(a) State the value of absolute zero on: (i) the Celsius temperature scale temperature = \_\_\_\_\_\_ $^{\circ}\text{C}$ (ii) the thermodynamic temperature scale. Give a unit with your answer. temperature = \_\_\_\_\_\_ unit \_\_\_\_\_\_ (b) A sample contains a fixed amount of gas. The gas has pressure $p$, volume $V$ and thermodynamic temperature $T$. Fig. 4.1 shows the variation of $pV$ with $kT$ for the sample, where $k$ is the Boltzmann constant. (i) State what is…

(a)(i) -273 °C · (a)(ii) 0 K · (b)(i) The gas behaves as an ideal gas. · (b)(ii) 3.4 × 10^22 · (b)(iii) 5.6 × 10^-2 mol · (c) 4.0 u

9702/43 Oct/Nov 2025 — Physics Solved Past Paper

Q1MediumMotion in a CircleOscillations

(a)

In terms of velocity and acceleration, describe uniform circular motion of an object.

2M Similar

DifficultyEasy

Worked solution

Answer

Speed is constant (magnitude of velocity constant).

Velocity changes direction continuously, so the object has an acceleration directed towards the centre of the circle (centripetal).

Final answer

Constant speed; acceleration towards centre of circle.

Detailed explanation

Background Concept

In uniform circular motion, an object moves in a circle at constant speed. Even though the speed is constant, the velocity is not constant because velocity is a vector: its direction changes continuously.

A changing velocity implies an acceleration. For circular motion, this acceleration is always directed towards the centre of the circle and is called centripetal acceleration.

Understanding the Question

The question asks you to describe uniform circular motion specifically in terms of:

what happens to the velocity
what happens to the acceleration

So you must mention both the constancy of speed and the inward acceleration.

Approach

Give two clear statements (one for velocity, one for acceleration):

speed (magnitude of velocity) is constant but velocity direction changes
acceleration exists and points towards the centre

Step-by-Step Reasoning

“Uniform” means the speed does not change.
In a circle, the direction of motion is always changing (tangent to the circle), so the velocity vector changes.
Therefore there must be acceleration.
That acceleration is towards the centre (perpendicular to the instantaneous velocity).

Key Takeaways

Constant speed does not mean constant velocity.
Uniform circular motion requires centripetal acceleration directed to the centre.

Common Mistakes

Saying “velocity is constant” (only speed is constant).
Saying “acceleration is zero” because speed is constant.

Things to Be Careful About

Always state the direction of the acceleration (towards the centre) to secure the mark.

Techniques used

state the defining features of uniform circular motionrelate constant speed to changing velocity directionidentify the direction of centripetal acceleration

(b)

Fig. 1.1

Fig. 1.1 shows the view from above of a polystyrene ball undergoing horizontal circular motion of radius $R$ .

The ball is illuminated by parallel light so that a shadow of the ball forms on a screen placed on the opposite side of the ball from the light source.

The line joining points O and P is perpendicular to the screen.

The angular speed of the circular motion is $\omega$ .

3M

(i)

State an expression, in terms of $R$ and $\omega$ , for the speed $v$ of the ball.

$v$ = ______

1M Similar

(ii)

Determine an expression, in terms of $v$ and $\omega$ , for the centripetal acceleration of the ball.

centripetal acceleration = ______

2M Similar

(c)

The ball in (b) is in the position shown in Fig. 1.1, such that line OB is at an angle $\theta$ to the line OP.

4M

(i)

Determine an expression, in terms of $R$ and $\theta$ , for the displacement $x$ of the shadow from P.

$x$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

x = R\sin\theta

Final answer

x = R\sin\theta

Detailed explanation

Background Concept

A shadow formed by parallel light rays is effectively a projection. If a point moves in a circle, its projected position along a line is a component of the radius vector.

In a circle of radius $R$ , if the radius makes an angle $\theta$ to a reference line, then the perpendicular component is found using trigonometry (sine/cosine depending on which side is opposite/adjacent to $\theta$ ).

Understanding the Question

The ball is at point $B$ on the circle. The line $OP$ is perpendicular to the screen, and $OB$ makes an angle $\theta$ to $OP$ . The shadow’s displacement $x$ along the screen from $P$ is the sideways (perpendicular-to- $OP$ ) component of $OB$ .

Approach

Treat $OB$ as a radius vector of length $R$ . The displacement $x$ is the component of $R$ opposite the angle $\theta$ , so use $\sin\theta$ .

Step-by-Step Reasoning

Consider the right triangle formed by dropping a perpendicular from $B$ to the line through $O$ in the direction of $OP$ .
The radius $OB$ is the hypotenuse and has magnitude $R$ .
The displacement $x$ on the screen corresponds to the component of $OB$ perpendicular to $OP$ , which is opposite the angle $\theta$ .

Thus

\sin\theta = \frac{x}{R}

so

x = R\sin\theta

Key Takeaways

The SHM displacement is a projection (component) of circular motion.
Identify whether the required component is opposite or adjacent to the angle.

Common Mistakes

Writing $x = R\cos\theta$ by choosing the wrong component.
Using $\theta$ in degrees inconsistently (here it is in radians because $\omega t$ is in radians).

Things to Be Careful About

Make sure the angle $\theta$ is defined between $OB$ and $OP$ (as stated), and choose sine/cosine accordingly.

Techniques used

use right-triangle trigonometry to find a projectionrelate displacement on screen to a component of radius

(ii)

The value of $\theta$ is zero at time $t = 0$ .

State an expression for $\theta$ in terms of $\omega$ and $t$ .

$\theta$ = ______

1M Similar

(iii)

Use your answers in (c)(i) and (c)(ii) to show that $x$ is given by

x = R \sin \omega t

1M Similar

(iv)

Explain, with reference to the equation in (c)(iii), why the motion of the shadow of the ball on the screen may be modelled as simple harmonic.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The equation

x = R\sin(\omega t)

has the standard SHM form $x=x_0\sin(\omega t)$ , so the shadow executes SHM with amplitude $R$ and angular frequency $\omega$ .

Final answer

Because x = R sin(\omega t) matches x = x0 sin(\omega t), so SHM.

Detailed explanation

Background Concept

A particle in simple harmonic motion has displacement that varies sinusoidally with time. A common form is

x = x_0\sin(\omega t)

where:

$x_0$ is the amplitude (maximum displacement)
$\omega$ is the angular frequency

Equivalently, SHM can be defined by the acceleration condition

a = -\omega^2 x

which means acceleration is proportional to displacement and opposite in direction.

Understanding the Question

You have obtained for the shadow:

x = R\sin(\omega t)

You must explain why this means the shadow’s motion can be modelled as SHM, referring to this equation.

Approach

Compare the equation directly with the standard SHM displacement equation $x=x_0\sin(\omega t)$ and identify what plays the roles of $x_0$ and $\omega$ .

Step-by-Step Reasoning

The equation $x = R\sin(\omega t)$ is a sinusoidal function of time.
This matches the SHM form $x=x_0\sin(\omega t)$ .
Therefore the shadow’s motion is simple harmonic with:
- amplitude $x_0 = R$
- angular frequency $\omega$ (same symbol in this question)

(You could also differentiate twice to show $a = -\omega^2 x$ , but the question only asks for reference to the displacement equation.)

Key Takeaways

If displacement varies as a sine or cosine with constant amplitude and constant $\omega$ , it is SHM.
The amplitude and angular frequency can be read directly from $x=x_0\sin(\omega t)$ .

Common Mistakes

Saying “it is periodic therefore SHM” (not every periodic motion is SHM).
Mixing up angular frequency $\omega$ with frequency $f$ .

Things to Be Careful About

SHM is a model: you justify it by matching the mathematical form (or the acceleration condition).
Here the same $\omega$ arises from the circular motion and becomes the SHM angular frequency.

Techniques used

compare a displacement-time equation with the standard SHM formidentify amplitude and angular frequency from an equation

(d)

The circular motion of the ball in Fig. 1.1 has a diameter of $0.46\ \text{m}$ and an angular speed of $1.9\ \text{rad s}^{-1}$ .

For the simple harmonic motion of the shadow of the ball in Fig. 1.1, calculate:

5M

(i)

the amplitude

amplitude = ______ $\text{m}$

1M Similar

(ii)

the period

period = ______ $\text{s}$

2M Similar

(iii)

the maximum acceleration.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For SHM,

a = -\omega^2 x \Rightarrow a_{\max} = \omega^2 x_0

Here $x_0=R=0.23\ \text{m}$ and $\omega=1.9\ \text{rad s}^{-1}$ :

a_{\max} = (1.9)^2(0.23) = 0.83\ \text{m s}^{-2}

Answer

Maximum acceleration $= 0.83\ \text{m s}^{-2}$

Final answer

0.83 m s^-2

Detailed explanation

Background Concept

In SHM,

a = -\omega^2 x

The magnitude of acceleration is greatest when $|x|$ is greatest, i.e. at the extremes where $x=\pm x_0$ . Hence

a_{\max} = \omega^2 x_0

Understanding the Question

You have:

amplitude $x_0 = R$ from the circular radius
angular frequency $\omega = 1.9\ \text{rad s}^{-1}$

You must find the maximum acceleration of the shadow in its SHM.

Approach

Use $a_{\max}=\omega^2 x_0$ , with $x_0=0.23\ \text{m}$ and $\omega=1.9\ \text{rad s}^{-1}$ .

Step-by-Step Reasoning

Identify amplitude:

x_0 = R = 0.23\ \text{m}

Apply maximum acceleration formula:

a_{\max} = \omega^2 x_0

Substitute:

a_{\max} = (1.9)^2(0.23)

Compute:

(1.9)^2 = 3.61

so

a_{\max} = 3.61 \times 0.23 = 0.8303\ \text{m s}^{-2} \approx 0.83\ \text{m s}^{-2}

Key Takeaways

In SHM, the maximum acceleration depends on both $\omega$ and the amplitude: $a_{\max}=\omega^2 x_0$ .
Maximum acceleration occurs at maximum displacement.

Common Mistakes

Using $a_{\max} = \omega x_0$ (missing a factor of $\omega$ ).
Using diameter instead of radius for the amplitude.

Things to Be Careful About

Units: $\omega^2 x_0$ gives $\text{s}^{-2}\times\text{m}=\text{m s}^{-2}$ .
Significant figures: the inputs are 2 s.f., so quote $a_{\max}$ to 2 s.f.

Techniques used

use the SHM maximum acceleration relationsubstitute amplitude and angular frequency into a_max = omega^2 x0

(e)

On Fig. 1.1, draw, and label with the letter A, the position of the shadow on the screen when the shadow has its maximum positive acceleration.

1M Similar

DifficultyMedium

Worked solution

Answer

Maximum positive acceleration occurs when $x$ is maximum and negative (since $a=-\omega^2 x$ ), i.e. at the left-hand extreme of the motion on the screen:

x=-R

diagram-e1

Final answer

Position at x = -R (left-hand extreme from P).

Detailed explanation

Background Concept

For SHM,

a = -\omega^2 x

This equation contains the key idea:

acceleration is proportional to displacement
acceleration is opposite in direction to displacement (minus sign)

The acceleration has maximum magnitude when $|x|$ is maximum, i.e. at $x=\pm x_0$ .

Understanding the Question

The shadow’s displacement on the screen is $x$ measured from $P$ (with the positive direction as shown on the diagram). You must indicate where the shadow is when its acceleration is maximum and positive.

Approach

Use

a = -\omega^2 x

To make $a$ positive and as large as possible:

you need $x$ to be as negative as possible
i.e. at the extreme $x=-x_0=-R$

Step-by-Step Reasoning

Maximum acceleration magnitude occurs at the extremes of motion: $x=\pm R$ .
At $x=+R$ :

a = -\omega^2(+R) < 0

so acceleration is negative.

At $x=-R$ :

a = -\omega^2(-R) = +\omega^2 R

which is maximum and positive.

Therefore point $A$ should be placed on the screen at the left-hand extreme of the shadow’s motion, a distance $R$ from $P$ in the negative $x$ direction.

diagram-e1

Key Takeaways

In SHM, maximum positive acceleration corresponds to maximum negative displacement.
Always use the sign in $a=-\omega^2 x$ .

Common Mistakes

Placing the point at $x=+R$ (that gives maximum acceleration but negative).
Placing the point at $x=0$ (acceleration is zero there).

Things to Be Careful About

Check the diagram’s sign convention for $x$ (right vs left of $P$ ).
The question asks for maximum positive acceleration, not maximum acceleration magnitude.

Techniques used

use the SHM relation a = -omega^2 x to link acceleration to displacementidentify the position corresponding to maximum positive accelerationinterpret sign convention for displacement on the screen

Q2MediumThermodynamicsTemperature

(a)

State two ways in which the first law of thermodynamics describes that the internal energy of a system may be changed.

1 ______

2 ______

2M Similar

DifficultyEasy

Worked solution

Answer

By transfer of thermal energy (heating/cooling), i.e. heat $q$ transferred.
By work done on/by the system, i.e. work $W$ done (e.g. compression/expansion).

Final answer

Thermal energy transfer (heat) and work done on/by the system.

Detailed explanation

Background Concept

The first law of thermodynamics is an energy conservation statement for a thermodynamic system. It is commonly written as

\Delta U = q + W

where:

$\Delta U$ is the change in internal energy of the system,
$q$ is the thermal energy (heat) transferred to the system (positive if the system gains thermal energy),
$W$ is the work done on the system (positive if work is done on the system).

So the internal energy can change in only two ways in this model: by heat transfer and by mechanical work.

Understanding the Question

The question asks for two ways that the internal energy of a system may be changed, according to the first law. So we simply name the two terms that can contribute to $\Delta U$ .

Approach

Recall the first law equation (or its verbal statement) and extract the two mechanisms that change $U$ .

Step-by-Step Reasoning

From

\Delta U = q + W,

the internal energy changes if either:

$q \neq 0$ (there is heat/thermal energy transfer into or out of the system), and/or
$W \neq 0$ (work is done on the system or by the system).

These are the two required ways.

Key Takeaways

The first law partitions changes in internal energy into heat transfer and work done.
In exam questions, “ways internal energy changes” almost always means “heat” and “work”.

Common Mistakes

Stating only “temperature change” (temperature is an effect, not a mechanism in the first law).
Listing examples (e.g. “friction”) without linking them to work done or heat transfer.

Things to Be Careful About

Sign convention can vary between texts, but here the key mark-worthy idea is the two mechanisms: thermal transfer and work.
Don’t confuse “internal energy” with “heat”; heat is energy in transit, internal energy is energy stored in the system.

Techniques used

state the first law of thermodynamics in wordsidentify the two energy-transfer mechanisms in the first law

(b)

6M

(i)

Use the first law of thermodynamics to explain why a bicycle pump gets hot when it is used to pump up a tyre quickly.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Using the pump, the gas is compressed so work is done on the air ( $W>0$ ).

Pumped quickly, there is little time for heat transfer so $q \approx 0$ .

From the first law,

\Delta U = q + W \approx W > 0

so the internal energy increases and the air temperature rises, heating the pump.

Final answer

Rapid compression means work is done on the air with negligible heat transfer, so internal energy and hence temperature increase, making the pump hot.

Detailed explanation

Background Concept

For a gas, internal energy $U$ is associated mainly with the random kinetic energy of its molecules (and, depending on the model, possibly other molecular energy modes). A rise in temperature corresponds to an increase in the average random kinetic energy.

The first law of thermodynamics relates internal energy change to heat transfer and work:

\Delta U = q + W

If a gas is compressed, the surroundings do work on it. For a rapid process, there is often insufficient time for significant thermal energy transfer to the surroundings, so the process is approximately adiabatic ( $q \approx 0$ ).

Understanding the Question

A bicycle pump is used to inflate a tyre quickly. You observe the pump becoming hot. The question asks you to explain this using the first law, so you must identify what happens to $q$ , $W$ and hence $\Delta U$ during rapid compression.

Approach

Identify that pumping compresses the air: work is done on the gas.
Recognise “quickly” implies little heat transfer during the compression.
Use $\Delta U = q + W$ to deduce that $\Delta U$ increases.
Link increased internal energy to increased temperature, and thus the pump warms.

Step-by-Step Reasoning

Compression means work is done on the gas.
When you push the pump handle in, you reduce the volume of the air inside. A force acts through a distance, so mechanical work is done on the gas. In first-law sign convention used here, that makes $W$ positive.
Rapid pumping implies negligible heat transfer during the compression.
Heat transfer needs time for thermal conduction through the pump walls and for convection to the air outside. If the compression happens quickly, the air does not have time to lose much energy as heat while it is being compressed, so $q \approx 0$ during that part of the cycle.
Apply the first law.

\Delta U = q + W \approx 0 + W

Since $W>0$ , it follows that $\Delta U > 0$ .

Connect internal energy to temperature.
Increased internal energy means the molecules have greater average random kinetic energy, so the temperature of the compressed air rises. The hot air then transfers energy to the pump walls, making the pump feel hot.

Key Takeaways

Compression: work done on a gas increases its internal energy.
“Quickly” is a key word: it suggests an (approximately) adiabatic process, so $q$ is small.
Higher internal energy for a gas usually means higher temperature.

Common Mistakes

Saying “friction heats the pump” without mentioning work done on the gas and the first law (friction may exist, but the question asks specifically for a first-law explanation).
Claiming heat flows into the gas during rapid compression; in reality, the temperature rises first, then heat may flow out afterwards.

Things to Be Careful About

Use the correct direction of energy transfer: it is mechanical work input that raises $U$ .
Don’t assume constant temperature; that would require enough time for heat to leave during compression (slow pumping).

Techniques used

apply the first law of thermodynamics to a compressionuse the adiabatic approximation for a rapid processlink increased internal energy to a temperature rise

(ii)

With reference to molecular energies, explain why the temperature of water remains at $100\ ^{\circ}\text{C}$ when it vaporises in a kettle, even though it is being heated.

3M Similar

DifficultyMedium

Worked solution

Answer

Temperature depends on the average random kinetic energy of the molecules.

At $100\ ^{\circ}\text{C}$ during vaporisation, the energy supplied is used to increase molecular potential energy by overcoming intermolecular forces (latent heat), not to increase average kinetic energy.

So the average kinetic energy (and hence temperature) stays constant at $100\ ^{\circ}\text{C}$ until all the water has vaporised.

Final answer

During boiling, supplied energy increases molecular potential energy (overcoming intermolecular forces) rather than average kinetic energy, so temperature remains at 100°C until vaporisation is complete.

Detailed explanation

Background Concept

Temperature is a measure of the average random kinetic energy of molecules in a substance. If the average kinetic energy increases, the temperature rises.

Internal energy includes:

random kinetic energy of molecules (related to temperature), and
potential energy associated with intermolecular forces (how strongly molecules are bound together).

When a substance changes phase (e.g. liquid to gas), energy is required to separate molecules against attractive intermolecular forces. This energy is the latent heat. During the phase change at a fixed pressure, the temperature typically remains constant even though energy is still being supplied.

Understanding the Question

Water in a kettle reaches $100\ ^{\circ}\text{C}$ and starts to vaporise (boil). Even though the kettle continues to supply energy via heating, the temperature of the water stays at $100\ ^{\circ}\text{C}$ while boiling continues. The question asks for an explanation specifically in terms of molecular energies.

Approach

State what temperature corresponds to at the molecular level (average kinetic energy).
Explain what happens to the supplied energy during vaporisation (it goes into increasing potential energy by separating molecules).
Conclude that since average kinetic energy is not increasing, the temperature remains constant until the phase change is complete.

Step-by-Step Reasoning

Link temperature to kinetic energy.
In a liquid, molecules move randomly. The hotter the liquid, the greater the average random kinetic energy. Therefore, temperature is tied to average kinetic energy, not directly to total energy supplied.
What extra energy is needed to vaporise?
To become a gas, molecules must be separated much more than in a liquid. This requires energy to overcome attractive intermolecular forces. That energy increases the potential energy part of the internal energy.
Why does the temperature not rise during boiling?
While boiling at $100\ ^{\circ}\text{C}$ (at atmospheric pressure), the energy delivered by the heater is mainly used for this separation process (latent heat). Since it is not increasing the average kinetic energy, the temperature stays constant.
When will the temperature rise again?
Only after all the liquid water has changed into steam can additional heating increase the average kinetic energy of the steam, allowing its temperature to rise above $100\ ^{\circ}\text{C}$ .

Key Takeaways

Temperature depends on average molecular kinetic energy.
During a phase change, added energy can go into potential energy (breaking/intermolecular separation) instead of kinetic energy.
Latent heat explains constant temperature during boiling.

Common Mistakes

Saying “energy is used to break bonds” but not specifying that this increases potential energy and does not increase average kinetic energy (so the temperature stays constant).
Claiming the temperature remains constant because “heat is lost to the surroundings” (not the key reason; even in an ideal insulated case, boiling at constant pressure uses energy for phase change).

Things to Be Careful About

Use the correct energy language: intermolecular forces (between molecules), not necessarily “chemical bonds”.
Be explicit that average kinetic energy stays constant, hence temperature stays constant.
The constant boiling temperature assumes (approximately) constant external pressure (e.g. atmospheric pressure for a kettle).

Techniques used

distinguish between kinetic and potential parts of internal energylink temperature to average molecular kinetic energyexplain latent heat as energy for phase change at constant temperature

Q3MediumGravitational Fields

(a)

Define gravitational field at a point.

1M Similar

(b)

Fig. 3.1

Fig. 3.1 shows an isolated point mass of mass $M$ .

Point P is at distance $x$ from the point mass.

5M

(i)

By considering the force exerted by the point mass on a test mass of mass $m$ placed at P, derive an equation for the gravitational field strength $g$ at P, in terms of $M$ and $x$ . Identify any other symbols you use.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Gravitational force on test mass $m$ at distance $x$ is

F = \frac{GMm}{x^2}

Using $g = F/m$ ,

g = \frac{F}{m} = \frac{GM}{x^2}

( $G$ is the gravitational constant.)

Answer

g = \frac{GM}{x^2}

Final answer

g = GM/x^2

Detailed explanation

Background Concept

For a point mass $M$ , Newton’s law of gravitation gives the force on a mass $m$ a distance $r$ away as

F = \frac{GMm}{r^2}

The gravitational field strength $g$ at that point is defined by

g = \frac{F}{m}

So for any gravitational force that is proportional to $m$ , dividing by $m$ removes the test mass and leaves a property of the field alone.

Understanding the Question

Point $P$ is a distance $x$ from an isolated point mass $M$ . You place a test mass $m$ at $P$ and consider the gravitational force acting on it. You are asked to derive an expression for $g$ at $P$ in terms of $M$ and $x$ , and to identify any extra symbols used (here, $G$ ).

Approach

Write the gravitational force on $m$ due to $M$ using Newton’s law with separation $r = x$ .
Use the definition $g = F/m$ to eliminate $m$ .

Step-by-Step Reasoning

Separation of the masses is $x$ , so:

F = \frac{GMm}{x^2}

Field strength is force per unit mass:

g = \frac{F}{m}

Substitute for $F$ and cancel $m$ :

g = \frac{1}{m}\cdot \frac{GMm}{x^2} = \frac{GM}{x^2}

$G$ is the universal gravitational constant.

Key Takeaways

Use $F = GMm/r^2$ then $g = F/m$ .
For a point (or spherically symmetric) mass, $g \propto 1/r^2$ .

Common Mistakes

Forgetting to divide by $m$ , leaving $m$ in the final expression.
Using $r$ instead of the given symbol $x$ (or mixing them).
Writing $g = GM/x$ (missing the square).

Things to Be Careful About

The distance must be the centre-to-point distance: here it is directly given as $x$ .
$g$ is a vector (direction towards $M$ ); the formula here gives the magnitude.

Techniques used

apply Newton's law of gravitation to find the force on a test massuse the definition g = F/msubstitute separation distance into an inverse-square law

(ii)

On Fig. 3.1, draw an arrow to indicate the direction of the gravitational field at P.

1M Similar

(iii)

Point Q is at distance $\frac{x}{2}$ from the point mass, on the opposite side of the mass from P, as shown in Fig. 3.2.

Fig. 3.2

Compare the gravitational field at Q with that at P.

2M Similar

DifficultyMedium

Worked solution

Working

At $P$ (distance $x$ ):

g_P = \frac{GM}{x^2}

At $Q$ (distance $x/2$ ):

g_Q = \frac{GM}{\left(x/2\right)^2} = \frac{GM}{x^2/4} = 4\frac{GM}{x^2} = 4g_P

Directions: at both points the field is towards $M$ , so $g_Q$ is opposite in direction to $g_P$ (since $Q$ and $P$ are on opposite sides of $M$ ).

Answer

Field at $Q$ has magnitude $4$ times that at $P$ and is in the opposite direction.

Final answer

At Q: 4 times larger magnitude than at P, opposite direction.

Detailed explanation

Background Concept

For a point mass $M$ , the gravitational field strength at distance $r$ is

g = \frac{GM}{r^2}

So:

the magnitude varies with inverse square of distance,
the direction is always towards the mass $M$ .

Understanding the Question

Point $P$ is at distance $x$ to one side of $M$ . Point $Q$ is on the other side, but closer, at distance $x/2$ from $M$ . You must compare the gravitational field at $Q$ with that at $P$ (so both magnitude and direction).

Approach

Write $g_P$ using $r=x$ .
Write $g_Q$ using $r=x/2$ .
Form the ratio $g_Q/g_P$ to compare magnitudes.
Use “towards $M$ ” to compare directions.

Step-by-Step Reasoning

At $P$ :

g_P = \frac{GM}{x^2}

At $Q$ , the distance is halved, so substitute $r=x/2$ :

g_Q = \frac{GM}{\left(x/2\right)^2}

Square the half-distance:

\left(\frac{x}{2}\right)^2 = \frac{x^2}{4}

So:

g_Q = \frac{GM}{x^2/4} = 4\frac{GM}{x^2} = 4g_P

Directions:

At $P$ (to the right of $M$ ), “towards $M$ ” means left.
At $Q$ (to the left of $M$ ), “towards $M$ ” means right.

So the fields point in opposite directions.

Key Takeaways

Halving $r$ makes $g$ four times bigger because $g \propto 1/r^2$ .
Field direction is always towards the mass, so it flips on opposite sides.

Common Mistakes

Saying it is “twice” as strong (forgetting the square).
Comparing only magnitudes and forgetting direction.
Thinking the direction is the same at $Q$ and $P$ just because both are “towards $M$ ” (they are towards $M$ but from opposite sides).

Things to Be Careful About

Keep clear whether you are comparing magnitude or the vector field.
When distance changes by a factor $k$ , inverse-square quantities change by $k^2$ .

Techniques used

apply inverse-square dependence of field strength on distancecompare magnitudes using proportional reasoningstate the direction of the field relative to the mass

(c)

Two identical isolated uniform spheres X and Y each have radius $R$ . The centres of the spheres are separated by distance $L$ , as shown in Fig. 3.3.

Fig. 3.3

Point P lies on the line joining the centres of X and Y, and is at a variable displacement $x$ from the centre of sphere X.

The gravitational field strength at the surface of each sphere is $g_0$ .

On Fig. 3.4, sketch the variation with $x$ of the gravitational field $g$ at point P between $x = R$ and $x = L - R$ .

Fig. 3.4

3M Similar

DifficultyMedium

Worked solution

Let positive $g$ be towards sphere $Y$ .

For a point between the spheres at position $x$ from centre of $X$ :

g(x) = \frac{GM}{(L-x)^2} - \frac{GM}{x^2}

Hence $g$ is negative near $x=R$ , increases monotonically, crosses $g=0$ at $x=L/2$ , and is positive near $x=L-R$ (end values close to $\mp g_0$ ).

diagram-c

Answer

Sketch as shown: increasing curve from about $-g_0$ at $x=R$ through $0$ at $x=L/2$ to about $+g_0$ at $x=L-R$ .

Final answer

Sketch: monotonic increase from ~−g0 at x=R, through 0 at x=L/2, to ~+g0 at x=L−R.

Detailed explanation

Background Concept

A uniform sphere produces the same gravitational field as a point mass located at its centre, provided the point is outside the sphere. So for points on or outside the surface,

g = \frac{GM}{r^2}

where $r$ is the distance from the centre of that sphere.

When more than one mass contributes to the field, the total gravitational field is found by superposition (vector addition). Along a straight line, this becomes a signed addition depending on the chosen positive direction.

Understanding the Question

There are two identical spheres $X$ and $Y$ of radius $R$ , centres separated by $L$ . Point $P$ lies between them on the line of centres. The horizontal coordinate $x$ is measured from the centre of sphere $X$ .

You must sketch how the net gravitational field $g$ at $P$ varies as $x$ goes from $R$ (the inner surface of $X$ ) to $L-R$ (the inner surface of $Y$ ).

The graph axes already show reference values $g_0$ , $\tfrac{1}{2}g_0$ , $0$ , etc. Here $g_0$ is the magnitude of the field at the surface of each sphere due to that sphere alone:

g_0 = \frac{GM}{R^2}

Approach

Choose a sign convention: take positive $g$ to be towards $Y$ (to the right).
Write the field at position $x$ due to sphere $X$ and sphere $Y$ separately using inverse-square dependence.
Add them with signs (superposition).
Use symmetry: because the spheres are identical, at the midpoint $x=L/2$ the fields cancel, so $g=0$ there.
Use qualitative inverse-square behaviour to draw the curve: it must be negative near $X$ , positive near $Y$ , and cross zero at the midpoint.

Step-by-Step Reasoning

Take + direction from $X$ towards $Y$ .

1) Field due to sphere X

At a point between the spheres, the distance to the centre of $X$ is $x$ . The field due to $X$ points towards $X$ (to the left), so it is negative:

g_X = -\frac{GM}{x^2}

2) Field due to sphere Y

Distance from the point to the centre of $Y$ is $L-x$ . The field due to $Y$ points towards $Y$ (to the right), so it is positive:

g_Y = +\frac{GM}{(L-x)^2}

3) Net field (superposition)

g(x) = g_Y + g_X = \frac{GM}{(L-x)^2} - \frac{GM}{x^2}

4) Key features for the sketch

At $x=R$ $x = R$ (inner surface of $X$ $X$ ):
- $|g_X| = GM/R^2 = g_0$ (negative contribution).
- $g_Y$ is smaller because $L-R > R$ (so its inverse-square is smaller).
- Net $g$ is therefore slightly greater than $-g_0$ (still negative).
At $x=L-R$ $x = L - R$ (inner surface of $Y$ $Y$ ):
- $g_Y \approx +g_0$ .
- $g_X$ is smaller in magnitude.
- Net is slightly less than $+g_0$ (still positive).
At $x=L/2$ (midpoint): distances to both centres are equal, so magnitudes are equal and opposite:

\frac{GM}{(L-L/2)^2} = \frac{GM}{(L/2)^2}

So net field is zero:

g(L/2)=0

The curve must be monotonically increasing from left to right (becoming less negative, then zero, then positive), because as you move towards $Y$ , the pull from $Y$ increases while the pull from $X$ decreases.

5) Sketch

The final sketch should show an increasing curve starting near $-g_0$ at $x=R$ , crossing $0$ at $x=L/2$ , and ending near $+g_0$ at $x=L-R$ , with inverse-square-type curvature (more curved near the spheres).

diagram-c

Key Takeaways

Outside a uniform sphere, treat it as a point mass at its centre.
Net gravitational field is the vector sum (here, signed sum) of contributions.
For identical masses, the net field between them is zero at the midpoint.

Common Mistakes

Adding magnitudes only (forgetting opposite directions).
Putting the zero-field point somewhere other than $L/2$ for identical spheres.
Drawing a straight line between endpoints (ignores inverse-square behaviour).
Getting the signs reversed at the two ends.

Things to Be Careful About

$x$ is measured from the centre of $X$ , but the distance to $Y$ is $(L-x)$ .
The interval is restricted to $R \le x \le L-R$ (points on the inner surfaces), not inside the spheres.
Endpoints are not exactly $\pm g_0$ unless the other sphere is infinitely far away; the sketch should be close to but slightly inside those values.

Techniques used

apply the inverse-square dependence of gravitational field outside a sphereuse superposition to add fields from two masses along a lineuse symmetry to locate the point where the net field is zerosketch a qualitative graph with correct intercepts and curvature

Q4MediumTemperatureIdeal Gases

(a)

State the value of absolute zero on:

2M

(i)

the Celsius temperature scale

temperature = ______ $^{\circ}\text{C}$

1M Similar

(ii)

the thermodynamic temperature scale. Give a unit with your answer.

temperature = ______ unit ______

1M Similar

(b)

A sample contains a fixed amount of gas. The gas has pressure $p$ , volume $V$ and thermodynamic temperature $T$ .

Fig. 4.1 shows the variation of $pV$ with $kT$ for the sample, where $k$ is the Boltzmann constant.

Fig. 4.1

4M

(i)

State what is indicated about the nature of the gas from the variation shown in Fig. 4.1.

1M Similar

(ii)

Determine the number $N$ of molecules of the gas in the sample.

$N$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For an ideal gas,

pV = NkT \Rightarrow N = \frac{pV}{kT}.

From the graph, at $kT = 8.0 \times 10^{-21}\ \text{J}$ , $pV \approx 270\ \text{J}$ .

N = \frac{270}{8.0 \times 10^{-21}} = 3.375 \times 10^{22} \approx 3.4 \times 10^{22}.

Answer

$N = 3.4 \times 10^{22}$

Final answer

3.4 × 10^22

Detailed explanation

Background Concept

For an ideal gas, using the molecular form of the gas equation:

pV = NkT,

where:

$p$ is pressure,
$V$ is volume,
$N$ is the number of molecules,
$k$ is the Boltzmann constant,
$T$ is thermodynamic temperature.

If $N$ is fixed, then a graph of $pV$ against $kT$ should be a straight line through the origin with

\text{gradient} = N.

Understanding the Question

You are given a straight-line graph of $pV$ (vertical axis) against $kT$ (horizontal axis). You must find how many molecules $N$ are in the sample.

Approach

Use $pV = NkT$ . Since the axes already use $pV$ and $kT$ , you can obtain $N$ by either:

taking the gradient of the line ( $\Delta(pV)/\Delta(kT)$ ), or
taking one convenient point on the line and calculating $N = (pV)/(kT)$ .

Step-by-Step Reasoning

Choose a clear point on the line. The graph indicates that when

kT = 8.0 \times 10^{-21}\ \text{J},

then

pV \approx 270\ \text{J}.

Substitute into $pV = NkT$ and rearrange:

N = \frac{pV}{kT}.

Calculate:

N = \frac{270}{8.0 \times 10^{-21}} = 3.375 \times 10^{22}.

Quote appropriately:

N \approx 3.4 \times 10^{22}.

Key Takeaways

On a $pV$ vs $kT$ graph, the gradient equals $N$ .
You can use any point on the straight line through the origin to find the gradient ratio.

Common Mistakes

Forgetting that the horizontal axis is $kT$ , not $T$ .
Using $N = (kT)/(pV)$ (inverting the ratio).
Ignoring the $\times 10^{-21}$ scale factor on the axis.

Things to Be Careful About

Read the axis scale carefully: $kT$ is given in units of $10^{-21}\ \text{J}$ .
Use a point well away from the origin to reduce percentage reading uncertainty.
Give $N$ to 2 or 3 significant figures.

Techniques used

use the ideal gas equation in the form pV = NkTread a value from a graph and calculate the gradient ratiorearrange an equation to isolate the required quantity

(iii)

Use your answer in (b)(ii) to determine the amount $n$ of gas in the sample.

$n$ = ______ $\text{mol}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

n = \frac{N}{N_A} = \frac{3.4 \times 10^{22}}{6.02 \times 10^{23}} = 5.6 \times 10^{-2}.

Answer

$n = 5.6 \times 10^{-2}\ \text{mol}$

Final answer

5.6 × 10^-2 mol

Detailed explanation

Background Concept

The amount of substance $n$ (in moles) is related to the number of molecules $N$ by

N = nN_A,

where $N_A$ is the Avogadro constant ( $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ ).

So,

n = \frac{N}{N_A}.

Understanding the Question

You have already found the number of molecules in the sample in part (b)(ii). This part asks you to convert that into the amount of gas in moles.

Approach

Use $n = N/N_A$ and substitute your value of $N$ from (b)(ii).

Step-by-Step Reasoning

Using $N \approx 3.4 \times 10^{22}$ :

n = \frac{3.4 \times 10^{22}}{6.02 \times 10^{23}}.

Separate powers of ten and the numerical ratio:

n = \left(\frac{3.4}{6.02}\right) \times 10^{22-23} \approx 0.565 \times 10^{-1} = 5.65 \times 10^{-2}\ \text{mol}.

Rounded suitably:

n \approx 5.6 \times 10^{-2}\ \text{mol}.

Key Takeaways

$1\ \text{mol}$ contains $N_A$ molecules.
Convert molecules to moles by dividing by $N_A$ .

Common Mistakes

Multiplying by $N_A$ instead of dividing.
Using $N_A$ with wrong power of ten.
Omitting the unit $\text{mol}$ .

Things to Be Careful About

Keep standard form consistent: handle the $10^{22}/10^{23} = 10^{-1}$ step carefully.
Your final significant figures should match the precision of $N$ from the graph.

Techniques used

convert number of molecules to amount of substance using Avogadro constantsubstitute values and round to appropriate significant figures

(c)

The root-mean-square (r.m.s.) speed of the molecules of the gas is $1900\ \text{m s}^{-1}$ when $pV$ is equal to $270\ \text{J}$ .

Determine the mass, in u, of one molecule of the gas, where u is the unified atomic mass unit.

mass = ______ $\text{u}$

4M Similar

DifficultyMedium

Worked solution

Working

From kinetic theory,

pV = \frac{1}{3}Nm c_{\text{rms}}^{2} \Rightarrow m = \frac{3pV}{N c_{\text{rms}}^{2}}.

Using $pV = 270\ \text{J}$ , $c_{\text{rms}} = 1900\ \text{m s}^{-1}$ and $N = 3.4 \times 10^{22}$ :

m = \frac{3(270)}{(3.4 \times 10^{22})(1900)^2} = 6.6 \times 10^{-27}\ \text{kg}.

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

\text{mass in u} = \frac{6.6 \times 10^{-27}}{1.66 \times 10^{-27}} = 4.0\ \text{u}.

Answer

$4.0\ \text{u}$

Final answer

4.0 u

Detailed explanation

Background Concept

Kinetic theory links the macroscopic gas quantity $pV$ to microscopic molecular motion. For an ideal gas:

pV = \frac{1}{3} N m \langle c^2 \rangle,

and since $c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$ , this is often written as

pV = \frac{1}{3} N m c_{\text{rms}}^2.

Here:

$N$ is the number of molecules,
$m$ is the mass of one molecule (in kg),
$c_{\text{rms}}$ is the root-mean-square speed.

To express a molecular mass in unified atomic mass units (u), use

1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}.

Understanding the Question

You are told that when $pV = 270\ \text{J}$ , the r.m.s. speed is $1900\ \text{m s}^{-1}$ . Using the number of molecules $N$ you found in part (b)(ii), you must determine the mass of one molecule and then convert it into u.

Approach

Start with the kinetic theory equation $pV = \frac{1}{3}Nm c_{\text{rms}}^2$ .
Rearrange to make $m$ the subject.
Substitute $pV$ , $N$ , and $c_{\text{rms}}$ .
Convert kg to u by dividing by $1.66 \times 10^{-27}\ \text{kg}$ .

Step-by-Step Reasoning

Rearrange the kinetic theory expression:

pV = \frac{1}{3}Nm c_{\text{rms}}^{2}

Multiply both sides by $3$ and divide by $N c_{\text{rms}}^{2}$ :

m = \frac{3pV}{N c_{\text{rms}}^{2}}.

Substitute values. Using $pV = 270\ \text{J}$ , $c_{\text{rms}} = 1900\ \text{m s}^{-1}$ , and $N \approx 3.4 \times 10^{22}$ :

m = \frac{3(270)}{(3.4 \times 10^{22})(1900)^2}.

Compute $c_{\text{rms}}^2$ :

(1900)^2 = 3.61 \times 10^6\ \text{m}^2\text{s}^{-2}.

Denominator:

(3.4 \times 10^{22})(3.61 \times 10^6) = 1.23 \times 10^{29}.

So

m = \frac{810}{1.23 \times 10^{29}} \approx 6.6 \times 10^{-27}\ \text{kg}.

Convert to unified atomic mass units:

\text{mass in u} = \frac{6.6 \times 10^{-27}}{1.66 \times 10^{-27}} \approx 4.0\ \text{u}.

Key Takeaways

$pV$ is connected to molecular motion via $pV = \frac{1}{3}Nm c_{\text{rms}}^2$ .
You can find molecular mass if you know $pV$ , $N$ , and $c_{\text{rms}}$ .
Convert kg to u by dividing by $1.66 \times 10^{-27}\ \text{kg}$ .

Common Mistakes

Using $pV = NkT$ again (that equation cannot give $m$ without additional information).
Forgetting the factor of $\frac{1}{3}$ .
Using $c_{\text{rms}}$ instead of $c_{\text{rms}}^2$ .
Converting to u by multiplying instead of dividing by $1.66 \times 10^{-27}$ .

Things to Be Careful About

Keep track of powers of ten when combining $N$ and $c^2$ .
$pV$ in joules is consistent with the kinetic theory equation (since $1\ \text{J} = 1\ \text{Pa m}^3$ ).
Your answer depends on the value of $N$ from the graph; small differences in graph reading can slightly change the final mass, so quote to appropriate significant figures.

Techniques used

apply the kinetic theory relation pV = (1/3) N m c_rms^2rearrange to find the molecular massconvert mass in kilograms to unified atomic mass units

Q5MediumElectric Fields

(a)

Define electric potential at a point.

2M Similar

DifficultyEasy

Worked solution

Answer

Electric potential at a point is the work done per unit positive charge 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.

Detailed explanation

Background Concept

Electric potential $V$ is a scalar quantity that tells you the electrical potential energy per unit charge.

It is defined by

V = \frac{W}{q}

where $W$ is the work done by an external agent (or work done against the electric field) in bringing a test charge $q$ from a reference point to the point of interest.

In electrostatics, we usually choose the reference point to be infinity, where the potential is taken to be zero.

Understanding the Question

You are asked to define electric potential at a point. For full marks you must mention:

it is work done per unit charge,
the charge is a small positive test charge (so it does not disturb the field),
the reference position is infinity.

Approach

Write the standard definition in one sentence, including “per unit positive charge” and “from infinity”. Some mark schemes also expect the condition “no change in kinetic energy”.

Step-by-Step Reasoning

“Potential” refers to energy per charge, so you need “work done / charge”.
For a unique value, you must specify the reference point: take $V=0$ at infinity.
To avoid ambiguity about extra energy going into motion, you can add “with no change in kinetic energy”.

Key Takeaways

Electric potential is energy per unit charge.
The standard reference is infinity.
Potential is a scalar.

Common Mistakes

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

Things to Be Careful About

Use “work done per unit positive charge”; potential is defined independent of the sign of the test charge, but exam definitions usually specify positive.
If you include “no change in kinetic energy”, ensure it is clearly linked to the process of bringing the test charge in slowly.

Techniques used

state the definition of electric potential as work done per unit chargespecify the reference point at infinity

(b)

A hydrogen atom may be considered to consist of a proton and an electron separated by a distance of $120\ \text{pm}$ , as shown in Fig. 5.1.

Fig. 5.1

The two particles may be considered as point charges.

Point P lies on the line joining the electron and the proton and is at a variable distance $x$ from the proton.

8M

(i)

Show that the electric potential $V$ at point P when $x = 10\ \text{pm}$ is equal to $130\ \text{V}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At point $P$ (between charges):

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{+e}{x} + \frac{-e}{120\ \text{pm}-x}\right)

For $x=10\ \text{pm}$ , distances are $10\ \text{pm}$ from proton and $110\ \text{pm}$ from electron:

V = (8.99\times 10^9)(1.60\times 10^{-19})\left(\frac{1}{10\times 10^{-12}}-\frac{1}{110\times 10^{-12}}\right)

V = 1.44\times 10^{-9}\times 10^{12}\left(0.1-0.00909\right)=1.44\times 10^3\times 0.0909 \approx 1.31\times 10^2\ \text{V}

Answer

$V \approx 130\ \text{V}$

Final answer

130 V

Detailed explanation

Background Concept

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

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

Potential is a scalar, so for more than one charge you add the potentials:

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

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

Understanding the Question

A hydrogen atom is modelled as:

a proton of charge $+e$ ,
an electron of charge $-e$ ,
separated by $120\ \text{pm}$ .

Point $P$ lies on the line between them and is at distance $x$ from the proton. When $x=10\ \text{pm}$ , you must calculate the net potential at $P$ and show it is about $130\ \text{V}$ .

Approach

Write the potential at $P$ due to the proton and electron separately using $V=kQ/r$ .
Add them (superposition), paying attention to the sign of each charge.
Substitute $x=10\ \text{pm}$ , so the distance to the electron is $120-10=110\ \text{pm}$ .
Convert pm to m to use SI units.

Step-by-Step Reasoning

Use

V = k\left(\frac{+e}{x}+\frac{-e}{120\ \text{pm}-x}\right)

where $k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\ \text{N m}^2\ \text{C}^{-2}$ .

For $x=10\ \text{pm}$ :
- distance to proton $r_p=10\ \text{pm}=10\times 10^{-12}\ \text{m}$ ,
- distance to electron $r_e=110\ \text{pm}=110\times 10^{-12}\ \text{m}$ .
Substitute:

V = (8.99\times 10^9)(1.60\times 10^{-19})\left(\frac{1}{10\times 10^{-12}}-\frac{1}{110\times 10^{-12}}\right)

Evaluate the constant:

ke = (8.99\times 10^9)(1.60\times 10^{-19}) \approx 1.44\times 10^{-9}

Evaluate the bracket (in $\text{m}^{-1}$ ):

\frac{1}{10\times 10^{-12}}=1.0\times 10^{11},\qquad \frac{1}{110\times 10^{-12}}=9.09\times 10^{9}

So

\left(1.0\times 10^{11}-9.09\times 10^{9}\right)=9.09\times 10^{10}

Multiply:

V \approx (1.44\times 10^{-9})(9.09\times 10^{10}) \approx 1.31\times 10^2\ \text{V} \approx 130\ \text{V}

Key Takeaways

Potential from point charges uses $V=kQ/r$ .
Net potential is found by adding potentials (scalar superposition).
Signs of charges matter directly in the potential expression.

Common Mistakes

Adding magnitudes only and forgetting the electron gives negative potential.
Using $120+x$ instead of $120-x$ for the distance to the electron.
Forgetting to convert pm to m (gives an answer wrong by $10^{12}$ ).

Things to Be Careful About

The distance in $V=kQ/r$ must be in metres.
The question asks to “show” $130\ \text{V}$ , so an answer like $131\ \text{V}$ is fine if rounding is sensible.
Do not confuse potential $V$ with field strength $E$ (which varies as $1/r^2$ ).

Techniques used

use electric potential of a point chargeapply superposition of electric potential from multiple chargesconvert picometres to metres consistently

(ii)

Calculate, to two significant figures, $V$ when $x = 30\ \text{pm}$ .

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{+e}{x} + \frac{-e}{120\ \text{pm}-x}\right)

For $x=30\ \text{pm}$ , distances are $30\ \text{pm}$ and $90\ \text{pm}$ :

V=(8.99\times10^9)(1.60\times10^{-19})\left(\frac{1}{30\times10^{-12}}-\frac{1}{90\times10^{-12}}\right)

V = 1.44\times10^3\left(\frac{1}{30}-\frac{1}{90}\right)=1.44\times10^3\left(0.0333-0.0111\right)=3.2\times10^{1}\ \text{V}

Answer

$V = 32\ \text{V}$

Final answer

32 V

Detailed explanation

Background Concept

For point charges, electric potential is

V = k\frac{Q}{r}

and potentials add algebraically:

V_{\text{total}} = k\left(\frac{Q_1}{r_1}+\frac{Q_2}{r_2}\right)

Understanding the Question

The proton and electron are $120\ \text{pm}$ apart. Point $P$ lies between them at distance $x$ from the proton.

For $x=30\ \text{pm}$ :

distance to proton is $30\ \text{pm}$ ,
distance to electron is $120-30=90\ \text{pm}$ .

You must calculate the net potential and give it to two significant figures.

Approach

Use $V=k\left(\frac{+e}{x}-\frac{e}{120-x}\right)$ because the electron has charge $-e$ .
Substitute $x=30\ \text{pm}$ and $120-x=90\ \text{pm}$ .
Convert pm to m.
Round to $2\ \text{s.f.}$ .

Step-by-Step Reasoning

Write the expression:

V = k e\left(\frac{1}{x}-\frac{1}{120\ \text{pm}-x}\right)

Convert distances:

30\ \text{pm}=30\times 10^{-12}\ \text{m},\qquad 90\ \text{pm}=90\times 10^{-12}\ \text{m}

Substitute using $k=8.99\times 10^9$ and $e=1.60\times 10^{-19}$ :

V=(8.99\times10^9)(1.60\times10^{-19})\left(\frac{1}{30\times10^{-12}}-\frac{1}{90\times10^{-12}}\right)

A useful simplification is $ke\times 10^{12}\approx 1.44\times 10^3$ (because $1\ \text{pm}=10^{-12}\ \text{m}$ ):

V = 1.44\times 10^3\left(\frac{1}{30}-\frac{1}{90}\right)

Evaluate the bracket:

\frac{1}{30}-\frac{1}{90}=\frac{3-1}{90}=\frac{2}{90}=0.0222

So

V \approx (1.44\times 10^3)(0.0222)\approx 32\ \text{V}

Key Takeaways

Between equal and opposite charges, the potential depends on the difference of reciprocals of distances.
Symmetry gives checks: at $x=30\ \text{pm}$ the potential should be positive (closer to proton).

Common Mistakes

Using $+e$ for both charges (forgetting the electron is negative).
Using $120+x$ instead of $120-x$ .
Rounding too early or not giving $2\ \text{s.f.}$ .

Things to Be Careful About

Keep units consistent: convert pm to m, or use the factor $10^{12}$ carefully.
The sign of $V$ matters: here it must be positive because $P$ is closer to the proton than to the electron.

Techniques used

use electric potential of a point chargeapply superposition with correct signsround the final value to the required significant figures

(iii)

On Fig. 5.1, draw a cross ( $\times$ ) at one position, other than infinity, where the electric potential is zero.

1M Similar

DifficultyEasy

Worked solution

Working

For $V=0$ between the charges:

\frac{+e}{x} + \frac{-e}{120\ \text{pm}-x}=0 \;\Rightarrow\; \frac{1}{x}=\frac{1}{120\ \text{pm}-x}

So $x=60\ \text{pm}$ .

Answer

Place the $\times$ halfway between the proton and electron (at $x=60\ \text{pm}$ from the proton).

Final answer

At the midpoint between the charges: x = 60 pm from the proton.

Detailed explanation

Background Concept

Electric potential is a scalar and adds algebraically. For two charges $+e$ and $-e$ at distances $r_p$ and $r_e$ from a point,

V = k\left(\frac{+e}{r_p} + \frac{-e}{r_e}\right)=ke\left(\frac{1}{r_p}-\frac{1}{r_e}\right)

For $V=0$ you need the two contributions to be equal in magnitude and opposite in sign.

Understanding the Question

Point $P$ is on the line between a proton and an electron separated by $120\ \text{pm}$ . You must mark one finite position (not infinity) where the net potential is zero.

Approach

Set the total potential to zero and solve for $x$ . Because the charges are equal and opposite, $V=0$ occurs when the point is equidistant from both charges.

Step-by-Step Reasoning

Between the charges, the distances are:
- to proton: $x$
- to electron: $120\ \text{pm}-x$
Set the potential to zero:

ke\left(\frac{1}{x}-\frac{1}{120\ \text{pm}-x}\right)=0

Since $ke\neq 0$ :

\frac{1}{x}=\frac{1}{120\ \text{pm}-x}

Therefore

x = 120\ \text{pm}-x \Rightarrow 2x=120\ \text{pm} \Rightarrow x=60\ \text{pm}

So the point is exactly the midpoint.

Key Takeaways

For equal and opposite charges, $V=0$ occurs where the point is equally far from both charges.
Potential can be zero even though the electric field is not necessarily zero.

Common Mistakes

Saying “at infinity only” (there is also a finite point here).
Confusing zero potential with zero electric field (field is not zero at the midpoint for a dipole).

Things to Be Careful About

The question restricts you to the line joining the charges; off the line, there are other points where $V=0$ (a plane perpendicular to the line through the midpoint), but you must place the cross on Fig. 5.1, so the midpoint on the line is the correct response.

Techniques used

set the algebraic sum of potentials to zerouse symmetry of equal and opposite charges

(iv)

On Fig. 5.2, sketch the variation of $V$ with $x$ between $x = 10\ \text{pm}$ and $x = 110\ \text{pm}$ .

Fig. 5.2

3M Similar

DifficultyMedium

Worked solution

Answer

Use

V = ke\left(\frac{1}{x}-\frac{1}{120\ \text{pm}-x}\right)

Key points:

$x=10\ \text{pm}$ , $V\approx +130\ \text{V}$
$x=60\ \text{pm}$ , $V=0$
$x=110\ \text{pm}$ , $V\approx -130\ \text{V}$

Sketch a smooth, continuously decreasing curve from $+130\ \text{V}$ at $x=10\ \text{pm}$ crossing $V=0$ at $x=60\ \text{pm}$ and reaching $-130\ \text{V}$ at $x=110\ \text{pm}$ , becoming steeper near both ends.

diagram-1

Final answer

Smooth decreasing curve crossing V = 0 at x = 60 pm; V ≈ +130 V at 10 pm and V ≈ −130 V at 110 pm.

Detailed explanation

Background Concept

For a proton-electron pair separated by $d=120\ \text{pm}$ , the potential at a point between them a distance $x$ from the proton is

V(x)=k\left(\frac{+e}{x}+\frac{-e}{d-x}\right)=ke\left(\frac{1}{x}-\frac{1}{d-x}\right)

This function is not linear: each term varies as $1/r$ , so the graph is curved.

Important features:

If you are closer to the proton ( $x$ small), the positive term dominates and $V>0$ .
If you are closer to the electron ( $x$ close to $d$ ), the negative term dominates and $V<0$ .
$V=0$ when $x=d-x$ (midpoint).

Understanding the Question

You must sketch $V$ against $x$ for $10\ \text{pm} \le x \le 110\ \text{pm}$ .

So you do not include the positions of the charges themselves ( $x=0$ and $x=120\ \text{pm}$ ), but you are close enough to see the curve becoming steep near the ends.

Approach

Use the expression $V(x)=ke\left(\frac{1}{x}-\frac{1}{120-x}\right)$ .
Plot/mark a few key points (you already have $V$ at $x=10$ from part (i), and can use symmetry).
Identify the crossing at $x=60\ \text{pm}$ .
Use the behaviour near the ends: as $x\to 0$ , $1/x$ becomes very large so $V$ rises steeply; as $x\to 120$ , $1/(120-x)$ becomes very large so $V$ falls steeply.
Draw a smooth curve consistent with these features.

Step-by-Step Reasoning

Key calculated/known values:
- At $x=10\ \text{pm}$ : $V\approx +130\ \text{V}$ (given by part (i)).
- At $x=60\ \text{pm}$ : $V=0$ because distances to the charges are equal.
- By symmetry, at $x=110\ \text{pm}$ the situation is the reverse of $x=10\ \text{pm}$ , so

V(110\ \text{pm})=-V(10\ \text{pm})\approx -130\ \text{V}

The curve must be monotonically decreasing between the charges: as you move from the proton towards the electron, the positive contribution weakens ( $1/x$ falls) and the negative contribution strengthens in magnitude (because $1/(120-x)$ grows).
It should be steeper near $x=10\ \text{pm}$ and near $x=110\ \text{pm}$ because the $1/r$ dependence changes rapidly when $r$ is small.

A correct sketch therefore passes through about $(10,130)$ , crosses the axis at $(60,0)$ , and reaches about $(110,-130)$ with a smooth curved shape.

diagram-1

Key Takeaways

For point charges, potential varies as $1/r$ , so graphs are curved.
For equal and opposite charges, $V$ changes sign at the midpoint.
Symmetry is a powerful check: $V(x)=-V(d-x)$ .

Common Mistakes

Drawing a straight line between the endpoints (ignores the $1/r$ dependence).
Crossing $V=0$ at the wrong $x$ value (it must be $60\ \text{pm}$ ).
Sketching the wrong sign near the electron end (it must be negative near $x=110\ \text{pm}$ ).

Things to Be Careful About

The axes given are from $x=10$ to $110\ \text{pm}$ , so do not try to draw vertical asymptotes at $x=0$ or $120\ \text{pm}$ on this plot; instead show the curve becoming steeper as you approach the ends of the plotted region.
Make sure the curve is smooth and continuous with a single zero crossing within the interval.

Techniques used

use the functional form V = k e (1/x - 1/(d-x))identify intercept where the potential is zerosketch a curve using key computed points and end behaviour

Q6MediumCapacitance

(a)

Two parallel plate capacitors $C_1$ and $C_2$ are connected to a supply that has a potential difference (p.d.) $V_S$ . The capacitors may be connected in series or in parallel.

The supply provides charge $Q_S$ and the plates of the two capacitors acquire charges $Q_1$ and $Q_2$ respectively. The p.d.s across the plates of the capacitors are $V_1$ and $V_2$ respectively.

Complete Table 6.1 to indicate how $Q_S$ , $Q_1$ and $Q_2$ relate to each other, and how $V_S$ , $V_1$ and $V_2$ relate to each other, for series and parallel connections of the capacitors to the supply.

Table 6.1

	relationship between charges	relationship between p.d.s
series
parallel

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Series:

Charges: $Q_S = Q_1 = Q_2$
p.d.s: $V_S = V_1 + V_2$

Parallel:

Charges: $Q_S = Q_1 + Q_2$
p.d.s: $V_S = V_1 = V_2$

Final answer

Series: QS = Q1 = Q2 and VS = V1 + V2. Parallel: QS = Q1 + Q2 and VS = V1 = V2.

Detailed explanation

Background Concept

Capacitance is defined by

C = \frac{Q}{V}

where $Q$ is the charge stored on a capacitor and $V$ is the potential difference (p.d.) across it.

For combinations of capacitors:

In series, the same charge must pass through each capacitor (there is only one path for charge flow), so each capacitor ends up with the same magnitude of charge.
In parallel, each capacitor is connected directly across the supply, so each has the same p.d. as the supply. The supply provides the total charge needed for both capacitors.

Understanding the Question

You are told there are two capacitors $C_1$ and $C_2$ connected to a supply of p.d. $V_S$ . The supply delivers charge $Q_S$ . The capacitors end up with charges $Q_1$ and $Q_2$ and p.d.s $V_1$ and $V_2$ .

The table asks for the relationships between:

$Q_S$ , $Q_1$ , $Q_2$ for series and parallel
$V_S$ , $V_1$ , $V_2$ for series and parallel

Approach

Recall the standard network rules:

Series: one path for charge (\Rightarrow) equal charge on each; voltages add.
Parallel: same two nodes (\Rightarrow) equal voltage across each; charges add.

Then write the required equalities.

Step-by-Step Reasoning

Series:

Because there is only one route for charge flow, the same amount of charge is transferred onto each capacitor plate pair. Hence

Q_1 = Q_2

The supply has provided that same amount of charge to the series combination, so

Q_S = Q_1 = Q_2

The p.d. from the supply is shared across the two capacitors in series, so the total p.d. is the sum:

V_S = V_1 + V_2

Parallel:

Each capacitor is connected directly across the supply terminals, so each has the full supply p.d.:

V_1 = V_2 = V_S

The supply must provide the total charge stored on both capacitors, so:

Q_S = Q_1 + Q_2

Key Takeaways

Series capacitors: same charge, voltages add.
Parallel capacitors: same voltage, charges add.

Common Mistakes

Writing $V_S = V_1 = V_2$ for series (this is only true for parallel).
Writing $Q_S = Q_1 + Q_2$ for series (this is only true for parallel).
Forgetting that in series it is the charge that is the same, not necessarily the capacitance or voltage.

Things to Be Careful About

Use the correct equality symbol structure: series requires a sum for voltages; parallel requires a sum for charges.
$Q_1$ and $Q_2$ are equal in magnitude in series (opposite signs on facing plates), but the question is about stored charge magnitudes, so $Q_1 = Q_2$ is the expected relationship.

Techniques used

use the defining relation C = Q/Vapply charge conservation in series and parallel capacitor networksapply potential difference rules for series and parallel connections

(b)

An isolated capacitor of capacitance $470\ \mu\text{F}$ stores $19\ \text{mJ}$ of energy.

7M

(i)

Calculate the p.d. across the capacitor.

p.d. = ______ $\text{V}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

W = \tfrac{1}{2}CV^2 \;\Rightarrow\; V = \sqrt{\frac{2W}{C}}

C = 470\ \mu\text{F} = 470 \times 10^{-6}\ \text{F},\quad W = 19\ \text{mJ} = 19 \times 10^{-3}\ \text{J}

V = \sqrt{\frac{2(19 \times 10^{-3})}{470 \times 10^{-6}}} = 8.99\ \text{V}

Answer

$9.0\ \text{V}$

Final answer

9.0 V

Detailed explanation

Background Concept

The energy stored in a capacitor is

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

where $W$ is energy in joules, $C$ is capacitance in farads, and $V$ is the p.d. across the capacitor in volts.

This formula comes from the work done to move charge onto the plates as the voltage rises from $0$ to $V$ .

Understanding the Question

An isolated capacitor has:

$C = 470\ \mu\text{F}$
stored energy $W = 19\ \text{mJ}$

You must find the p.d. $V$ across it.

Approach

Convert $\mu\text{F}$ to $\text{F}$ and $\text{mJ}$ to $\text{J}$ , then rearrange

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

to make $V$ the subject.

Step-by-Step Reasoning

Convert units:

C = 470\ \mu\text{F} = 470 \times 10^{-6}\ \text{F}

W = 19\ \text{mJ} = 19 \times 10^{-3}\ \text{J}

Rearrange the energy equation:

W = \frac{1}{2}CV^2 \Rightarrow V^2 = \frac{2W}{C} \Rightarrow V = \sqrt{\frac{2W}{C}}

Substitute:

V = \sqrt{\frac{2(19 \times 10^{-3})}{470 \times 10^{-6}}}

Calculate inside the square root:

\frac{2(19 \times 10^{-3})}{470 \times 10^{-6}} = \frac{38 \times 10^{-3}}{470 \times 10^{-6}} \approx 80.85

So

V \approx \sqrt{80.85} \approx 8.99\ \text{V} \approx 9.0\ \text{V}

Key Takeaways

Use $W = \tfrac{1}{2}CV^2$ when energy and capacitance are given.
Always convert prefixes: $\mu = 10^{-6}$ , $\text{m} = 10^{-3}$ .

Common Mistakes

Using $W = CV^2$ (missing the factor $\tfrac{1}{2}$ ).
Forgetting to convert $\mu\text{F}$ to $\text{F}$ , giving an answer too small by a factor of $\sqrt{10^6}$ .
Leaving energy in mJ but treating it as J.

Things to Be Careful About

The square root step: calculate $\frac{2W}{C}$ first, then take the square root.
Significant figures: inputs are typically given to 2 s.f. (19 mJ), so $V$ to 2 s.f. ( $9.0\ \text{V}$ ) is appropriate.

Techniques used

use the capacitor energy equation W = 1/2 C V^2convert micro- and milli- units to SIrearrange an equation to solve for potential difference

(ii)

Calculate the charge on the capacitor.

charge = ______ $\text{C}$

2M Similar

(iii)

The capacitor is now connected in parallel with a capacitor of capacitance $180\ \mu\text{F}$ that is initially uncharged.

Determine the total energy, in mJ, now stored in the two capacitors.

energy = ______ $\text{mJ}$

3M Similar

DifficultyMedium

Worked solution

Working

Initial charge (from (ii)):

Q = 4.23 \times 10^{-3}\ \text{C}

Parallel combination:

C_T = 470\ \mu\text{F} + 180\ \mu\text{F} = 650 \times 10^{-6}\ \text{F}

Charge is conserved, so final common p.d.

V = \frac{Q}{C_T} = \frac{4.23 \times 10^{-3}}{650 \times 10^{-6}} = 6.50\ \text{V}

Total energy stored:

W = \tfrac{1}{2} C_T V^2 = \tfrac{1}{2}(650 \times 10^{-6})(6.50)^2 = 1.38 \times 10^{-2}\ \text{J}

W = 13.8\ \text{mJ}

Answer

$13.8\ \text{mJ}$

Final answer

13.8 mJ

Detailed explanation

Background Concept

When a charged capacitor is connected in parallel with an initially uncharged capacitor:

The final p.d. across both is the same (parallel connection).
Total charge is conserved (assuming no external supply and negligible leakage): charge redistributes until both capacitors have the same voltage.

For parallel capacitors, the total capacitance is

C_T = C_1 + C_2.

If total charge is $Q$ , the final common voltage is

V = \frac{Q}{C_T}.

Energy stored at the end is

W = \frac{1}{2}C_TV^2.

Note: the total energy decreases because energy is dissipated as thermal energy in the connecting wires during the rapid redistribution of charge.

Understanding the Question

You start with:

Capacitor 1: $C_1 = 470\ \mu\text{F}$ , initially charged (energy $19\ \text{mJ}$ ).
Capacitor 2: $C_2 = 180\ \mu\text{F}$ , initially uncharged.

They are connected in parallel with each other (not to a supply). You must find the total final energy stored in both capacitors.

Approach

Find the initial total charge available (this is the charge on the first capacitor just before connection).
In parallel, compute $C_T = C_1 + C_2$ .
Use charge conservation to find the final voltage $V = Q/C_T$ .
Use $W = \tfrac{1}{2} C_T V^2$ to find the final stored energy, then convert to mJ.

Step-by-Step Reasoning

Initial charge on the charged capacitor (from part (ii)) is approximately

Q = 4.23 \times 10^{-3}\ \text{C}.

This is the total charge in the isolated system before connection, so it is conserved after connection.

Total capacitance in parallel:

C_T = C_1 + C_2 = (470 + 180)\ \mu\text{F} = 650\ \mu\text{F} = 650 \times 10^{-6}\ \text{F}.

Final common voltage across both capacitors:

V = \frac{Q}{C_T} = \frac{4.23 \times 10^{-3}}{650 \times 10^{-6}} \approx 6.50\ \text{V}.

This is lower than the original $\approx 9.0\ \text{V}$ because the same total charge is now spread over a larger total capacitance.

Final total stored energy:

W = \frac{1}{2} C_T V^2 = \frac{1}{2}(650 \times 10^{-6})(6.50)^2\ \text{J}.

Compute:

(6.50)^2 = 42.25

W = 0.5 \times 650 \times 10^{-6} \times 42.25 = 1.37 \times 10^{-2}\ \text{J}.

Convert to mJ:

1.37 \times 10^{-2}\ \text{J} = 13.7\ \text{mJ} \approx 13.8\ \text{mJ}.

Key Takeaways

Parallel connection forces the same final voltage across both capacitors.
In an isolated connection, total charge is conserved.
Stored energy usually decreases when charge redistributes (dissipated in circuit resistance).

Common Mistakes

Assuming the voltage stays at $9.0\ \text{V}$ after connecting in parallel (it does not, because the second capacitor draws charge).
Adding energies directly without recalculating the new voltage.
Forgetting that $C_T = C_1 + C_2$ only applies to parallel, not series.

Things to Be Careful About

The system is described as isolated (no supply), so use charge conservation, not constant voltage.
Use SI units in calculations; convert back to mJ only at the end.
A lower final energy than the initial $19\ \text{mJ}$ is physically reasonable here.

Techniques used

apply conservation of charge when connecting capacitors in paralleluse equivalent capacitance for capacitors in parallelcalculate final potential difference from Q = CVcompute stored energy using W = 1/2 CV^2

Q7MediumMagnetic Fields

(a)

State Faraday’s law of electromagnetic induction.

2M Similar

DifficultyEasy

Worked solution

Answer

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

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

Final answer

The induced e.m.f. equals the rate of change of magnetic flux linkage: ε = − d(NΦ)/dt.

Detailed explanation

Background Concept

Faraday’s law links electromagnetic induction to changing magnetic flux. The magnetic flux through a surface is $\Phi = BA$ (more generally $\Phi = BA\cos\theta$ ), and the flux linkage for a coil of $N$ turns is $N\Phi$ .

Faraday’s law states that an e.m.f. is induced when the flux linkage changes, and its magnitude equals the rate of change of flux linkage.

Including the direction information from Lenz’s law gives the negative sign:

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

Understanding the Question

You are asked to state Faraday’s law, so you must give the correct wording (rate of change of flux linkage) and/or the correct equation. For 2 marks, examiners typically expect both the idea and the mathematical form including the minus sign.

Approach

Provide the statement in words and then write the standard equation for induced e.m.f. in terms of the time rate of change of flux linkage, with the negative sign.

Step-by-Step Reasoning

Flux linkage is $N\Phi$ .
Induced e.m.f. equals the time rate of change of this quantity.
The negative sign indicates the induced e.m.f. acts to oppose the change that produces it (Lenz’s law).

So:

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

Key Takeaways

Induction requires a change in flux linkage.
Faraday gives the magnitude; the negative sign (Lenz) gives the direction.

Common Mistakes

Writing $\varepsilon = -\frac{\mathrm{d}\Phi}{\mathrm{d}t}$ without mentioning flux linkage when $N \neq 1$ .
Omitting the minus sign.
Saying “rate of change of flux” but not indicating it is flux linkage $N\Phi$ .

Things to Be Careful About

Use “flux linkage” in the statement to be unambiguous.
Ensure the differential is with respect to time $t$ .

Techniques used

state Faraday’s law in terms of flux linkageexpress the law using a differential equation including the negative sign

(b)

An aircraft is flying horizontally at constant speed $v$ through the Earth’s magnetic field, as shown in Fig. 7.1.

Fig. 7.1

At the location of the aircraft, the vertical component of the Earth’s magnetic field is $38\ \mu\text{T}$ towards the ground.

The distance between the wingtips P and Q of the aircraft is $68\ \text{m}$ .

As the aircraft moves through the magnetic field, an electromotive force (e.m.f.) of $0.54\ \text{V}$ is induced between the wingtips P and Q.

9M

(i)

Calculate the magnetic flux cut by the wings of the aircraft in a time of $15\ \text{s}$ . Give a unit with your answer.

magnetic flux = ______ unit ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $\varepsilon = \frac{\Delta \Phi}{\Delta t}$ ,

\Delta \Phi = \varepsilon \Delta t = 0.54 \times 15 = 8.1

Answer

Magnetic flux cut $= 8.1\ \text{Wb}$ (= $8.1\ \text{V s}$ ).

Final answer

8.1 Wb

Detailed explanation

Background Concept

Faraday’s law for a single conductor cutting magnetic field lines can be used in the form

\varepsilon = \frac{\Delta \Phi}{\Delta t}

This works because an induced e.m.f. corresponds to how quickly magnetic flux is being “cut” (i.e. how quickly the flux linkage changes). Magnetic flux $\Phi$ has unit weber (Wb), and

1\ \text{Wb} = 1\ \text{V s}.

Understanding the Question

You are given that the e.m.f. between the wingtips is $0.54\ \text{V}$ . Over $15\ \text{s}$ , the question asks for the total magnetic flux cut in that time. That is the change in flux corresponding to that sustained e.m.f.

Approach

Treat the given e.m.f. as a constant rate of cutting flux: multiply e.m.f. by time to get the flux cut. Then quote the correct unit (Wb).

Step-by-Step Reasoning

Starting from

\varepsilon = \frac{\Delta \Phi}{\Delta t}

Rearrange:

\Delta \Phi = \varepsilon \Delta t

Substitute:

\Delta \Phi = 0.54 \times 15 = 8.1

The unit is

\text{V} \times \text{s} = \text{Wb}.

Key Takeaways

If $\varepsilon$ is constant, then $\Delta\Phi = \varepsilon t$ .
Always attach the correct derived unit: Wb (or equivalently V s).

Common Mistakes

Forgetting the unit, or writing $\text{T}$ (tesla) instead of Wb.
Dividing by time instead of multiplying.

Things to Be Careful About

The question says “flux cut in a time of $15\ \text{s}$ ” so it is a total over time, not a rate.
Keep consistent significant figures (here 2 s.f. is appropriate).

Techniques used

use induced e.m.f. as the rate of change of magnetic fluxrearrange to find change in flux from e.m.f. and timestate the correct unit for magnetic flux

(ii)

Determine the area of flux cut by the wings in a time of $15\ \text{s}$ .

area = ______ $\text{m}^2$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

A = \frac{\Phi}{B} = \frac{8.1}{38 \times 10^{-6}} = 2.13 \times 10^{5}

Answer

Area $= 2.1 \times 10^{5}\ \text{m}^2$ .

Final answer

2.1 × 10^5 m^2

Detailed explanation

Background Concept

For a uniform magnetic field perpendicular to the area swept out,

\Phi = BA

where:

$\Phi$ is magnetic flux (Wb),
$B$ is magnetic flux density (T),
$A$ is area (m $^2$ ).

(If the field is not perpendicular, $\Phi = BA\cos\theta$ .)

Understanding the Question

You have already found the flux cut in $15\ \text{s}$ . The field component given ( $38\ \mu\text{T}$ ) is the component relevant to the aircraft’s horizontal motion, so you can treat the swept area as perpendicular to $B$ . The question asks for the area of flux cut in that time, meaning the area $A$ such that $\Phi = BA$ .

Approach

Rearrange $\Phi = BA$ to $A = \Phi/B$ , convert $38\ \mu\text{T}$ to tesla, then calculate $A$ .

Step-by-Step Reasoning

Use

A = \frac{\Phi}{B}

Convert field strength:

38\ \mu\text{T} = 38 \times 10^{-6}\ \text{T}

Substitute $\Phi = 8.1\ \text{Wb}$ :

A = \frac{8.1}{38 \times 10^{-6}} = 2.13 \times 10^{5}\ \text{m}^2

Key Takeaways

Flux is proportional to area for a uniform perpendicular field.
Micro ( $\mu$ ) means $\times 10^{-6}$ .

Common Mistakes

Forgetting to convert $\mu\text{T}$ to T.
Using $A = B/\Phi$ instead of $\Phi/B$ .
Including an unnecessary $\cos\theta$ when the relevant component is already given.

Things to Be Careful About

Keep track of powers of ten: dividing by $10^{-6}$ makes the result $10^{6}$ times larger.
Quote the final area in $\text{m}^2$ as requested.

Techniques used

apply the flux definition \u03a6 = BA for a uniform fieldrearrange to find area from flux and flux densityconvert magnetic flux density to tesla

(iii)

Use your answer in (b)(ii) to determine the speed $v$ of the aircraft.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

In $15\ \text{s}$ , distance travelled $= vt$ .

A = (68)(vt) = 68 \times v \times 15

v = \frac{A}{68 \times 15} = \frac{2.13 \times 10^{5}}{68 \times 15} = 2.09 \times 10^{2}

Answer

v = 2.1 \times 10^{2}\ \text{m s}^{-1}

Final answer

2.1 × 10^2 m s^-1

Detailed explanation

Background Concept

As an aircraft moves forward, its wings sweep out a rectangular area in space. If the wingspan is $L$ and the aircraft travels a distance $s$ in time $t$ , then the swept area is

A = Ls = L(vt).

This geometric area is the area used in $\Phi = BA$ when the field is perpendicular to the area.

Understanding the Question

From (b)(ii) you have the area of flux cut in $15\ \text{s}$ . The wingtip separation is $68\ \text{m}$ , which is the rectangle width, and the forward distance travelled in $15\ \text{s}$ is $vt$ , which is the rectangle length. You are asked to find $v$ .

Approach

Use the rectangle area relationship $A = 68 \times (v \times 15)$ and rearrange to solve for $v$ .

Step-by-Step Reasoning

Distance travelled in $15\ \text{s}$ :

s = vt

Swept area:

A = Ls = 68(v \times 15)

Rearrange:

v = \frac{A}{68 \times 15}

Substitute $A = 2.13 \times 10^5\ \text{m}^2$ :

v = \frac{2.13 \times 10^{5}}{1020} = 2.09 \times 10^{2}\ \text{m s}^{-1}

Key Takeaways

Converting induction information into speed often uses geometry of the “swept area”.
Always identify which length corresponds to wingspan and which corresponds to distance travelled.

Common Mistakes

Using $A = 68/ (vt)$ instead of $A = 68(vt)$ .
Forgetting that the time is $15\ \text{s}$ (using $t=1\ \text{s}$ ).
Mixing up units (e.g. using $68\ \text{cm}$ ).

Things to Be Careful About

Dimensional check: $A/(Lt)$ has units $\text{m}^2/(\text{m s}) = \text{m s}^{-1}$ .
Use a sensible number of significant figures consistent with the data.

Techniques used

relate swept area to distance traveled and wingspanrearrange to solve for speedcheck dimensional consistency

(iv)

Use Lenz’s law of electromagnetic induction to explain which of the wingtips P and Q is at the higher induced potential.

3M Similar

DifficultyMedium

Worked solution

Answer

If the circuit were completed, the induced current must produce a magnetic force opposing the aircraft’s motion (Lenz’s law).

For $\vec{v}$ to the right and $\vec{B}$ into the page (vertical downwards), the current in the wing must be from $Q$ to $P$ so that $\vec{F} = I\vec{L} \times \vec{B}$ is to the left.

Hence conventional current is $Q \to P$ , so $Q$ is at the higher potential (positive) and electrons collect at $P$ .

Final answer

Q is at the higher induced potential.

Detailed explanation

Background Concept

Two linked ideas are needed here:

Magnetic force on moving charge
A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force

\vec{F} = q(\vec{v} \times \vec{B}).

This causes charges in a moving conductor to separate, creating a potential difference.

Lenz’s law (direction of induction)
The induced e.m.f. (and induced current if a complete circuit exists) acts in such a direction as to oppose the change that produces it. In motional induction, that often means the induced current would create a magnetic force that opposes the motion causing the induction.

Understanding the Question

From the diagram: the aircraft moves to the right. The relevant component of Earth’s field is vertically downward, which is shown as crosses in the “view from above” (field into the page). The wingtips are labelled $P$ (upper wingtip on the page) and $Q$ (lower wingtip on the page). You must decide which wingtip ends up at higher potential.

Approach

Use the fact that if the wing formed part of a circuit, the induced current must oppose the aircraft’s motion (Lenz’s law).
Choose the current direction along the wing that would give a magnetic force on the wing opposite to the velocity using $\vec{F} = I\vec{L} \times \vec{B}$ (or Fleming’s left-hand rule).
Convert current direction into polarity: conventional current flows from higher to lower potential.

Step-by-Step Reasoning

diagram-iv-1

Take the “view from above”:

Velocity $\vec{v}$ is to the right.
Magnetic field $\vec{B}$ is into the page.

If there were an induced current in the wing, Lenz’s law says the magnetic force due to that current must oppose the motion (force to the left).

Let the current flow from $Q$ to $P$ (up the page). Then $\vec{L}$ is upward.
Using

\vec{F} = I\vec{L} \times \vec{B},

upward $\times$ into the page gives a force to the left, which opposes the motion. So this is the correct current direction.

Conventional current therefore is $Q \to P$ , meaning positive charge is at $Q$ and negative charge at $P$ . Hence $Q$ is at the higher potential.

(Equivalently, using $q(\vec{v} \times \vec{B})$ : with $\vec{v}$ right and $\vec{B}$ into page, $\vec{v} \times \vec{B}$ points down the page, so positive charges are driven toward $Q$ and electrons toward $P$ .)

Key Takeaways

Lenz’s law can be used as a quick direction check: induced current must oppose the motion producing it.
Use either $\vec{F}=q(\vec{v}\times\vec{B})$ for charge separation, or $\vec{F}=I\vec{L}\times\vec{B}$ for force on a current-carrying conductor.
Higher potential is where positive charge accumulates.

Common Mistakes

Reversing the $\vec{v} \times \vec{B}$ direction.
Forgetting that conventional current direction is opposite to electron motion.
Stating $P$ or $Q$ without any Lenz’s law argument (often loses explanation marks).

Things to Be Careful About

Be consistent with the field direction: crosses mean into the page.
Clearly link Lenz’s law to “force opposing motion” when using the conductor-force method.
If using Fleming’s left-hand rule, label field, current, and force directions explicitly.

Techniques used

apply Lenz’s law to determine the direction of the induced effectuse Fleming’s left-hand rule / vector product to relate current, field and forcededuce charge separation and hence polarity from magnetic force on charges

Q8MediumQuantum PhysicsNuclear Physics

(a)

State what is meant by a photon.

2M Similar

(b)

A stationary nucleus of uranium-238 ( $^{238}_{92}\text{U}$ ) undergoes alpha decay to produce a nucleus of thorium-234 ( $^{234}_{90}\text{Th}$ ). The kinetic energy of the emitted alpha particle is $4.200\ \text{MeV}$ . A gamma-ray photon is also emitted during the decay.

Assume that the rebound kinetic energy of the thorium nucleus is negligible.

Table 8.1 shows the masses of the nuclides involved in the decay reaction. The mass of the uranium-238 nuclide is missing.

Table 8.1

nuclide	nuclide mass / u
$^4_2\alpha$	4.000407
$^{234}_{90}\text{Th}$	233.915174
$^{238}_{92}\text{U}$

The total energy released in the decay of the nucleus of uranium-238 is $4.274\ \text{MeV}$ .

7M

(i)

Calculate the mass, in u, of the uranium-238 nuclide. Give your answer to five decimal places.

mass = ______ $\text{u}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Total energy released

Q = \left(m_{\text{U}} - m_{\text{Th}} - m_{\alpha}\right)c^2

Using $1\ \text{u}\,c^2 = 931.5\ \text{MeV}$ ,

m_{\text{U}} = m_{\text{Th}} + m_{\alpha} + \frac{Q}{931.5}

m_{\text{U}} = 233.915174 + 4.000407 + \frac{4.274}{931.5}

m_{\text{U}} = 237.920169\ \text{u}

Answer

m_{\text{U}} = 237.92017\ \text{u}

Final answer

237.92017 u

Detailed explanation

Background Concept

In a nuclear decay, the energy released (the $Q$ -value) comes from a decrease in total rest mass of the particles involved:

Q = \Delta mc^2 = \left(m_{\text{initial}} - m_{\text{final}}\right)c^2

If we work in atomic mass units $\text{u}$ , a very useful conversion is:

1\ \text{u}\,c^2 = 931.5\ \text{MeV}

So a mass change in $\text{u}$ can be obtained by dividing the energy in MeV by $931.5$ .

Understanding the Question

We are told uranium-238 decays to thorium-234 plus an alpha particle, and the total energy released is $4.274\ \text{MeV}$ . The table gives $m_{\alpha}$ and $m_{\text{Th}}$ , but $m_{\text{U}}$ is missing.

The statement about recoil being negligible is not needed for part (i) because $Q$ depends on rest masses only.

Approach

Use the definition of $Q$ :

initial mass = $m_{\text{U}}$
final mass = $m_{\text{Th}} + m_{\alpha}$

Rearrange to solve for $m_{\text{U}}$ and convert $Q$ into a mass in $\text{u}$ .

Step-by-Step Reasoning

Write the $Q$ -value relationship:

Q = \left(m_{\text{U}} - m_{\text{Th}} - m_{\alpha}\right)c^2

Convert energy to mass in $\text{u}$ :

\Delta m = \frac{Q}{931.5}\ \text{u}

Add this mass difference to the total final mass:

m_{\text{U}} = m_{\text{Th}} + m_{\alpha} + \Delta m

Substitute the given values:

m_{\text{Th}} + m_{\alpha} = 233.915174 + 4.000407 = 237.915581\ \text{u}

\Delta m = \frac{4.274}{931.5} = 0.004588\ \text{u}

m_{\text{U}} = 237.915581 + 0.004588 = 237.920169\ \text{u}

Round to five decimal places as requested.

Key Takeaways

Energy released in nuclear processes comes from mass decrease: $Q = \Delta mc^2$ .
With $Q$ in MeV and mass in u, use $1\ \text{u}c^2 = 931.5\ \text{MeV}$ .

Common Mistakes

Subtracting instead of adding $\frac{Q}{931.5}$ when finding $m_{\text{U}}$ .
Forgetting that $Q$ is the total released energy, not just the alpha particle kinetic energy.
Rounding too early and losing the required 5 d.p.

Things to Be Careful About

Quote the uranium mass to five decimal places exactly.
Keep masses in $\text{u}$ consistently; only convert the energy term.

Techniques used

use Q-value as mass difference times c^2convert energy in MeV to mass in u using 1 u c^2 = 931.5 MeVrearrange to find the unknown nuclide mass

(ii)

Determine a value for the wavelength of the gamma radiation emitted during the decay of the uranium-238 nucleus.

wavelength = ______ $\text{m}$

3M Similar

DifficultyMedium

Worked solution

Working

Energy of gamma photon (recoil KE negligible):

E_\gamma = 4.274 - 4.200 = 0.074\ \text{MeV}

E_\gamma = 0.074 \times 10^6 \times 1.602 \times 10^{-19} = 1.19 \times 10^{-14}\ \text{J}

\lambda = \frac{hc}{E_\gamma} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{1.19 \times 10^{-14}}

\lambda = 1.68 \times 10^{-11}\ \text{m}

Answer

\lambda = 1.68 \times 10^{-11}\ \text{m}

Final answer

1.68 × 10^-11 m

Detailed explanation

Background Concept

The total energy released in a decay ( $Q$ ) is shared between the kinetic energies of the decay products and any emitted photons. If a gamma photon is emitted, its energy is

E_\gamma = hf = \frac{hc}{\lambda}

So once you know $E_\gamma$ , you can find the wavelength using

\lambda = \frac{hc}{E_\gamma}

Also, converting from electronvolts:

1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}

Understanding the Question

We are given:

total energy released: $Q = 4.274\ \text{MeV}$
kinetic energy of alpha particle: $E_{k,\alpha} = 4.200\ \text{MeV}$
recoil kinetic energy of thorium is assumed negligible

So essentially, the remaining energy must be the gamma photon energy. Then we convert that energy to a wavelength.

Approach

Use energy conservation: $Q = E_{k,\alpha} + E_{k,\text{Th}} + E_\gamma$ .
With $E_{k,\text{Th}} \approx 0$ , find $E_\gamma = Q - E_{k,\alpha}$ .
Convert MeV to J.
Use $\lambda = hc/E_\gamma$ .

Step-by-Step Reasoning

Photon energy:

E_\gamma = 4.274 - 4.200 = 0.074\ \text{MeV}

Convert to joules:

0.074\ \text{MeV} = 0.074 \times 10^6\ \text{eV}

E_\gamma = 0.074 \times 10^6 \times 1.602 \times 10^{-19} = 1.185 \times 10^{-14}\ \text{J}

Use $\lambda = hc/E_\gamma$ :

\lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{1.185 \times 10^{-14}}

Compute the powers of ten: $10^{-34} \times 10^{8} / 10^{-14} = 10^{-12}$ , and the numbers give about $1.68$ :

\lambda \approx 1.68 \times 10^{-11}\ \text{m}

This is in the gamma-ray region (very short wavelength), which checks the plausibility.

Key Takeaways

The gamma photon takes the “leftover” energy after kinetic energies of particles are accounted for.
Convert MeV to J carefully before using $\lambda = hc/E$ .

Common Mistakes

Using $E_\gamma = 4.274\ \text{MeV}$ (forgetting the alpha already has $4.200\ \text{MeV}$ ).
Forgetting $10^6$ when converting MeV to eV.
Using $\lambda = E/hc$ instead of $\lambda = hc/E$ .

Things to Be Careful About

The assumption “recoil negligible” is what allows $E_\gamma = Q - E_{k,\alpha}$ .
Keep $h$ and $c$ to appropriate significant figures; final answer typically 2–3 s.f.

Techniques used

use energy conservation to find the photon energy from the Q-valueconvert photon energy from MeV to joulesapply E = hc/lambda to determine wavelength

(iii)

In practice, the rebound kinetic energy of the thorium nucleus is not negligible.

Explain, without further calculation, how your answer in (b)(ii) compares with the true wavelength of gamma radiation emitted during the decay of the uranium-238 nucleus.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

If the thorium nucleus has non-negligible recoil kinetic energy, less energy is available for the photon, so $E_\gamma$ is smaller.

Since

\lambda = \frac{hc}{E_\gamma}

the true wavelength is greater than the value found in (b)(ii).

Final answer

True wavelength is greater than the calculated value in (b)(ii).

Detailed explanation

Background Concept

In any decay, the released energy $Q$ is shared between:

kinetic energy of emitted particles,
kinetic energy of the recoiling daughter nucleus,
energy of any emitted gamma photons.

For a photon,

E_\gamma = \frac{hc}{\lambda}

So if the photon gets less energy, its wavelength must increase.

Understanding the Question

Part (b)(ii) assumed the thorium recoil kinetic energy was negligible, so essentially all energy not in the alpha’s kinetic energy went to the gamma photon. This part asks how the answer changes when that assumption is not valid.

Approach

Use energy sharing:

adding a non-zero recoil kinetic energy means the photon energy must be smaller than before.
then use the inverse relationship between wavelength and energy.

Step-by-Step Reasoning

True energy balance is

Q = E_{k,\alpha} + E_{k,\text{Th}} + E_\gamma

If $E_{k,\text{Th}}$ is not negligible, then compared to the earlier calculation, some of the “remaining” energy goes into $E_{k,\text{Th}}$ .
Therefore $E_\gamma$ is less than $Q - E_{k,\alpha}$ .
Since

\lambda = \frac{hc}{E_\gamma}

smaller $E_\gamma$ gives a larger $\lambda$ .

Key Takeaways

Recoil reduces photon energy.
Lower photon energy means longer wavelength.

Common Mistakes

Saying the wavelength becomes smaller (this would correspond to a higher photon energy).
Claiming the energy released $Q$ changes (it does not; only the partition changes).

Things to Be Careful About

The relationship is inverse: $E \propto 1/\lambda$ .
The question asks for comparison only, so no calculation is required.

Techniques used

apply energy conservation including recoil kinetic energyuse inverse relationship between photon energy and wavelength

(c)

Gamma radiation emitted during the decay of a sample of uranium-238 has a single wavelength.

Nuclei of cobalt-60 ( $^{60}_{27}\text{Co}$ ) decay by beta emission, and also emit gamma radiation in the process.

Suggest why there is not a single wavelength for the gamma radiation emitted during the decay of a sample of cobalt-60.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Gamma rays are emitted when the daughter nucleus drops between discrete nuclear energy levels. In $^{60}_{27}\text{Co}$ beta decay can leave the $^{60}\text{Ni}$ nucleus in excited states and it can de-excite via more than one transition (often a cascade), so photons of different energies (hence different wavelengths) are emitted.

Final answer

Because beta decay can populate different excited nuclear states (and/or a cascade of de-excitations), there are multiple gamma transitions so multiple wavelengths.

Detailed explanation

Background Concept

Gamma emission comes from a nucleus changing from a higher energy state to a lower energy state. Because nuclear energy levels are discrete, each transition has a specific energy difference $\Delta E$ , giving a photon of energy

E_\gamma = \Delta E = hf = \frac{hc}{\lambda}

If there is more than one possible transition (different excited states, or a multi-step cascade), then there will be more than one possible $\Delta E$ , so multiple gamma photon energies and wavelengths.

Understanding the Question

The question contrasts uranium-238 (given as producing gamma radiation of a single wavelength) with cobalt-60, which beta decays and also emits gamma radiation. It asks why cobalt-60 does not give a single gamma wavelength.

Approach

Explain in terms of nuclear energy-level structure:

beta decay can populate excited states of the daughter nucleus in different ways,
the daughter can de-excite through more than one transition (including cascades),
therefore multiple photon energies occur.

Step-by-Step Reasoning

In alpha decay of a particular nuclide, you may have effectively one dominant gamma transition, so a single photon energy is observed.
For $^{60}_{27}\text{Co}$ beta decay, the daughter nucleus ( $^{60}\text{Ni}$ ) can be left in excited states.
The nucleus may return to the ground state via more than one allowed transition, or via a two-step cascade (excited level $\to$ intermediate level $\to$ ground level).
Each transition corresponds to a different energy gap $\Delta E$ , so different gamma photon energies and hence different wavelengths.

Key Takeaways

Gamma photon energy is set by nuclear energy-level differences.
Multiple possible transitions/cascades lead to multiple gamma wavelengths.

Common Mistakes

Saying “beta particles have different energies so gammas do too” without mentioning nuclear energy levels/transition pathways.
Claiming gamma rays have a continuous spectrum (gamma lines are discrete; there are just multiple lines).

Things to Be Careful About

The mark is usually for: (1) reference to excited nuclear states/energy levels and (2) more than one transition (or cascade) leading to different $\lambda$ .

Techniques used

relate gamma emission to transitions between nuclear energy levelsexplain multiple photon energies via multiple transitions or cascades

Q9MediumAstronomy and Cosmology

(a)

State Wien’s displacement law.

2M Similar

DifficultyEasy

Worked solution

Answer

Wien’s displacement law:

\lambda_{\text{max}} T = b

where $b \approx 2.90 \times 10^{-3}\ \text{m K}$ .

Final answer

$\lambda_{\text{max}}T=2.90\times 10^{-3}\ \text{m K}$

Detailed explanation

Background Concept

A hot object emits a continuous spectrum of electromagnetic radiation. The spectrum has a peak at a particular wavelength $\lambda_{\text{max}}$ (the wavelength where the intensity is greatest). Wien’s displacement law links the position of this peak to the absolute temperature $T$ of the emitting surface.

It states that the product of the peak wavelength and absolute temperature is a constant:

\lambda_{\text{max}} T = b

where $b$ is the Wien constant, approximately $2.90 \times 10^{-3}\ \text{m K}$ .

Understanding the Question

You are asked to state Wien’s displacement law. That means give the relationship (ideally as an equation) between peak wavelength and temperature, and (for full credit) include the constant value and units.

Approach

Write the equation $\lambda_{\text{max}}T = b$ and quote the standard value of $b$ with correct units.

Step-by-Step Reasoning

Identify the relevant peak wavelength: $\lambda_{\text{max}}$ .
Use the law that links it to temperature:

\lambda_{\text{max}} T = b

Quote the constant:

b \approx 2.90 \times 10^{-3}\ \text{m K}

Key Takeaways

Hotter objects peak at shorter wavelengths.
The quantitative link is $\lambda_{\text{max}}T = 2.90 \times 10^{-3}\ \text{m K}$ .

Common Mistakes

Writing $\lambda_{\text{max}} \propto T$ (wrong direction; actually $\lambda_{\text{max}} \propto 1/T$ ).
Omitting the constant or giving it without units.
Using frequency instead of wavelength without stating the correct form.

Things to Be Careful About

$T$ must be in kelvin.
$\lambda_{\text{max}}$ must be in metres if using $b$ in $\text{m K}$ .

Techniques used

recall and state Wien’s displacement law in equation formquote the Wien constant with appropriate units

(b)

Fig. 9.1 shows the variation with $d^{-2}$ of the radiant flux intensity $F$ observed from a star X, where $d$ is the distance of the observer from the star. Fig. 9.2 shows the variation with wavelength $\lambda$ of the rates of emission $P$ of radiation by star X and the Sun.

Fig. 9.1

Fig. 9.2

The surface temperature of the Sun is $5770\ \text{K}$ .

State three conclusions about star X that can be drawn from this data. The conclusions may be qualitative or quantitative. Use the space for any working.

1 ______

2 ______

3 ______

3M Similar

DifficultyMedium

Worked solution

Working

From Fig. 9.1, straight line through origin for $F$ against $d^{-2}$ , so

F \propto d^{-2}\quad (F \propto \tfrac{1}{d^2}).

Gradient using the point $(d^{-2},F)=(3.0\times 10^{-23},\ 6.0\times 10^{3})$ :

\text{gradient}=\frac{6.0\times 10^{3}}{3.0\times 10^{-23}}=2.0\times 10^{26}\ \text{W}.

Using $F=\dfrac{L}{4\pi d^2}=(\dfrac{L}{4\pi})d^{-2}$ ,

\frac{L}{4\pi}=2.0\times 10^{26}\ \Rightarrow\ L=4\pi(2.0\times 10^{26})\approx 2.5\times 10^{27}\ \text{W}.

From Fig. 9.2, $\lambda_{\text{max,X}}\approx 4.0\times 10^{-7}\ \text{m}$ and $\lambda_{\text{max,Sun}}\approx 5.0\times 10^{-7}\ \text{m}$ .

Using $\lambda_{\text{max}}T=\text{constant}$ ,

T_X=T_{\odot}\frac{\lambda_{\text{max,Sun}}}{\lambda_{\text{max,X}}}=5770\times \frac{5.0}{4.0}\approx 7.2\times 10^{3}\ \text{K}.

Answer (three conclusions)

$F \propto d^{-2}$ (inverse-square dependence).
Luminosity of star X: $L \approx 2.5\times 10^{27}\ \text{W}$ .
Star X has higher surface temperature than the Sun, $T_X \approx 7.2\times 10^{3}\ \text{K}$ (shorter $\lambda_{\text{max}}$ ).

Final answer

F ∝ d⁻². 2. L ≈ 2.5 × 10^27 W. 3. T_X ≈ 7.2 × 10^3 K (hotter than Sun).

Detailed explanation

Background Concept

Two key ideas are being tested:

Inverse-square law for radiation
If a star emits power (luminosity) $L$ equally in all directions, then at distance $d$ that power is spread over the surface area of a sphere, $4\pi d^2$ . The radiant flux intensity (power received per unit area) is

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

So $F \propto 1/d^2$ , and if you plot $F$ against $d^{-2}$ you should get a straight line through the origin.

Wien’s displacement law
A hotter black-body spectrum peaks at a shorter wavelength. Quantitatively,

\lambda_{\text{max}}T=b.

So for two stars,

\frac{T_1}{T_2}=\frac{\lambda_{\text{max},2}}{\lambda_{\text{max},1}}.

Understanding the Question

You are given two graphs:

Fig. 9.1: $F$ plotted against $d^{-2}$ for star X.
Fig. 9.2: spectral curves $P$ vs $\lambda$ for star X and for the Sun.

You must make three conclusions about star X using information from these graphs. These can be qualitative (e.g. “hotter than the Sun”) or quantitative (e.g. “ $L \approx ...$ ”).

Approach

Use Fig. 9.1 to identify the dependence of $F$ on $d$ , and use the gradient to find $L$ via $F=(L/4\pi)d^{-2}$ .
Use Fig. 9.2 to read off the peak wavelengths and apply Wien’s law to estimate $T_X$ compared with $T_{\odot}=5770\ \text{K}$ .
Choose three clear, distinct conclusions (relationship + luminosity + temperature is a natural set).

Step-by-Step Reasoning

1) Relationship between $F$ and $d$

Fig. 9.1 is a straight line through the origin when plotting $F$ against $d^{-2}$ . That means

F \propto d^{-2} \Rightarrow F \propto \frac{1}{d^2}.

This is exactly what is expected for isotropic emission with negligible absorption between star and observer.

2) Luminosity of star X from the gradient

For isotropic emission,

F=\frac{L}{4\pi d^2} = \left(\frac{L}{4\pi}\right)d^{-2}.

So, on a graph of $F$ vs $d^{-2}$ :

gradient $= L/(4\pi)$ .

Using the given point on the line ( $d^{-2}=3\times 10^{-23}\ \text{m}^{-2}$ , $F=6\times 10^{3}\ \text{W m}^{-2}$ ):

\text{gradient}=\frac{6\times 10^{3}}{3\times 10^{-23}}=2\times 10^{26}\ \text{W}.

Then

L=4\pi(2\times 10^{26}) \approx 2.5\times 10^{27}\ \text{W}.

(Notice the unit: gradient has units of watts because $(\text{W m}^{-2})/(\text{m}^{-2})=\text{W}$ .)

3) Surface temperature of star X from the peak wavelength

From Fig. 9.2 the peak of star X occurs at a shorter wavelength than the Sun, so star X is hotter.

Reading approximate peaks:

Sun: $\lambda_{\text{max,Sun}}\approx 5\times 10^{-7}\ \text{m}$
Star X: $\lambda_{\text{max,X}}\approx 4\times 10^{-7}\ \text{m}$

Using Wien’s law in ratio form:

T_X = T_{\odot}\frac{\lambda_{\text{max,Sun}}}{\lambda_{\text{max,X}}} =5770\times \frac{5}{4}\approx 7.2\times 10^{3}\ \text{K}.

So a clear conclusion is $T_X \approx 7200\ \text{K}$ (hotter than the Sun).

Key Takeaways

A straight line of $F$ vs $d^{-2}$ implies the inverse-square law and lets you extract luminosity from the gradient.
Wien’s law lets you infer temperature from the peak wavelength, and compare temperatures using ratios.
Good “conclusions” are distinct: dependence, luminosity, and temperature are different properties.

Common Mistakes

Using $F = L/(4\pi d)$ instead of $F = L/(4\pi d^2)$ .
Treating the gradient as $L$ directly and forgetting the factor $4\pi$ .
Forgetting to convert the axis scale factors (e.g. $10^{3}$ and $10^{-23}$ ) when calculating the gradient.
Thinking a shorter peak wavelength means cooler (it means hotter).

Things to Be Careful About

Read approximate peak wavelengths from the graph; the final temperature should be given as an estimate.
When you quote three conclusions, make them clearly separate and unambiguous (one per line, as the question layout suggests).
Units: luminosity in $\text{W}$ , flux in $\text{W m}^{-2}$ , wavelength in $\text{m}$ , temperature in $\text{K}$ .

Techniques used

interpret a straight-line graph to infer proportionalityuse the inverse-square relationship between flux and distancedetermine a gradient from graph datarelate graph gradient to stellar luminosityuse Wien’s displacement law to estimate surface temperature from peak wavelength

(c)

Star X is in a galaxy that is moving away from the Earth.

Suggest, with a reason, how the line for star X in Fig. 9.2 would appear differently if it had been obtained from data measured on the Earth.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The spectrum for star X measured on Earth would be shifted to larger wavelengths (peak moves to the right), because light from a receding galaxy is red-shifted (Doppler/cosmological redshift stretches $\lambda$ ).

Final answer

Shift to longer wavelength (peak to the right) due to redshift from recession.

Detailed explanation

Background Concept

When a source of waves moves away from an observer, the observed wavelength increases. For light from a receding galaxy this is called redshift. In simplest terms:

observed wavelength is larger than emitted wavelength
equivalently, observed frequency is smaller

For spectral lines, this appears as lines shifted toward the red end. For a continuous black-body spectrum, the whole curve shifts so that the peak occurs at a larger wavelength.

Understanding the Question

Fig. 9.2 shows the emission of star X as it would be at the star (or at least not including the effect of the galaxy’s recession). The question asks: if the data were measured on Earth (with the galaxy moving away), how would the plotted curve for star X look different?

You need two things for 2 marks:

a correct suggested change to the curve,
a reason linked to recession/redshift.

Approach

Use the idea “receding source (\Rightarrow) redshift (\Rightarrow) wavelengths increase”. Then translate that into what happens on a $P$ vs $\lambda$ graph: the same shape but shifted to the right so $\lambda_{\text{max}}$ is larger.

Step-by-Step Reasoning

Because the galaxy is moving away, the light received at Earth is redshifted.
Redshift means each wavelength is multiplied by a factor greater than 1:

\lambda_{\text{obs}} > \lambda_{\text{emit}}.

Therefore every feature of the spectrum (including the peak) appears at a larger wavelength.

diagram-c1

So on Fig. 9.2 the curve for star X would be displaced to the right (toward longer $\lambda$ ). The overall shape is essentially the same; the key mark-worthy change is the shift in $\lambda_{\text{max}}$ .

Key Takeaways

Recession of a galaxy causes redshift.
Redshift shifts all wavelengths to larger values, so the peak wavelength increases.

Common Mistakes

Saying the peak shifts to smaller wavelength (that would correspond to blueshift, approaching source).
Only stating “redshift happens” without describing what happens on the graph.
Confusing a shift along the wavelength axis with a change in the height of the curve (the question is primarily about wavelength shift).

Things to Be Careful About

The graph is $P$ against $\lambda$ : redshift is most directly shown as a horizontal shift to higher $\lambda$ .
Use correct wording: “peak moves to the right / to longer wavelength”.

Techniques used

apply Doppler redshift ideas to electromagnetic spectrapredict how a spectral curve changes under wavelength stretchingjustify a graphical change with physical reasoning about recession

Q10Medium-EasyMedical Physics

(a)

Define specific acoustic impedance.

2M Similar

DifficultyEasy

Worked solution

Answer

Specific acoustic impedance $Z$ is the ratio of acoustic pressure to particle (oscillation) speed:

Z = \frac{p}{u}

Equivalently,

Z = \rho c

Unit: $\text{kg m}^{-2} \text{s}^{-1}$ .

Final answer

Z = p/u = ρc (unit: kg m^-2 s^-1)

Detailed explanation

Background Concept

When a sound (or ultrasound) wave travels through a medium, it produces:

a small alternating pressure change $p$ (acoustic pressure), and
a small alternating particle speed $u$ (the speed of oscillation of the medium’s particles, not the wave speed).

The specific acoustic impedance $Z$ is a property of the medium that describes how much pressure change is needed to produce a given particle speed. It is closely related to how strongly waves reflect at boundaries.

Two equivalent expressions used at A level are:

Z = \frac{p}{u}

and, for a plane progressive wave,

Z = \rho c

where $\rho$ is the density of the medium and $c$ is the speed of sound in that medium.

Understanding the Question

You are asked to define specific acoustic impedance, so you need a clear statement of what it is, usually with a formula, and (for full credit) the standard equivalent form or the unit.

Approach

State the definition as a ratio ( $p/u$ ), then give the commonly used equivalent expression ( $\rho c$ ) and the unit.

Step-by-Step Reasoning

By definition, specific acoustic impedance compares wave pressure to particle speed, so write:

Z = \frac{p}{u}

For a plane wave in a medium, the relationship between pressure and particle speed leads to:

Z = \rho c

Check the unit using $Z=\rho c$ :
$\rho$ has unit $\text{kg m}^{-3}$ and $c$ has unit $\text{m s}^{-1}$ , so
$Z$ has unit $\text{kg m}^{-2} \text{s}^{-1}$ .

Key Takeaways

Specific acoustic impedance is a medium property.
Useful forms: $Z = p/u$ and $Z = \rho c$ .
Unit: $\text{kg m}^{-2} \text{s}^{-1}$ .

Common Mistakes

Saying “impedance is resistance to sound” without defining it quantitatively.
Confusing particle speed $u$ with wave speed $c$ .
Writing $Z = p/v$ and using $v$ ambiguously.

Things to Be Careful About

Use the correct symbols: $p$ for acoustic pressure, $u$ for particle speed, $c$ for wave speed.
Include the unit if you are aiming for full definition marks.

Techniques used

state a definition using the ratio of acoustic pressure to particle velocityuse the equivalent expression as the product of density and wave speedquote the correct SI unit for the defined quantity

(b)

Explain how ultrasound waves are detected by a piezoelectric crystal.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Incident ultrasound produces alternating pressure on the piezoelectric crystal so it alternately compresses and expands.

This deformation causes charge separation in the crystal, producing an alternating potential difference across electrodes at the same frequency as the ultrasound (detected as a voltage signal).

Final answer

Ultrasound causes alternating compression/expansion of the piezoelectric crystal, producing alternating charge separation and hence an a.c. p.d. across electrodes that is measured as the signal.

Detailed explanation

Background Concept

A piezoelectric material produces an electrical effect when it is mechanically stressed:

compression/extension changes the positions of ions in the crystal lattice,
this creates a separation of charge, giving a potential difference across opposite faces.

This works in reverse too: applying an alternating p.d. makes the crystal vibrate and generate ultrasound.

Understanding the Question

The question asks specifically about detection. So you must explain how an incoming ultrasound wave is converted into an electrical signal by the piezoelectric crystal.

Approach

Describe the chain:

ultrasound exerts alternating pressure,
crystal deforms at the wave frequency,
deformation creates alternating charge separation,
measurable alternating voltage/current appears at the electrodes.

Step-by-Step Reasoning

Ultrasound is a longitudinal wave, so it consists of compressions and rarefactions (pressure increases and decreases).
When the wave reaches the crystal, the changing pressure applies a changing force on the crystal faces.
The crystal therefore alternately compresses and expands in step with the pressure oscillations.
Due to the piezoelectric effect, this mechanical strain causes electric dipoles within the crystal to align/shift, producing opposite charges on opposite faces.
If electrodes are connected to these faces, an alternating potential difference is produced and can be amplified/recorded. The frequency matches the ultrasound frequency, so it is a faithful detection of the incoming wave.

Key Takeaways

Detection is mechanical-to-electrical energy conversion.
Alternating pressure from ultrasound produces an alternating voltage signal.

Common Mistakes

Describing only the generation mode (applying a.c. voltage makes ultrasound) without mentioning incoming pressure producing voltage.
Saying “the crystal vibrates” but not linking this to charge separation / p.d.

Things to Be Careful About

Make clear it is the changing pressure (compressions/rarefactions) that produces an alternating electrical signal.
Mention electrodes/potential difference to show it is detectable/measureable.

Techniques used

describe the effect of alternating pressure on the crystal shapelink mechanical deformation to charge separation (piezoelectric effect)state that the resulting alternating p.d. can be measured as the detected signal

(c)

Table 10.1 shows the specific acoustic impedance $Z$ for body tissue, water and steel.

Table 10.1

material	$Z / \text{kg m}^{-2} \text{s}^{-1}$
body tissue	$1.38 \times 10^6$
water	$1.48 \times 10^6$
steel	$4.04 \times 10^7$

4M

(i)

Calculate the intensity reflection coefficient for ultrasound incident on a water–steel boundary.

intensity reflection coefficient = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For normal incidence,

R = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2

Water: $Z_1 = 1.48 \times 10^6\ \text{kg m}^{-2} \text{s}^{-1}$ , steel: $Z_2 = 4.04 \times 10^7\ \text{kg m}^{-2} \text{s}^{-1}$ .

R = \left( \frac{4.04\times 10^7 - 1.48\times 10^6}{4.04\times 10^7 + 1.48\times 10^6} \right)^2 = \left( \frac{3.892\times 10^7}{4.188\times 10^7} \right)^2 \approx (0.929)^2 \approx 0.86

Answer

Intensity reflection coefficient $\approx 0.86$ .

Final answer

0.86

Detailed explanation

Background Concept

When a wave meets a boundary between two media, part of the wave is reflected and part is transmitted. For ultrasound at normal incidence, the fraction of intensity reflected depends on the specific acoustic impedances $Z_1$ and $Z_2$ .

The (intensity) reflection coefficient $R$ is

R = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2

Key idea: a large mismatch in $Z$ gives a value of $R$ close to $1$ (strong reflection).

Understanding the Question

You are given $Z$ values for water and steel and asked for the intensity reflection coefficient when ultrasound hits a water–steel boundary. The boundary is a big impedance mismatch, so you should expect a large reflected fraction.

Given:

$Z_{\text{water}} = 1.48 \times 10^6\ \text{kg m}^{-2} \text{s}^{-1}$
$Z_{\text{steel}} = 4.04 \times 10^7\ \text{kg m}^{-2} \text{s}^{-1}$

Unknown: $R$ (a dimensionless fraction between $0$ and $1$ ).

Approach

Use the standard formula for $R$ at normal incidence, substitute $Z_1$ and $Z_2$ , then square the ratio. Round sensibly (typically 2 s.f.).

Step-by-Step Reasoning

Write the reflection coefficient formula:

R = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2

Assign the media. Either order works because the expression is squared, but keep consistent:
$Z_1 = Z_{\text{water}}$ , $Z_2 = Z_{\text{steel}}$ .
Calculate difference and sum:

Z_2 - Z_1 = 4.04\times 10^7 - 1.48\times 10^6 = 3.892\times 10^7

Z_2 + Z_1 = 4.04\times 10^7 + 1.48\times 10^6 = 4.188\times 10^7

Form the ratio and square:

\frac{Z_2 - Z_1}{Z_2 + Z_1} = \frac{3.892}{4.188} \approx 0.929

R \approx (0.929)^2 \approx 0.864 \approx 0.86

Interpretation: about $86\%$ of the intensity is reflected at a water–steel boundary.

Key Takeaways

Use $R=\left(\frac{Z_2-Z_1}{Z_2+Z_1}\right)^2$ for normal incidence.
Big impedance mismatch (steel vs water) gives strong reflection.

Common Mistakes

Forgetting to square the ratio (that would give the amplitude reflection coefficient, not intensity).
Mixing up powers of ten when subtracting/adding $Z$ values.
Giving a value greater than $1$ (reflection coefficient must be between $0$ and $1$ ).

Things to Be Careful About

Ensure both impedances are in the same units (they are).
Keep enough significant figures during the ratio before squaring, then round at the end.
The result is dimensionless (no unit).

Techniques used

select the intensity reflection coefficient formula in terms of impedancessubstitute impedance values with correct powers of tensquare the amplitude reflection coefficient to obtain an intensity ratio

(ii)

Explain, without calculation, what is likely to happen when ultrasound is incident on a body tissue–water boundary.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

$Z$ for body tissue and water are very similar, so there is little impedance mismatch.

Therefore only a small fraction of the ultrasound intensity is reflected at the boundary and most is transmitted into the water.

Final answer

Very little reflection; most ultrasound transmitted (tissue and water impedances are similar).

Detailed explanation

Background Concept

Reflection at a boundary happens because the wave must satisfy boundary conditions in two different media. The amount reflected depends mainly on how different the acoustic impedances are.

If $Z_1 \approx Z_2$ , the wave “matches” well across the boundary, so reflection is small and transmission is large.
If $Z_1$ and $Z_2$ differ greatly, reflection is large.

This is why ultrasound gel (similar impedance to skin/tissue) is used: it reduces reflection at the skin surface and allows more energy into the body.

Understanding the Question

You are not asked to calculate. You are asked what is likely to happen at a body tissue–water boundary. From Table 10.1:

body tissue: $1.38 \times 10^6$
water: $1.48 \times 10^6$

These are close, so you infer a small reflection coefficient.

Approach

Compare the numerical sizes of the two impedances and use the rule “small mismatch → small reflection”. Then state what happens to the incident ultrasound intensity (mostly transmitted, small reflected echo).

Step-by-Step Reasoning

Look at the table values: $Z_{\text{tissue}}$ and $Z_{\text{water}}$ differ by only about $0.10 \times 10^6$ on a base of about $1.4 \times 10^6$ (a small percentage difference).
Because the impedances are similar, the boundary does not strongly oppose the wave motion in either direction.
Therefore the reflected wave is weak (small reflected intensity).
Most of the ultrasound energy continues across the boundary into the water (large transmitted intensity).

Key Takeaways

Reflection depends on impedance mismatch.
Tissue and water are well matched, so reflection is minimal.

Common Mistakes

Claiming “most is reflected” just because there is a boundary.
Confusing this with the water–steel case, where the mismatch is huge.

Things to Be Careful About

Use the table: your justification must explicitly come from “ $Z$ values are similar”.
Keep the answer qualitative (the question explicitly says “without calculation”).

Techniques used

compare the relative size of the two impedances qualitativelyinfer the size of the reflection coefficient from impedance mismatchstate the consequence for reflected and transmitted intensity

Physics 9702/43 — October/November 2025

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