$ the block is pulled down and released, so it starts at the lowest height with $v=0$ .

Approach

To sketch correctly, you need four features:

correct midline ( $6.5\ \text{cm}$ ),
correct amplitude (reach $3.5$ and $9.5\ \text{cm}$ ),
correct period spacing (about $2\ \text{s}$ per full cycle),
correct phase: start at minimum at $t=0$ and then rise.

Step-by-Step Reasoning

diagram-4-b-iv

Set the midline

From the extreme heights $3.5$ and $9.5\ \text{cm}$ , the equilibrium is

\frac{3.5+9.5}{2} = 6.5\ \text{cm}

Draw a faint central line at $h=6.5\ \text{cm}$ .

Set the amplitude

Amplitude $A=3.0\ \text{cm}$ , so the curve must touch $h=3.5\ \text{cm}$ and $h=9.5\ \text{cm}$ .

Set the period

Using $T = 2\pi/\omega$ with $\omega=3.2\ \text{rad s}^{-1}$ gives $T\approx 2.0\ \text{s}$ .
So from $t=0$ to $t=6\ \text{s}$ you should draw about three full cycles.

Set the starting point and direction

“Pulled down and released at $t=0$ $t = 0$ ” means:
- start at the bottom turning point: $h(0)=3.5\ \text{cm}$ ,
- released from rest so initial gradient is zero,
- then it moves upward, so immediately after $t=0$ the curve rises.

Mark key times (helps accuracy of the sketch)

Quarter-period steps are useful:
- at $t=T/4 \approx 0.50\ \text{s}$ : passes equilibrium ( $h=6.5\ \text{cm}$ ) with maximum positive slope,
- at $t=T/2 \approx 1.0\ \text{s}$ : reaches maximum $h=9.5\ \text{cm}$ ,
- at $t=3T/4 \approx 1.5\ \text{s}$ : crosses equilibrium downward,
- at $t=T \approx 2.0\ \text{s}$ : back to minimum.

Repeat this pattern up to $6.0\ \text{s}$ .

Key Takeaways

SHM position-time graphs are sinusoidal.
The centre line is the equilibrium value; amplitude is measured from that line.
The wording at $t=0$ sets the phase (minimum/maximum/crossing equilibrium).

Common Mistakes

Starting at the equilibrium position at $t=0$ even though the block is released from a pulled-down position.
Drawing a triangle wave or non-sinusoidal curve.
Using the wrong midline (e.g. centring at $h=0$ instead of $h=6.5\ \text{cm}$ ).
Incorrect period spacing (too few or too many cycles in 6 s).

Things to Be Careful About

Turning points must have zero gradient.
Ensure the maximum and minimum heights match the graph values ( $9.5$ and $3.5\ \text{cm}$ ).
A sketch should use the graph grid sensibly: clear sinusoid, consistent period, consistent amplitude.

Techniques used

identify initial phase from the description of motionuse amplitude and equilibrium position to set vertical scaleuse the period to set horizontal spacing of oscillationssketch a sinusoidal displacement-time graph

Q5MediumElectric Fields

(a)

State the relationship between electric field and electric potential.

2M Similar

DifficultyEasy

Worked solution

Answer

Electric field strength is the negative potential gradient:

\vec{E} = -\nabla V

(along a line)

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

Final answer

$\vec{E}=-\nabla V$ (so along a line $E=-\mathrm{d}V/\mathrm{d}x$ )

Detailed explanation

Background Concept

Electric potential $V$ at a point is the potential energy per unit charge. The electric field strength $\vec{E}$ describes the force per unit positive charge. These are linked because a charge moving a small distance in an electric field changes its potential energy.

The electric field points in the direction of decreasing potential, so the field is related to the negative spatial rate of change (gradient) of potential.

Understanding the Question

You are asked to state the mathematical relationship between electric field and electric potential (no numbers, just the defining equation).

Approach

Recall that $\vec{E}$ is the negative gradient of $V$ . If only one dimension is involved (e.g. along $x$ ), replace the gradient with a derivative with respect to distance.

Step-by-Step Reasoning

In vector form:

\vec{E} = -\nabla V

Along a straight line coordinate $x$ :

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

The minus sign encodes that $V$ decreases in the direction of $\vec{E}$ .

Key Takeaways

$\vec{E}$ is a vector; $V$ is a scalar.
The field direction is the direction of steepest decrease of potential.

Common Mistakes

Writing $\vec{E} = +\nabla V$ (wrong sign).
Confusing $E = V/d$ (only for a uniform field between parallel plates) with the general relationship.

Things to Be Careful About

Use a derivative/gradient (rate of change with position), not a simple ratio unless the field is uniform.
Include the negative sign.

Techniques used

state the relationship between field strength and potential gradientuse the sign convention linking field direction to 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 $P$ .

3M Similar

DifficultyMedium

Worked solution

Answer

For $V_{\text{total}}(P)=0$ , the potentials from the two spheres must be equal and opposite, so the charges must be of opposite sign.

At a point between opposite charges, the electric fields due to each charge are in the same direction (from $+$ to $-$ ), so they add and cannot give zero resultant field.

Therefore $V_{\text{total}}=0$ and $\vec{E}_{\text{resultant}}=\vec{0}$ cannot both occur at $P$ .

Final answer

To have $V_{\text{total}}=0$ the charges must be opposite signs, but then the fields at a point between them are in the same direction and cannot cancel, so $\vec{E}$ cannot be zero.

Detailed explanation

Background Concept

For point charges (or charged spheres treated as point charges), the electric potential at a point is

V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}

Potential is a scalar, so total potential is the algebraic sum:

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

Electric field strength is a vector:

\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\,\hat{r}

and fields add by vector addition:

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

Direction rule: field points away from a positive charge and towards a negative charge.

Understanding the Question

Two charged insulating spheres are near each other, and $P$ lies on the line between their centres. You must explain (using physics reasoning, not calculations) why it is impossible for both:

the total potential at $P$ to be zero, and
the resultant electric field at $P$ to be zero,
at the same time.

Approach

Consider what charge signs are required for $V_{\text{total}}(P)=0$ , then examine what that implies for the directions of the two electric fields at a point between the spheres. Use the fact that potentials add as scalars but fields add as vectors.

Step-by-Step Reasoning

For $V_{\text{total}}(P)=0$ with two charges, we need

V_X + V_Y = 0

So one contribution must be positive and the other negative. That requires opposite signs of charge (one sphere positive, the other negative).

Now look at the electric field directions at a point between opposite charges.

diagram-b-1

At point $P$ :

the field due to the positive charge points away from it (towards the other sphere),
the field due to the negative charge points towards it (also towards the other sphere),
so both fields point in the same direction along the line.

Since the two field vectors at $P$ point the same way, they cannot cancel to give zero resultant field. They add to a non-zero resultant.

Therefore it is not possible for $V_{\text{total}}=0$ and $\vec{E}=\vec{0}$ simultaneously at $P$ .

(Conversely, if you tried to make $\vec{E}=\vec{0}$ between two charges, you would need like charges so the fields oppose there, but then both potentials would have the same sign and could not sum to zero.)

Key Takeaways

Electric potential is a scalar: add with signs.
Electric field is a vector: direction matters.
Between opposite charges, fields point in the same direction, so they cannot cancel.

Common Mistakes

Thinking that if potentials cancel then fields must cancel (they are related but not the same quantity).
Forgetting field direction (away from $+$ , towards $-$ ).

Things to Be Careful About

The statement depends on $P$ being on the line between the charges; that fixes the field directions.
“Zero potential” is relative to infinity; it can occur by cancellation of positive and negative contributions, not by each being zero.

Techniques used

use superposition of electric potential as a scalar sumuse superposition of electric field as a vector sumreason about field directions for like and unlike charges

(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 $2x$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For $V(P)=0$ , the charges must be opposite in sign, so

\frac{1}{4\pi\epsilon_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 due to a point charge is

V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}

Potential is a scalar, so for multiple charges you add the potentials (including sign of charge):

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

A positive charge gives positive potential; a negative charge gives negative potential.

Understanding the Question

Sphere X has charge magnitude $Q$ and sphere Y has charge magnitude $2Q$ . Point $P$ lies on the line between them, at distance $x$ from X and distance $y$ from Y. You are told that the electric potential at $P$ is zero, and you must show that this implies $y=2x$ .

Approach

Write $V(P)=V_X+V_Y$ using $V = (1/4\pi\epsilon_0)(Q/r)$ . For the sum to be zero with non-zero charges, one term must be positive and the other negative, so take the charges to be opposite sign (e.g. $+Q$ and $-2Q$ ). Then solve the resulting equation for $y$ in terms of $x$ .

Step-by-Step Reasoning

Let X be $+Q$ and Y be $-2Q$ (the opposite assignment gives the same distance result).

Potential at $P$ :

V(P)=\frac{1}{4\pi\epsilon_0}\left(\frac{Q}{x}+\frac{-2Q}{y}\right)

Given $V(P)=0$ :

\frac{1}{4\pi\epsilon_0}\left(\frac{Q}{x}-\frac{2Q}{y}\right)=0

The constant factor is non-zero, so the bracket must be zero:

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

Cancel $Q$ :

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

Rearrange:

y=2x

Key Takeaways

To get zero total potential from two charges, they must be opposite sign.
Potential varies as $1/r$ , so the larger-magnitude charge must be further away to cancel.

Common Mistakes

Adding magnitudes instead of signed potentials (you must include a minus sign for a negative charge).
Using $1/r^2$ (that is for field, not potential).

Things to Be Careful About

The relationship uses distances from the charge centres because the spheres are treated as point charges.
Either sign assignment gives the same distance ratio; what matters is opposite sign for cancellation.

Techniques used

use the point-charge potential equationapply superposition of potentials at a pointsolve an algebraic ratio from a zero-sum condition

(ii)

State an expression, in terms of $Q$ , $x$ and the permittivity of free space $\epsilon_0$ , for the electric field strength $E_X$ at $P$ due to sphere X.

$E_X$ = ______

1M Similar

(iii)

Determine an expression, in terms of $Q$ , $x$ and $\epsilon_0$ , for the resultant electric field strength $E$ at point $P$ due to the two spheres.

$E$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

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

Field magnitudes at $P$ :

E_X = \frac{1}{4\pi\epsilon_0}\frac{Q}{x^2}

E_Y = \frac{1}{4\pi\epsilon_0}\frac{2Q}{y^2}

The charges are opposite sign, so at $P$ the fields are in the same direction and add:

E = E_X + E_Y

Substitute $y=2x$ :

E = \frac{1}{4\pi\epsilon_0}\left(\frac{Q}{x^2} + \frac{2Q}{(2x)^2}\right) = \frac{1}{4\pi\epsilon_0}\left(\frac{Q}{x^2} + \frac{Q}{2x^2}\right) = \frac{3Q}{8\pi\epsilon_0 x^2}

Answer

E = \frac{3Q}{8\pi\epsilon_0 x^2}

(direction towards the negative sphere)

Final answer

$E=\dfrac{3Q}{8\pi\epsilon_0 x^2}$

Detailed explanation

Background Concept

For a point charge, electric field magnitude is

E = \frac{1}{4\pi\epsilon_0}\frac{|Q|}{r^2}

Electric field is a vector. On a straight line, you can add fields algebraically provided you choose a direction as positive and assign signs based on direction.

Also, if the total potential at $P$ is zero for two non-zero charges, the charges must be of opposite sign. Between opposite charges, both field contributions at a point between them point from the positive charge towards the negative charge.

Understanding the Question

You already have that the potential at $P$ is zero and from part (i) this gives $y=2x$ . Now you must find an expression for the resultant electric field strength at $P$ due to both spheres (treated as point charges).

Approach

Write the field at $P$ due to X and due to Y using the inverse-square law.
Decide whether they add or subtract by considering charge signs/directions.
Use $y=2x$ to eliminate $y$ and simplify to a single expression in $Q$ , $x$ , $\epsilon_0$ .

Step-by-Step Reasoning

Field due to X at distance $x$ :

E_X = \frac{1}{4\pi\epsilon_0}\frac{Q}{x^2}

Field due to Y at distance $y$ with charge magnitude $2Q$ :

E_Y = \frac{1}{4\pi\epsilon_0}\frac{2Q}{y^2}

Direction check:

Since $V(P)=0$ , the charges must be opposite sign.
At a point between opposite charges, both fields point in the same direction (towards the negative charge), so the resultant magnitude is the sum:

E = E_X + E_Y

Substitute $y=2x$ :

E = \frac{1}{4\pi\epsilon_0}\left(\frac{Q}{x^2} + \frac{2Q}{(2x)^2}\right)

Compute $(2x)^2 = 4x^2$ :

E = \frac{1}{4\pi\epsilon_0}\left(\frac{Q}{x^2} + \frac{2Q}{4x^2}\right) = \frac{1}{4\pi\epsilon_0}\left(\frac{Q}{x^2} + \frac{Q}{2x^2}\right)

Add the fractions:

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

So:

E = \frac{1}{4\pi\epsilon_0}\cdot \frac{3Q}{2x^2} = \frac{3Q}{8\pi\epsilon_0 x^2}

Direction: along the line from the positive sphere to the negative sphere.

Key Takeaways

Zero potential implies opposite charge signs.
Between opposite charges, fields reinforce (same direction), even though potentials can cancel.
Use $1/r^2$ for field and $1/r$ for potential.

Common Mistakes

Subtracting fields here (they would subtract between like charges, not unlike charges).
Using $y=2x$ but then forgetting to square it when substituting into $E_Y$ .
Mixing up $2Q/y^2$ with $(2Q/y)^2$ .

Things to Be Careful About

The expression requested is for field strength at $P$ ; if the exam expects a direction statement, include it (towards the negative sphere).
Keep algebra tidy: $(2x)^2=4x^2$ , not $2x^2$ .

Techniques used

use the point-charge electric field equationuse the distance relationship found from zero potentialadd electric field contributions along a line using directions

Q6MediumAlternating CurrentsCapacitance

(a)

3M

(i)

State what is meant by rectification of an alternating voltage.

1M Similar

(ii)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Half-wave rectification: only one half-cycle of the input produces an output; the other half-cycle is blocked ((V_{\text{out}} \approx 0)).
Full-wave rectification: both half-cycles produce an output of the same polarity (so pulses occur every half-cycle; output frequency is doubled).

Final answer

Half-wave uses only one half-cycle; full-wave uses both half-cycles to give output of one polarity (pulses every half-cycle, frequency doubled).

Detailed explanation

Background Concept

A diode conducts when forward biased and blocks current when reverse biased. Rectifier circuits use this to ensure the load voltage does not reverse polarity.

In half-wave rectification, the circuit allows current through the load for only one polarity of the input (one half-cycle).
In full-wave rectification, the circuit arranges that whichever polarity the input has, current through the load is in the same direction (both half-cycles are used).

Understanding the Question

You must state the difference between half-wave and full-wave rectification. For 2 marks, you should give two contrasting points (e.g. “one half-cycle vs both half-cycles” and “pulse frequency is the same vs doubled / smoother output”).

Approach

Describe what happens to the output during the positive and negative half-cycles for each type.

Step-by-Step Reasoning

Half-wave: during (say) the positive half-cycle the diode is forward biased so the load gets a pulse; during the negative half-cycle the diode is reverse biased so the load gets almost no voltage.
Full-wave: the circuit (e.g. bridge rectifier) effectively swaps connections on alternate half-cycles so the load always has the same polarity across it.
Since full-wave produces a pulse every half-cycle, the ripple has double the frequency compared to half-wave, making smoothing easier.

Key Takeaways

Half-wave: uses one half-cycle only.
Full-wave: uses both half-cycles and gives higher ripple frequency and typically smoother output.

Common Mistakes

Mixing up “full-wave” with “higher voltage”: the key point is both half-cycles contribute.
Forgetting to state that the output polarity is the same for both half-cycles in full-wave.

Things to Be Careful About

Use the word “half-cycle” and mention polarity/direction through the load; vague statements like “full-wave is better” are not creditworthy.

Techniques used

compare the output waveform for half-wave and full-wave rectificationstate how many half-cycles contribute to the outputrelate rectification type to output pulse frequency

(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

2M Similar

DifficultyMedium-Easy

Worked solution

diagram-6b1

Answer

Place a diode in series between the top input terminal and the top rail of the (R\parallel C) combination, oriented to conduct when the top input is positive (anode to input, cathode to the (R/C) node). Connect the bottom input terminal directly to the bottom rail.

Final answer

Single series diode between input and R/C node (anode to input, cathode to output node); return directly to the other input terminal.

Detailed explanation

Background Concept

A half-wave rectifier uses a single diode so that current can flow through the load for one input polarity only.

A smoothing capacitor is usually placed across the load (in parallel) so it can charge when the diode conducts and then discharge through the load when the diode is off.

Understanding the Question

You are given an incomplete circuit: an a.c. input (V_{IN}) on the left and, on the right, a resistor (R) with capacitor (C) in parallel, with (V_{OUT}) measured across (R).

To get half-wave rectification, you must add a single diode in the correct place and orientation so that only one half-cycle charges the capacitor and provides output across (R).

Approach

Put the diode in series between the source and the load node.
Choose its orientation so that it is forward biased on the half-cycle you want to pass (typically when the top input terminal is positive relative to the bottom).

Step-by-Step Reasoning

diagram-6b1

The top of the (R\parallel C) network is the output “positive” node, and the bottom is the return.
Put the diode in the top lead from the input to this output node.
Orient the diode so that when (V_{IN}) is positive, the diode is forward biased and current flows into the (R\parallel C) network, charging (C) and producing (V_{OUT}).
When (V_{IN}) becomes negative, the diode is reverse biased so the input is effectively disconnected; the capacitor then discharges only through (R), maintaining a positive (V_{OUT}) for a short time.

Key Takeaways

Half-wave rectification can be achieved with one diode in series.
The diode orientation determines which half-cycle appears at the output.
The capacitor must be in parallel with the load to smooth the output.

Common Mistakes

Putting the diode across the load (would not rectify the supply to the load correctly).
Reversing the diode so that the output is negative pulses instead of positive (may be marked wrong if inconsistent with (V_{OUT}) definition).
Putting the capacitor in series with the load (prevents correct smoothing behaviour).

Things to Be Careful About

The output (V_{OUT}) is measured across (R), so the diode must be placed so that the voltage across (R) becomes unidirectional.
In diagrams, show the diode symbol with a clear bar (cathode) so the direction of conduction is unambiguous.

Techniques used

identify a half-wave rectifier circuit using a single diodechoose diode orientation to conduct on the desired half-cycleconnect the load and smoothing capacitor correctly

(ii)

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

1M Similar

(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 $570\ \mu\text{F}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Maximum capacitor voltage (V = 12\ \text{V}).

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

C = \frac{2W}{V^2} = \frac{2(0.041)}{12^2} = 5.69 \times 10^{-4}\ \text{F}

C = 5.69 \times 10^{-4}\ \text{F} = 5.69 \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. The energy stored when the potential difference across it is (V) is

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

where:

(W) is energy in joules (J),
(C) is capacitance in farads (F),
(V) is potential difference in volts (V).

In a smoothing circuit, the capacitor charges up to approximately the peak of the rectified input, so the maximum energy corresponds to the maximum capacitor voltage.

Understanding the Question

From the graphs:

(V_{IN}) is a square wave between (+12\ \text{V}) and (-12\ \text{V}), period (0.02\ \text{s}).
(V_{OUT}) is held at (12\ \text{V}) during the conducting interval.

You are told the maximum energy stored in the capacitor is (0.041\ \text{J}). You must show that this implies (C \approx 570\ \mu\text{F}).

Approach

Take the maximum capacitor voltage from the output waveform: (V_{\max} = 12\ \text{V}).
Rearrange (W = \tfrac{1}{2}CV^2) to find (C).
Convert farads to microfarads using (1\ \mu\text{F} = 10^{-6}\ \text{F}).

Step-by-Step Reasoning

Identify (V_{\max}): the capacitor is across the output, and the output reaches (12\ \text{V}), so (V_{\max} = 12\ \text{V}).
Rearrange:

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

Substitute (W = 0.041\ \text{J}), (V = 12\ \text{V}):

C = \frac{2(0.041)}{12^2} = \frac{0.082}{144} = 5.69 \times 10^{-4}\ \text{F}

Convert to (\mu\text{F}):

5.69 \times 10^{-4}\ \text{F} = 5.69 \times 10^{-4} \times 10^{6}\ \mu\text{F} = 5.69 \times 10^{2}\ \mu\text{F} \approx 570\ \mu\text{F}

Key Takeaways

Use (W = \tfrac{1}{2}CV^2) for capacitor energy.
Maximum stored energy corresponds to maximum voltage across the capacitor.
Be fluent with (\mu\text{F}\leftrightarrow\text{F}) conversions.

Common Mistakes

Using (V = 24\ \text{V}) (peak-to-peak) instead of the maximum capacitor voltage (12\ \text{V}).
Forgetting the (\tfrac{1}{2}) in the energy formula.
Incorrect conversion to microfarads (missing factor (10^6)).

Things to Be Careful About

(V) in (\tfrac{1}{2}CV^2) is the capacitor p.d., not the input peak-to-peak value.
Keep units consistent: joules, volts, farads.

Techniques used

use capacitor energy equation to relate energy, capacitance and voltageread the peak output voltage from the waveformconvert between farads and microfarads

(ii)

Determine the resistance of $R$ .

resistance = ______ $\Omega$

3M Similar

DifficultyMedium

Worked solution

Working

During discharge (from (t = 0.01\ \text{s}) to (0.02\ \text{s})), (V) falls from (12\ \text{V}) to (8\ \text{V}) in (\Delta t = 0.010\ \text{s}).

V = V_0 e^{-t/RC}

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

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

Using (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}} = 4.33 \times 10^{1}\ \Omega \approx 43\ \Omega

Answer

(R \approx 43\ \Omega).

Final answer

43 \Omega

Detailed explanation

Background Concept

When a capacitor discharges through a resistor (with no external supply connected), the voltage across the capacitor decays exponentially:

V = V_0 e^{-t/RC}

where:

(V_0) is the initial capacitor voltage at the start of the discharge,
(t) is time since discharge began,
(RC) is the time constant (\tau).

In a rectifier with smoothing capacitor, the capacitor charges rapidly when the diode conducts and discharges through the load resistor when the diode is reverse biased, causing the output to fall between charging intervals.

Understanding the Question

From the output graph (V_{OUT}):

It remains at (12\ \text{V}) up to (t = 0.01\ \text{s}).
Then it drops to (8\ \text{V}) by (t = 0.02\ \text{s}).

So, during the non-conducting half-cycle, the capacitor discharges for (0.010\ \text{s}), and the voltage falls from (12\ \text{V}) to (8\ \text{V}).

You have already found (C = 570\ \mu\text{F}), and you must determine (R).

Approach

Treat the interval where (V_{OUT}) falls as a capacitor discharge through (R).
Use (V = V_0 e^{-t/RC}) with (V_0 = 12\ \text{V}), (V = 8\ \text{V}), (t = 0.010\ \text{s}).
Solve for (RC), then divide by (C) to get (R).

Step-by-Step Reasoning

Identify the discharge time:
- Discharge starts when the diode stops conducting: at (t = 0.01\ \text{s}).
- Next charging begins at (t = 0.02\ \text{s}).
- So (t = 0.02 - 0.01 = 0.010\ \text{s}).
Identify (V_0) and (V):
- (V_0 = 12\ \text{V}) at the start of discharge.
- (V = 8\ \text{V}) after (0.010\ \text{s}).
Apply exponential decay:

\frac{V}{V_0} = e^{-t/RC}

so

\frac{8}{12} = e^{-0.010/(RC)}

Take natural logarithms:

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

and since (\ln(2/3) = -\ln(3/2)),

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

Use (R = (RC)/C) with (C = 570\ \mu\text{F} = 5.70 \times 10^{-4}\ \text{F}):

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

Note: although the graph shows a straight-line fall, in a real RC discharge the curve is exponential; exam questions typically still expect the exponential discharge equation.

Key Takeaways

Use (V = V_0 e^{-t/RC}) for capacitor discharge.
The time constant is (\tau = RC).
Read (t), (V_0), and (V) carefully from the waveform.

Common Mistakes

Using the full period (0.02\ \text{s}) as the discharge time instead of the discharge interval (0.01\ \text{s}).
Swapping the ratio (V/V_0) (e.g. using (12/8)) leading to a negative time constant.
Forgetting to convert (570\ \mu\text{F}) to farads.

Things to Be Careful About

(t) in the exponential formula is time since discharge began, not absolute clock time.
Always keep (RC) in seconds: (\ln) is dimensionless, so the numerator must be in seconds.
Quote (R) to a sensible number of significant figures consistent with the data (typically 2–3 s.f.).

Techniques used

read discharge interval and voltages from the output waveformapply the exponential discharge equation for a capacitorrearrange to determine the time constant and then resistance

Q7Medium-EasyMagnetic Fields

(a)

Define magnetic flux density.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

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

B = \frac{F}{IL}

Final answer

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

Detailed explanation

Background Concept

Magnetic flux density $B$ describes the “strength” of a magnetic field in terms of the force it can exert on moving charges, or equivalently on a current in a wire.

For a straight wire of length $L$ carrying current $I$ in a uniform magnetic field, the magnetic force is

F = BIL\sin\theta

where $\theta$ is the angle between the wire (direction of conventional current) and the magnetic field direction.

Understanding the Question

You are asked for a definition of magnetic flux density. In Cambridge mark schemes this is usually the operational definition based on the force on a current-carrying conductor. For full credit you must include:

force per unit current,
per unit length,
when the conductor is perpendicular to the field.

Approach

Start from the standard force equation on a current-carrying conductor. Set $\theta = 90^\circ$ so $\sin\theta = 1$ , then rearrange to express $B$ as a ratio involving $F$ , $I$ and $L$ . State this in words.

Step-by-Step Reasoning

Use

F = BIL\sin\theta

For the definition, the conductor is at right angles to the field: $\theta = 90^\circ$ , so $\sin\theta = 1$ .

F = BIL

Rearrange:

B = \frac{F}{IL}

Put into words: $B$ is the force per unit current per unit length on a conductor at $90^\circ$ to the field.

Key Takeaways

Magnetic flux density can be defined via a measurable force on a current.
The perpendicular condition matters because of the $\sin\theta$ factor.

Common Mistakes

Missing “per unit length” (writing $B = F/I$ ).
Forgetting the perpendicular condition (the definition uses maximum force).
Confusing $B$ with magnetic flux $\Phi$ .

Things to Be Careful About

The definition is tied to $\theta = 90^\circ$ ; otherwise $B = F/(IL\sin\theta)$ , which is not the standard definition statement.
Use conventional current direction when relating to force rules (e.g. Fleming’s left-hand rule).

Techniques used

recall the operational definition of magnetic flux densityapply the force equation for a current-carrying conductor in a magnetic fieldstate the required perpendicular condition 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.

3M Similar

(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.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Each wire produces a magnetic field around it.

Wire X lies in the magnetic field produced by wire Y (and vice versa), so a current-carrying wire in a magnetic field experiences a force (e.g. $F = BIL$ ). Hence the wires exert forces on each other.

Final answer

Each wire produces a magnetic field; the other wire’s current in this field experiences a force, so the wires exert forces on each other.

Detailed explanation

Background Concept

A current in a wire creates a magnetic field around the wire. If another current-carrying conductor is placed in this magnetic field, it experiences a force.

The magnitude for a straight conductor in a magnetic field is

F = BIL\sin\theta

where $\theta$ is the angle between current direction and the magnetic field.

In the special case of two long parallel wires, each wire is inside the magnetic field produced by the other, so forces arise on both wires.

Understanding the Question

Two parallel wires X and Y both carry current into the page. You must explain why there is a magnetic force between them (not necessarily calculate it). The essential points are:

a current produces a magnetic field,
a current in an external magnetic field experiences a force,
so each wire is forced by the field of the other.

Approach

Use a cause-and-effect chain: current $\Rightarrow$ magnetic field; other wire is in that field; magnetic field acting on a current $\Rightarrow$ force.

Step-by-Step Reasoning

Current in wire Y produces a magnetic field around Y.
Wire X is close by, so it is within that magnetic field.
Since X carries current, it experiences a force due to the magnetic field from Y.
Similarly, current in X produces a magnetic field that exerts a force on wire Y.
Therefore, the two wires exert magnetic forces on each other.

Key Takeaways

Magnetic forces between wires come from “field of one acting on current in the other”.

Common Mistakes

Saying “they attract because of magnetism” with no mechanism.
Mentioning electric forces instead of magnetic forces.

Things to Be Careful About

You do not need a calculation; just the correct physical reason.
Make clear it’s the other wire’s magnetic field that causes the force.

Techniques used

state that a current produces a magnetic field around a conductorapply the idea that a current-carrying wire in a magnetic field experiences a forcelink mutual forces to each wire being in the other wire's magnetic field

(ii)

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

1M Similar

(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.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The force on wire Y is equal in magnitude to the force on wire X (and opposite in direction).

Final answer

Same magnitude as on X (opposite direction).

Detailed explanation

Background Concept

When two objects interact, the forces they exert on each other are an action–reaction pair:

equal magnitude,
opposite direction,
acting on different bodies.

For two long parallel wires, the magnitude per unit length is

\frac{F}{L} = \frac{\mu_0 I_X I_Y}{2\pi r}

which is symmetric in $I_X$ and $I_Y$ , so each wire experiences the same magnitude force.

Understanding the Question

You are told $I_X = 2 I_Y$ . You are asked how the force on Y compares with the force on X.

Even though the currents differ, the interaction is mutual: the force of X on Y equals the force of Y on X in magnitude.

Approach

Use Newton’s third law (or the symmetry of the force formula) to compare forces.

Step-by-Step Reasoning

Wire X produces a magnetic field that acts on current in wire Y, producing a force on Y.
Wire Y produces a magnetic field that acts on current in wire X, producing a force on X.
These are forces of interaction between the same two bodies, so by Newton’s third law:
- $|F_{X\to Y}| = |F_{Y\to X}|$ .
Therefore, force on Y has the same magnitude as force on X, but acts in the opposite direction.

Key Takeaways

Forces between the two wires are always equal and opposite as an interaction pair.
Different currents do not break Newton’s third law.

Common Mistakes

Saying the force on X is double because $I_X$ is double (missing that the product $I_X I_Y$ applies to both, and that forces are equal and opposite).
Claiming both forces act in the same direction.

Things to Be Careful About

The question asks a comparison (not a calculation), so a clear statement about magnitude (and ideally direction) is enough.
If you mention direction: they are opposite because they act on different wires.

Techniques used

use Newton's third law for interaction forcescompare forces on two interacting current-carrying wiresstate relative magnitudes and directions

(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.

1M Similar

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.

1M Similar

(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 = ______ $\text{J}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At threshold, $E_{\text{photon}} = \phi$ so

\phi = hf_0

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

Answer

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

Final answer

5.8 × 10^-19 J

Detailed explanation

Background Concept

In the photoelectric effect, the minimum energy needed to remove an electron from the metal surface is the work function energy $\phi$ . Emission just begins when the photon energy equals the work function:

hf_0 = \phi

where $f_0$ is the threshold frequency.

Understanding the Question

You are told that electrons are emitted only if the frequency is at least $8.8 \times 10^{14}\ \text{Hz}$ . That is the threshold frequency $f_0$ . You must calculate the work function energy of magnesium in joules.

Approach

Use the threshold condition $\phi = hf_0$ . Substitute the given $f_0$ and Planck constant $h$ .

Step-by-Step Reasoning

At threshold, the emitted electrons have zero maximum kinetic energy, so all the photon energy goes into overcoming the work function:

\phi = hf_0

Substitute values:

\phi = (6.63 \times 10^{-34}\ \text{J s})(8.8 \times 10^{14}\ \text{s}^{-1})

Multiply the numbers and powers of ten:

\phi = 5.83 \times 10^{-19}\ \text{J}

Quote to appropriate significant figures (typically 2 s.f. from $8.8$ ):

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

Key Takeaways

Threshold frequency $f_0$ is linked to work function by $\phi = hf_0$ .
At threshold, $E_{\text{max}} = 0$ .

Common Mistakes

Using $\phi = h(f-f_0)$ (that expression gives $E_{\text{max}}$ , not $\phi$ ).
Forgetting units or writing $\text{eV}$ when the question demands joules.
Errors with powers of ten when multiplying $10^{-34}$ by $10^{14}$ .

Things to Be Careful About

$\text{Hz} = \text{s}^{-1}$ , so $\text{J s} \times \text{s}^{-1} = \text{J}$ .
Use the given threshold frequency ( $8.8 \times 10^{14}\ \text{Hz}$ ), not the lower limit ( $8.0 \times 10^{14}\ \text{Hz}$ ) mentioned later in the question.

Techniques used

use the threshold frequency condition for photoemissionapply the photon energy equationsubstitute numerical values with correct powers of ten

(ii)

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

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

3M Similar

DifficultyMedium

Worked solution

Working

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

E_{\text{max}} = (6.63 \times 10^{-34})\left[(11.0-8.8)\times 10^{14}\right] = 1.46 \times 10^{-19}\ \text{J}

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

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

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

Final answer

5.7 × 10^5 m s^-1

Detailed explanation

Background Concept

Einstein’s photoelectric equation relates photon energy to the work function and the maximum kinetic energy of emitted electrons:

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

At threshold frequency $f_0$ , $E_{\text{max}} = 0$ , so $\phi = hf_0$ . Combining gives

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

The maximum kinetic energy is also

E_{\text{max}} = \frac{1}{2}mv_{\text{max}}^2

with $m$ the electron mass.

Understanding the Question

The incident frequency is $11 \times 10^{14}\ \text{Hz}$ , which is above the threshold $8.8 \times 10^{14}\ \text{Hz}$ . So electrons are emitted, and you must find the maximum speed (not the average speed). That comes from the maximum kinetic energy.

Approach

Use $E_{\text{max}} = h(f-f_0)$ to find the maximum kinetic energy.
Convert kinetic energy to speed using $E_{\text{max}} = \tfrac{1}{2}mv^2$ .

Step-by-Step Reasoning

Find the frequency difference:

\Delta f = f-f_0 = (11.0-8.8)\times 10^{14} = 2.2 \times 10^{14}\ \text{Hz}

Compute maximum kinetic energy:

E_{\text{max}} = h\Delta f = (6.63 \times 10^{-34})(2.2 \times 10^{14})

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

Use kinetic energy to find speed (electron mass $m = 9.11 \times 10^{-31}\ \text{kg}$ ):

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

v_{\text{max}} = 5.66 \times 10^{5}\ \text{m s}^{-1} \approx 5.7 \times 10^{5}\ \text{m s}^{-1}

Key Takeaways

For photoemission, $E_{\text{max}}$ depends on frequency difference: $E_{\text{max}} = h(f-f_0)$ .
Convert from kinetic energy to speed with $E_k = \tfrac{1}{2}mv^2$ .

Common Mistakes

Using $E_{\text{max}} = hf$ (forgetting to subtract the work function / threshold term).
Using the wrong mass (e.g. proton mass) or forgetting to use $m = 9.11 \times 10^{-31}\ \text{kg}$ .
Not taking the square root when solving for $v$ .

Things to Be Careful About

Ensure $f$ and $f_0$ are in Hz and written with consistent powers of ten before subtracting.
Quote speed to 2 or 3 significant figures, matching the input data.
The question asks for maximum speed, so you must use $E_{\text{max}}$ (not average kinetic energy).

Techniques used

apply the photoelectric equation to find maximum kinetic energycompute kinetic energy from frequency differenceuse 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

3M Similar

DifficultyMedium

Worked solution

Working

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

So $E_{\text{max}}=0$ for $f \le f_0$ and a straight line for $f \ge f_0$ .

At $f_0 = 8.8 \times 10^{14}\ \text{Hz}$ , $E_{\text{max}}=0$ .

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

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

Answer

diagram-8c

Final answer

Straight line: E_MAX = 0 up to f = 8.8×10^14 Hz, then linear increase; passes through (8.8,0) and (11.0,1.46) on the given scaled axes.

Detailed explanation

Background Concept

From Einstein’s photoelectric equation,

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

and using $\phi = hf_0$ , we get

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

This tells us two key graphical features:

There is a threshold frequency $f_0$ below which no electrons are emitted.
For frequencies above $f_0$ , $E_{\text{max}}$ increases linearly with $f$ with gradient $h$ .

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 of emitted electrons changes.

Given earlier, the threshold frequency is $f_0 = 8.8 \times 10^{14}\ \text{Hz}$ . So for part of the x-range (from $8.0$ to $8.8$ in units of $10^{14}\ \text{Hz}$ ) there should be no emission.

The axes are scaled as:

horizontal axis: $f / 10^{14}\ \text{Hz}$ (so you plot numbers like $8.8$ and $11.0$ )
vertical axis: $E_{\text{max}} / 10^{-19}\ \text{J}$ (so you plot numbers like $1.46$ for $1.46\times 10^{-19}\ \text{J}$ ).

Approach

Mark the threshold point where $E_{\text{max}}=0$ : at $f=8.8 \times 10^{14}\ \text{Hz}$ .
For $f < f_0$ , show no emission (so $E_{\text{max}}=0$ on the graph).
For $f > f_0$ , draw a straight line with positive slope. Use a second point (e.g. at $11 \times 10^{14}\ \text{Hz}$ ) to set the line correctly.

Step-by-Step Reasoning

Threshold point:

f_0 = 8.8 \times 10^{14}\ \text{Hz} \Rightarrow \left(\frac{f}{10^{14}\ \text{Hz}} = 8.8,\ \frac{E_{\text{max}}}{10^{-19}\ \text{J}} = 0\right)

Find a second point at $f=11 \times 10^{14}\ \text{Hz}$ :

E_{\text{max}} = h(f-f_0) = (6.63\times 10^{-34})\left[(11.0-8.8)\times 10^{14}\right]

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

So on the scaled axes this is the point $(11.0,\ 1.46)$ .

Sketch:

From $f=8.0$ to $f=8.8$ , draw the graph along the horizontal axis at $E_{\text{max}}=0$ .
From $f=8.8$ upward, draw a straight line through $(8.8,0)$ and $(11.0,1.46)$ .

diagram-8c

Key Takeaways

The photoelectric $E_{\text{max}}$ vs $f$ graph is a straight line with x-intercept at $f_0$ .
Below $f_0$ , there is no photoemission (so $E_{\text{max}}$ is zero on such exam sketches).

Common Mistakes

Drawing the straight line starting at $f=0$ (incorrect: threshold is not at the origin).
Giving a non-zero $E_{\text{max}}$ at $f=8.0 \times 10^{14}\ \text{Hz}$ even though $f < f_0$ .
Drawing a curve rather than a straight line for $f>f_0$ .

Things to Be Careful About

Use the axis scaling correctly: plot $8.8$ and $11.0$ (not $8.8\times 10^{14}$ and $11\times 10^{14}$ ) and plot $1.46$ (not $1.46\times 10^{-19}$ ).
Make sure the line hits $E_{\text{max}}=0$ exactly at $f=8.8$ .
If you extend the straight line back, it should intersect the frequency axis at $f=f_0$ (that is the key physical meaning of the threshold).

Techniques used

use the photoelectric equation to relate kinetic energy to frequencyidentify the threshold frequency on a graphsketch a straight-line relationship with correct intercept and gradient

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.

1M Similar

(ii)

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Half-life:

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 × 10^-4 s^-1

Detailed explanation

Background Concept

Radioactive decay is random, but for a large number of nuclei the decay rate is predictable. The decay constant (\lambda) is the probability per unit time that a nucleus decays.

Half-life (T_{1/2}) is the time for the number of undecayed nuclei (or activity) to fall to half. They are related by:

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

so

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

Understanding the Question

You are given the half-life of (^{18}_{9}\text{F}) as 110 minutes and asked to show that (\lambda = 1.05 \times 10^{-4}\ \text{s}^{-1}). So you must:

convert minutes to seconds, then
apply (\lambda = \ln 2 / T_{1/2}).

Approach

Convert (110\ \text{min}) to seconds because (\lambda) is required in (\text{s}^{-1}).
Substitute into (\lambda = \ln 2 / T_{1/2}).

Step-by-Step Reasoning

Unit conversion:

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

Use the half-life relation:

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

Substitute numbers:

\lambda = \frac{0.693}{6600} = 1.05 \times 10^{-4}\ \text{s}^{-1}.

This matches the value required.

Key Takeaways

Always convert (T_{1/2}) into seconds if (\lambda) is needed in (\text{s}^{-1}).
(\lambda) and (T_{1/2}) are connected by (\ln 2).

Common Mistakes

Forgetting to multiply by 60 (leaving (T_{1/2}) in minutes gives (\lambda) in (\text{min}^{-1})).
Using (\log_{10}) instead of (\ln).
Writing (\lambda = 2/T_{1/2}) or similar (incorrect).

Things to Be Careful About

(\ln 2 = 0.693) (not (0.301), which is (\log_{10} 2)).
Quote the final (\lambda) to appropriate significant figures and in standard form.

Techniques used

convert half-life into secondsuse the relation between decay constant and half-lifesubstitute values and evaluate in standard form

(iii)

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

activity = ______ $\text{Bq}$

3M Similar

DifficultyMedium

Worked solution

Working

Molar mass of (^{18}\text{F}): (M = 18\ \text{g mol}^{-1} = 18 \times 10^{-3}\ \text{kg mol}^{-1}).

n = \frac{m}{M} = \frac{2.1 \times 10^{-12}}{18 \times 10^{-3}} = 1.17 \times 10^{-10}\ \text{mol}

N = nN_A = (1.17 \times 10^{-10})(6.02 \times 10^{23}) = 7.0 \times 10^{13}

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

A = \lambda N = (1.05 \times 10^{-4})(7.0 \times 10^{13}) = 7.4 \times 10^{9}\ \text{Bq}

Answer

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

Final answer

7.4 × 10^9 Bq

Detailed explanation

Background Concept

Activity (A) is the number of decays per second, measured in becquerel ((\text{Bq})). It is related to the number of undecayed nuclei (N) by:

A = \lambda N

where (\lambda) is the decay constant in (\text{s}^{-1}).

To find (N) from a mass, use moles:

n = \frac{m}{M}, \qquad N = nN_A

where:

(m) is mass (in kg here)
(M) is molar mass (kg mol(^{-1}))
(N_A) is Avogadro constant (6.02 \times 10^{23}\ \text{mol}^{-1}).

Understanding the Question

You are given a small mass of fluorine-18: (2.1 \times 10^{-12}\ \text{kg}). You must find its activity, using the decay constant found earlier ((1.05 \times 10^{-4}\ \text{s}^{-1})).

So you need:

number of nuclei (N) in that mass
apply (A = \lambda N).

Approach

Convert the mass into moles using the molar mass of (^{18}\text{F}) (18 g per mole).
Convert moles into number of nuclei with (N_A).
Multiply by (\lambda) to get activity.

Step-by-Step Reasoning

Use (M = 18\ \text{g mol}^{-1} = 18 \times 10^{-3}\ \text{kg mol}^{-1}).
Find moles:

n = \frac{m}{M} = \frac{2.1 \times 10^{-12}}{18 \times 10^{-3}} = 1.17 \times 10^{-10}\ \text{mol}.

Find number of nuclei:

N = nN_A = (1.17 \times 10^{-10})(6.02 \times 10^{23}) \approx 7.0 \times 10^{13}.

Use (A = \lambda N):

A = (1.05 \times 10^{-4})(7.0 \times 10^{13}) \approx 7.4 \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 (\lambda) and how many nuclei are present.
To get (N) from mass, you must go via moles and Avogadro’s constant.

Common Mistakes

Using (M = 18\ \text{kg mol}^{-1}) instead of (18 \times 10^{-3}\ \text{kg mol}^{-1}) (factor of (10^3) error).
Forgetting (A = \lambda N) and using (A = N/\lambda) (inverted).
Confusing (^{18}\text{F}) with fluorine’s relative atomic mass (19) (would give a small but mark-losing difference).

Things to Be Careful About

Keep units consistent: kg with kg mol(^{-1}), so (n) comes out in mol.
Quote the activity to 2–3 significant figures and include (\text{Bq}).
Ensure you use (\lambda) in (\text{s}^{-1}), not (\text{min}^{-1}).

Techniques used

convert mass to amount of substance using molar masscalculate number of nuclei using Avogadro's constantuse A = lambda N to determine activity

(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.

3M Similar

DifficultyMedium

Worked solution

Answer

The emitted positron interacts with an electron and they annihilate.

Two (\gamma)-photons (each about (511\ \text{keV})) are produced and travel in opposite directions.

Detectors outside the body detect the two (\gamma) photons in coincidence; the line along which they are detected locates the decay position, allowing an image of tracer distribution to be built up.

Final answer

Positron annihilation with electron produces two opposite 511 keV gamma photons; coincidence detection locates decay position to form an image.

Detailed explanation

Background Concept

In PET (positron emission tomography), a radioactive tracer emits positrons. A positron is antimatter, so when it meets an electron, they can undergo annihilation:

their mass-energy is converted into electromagnetic energy,
typically producing two gamma photons.

Because momentum must be conserved, the two photons are emitted approximately back-to-back (opposite directions). Each photon has energy (511\ \text{keV}), corresponding to the electron/positron rest energy (m_ec^2).

The PET scanner has detectors arranged around the body. If two detectors register gamma photons at the same time (a coincidence event), the source lies somewhere along the straight line joining those detectors.

Understanding the Question

You are asked to describe how a (\beta^+) particle (positron) interacting with an electron in the body enables image formation. So the key steps are:

positron meets electron
annihilation produces gamma photons
detection system uses those photons to locate where the decay occurred
many events build up an image of tracer concentration.

Approach

Describe the physical chain from decay to image:

(\beta^+) emission → positron travels a short distance → annihilation → two gamma photons → coincidence detection → reconstruction of tracer distribution.

Step-by-Step Reasoning

Emission and slowing down: The fluorine-18 nucleus emits a positron. In tissue, the positron quickly loses kinetic energy through collisions.
Annihilation: The positron meets an electron and they annihilate.
Gamma production: The annihilation produces two gamma photons, each with energy about (511\ \text{keV}).
Opposite directions: The photons travel in opposite directions (approximately (180^{\circ}) apart) so that momentum is conserved.
Detection and localisation: Detectors around the body detect pairs of gamma photons arriving at the same time. The pair defines a line of response, indicating where the annihilation happened.
Image formation: Repeating for many decays allows a computer to map where annihilations occur most often, producing an image of tracer distribution.

Key Takeaways

PET imaging uses annihilation gamma rays, not the positron directly.
Two opposite gamma photons + coincidence timing gives positional information.

Common Mistakes

Saying one gamma photon is produced (typically two are expected for PET).
Not linking detection to localisation (must explain how an image is built).
Confusing PET with gamma-camera imaging that detects gamma rays emitted directly by the nucleus.

Things to Be Careful About

The key credited ideas are: annihilation, two gamma photons, opposite directions, and coincidence detection leading to location/image.
“Opposite directions” may be described as (180^{\circ}) apart or “back-to-back”.

Techniques used

describe positron-electron annihilationapply conservation ideas to gamma emission directionsexplain coincidence detection used to locate the tracer

(ii)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

110 min is long enough for the tracer to be prepared, administered and for measurements to be taken with sufficient activity.

It is short enough that the activity falls rapidly after the scan, so the patient’s radiation dose is reduced.

Final answer

Long enough for preparation/scan with detectable activity; short enough to reduce dose as it decays soon after.

Detailed explanation

Background Concept

A tracer must be radioactive enough to give a detectable signal, but not so long-lived that it keeps irradiating the patient unnecessarily.

The half-life controls how quickly the activity decreases:

long half-life → activity persists for a long time (higher total dose)
short half-life → activity drops quickly (lower dose) but may become too weak to detect.

Understanding the Question

The question asks why a half-life of 110 minutes is suitable for a medical tracer (like (^{18}\text{F}) used in PET). You need two sensible reasons, typically one practical and one safety-related.

Approach

Give two balanced points:

practical usability during the scanning time
reduced dose because it decays away reasonably quickly.

Step-by-Step Reasoning

Practical point: A PET procedure involves producing the tracer, transporting it (possibly within a hospital), injecting it, waiting for uptake, and then scanning. A half-life of 110 minutes means the activity does not halve too quickly during these steps, so the signal remains strong enough to detect.
Safety point: Because 110 minutes is not very long, the activity drops substantially within a few hours. This reduces the time the patient remains radioactive and lowers the radiation dose after diagnosis.

Key Takeaways

Suitable tracer half-life = compromise between detectability and minimising dose.
(^{18}\text{F}) (110 min) is short-lived enough for safety but long-lived enough for imaging logistics.

Common Mistakes

Saying only “short half-life is good” without mentioning that it must still be long enough for measurement.
Claiming that a long half-life is always safer (it is usually the opposite for dose accumulation).

Things to Be Careful About

The mark scheme usually expects two distinct reasons; don’t repeat the same idea in different words.
Avoid vague statements like “it is convenient”; specify what it is convenient for (time to prepare/inject/scan).

Techniques used

relate half-life to usable measurement timerelate half-life to patient radiation dosebalance practical and safety considerations

Q10Medium-EasyAstronomy and Cosmology

(a)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Redshift means spectral lines are shifted to longer wavelength, showing (Doppler) that the source is moving away.

Most galaxies in all directions show redshift, so they are receding from us.

More distant galaxies have larger redshift (greater recession speed), consistent with space itself expanding.

Final answer

Redshift shows galaxies are receding (Doppler shift to longer wavelength); since this occurs in all directions and increases with distance, it implies the Universe is expanding.

Detailed explanation

Background Concept

Redshift is an increase in the observed wavelength of a spectral line compared with its emitted (laboratory) wavelength:

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

For relatively small speeds compared with the speed of light $c$ , the Doppler interpretation gives

\frac{v}{c} \approx \frac{\Delta \lambda}{\lambda}

So a positive $\Delta \lambda$ (redshift) corresponds to recession (moving away).

In cosmology, the key observation is not just that some objects are redshifted, but that almost all distant galaxies are redshifted, and the redshift increases systematically with distance.

Understanding the Question

You are asked to explain how observing redshift in light from galaxies leads scientists to conclude that the Universe is expanding.

So you need to connect:

spectral lines shifted to longer wavelengths,
interpretation as recession,
the pattern across many galaxies (in all directions, and stronger for more distant galaxies),
to the idea of an expanding Universe.

Approach

Make three clear points (typical 3-mark structure):

Define what redshift means in terms of wavelength.
State the Doppler/recession interpretation (galaxy moving away).
Use the global trend (many galaxies, in all directions; larger redshift for larger distance) to infer expansion of space.

Step-by-Step Reasoning

Observation: When we compare known spectral lines (e.g. from hydrogen) with those from a galaxy, the galaxy’s lines appear at a longer wavelength. That is a redshift.
Physical meaning: For waves, a longer observed wavelength occurs if the source is receding, which is a Doppler effect. So the galaxy is moving away from the Earth.
Cosmological inference: When this is seen for very many galaxies in every direction, it indicates that galaxies are generally receding from each other. Furthermore, the fact that more distant galaxies have larger redshift means they recede faster, matching Hubble’s law ( $v \propto d$ ). This pattern is naturally explained if the scale of the Universe is increasing with time: the Universe is expanding.

Key Takeaways

Redshift = longer observed wavelength.
Redshift usually indicates recession.
The universal, distance-dependent recession of galaxies implies expansion.

Common Mistakes

Saying “redshift means the galaxy is moving faster” without stating it is moving away.
Mentioning Doppler shift but not connecting it to the pattern for many galaxies.
Claiming it proves “the Earth is at the centre”; expansion does not require a central point.

Things to Be Careful About

Use the language of recession and increasing redshift with distance.
Keep the explanation qualitative; no calculation is required here.

Techniques used

relate spectral-line redshift to Doppler recessionlink recession in all directions to expansion of spacestate the distance-dependence of redshift/recession speed

(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

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using

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

d = \sqrt{\frac{L}{4\pi F}} = \sqrt{\frac{1.90 \times 10^{36}}{4\pi (8.42 \times 10^{-16})}}

d = 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 $F$ is the power received per unit area (in $\text{W m}^{-2}$ ).

If the source radiates uniformly in all directions, the power spreads out over the surface area of a sphere of radius $d$ :

\text{area} = 4\pi d^2

So the flux at distance $d$ is

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

This is an inverse-square law: doubling $d$ makes $F$ four times smaller.

Understanding the Question

You are given the galaxy’s luminosity $L = 1.90 \times 10^{36}\ \text{W}$ and the observed flux on Earth $F = 8.42 \times 10^{-16}\ \text{W m}^{-2}$ . You must determine the distance $d$ .

Approach

Rearrange the inverse-square relation for $d$ :

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

Then substitute the values (already in SI units).

Step-by-Step Reasoning

Start with

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

Rearrange:

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

Substitute:

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

Compute the denominator first: $4\pi(8.42\times 10^{-16}) \approx 1.06 \times 10^{-14}$ .

Then

\frac{1.90 \times 10^{36}}{1.06 \times 10^{-14}} \approx 1.80 \times 10^{50}

Square root:

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

Key Takeaways

Use $F = L/(4\pi d^2)$ for isotropic emission.
Rearranging introduces a square root.
Keep units in $\text{W}$ and $\text{W m}^{-2}$ to get $d$ in $\text{m}$ .

Common Mistakes

Forgetting the $4\pi$ factor.
Using $F = L/(\pi d^2)$ or $F = L/d^2$ .
Not taking the square root at the end.

Things to Be Careful About

Powers of ten: dividing by $10^{-16}$ increases the power by $10^{16}$ .
Quote the final distance in standard form and with unit $\text{m}$ .

Techniques used

apply the inverse-square law for radiant flux intensityrearrange for distanceuse consistent SI units and standard form

(ii)

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

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\frac{v}{c} \approx \frac{\Delta \lambda}{\lambda} = \frac{726-658}{658}

\frac{v}{c} = 0.103 qquad\Rightarrow\qquad v = 0.103 \times 3.00 \times 10^8

v = 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 spectral line has a known emitted wavelength $\lambda_{\text{em}}$ . If the source is moving relative to the observer, the observed wavelength $\lambda_{\text{obs}}$ is shifted.

The redshift is

z = \frac{\lambda_{\text{obs}} - \lambda_{\text{em}}}{\lambda_{\text{em}}}

For speeds small compared with $c$ , the Doppler approximation gives

\frac{v}{c} \approx z = \frac{\Delta \lambda}{\lambda}

A positive $\Delta \lambda$ (redshift) means recession.

Understanding the Question

Line X is emitted at $658\ \text{nm}$ but observed at $726\ \text{nm}$ . You are asked to determine the galaxy’s speed relative to Earth.

Approach

Find the fractional shift $\Delta \lambda/\lambda$ .
Multiply by $c$ to get $v$ .

Step-by-Step Reasoning

Compute the wavelength change:

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

Compute the fractional change:

\frac{\Delta \lambda}{\lambda} = \frac{68}{658} = 0.103

Use $v/c \approx \Delta \lambda/\lambda$ :

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

Key Takeaways

Redshift comes from comparing observed and emitted wavelengths.
For small redshift, $v \approx cz$ .
Redshift implies motion away from the observer.

Common Mistakes

Using $\lambda_{\text{obs}}$ in the denominator instead of $\lambda_{\text{em}}$ .
Forgetting to subtract wavelengths (using $726/658$ directly without minus 1).
Mixing nm and m unnecessarily (the ratio is unitless, so nm cancels).

Things to Be Careful About

This uses the non-relativistic approximation; at higher redshift you would need a relativistic formula. For $z \approx 0.10$ this approximation is typically accepted at this level.
Quote the answer in $\text{m s}^{-1}$ and standard form.

Techniques used

calculate redshift using wavelength changeapply the (non-relativistic) Doppler approximation v/c ≈ Δλ/λuse standard form for the final speed

(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

2M Similar

(ii)

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

gradient = ______

1M Similar

Physics 9702/41 — October/November 2024

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question