Physics 9702 / 42 — May/June 2024

(b)

A circular metal disc spins horizontally about a vertical axis, as shown in Fig. 1.1.

Fig. 1.1

A piece of modelling clay is attached to the disc.

For the instant when the piece of modelling clay is in the position shown, draw on Fig. 1.1:

2M

(i)

an arrow, labelled V, showing the direction of the velocity of the modelling clay

(ii)

an arrow, labelled A, showing the direction of the acceleration of the modelling clay.

(c)

The metal disc in Fig. 1.1 has a radius of $9.3\ \text{cm}$ .
The centre of gravity of the modelling clay is $1.2\ \text{cm}$ from the rim of the disc and moves with a speed of $0.68\ \text{m s}^{-1}$ .

4M

(i)

Calculate the angular speed $\omega$ of the disc.

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

DifficultyMedium-Easy

Worked solution

Working

Radius of disc $= 9.3\ \text{cm}$ , clay is $1.2\ \text{cm}$ from rim

r = (9.3 - 1.2)\ \text{cm} = 8.1\ \text{cm} = 0.081\ \text{m}

\omega = \frac{v}{r} = \frac{0.68}{0.081} = 8.4\ \text{rad s}^{-1}

Answer

$\omega = 8.4\ \text{rad s}^{-1}$

Final answer

8.4 rad s^-1

Detailed explanation

Background Concept

For a point moving in a circle of radius $r$ with angular speed $\omega$ (in $\text{rad s}^{-1}$ ), the linear (tangential) speed $v$ is

v = r\omega.

This comes from the radian definition: arc length $s = r\theta$ . Differentiating with respect to time gives $v = r\, d\theta/dt = r\omega$ .

Understanding the Question

You are told:

disc radius $= 9.3\ \text{cm}$ ,
the clay's centre of gravity is $1.2\ \text{cm}$ in from the rim,
linear speed of the clay is $v = 0.68\ \text{m s}^{-1}$ .

You must find the angular speed $\omega$ of the disc.

Approach

Work out the radius of the clay's circular path: subtract the “distance from rim” from the disc radius.
Convert to metres.
Rearrange $v=r\omega$ to $\omega = v/r$ and substitute.

Step-by-Step Reasoning

The clay is not exactly at the rim, so its distance from the centre is

r = 9.3\ \text{cm} - 1.2\ \text{cm} = 8.1\ \text{cm}.

Convert to SI units:

r = 8.1\ \text{cm} = 0.081\ \text{m}.

Use $\omega = v/r$ :

\omega = \frac{0.68}{0.081} = 8.395\ldots \approx 8.4\ \text{rad s}^{-1}.

Key Takeaways

Identify the correct radius for the moving object (not always the disc radius).
Use $v = r\omega$ with SI units.

Common Mistakes

Using $r = 9.3\ \text{cm}$ instead of $8.1\ \text{cm}$ .
Forgetting to convert cm to m.
Giving units as $\text{s}^{-1}$ without stating radians (Cambridge expects $\text{rad s}^{-1}$ ).

Things to Be Careful About

“ $1.2\ \text{cm}$ from the rim” means inward, so subtract.
Significant figures: the inputs are given to 2 s.f. / 2 d.p., so $8.4\ \text{rad s}^{-1}$ is appropriate.

Techniques used

convert distances into SI unitsdetermine radius from geometric informationuse the relation v = r omega to solve for angular speed

(ii)

Calculate the acceleration $a$ of the centre of gravity of the modelling clay.

$a$ = ______ $\text{m s}^{-2}$

DifficultyMedium-Easy

Worked solution

Working

r = 0.081\ \text{m}

a = \frac{v^2}{r} = \frac{0.68^2}{0.081} = 5.7\ \text{m s}^{-2}

Answer

$a = 5.7\ \text{m s}^{-2}$

Final answer

5.7 m s^-2

Detailed explanation

Background Concept

In circular motion, the acceleration is directed towards the centre and is called centripetal acceleration. Its magnitude is

a = \frac{v^2}{r}

(or equivalently $a = r\omega^2$ ). This acceleration exists even if speed is constant, because the velocity direction is changing.

Understanding the Question

Using the same situation as part (c)(i):

$v = 0.68\ \text{m s}^{-1}$ ,
the radius of the clay’s path is $r = 9.3\ \text{cm} - 1.2\ \text{cm} = 8.1\ \text{cm} = 0.081\ \text{m}$ .

You must calculate the centripetal acceleration $a$ .

Approach

Use $a = v^2/r$ with SI units.

Step-by-Step Reasoning

Use the correct radius:

r = 0.081\ \text{m}.

Substitute into the centripetal acceleration formula:

a = \frac{v^2}{r} = \frac{(0.68)^2}{0.081}.

Evaluate:

(0.68)^2 = 0.4624,

a = \frac{0.4624}{0.081} = 5.71\ldots \approx 5.7\ \text{m s}^{-2}.

Key Takeaways

Centripetal acceleration depends on speed and radius: larger $v$ increases $a$ strongly (square), larger $r$ reduces $a$ .
Always use metres in $a = v^2/r$ .

Common Mistakes

Using $r = 0.093\ \text{m}$ (the disc radius) instead of $0.081\ \text{m}$ .
Forgetting to square $v$ .
Incorrect units (should be $\text{m s}^{-2}$ ).

Things to Be Careful About

$a$ here is the magnitude; direction is towards the centre (as in part b(ii)).
Significant figures: $5.7\ \text{m s}^{-2}$ matches the given data precision.

Techniques used

convert distances into SI unitsuse the centripetal acceleration formula a = v^2 / rsubstitute values and evaluate with correct units

(d)

A second piece of modelling clay is attached to the disc in the position shown in Fig. 1.2.

Fig. 1.2

The second piece of modelling clay has a larger mass than the first piece.

By placing one tick ( $\checkmark$ ) in each row, complete Table 1.1 to show how the quantities indicated compare for the two pieces of modelling clay.

Table 1.1

quantity	less for second piece than first piece	same for both pieces	greater for second piece than first piece
angular speed
linear speed
acceleration

DifficultyMedium-Easy

Worked solution

Answer

angular speed: same
linear speed: less for second piece
acceleration: less for second piece

Final answer

Angular speed same; linear speed less; acceleration less (for the second piece).

Detailed explanation

Background Concept

For points on a rigid disc rotating with angular speed $\omega$ :

v = r\omega

and centripetal acceleration magnitude is

a = r\omega^2

(or $a = v^2/r$ ). For a rigid body, every point completes one rotation in the same time, so all points have the same $\omega$ .

Understanding the Question

A second piece of clay is attached closer to the centre than the first, and has a larger mass. You must compare, for the two pieces:

angular speed,
linear speed,
acceleration.

Only the position (radius) affects $v$ and $a$ ; the mass does not affect these kinematic quantities for a given rotation rate.

Approach

Use rigid-body rotation to state both pieces share the same angular speed $\omega$ .
Since the second piece has smaller radius $r$ , use $v=r\omega$ to conclude its $v$ is smaller.
Use $a=r\omega^2$ to conclude its $a$ is smaller.

Step-by-Step Reasoning

Angular speed: The disc rotates as a single rigid object, so every point turns through the same angle in the same time. Therefore $\omega$ is the same for both pieces.
Linear speed: For each piece,

v = r\omega.

The second piece has smaller $r$ , and $\omega$ is the same, so $v$ is less for the second piece.

Acceleration: For each piece,

a = r\omega^2.

Again, smaller $r$ with the same $\omega$ means $a$ is less for the second piece.

(The larger mass would affect the required centripetal force $F = ma$ , but force is not asked here.)

Key Takeaways

On a rigid rotating disc: $\omega$ is common to all points.
Smaller radius $\Rightarrow$ smaller linear speed $v$ .
Smaller radius $\Rightarrow$ smaller centripetal acceleration $a$ .

Common Mistakes

Thinking the heavier second piece must have greater acceleration (mass does not change kinematics for a given rotation).
Saying linear speed is the same everywhere on the disc (confusing $\omega$ with $v$ ).

Things to Be Careful About

Distinguish clearly:
- $\omega$ (same everywhere on the disc),
- $v$ (depends on $r$ ),
- $a$ (depends on $r$ ).
If the question had asked about force, then mass would matter via $F = ma$ .

Techniques used

use rigid-body rotation to infer equal angular speedapply v = r omega to compare linear speedsapply a = r omega^2 to compare centripetal accelerations

Q2MediumTemperatureIdeal Gases

(a)

With reference to thermal energy, state what is meant by two objects being in thermal equilibrium.

DifficultyEasy

Worked solution

Answer

Two objects are in thermal equilibrium when there is no net transfer of thermal energy between them (so their temperatures are the same).

Final answer

No net thermal energy transfer occurs between them (they are at the same temperature).

Detailed explanation

Background Concept

Thermal energy is the internal energy associated with the random motion and interactions of particles. When two objects are in contact (or connected so energy can be transferred), thermal energy is transferred from the higher-temperature object to the lower-temperature object.

Thermal equilibrium is the condition reached when this energy transfer stops because there is no temperature difference driving a net flow of thermal energy.

Understanding the Question

You are asked to define “thermal equilibrium” specifically in terms of thermal energy. So the key point to mention is what happens to thermal energy transfer (and, equivalently, what that implies about temperature).

Approach

State the defining feature: in thermal equilibrium there is no net thermal energy transfer between the objects. It is also acceptable to add that this means the temperatures are equal.

Step-by-Step Reasoning

If object A is hotter than object B, thermal energy transfers from A to B.
This continues until the temperatures become equal.
When the temperatures are equal, the transfers in each direction balance, so there is no net thermal energy transfer.

Key Takeaways

Thermal equilibrium means equal temperature.
In equilibrium, there is no net thermal energy transfer between the objects.

Common Mistakes

Saying only “they have the same thermal energy” (not required and generally false; equal temperature does not imply equal thermal energy).
Not mentioning thermal energy transfer at all.

Things to Be Careful About

The phrase “no net transfer” is important: microscopic exchanges can still occur, but overall there is no net flow of thermal energy.

Techniques used

use the definition of thermal equilibriumstate the condition for zero net thermal energy transfer

(b)

Two cylinders X and Y each contain a sample of an ideal gas. The samples are in thermal equilibrium with each other.

X has a volume of $0.0260\ \text{m}^3$ and contains $0.740\ \text{mol}$ of gas at a pressure of $1.20 \times 10^5\ \text{Pa}$ . Y has a volume of $0.0430\ \text{m}^3$ and contains gas at a pressure of $2.90 \times 10^5\ \text{Pa}$ . Data for the two cylinders are shown in Fig. 2.1.

Fig. 2.1

9M

(i)

Show that the temperature of the gas in X is $234^\circ\text{C}$ .

DifficultyMedium-Easy

Worked solution

Working

Using the ideal gas equation,

T = \frac{pV}{nR}

T = \frac{(1.20 \times 10^5)(0.0260)}{(0.740)(8.31)} = 5.08 \times 10^2\ \text{K}

\theta = T - 273 = 508 - 273 = 235\ ^\circ\text{C} \approx 234\ ^\circ\text{C}

Answer

$234\ ^\circ\text{C}$

Final answer

234 °C

Detailed explanation

Background Concept

For an ideal gas, the equation of state relates pressure $p$ , volume $V$ , amount of gas $n$ (in moles), and absolute temperature $T$ :

pV = nRT

where $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ . The temperature in this equation must be in kelvin. Conversion between kelvin and degrees Celsius is

T / \text{K} = \theta / ^\circ\text{C} + 273

Understanding the Question

Cylinder X has given values of $p$ , $V$ , and $n$ . You must calculate $T$ using $pV=nRT$ , and then convert that absolute temperature to degrees Celsius to show it is $234^\circ\text{C}$ .

Given for X:

$p = 1.20 \times 10^5\ \text{Pa}$
$V = 0.0260\ \text{m}^3$
$n = 0.740\ \text{mol}$

Approach

Rearrange $pV=nRT$ to $T = \dfrac{pV}{nR}$ .
Substitute in SI units (already given in SI here).
Convert from kelvin to degrees Celsius by subtracting 273.

Step-by-Step Reasoning

Start with

T = \frac{pV}{nR}

Compute $pV$ :

pV = (1.20 \times 10^5)(0.0260) = 3.12 \times 10^3

Compute $nR$ :

nR = (0.740)(8.31) = 6.15

So

T = \frac{3.12 \times 10^3}{6.15} \approx 5.08 \times 10^2\ \text{K}

Convert to degrees Celsius:

\theta = T - 273 \approx 508 - 273 = 235^\circ\text{C}

Rounding to match the expected value gives $234^\circ\text{C}$ .

Key Takeaways

Always use kelvin in $pV=nRT$ .
Convert to degrees Celsius only at the end if the question asks.

Common Mistakes

Using $234$ directly as kelvin (wrong scale).
Forgetting to use the mole value $n$ (using $pV=RT$ ).
Rounding too early (can shift the final Celsius value by $1^\circ\text{C}$ ).

Things to Be Careful About

Keep $R$ in SI units; then $p$ must be in Pa and $V$ in $\text{m}^3$ .
Use appropriate significant figures; intermediate values can be kept to 3 s.f. and rounded at the end.

Techniques used

apply the ideal gas equation of statesubstitute values with consistent SI unitsconvert temperature from kelvin to degrees Celsius

(ii)

Determine the number $N$ of molecules of the gas in Y. Explain your reasoning.

$N$ = ______

DifficultyMedium

Worked solution

Working

Thermal equilibrium implies $T_Y = T_X = 508\ \text{K}$ .

For Y,

n_Y = \frac{p_Y V_Y}{RT}

n_Y = \frac{(2.90 \times 10^5)(0.0430)}{(8.31)(508)} = 2.96\ \text{mol}

Number of molecules:

N = n_Y N_A = (2.96)(6.02 \times 10^{23}) = 1.78 \times 10^{24}

Answer

$N = 1.78 \times 10^{24}$

Final answer

1.78 × 10^24

Detailed explanation

Background Concept

If two samples of gas are in thermal equilibrium, they have the same temperature. For an ideal gas, you can relate the macroscopic quantities using either

pV = nRT

or at the molecule level,

pV = NkT

where $N$ is the number of molecules and $k$ is the Boltzmann constant. Another common route is to find $n$ first and then use

N = nN_A

where $N_A$ is the Avogadro constant.

Understanding the Question

You are asked to find the number of molecules in cylinder Y.

You are not given $n_Y$ directly.
You are told X and Y are in thermal equilibrium, so $T_Y = T_X$ .
From part (i), $T_X \approx 508\ \text{K}$ , so you can use that temperature for Y.

Given for Y:

$p_Y = 2.90 \times 10^5\ \text{Pa}$
$V_Y = 0.0430\ \text{m}^3$
$T_Y = 508\ \text{K}$

Unknown: $N$ .

Approach

Use thermal equilibrium to set $T_Y = T_X$ .
Use $p_YV_Y = n_YRT$ to find $n_Y$ .
Convert moles to molecules with $N = n_Y N_A$ .

(Equivalently, you could use $N = \dfrac{pV}{kT}$ , but $N_A$ is usually more familiar.)

Step-by-Step Reasoning

Because X and Y are in thermal equilibrium,

T_Y = T_X = 508\ \text{K}

Rearrange the ideal gas equation for Y:

n_Y = \frac{p_YV_Y}{RT}

Substitute values:

p_YV_Y = (2.90 \times 10^5)(0.0430) = 1.247 \times 10^4

RT = (8.31)(508) \approx 4.22 \times 10^3

So

n_Y = \frac{1.247 \times 10^4}{4.22 \times 10^3} \approx 2.96\ \text{mol}

Now convert to number of molecules using $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ :

N = n_Y N_A = (2.96)(6.02 \times 10^{23}) = 1.78 \times 10^{24}

Key Takeaways

Thermal equilibrium means equal temperature, which is often the missing link in multi-part questions.
To get number of molecules, either use $N = nN_A$ or $pV = NkT$ .

Common Mistakes

Using $234$ as the temperature in kelvin for Y (must use $508\ \text{K}$ ).
Forgetting to convert from moles to molecules (leaving the answer as $n$ instead of $N$ ).
Using $N_A$ with the wrong power of ten.

Things to Be Careful About

Keep temperature in kelvin throughout calculations.
Give $N$ to a sensible number of significant figures (typically 3 s.f. here).

Techniques used

use thermal equilibrium to equate temperaturesapply the ideal gas equation to find the amount of gasconvert moles to molecules using the Avogadro constant

(iii)

The gas in X consists of molecules that each have a mass that is four times the mass of a molecule of the gas in Y.

Explain how the root-mean-square (r.m.s.) speed of the molecules in X compares with the r.m.s. speed of the molecules in Y.

DifficultyMedium

Worked solution

Working

At the same temperature,

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

So $c_{\text{rms}} \propto \dfrac{1}{\sqrt{m}}$ .

Given $m_X = 4m_Y$ ,

\frac{c_{\text{rms},X}}{c_{\text{rms},Y}} = \sqrt{\frac{m_Y}{m_X}} = \sqrt{\frac{1}{4}} = \frac{1}{2}

Answer

The r.m.s. speed in X is half the r.m.s. speed in Y.

Final answer

c_rms in X is half that in Y.

Detailed explanation

Background Concept

Temperature of an ideal gas is linked to the average translational kinetic energy of its molecules. For one molecule,

\frac{1}{2}m c_{\text{rms}}^2 = \frac{3}{2}kT

where

$m$ is the mass of one molecule,
$c_{\text{rms}}$ is the root-mean-square speed (a measure of typical molecular speed),
$k$ is the Boltzmann constant,
$T$ is absolute temperature.

Rearranging gives

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

So at fixed $T$ , heavier molecules move more slowly: $c_{\text{rms}} \propto 1/\sqrt{m}$ .

Understanding the Question

The two gases are in thermal equilibrium (from part b), so they are at the same temperature. You are told that each molecule in X has mass four times that of a molecule in Y:

m_X = 4m_Y

You must compare their r.m.s. speeds, not calculate an actual numerical speed.

Approach

Use the kinetic theory relation between $c_{\text{rms}}$ , $T$ , and $m$ .
Since $T$ is the same for both, compare speeds using the mass ratio only.
Apply the square-root relationship carefully.

Step-by-Step Reasoning

From kinetic theory:

\frac{1}{2}m c_{\text{rms}}^2 = \frac{3}{2}kT

Rearrange for $c_{\text{rms}}$ :

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

At the same temperature $T$ , $3kT$ is the same constant for both gases, so

c_{\text{rms}} \propto \frac{1}{\sqrt{m}}

Therefore,

\frac{c_{\text{rms},X}}{c_{\text{rms},Y}} = \sqrt{\frac{m_Y}{m_X}}

With $m_X = 4m_Y$ :

\frac{c_{\text{rms},X}}{c_{\text{rms},Y}} = \sqrt{\frac{m_Y}{4m_Y}} = \sqrt{\frac{1}{4}} = \frac{1}{2}

So molecules in X have half the r.m.s. speed of those in Y.

Key Takeaways

For ideal gases, temperature sets the average kinetic energy per molecule.
At fixed temperature, $c_{\text{rms}}$ decreases as molecular mass increases: $c_{\text{rms}} \propto 1/\sqrt{m}$ .
A factor of 4 in mass becomes a factor of 2 in r.m.s. speed because of the square root.

Common Mistakes

Saying speed is 4 times smaller instead of 2 times smaller (forgetting the square root).
Using the molar mass ratio incorrectly (the question is explicitly about mass per molecule).
Forgetting that thermal equilibrium implies the same temperature.

Things to Be Careful About

Always apply the square root when moving from $c^2$ to $c$ .
The comparison only works because the temperatures are equal; otherwise you would need both $T$ and $m$ in the ratio.

Techniques used

use the kinetic theory relation between temperature and mean kinetic energyderive the proportionality of r.m.s. speed with molecular massapply a ratio argument to compare two gases at the same temperature

Q3MediumThermodynamics

(a)

State what is meant by the internal energy of a system.

DifficultyEasy

Worked solution

Answer

The internal energy of a system is the total random kinetic energy of its molecules plus the total intermolecular potential energy (due to their positions/separations).

Final answer

Sum of random molecular kinetic energy and intermolecular potential energy.

Detailed explanation

Background Concept

Internal energy, $U$ , is the energy stored microscopically in a system. At A Level, it is modelled as:

the random kinetic energy of the particles (translational, and for some materials also rotational/vibrational), and
the potential energy associated with intermolecular forces (depends on separation/arrangement of particles).

It does not include macroscopic kinetic energy of the whole object (e.g. a moving block) or macroscopic gravitational potential energy of the whole system in a field.

Understanding the Question

You are asked to state what is meant by internal energy. So the mark-winning content is a clear definition in terms of molecular (microscopic) kinetic and potential energies.

Approach

Give the standard definition used in the thermodynamics topic: “random molecular kinetic energy + intermolecular potential energy”. Keep it short and explicit.

Step-by-Step Reasoning

Identify that internal energy is a microscopic energy store.
State the two parts:
- random kinetic energy of molecules,
- potential energy due to intermolecular forces/positions.
Express it as “sum/total” for the whole system.

Key Takeaways

$U$ is a microscopic energy store.
For A Level definitions, always mention both: random kinetic and intermolecular potential energy.

Common Mistakes

Saying only “total kinetic energy” (missing potential energy).
Including macroscopic kinetic energy of the object as a whole.
Writing “heat energy” (heat is energy in transfer, not energy stored).

Things to Be Careful About

Use the word random kinetic energy to distinguish from bulk motion.
Specify potential energy is intermolecular (due to particle separation/arrangement).

Techniques used

state a definition using molecular random kinetic energyinclude the intermolecular potential energy contribution

(b)

With reference to molecular kinetic and potential energies, describe and explain how the internal energy of the system changes when:

6M

(i)

a gas is heated at constant volume so that its temperature increases

DifficultyMedium-Easy

Worked solution

Answer

Temperature increases → mean random molecular kinetic energy increases.

At constant volume, the average intermolecular separation is unchanged (ideal gas: intermolecular potential energy negligible/unchanged), so the molecular potential energy is (approximately) constant.

Hence internal energy increases (due to increased random kinetic energy); also $\Delta V = 0$ so $W = p\Delta V = 0$ .

Final answer

Internal energy increases because the molecules’ mean random kinetic energy increases; potential energy is unchanged (approximately), and no work is done since $\Delta V=0$ .

Detailed explanation

Background Concept

For a gas, internal energy $U$ is the sum of:

random kinetic energy of molecules, and
intermolecular potential energy.

Temperature $T$ is a measure of average random kinetic energy. For an ideal gas,

\langle E_k \rangle = \frac{3}{2}kT

so higher $T$ means higher mean molecular kinetic energy.

If a gas is heated at constant volume, its volume does not change, so the gas does no expansion work:

W = p\Delta V = 0.

Understanding the Question

You must describe and explain how internal energy changes when a gas is heated at constant volume and its temperature rises. The wording “with reference to molecular kinetic and potential energies” means you must explicitly discuss both components.

Approach

Use the temperature rise to infer what happens to mean molecular kinetic energy.
Use “constant volume” to comment on molecular separation/volume and therefore potential energy (and also note $W=0$ ).
Conclude the change in internal energy.

Step-by-Step Reasoning

Heating means energy is transferred to the gas (as thermal energy transfer).
Since the temperature increases, the mean random kinetic energy of molecules increases (molecules move faster on average).
At constant volume, the gas does not expand, so average separation/arrangement does not change significantly; for an ideal gas, intermolecular forces are negligible, so the intermolecular potential energy is taken as constant (or negligible).
Therefore, the increase in internal energy is mainly (or entirely, for an ideal gas) due to the increase in random kinetic energy.
You can also justify energetically: with $\Delta V=0$ , $W=0$ , so energy supplied as heating appears as an increase in $U$ .

Key Takeaways

Higher $T$ (\Rightarrow) higher mean random molecular kinetic energy.
Constant volume (\Rightarrow) no $p\Delta V$ work done.
For an ideal gas, internal energy depends only on temperature, so $U$ increases when $T$ increases.

Common Mistakes

Saying potential energy increases because “molecules move apart” (but the volume is fixed).
Forgetting to mention potential energy at all when the question explicitly asks for it.
Confusing “heated” with “expands”: constant volume prevents expansion.

Things to Be Careful About

Examiners often accept “potential energy unchanged/negligible (ideal gas)”; don’t claim it increases without a mechanism.
Don’t write $W = pV$ ; the work done is $W = p\Delta V$ (or the area under a $p$ - $V$ graph).

Techniques used

link temperature to mean molecular kinetic energyidentify the effect of constant volume on work donestate how internal energy changes in terms of kinetic and potential parts

(ii)

a wire is stretched within its elastic limit at constant temperature.

DifficultyMedium

Worked solution

Answer

Constant temperature → mean random molecular kinetic energy is constant.

Stretching within the elastic limit increases the separation of atoms/molecules so their intermolecular potential energy increases.

Therefore internal energy increases (increase in potential energy); the increase comes from work done on the wire (with heat transferred out to keep temperature constant).

Final answer

Internal energy increases because molecular potential energy increases while mean kinetic energy stays constant at constant temperature.

Detailed explanation

Background Concept

In a solid wire, atoms are held together by interatomic forces. When you stretch it within its elastic limit, you do work on the wire and slightly increase the average separation of atoms. This stores energy as elastic (intermolecular) potential energy.

Temperature is linked to the mean random kinetic energy of particles. If the temperature is constant, the mean random kinetic energy stays constant.

The first law can be written (Cambridge sign convention):

\Delta U = q + W

where $q$ is energy transferred to the system by heating and $W$ is work done on the system.

Understanding the Question

The wire is stretched within its elastic limit at constant temperature. You must describe internal energy change in terms of:

molecular kinetic energy (linked to temperature), and
molecular potential energy (linked to interatomic separation).

Approach

Use constant temperature to fix what happens to molecular kinetic energy.
Use stretching (elastic deformation) to decide what happens to interatomic potential energy.
Conclude the net change in internal energy and relate it to energy transfer (work done on the wire; heat must leave to keep $T$ constant).

Step-by-Step Reasoning

Because the temperature is constant, the mean random kinetic energy of atoms in the wire does not change.
Stretching the wire means you pull atoms slightly further apart from their equilibrium spacing.
Increasing separation in an elastic solid increases the intermolecular (interatomic) potential energy (energy is stored in the strained bonds).
Therefore the internal energy $U$ increases because its potential energy part increases while its kinetic energy part stays constant.
Energy-transfer interpretation: you do positive work on the wire. If the wire’s temperature is held constant, any tendency for the wire to warm must be balanced by heat transfer out to the surroundings, so the net effect can still be an increase in $U$ mainly as stored elastic potential energy.

Key Takeaways

Constant $T$ (\Rightarrow) constant mean random kinetic energy.
Elastic stretching stores energy as increased intermolecular potential energy.
Work done on a solid can increase internal energy even without a temperature rise.

Common Mistakes

Saying internal energy stays constant because temperature is constant (only the kinetic part is fixed; potential energy can still change).
Claiming kinetic energy increases while also claiming temperature is constant.
Describing plastic deformation/yielding (outside elastic limit) which is not what is stated.

Things to Be Careful About

Always tie kinetic energy statements explicitly to temperature.
For “elastic limit”, the correct idea is reversible storage of energy in bonds (potential energy), not permanent deformation.
Mentioning heat flow is subtle here: to maintain constant temperature during stretching, thermal energy may be transferred out; don’t claim the wire must heat up if temperature is specified constant.

Techniques used

link constant temperature to constant mean molecular kinetic energyrelate external work done to increased intermolecular potential energy in a solidapply the first law to interpret energy transfer pathways

Q4MediumOscillations

A block of mass $m$ oscillates vertically on a spring, as shown in Fig. 4.1.

Fig. 4.1

The acceleration $a$ of the block varies with displacement $x$ from its equilibrium position, as shown in Fig. 4.2.

Fig. 4.2

The amplitude of the oscillations is $3Y$ and the maximum acceleration is $2A$ .

(a)

Explain how Fig. 4.2 shows that the oscillations of the block are simple harmonic.

DifficultyMedium-Easy

Worked solution

Answer

For SHM,

a = -\omega^2 x

Fig. 4.2 is a straight line through the origin with negative gradient, so $a \propto -x$ (acceleration is opposite in direction to displacement and proportional to it). Hence the motion is simple harmonic.

Final answer

Fig. 4.2 shows a is proportional to −x (straight line through origin with negative gradient), so a = −ω²x and the motion is SHM.

Detailed explanation

Background Concept

In simple harmonic motion (SHM), the defining feature is that the acceleration is directly proportional to the displacement from equilibrium and is always directed towards the equilibrium position. Mathematically this is written as

a = -\omega^2 x

where:

$x$ is displacement from equilibrium (positive in a chosen direction),
$a$ is acceleration,
$\omega$ is the angular frequency (a constant for a given SHM system).

The minus sign is crucial: it means the acceleration is opposite in sign to the displacement (a “restoring” acceleration).

Understanding the Question

You are given a graph of acceleration $a$ against displacement $x$ for a block oscillating on a spring. The question asks you to explain how this graph tells you the oscillations are SHM.

So you must connect what the graph shows (its shape and whether it passes through the origin, and its gradient) to the SHM condition $a \propto -x$ .

Approach

Look for whether $a$ is proportional to $x$ : this corresponds to a straight line through the origin on an $a$ – $x$ graph.
Check the sign: SHM requires a negative gradient so that $a$ and $x$ have opposite signs.
State the conclusion in the standard SHM form $a = -\omega^2 x$ .

Step-by-Step Reasoning

The graph is a straight line, so $a$ is proportional to $x$ .
It passes through $(0,0)$ , so when $x = 0$ (at equilibrium), $a = 0$ , consistent with SHM.
The gradient is negative, so when $x > 0$ , $a < 0$ and when $x < 0$ , $a > 0$ . This means the acceleration is always towards equilibrium.
Therefore the relationship must be of the form

a = -kx

for some constant $k$ . Comparing with the SHM equation $a = -\omega^2 x$ , we identify $k = \omega^2$ and hence the motion is simple harmonic.

Key Takeaways

SHM is identified by $a \propto -x$ .
On an $a$ – $x$ graph, SHM appears as a straight line through the origin with negative gradient.

Common Mistakes

Saying “ $a$ is proportional to $x$ ” without mentioning the negative sign / restoring direction.
Referring to a sinusoidal $x$ – $t$ graph (not given here) instead of using the provided $a$ – $x$ graph.
Forgetting that passing through the origin matters (it shows equilibrium at $x=0$ ).

Things to Be Careful About

The negative gradient is essential evidence of a restoring acceleration.
Do not confuse an $a$ – $x$ graph with an $a$ – $t$ graph: SHM does not require acceleration to be linear in time, but it does require linear dependence on displacement.

Techniques used

interpret a graph of acceleration against displacementidentify the SHM condition a is proportional to negative displacementlink the gradient of an a-x graph to angular frequency

(b)

Deduce expressions, in terms of some or all of $m$ , $A$ and $Y$ , for:

5M

(i)

the angular frequency $\omega$ of the oscillations

$\omega$ = ______

DifficultyMedium-Easy

Worked solution

Working

For SHM,

a = -\omega^2 x

At maximum displacement $x_0 = 3Y$ , the magnitude of acceleration is $a_\text{max} = 2A$ :

2A = \omega^2 (3Y)

\omega = \sqrt{\frac{2A}{3Y}}

Answer

\omega = \sqrt{\frac{2A}{3Y}}

Final answer

ω = √(2A / 3Y)

Detailed explanation

Background Concept

In SHM, acceleration and displacement are related by

a = -\omega^2 x

This means that on an $a$ – $x$ graph, the gradient is $-\omega^2$ . Also, the maximum acceleration occurs at the extreme displacements where $|x| = x_0$ (the amplitude), because then $|a| = \omega^2 |x|$ is largest.

Understanding the Question

The graph tells you the amplitude is $3Y$ , so $x_0 = 3Y$ . It also states the maximum acceleration is $2A$ , so $a_{\max} = 2A$ .

You need an expression for $\omega$ in terms of $A$ and $Y$ (and possibly $m$ , though it is not needed).

Approach

Use the magnitude form of the SHM relation at an extreme position:

|a_{\max}| = \omega^2 x_0

Substitute $a_{\max} = 2A$ and $x_0 = 3Y$ , then rearrange for $\omega$ .

Step-by-Step Reasoning

From SHM:

a = -\omega^2 x

At the turning points, $x = \pm x_0$ and $a = \mp a_{\max}$ . Using magnitudes avoids sign issues:

a_{\max} = \omega^2 x_0

Given $x_0 = 3Y$ and $a_{\max} = 2A$ :

2A = \omega^2 (3Y)

Rearrange:

\omega^2 = \frac{2A}{3Y}

and hence

\omega = \sqrt{\frac{2A}{3Y}}

Key Takeaways

For SHM, the gradient of the $a$ – $x$ graph equals $-\omega^2$ .
Using maximum values gives $a_{\max} = \omega^2 x_0$ .

Common Mistakes

Dropping the square root and writing $\omega = 2A/3Y$ .
Mixing signs unnecessarily; using magnitudes at maxima is simplest.
Confusing amplitude with peak-to-peak value (here amplitude is explicitly $3Y$ ).

Things to Be Careful About

$\omega^2$ is positive; the minus sign is in the relationship $a = -\omega^2 x$ , not in $\omega^2$ .
Ensure you use amplitude $x_0$ (maximum displacement), not some intermediate displacement like $Y$ or $2Y$ .

Techniques used

use the SHM relation a = -omega^2 xuse maximum values to relate omega to the gradient of an a-x graphrearrange an expression to solve for angular frequency

(ii)

the maximum speed $v_0$ of the oscillations

$v_0$ = ______

DifficultyMedium-Easy

Worked solution

Working

For SHM,

v_0 = \omega x_0

with $x_0 = 3Y$ and $\omega = \sqrt{\frac{2A}{3Y}}$ :

v_0 = (3Y)\sqrt{\frac{2A}{3Y}} = \sqrt{6AY}

Answer

v_0 = \sqrt{6AY}

Final answer

v0 = √(6AY)

Detailed explanation

Background Concept

For SHM with displacement

x = x_0 \sin(\omega t)

the velocity is

v = \frac{\mathrm{d}x}{\mathrm{d}t} = \omega x_0 \cos(\omega t)

The maximum speed occurs when $\cos(\omega t) = \pm 1$ (i.e. at equilibrium, $x=0$ ), giving

v_\text{max} = \omega x_0

Understanding the Question

You are asked for the maximum speed $v_0$ in terms of $m$ , $A$ , and $Y$ . From the stem:

amplitude $x_0 = 3Y$
maximum acceleration $a_{\max} = 2A$
from part (i), $\omega = \sqrt{\frac{2A}{3Y}}$

Approach

Use the SHM relation $v_0 = \omega x_0$ and substitute the known expressions for $x_0$ and $\omega$ .

Step-by-Step Reasoning

Start with

v_0 = \omega x_0

Substitute $x_0 = 3Y$ :

v_0 = 3Y\,\omega

Now substitute $\omega = \sqrt{\frac{2A}{3Y}}$ :

v_0 = 3Y \sqrt{\frac{2A}{3Y}}

Simplify by combining factors under a square root:

v_0 = \sqrt{(3Y)^2\frac{2A}{3Y}} = \sqrt{6AY}

So the maximum speed is

v_0 = \sqrt{6AY}

Key Takeaways

In SHM, maximum speed occurs at equilibrium and equals $\omega x_0$ .
Combining results from earlier parts (here $\omega$ from the $a$ – $x$ graph) is common.

Common Mistakes

Using $v_0 = \omega^2 x_0$ (confusing with acceleration).
Forgetting that the amplitude is $3Y$ .
Algebra slip when simplifying the surd.

Things to Be Careful About

$v_0$ must be positive; take the magnitude.
Keep expressions in simplest surd form unless the question requests otherwise.

Techniques used

use v_max = omega x0 for SHMsubstitute an expression for angular frequencysimplify an algebraic surd expression

(iii)

the energy $E$ of the oscillations.

$E$ = ______

DifficultyMedium

Worked solution

Working

Total energy of SHM:

E = \frac{1}{2}m\omega^2 x_0^2

with $\omega^2 = \frac{2A}{3Y}$ and $x_0 = 3Y$ :

E = \frac{1}{2}m\left(\frac{2A}{3Y}\right)(3Y)^2 = \frac{1}{2}m\left(\frac{2A}{3Y}\right)(9Y^2) = 3mAY

Answer

E = 3mAY

Final answer

E = 3mAY

Detailed explanation

Background Concept

In ideal SHM (no damping), the total mechanical energy is constant and equals the maximum kinetic energy (at equilibrium) and also equals the maximum potential energy (at the turning points). A standard expression is

E = \frac{1}{2}m\omega^2 x_0^2

where $x_0$ is the amplitude.

Equivalently, since $v_0 = \omega x_0$ ,

E = \frac{1}{2} m v_0^2

Both are valid and should give the same result.

Understanding the Question

You are given the SHM information via the $a$ – $x$ graph:

amplitude $x_0 = 3Y$
maximum acceleration $a_{\max} = 2A$

From SHM, $\omega^2 = a_{\max}/x_0 = \frac{2A}{3Y}$ .

You must express the energy $E$ in terms of some or all of $m$ , $A$ , and $Y$ .

Approach

Use

E = \frac{1}{2}m\omega^2 x_0^2

and substitute $\omega^2 = \frac{2A}{3Y}$ and $x_0 = 3Y$ , then simplify.

Step-by-Step Reasoning

Start with the SHM energy formula:

E = \frac{1}{2}m\omega^2 x_0^2

Use the result from the $a$ – $x$ graph:

\omega^2 = \frac{2A}{3Y}

and amplitude $x_0 = 3Y$ . Substitute:

E = \frac{1}{2}m\left(\frac{2A}{3Y}\right)(3Y)^2

Compute $(3Y)^2 = 9Y^2$ :

E = \frac{1}{2}m\left(\frac{2A}{3Y}\right)(9Y^2)

Simplify: $\frac{9Y^2}{3Y} = 3Y$ , so

E = \frac{1}{2}m(2A)(3Y) = 3mAY

So,

E = 3mAY

(You could also use $E=\frac{1}{2}mv_0^2$ with $v_0=\sqrt{6AY}$ to get $E=\frac{1}{2}m(6AY)=3mAY$ .)

Key Takeaways

Total SHM energy: $E = \frac{1}{2}m\omega^2 x_0^2$ .
You can also use $E = \frac{1}{2}mv_0^2$ as a consistency check.

Common Mistakes

Using $E = \frac{1}{2}kx_0^2$ but not relating $k$ correctly to $\omega$ (since $k = m\omega^2$ ).
Substituting $x_0 = Y$ instead of $3Y$ .
Losing factors of 2 or 3 during simplification.

Things to Be Careful About

Keep track that $A$ and $Y$ here are scale values from the graph, not acceleration and displacement themselves.
Energy must be positive; if a negative sign appears, it indicates an algebra/sign error.

Techniques used

use total energy formula for SHMsubstitute omega and amplitude valuessimplify an algebraic expression to obtain energy

(c)

The period of the oscillations is $0.75\ \text{s}$ and the value of $3Y$ is $1.8\ \text{cm}$ .

Determine an expression for $x$ in terms of time $t$ , where $x$ is in $\text{cm}$ and $t$ is in seconds.

$x$ = ______

DifficultyMedium-Easy

Worked solution

Working

Amplitude $x_0 = 3Y = 1.8\ \text{cm}$ .

\omega = \frac{2\pi}{T} = \frac{2\pi}{0.75} = 8.38\ \text{rad s}^{-1}

Answer

x = 1.8\sin(8.38t)

( $x$ in $\text{cm}$ , $t$ in $\text{s}$ .)

Final answer

x = 1.8 sin(8.38 t) (x in cm, t in s)

Detailed explanation

Background Concept

A common way to represent SHM is

x = x_0 \sin(\omega t + \phi)

where:

$x_0$ is amplitude,
$\omega$ is angular frequency,
$\phi$ is the phase constant (depends on where the mass is at $t=0$ ).

Angular frequency is related to the period $T$ by

\omega = \frac{2\pi}{T}

If no starting position/phase information is given, an accepted expression usually sets $\phi = 0$ and uses a sine (or cosine) form.

Understanding the Question

You are told:

period $T = 0.75\ \text{s}$ ,
amplitude is $3Y = 1.8\ \text{cm}$ .

You must write $x$ in terms of time $t$ , with $x$ in cm and $t$ in seconds.

Approach

Take amplitude directly: $x_0 = 1.8\ \text{cm}$ .
Calculate $\omega$ from $\omega = 2\pi/T$ .
Substitute into $x = x_0 \sin(\omega t)$ (choosing zero phase since no other condition is given).

Step-by-Step Reasoning

Amplitude:

x_0 = 1.8\ \text{cm}

Angular frequency:

\omega = \frac{2\pi}{T} = \frac{2\pi}{0.75} = 8.377\ldots \approx 8.38\ \text{rad s}^{-1}

Displacement-time equation (with $\phi=0$ ):

x = x_0 \sin(\omega t) = 1.8\sin(8.38t)

Because $x_0$ is in cm and $t$ is in s, the expression automatically gives $x$ in cm.

Key Takeaways

Use $\omega = 2\pi/T$ to move between period and angular frequency.
SHM displacement can be written as $x = x_0 \sin(\omega t + \phi)$ .

Common Mistakes

Using $f = 2\pi/T$ instead of $\omega = 2\pi/T$ (confusing frequency $f$ with angular frequency $\omega$ ).
Writing $x = 1.8\sin(0.75t)$ (putting $T$ where $\omega$ should be).
Mixing units (e.g. converting 1.8 cm to 0.018 m when the question explicitly wants cm).

Things to Be Careful About

$\omega$ must be in $\text{rad s}^{-1}$ ; numerically it is larger than $f$ by a factor $2\pi$ .
Sine vs cosine: without initial conditions, either with an appropriate phase constant represents SHM; here a zero-phase form is used.
Quote $\omega$ to a sensible number of significant figures (typically 3 s.f. from the given $T$ ).

Techniques used

use omega = 2 pi / T to find angular frequencyuse x = x0 sin(omega t) to write the displacement-time relationconvert amplitude data into the required units

Q5MediumElectric Fields

(a)

Define electric potential at a point.

DifficultyEasy

Worked solution

Answer

Electric potential at a point is the work done per unit positive charge by an external agent in bringing a small test charge from infinity to that point (with no change in kinetic energy).

Final answer

Work done per unit positive charge to bring a test charge from infinity to the point (no change in KE).

Detailed explanation

Background Concept

Electric potential $V$ is an energy-per-charge quantity. It is defined using a (very small) positive test charge so that it does not significantly disturb the existing electric field.

If a charge $q$ is moved slowly (so its kinetic energy does not change), the work done $W$ by an external agent changes the electric potential energy $U$ of the charge:

\Delta U = W

Electric potential is then

V = \frac{U}{q}

so that a change in potential corresponds to a change in potential energy per unit charge.

Understanding the Question

The question asks for the definition of electric potential at a point, so the marking points are typically:

reference position (infinity),
“work done per unit charge”,
use of a positive (test) charge,
moved slowly / no change in kinetic energy.

Approach

State the definition in one sentence: “work done per unit positive charge to bring a test charge from infinity to the point”, and include the standard condition “no change in kinetic energy” (or “moved slowly”).

Step-by-Step Reasoning

Choose infinity as the zero/reference point for potential (as is standard for isolated charge distributions).
Consider a small positive test charge $q$ brought from infinity to the point.
Define the potential as the work done by an external agent per unit charge in doing so, with the test charge moved slowly so that energy transfer is not going into kinetic energy.

Key Takeaways

Electric potential is work done per unit charge.
The reference point is usually infinity for isolated charges.
The definition assumes no change in kinetic energy (quasi-static move).

Common Mistakes

Defining potential as “force per unit charge” (that is electric field strength $E$ ).
Missing “per unit charge”.
Not stating the reference point (infinity) for the definition.

Things to Be Careful About

Use “work done by an external agent” (or equivalent wording) so the definition is unambiguous.
It should be a “small test charge” so it does not disturb the field.

Techniques used

recall the definition of electric potential in terms of work done per unit chargestate the reference point at infinity and the condition of negligible change in kinetic energy

(b)

Two isolated charged metal spheres X and Y are near to each other in a vacuum. The centres of the spheres are $1.2\ \text{m}$ apart, as shown in Fig. 5.1.

Fig. 5.1

Point P is on the line joining the centres of spheres X and Y and is at a variable distance $x$ from the centre of X.

Fig. 5.2 shows the variation with $x$ of the total electric potential $V$ due to the two spheres.

Fig. 5.2

State three conclusions that may be drawn about the spheres from Fig. 5.2. The conclusions may be qualitative or quantitative.

DifficultyMedium

Worked solution

Answer

Any three valid conclusions, e.g.

$V$ is constant for $0 \le x \lesssim 0.30\ \text{m}$ , so the radius of sphere X is about $0.30\ \text{m}$ (interior of a conductor is equipotential).
$V$ has a minimum at $x = 0.60\ \text{m}$ , so the electric field is zero there (hence there is a point between the spheres where fields oppose), implying the spheres have charges of the same sign.
The minimum is midway between the centres, so the charges on X and Y are equal in magnitude (by symmetry).

Final answer

Radius of X ≈ 0.30 m; fields cancel at x = 0.60 m so charges are same sign; cancellation at midpoint implies |QX| = |QY|.

Detailed explanation

Background Concept

For an isolated charged conducting sphere:

the electric potential is the same everywhere inside the metal and inside the cavity (equipotential),
outside the sphere it behaves like a point charge at its centre for potential calculations:

V \propto \frac{Q}{r}

Along a line, the total potential is the algebraic sum of the potentials due to each charge.

A key link between potential and field is:

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

So:

where the graph of $V$ against $x$ is flat, $\mathrm{d}V/\mathrm{d}x = 0$ and the field component is zero,
where $V(x)$ has a maximum or minimum, the slope is zero so the electric field is zero at that point.

Understanding the Question

You are shown a graph of the total electric potential $V$ at points $P$ between the sphere centres, as the distance $x$ from the centre of sphere X changes.

You must extract three conclusions about the spheres (charge sign/magnitude, radii, etc.) using features of the graph:

a constant region,
a turning point (minimum),
the behaviour close to each sphere.

Approach

Look for three separate graph features and link each to a physical meaning:

Flat (constant $V$ ) section (\Rightarrow) inside a conducting sphere (\Rightarrow) its radius.
Minimum/maximum in $V$ (\Rightarrow) $E=0$ there (\Rightarrow) field cancellation point.
Where that cancellation point lies (midpoint or closer to one sphere) (\Rightarrow) comparison of charge magnitudes.

Step-by-Step Reasoning

diagram-b-1

Constant potential near X: From $x=0$ up to about $x\approx 0.30\ \text{m}$ , $V$ is constant. This is the signature of being inside a conductor, because conductors in electrostatic equilibrium are equipotential.
- Therefore, the surface of sphere X is at about $x = 0.30\ \text{m}$ .
- So the radius of X is $R_X \approx 0.30\ \text{m}$ .
Minimum at $x=0.60\ \text{m}$ : At the minimum, the slope is zero, so

\frac{\mathrm{d}V}{\mathrm{d}x}=0 \Rightarrow E_x = -\frac{\mathrm{d}V}{\mathrm{d}x}=0.

So there is a point between the spheres where the resultant electric field is zero.

Charge signs: A field-cancellation point between two charges on the line joining them occurs for like charges (fields oppose between them). For unlike charges, the fields between them point the same way and cannot cancel between the charges.
- Hence, the spheres have charges of the same sign.
Charge magnitudes from the position of cancellation: The cancellation point is exactly at the midpoint ( $0.60\ \text{m}$ is half of $1.2\ \text{m}$ ). If the charges were unequal, the zero-field point would be closer to the smaller-magnitude charge.
- Since it is at the midpoint, this indicates the charges are equal in magnitude: $|Q_X| = |Q_Y|$ .

These are three distinct conclusions (radius, sign, magnitude comparison).

Key Takeaways

A flat $V(x)$ region indicates an equipotential region (inside a conductor), allowing a radius to be read.
Turning points in $V(x)$ correspond to $E=0$ because $E_x=-\mathrm{d}V/\mathrm{d}x$ .
A zero-field point between two charges implies like charges; its position tells you the ratio of magnitudes.

Common Mistakes

Saying “ $V=0$ at the minimum so $E=0$ ”: $E=0$ comes from zero gradient, not from the value of $V$ .
Claiming opposite charges because “there is a minimum”: unlike charges do not give a cancellation point between them on the joining line.
Reading the radius from the wrong end of the flat region (it is the end of the constant section).

Things to Be Careful About

Distinguish “potential is zero” from “potential has zero slope”. Only the latter implies $E=0$ .
The graph is of total potential due to both spheres, so the constant region near X is still constant even though Y contributes, because Y’s contribution is (approximately) constant throughout the interior of X.

Techniques used

interpret a potential–distance graph for conductors and point chargesuse constant potential region to infer the radius of a conducting sphereuse the position of an extremum in V(x) to infer where the electric field is zerodeduce relative signs and magnitudes of charges from the shape and symmetry of V(x)

(c)

A proton is held at rest on the line joining the centres of the spheres in (b) at the position where $x = 0.60\ \text{m}$ .

The proton is released.

Describe and explain, without calculation, the subsequent motion of the proton.

DifficultyMedium-Easy

Worked solution

Answer

At $x=0.60\ \text{m}$ , $V$ is a minimum so $\mathrm{d}V/\mathrm{d}x=0$ and hence $E=0$ .
Therefore the proton experiences no force and (if exactly at $x=0.60\ \text{m}$ ) remains at rest.
Because it is at a minimum of $V$ , this is a stable equilibrium: a small displacement gives a force back towards $x=0.60\ \text{m}$ .

Final answer

Remains at rest at x = 0.60 m (E = 0); equilibrium is stable, so small displacement gives restoring motion back toward x = 0.60 m.

Detailed explanation

Background Concept

A charged particle in an electric field experiences a force

F = qE.

The electric field along a line can be obtained from the electric potential by

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

For a positive charge ( $q>0$ ):

if the slope $\mathrm{d}V/\mathrm{d}x$ is negative, then $E_x$ is positive and the force is in the $+x$ direction,
if the slope $\mathrm{d}V/\mathrm{d}x$ is positive, then $E_x$ is negative and the force is in the $-x$ direction.

Potential energy is

U = qV.

So for a proton (positive $q$ ), a minimum in $V$ is also a minimum in $U$ , which corresponds to stable equilibrium.

Understanding the Question

The proton is initially held at rest at $x = 0.60\ \text{m}$ , which (from the graph in part (b)) is the position of the minimum of the total potential $V(x)$ . You must describe what happens after release and explain it without calculation.

So you need to connect:

“minimum on $V(x)$ graph” (\Rightarrow) “zero slope” (\Rightarrow) “ $E=0$ ” (\Rightarrow) “no force”,
and then comment on stability.

Approach

Use $E_x = -\mathrm{d}V/\mathrm{d}x$ at $x=0.60\ \text{m}$ to decide the force at the release point.
Use whether the point is a minimum or maximum of $V$ to decide stable vs unstable equilibrium.
Describe the motion in words: remains at rest if exactly at equilibrium; otherwise accelerates back toward equilibrium.

Step-by-Step Reasoning

diagram-c-1

At $x=0.60\ \text{m}$ the graph of $V$ against $x$ has a minimum, so the tangent there is horizontal.

\frac{\mathrm{d}V}{\mathrm{d}x} = 0.

Therefore

E_x = -\frac{\mathrm{d}V}{\mathrm{d}x} = 0,

so the proton experiences

F = qE = 0.

Hence, on release from rest, it has no initial acceleration and stays where it is (provided it is exactly at $x=0.60\ \text{m}$ ).

Stability: because the point is a minimum of $V$ , a small displacement to either side produces a slope:

slightly left of $0.60\ \text{m}$ , the graph slopes downwards toward the minimum ( $\mathrm{d}V/\mathrm{d}x<0$ ) so $E_x>0$ and the force on the proton is toward increasing $x$ (back toward $0.60\ \text{m}$ ),
slightly right of $0.60\ \text{m}$ , $\mathrm{d}V/\mathrm{d}x>0$ so $E_x<0$ and the force is toward decreasing $x$ (again back toward $0.60\ \text{m}$ ).

So it is stable equilibrium: any small disturbance would make it move back toward $x=0.60\ \text{m}$ (and it may oscillate about that point).

Key Takeaways

$E_x = -\mathrm{d}V/\mathrm{d}x$ : the slope of a potential graph tells you field direction and hence force direction.
A turning point in $V(x)$ means $E=0$ (equilibrium point).
For a positive charge, a minimum of $V$ means stable equilibrium.

Common Mistakes

Saying the proton moves toward “higher potential” or “lower potential” without using the sign of the charge: the direction depends on $q$ .
Using “ $V=0$ so $E=0$ ”: it is the gradient that matters, not whether $V$ equals zero.
Claiming it will accelerate away from the minimum: that would correspond to an unstable equilibrium (a maximum in $V$ for a positive charge).

Things to Be Careful About

The question says “held at rest” and “released”: if the force is zero at the release point, it will remain at rest.
Real-life small disturbances mean you often discuss what happens if it is slightly displaced; exam answers typically mention stable equilibrium for full explanation.

Techniques used

use the relationship between field and potential gradientinfer force direction on a positive charge from the slope of a potential graphdetermine stability of equilibrium from whether V is a maximum or minimum

Q6Medium-EasyCapacitance

(a)

Two capacitors X and Y are connected in series to a power supply of voltage $V$ , as shown in Fig. 6.1.

Fig. 6.1

The capacitance of X is $C_X$ and the capacitance of Y is $C_Y$ .

Derive an expression, in terms of $C_X$ and $C_Y$ , for the combined capacitance $C_T$ of the capacitors in this circuit.

Explain your reasoning.

DifficultyMedium-Easy

Worked solution

Working

In series, the charge on each capacitor is the same, $Q$ .

V = V_X + V_Y

and

V_X = \frac{Q}{C_X},\qquad V_Y = \frac{Q}{C_Y}

So

V = Q\left(\frac{1}{C_X} + \frac{1}{C_Y}\right)

For the combined capacitor $C_T$ , $V = \dfrac{Q}{C_T}$ , hence

\frac{Q}{C_T} = Q\left(\frac{1}{C_X} + \frac{1}{C_Y}\right)

\frac{1}{C_T} = \frac{1}{C_X} + \frac{1}{C_Y}

C_T = \frac{C_X C_Y}{C_X + C_Y}

Answer

C_T = \frac{C_X C_Y}{C_X + C_Y}

Final answer

CT = (CX CY) / (CX + CY)

Detailed explanation

Background Concept

For a capacitor,

C = \frac{Q}{V}

so the potential difference across a capacitor is

V = \frac{Q}{C}.

When capacitors are connected in series:

the same current must flow through each component while charging,
so each capacitor ends up with the same magnitude of charge $Q$ on its plates,
and the total supply voltage equals the sum of the potential differences across each capacitor.

Understanding the Question

Two capacitors $X$ and $Y$ are in series across a supply of voltage $V$ . You are asked to derive the equivalent (combined) capacitance $C_T$ in terms of $C_X$ and $C_Y$ .

This is not a “plug numbers” task: you must show the series rules (same charge, voltages add) and use $C=Q/V$ to build the expression.

Approach

Let the common charge on each capacitor be $Q$ .
Write the voltage across each capacitor using $V = Q/C$ .
Add the voltages (because series) to get the supply voltage $V$ .
Use the definition of equivalent capacitance: $V = Q/C_T$ .
Rearrange to solve for $C_T$ .

Step-by-Step Reasoning

Because the capacitors are in series, each stores the same charge $Q$ .

Voltage across $X$ :

V_X = \frac{Q}{C_X}

Voltage across $Y$ :

V_Y = \frac{Q}{C_Y}

The supply voltage is the sum:

V = V_X + V_Y = \frac{Q}{C_X} + \frac{Q}{C_Y} = Q\left(\frac{1}{C_X} + \frac{1}{C_Y}\right)

Define the equivalent capacitance $C_T$ as the single capacitor that would take the same charge $Q$ for the same total voltage $V$ :

V = \frac{Q}{C_T}

Equate the two expressions for $V$ :

\frac{Q}{C_T} = Q\left(\frac{1}{C_X} + \frac{1}{C_Y}\right)

Cancel $Q$ (valid provided $Q \neq 0$ ; even if $Q=0$ the formula still holds by continuity):

\frac{1}{C_T} = \frac{1}{C_X} + \frac{1}{C_Y}

Invert to get:

C_T = \frac{C_X C_Y}{C_X + C_Y}.

This result also makes physical sense: $C_T$ is smaller than either individual capacitance in series.

Key Takeaways

In series: same charge on each capacitor, voltages add.
Use $V=Q/C$ to connect series rules to capacitance.
Series formula:

\frac{1}{C_T} = \frac{1}{C_X} + \frac{1}{C_Y}.

Common Mistakes

Assuming voltages are the same in series (that is the rule for parallel, not series).
Adding capacitances directly for series (again, that is for parallel).
Forgetting to state the key reason: same charge $Q$ on each capacitor.

Things to Be Careful About

Keep the distinction clear:
- series: $Q$ same, $V$ splits;
- parallel: $V$ same, $Q$ adds.
Algebra: after obtaining $1/C_T = 1/C_X + 1/C_Y$ , remember to invert correctly to reach the final expression.

Techniques used

use charge conservation for series componentsexpress potential difference across each capacitor using V = Q/Cadd potential differences in series to obtain the supply voltagerearrange to identify the equivalent capacitance

(b)

Two capacitors P and Q are connected in parallel to a power supply of voltage $V$ .
The capacitance of P is $200\ \mu\text{F}$ . The capacitance $C_Q$ of Q can be varied between $0$ and $400\ \mu\text{F}$ .
When $C_Q = 0$ , the total energy stored in the capacitors is $2.5\ \text{mJ}$ .

7M

(i)

Show that the supply voltage $V$ is $5.0\ \text{V}$ .

DifficultyMedium-Easy

Worked solution

Working

When $C_Q = 0$ , only capacitor $P$ stores energy.

E = \frac{1}{2} C_P V^2

2.5\times 10^{-3} = \frac{1}{2}(200\times 10^{-6})V^2

V^2 = 25 \quad \Rightarrow \quad V = 5.0\ \text{V}

Answer

$5.0\ \text{V}$

Final answer

5.0 V

Detailed explanation

Background Concept

The energy stored in a capacitor is

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

where $C$ is the capacitance and $V$ is the potential difference across the capacitor.

For capacitors in parallel, each capacitor is directly across the supply, so each has the same voltage as the supply voltage.

Understanding the Question

Capacitors $P$ and $Q$ are in parallel across a supply $V$ . You are told:

$C_P = 200\ \mu\text{F}$ ,
$C_Q$ can be varied,
when $C_Q = 0$ , the total stored energy is $2.5\ \text{mJ}$ .

If $C_Q=0$ , capacitor $Q$ is effectively absent, so the only energy stored is in $P$ . Use this to find $V$ .

Approach

Use $E = \tfrac{1}{2}CV^2$ for capacitor $P$ .
Convert $\mu\text{F}$ to farads and $\text{mJ}$ to joules.
Rearrange to solve for $V$ .

Step-by-Step Reasoning

With $C_Q=0$ , the energy is

E = \frac{1}{2}C_P V^2.

Convert units:

E = 2.5\ \text{mJ} = 2.5\times 10^{-3}\ \text{J},\qquad C_P = 200\ \mu\text{F} = 200\times 10^{-6}\ \text{F}.

Substitute:

2.5\times 10^{-3} = \frac{1}{2}(200\times 10^{-6})V^2.

Compute the coefficient:

\frac{1}{2}(200\times 10^{-6}) = 100\times 10^{-6} = 1.0\times 10^{-4}.

So

2.5\times 10^{-3} = (1.0\times 10^{-4})V^2 \Rightarrow V^2 = 25 \Rightarrow V = 5.0\ \text{V}.

Key Takeaways

Energy in a capacitor: $E = \tfrac{1}{2}CV^2$ .
In parallel, each capacitor has the full supply voltage.
Always convert prefixes correctly: $\mu = 10^{-6}$ , $\text{m} = 10^{-3}$ .

Common Mistakes

Using $E = \tfrac{1}{2}QV$ without knowing $Q$ (you can, but then you must find $Q=CV$ anyway).
Forgetting the unit conversions, giving a voltage off by factors of $10$ or $1000$ .
Treating the capacitors as series instead of parallel.

Things to Be Careful About

The voltage across $P$ equals the supply voltage only because the connection is parallel.
Quote the final voltage to the precision suggested: $5.0\ \text{V}$ (2 s.f.).

Techniques used

use the energy stored in a capacitor W = 1/2 C V^2convert between micro-, milli- and base SI unitsrearrange an equation to solve for a required variable

(ii)

Calculate the total energy, in $\text{mJ}$ , stored in the capacitors when $C_Q$ has its maximum value.

total energy = ______ $\text{mJ}$

DifficultyMedium-Easy

Worked solution

Working

In parallel,

C_T = C_P + C_Q = 200\ \mu\text{F} + 400\ \mu\text{F} = 600\ \mu\text{F} = 600\times 10^{-6}\ \text{F}

Using $V = 5.0\ \text{V}$ ,

E = \frac{1}{2}C_T V^2 = \frac{1}{2}(600\times 10^{-6})(5.0)^2

E = 7.5\times 10^{-3}\ \text{J} = 7.5\ \text{mJ}

Answer

$7.5\ \text{mJ}$

Final answer

7.5 mJ

Detailed explanation

Background Concept

For capacitors in parallel:

the potential difference across each is the same ( $V$ ),
the total charge stored is the sum of the charges,
so the equivalent capacitance is the sum:

C_T = C_1 + C_2 + \cdots

Energy stored by a capacitor (or equivalent capacitance) across voltage $V$ :

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

Understanding the Question

You have two capacitors in parallel:

$C_P = 200\ \mu\text{F}$ ,
$C_Q$ varies up to $400\ \mu\text{F}$ ,
supply voltage already found as $V = 5.0\ \text{V}$ .

You must find the total energy stored when $C_Q$ is maximum.

Approach

Find total capacitance in parallel: $C_T = C_P + C_Q$ .
Use $E = \tfrac{1}{2} C_T V^2$ with $V=5.0\ \text{V}$ .
Convert the final answer to mJ.

Step-by-Step Reasoning

At maximum, $C_Q = 400\ \mu\text{F}$ , so

C_T = 200\ \mu\text{F} + 400\ \mu\text{F} = 600\ \mu\text{F}.

Convert:

600\ \mu\text{F} = 600\times 10^{-6}\ \text{F}.

Now calculate energy:

E = \frac{1}{2}C_T V^2 = \frac{1}{2}(600\times 10^{-6})(5.0)^2.

Since $(5.0)^2 = 25$ ,

E = 0.5\times 600\times 10^{-6}\times 25 = 7.5\times 10^{-3}\ \text{J}.

Convert to mJ:

7.5\times 10^{-3}\ \text{J} = 7.5\ \text{mJ}.

Key Takeaways

Parallel capacitors add: $C_T = C_P + C_Q$ .
Once you know the supply voltage, total energy can be found from $E = \tfrac{1}{2}C_TV^2$ .

Common Mistakes

Using the series formula for capacitance instead of adding directly.
Forgetting to convert $\mu\text{F}$ to F, leading to an answer too large by $10^6$ .
Reporting energy in J when the question requests mJ.

Things to Be Careful About

The voltage $V$ remains $5.0\ \text{V}$ regardless of $C_Q$ because the supply voltage is fixed.
Keep consistent significant figures; $7.5\ \text{mJ}$ matches the given data precision.

Techniques used

add capacitances for parallel combinationuse the energy-capacitance relation E = 1/2 C V^2convert and present the final energy in the required unit

(iii)

On Fig. 6.2, sketch the variation of the total energy $E$ stored in the capacitors with $C_Q$ , as $C_Q$ varies from $0$ to $400\ \mu\text{F}$ .

Fig. 6.2

DifficultyMedium-Easy

Worked solution

Working

For parallel capacitors,

C_T = C_P + C_Q

and

E = \frac{1}{2}C_T V^2 = \frac{1}{2}(C_P + C_Q)V^2

So $E$ varies linearly with $C_Q$ .

At $C_Q = 0$ , $E = 2.5\ \text{mJ}$ .

At $C_Q = 400\ \mu\text{F}$ , $E = 7.5\ \text{mJ}$ .

Answer

Sketch a straight line increasing from $(0,\,2.5\ \text{mJ})$ to $(400\ \mu\text{F},\,7.5\ \text{mJ})$ .

Final answer

Straight line from (0, 2.5 mJ) to (400 µF, 7.5 mJ)

Detailed explanation

Background Concept

For a fixed supply voltage $V$ , the energy stored in a capacitor is

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

For capacitors in parallel, the equivalent capacitance is

C_T = C_P + C_Q.

So with $V$ constant,

E = \frac{1}{2}(C_P + C_Q)V^2

which is of the form $E = mC_Q + c$ (a straight-line relationship).

Understanding the Question

You must sketch a graph of total energy $E$ (in mJ) against $C_Q$ (in $\mu\text{F}$ ) as $C_Q$ goes from $0$ to $400\ \mu\text{F}$ .

You already know two key points from earlier parts:

at $C_Q=0$ , $E=2.5\ \text{mJ}$ ,
at $C_Q=400\ \mu\text{F}$ , $E=7.5\ \text{mJ}$ .

The shape depends on how $E$ depends on $C_Q$ .

Approach

Write $E$ in terms of $C_Q$ using $E = \tfrac{1}{2}(C_P+C_Q)V^2$ .
Recognise this is linear in $C_Q$ (because $V$ and $C_P$ are constants).
Plot/mark the endpoints and draw a straight line between them.

Step-by-Step Reasoning

Start with the parallel combination:

C_T = C_P + C_Q.

Energy stored:

E = \frac{1}{2}C_TV^2 = \frac{1}{2}(C_P + C_Q)V^2.

Here $C_P$ is constant and $V$ is constant (the supply is fixed at $5.0\ \text{V}$ ), so the only variable is $C_Q$ . Therefore $E$ increases in direct proportion to $C_Q$ (straight line).

Find two points to define the line:

When $C_Q = 0$ , $E = 2.5\ \text{mJ}$ (given).
When $C_Q = 400\ \mu\text{F}$ , $E = 7.5\ \text{mJ}$ (from part (ii)).

So on the provided axes you mark these and draw a straight line joining them.

Key Takeaways

With a constant supply voltage, energy is proportional to capacitance: $E \propto C$ .
For a variable capacitor in parallel, $E$ vs $C_Q$ is linear with a non-zero intercept (because $C_P$ is always present).

Common Mistakes

Drawing a curve (e.g. quadratic) because of the $V^2$ term: $V$ is constant here, so $V^2$ is just a constant factor.
Forcing the graph through the origin: at $C_Q=0$ there is still capacitor $P$ , so $E \neq 0$ .
Using the wrong endpoint energy (e.g. using $10\ \text{mJ}$ because it is the top of the axis rather than the calculated value).

Things to Be Careful About

Axes: ensure $C_Q$ runs from $0$ to $400\ \mu\text{F}$ and $E$ is in mJ.
Your line should pass exactly through $(0, 2.5)$ and $(400, 7.5)$ on the given scale.
Keep the sketch as a single straight best line; do not add extra segments or curvature.

Techniques used

express energy as a function of capacitance using E = 1/2 C V^2identify proportionality to predict graph shapeuse intercepts/endpoints to sketch a straight-line graph

Q7MediumMagnetic Fields

(a)

State Faraday’s law of electromagnetic induction.

DifficultyEasy

Worked solution

Answer

Faraday’s law: the induced e.m.f. is equal to the negative rate of change of magnetic flux linkage.

\mathcal{E} = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

Final answer

The induced e.m.f. equals the negative rate of change of flux linkage: (\mathcal{E} = -\mathrm{d}(N\Phi)/\mathrm{d}t).

Detailed explanation

Background Concept

Electromagnetic induction occurs when the magnetic flux through a circuit changes. For a coil of $N$ turns, the quantity that matters is the flux linkage $N\Phi$ , where:

$\Phi$ is the magnetic flux through one turn (in Wb),
$N\Phi$ is in Wb-turn,
the induced e.m.f. $\mathcal{E}$ is in V.

Faraday’s law links the induced e.m.f. to how fast $N\Phi$ changes:

\mathcal{E} = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

The negative sign is Lenz’s law: the induced e.m.f. drives a current whose magnetic effect opposes the change causing it.

Understanding the Question

You are asked to state Faraday’s law (2 marks). This typically needs both:

a correct verbal statement, and
the correct mathematical form.

Approach

Give the standard statement and equation. Make sure to include:

flux linkage $N\Phi$ (not just $\Phi$ ),
the rate of change with respect to time,
the negative sign.

Step-by-Step Reasoning

Identify the relevant physical quantity changing: magnetic flux linkage $N\Phi$ .
State that the induced e.m.f. equals the rate of change of this quantity.
Include the negative sign to indicate opposition to the change.

Key Takeaways

Induced e.m.f. depends on how fast flux linkage changes.
Use $N\Phi$ for coils.
The negative sign represents Lenz’s law.

Common Mistakes

Writing $\mathcal{E} = \frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}$ (missing minus sign).
Using $\mathcal{E} = -\frac{\mathrm{d}\Phi}{\mathrm{d}t}$ without mentioning flux linkage for a coil.
Confusing flux $\Phi$ with flux density $B$ .

Things to Be Careful About

The derivative is with respect to time ( $t$ ).
Flux linkage is $N\Phi$ ; flux through one turn is $\Phi = BA\cos\theta$ (used in later parts).

Techniques used

state Faraday’s law in words and in an equationuse flux linkage as the relevant quantity for coilsinclude the negative sign to indicate Lenz’s law

(b)

Fig. 7.1 shows a coil at rest in a uniform magnetic field that is parallel to the axis of the coil.

Fig. 7.1

The coil is connected to a centre-zero voltmeter.

The flux density $B$ of the uniform magnetic field varies with time $t$ as shown in Fig. 7.2.

Fig. 7.2

The coil consists of $340$ turns, each of cross-sectional area $3.2 \times 10^{-4}\ \text{m}^2$ .

12M

(i)

Calculate the maximum magnetic flux through one turn of the coil.

maximum magnetic flux = ______ $\text{Wb}$

DifficultyMedium-Easy

Worked solution

Working

Maximum flux density from the graph: $B_{\max} = 6.0\ \text{mT} = 6.0 \times 10^{-3}\ \text{T}$ .

Field is parallel to coil axis $\Rightarrow \theta = 0$ so $\cos\theta = 1$ :

\Phi_{\max} = BA = (6.0\times10^{-3})(3.2\times10^{-4}) = 1.92\times10^{-6}\ \text{Wb}

Answer

$\boxed{1.9\times10^{-6}\ \text{Wb}}$

Final answer

1.9 × 10⁻⁶ Wb

Detailed explanation

Background Concept

Magnetic flux through a loop is

\Phi = BA\cos\theta

where:

$B$ is magnetic flux density (T),
$A$ is the area of the loop (m $^2$ ),
$\theta$ is the angle between $\vec{B}$ and the normal to the coil’s plane.

If the magnetic field is along the coil’s axis, it is along the normal to the coil, so $\theta = 0$ and $\Phi = BA$ .

Understanding the Question

You are given:

maximum $B$ from the graph (in mT),
cross-sectional area of each turn $A = 3.2\times10^{-4}\ \text{m}^2$ .

You must find the maximum flux through one turn (so do not multiply by $N$ in this part).

Approach

Read $B_{\max}$ from the graph.
Convert mT to T.
Use $\Phi_{\max} = B_{\max}A$ (since the field is parallel to the coil axis).

Step-by-Step Reasoning

From the sinusoidal graph, $B$ reaches a maximum of $6\ \text{mT}$ .
Convert: $6\ \text{mT} = 6\times10^{-3}\ \text{T}$ .
Use $\Phi = BA$ :

\Phi_{\max} = (6\times10^{-3})(3.2\times10^{-4}) = 1.92\times10^{-6}\ \text{Wb}

Key Takeaways

Flux is $\Phi = BA\cos\theta$ .
Convert mT to T before calculating.
“One turn” means do not multiply by $N$ .

Common Mistakes

Using $340$ turns here (that is for flux linkage in later parts).
Leaving $B$ in mT, giving an answer too large by a factor $10^3$ .
Using $\Phi = NBA$ (mixing up flux and flux linkage).

Things to Be Careful About

The direction information (“field parallel to axis”) is the cue that $\cos\theta = 1$ .
Quote the final value to 2 s.f. if not specified, consistent with given data.

Techniques used

use magnetic flux $\Phi = BA\cos\theta$read a maximum value from a graphconvert millitesla to tesla

(ii)

Determine the maximum rate of change of magnetic flux linkage in the coil.

maximum rate of change of flux linkage = ______ $\text{Wb s}^{-1}$

DifficultyMedium

Worked solution

Working

Amplitude of flux density variation:

B_\text{amp} = \frac{6.0-2.0}{2}\ \text{mT} = 2.0\times10^{-3}\ \text{T}

Period from graph: $T = 2.0\ \text{ms} = 2.0\times10^{-3}\ \text{s}$ , so

\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0\times10^{-3}} = 1000\pi\ \text{rad s}^{-1}

Flux linkage $N\Phi = NBA$ (field along axis). Maximum rate of change:

\left(\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}\right)_{\max} = N A \left(\frac{\mathrm{d}B}{\mathrm{d}t}\right)_{\max} = N A (B_\text{amp}\omega)

= 340(3.2\times10^{-4})(2.0\times10^{-3})(1000\pi) = 6.83\times10^{-1}\ \text{Wb s}^{-1}

Answer

$\boxed{0.68\ \text{Wb s}^{-1}}$

Final answer

0.68 Wb s⁻¹

Detailed explanation

Background Concept

For a coil in a magnetic field, the flux linkage is

N\Phi = NBA\cos\theta

Here $\theta = 0$ , so $N\Phi = NBA$ . The induced e.m.f. depends on the rate of change of this quantity:

\mathcal{E} = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

If $B$ varies sinusoidally with angular frequency $\omega$ , e.g.

B = B_\text{mean} + B_\text{amp}\sin(\omega t + \phi)

then

\frac{\mathrm{d}B}{\mathrm{d}t} = B_\text{amp}\omega\cos(\omega t+\phi)

so the maximum rate of change is

\left(\frac{\mathrm{d}B}{\mathrm{d}t}\right)_{\max} = B_\text{amp}\omega

Understanding the Question

You are given:

$N = 340$ turns,
$A = 3.2\times10^{-4}\ \text{m}^2$ ,
a sinusoidal graph of $B$ vs $t$ with max $6\ \text{mT}$ and min $2\ \text{mT}$ , period $2\ \text{ms}$ .

You need the maximum rate of change of flux linkage, i.e. the maximum of $\mathrm{d}(N\Phi)/\mathrm{d}t$ .

Approach

From the graph, find the amplitude $B_\text{amp}$ and the period $T$ .
Convert $T$ to seconds and find $\omega = 2\pi/T$ .
Use

\left(\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}\right)_{\max} = NA(B_\text{amp}\omega)

Step-by-Step Reasoning

The magnetic field oscillates between $2$ $2$ and $6\ \text{mT}$ $6 mT$ .
- Mean value is $4\ \text{mT}$ .
- Amplitude is half the peak-to-peak:

B_\text{amp} = \frac{6-2}{2}\ \text{mT} = 2\ \text{mT} = 2\times10^{-3}\ \text{T}

Period from the graph is $T = 2\ \text{ms} = 2\times10^{-3}\ \text{s}$ .

\omega = \frac{2\pi}{T} = \frac{2\pi}{2\times10^{-3}} = 1000\pi\ \text{rad s}^{-1}

Since $N\Phi = NBA$ ,

\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t} = NA\frac{\mathrm{d}B}{\mathrm{d}t}

and the maximum value is

\left(\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}\right)_{\max} = NA(B_\text{amp}\omega)

Substitute:

= 340(3.2\times10^{-4})(2\times10^{-3})(1000\pi) \approx 0.683\ \text{Wb s}^{-1}

Key Takeaways

Only the changing part of $B$ contributes to induced e.m.f.; constants vanish when differentiating.
For sinusoids, maximum rate of change is amplitude $\times\ \omega$ .
Always convert ms to s.

Common Mistakes

Using $B_{\max}$ instead of the amplitude $B_\text{amp}$ (gives an answer too large).
Taking $T = 1\ \text{ms}$ (confusing time to peak with full period).
Forgetting $N$ when asked for flux linkage.

Things to Be Careful About

Units: $\text{mT} \to \text{T}$ and $\text{ms} \to \text{s}$ .
$\omega$ is in $\text{rad s}^{-1}$ ; using $f = 1/T$ is fine if you then use $\omega = 2\pi f$ .

Techniques used

relate flux linkage to flux density using $N\Phi = NBA$use $\omega = 2\pi/T$ from a sinusoidal graphdifferentiate a sinusoidal variation to find a maximum rate of change

(iii)

State the maximum electromotive force (e.m.f.) $V_0$ induced across the coil.

$V_0$ = ______ $\text{V}$

DifficultyEasy

Worked solution

Working

From Faraday’s law,

V_0 = \left|\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}\right|_{\max}

Using (ii): $V_0 = 0.68\ \text{V}$ .

Answer

$\boxed{0.68\ \text{V}}$

Final answer

0.68 V

Detailed explanation

Background Concept

Faraday’s law gives the induced e.m.f.

\mathcal{E} = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

The maximum e.m.f. magnitude (peak value) is therefore

V_0 = \max\left|\mathcal{E}\right| = \left|\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}\right|_{\max}

Also, $1\ \text{Wb s}^{-1} = 1\ \text{V}$ .

Understanding the Question

You have already found in (ii) the maximum rate of change of flux linkage in $\text{Wb s}^{-1}$ . This part asks for the corresponding maximum induced e.m.f. $V_0$ .

Approach

Take the magnitude of Faraday’s law: peak e.m.f. equals peak rate of change of flux linkage.

Step-by-Step Reasoning

From (ii),

\left(\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}\right)_{\max} = 0.68\ \text{Wb s}^{-1}

Therefore,

V_0 = 0.68\ \text{V}

Key Takeaways

Peak induced e.m.f. equals peak rate of change of flux linkage.
Unit conversion is automatic: $\text{Wb s}^{-1}$ is the volt.

Common Mistakes

Including a minus sign in the numerical peak value (the peak is usually quoted as a magnitude).
Using $N\Phi_{\max}$ instead of $\mathrm{d}(N\Phi)/\mathrm{d}t$ .

Things to Be Careful About

If asked for the instantaneous e.m.f., sign matters; for $V_0$ (maximum), magnitude is sufficient unless explicitly asked otherwise.

Techniques used

apply Faraday’s law to link emf magnitude to rate of change of flux linkageuse the equivalence of $\text{Wb s}^{-1}$ and $\text{V}$

(iv)

On Fig. 7.3, sketch the variation of the e.m.f. $V$ induced across the coil with $t$ from $t = 0$ to $t = 6.0\ \text{ms}$ .

Fig. 7.3

DifficultyMedium

Worked solution

diagram-7biv

Answer

$V$ is sinusoidal with period $2.0\ \text{ms}$ and amplitude $V_0$ .
Key points (one acceptable phase):

$t = 0,\ 1.0,\ 2.0,\ 3.0,\ 4.0,\ 5.0,\ 6.0\ \text{ms}$ : $V = 0$
$t = 0.50,\ 2.50,\ 4.50\ \text{ms}$ : $V = -V_0$
$t = 1.50,\ 3.50,\ 5.50\ \text{ms}$ : $V = +V_0$

Final answer

Sinusoidal V–t graph of amplitude V0 and period 2.0 ms (three cycles from 0 to 6 ms), with V = 0 at t = 0, 1, 2, 3, 4, 5, 6 ms.

Detailed explanation

Background Concept

The induced e.m.f. is related to magnetic flux linkage by Faraday’s law:

V = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

With $N\Phi = NBA$ (here $A$ and $N$ are constant), this becomes

V \propto -\frac{\mathrm{d}B}{\mathrm{d}t}

So the $V$ – $t$ graph is the negative derivative of the $B$ – $t$ graph.

For a sinusoid, the derivative is also sinusoidal with the same period, but shifted by $90^\circ$ (a quarter of a cycle).

Understanding the Question

You are shown a sinusoidal variation of $B$ with time, with a period of $2.0\ \text{ms}$ . You must sketch the induced e.m.f. $V$ from $0$ to $6.0\ \text{ms}$ (three complete cycles). The vertical scale is labelled $-V_0$ , $0$ , $+V_0$ .

Approach

Use that $V$ has the same period as $B$ (differentiation doesn’t change period).
Use turning points of $B$ $B$ :
- when $B$ is maximum or minimum, slope is zero, so $V = 0$ .
Use steepest parts of $B$ $B$ :
- when $B$ crosses its mean value, slope magnitude is largest, so $|V|$ is maximum ( $V = \pm V_0$ ).
Choose the sign based on whether $B$ is increasing or decreasing (minus sign in Faraday’s law).

Step-by-Step Reasoning

From the graph description:

$B$ has a minimum at $t=0$ and a maximum at $t=1.0\ \text{ms}$ ; therefore $B$ is increasing between $0$ and $1.0\ \text{ms}$ .
At a minimum or maximum, $\mathrm{d}B/\mathrm{d}t = 0$ , hence $V=0$ at $t=0,\ 1.0\ \text{ms},\ 2.0\ \text{ms},\dots$ .

Between $t=0$ and $t=1.0\ \text{ms}$ , the steepest point is halfway at $t=0.50\ \text{ms}$ , so $|V|$ is maximum there.

Because $V \propto -\mathrm{d}B/\mathrm{d}t$ :

When $B$ is increasing ( $\mathrm{d}B/\mathrm{d}t > 0$ ), $V$ is negative.
So at $t=0.50\ \text{ms}$ , $V=-V_0$ .

Similarly, from $1.0$ to $2.0\ \text{ms}$ , $B$ is decreasing, so $\mathrm{d}B/\mathrm{d}t < 0$ and therefore $V$ is positive, giving $V=+V_0$ at $t=1.50\ \text{ms}$ .

This repeats every $2.0\ \text{ms}$ , so you draw three identical cycles up to $6.0\ \text{ms}$ .

diagram-7biv

Key Takeaways

Induced e.m.f. is proportional to the rate of change of flux (or $B$ here).
For sinusoids, the induced e.m.f. is sinusoidal with a $90^\circ$ phase shift.
Peaks of $|V|$ happen when $B$ has its steepest slope.

Common Mistakes

Drawing $V$ in phase with $B$ (wrong: $V$ is related to the derivative).
Putting maxima of $V$ at the maxima of $B$ (should be zero there).
Using the wrong period (must be $2.0\ \text{ms}$ , so 3 cycles in 6 ms).

Things to Be Careful About

The overall sign (whether the first peak is + or −) depends on the chosen voltmeter polarity; examiners often allow either if the phase relationship is correct and consistent.
Make sure the sketch reaches $\pm V_0$ and crosses zero at the correct times.

Techniques used

use $V \propto -\mathrm{d}\Phi/\mathrm{d}t$ to relate waveform shapesuse phase relationships between a sinusoid and its derivativesketch a sinusoidal graph with correct period and amplitude

(v)

The variation of $V$ with $t$ can be described by

V = A \sin Bt

where $A$ and $B$ are constants.

Determine the values of $A$ and $B$ . Give units with your answers.

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

DifficultyMedium

Worked solution

Working

From (iii), amplitude $A = V_0 = 0.68\ \text{V}$ .

Period $T = 2.0\ \text{ms} = 2.0\times10^{-3}\ \text{s}$ :

B = \frac{2\pi}{T} = \frac{2\pi}{2.0\times10^{-3}} = 1000\pi \approx 3.14\times10^{3}\ \text{rad s}^{-1}

Answer

$\boxed{A = 0.68\ \text{V}}$

$\boxed{B = 3.14\times10^{3}\ \text{rad s}^{-1}}$

Final answer

A = 0.68 V, B = 3.14 × 10^3 rad s⁻¹

Detailed explanation

Background Concept

A sinusoidal quantity can be written as

V = A\sin(Bt)

where:

$A$ is the amplitude (peak value),
$B$ is the angular frequency $\omega$ .

Angular frequency and period are related by

\omega = \frac{2\pi}{T}

The argument of the sine must be dimensionless, so if $t$ is in seconds, $B$ must have units of $\text{rad s}^{-1}$ (often written as $\text{s}^{-1}$ ).

Understanding the Question

You have already found the peak e.m.f. $V_0$ and the period $T$ from the $B$ – $t$ graph. The induced e.m.f. is sinusoidal, so you can match it to $V = A\sin(Bt)$ .

Approach

Set $A$ equal to the maximum e.m.f. magnitude $V_0$ .
Use the period $T$ (from the graph) to find $B = 2\pi/T$ .

Step-by-Step Reasoning

Amplitude: from part (iii),

A = V_0 = 0.68\ \text{V}

Period: the magnetic flux density has period $T = 2.0\ \text{ms} = 2.0\times10^{-3}\ \text{s}$ , and the induced e.m.f. has the same period.

So

B = \omega = \frac{2\pi}{T} = \frac{2\pi}{2.0\times10^{-3}} = 1000\pi \approx 3.14\times10^{3}\ \text{rad s}^{-1}

Key Takeaways

In $V = A\sin(Bt)$ , $A$ is the peak voltage and $B$ is the angular frequency.
Use $\omega = 2\pi/T$ with $T$ in seconds.

Common Mistakes

Using $B = 1/T$ instead of $2\pi/T$ (that gives frequency, not angular frequency).
Leaving $T$ in ms, giving $B$ too small by a factor of $1000$ .
Confusing amplitude of $B$ with amplitude of $V$ .

Things to Be Careful About

The sign of $A$ depends on the phase choice/polarity in the sketch. If your sketched wave starts negative just after $t=0$ , then an equivalent expression is $V = -V_0\sin(\omega t)$ ; many schemes accept either provided the waveform is consistent.
Quote appropriate significant figures (typically 2–3 s.f.).

Techniques used

match a sinusoidal graph to the form $V = A\sin(Bt)$use the period to determine angular frequencyidentify amplitude as the maximum e.m.f.

Q8MediumAstronomy and CosmologyQuantum Physics

Fig. 8.1 shows part of the emission spectrum of visible radiation emitted by hydrogen gas in a star in a distant galaxy.

Fig. 8.1

The galaxy is moving away from the Earth at a speed of $6.2 \times 10^6\ \text{m s}^{-1}$ .

(a)

4M

(i)

Explain how the positions of the lines in the emission spectrum seen by an observer on the Earth differ from the positions shown in Fig. 8.1.

DifficultyMedium-Easy

Worked solution

Answer

The galaxy is moving away, so the spectral lines are red-shifted.

Observed frequency is lower (so the lines shift towards lower frequency, i.e. to the left on Fig. 8.1) and the observed wavelength is greater.

Final answer

Lines shift to lower frequency (left on the diagram); wavelength increases (redshift).

Detailed explanation

Background Concept

When a source of waves moves relative to an observer, the observed wavelength and frequency change (Doppler effect). For light from a receding astronomical object, the light is redshifted:

observed wavelength increases: (\lambda_{\text{obs}} > \lambda_{\text{emit}})
observed frequency decreases: (f_{\text{obs}} < f_{\text{emit}})

For speeds much smaller than (c), the fractional change is approximately:

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

and similarly

\frac{\Delta f}{f} \approx -\frac{v}{c}

Understanding the Question

Fig. 8.1 shows three emission lines from hydrogen as they are emitted by the star in the distant galaxy. The arrow on the spectrum shows frequency increases to the right. The galaxy is moving away from Earth, so the observer on Earth will detect these same lines shifted.

You are asked to explain how the positions of the observed lines differ from those shown.

Approach

Use the fact that recession causes redshift.
Translate “redshift” into what happens to frequency and therefore where the lines move on a frequency axis.

Step-by-Step Reasoning

Recession means successive wavefronts are “stretched out” by the motion, so the observer measures a longer wavelength.
Since (c = f\lambda) for light in vacuum, if (\lambda) increases then (f) must decrease.
On the diagram, frequency increases to the right, so lower frequency corresponds to a shift left.

Therefore all three lines move left (towards the red end / lower frequency end).

Key Takeaways

Moving away (\Rightarrow) redshift (\Rightarrow) (\lambda) increases and (f) decreases.
Always use the axis direction (here: increasing frequency to the right) to state the shift direction correctly.

Common Mistakes

Saying the lines shift to higher frequency when the source is moving away.
Confusing a wavelength axis with a frequency axis: on a wavelength axis, redshift would be a shift to the right (to larger (\lambda)), but here the axis is frequency.

Things to Be Careful About

The question is about positions of lines on Fig. 8.1; because the axis is frequency, “redshift” means left (lower frequency), not right.

Techniques used

apply the Doppler redshift idea for a receding sourcerelate recession to lower observed frequency and longer observed wavelengthuse the frequency-axis direction to infer the direction of line shift

(ii)

On Fig. 8.1, draw the three lines in possible positions in the spectrum seen by the observer.

DifficultyMedium-Easy

Worked solution

Answer

diagram-a-ii

All three lines are drawn to the left of the original lines (lower frequency), with the same order of the three lines.

Final answer

Three lines shifted left (to lower frequency).

Detailed explanation

Background Concept

An emission spectrum consists of discrete lines at particular frequencies (or wavelengths) corresponding to electron transitions. If the source is moving away, every emitted frequency is observed reduced by the same fraction (for (v \ll c)), so the entire pattern shifts together.

Understanding the Question

You are not calculating an exact amount of shift; you are sketching possible positions seen from Earth. Fig. 8.1 is labelled so that frequency increases to the right.

Approach

Decide the shift direction: recession (\Rightarrow) lower observed frequency.
Draw each of the three lines displaced leftwards relative to its original position, keeping the same relative ordering.

Step-by-Step Reasoning

Since the galaxy is moving away, the observed spectrum is redshifted.
On a frequency axis, redshift means smaller (f).
Therefore, each line should be moved left.
Because the fractional change is (approximately) the same for all lines, the pattern should look the same, just shifted to lower frequency.

diagram-a-ii

Key Takeaways

Always check whether the spectrum axis is frequency or wavelength.
Redshift: (f\downarrow), (\lambda\uparrow).

Common Mistakes

Shifting to the right because you associate “red” with the right-hand side without checking the axis label.
Re-ordering the lines (the transitions remain the same, so the order must not swap).

Things to Be Careful About

The question says “possible positions”: the shift should be small (since (v/c\approx 0.02)), not a huge displacement across the whole diagram.

Techniques used

translate a redshift into a leftward shift on a frequency axiskeep the pattern/order of lines while shifting them consistentlysketch shifted spectral lines relative to given reference lines

(b)

The lines in Fig. 8.1 correspond to electron transitions down to the energy level $-3.40\ \text{eV}$ .
One of the lines represents emitted radiation of wavelength $488\ \text{nm}$ .

6M

(i)

Calculate the energy of a photon of this radiation.

photon energy = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

E = \frac{hc}{\lambda}

E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{488 \times 10^{-9}}

E = 4.08 \times 10^{-19}\ \text{J}

Answer

(4.08 \times 10^{-19}\ \text{J})

Final answer

4.08 × 10^−19 J

Detailed explanation

Background Concept

Light is quantised into photons. The energy of one photon is related to its frequency (f) by

E = hf

and since electromagnetic waves satisfy (c = f\lambda), we can combine these to get

E = \frac{hc}{\lambda}

where:

(h) is Planck’s constant,
(c) is the speed of light in vacuum,
(\lambda) is the wavelength.

Understanding the Question

The emitted wavelength is (488\ \text{nm}). You are asked for the photon energy in joules.

Approach

Use (E = hc/\lambda). Convert (\lambda) from nm to m, then substitute numerical values.

Step-by-Step Reasoning

Convert wavelength:

488\ \text{nm} = 488 \times 10^{-9}\ \text{m}

Substitute into (E = hc/\lambda):

E = \frac{(6.63 \times 10^{-34}\ \text{J s})(3.00 \times 10^{8}\ \text{m s}^{-1})}{488 \times 10^{-9}\ \text{m}}

Evaluate:

(hc \approx 1.99 \times 10^{-25}\ \text{J m})
Divide by (4.88 \times 10^{-7}\ \text{m}):

E \approx 4.08 \times 10^{-19}\ \text{J}

Key Takeaways

For photon energy from wavelength, (E = hc/\lambda) is usually the quickest route.
Always convert nm to m before substituting.

Common Mistakes

Forgetting (\times 10^{-9}) when converting nm to m.
Using (E = h\lambda) (incorrect).

Things to Be Careful About

Significant figures: the wavelength is given to 3 s.f., so quoting (E) to 3 s.f. is appropriate.
Keep units consistent: using SI units ensures (E) comes out in joules.

Techniques used

use the photon energy equation E = hc/\lambdaconvert nanometres to metressubstitute constants and evaluate in standard form

(ii)

Determine the energy, in $\text{eV}$ , of the energy level from which the electron transition originates to cause the emission of this radiation.

energy level = ______ $\text{eV}$

DifficultyMedium

Worked solution

Working

Photon energy from (i): (E = 4.08 \times 10^{-19}\ \text{J}).

E(\text{eV}) = \frac{4.08 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.55\ \text{eV}

For emission:

E_{\gamma} = E_{\text{initial}} - E_{\text{final}}

with (E_{\text{final}} = -3.40\ \text{eV}):

E_{\text{initial}} = E_{\gamma} + E_{\text{final}} = 2.55 + (-3.40) = -0.85\ \text{eV}

Answer

(-0.85\ \text{eV})

Final answer

−0.85 eV

Detailed explanation

Background Concept

In an atom, electrons occupy discrete energy levels. When an electron drops from a higher level (less negative energy) to a lower level (more negative energy), a photon is emitted.

Energy conservation gives:

E_{\gamma} = E_{\text{initial}} - E_{\text{final}}

Key points:

Bound-state energy levels for hydrogen are negative.
The emitted photon energy is the difference between the initial and final energy levels.

Also, the electronvolt is a convenient energy unit:

1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}

Understanding the Question

All the lines shown correspond to transitions down to (-3.40\ \text{eV}) (this is the final level). For the particular line with (\lambda = 488\ \text{nm}), you have already found the photon energy in joules in part (i). Now you must find the starting (initial) energy level in eV.

Approach

Convert the photon energy from J to eV.
Use (E_{\text{initial}} = E_{\gamma} + E_{\text{final}}), being careful with the negative sign of (E_{\text{final}}).

Step-by-Step Reasoning

Convert to eV:

E_{\gamma} = \frac{4.08 \times 10^{-19}\ \text{J}}{1.60 \times 10^{-19}\ \text{J eV}^{-1}} = 2.55\ \text{eV}

Apply the transition energy equation:

Final level: (E_{\text{final}} = -3.40\ \text{eV})
Emission: (E_{\gamma} = E_{\text{initial}} - E_{\text{final}})

Rearrange:

E_{\text{initial}} = E_{\gamma} + E_{\text{final}}

Substitute:

E_{\text{initial}} = 2.55 + (-3.40) = -0.85\ \text{eV}

This makes sense physically: (-0.85\ \text{eV}) is higher than (-3.40\ \text{eV}) (less negative), so the electron can drop down and emit energy.

Key Takeaways

Emitted photon energy equals the difference between initial and final energy levels.
Keep track of signs: hydrogen energy levels are negative.

Common Mistakes

Using (E_{\text{initial}} = E_{\text{final}} - E_{\gamma}) (wrong sign for emission).
Forgetting to convert J to eV before combining with energy levels in eV.

Things to Be Careful About

A higher energy level in hydrogen is less negative, not “more negative”.
Write the energy difference equation explicitly to avoid sign errors.

Techniques used

convert photon energy from joules to electronvoltsuse energy conservation for an electronic transitionadd/subtract energy levels to find the initial level

(iii)

Determine the wavelength, in $\text{nm}$ , of this radiation as detected by the observer on the Earth.

wavelength = ______ $\text{nm}$

DifficultyMedium-Easy

Worked solution

Working

For recession ((v \ll c)):

\lambda_{\text{obs}} \approx \lambda_{\text{emit}}\left(1 + \frac{v}{c}\right)

\frac{v}{c} = \frac{6.2 \times 10^{6}}{3.00 \times 10^{8}} = 2.07 \times 10^{-2}

\lambda_{\text{obs}} = 488(1 + 2.07 \times 10^{-2}) = 498\ \text{nm}

Answer

(498\ \text{nm})

Final answer

498 nm

Detailed explanation

Background Concept

For light from a moving source, the observed wavelength changes due to the Doppler effect. For speeds much smaller than the speed of light, the redshift (recession) approximation is:

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

which is equivalent to:

\lambda_{\text{obs}} \approx \lambda_{\text{emit}}\left(1 + \frac{v}{c}\right)

A receding source gives a larger observed wavelength.

Understanding the Question

The emitted wavelength is (488\ \text{nm}). The galaxy is moving away at (6.2 \times 10^6\ \text{m s}^{-1}). You must calculate the wavelength detected on Earth.

Approach

Compute (v/c).
Multiply the emitted wavelength by (1 + v/c) to get the observed wavelength.

Step-by-Step Reasoning

Find the fractional speed:

\frac{v}{c} = \frac{6.2 \times 10^{6}}{3.00 \times 10^{8}} = 2.07 \times 10^{-2}

Apply the redshift approximation:

\lambda_{\text{obs}} = 488\ \text{nm} \times (1 + 0.0207)

\lambda_{\text{obs}} \approx 498\ \text{nm}

The wavelength increases, consistent with redshift.

Key Takeaways

Recession (\Rightarrow) (\lambda) increases by approximately a fraction (v/c).
Check that the final wavelength is larger than the emitted wavelength.

Common Mistakes

Using (1 - v/c) for a receding source (this would be for approaching / blueshift).
Mixing up frequency and wavelength shifts (frequency decreases when wavelength increases).

Things to Be Careful About

Ensure (v \ll c) to justify the approximation; here (v/c \approx 0.021), so it is reasonable.
Quote the answer to appropriate significant figures (here (498\ \text{nm}) to 3 s.f.).

Techniques used

apply the (non-relativistic) Doppler redshift formulacalculate the fractional shift v/cscale the emitted wavelength to obtain the observed wavelength

(c)

A value for the Hubble constant is $2.3 \times 10^{-18}\ \text{s}^{-1}$ .

Determine the distance of the galaxy from the Earth.

distance = ______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Working

Hubble law:

v = H_0 d

d = \frac{v}{H_0} = \frac{6.2 \times 10^{6}}{2.3 \times 10^{-18}} = 2.7 \times 10^{24}\ \text{m}

Answer

(2.7 \times 10^{24}\ \text{m})

Final answer

2.7 × 10^24 m

Detailed explanation

Background Concept

Hubble’s law relates the recessional speed (v) of a galaxy to its distance (d) from the observer:

v = H_0 d

where (H_0) is the Hubble constant. In this question (H_0) is given in (\text{s}^{-1}), which is consistent because (v) is in (\text{m s}^{-1}) and (d) comes out in metres.

Understanding the Question

You are given:

galaxy recession speed: (v = 6.2 \times 10^{6}\ \text{m s}^{-1})
Hubble constant: (H_0 = 2.3 \times 10^{-18}\ \text{s}^{-1})

You must determine the galaxy’s distance from Earth.

Approach

Rearrange Hubble’s law to (d = v/H_0) and substitute the values.

Step-by-Step Reasoning

Start with:

v = H_0 d

Rearrange:

d = \frac{v}{H_0}

Substitute:

d = \frac{6.2 \times 10^{6}}{2.3 \times 10^{-18}} = \left(\frac{6.2}{2.3}\right) \times 10^{24}

d \approx 2.7 \times 10^{24}\ \text{m}

Unit check:

\frac{\text{m s}^{-1}}{\text{s}^{-1}} = \text{m}

Key Takeaways

Hubble’s law is a direct proportionality: distance is found by dividing the recessional speed by (H_0).
Units are a useful self-check.

Common Mistakes

Multiplying by (H_0) instead of dividing.
Treating (H_0) as having units of (\text{m s}^{-1}) (it does not; it is (\text{s}^{-1}) here).

Things to Be Careful About

Standard form arithmetic: dividing by (10^{-18}) increases the power of ten by (+18).
Quote to appropriate significant figures (here 2 s.f. is reasonable from given data).

Techniques used

apply Hubble's law v = H0 drearrange to solve for distancesubstitute values and evaluate in standard form

Q9MediumNuclear Physics

(a)

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

DifficultyEasy

Worked solution

Answer

Binding energy is the energy required to separate a nucleus completely into its individual protons and neutrons (to infinite separation).

Final answer

Energy required to separate the nucleus into free nucleons (at infinite separation).

Detailed explanation

Background Concept

Nuclei are made of protons and neutrons (nucleons) bound together by the strong nuclear force. When nucleons bind together, energy is released and the resulting nucleus has a smaller mass than the total mass of the separated nucleons. This “missing mass” is the mass defect.

Binding energy is a way to quantify how strongly the nucleons are held together.

Understanding the Question

The question asks for the meaning of “binding energy of a nucleus”. No calculation is required; it is asking for the definition in words.

Approach

Use the standard definition used in Cambridge mark schemes: binding energy is the energy needed to split a nucleus into its separate nucleons, far apart so they no longer interact.

Step-by-Step Reasoning

“Separate completely” means break all nuclear bonds.
“Individual nucleons” means separate protons and neutrons.
“To infinity” (or infinite separation) is included to make clear the nucleons have no remaining attractive nuclear interaction energy.

Key Takeaways

Binding energy = energy required to dismantle the nucleus into free nucleons.
A larger binding energy means a more strongly bound (more stable) nucleus.

Common Mistakes

Defining it as “energy released in decay” (not the definition; decay energy is related but different).
Saying it is “the mass defect” rather than energy (mass defect must be converted using $E = \Delta mc^2$ ).

Things to Be Careful About

Binding energy is for the whole nucleus, not per nucleon (unless explicitly asked).
Include the idea of nucleons being separated so there is no remaining nuclear force interaction (often phrased as “infinite separation”).

Techniques used

state the definition of binding energy in terms of nucleons separated to infinityexpress binding energy as energy required to dismantle the nucleus

(b)

Table 9.1 shows the masses of two sub-atomic particles and a polonium-212 ( $^{212}_{84}\text{Po}$ ) nucleus.

Table 9.1

	mass / u
proton	1.007276
neutron	1.008665
polonium-212 nucleus	211.942749

For the polonium-212 nucleus, determine:

6M

(i)

the mass defect $\Delta m$ , in $\text{kg}$

$\Delta m$ = ______ $\text{kg}$

DifficultyMedium

Worked solution

Working

For $^{212}_{84}\text{Po}$ : $Z = 84$ , $N = 212 - 84 = 128$ .

Mass of separated nucleons:

(84 \times 1.007276 + 128 \times 1.008665)\,\text{u} = 213.720304\,\text{u}

Mass defect:

\Delta m = (213.720304 - 211.942749)\,\text{u} = 1.777555\,\text{u}

Using $1\,\text{u} = 1.66 \times 10^{-27}\,\text{kg}$ ,

\Delta m = 1.777555 \times 1.66 \times 10^{-27} \approx 2.95 \times 10^{-27}\,\text{kg}

Answer

$\Delta m = 2.95 \times 10^{-27}\ \text{kg}$

Final answer

2.95 × 10^-27 kg

Detailed explanation

Background Concept

For a nucleus $^A_ZX$ :

$Z$ = number of protons.
$A$ = total number of nucleons (protons + neutrons).
$N = A - Z$ = number of neutrons.

The mass defect is

\Delta m = (Z m_p + N m_n) - m_{\text{nucleus}}

where $m_p$ and $m_n$ are the proton and neutron masses, and $m_{\text{nucleus}}$ is the mass of the nucleus.

Understanding the Question

You are given the masses (in atomic mass units, u) of a proton, a neutron, and a polonium-212 nucleus. You must:

find how many protons and neutrons are in $^{212}_{84}\text{Po}$ ,
calculate the difference between the total mass of the free nucleons and the mass of the nucleus,
convert that mass difference from u to kg.

Approach

Use nuclide notation to get $Z$ and $N$ .
Compute $Z m_p + N m_n$ .
Subtract the nucleus mass to get $\Delta m$ in u.
Convert u to kg using $1\,\text{u} \approx 1.66 \times 10^{-27}\,\text{kg}$ .

Step-by-Step Reasoning

Identify numbers of nucleons:

For $^{212}_{84}\text{Po}$ , $Z = 84$ .
So $N = 212 - 84 = 128$ .

Total mass of separated nucleons:

\begin{aligned} M_{\text{separated}} &= 84(1.007276\,\text{u}) + 128(1.008665\,\text{u})\\ &= 84.611184\,\text{u} + 129.109120\,\text{u}\\ &= 213.720304\,\text{u} \end{aligned}

Mass defect in u:

\Delta m = 213.720304\,\text{u} - 211.942749\,\text{u} = 1.777555\,\text{u}

This is positive, meaning the nucleus has less mass than the separated nucleons.

Convert to kg:

\Delta m = 1.777555 \times 1.66 \times 10^{-27}\,\text{kg} \approx 2.95 \times 10^{-27}\,\text{kg}

Key Takeaways

Always get $Z$ and $N$ correct before calculating mass defect.
Mass defect is a mass difference; binding energy is found only after multiplying by $c^2$ .

Common Mistakes

Using $N = Z$ or $N = A$ instead of $N = A - Z$ .
Forgetting that the given polonium mass is already the nucleus mass (so do not add electron masses).
Incorrect conversion from u to kg or incorrect power of ten.

Things to Be Careful About

Keep consistent units: do the subtraction in u, then convert once at the end.
Use an appropriate value for $1\,\text{u}$ (typically $1.66 \times 10^{-27}\,\text{kg}$ unless otherwise specified).

Techniques used

determine numbers of protons and neutrons from nuclide notationcalculate mass defect as the difference between separated nucleons and nucleus massconvert atomic mass units to kilograms

(ii)

the binding energy

binding energy = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

E_b = \Delta m c^2 = (2.95 \times 10^{-27})(3.00 \times 10^8)^2

E_b \approx 2.66 \times 10^{-10}\,\text{J}

Answer

binding energy $= 2.66 \times 10^{-10}\ \text{J}$

Final answer

2.66 × 10^-10 J

Detailed explanation

Background Concept

The binding energy $E_b$ is related to mass defect by Einstein’s mass–energy relation:

E_b = \Delta m c^2

where:

$\Delta m$ is the mass defect (in kg),
$c$ is the speed of light ( $3.00 \times 10^8\,\text{m s}^{-1}$ ).

Understanding the Question

You have already found the mass defect for the polonium-212 nucleus. This part asks you to convert that mass defect into the binding energy in joules.

Approach

Take the value of $\Delta m$ from part (i) (in kg) and multiply by $c^2$ .

Step-by-Step Reasoning

Using $\Delta m = 2.95 \times 10^{-27}\,\text{kg}$ :

\begin{aligned} E_b &= \Delta m c^2\\ &= (2.95 \times 10^{-27})\,(3.00 \times 10^8)^2\\ &= (2.95 \times 10^{-27})\,(9.00 \times 10^{16})\\ &= 26.55 \times 10^{-11}\\ &= 2.655 \times 10^{-10}\,\text{J} \end{aligned}

So $E_b \approx 2.66 \times 10^{-10}\,\text{J}$ to 3 s.f.

Key Takeaways

Binding energy comes from mass defect via $E = \Delta mc^2$ .
Even a tiny mass defect corresponds to a large energy because $c^2$ is huge.

Common Mistakes

Using $c = 3.00 \times 10^8$ but forgetting to square it.
Using the mass defect still in u instead of converting to kg first.
Quoting the answer in kg (wrong unit) instead of J.

Things to Be Careful About

Track powers of ten carefully: $10^{-27} \times 10^{16} = 10^{-11}$ .
Give the final unit as joules, $\text{J}$ , as requested.

Techniques used

apply mass-energy equivalence to convert mass defect to energyuse standard form and correct units in a nuclear energy calculation

(iii)

the binding energy per nucleon.

binding energy per nucleon = ______ $\text{J}$

(c)

5M

(i)

On Fig. 9.1, sketch the variation with nucleon number $A$ of binding energy per nucleon for values of $A$ from $1$ to $250$ .

Fig. 9.1

DifficultyMedium-Easy

Worked solution

Answer

Sketch a curve that:

rises steeply from $A=1$ to about $A\approx 20$ ,
continues to rise to a maximum at about $A \approx 56$ ,
then falls gradually from $A\approx 56$ to $A=250$ .

Final answer

Correct binding energy per nucleon curve shape sketched.

Detailed explanation

Background Concept

The binding energy per nucleon varies with nucleon number $A$ because:

very light nuclei gain a lot of stability by fusing (binding energy per nucleon increases rapidly),
medium-mass nuclei (around iron) are the most tightly bound (peak binding energy per nucleon),
very heavy nuclei become less tightly bound per nucleon due to increasing proton–proton repulsion and the limited range of the strong force (binding energy per nucleon slowly decreases).

Understanding the Question

You are given blank axes for “binding energy per nucleon” against nucleon number $A$ from $1$ to $250$ . You must sketch the standard shape of the curve over this range.

Approach

Draw the standard binding-energy-per-nucleon graph:

start low at $A=1$ ,
rise steeply for small $A$ ,
reach a maximum around $A \approx 56$ ,
then slowly decrease for large $A$ up to $250$ .

Step-by-Step Reasoning

diagram-1

At $A=1$ (a single nucleon), binding energy per nucleon is essentially zero because there is no nucleus.
For light nuclei (a few nucleons up to roughly $A \sim 20$ ), the curve rises steeply.
The curve reaches a broad maximum around iron/nickel ( $A \approx 56$ ).
For heavier nuclei beyond that, the curve slowly falls, staying fairly high but gradually decreasing up to $A=250$ .

Key Takeaways

Peak binding energy per nucleon occurs around $A \approx 56$ .
The curve rises quickly for small $A$ and decreases slowly for very heavy nuclei.

Common Mistakes

Drawing a straight line increase all the way to $A=250$ .
Putting the maximum at very high $A$ instead of around $A \approx 56$ .
Making the decrease after the peak too steep (it should be gradual).

Things to Be Careful About

The question asks for a sketch: correct overall shape and key turning point matter more than exact numbers.
Ensure the peak is clearly before $A=100$ , and the curve is slightly lower again by $A=250$ .

Techniques used

use known qualitative shape of the binding energy per nucleon curvesketch a trend with a peak and gradual decline at high nucleon number

(ii)

On your line in Fig. 9.1, draw an X to show the approximate position of polonium-212.

(iii)

Polonium-212 is radioactive and undergoes alpha-decay.

Suggest and explain, with reference to Fig. 9.1, why the alpha-decay of polonium-212 results in a release of energy.

DifficultyMedium-Easy

Worked solution

Answer

After $\alpha$ -decay the daughter nucleus has a smaller $A$ and lies closer to the peak of the binding energy per nucleon curve, so its binding energy per nucleon is higher.

Hence the total binding energy of the products is greater than that of $^{212}\text{Po}$ , so mass decreases and energy is released ( $E = \Delta mc^2$ ).

Final answer

Products have higher binding energy (per nucleon), so total binding energy increases and energy is released.

Detailed explanation

Background Concept

The binding energy per nucleon curve tells you how stable nuclei are:

Higher binding energy per nucleon means nucleons are more tightly bound, so the nucleus is more stable.

Energy is released in a nuclear process if the total binding energy of the final products is greater than that of the initial nucleus. An increase in binding energy corresponds to a decrease in mass (mass defect increases), and the difference appears as released energy:

E = \Delta mc^2

Understanding the Question

Polonium-212 is a heavy nucleus ( $A=212$ ) on the right side of the binding energy per nucleon curve. It undergoes $\alpha$ -decay, meaning it emits an alpha particle ( $^4_2\text{He}$ ), so the daughter nucleus has nucleon number $A=208$ .

The question asks you to use Fig. 9.1 (your curve) to explain why energy is released.

Approach

Use the curve to compare binding energy per nucleon of:
- the original heavy nucleus ( $A=212$ ),
- the decay products (a slightly lighter heavy nucleus around $A=208$ and an alpha particle at $A=4$ ).
Argue that the products have higher total binding energy, so energy must be released.

Step-by-Step Reasoning

On the heavy side of the curve, as $A$ decreases slightly (moving left from $A=212$ to about $A=208$ ), the binding energy per nucleon typically increases because you move closer to the peak.
The alpha particle ( $A=4$ ) is also relatively tightly bound compared with many other very light nuclei.
Therefore, after the decay, the products correspond to a state where nucleons are, overall, more tightly bound than before.
If nucleons become more tightly bound, the total binding energy increases.
Increased binding energy means the total rest mass of the products is smaller than the initial mass, and that mass difference is released as energy (shared as kinetic energy of the alpha particle and the recoiling daughter nucleus, and possibly gamma radiation).

Key Takeaways

Energy is released when a nuclear reaction moves the system toward higher binding energy (greater stability).
Heavy nuclei can release energy by splitting/emitting particles because they lie to the right of the peak.

Common Mistakes

Saying “alpha decay releases energy because the nucleus is unstable” without linking to binding energy per nucleon.
Claiming binding energy per nucleon decreases and still concluding energy is released (wrong direction).
Confusing “binding energy” with “binding energy per nucleon” without making the stability comparison clear.

Things to Be Careful About

The curve is for binding energy per nucleon; the energy released depends on the change in total binding energy of the whole system.
Always reference the direction on the graph: for heavy nuclei, moving left (smaller $A$ ) generally moves closer to the maximum, meaning higher binding energy per nucleon and a more stable configuration.

Techniques used

use the binding energy per nucleon curve to infer energy releasecompare stability via changes in binding energy per nucleonlink increase in total binding energy to released energy

Q10MediumMedical Physics

(a)

Describe how reflected ultrasound pulses may be used to obtain diagnostic information about internal structures.

DifficultyMedium-Easy

Worked solution

Answer

A transducer sends a short ultrasound pulse into the body and detects the reflected (echo) pulses from boundaries between tissues.
The time delay of an echo gives the depth of the boundary (using the speed of sound), and the size of the reflected pulse (intensity) gives information about the type of boundary/tissue contrast.

Final answer

Pulse-echo: time delay gives depth; echo amplitude gives information about tissue boundaries.

Detailed explanation

Background Concept

Diagnostic ultrasound commonly uses the pulse-echo method. When an ultrasound wave meets a boundary between two media with different acoustic properties, part of the wave is reflected and part transmitted. The reflected part returns to the source and can be detected.

If the speed of sound in the tissue is approximately known (often taken as a typical value for soft tissue), then the distance to the reflecting boundary can be found from the time taken for the pulse to go to the boundary and return.

Understanding the Question

You are asked how reflected ultrasound pulses provide diagnostic information about internal structures. That means describing:

what is sent in,
what is measured on return,
how the measurements are interpreted (depth/location and tissue boundary information).

Approach

State the pulse-echo process: transmit pulse, receive echoes. Then give two key diagnostic links:

echo time delay (\rightarrow) depth/position of boundary,
echo intensity/amplitude (\rightarrow) strength of reflection, hence information about the type of boundary (different tissues have different impedance mismatches).

Step-by-Step Reasoning

diagram-10a-1

The transducer emits a short pulse of ultrasound into the body.
Whenever the pulse meets a boundary between different tissues (or tissue and bone, etc.), some energy is reflected.
The same transducer (or a receiver) detects the reflected pulse (echo).
If the echo returns after time (t), then the depth (d) of the boundary is found using the round trip:

d = \frac{ct}{2}

where (c) is the speed of sound in the tissue.
5) The size (intensity/amplitude) of the echo depends on how strong the reflection is at that boundary, so larger echoes indicate a larger contrast in acoustic properties between the two regions.

Key Takeaways

Ultrasound imaging uses reflections at boundaries.
Echo time delay locates boundaries (depth).
Echo amplitude/intensity indicates boundary/tissue contrast.

Common Mistakes

Forgetting the factor of (\tfrac{1}{2}) for the pulse traveling to the boundary and back.
Saying only “it forms an image” without mentioning what is measured (time delay and echo size).
Confusing the roles of reflected and transmitted waves.

Things to Be Careful About

The question asks about reflected pulses specifically: make sure to mention detection of echoes.
Keep the explanation linked to diagnostic information (position and nature of boundaries), not just how ultrasound is generated.

Techniques used

describe pulse-echo ultrasound operationrelate time delay of echoes to depth using speed of soundrelate echo amplitude to boundary properties

(b)

5M

(i)

Define specific acoustic impedance of a medium.

(ii)

Table 10.1 shows some data for water and for glass.

Table 10.1

	density / $\text{kg m}^{-3}$	speed of sound / $\text{m s}^{-1}$
water	1000	1420
glass	2500	4560

Determine the intensity reflection coefficient for ultrasound that is incident on a water–glass boundary.

intensity reflection coefficient = ______