(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 (obeys pV = NkT). · (b)(ii) 3.4 × 10^22 · (b)(iii) 5.6 × 10^-2 mol · (c) 4.0 u

9702/41 Oct/Nov 2025 — Physics Solved Past Paper

Q1Medium-EasyMotion in a CircleOscillations

(a)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Speed (magnitude of velocity) is constant but the velocity direction changes continuously (tangent to the circle).

The acceleration is centripetal: directed towards the centre (perpendicular to the velocity).

Final answer

Speed is constant but velocity direction changes; acceleration is towards the centre (perpendicular to velocity).

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 changing because velocity is a vector (it has direction as well as magnitude).

A change in velocity implies an acceleration. For circular motion, this acceleration is called centripetal acceleration and it always points towards the centre of the circle, continually changing the direction of the velocity so that the object stays on a circular path.

Understanding the Question

You are asked to describe uniform circular motion specifically in terms of velocity and acceleration. So you must mention (1) what happens to the velocity, and (2) what the acceleration is like (direction and nature).

Approach

Give two crisp statements:

Velocity: magnitude constant, direction changing (tangent to circle).
Acceleration: towards centre, perpendicular to velocity (centripetal).

Step-by-Step Reasoning

Because the motion is uniform, the speed stays the same.
But as the object goes around the circle, the direction of motion changes at every point, so the velocity vector changes.
Therefore there must be acceleration.
The acceleration must be directed towards the centre to bend the path into a circle, and it is perpendicular to the instantaneous velocity.

Key Takeaways

Uniform circular motion: constant speed, changing velocity.
Centripetal acceleration points towards the centre and is perpendicular to velocity.

Common Mistakes

Saying “velocity is constant” (only the speed is constant).
Missing the direction of acceleration (must be towards the centre).

Things to Be Careful About

Use vector language correctly: velocity changes in direction, not magnitude.
Acceleration is not zero even though speed is constant.

Techniques used

state how velocity changes in uniform circular motionstate the direction and nature of centripetal acceleration

(b)

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

Fig. 1.1

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

When an object moves in a circle, its position can be described by an angle $\theta$ measured from some reference line. If you take the projection of the radius onto a line perpendicular to that reference, you get a sinusoidal relationship (involving $\sin$ or $\cos$ depending on how $\theta$ is defined).

Understanding the Question

The ball is at point $B$ on a circle of radius $R$ centred at $O$ . The line $OP$ is perpendicular to the screen (and parallel to the light rays). The shadow position on the screen is the projection of $B$ along $OP$ onto the screen, so the displacement $x$ of the shadow from $P$ is the component of $OB$ parallel to the screen.

Angle $\theta$ is between $OP$ and $OB$ . Therefore, the component of $OB$ perpendicular to $OP$ (i.e. along the screen) is $R\sin\theta$ .

Approach

Use right-triangle trigonometry on the radius vector $OB$ : the displacement along the screen is the opposite side to angle $\theta$ .

Step-by-Step Reasoning

Radius magnitude: $|OB| = R$ .
Angle between $OB$ and $OP$ is $\theta$ .
Component of $OB$ perpendicular to $OP$ is

x = R\sin\theta

Key Takeaways

The shadow displacement is a projection of circular motion, giving a sine (or cosine) dependence.

Common Mistakes

Using $x = R\cos\theta$ when the geometry given defines $\theta$ such that the opposite component is required.
Forgetting that $x$ is a displacement from $P$ , not the arc length.

Things to Be Careful About

Ensure $\theta$ is the angle between $OP$ and $OB$ as stated, not between $OB$ and the screen.
Sign convention: the expression gives magnitude; direction depends on which side of $P$ is defined positive in the diagram.

Techniques used

project circular motion onto a diameter using trigonometryrelate displacement to radius and angular position

(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

DifficultyMedium-Easy

Worked solution

Working

From (c)(i):

x = R\sin\theta

From (c)(ii):

\theta = \omega t

Substitute:

x = R\sin(\omega t)

Answer

x = R\sin\omega t

Final answer

x = R\sin\omega t

Detailed explanation

Background Concept

The projection of uniform circular motion onto a diameter gives simple harmonic motion. Mathematically, if a point goes around a circle with constant angular speed, its horizontal/vertical projection varies as $\sin\theta$ or $\cos\theta$ . Since $\theta$ increases linearly with time ( $\theta = \omega t$ ), the projection becomes a sine/cosine function of time.

Understanding the Question

You already found:

a geometric link between the shadow displacement $x$ and the angle $\theta$ ;
a time dependence for $\theta$ .

You must combine them to show $x$ depends on time as $R\sin\omega t$ .

Approach

Take the expression for $x(\theta)$ and substitute $\theta(t)$ to obtain $x(t)$ .

Step-by-Step Reasoning

From the geometry,

x = R\sin\theta

For constant angular speed with $\theta=0$ at $t=0$ ,

\theta = \omega t

Substitute $\theta = \omega t$ into $x = R\sin\theta$ :

x = R\sin(\omega t)

which is exactly the required result.

Key Takeaways

Combining a projection ( $\sin\theta$ ) with $\theta = \omega t$ gives sinusoidal motion in time.

Common Mistakes

Writing $x = R\sin\theta = R\sin\omega \times t$ (incorrectly separating the sine argument).
Dropping brackets: must be $\sin(\omega t)$ .

Things to Be Careful About

Ensure the sine argument is dimensionless: $\omega t$ is in radians.
Use the same symbol $\omega$ consistently (it is angular speed for both the circle and the SHM model here).

Techniques used

substitute an angular-position expression into a displacement projectionuse a trigonometric form to identify sinusoidal time dependence

(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)

is of the standard SHM form $x = x_0\sin(\omega t)$ , so the shadow’s displacement varies sinusoidally with time (amplitude $R$ , angular frequency $\omega$ ) and can be modelled as SHM.

Final answer

Because x = R sin(ωt) matches the SHM form x = x0 sin(ωt) (sinusoidal with amplitude R and angular frequency ω).

Detailed explanation

Background Concept

Simple harmonic motion (SHM) can be recognised in two equivalent ways:

The displacement varies sinusoidally with time:

x = x_0\sin(\omega t) \quad \text{or} \quad x = x_0\cos(\omega t)

The acceleration is proportional to and opposite the displacement:

a = -\omega^2 x

Either of these is enough to justify that a motion is SHM (provided the motion is along a line about an equilibrium position).

Understanding the Question

You have obtained

x = R\sin(\omega t)

The question asks you to explain why the shadow motion on the screen may be modelled as SHM, referring to this equation.

Approach

Compare the given equation directly to the standard SHM displacement equation and identify the parameters (amplitude and angular frequency). Optionally note that it implies $a \propto -x$ .

Step-by-Step Reasoning

Standard SHM form is $x = x_0\sin(\omega t)$ .
Your equation is exactly that with $x_0 = R$ .
Therefore the shadow oscillates back and forth about $x=0$ with sinusoidal variation in time, i.e. it is SHM.

(If you differentiate twice: $x = R\sin(\omega t) \Rightarrow a = \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\omega^2 R\sin(\omega t) = -\omega^2 x$ , which is the defining acceleration condition for SHM.)

Key Takeaways

Recognise SHM by matching $x(t)$ to $x_0\sin(\omega t)$ / $x_0\cos(\omega t)$ .
Projection of uniform circular motion produces SHM.

Common Mistakes

Saying “it is SHM because it is a sine graph” without linking to the standard form and identifying the equilibrium position.
Mixing up $\omega$ (angular frequency) with $f$ or $T$ .

Things to Be Careful About

The model is for the shadow displacement along the screen, not the ball’s circular motion itself.
State clearly the connection to the SHM equation (that is what the mark is for).

Techniques used

compare a displacement-time equation with the standard SHM formidentify amplitude and angular frequency from the 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_{\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, acceleration is related to displacement by

a = -\omega^2 x

The magnitude of acceleration is largest when $|x|$ is largest, i.e. at the extremes $x = \pm x_0$ (where $x_0$ is amplitude). Therefore,

a_{\max} = \omega^2 x_0

Understanding the Question

You have angular frequency $\omega = 1.9\ \text{rad s}^{-1}$ and the amplitude is the radius $R = 0.23\ \text{m}$ . You must find the maximum acceleration of the shadow in SHM.

Approach

Use $a_{\max} = \omega^2 x_0$ with $x_0 = 0.23\ \text{m}$ .

Step-by-Step Reasoning

Identify amplitude:

x_0 = R = 0.23\ \text{m}

Apply the formula:

a_{\max} = \omega^2 x_0 = (1.9)^2(0.23)

Calculate:

(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, maximum acceleration occurs at maximum displacement.
Use $a_{\max} = \omega^2 x_0$ .

Common Mistakes

Using $a_{\max} = \omega x_0$ (missing one power of $\omega$ ).
Using diameter as $x_0$ .

Things to Be Careful About

Units: rad is dimensionless, so $\omega^2 x_0$ gives $\text{m s}^{-2}$ .
Use the amplitude for the shadow, which equals the radius $R$ , not the diameter.

Techniques used

use the SHM relation between maximum acceleration, angular frequency and amplitudesubstitute numerical values and calculate

(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-Easy

Worked solution

diagram-1

Answer

Maximum positive acceleration occurs at the most negative displacement: $x = -R$ (left of $P$ on the screen). Mark this point as $A$ .

Final answer

Point A at x = −R (left of P on the screen).

Detailed explanation

Background Concept

For SHM,

a = -\omega^2 x

So:

acceleration is always directed towards equilibrium ( $x=0$ ), because it has the opposite sign to $x$ ;
the acceleration magnitude is

|a| = \omega^2 |x|

and is therefore maximum when $|x|$ is maximum, i.e. at $x = \pm x_0$ .

Understanding the Question

The shadow performs SHM along the screen with displacement $x$ measured from $P$ (as shown on the diagram). You are asked to mark the position of the shadow when it has maximum positive acceleration.

Approach

Maximum magnitude acceleration happens at the ends of the motion ( $x = \pm x_0$ ).
Use $a = -\omega^2 x$ to decide which end gives positive acceleration.

Step-by-Step Reasoning

At the extremes, $x = \pm R$ .
If $x = +R$ then

a = -\omega^2(+R)

so $a$ is negative.

If $x = -R$ then

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

so $a$ is positive and its magnitude is maximum.

Therefore the shadow must be at the left-hand extreme (negative $x$ direction from $P$ , given the diagram’s arrow for positive $x$ ).

diagram-1

Key Takeaways

Maximum acceleration in SHM occurs at maximum displacement.
The sign comes from $a = -\omega^2 x$ (opposite to displacement).

Common Mistakes

Choosing $x = +R$ because it is “maximum” displacement, without considering the sign of acceleration.
Confusing maximum acceleration with maximum speed (maximum speed occurs at $x=0$ ).

Things to Be Careful About

Always use the diagram’s sign convention for $x$ (here positive $x$ is to the right from $P$ ).
“Maximum positive acceleration” means largest acceleration in the +x direction, not largest magnitude regardless of direction.

Techniques used

use the SHM relation a = -omega^2 x to link acceleration direction to displacementidentify the position corresponding to maximum magnitude acceleration

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 to/from the system (heating/cooling), $q$ .
By work done on/by the system, $W$ (e.g. compression/expansion).

Final answer

By heating (thermal energy transfer) and by doing work (work done on/by the system).

Detailed explanation

Background Concept

The first law of thermodynamics is an energy-conservation statement for a thermodynamic system:

\Delta U = q + W

where:

$\Delta U$ is the change in internal energy of the system,
$q$ is the thermal energy transferred to the system (positive when energy is supplied as heat),
$W$ is the work done on the system (positive when the surroundings do work on the system, e.g. compression).

So internal energy can be changed by (i) heating/cooling and (ii) doing mechanical/electrical work on the system.

Understanding the Question

The question asks for two distinct ways, according to the first law, that $U$ can change. These correspond directly to the two terms on the right-hand side of $\Delta U = q + W$ .

Approach

Read the first law and identify the two mechanisms of energy transfer into/out of the system: heat transfer $q$ and work transfer $W$ . State each clearly.

Step-by-Step Reasoning

From $\Delta U = q + W$ , one contribution is $q$ : supplying thermal energy increases $U$ (removing thermal energy decreases $U$ ).
The other contribution is $W$ : work done on the system increases $U$ (work done by the system tends to decrease $U$ ).

Key Takeaways

The first law gives two routes to change internal energy: heat transfer and work transfer.
These are the only two pathways in the first law expression.

Common Mistakes

Stating only “temperature change” instead of the mechanism (heat transfer or work).
Mixing up sign conventions without making it clear that the mechanism is work done on/by the system.

Things to Be Careful About

Use the language of energy transfer: “thermal energy transferred” and “work done”.
Ensure the two ways are distinct (do not give two examples of work, for instance).

Techniques used

state the first law of thermodynamics in wordsidentify the two energy-transfer pathways 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 quickly means there is little time for heat transfer, so $q \approx 0$ .

When the gas is compressed, work is done on the gas ( $W>0$ ), so by

\Delta U = q + W

$\Delta U \approx W > 0$ and the internal energy increases.

The temperature of the air rises, and heat is then transferred from the hot air to the pump so the pump becomes hot.

Final answer

Rapid compression is approximately adiabatic (q ≈ 0); work done on the air increases internal energy so temperature rises and the pump warms.

Detailed explanation

Background Concept

For a gas, internal energy $U$ is mainly the random kinetic energy (and, for real gases, some potential energy) of its molecules. A higher temperature corresponds to a higher mean random kinetic energy.

The first law relates changes in $U$ to energy transfers:

\Delta U = q + W

If you compress a gas, the surroundings do work on it, so $W>0$ .
If the process happens so quickly that little heat can flow in or out, then $q \approx 0$ (adiabatic-like behaviour).

Understanding the Question

A bicycle pump is used to inflate a tyre quickly. The observation is that the pump gets hot. The question asks for an explanation specifically using the first law: identify $q$ and $W$ and deduce what happens to $\Delta U$ and temperature.

Approach

Decide whether heat transfer $q$ is significant during the rapid pumping.
Identify the sign of $W$ during compression.
Use $\Delta U = q + W$ to conclude what happens to $U$ .
Link increased internal energy to increased temperature of the gas and then to heating of the pump cylinder.

Step-by-Step Reasoning

Rapid process implies small heat transfer: Pumping quickly means the compression happens over a short time, so there is not much time for thermal energy to flow from the gas to the surroundings. Therefore, take $q \approx 0$ during the compression stroke.
Compression means work done on the gas: You push the handle in, decreasing the gas volume. The pump (surroundings) does work on the gas, so $W>0$ .
Apply the first law:

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

So $\Delta U > 0$ .
4. Interpret $\Delta U>0$ : The internal energy of the air increases. For a gas, this corresponds to an increase in the average random kinetic energy of molecules, so the gas temperature rises.
5. Why the pump feels hot: The hotter compressed air is in contact with the pump walls. After (or during) repeated strokes, thermal energy transfers from the hot gas to the metal/plastic cylinder, so the pump itself becomes warm/hot.

Key Takeaways

Fast compression is approximately adiabatic: $q \approx 0$ .
Compression implies $W>0$ (work done on the gas), so $\Delta U$ increases.
Increased internal energy of a gas means higher temperature, so objects in contact warm up.

Common Mistakes

Saying “pressure increases so temperature increases” without using the first law or mentioning work done.
Claiming heat is transferred into the gas from outside during rapid pumping (the key point is that heat transfer is minimal during the rapid compression).
Mixing sign conventions and concluding $\Delta U$ decreases during compression.

Things to Be Careful About

The mark-worthy idea is the energy pathway: temperature rises because mechanical work is done on the gas, not because heat is supplied.
Use wording like “little time for heat transfer” rather than claiming $q$ is exactly zero.

Techniques used

apply the first law of thermodynamics to a physical processidentify when heat transfer is negligible (adiabatic process)link internal energy increase to 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 is related to the average random kinetic energy of the molecules.

During vaporisation at $100\ ^{\circ}\text{C}$ , the energy supplied is used to increase the (intermolecular) potential energy as molecules separate / to break intermolecular bonds, not to increase their average kinetic energy.

Since the average kinetic energy does not increase, the temperature remains at $100\ ^{\circ}\text{C}$ until all the water has vaporised.

Final answer

Energy supplied during boiling increases intermolecular potential energy (separation) rather than average kinetic energy, so temperature stays at 100 °C until vaporisation is complete.

Detailed explanation

Background Concept

Temperature is a measure of the average random kinetic energy of molecules. In simple terms:

higher $T$ (\Rightarrow) molecules have greater average kinetic energy,
constant $T$ (\Rightarrow) average kinetic energy stays constant.

Internal energy $U$ is the total microscopic energy of the molecules:

random kinetic energy of molecules,
potential energy due to intermolecular forces.

When a substance is heated without changing state, the added energy mainly increases kinetic energy, so temperature rises.

During a change of state (e.g. liquid to gas), energy is required to overcome intermolecular attractions and separate molecules; this increases potential energy. That energy is the latent heat.

Understanding the Question

Water in a kettle boils at $100\ ^{\circ}\text{C}$ (at atmospheric pressure). The kettle continues to supply energy, yet the temperature reading stays at $100\ ^{\circ}\text{C}$ while boiling occurs. The question asks for a molecular-energy explanation: what happens to the supplied energy if temperature (average kinetic energy) does not increase.

Approach

State what temperature represents at molecular level (average kinetic energy).
Describe what must happen to molecules during vaporisation (they separate and escape the liquid).
Explain that the supplied energy increases intermolecular potential energy (latent heat), not the average kinetic energy.
Conclude that temperature remains constant until the phase change finishes.

Step-by-Step Reasoning

Link temperature to kinetic energy: For water molecules, temperature corresponds to their average random kinetic energy.
What vaporisation requires: In the liquid, molecules are close together and held by intermolecular attractions. To become vapour, molecules must separate significantly and do work against these attractive forces.
Where the heating energy goes: The energy supplied by the kettle during boiling is used to increase the molecules' potential energy (often described as "breaking" or overcoming intermolecular bonds) so molecules can escape into the gas phase.
Why temperature stays at $100\ ^{\circ}\text{C}$ : Because the energy is not increasing the average kinetic energy, the average kinetic energy remains (approximately) constant, so the temperature remains at $100\ ^{\circ}\text{C}$ during the phase change.
After all water has boiled: Once all liquid has become gas, further heating will then increase the kinetic energy of the steam, so its temperature can rise above $100\ ^{\circ}\text{C}$ .

Key Takeaways

Temperature depends on average molecular kinetic energy.
During boiling, added energy becomes latent heat: it increases potential energy by separating molecules.
Therefore temperature stays constant during the phase change.

Common Mistakes

Saying "the molecules stop gaining energy" (they do gain energy, but as potential energy).
Claiming temperature stays constant because "heat is used to evaporate" without mentioning kinetic vs potential energy.
Confusing latent heat with specific heat capacity (temperature rise) processes.

Things to Be Careful About

Mention molecular energies explicitly: kinetic energy stays constant on average; potential energy increases.
State the condition: this constant temperature is while vaporisation is occurring at constant pressure (e.g. boiling point at atmospheric pressure).

Techniques used

relate temperature to average molecular kinetic energydistinguish kinetic-energy change from potential-energy change during a phase changeexplain a constant-temperature process using energy distribution at molecular level

Q3MediumGravitational Fields

(a)

Define gravitational field at a point.

1M Similar

DifficultyEasy

Worked solution

Answer

Gravitational field (strength) at a point is the force per unit mass on a small test mass placed at that point (direction is the direction of this force).

Final answer

Force per unit mass on a small test mass at that point.

Detailed explanation

Background Concept

A (gravitational) field describes the effect a mass has on other masses around it without contact. At any point in space we define the gravitational field strength $g$ so that it tells us how much force a small test mass would experience.

By definition,

g = \frac{F}{m}

where $F$ is the gravitational force on a test mass $m$ placed at that point. The direction of $g$ is the same as the direction of the force on a positive test mass (i.e. towards the attracting mass).

Understanding the Question

You are asked to define gravitational field at a point. For full credit, you must include the key idea “force per unit mass” and make clear it refers to a small test mass placed at that point.

Approach

Recall the standard definition used in A Level: field strength equals gravitational force per unit mass. State it clearly (and optionally mention direction).

Step-by-Step Reasoning

Place a small test mass $m$ at the point.
Let the gravitational force on it be $F$ .
Define gravitational field strength:

g = \frac{F}{m}

Since gravity is attractive, the direction is towards the mass producing the field.

Key Takeaways

Gravitational field strength is defined locally at a point.
It is force per unit mass, with direction given by the force on a test mass.

Common Mistakes

Saying “force” only, without “per unit mass”.
Confusing gravitational field strength $g$ with gravitational potential.

Things to Be Careful About

Use “test mass” language: the field exists whether or not the test mass is present, but the definition uses it conceptually.
Direction matters if the definition asked for “field” as a vector; stating “towards the mass” is safe.

Techniques used

use the definition of field strength as force per unit massstate the direction of the field as the direction of the force on a test mass

(b)

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

Fig. 3.1

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

Force on test mass $m$ at distance $x$ :

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

Gravitational field strength

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 two point masses $M$ and $m$ separated by distance $r$ , Newton’s law of gravitation gives the attractive force:

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

where $G$ is the gravitational constant.

Gravitational field strength $g$ at a point is defined by

g = \frac{F}{m}

so if you know the gravitational force on a test mass $m$ at that point, dividing by $m$ gives the field strength due to the source mass.

Understanding the Question

A point mass $M$ is isolated. Point $P$ is a distance $x$ from $M$ . You place a test mass $m$ at $P$ and use the force on $m$ to derive $g$ at $P$ in terms of $M$ and $x$ . You must also identify any extra symbols (here, $G$ ).

Approach

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

Step-by-Step Reasoning

The separation between $M$ and the test mass at $P$ is $x$ , so

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

By definition of field strength,

g = \frac{F}{m}

Substitute $F$ :

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

Identify symbol: $G$ is the gravitational constant.

Key Takeaways

For a point mass (or outside a spherical mass), gravitational field strength follows an inverse-square law.
The test mass cancels: $g$ depends only on the source mass and distance.

Common Mistakes

Forgetting to divide by $m$ , giving $F$ instead of $g$ .
Using $x$ instead of $x^2$ in the denominator.
Omitting $G$ or not identifying it.

Things to Be Careful About

$g$ is a vector; this part asked for an equation for its strength (magnitude). Direction is asked in part (ii).
The formula is valid for a point mass and for points outside a spherically symmetric mass (treated as if all mass is at the centre).

Techniques used

apply Newton's law of gravitation to find the force on a test massuse field strength as force per unit masssubstitute 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-Easy

Worked solution

Working

At $P$ :

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

At $Q$ , distance from $M$ is $x/2$ :

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

Direction at both points is towards $M$ , so directions at $P$ and $Q$ are opposite.

Answer

Magnitude at $Q$ is $4$ times that at $P$ , and the direction is opposite to that at $P$ (both point towards $M$ ).

Final answer

At Q, g is four times larger than at P and in the opposite direction (towards M).

Detailed explanation

Background Concept

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

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

This is an inverse-square law: halving $r$ makes $g$ four times bigger.

Direction: the field always points towards the mass $M$ .

Understanding the Question

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

Approach

Write $g_P$ using $r=x$ .
Write $g_Q$ using $r=x/2$ .
Take the ratio to compare magnitudes.
Use “towards $M$ ” to compare directions for points on opposite sides.

Step-by-Step Reasoning

At $P$ :

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

At $Q$ , the distance from $M$ is $x/2$ , so:

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

Square the distance:

(x/2)^2 = \frac{x^2}{4}

Substitute:

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

Direction: at $P$ the field points towards $M$ (towards the left if $M$ is left of $P$ ), while at $Q$ it points towards $M$ (towards the right if $M$ is right of $Q$ ). So the directions at $P$ and $Q$ are opposite.

Key Takeaways

Inverse-square laws: change in distance is squared in the effect.
Always comment on direction when comparing fields on opposite sides of a source.

Common Mistakes

Saying the field at $Q$ is “twice” rather than “four times” (forgetting to square).
Saying the direction is the same as at $P$ (it is towards $M$ in both cases, but since the points are on opposite sides, that means opposite directions along the line).

Things to Be Careful About

Be explicit about what “same direction” means: the physically correct statement is “towards $M$ ”; when comparing two points on opposite sides, convert that into “opposite directions relative to the line.”

Techniques used

use inverse-square dependence of field strength on distancecompare magnitudes by taking a ratiostate relative directions using attraction towards 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

Answer

Take positive $g$ towards sphere $Y$ .

Between the spheres,

at $x = R$ , resultant $g$ is approximately $-g_0$ (towards $X$ ),
at $x = L/2$ , resultant $g = 0$ (equal and opposite fields),
at $x = L - R$ , resultant $g$ is approximately $+g_0$ (towards $Y$ ).

Sketch a smooth curve increasing from near $-g_0$ at $x=R$ to $0$ at $x=L/2$ and then to near $+g_0$ at $x=L-R$ , becoming steeper near each surface (inverse-square behaviour).

Final answer

Sketch: g rises from about −g0 at x=R to 0 at x=L/2 then to about +g0 at x=L−R, steeper near surfaces.

Detailed explanation

Background Concept

Outside a uniform sphere, the gravitational field is the same as if all its mass were concentrated at its centre:

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

where $r$ is the distance from the centre.

When more than one mass produces a field at the same point, the resultant gravitational field is found by superposition (vector addition):

\vec{g}_{\text{resultant}} = \vec{g}_X + \vec{g}_Y

Because the two spheres are identical and arranged symmetrically, the midpoint between their centres is a key symmetry point.

Understanding the Question

Two identical spheres $X$ and $Y$ (radius $R$ ) have centres separated by $L$ . A point $P$ lies between them on the line of centres.

The horizontal coordinate is $x$ , measured from the centre of sphere $X$ .
You are asked to sketch how the resultant gravitational field $g$ at $P$ varies as $x$ goes from $R$ (just outside sphere $X$ ) to $L-R$ (just outside sphere $Y$ ).
The field strength at the surface of an isolated sphere is given as $g_0$ , so the field due to one sphere alone at distance $R$ from its centre is $g_0$ .

The provided graph axis includes positive and negative values, so you must choose a sign convention and be consistent.

Approach

Choose a positive direction along the line (natural choice: from $X$ towards $Y$ , i.e. increasing $x$ ).
Write the field contribution from each sphere at a general point $x$ $x$ :
- Distance from centre of $X$ is $x$ .
- Distance from centre of $Y$ is $L-x$ .
Use directions:
- Field due to $X$ points towards $X$ (negative direction).
- Field due to $Y$ points towards $Y$ (positive direction).
Use symmetry to locate where resultant field is zero: for identical spheres it must be at $x=L/2$ .
Use endpoint values (near $x=R$ and $x=L-R$ ) and the inverse-square idea to sketch curvature (steeper near surfaces, flatter in the middle).

Step-by-Step Reasoning

Let positive $g$ be towards $Y$ (to the right, increasing $x$ ).

Field due to sphere $X$ at point $P$ (distance $x$ from its centre) has magnitude $GM/x^2$ but points towards $X$ (negative):

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

Field due to sphere $Y$ at point $P$ (distance $L-x$ from its centre) points towards $Y$ (positive):

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

Resultant:

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

Now use key points for the sketch:

At the midpoint $x = L/2$ :
$g = \frac{GM}{(L-L/2)^2} - \frac{GM}{(L/2)^2} = \frac{GM}{(L/2)^2} - \frac{GM}{(L/2)^2} = 0$
So the curve must cross the horizontal axis at $x=L/2$ .
Near sphere $X$ , at $x=R$ :
- The contribution from $X$ alone has magnitude $g_0$ (given).
- The contribution from $Y$ is smaller because it is farther away (distance $L-R$ ).
  So the resultant is negative and close to $-g_0$ .
Near sphere $Y$ , at $x=L-R$ :
By symmetry the resultant is positive and close to $+g_0$ .
Shape/curvature: because each contribution varies as $1/r^2$ , the magnitude changes rapidly when you are close to a sphere and more slowly when you are far away. So the sketch should be steeper near $x=R$ and near $x=L-R$ , and flatter near the middle.

A clear qualitative sketch is therefore:

diagram-1

Key Takeaways

Outside a sphere, treat it as a point mass at its centre.
Resultant gravitational field is found by vector superposition.
For two identical masses, the zero-field point on the line joining them is at the midpoint.

Common Mistakes

Adding magnitudes without directions (forgetting one field is opposite in direction).
Putting the zero-field point somewhere other than $L/2$ despite symmetry.
Sketching straight lines instead of showing curvature consistent with inverse-square behaviour.

Things to Be Careful About

State or imply a sign convention (positive towards $Y$ is consistent with the graph running from $x=R$ to $x=L-R$ ).
The values at the surfaces are described as “approximately” $\pm g_0$ because the other sphere still contributes; the important marking points are: negative near $x=R$ , zero at $L/2$ , positive near $x=L-R$ , and the correct curved shape.

Techniques used

use the principle of superposition to add gravitational fieldsapply the inverse-square variation of field strength with distanceuse symmetry to locate the zero-field pointsketch a curve using limiting values and qualitative curvature

Q4MediumTemperatureIdeal Gases

(a)

State the value of absolute zero on:

2M

(i)

the Celsius temperature scale

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

1M Similar

DifficultyEasy

Worked solution

Answer

$-273\ ^{\circ}\text{C}$

Final answer

-273 °C

Detailed explanation

Background Concept

Absolute zero is the lowest possible thermodynamic temperature, corresponding (in the ideal-gas model) to zero random translational kinetic energy per molecule.

The Celsius scale is defined relative to the kelvin (thermodynamic) scale by

T / \text{K} = \theta / ^{\circ}\text{C} + 273.15

where $T$ is the thermodynamic temperature and $\theta$ is the Celsius temperature.

Understanding the Question

You are asked to state the value of absolute zero on the Celsius scale (i.e. what Celsius temperature corresponds to $T = 0\ \text{K}$ ).

Approach

Use the fixed offset between the kelvin and Celsius scales: kelvin temperature is Celsius temperature plus about $273$ .

Step-by-Step Reasoning

At absolute zero, $T = 0\ \text{K}$ .

Using

T = \theta + 273.15

set $T = 0$ :

0 = \theta + 273.15 \;\Rightarrow\; \theta = -273.15\ ^{\circ}\text{C}

In exams this is usually quoted as $-273\ ^{\circ}\text{C}$ .

Key Takeaways

The kelvin and Celsius scales have the same size degree, but different zero points.
Absolute zero corresponds to $0\ \text{K} = -273\ ^{\circ}\text{C}$ (approximately).

Common Mistakes

Writing $+273\ ^{\circ}\text{C}$ instead of a negative value.
Confusing $0\ ^{\circ}\text{C}$ (freezing point of water) with absolute zero.

Things to Be Careful About

The more accurate value is $-273.15\ ^{\circ}\text{C}$ , but many mark schemes accept $-273\ ^{\circ}\text{C}$ .
Keep the unit as $^{\circ}\text{C}$ for the Celsius scale.

Techniques used

recall the definition of absolute zeroconvert between Celsius and kelvin zero points

(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

DifficultyMedium-Easy

Worked solution

Answer

$pV$ is proportional to $kT$ (straight line through origin), so the gas behaves as an ideal gas and obeys $pV = NkT$ .

Final answer

The gas behaves as an ideal gas (obeys pV = NkT).

Detailed explanation

Background Concept

For an ideal gas,

pV = NkT

where:

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

If $N$ is fixed, then $pV$ is directly proportional to $T$ , and also directly proportional to $kT$ .

Understanding the Question

A fixed amount of gas is used, so $N$ is constant. The graph shows how $pV$ varies with $kT$ . You must state what this tells you about the nature of the gas.

Approach

Recognise that a straight line through the origin means a direct proportionality: $pV \propto kT$ . Compare this with the ideal-gas equation $pV = NkT$ .

Step-by-Step Reasoning

From the graph:

It is a straight line.
It passes through $(0,0)$ .

Therefore,

pV \propto kT

For an ideal gas,

pV = NkT

which is exactly a direct proportionality between $pV$ and $kT$ for constant $N$ .

So the gas is behaving ideally (the ideal-gas model fits the data).

Key Takeaways

A straight line through the origin indicates direct proportionality.
$pV$ vs $kT$ being linear through the origin is evidence for $pV = NkT$ (ideal gas behaviour).

Common Mistakes

Saying only “it is a straight line” without stating what that implies physically.
Confusing $pV = NkT$ with $pV = nRT$ without connecting to the meaning of the graph.

Things to Be Careful About

The conclusion is about the model (ideal gas) fitting the data, not about the gas being a particular chemical substance.
The phrase “fixed amount of gas” implies $N$ is constant, which is essential for interpreting the slope as a constant.

Techniques used

interpret a straight-line graph through the originlink proportionality to the ideal-gas equation

(ii)

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

$N$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

For an ideal gas,

pV = NkT

So for a graph of $pV$ against $kT$ , gradient $= N$ .

Using the point $kT = 8.0 \times 10^{-21}\ \text{J}$ , $pV = 270\ \text{J}$ :

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,

pV = NkT

If you plot $pV$ (y-axis) against $kT$ (x-axis), then the equation matches the straight-line form:

\text{y} = N\,\text{x}

So:

the graph should be a straight line through the origin,
the gradient (slope) of the line is the number of molecules $N$ .

Understanding the Question

You are given a straight-line graph of $pV$ versus $kT$ for a fixed sample. You must determine $N$ by finding the gradient.

A specific point is readable from the graph: when $kT = 8 \times 10^{-21}\ \text{J}$ , then $pV \approx 270\ \text{J}$ .

Approach

Use the ideal-gas relation $pV = NkT$ .
Recognise $N$ is the gradient of the $pV$ vs $kT$ graph.
Calculate gradient using a point on the line (or a large triangle):

N = \frac{\Delta (pV)}{\Delta (kT)}

Step-by-Step Reasoning

From $pV = NkT$ ,

N = \frac{pV}{kT}

Use the values from the graph:

$pV = 270\ \text{J}$ ,
$kT = 8.0 \times 10^{-21}\ \text{J}$ .

Then

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

Compute:

\frac{270}{8.0} = 33.75

and dividing by $10^{-21}$ multiplies by $10^{21}$ :

N = 33.75 \times 10^{21} = 3.375 \times 10^{22}

To sensible significant figures (limited by reading the graph),

N \approx 3.4 \times 10^{22}

The unit cancels (J/J), so $N$ is dimensionless, as it should be.

Key Takeaways

For $pV$ vs $kT$ , the gradient equals $N$ because $pV = NkT$ .
Always include the $\times 10^{-21}$ factor from the x-axis label when using graph readings.

Common Mistakes

Forgetting that the x-axis is in $10^{-21}\ \text{J}$ and using $8$ instead of $8 \times 10^{-21}$ .
Using $kT/pV$ (inverting the gradient).
Giving $N$ a unit (it has no unit).

Things to Be Careful About

Use a clear point on the straight line (or a large triangle) to reduce percentage uncertainty.
Quote an appropriate number of significant figures; readings from a graph typically justify 2 s.f.

Techniques used

use the gradient of a straight-line graphapply pV = NkT to relate slope to number of moleculeshandle powers of ten from axis scales

(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}

Using $N = 3.4 \times 10^{22}$ and $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ :

n = \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 Avogadro’s constant $N_A$ :

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

where $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ .

Understanding the Question

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

Approach

Apply

n = \frac{N}{N_A}

and keep the unit as $\text{mol}$ .

Step-by-Step Reasoning

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

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

Separate the numbers and powers of ten:

n = \left(\frac{3.4}{6.02}\right) \times 10^{22-23}

\frac{3.4}{6.02} \approx 0.564 \quad \text{and} \quad 10^{-1} = 0.1

So

n \approx 0.564 \times 10^{-1} = 5.64 \times 10^{-2}\ \text{mol}

Rounded:

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

Key Takeaways

Converting molecules to moles uses $n = N/N_A$ .
$N_A$ has units $\text{mol}^{-1}$ , so $n$ comes out in $\text{mol}$ .

Common Mistakes

Multiplying by $N_A$ instead of dividing.
Forgetting the unit $\text{mol}$ .

Things to Be Careful About

Keep powers of ten organised: $10^{22}/10^{23} = 10^{-1}$ .
Your $n$ should be much less than $1\ \text{mol}$ because $N$ is much less than $N_A$ .

Techniques used

use n = N/NA to convert molecules to moleshandle standard form arithmetic

(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} N m c_{\text{rms}}^2

So

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

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

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}} \approx 4.0

Answer

$4.0\ \text{u}$

Final answer

4.0 u

Detailed explanation

Background Concept

In the kinetic theory of gases, pressure arises from molecules colliding with the container walls. One important result is

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

For a gas where $c_{\text{rms}}$ is the root-mean-square speed,

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

so

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

Here:

$N$ is the number of molecules,
$m$ is the mass of one molecule,
$c_{\text{rms}}$ is the r.m.s. speed.

The unified atomic mass unit is

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 your previously found $N$ , you must calculate the mass of one molecule and express it in u.

Approach

Start from the kinetic theory equation linking $pV$ to $N$ , $m$ and $c_{\text{rms}}$ .
Rearrange it to make $m$ the subject.
Substitute the given values to obtain $m$ in kg.
Convert kg to u using $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$ .

Step-by-Step Reasoning

Use

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

Rearrange for $m$ :

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

Substitute:

$pV = 270\ \text{J}$ ,
$N = 3.4 \times 10^{22}$ ,
$c_{\text{rms}} = 1900\ \text{m s}^{-1}$ so $c_{\text{rms}}^2 = (1900)^2 = 3.61 \times 10^6\ \text{m}^2\text{s}^{-2}$ .

Then

m = \frac{3(270)}{(3.4 \times 10^{22})(3.61 \times 10^6)}

Combine the denominator:

(3.4 \times 10^{22})(3.61 \times 10^6) = 12.3 \times 10^{28} = 1.23 \times 10^{29}

So

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

Convert to u:

m(\text{u}) = \frac{6.6 \times 10^{-27}}{1.66 \times 10^{-27}} \approx 4.0\ \text{u}

Key Takeaways

The kinetic theory equation $pV = \tfrac{1}{3}Nmc_{\text{rms}}^2$ links macroscopic gas behaviour to molecular properties.
$pV$ in joules is consistent with kinetic energy units, making the equation dimensionally sensible.
Converting to u is a simple division by $1.66 \times 10^{-27}\ \text{kg}$ .

Common Mistakes

Using $pV = NkT$ instead of the kinetic theory equation (that would not involve $c_{\text{rms}}$ ).
Forgetting to square the r.m.s. speed.
Using $\tfrac{1}{2}mc^2$ directly without the factor and the $N$ relationship.
Not converting to u, or using the wrong value for $1\ \text{u}$ .

Things to Be Careful About

Use the same $N$ as in (b)(ii); if it was rounded, your final mass will be correspondingly approximate (error carried forward is usually allowed).
Keep track of powers of ten when multiplying $N$ by $c^2$ .
Quote the final value with a sensible number of significant figures (typically 2 s.f. from graph-derived $N$ ).

Techniques used

apply the kinetic theory relation pV = (1/3) N m c_rms^2rearrange to solve for molecular massconvert mass in kg to unified atomic mass unit

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 positive test charge from infinity to that point (with no change in kinetic energy).

Final answer

Work done per unit positive charge in bringing a test charge from infinity to the point (no change in kinetic energy).

Detailed explanation

Background Concept

Electric potential $V$ at a point is a scalar quantity defined via energy (or work) per unit charge.

If an external agent brings a small positive test charge $q$ from infinity to a point in an electric field, then the work done $W$ by the external agent is related to the potential difference by

V = \frac{W}{q}

When we use the definition “from infinity”, we are choosing the zero of potential to be $V = 0$ at infinity.

Understanding the Question

You are asked to define electric potential at a point. That means you must give the standard wording: what is done (work done), per what (per unit charge), and from where (from infinity) to where (the point).

Approach

Use the formal definition: electric potential at a point equals work done per unit positive charge to bring a test charge from infinity to that point, typically specifying “no change in kinetic energy” (or “work done against the field”).

Step-by-Step Reasoning

Start with the idea that potential measures energy per coulomb.
Use a small positive test charge so it does not disturb the existing field.
Specify the reference point “infinity” where the potential is taken as zero.
State the work done per unit charge to bring it to the point.

Key Takeaways

$V$ is work done per unit charge.
The phrase “from infinity” fixes the reference of zero potential.
Potential is a scalar (no direction).

Common Mistakes

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

Things to Be Careful About

Use “work done by an external agent” or “work done against the field” (either is acceptable if clearly per unit charge).
Include “positive test charge” to avoid sign confusion.

Techniques used

state the definition of electric potential in terms of work done per unit chargereference moving a positive test charge from infinity without change in kinetic energy

(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

Potential at $P$ due to a point charge:

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

Here $r_p = x$ from the proton, and $r_e = (120 - x)$ from the electron (with $x$ in pm).

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

For $x = 10\ \text{pm}$ :

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 \approx 1.31\times 10^{2}\ \text{V} \approx 130\ \text{V}

Answer

$V = 130\ \text{V}$

Final answer

130 V

Detailed explanation

Background Concept

For a point charge $Q$ , the electric potential at a distance $r$ (taking $V=0$ at infinity) is

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

Potential is a scalar, so when there are multiple charges, the total potential is the algebraic sum (superposition):

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

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

Understanding the Question

A proton ( $+e$ ) and an electron ( $-e$ ) are separated by $120\ \text{pm}$ . Point $P$ lies on the line between them, at distance $x$ from the proton. When $x=10\ \text{pm}$ , you must show the potential at $P$ is $130\ \text{V}$ .

Distances from $P$ :

to proton: $r_p = x$
to electron: $r_e = 120 - x$

Approach

Write the total potential as the sum of the potentials from the proton and electron.
Use $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$ for each charge.
Substitute $x=10\ \text{pm}$ and convert pm to m.
Keep the sign of the electron’s contribution negative.

Step-by-Step Reasoning

Total potential:

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

Now substitute $x=10\ \text{pm}$ :

$x = 10\times 10^{-12}\ \text{m}$
$120-x = 110\ \text{pm} = 110\times 10^{-12}\ \text{m}$

So

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 bracket:

\frac{1}{10\times 10^{-12}} = 1.0\times 10^{11} \quad\text{and}\quad \frac{1}{110\times 10^{-12}} \approx 9.09\times 10^{9}

Difference:

\approx 9.09\times 10^{10}

Multiply by $(8.99\times 10^9)(1.60\times 10^{-19}) \approx 1.44\times 10^{-9}$ :

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

Use superposition: add potentials as scalars.
The electron contributes a negative term.
Convert $\text{pm}$ to $\text{m}$ carefully.

Common Mistakes

Adding magnitudes instead of using the negative sign for the electron.
Using separation $120\ \text{pm}$ as a distance to $P$ (you need $x$ and $120-x$ ).
Forgetting the $10^{-12}$ conversion from pm to m.

Things to Be Careful About

The point is between the charges here, so $r_e = 120-x$ (not $120+x$ ).
Keep enough significant figures during working so the final rounding to $130\ \text{V}$ is justified.

Techniques used

use superposition of electric potential from point chargessubstitute distances into the point-charge potential equationconvert picometres to metres

(ii)

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

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For $x = 30\ \text{pm}$ , distances are $r_p = 30\ \text{pm}$ and $r_e = 90\ \text{pm}$ .

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{e}{30\times 10^{-12}} - \frac{e}{90\times 10^{-12}}\right)

V = (8.99\times 10^9)(1.60\times 10^{-19})\left(\frac{1}{30\times 10^{-12}} - \frac{1}{90\times 10^{-12}}\right)

V \approx 31.9\ \text{V} \approx 32\ \text{V}\ (2\ \text{s.f.})

Answer

$V = 32\ \text{V}$

Final answer

32 V

Detailed explanation

Background Concept

The potential at a point due to multiple charges is found by adding their potentials (with signs):

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

For a proton–electron pair, at a point between them, the total potential is

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

Understanding the Question

You must calculate the potential at point $P$ when it is $x=30\ \text{pm}$ from the proton along the line joining the charges (separation $120\ \text{pm}$ ). The answer must be to two significant figures.

So the distances are:

proton to $P$ : $30\ \text{pm}$
electron to $P$ : $120-30=90\ \text{pm}$

Approach

Write the same total potential expression as in (i).
Substitute $x=30\ \text{pm}$ and $120-x=90\ \text{pm}$ .
Convert pm to m inside the calculation.
Round the final number to 2 s.f.

Step-by-Step Reasoning

Start:

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

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

V = (8.99\times 10^9)(1.60\times 10^{-19})\left(\frac{1}{30\times 10^{-12}} - \frac{1}{90\times 10^{-12}}\right)

Compute the bracket:

\frac{1}{30\times 10^{-12}} = 3.33\times 10^{10}, \quad \frac{1}{90\times 10^{-12}} = 1.11\times 10^{10}

Difference:

2.22\times 10^{10}

Multiply by $(8.99\times 10^9)(1.60\times 10^{-19}) \approx 1.44\times 10^{-9}$ :

V \approx (1.44\times 10^{-9})(2.22\times 10^{10}) \approx 31.9\ \text{V}

To 2 s.f.:

V = 32\ \text{V}

Key Takeaways

Between charges: use $x$ and $120-x$ as the two distances.
Electron term is negative.
Rounding instruction (2 s.f.) is part of the mark.

Common Mistakes

Using $120+x$ instead of $120-x$ .
Forgetting to convert pm to m.
Rounding too early and losing accuracy.

Things to Be Careful About

The potential can be small even near charges because the positive and negative contributions partially cancel.
Write units ( $\text{V}$ ) and apply the requested significant figures.

Techniques used

apply superposition of electric potentialhandle sign of contributions from positive and negative chargesround to a stated number of 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

DifficultyMedium-Easy

Worked solution

Working

For $V=0$ between the charges:

\frac{e}{x} - \frac{e}{120-x} = 0

So $x = 120-x \Rightarrow x = 60\ \text{pm}$ .

Answer

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

Final answer

Midpoint between the charges, x = 60 pm from the proton.

Detailed explanation

Background Concept

Electric potential from charges adds as a scalar:

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q_1}{r_1} + \frac{Q_2}{r_2}\right)

For equal and opposite charges ( $+e$ and $-e$ ), the potential can be zero where the positive and negative contributions are equal in magnitude.

Understanding the Question

You need to mark one position (not infinity) where the total potential due to the proton and electron is zero. The point must lie on the line joining them (as in the diagram).

Approach

Set the total potential expression to zero and solve for $x$ (distance from the proton). Between the charges the distances are $x$ and $120-x$ .

Step-by-Step Reasoning

Between the charges:

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

For $V=0$ :

\frac{e}{x} = \frac{e}{120-x}

Cancel $e$ :

\frac{1}{x} = \frac{1}{120-x} \Rightarrow x = 120-x

So

x = 60\ \text{pm}

This is exactly the midpoint, which makes sense by symmetry: you are equally far from $+e$ and $-e$ , so their potentials cancel.

Key Takeaways

For $+e$ and $-e$ , $V=0$ where the distances to each charge are equal.
Along the line segment between them, that is the midpoint.

Common Mistakes

Stating “at infinity” (the question explicitly says other than infinity).
Choosing a point closer to one charge than the other (then magnitudes cannot cancel).

Things to Be Careful About

The point $V=0$ here is a point on the line; in 3D it corresponds to a plane perpendicular to the line at the midpoint (but you only need one point on the given line).

Techniques used

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

Sketch a smooth decreasing curve for

V \propto \left(\frac{1}{x} - \frac{1}{120-x}\right)

with the following features:

passes through $(10\ \text{pm},\ +130\ \text{V})$
crosses $V=0$ at $x=60\ \text{pm}$
passes through $(110\ \text{pm},\ -130\ \text{V})$
non-linear (steeper near the ends $x=10$ and $x=110$ , flatter near $x=60$ ).

Final answer

Decreasing non-linear curve: +130 V at 10 pm, crosses 0 at 60 pm, −130 V at 110 pm (steeper near ends).

Detailed explanation

Background Concept

For a point charge, potential varies as $1/r$ . For multiple charges, potentials add:

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

Because of the $1/r$ dependence, the graph is generally curved, not a straight line.

Also note the symmetry for equal and opposite charges: if you measure displacement from the midpoint, the potential changes sign.

Understanding the Question

You must sketch $V$ against $x$ for $10\ \text{pm} \le x \le 110\ \text{pm}$ , where $x$ is the distance from the proton to point $P$ along the line joining the charges (total separation $120\ \text{pm}$ ).

You already found/verified:

$V(10\ \text{pm}) = +130\ \text{V}$
$V(60\ \text{pm}) = 0$ (from part iii)

By symmetry you can also infer what happens near $x=110\ \text{pm}$ (close to the electron).

Approach

To sketch correctly, anchor the curve using key physics features:

Use the formula to know it is not linear ( $1/x$ terms).
Plot key points: $x=10$ gives +130 V, $x=60$ gives 0.
Use symmetry: exchanging $x$ with $120-x$ swaps the roles of proton/electron and flips the sign, so $V(110) = -V(10) = -130\ \text{V}$ .
Consider shape: near the proton end ( $x$ small) the positive $1/x$ term dominates, so $V$ is large positive; near the electron end ( $120-x$ small) the negative term dominates, so $V$ is large negative.

Step-by-Step Reasoning

Start from

V(x) = k e\left(\frac{1}{x} - \frac{1}{120-x}\right) \quad\text{where } k = \frac{1}{4\pi\varepsilon_0}

Key points:

At $x=10\ \text{pm}$ , you already showed $V=+130\ \text{V}$ .
At $x=60\ \text{pm}$ , distances to both charges are equal ( $60\ \text{pm}$ ), so the two terms cancel and $V=0$ .
At $x=110\ \text{pm}$ :

V(110) = k e\left(\frac{1}{110} - \frac{1}{10}\right) = -k e\left(\frac{1}{10} - \frac{1}{110}\right) = -130\ \text{V}

So the curve must pass through $(110, -130)$ .

Shape / curvature:

As $x$ increases from $10$ toward $60$ , $1/x$ decreases and $1/(120-x)$ increases, so $V$ falls.
Near $x=60$ the two terms are close in value, so changes in $V$ are less dramatic (the curve looks flatter).
Moving toward $x=110$ , the negative term becomes much larger in magnitude, so the curve becomes more steeply negative.

Therefore: a smooth monotonic decreasing curve, crossing the axis at $x=60\ \text{pm}$ , with steeper gradients nearer the ends and a gentler slope around the midpoint.

Key Takeaways

Potential between $+e$ and $-e$ varies like $\frac{1}{x} - \frac{1}{120-x}$ .
Zero potential occurs where distances are equal (midpoint).
The graph is non-linear due to the $1/r$ dependence.

Common Mistakes

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

Things to Be Careful About

The question only asks for $x=10$ to $110\ \text{pm}$ , so you do not need to show the divergences at $x=0$ or $x=120\ \text{pm}$ , but the curve should still look steeper toward the ends of the given interval.
Ensure the plotted values are consistent with the provided axes scale (about +130 V at $x=10$ and about -130 V at $x=110$ ).

Techniques used

use the functional form of potential for two point chargesidentify key points and symmetry to sketch a curveuse sign changes and curvature to shape a qualitative graph

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

For any capacitor,

C = \frac{Q}{V}

where $C$ is capacitance, $Q$ is the charge stored (magnitude on one plate), and $V$ is the potential difference (p.d.) across the plates.

When capacitors are connected:

In series, the same charge must pass through each component, so each capacitor ends up with the same magnitude of charge.
In parallel, components share the same two connection points, so each has the same p.d. across it.

Understanding the Question

You have two capacitors connected to a supply of p.d. $V_S$ . The supply delivers a total charge $Q_S$ to the combination. The capacitors acquire charges $Q_1$ and $Q_2$ and have p.d.s $V_1$ and $V_2$ .

You are asked to fill a table stating:

how the charges relate ( $Q_S$ , $Q_1$ , $Q_2$ )
how the p.d.s relate ( $V_S$ , $V_1$ , $V_2$ )

for both series and parallel connections.

Approach

Use the standard circuit rules:

Series: same charge on each capacitor; supply p.d. is shared so p.d.s add.
Parallel: same p.d. on each capacitor; supply charge is shared so charges add.

Step-by-Step Reasoning

Series connection

There is only one path for charge to move. Any charge transferred from the supply must appear on each capacitor in turn, so:

Q_1 = Q_2

The charge delivered by the supply to the series combination is the same magnitude as the charge on each capacitor:

Q_S = Q_1 = Q_2

The supply p.d. is split across the two capacitors, so total p.d. is the sum:

V_S = V_1 + V_2

Parallel connection

Each capacitor is connected directly across the same two supply terminals, so the p.d. across each equals the supply p.d.:

V_1 = V_2 = V_S

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

Q_S = Q_1 + Q_2

Key Takeaways

Series capacitors: same $Q$ , $V$ adds.
Parallel capacitors: same $V$ , $Q$ adds.

Common Mistakes

Swapping the rules (e.g. writing $V_S = V_1 = V_2$ for series).
Writing $Q_S = Q_1 + Q_2$ for series (this is for parallel).
Forgetting that p.d.s in series add to the supply.

Things to Be Careful About

These are relationships between magnitudes of charges on plates (one plate has $+Q$ , the other $-Q$ ).
The question uses $Q_S$ as “charge provided by the supply”, so in parallel it must equal the sum of the capacitor charges.

Techniques used

use the defining relation between charge, p.d. and capacitanceapply charge conservation rules for series and parallel circuitsapply the rule for how potential differences combine in series and parallel

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

W = 19\ \text{mJ} = 0.019\ \text{J},\quad C = 470\ \mu\text{F} = 470\times 10^{-6}\ \text{F}

V = \sqrt{\frac{2(0.019)}{470\times 10^{-6}}} = 8.99\ \text{V} \approx 9.0\ \text{V}

Answer

$9.0\ \text{V}$

Final answer

9.0 V

Detailed explanation

Background Concept

Energy stored in a capacitor can be written in several equivalent forms:

W = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}

The form $W = \frac{1}{2}CV^2$ is most useful when you know $C$ and want $V$ .

Understanding the Question

A single isolated capacitor has:

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

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

Approach

Use

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

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

Step-by-Step Reasoning

Convert units:

W = 19\ \text{mJ} = 19\times 10^{-3}\ \text{J} = 0.019\ \text{J}

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

Rearrange:

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

Substitute:

V = \sqrt{\frac{2(0.019)}{470\times 10^{-6}}} = 8.99\ \text{V}

So $V \approx 9.0\ \text{V}$ (to 2 s.f., matching the given energy).

Key Takeaways

Choose the energy expression that matches the known quantities.
Always convert prefixes (milli-, micro-) into SI before substituting.

Common Mistakes

Using $W = CV^2$ (missing the factor $\tfrac{1}{2}$ ).
Forgetting to convert $\mu\text{F}$ to $\text{F}$ or $\text{mJ}$ to $\text{J}$ .

Things to Be Careful About

Check reasonableness: a large capacitance like $470\ \mu\text{F}$ storing only $19\ \text{mJ}$ should have a modest voltage (a few volts to a few tens of volts), not hundreds of volts.
Quote the final voltage to a sensible number of significant figures.

Techniques used

use the capacitor energy equationrearrange to make potential difference the subjectconvert between milli- and base SI units

(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 on $470\ \mu\text{F}$ capacitor:

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

After connecting in parallel with $180\ \mu\text{F}$ , common p.d. $V_f$ :

C_{\text{tot}} = (470+180)\ \mu\text{F} = 650\times 10^{-6}\ \text{F}

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

Total energy stored:

W_f = \frac{1}{2}C_{\text{tot}}V_f^2

W_f = \frac{1}{2}(650\times 10^{-6})(6.50)^2 = 1.37\times 10^{-2}\ \text{J} = 13.7\ \text{mJ}

Answer

$14\ \text{mJ}$

Final answer

14 mJ

Detailed explanation

Background Concept

When a charged capacitor is connected in parallel with another capacitor:

The final p.d. across both capacitors becomes the same (parallel rule).
If the system is isolated (no battery connected), total charge is conserved.

Energy in capacitors is not necessarily conserved during redistribution: energy can be dissipated as heat in the connecting wires due to current flow.

Understanding the Question

You start with:

Capacitor $C_1 = 470\ \mu\text{F}$ charged (it stored $19\ \text{mJ}$ initially).
Capacitor $C_2 = 180\ \mu\text{F}$ initially uncharged.

They are then connected in parallel, with no mention of a supply, so the total charge initially on $C_1$ is shared between both capacitors.

You must find the total final energy stored in the two capacitors.

Approach

Find the initial total charge available (this is the charge from part (ii)).
In parallel, the final p.d. $V_f$ is common, and

Q_{\text{total}} = (C_1 + C_2)V_f

So compute $V_f$ .
3. Then compute total energy using the equivalent capacitance:

W_f = \frac{1}{2}(C_1 + C_2)V_f^2

Step-by-Step Reasoning

Initial charge on the charged capacitor (from earlier parts):

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

In parallel, capacitances add:

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

Charge conservation gives:

Q_{\text{total}} = C_{\text{tot}}V_f

so

V_f = \frac{Q_{\text{total}}}{C_{\text{tot}}} = \frac{4.23\times 10^{-3}}{650\times 10^{-6}} = 6.50\ \text{V}

This is smaller than the original $9.0\ \text{V}$ , as expected because adding an uncharged capacitor increases the total capacitance and shares the charge.

Total final energy:

W_f = \frac{1}{2}C_{\text{tot}}V_f^2 = \frac{1}{2}(650\times 10^{-6})(6.50)^2

W_f = 1.37\times 10^{-2}\ \text{J} = 13.7\ \text{mJ} \approx 14\ \text{mJ}

Notice $13.7\ \text{mJ}$ is less than the original $19\ \text{mJ}$ : the “missing” energy is dissipated as heat during the brief current flow when the capacitors are connected.

Key Takeaways

In parallel: same final voltage; capacitances add.
For an isolated connection: total charge is conserved.
Energy usually decreases during charge sharing (dissipation in the circuit).

Common Mistakes

Assuming energy is conserved and answering $19\ \text{mJ}$ .
Using $V_f = 9.0\ \text{V}$ (forgetting that voltage changes when connected to an uncharged capacitor without a supply).
Calculating energy in only one capacitor instead of the total.

Things to Be Careful About

The phrase “initially uncharged” means $Q_2 = 0$ before connection.
Use $C_{\text{tot}} = C_1 + C_2$ only because the connection is parallel.
Keep units consistent: convert $\mu\text{F}$ to $\text{F}$ before using energy equations.

Techniques used

apply conservation of charge when connecting an isolated charged capacitor to an uncharged oneuse the parallel rule that potential differences become equal after connectioncalculate new common voltage using total capacitancecalculate total stored energy from equivalent capacitance

Q7MediumMagnetic Fields

(a)

State Faraday’s law of electromagnetic induction.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

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

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

Detailed explanation

Background Concept

Magnetic flux $\Phi$ through a surface is

\Phi = BA\cos\theta

where $B$ is magnetic flux density, $A$ is area, and $\theta$ is the angle between $\vec B$ and the normal to the area.

Flux linkage is $N\Phi$ for a coil of $N$ turns.

Faraday’s law states that an e.m.f. is induced when the magnetic flux linkage through a circuit changes, and its magnitude equals the rate of change of flux linkage. The negative sign in the equation (often associated with Lenz’s law) indicates the induced e.m.f. acts to oppose the change producing it.

Understanding the Question

This part asks for a statement of Faraday’s law (2 marks), so you should give (i) the relationship in words and (ii) the mathematical expression including flux linkage $N\Phi$ and the negative sign.

Approach

Provide the standard Cambridge A Level form:

a clear statement about “rate of change of flux linkage”, and
the equation $\varepsilon = -\mathrm{d}(N\Phi)/\mathrm{d}t$ .

Step-by-Step Reasoning

Identify the quantity that must change: magnetic flux linkage $N\Phi$ .
State proportionality/equality: induced e.m.f. equals the rate of change of $N\Phi$ .
Write the calculus form and include the negative sign to show opposition to the change.

Key Takeaways

Induction requires a change in flux linkage, not merely the presence of a magnetic field.
The core formula is $\varepsilon = -\mathrm{d}(N\Phi)/\mathrm{d}t$ .

Common Mistakes

Writing $\varepsilon = -\mathrm{d}\Phi/\mathrm{d}t$ without the flux linkage $N\Phi$ .
Omitting the negative sign.
Stating “flux” rather than “flux linkage” when the question asks for Faraday’s law.

Things to Be Careful About

Use “rate of change” wording (not just “change”).
Include $N$ explicitly (even if later questions may have $N=1$ ).
Keep the negative sign: it is often required for full credit.

Techniques used

state Faraday’s law in wordswrite Faraday’s law using flux linkage and a 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 Faraday’s law for one turn:

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

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

Answer

Magnetic flux cut $= 8.1\ \text{Wb}$

Final answer

8.1 Wb

Detailed explanation

Background Concept

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

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

If we are dealing with a single conductor effectively cutting flux (equivalent to $N=1$ ), the magnitude relation becomes

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

so the flux cut in a time interval is $\Delta\Phi = \varepsilon\Delta t$ .

Understanding the Question

You are told an e.m.f. of $0.54\ \text{V}$ is induced between the wingtips. Over $15\ \text{s}$ , the question asks for the magnetic flux cut (i.e. the total change in flux linkage for $N=1$ ) and to give the correct unit.

Approach

Treat the given e.m.f. as a constant average e.m.f. over the $15\ \text{s}$ interval and use

\Delta\Phi = \varepsilon\Delta t.

Step-by-Step Reasoning

Start from

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

Rearrange:

\Delta\Phi = \varepsilon\Delta t.

Substitute values:

\Delta\Phi = 0.54 \times 15 = 8.1.

The unit of magnetic flux is the weber (Wb). (Equivalently, $1\ \text{Wb} = 1\ \text{V s}$ .)

Key Takeaways

Flux cut over time is the time integral of e.m.f.; for constant e.m.f., $\Delta\Phi = \varepsilon t$ .
Remember the unit: $\text{Wb}$ .

Common Mistakes

Forgetting the unit, or writing $\text{T m}^2$ without recognising it is a weber.
Dividing by time instead of multiplying.
Using $N\Phi$ without being clear what $N$ is (here effectively $N=1$ ).

Things to Be Careful About

Use magnitude: the negative sign is about direction/polarity, not needed for a scalar “flux cut” value.
Ensure time is in seconds and e.m.f. in volts so that $\text{V s} = \text{Wb}$ .

Techniques used

apply Faraday’s law in the form \varepsilon = \Delta\Phi/\Delta trearrange to find magnetic flux from e.m.f. and time

(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

Vertical field component:

B = 38\ \mu\text{T} = 3.8 \times 10^{-5}\ \text{T}

\Phi = BA \;\Rightarrow\; A = \frac{\Phi}{B} = \frac{8.1}{3.8 \times 10^{-5}} = 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

Magnetic flux through an area is

\Phi = BA\cos\theta.

In this question, the aircraft is cutting the vertical component of Earth’s field, so we take the relevant area as perpendicular to that vertical field component (so $\cos\theta = 1$ ), giving $\Phi = BA$ .

Understanding the Question

From part (b)(i) you have the flux cut in $15\ \text{s}$ . You are told the vertical component of the Earth’s field is $38\ \mu\text{T}$ . The question asks for the area of flux cut in that same time.

Approach

Use

A = \frac{\Phi}{B}

and be careful to convert microtesla to tesla.

Step-by-Step Reasoning

Convert the magnetic flux density:

38\ \mu\text{T} = 38 \times 10^{-6}\ \text{T} = 3.8 \times 10^{-5}\ \text{T}.

Use $\Phi = BA$ (field perpendicular to area):

A = \frac{\Phi}{B}.

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

A = \frac{8.1}{3.8 \times 10^{-5}} = 2.13 \times 10^{5}\ \text{m}^2.

Quote to suitable significant figures (limited by 2 s.f. in $38\ \mu\text{T}$ ):

A \approx 2.1 \times 10^{5}\ \text{m}^2.

Key Takeaways

For flux cutting problems, $\Phi$ is linked to area by $\Phi = BA$ when the area is normal to $\vec B$ .
Always convert $\mu\text{T}$ to $\text{T}$ .

Common Mistakes

Using $38\ \mu\text{T}$ as $38\ \text{T}$ (missing the $10^{-6}$ factor).
Forgetting the $\cos\theta$ factor conceptually and becoming unsure whether to use $\Phi=BA$ .
Giving the area unit as $\text{Wb}$ or $\text{T}$ .

Things to Be Careful About

The vertical component is already provided; do not attempt to resolve the field further.
Keep track of powers of ten when dividing by $10^{-5}$ (it increases the result by $10^{5}$ ).

Techniques used

use \Phi = BA to connect flux and areaconvert microtesla to tesla and rearrange for area

(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 $15\ \text{s}$ the area swept out is

A = (68)(v)(15)

v = \frac{A}{68 \times 15} = \frac{2.13 \times 10^{5}}{1020} = 2.09 \times 10^{2}

Answer

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

Final answer

2.1 × 10^2 m s^-1

Detailed explanation

Background Concept

When a straight conductor of length $L$ moves a distance $x$ perpendicular to its length, it “sweeps out” an area

A = Lx.

If it moves at constant speed $v$ for time $t$ , then $x = vt$ , so

A = Lvt.

This area is the area used in $\Phi = BA$ for flux cutting when the magnetic field is perpendicular to the swept area.

Understanding the Question

You have already found the area of flux cut in $15\ \text{s}$ in part (b)(ii). The wingspan (distance between wingtips P and Q) is $68\ \text{m}$ . The aircraft travels forward a distance $vt$ in $15\ \text{s}$ . The question asks you to use these to find the aircraft speed $v$ .

Approach

Model the swept area as a rectangle:

one side is wingspan $L = 68\ \text{m}$ ,
the other side is distance travelled in $15\ \text{s}$ , i.e. $vt$ .

So $A = Lvt$ and rearrange to $v = A/(Lt)$ .

Step-by-Step Reasoning

Write the area relation:

A = Lvt.

Substitute $A = 2.13 \times 10^{5}\ \text{m}^2$ , $L=68\ \text{m}$ , $t=15\ \text{s}$ :

v = \frac{2.13 \times 10^{5}}{68 \times 15}.

Calculate the denominator:

68 \times 15 = 1020.

Evaluate:

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

Quote to suitable significant figures:

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

Key Takeaways

The “area of flux cut” here is simply the swept area: wingspan $\times$ distance travelled.
Convert the physical picture into $A=Lvt$ .

Common Mistakes

Using $A = 68v$ (forgetting the time factor).
Using $A = 68/(vt)$ (incorrect rearrangement).
Using $t=15$ minutes instead of seconds.

Things to Be Careful About

Check units: $\text{m}^2/(\text{m}\,\text{s}) = \text{m s}^{-1}$ , so the formula is dimensionally consistent.
Use the area from (b)(ii) consistently (carry forward is normally allowed but arithmetic must be correct).

Techniques used

relate swept area to distance travelled and wingspanrearrange A = Lvt to determine speed

(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

As the aircraft moves, charge carriers in the wings experience a magnetic force. The induced effect must oppose the motion (Lenz’s law), so the induced current in the wings must be such that the magnetic force on the wing is opposite to $v$ .

With $\vec v$ to the right and $\vec B$ vertically downwards (into the page in the view from above), the required current is from Q to P, giving a force to the left.

Hence wingtip P is at the higher induced potential (P is positive relative to Q).

Final answer

P is at higher potential.

Detailed explanation

Background Concept

When a conductor moves through a magnetic field, its charge carriers have velocity $\vec v$ and experience a magnetic force

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

This pushes positive and negative charges to opposite ends of the conductor, creating a potential difference (a motional e.m.f.).

Lenz’s law gives the direction: the induced e.m.f./current acts to oppose the change that produces it. In a motional induction situation, that often means the induced current produces a magnetic force that opposes the motion (a magnetic “drag”).

Understanding the Question

You are shown the aircraft moving horizontally. The vertical component of Earth’s magnetic field is towards the ground. The wingtips are P and Q. The question asks which wingtip is at higher potential and requires an explanation using Lenz’s law.

From the view from above, the vertical downward field is represented as crosses (field into the page). The aircraft moves to the right.

Approach

Use Lenz’s law: induced current must create a force opposing the aircraft’s motion.
Use the motor/generator force rule $\vec F = I(\vec L \times \vec B)$ (or Fleming’s right-hand rule) to decide which direction of current along the wing gives a force opposite to $\vec v$ .
Once current direction along the wing is known, infer which tip is at higher potential (the tip where positive charge accumulates is at higher potential).

Step-by-Step Reasoning

diagram-biv-1

Take the view from above:

aircraft velocity $\vec v$ is to the right.
magnetic field $\vec B$ is vertically downward, i.e. into the page.

Lenz’s law requirement: the induced current must produce a magnetic force on the wing to the left (opposing the motion to the right).
Consider current along the wing (from one tip to the other). If conventional current is from Q to P (up the page), then using

\vec F = I(\vec L \times \vec B),

with $\vec L$ upwards and $\vec B$ into the page, $\vec L \times \vec B$ points to the left. So that current direction produces a force opposing the motion, which matches Lenz’s law.

Therefore, the induced current in the wing itself is from Q to P.

In a source of e.m.f., conventional current inside the source goes from lower potential to higher potential (from negative to positive). Since current in the wing is Q to P, P must be at the higher potential.

Equivalently, using $\vec v \times \vec B$ : with $\vec v$ right and $\vec B$ into the page, $\vec v \times \vec B$ is upwards, so positive charges are driven towards P, making P positive and hence at higher potential.

Key Takeaways

Lenz’s law is about opposition: induced effects oppose the cause (here, the motion through the field).
Use direction rules ( $\vec v \times \vec B$ or $\vec F = I\vec L \times \vec B$ ) carefully to link motion, field direction, current direction, and polarity.

Common Mistakes

Saying “P is higher” without any reasoning based on Lenz’s law.
Mixing up field direction (crosses mean into the page).
Assuming conventional current always flows from high to low potential inside the moving conductor; in the induced source region it goes from negative to positive.
Using Fleming’s left-hand rule (motor) without recognising it is still valid for finding force direction once current direction is known.

Things to Be Careful About

Be explicit about which view you are using (above view is easiest because the downward field becomes “into the page”).
Keep the distinction clear:
- charge separation sets up a potential difference,
- if a circuit were completed, current in the external circuit would flow from P to Q, but within the wing (the source region) it is Q to P.
The question asks “higher induced potential”, so you must name P or Q and indicate polarity (e.g. “P is positive relative to Q”).

Techniques used

use the magnetic force on moving charges to infer charge separationapply Lenz’s law to determine the direction of induced current and polarityuse Fleming’s right-hand rule or the v × B rule to determine the direction

Q8Medium-EasyQuantum 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

Energy released:

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

So

m_{\text{U}} = m_{\text{Th}} + m_{\alpha} + \frac{Q}{c^2}

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

\frac{Q}{c^2} = \frac{4.274}{931.5}\ \text{u} = 0.0045883\ \text{u}

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

m_{\text{U}} = 237.915581 + 0.0045883 = 237.9201693\ \text{u}

Answer

m_{\,^{238}_{92}\text{U}} = 237.92017\ \text{u}

Final answer

237.92017 u

Detailed explanation

Background Concept

In nuclear reactions/decays, the total mass of the products can be less than the mass of the reactant(s). The “missing” mass (mass defect) appears as energy released:

Q = \Delta m\,c^2

For a decay

\text{parent} \rightarrow \text{daughter} + \alpha + \gamma

the energy release $Q$ corresponds to the mass difference between parent and all products. Often we use the conversion

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

so that MeV energy releases can be converted directly to a mass in u.

Understanding the Question

You are told:

total energy released $Q = 4.274\ \text{MeV}$ ,
nuclide masses: $m_{\alpha}=4.000407\ \text{u}$ and $m_{\text{Th}}=233.915174\ \text{u}$ ,
the uranium-238 nuclide mass is missing.

The question asks for $m_{\,^{238}_{92}\text{U}}$ in u, to five decimal places.

Approach

Use mass-energy equivalence with the definition of energy released:

Write $Q = (m_{\text{U}} - m_{\text{Th}} - m_{\alpha})c^2$ .
Rearrange for $m_{\text{U}}$ .
Convert $Q$ (MeV) to an equivalent mass in u using $1\ \text{u}c^2 = 931.5\ \text{MeV}$ .
Add masses and round to 5 d.p.

Step-by-Step Reasoning

Mass difference for this decay (ignoring any later recoil detail, since $Q$ is given as the total release):

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

Rearrange:

m_{\text{U}} = m_{\text{Th}} + m_{\alpha} + \frac{Q}{c^2}

Convert $Q$ to u:

\frac{Q}{c^2} = \frac{4.274\ \text{MeV}}{931.5\ \text{MeV per u}} = 0.0045883\ \text{u}

Add the given product masses:

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

Hence:

m_{\text{U}} = 237.915581 + 0.0045883 = 237.9201693\ \text{u}

Rounded to five decimal places:

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

Key Takeaways

Total energy release in a nuclear decay is found from mass defect: $Q=\Delta m c^2$ .
Use $1\ \text{u}c^2 = 931.5\ \text{MeV}$ to convert MeV (\leftrightarrow) u efficiently.

Common Mistakes

Using the electron mass unit conversion (eV to kg) unnecessarily and making power-of-ten errors.
Subtracting in the wrong direction (giving a negative mass defect).
Rounding too early and losing accuracy before the final 5 d.p.

Things to Be Careful About

The uranium mass must be slightly larger than $m_{\text{Th}}+m_{\alpha}$ because energy is released.
Keep enough digits in the intermediate steps to ensure correct rounding to five decimal places.

Techniques used

apply mass-energy equivalence to a nuclear decayrelate energy released to mass defectconvert between MeV and atomic mass unit using 1 u c^2

(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-Easy

Worked solution

Working

With recoil of thorium negligible:

Q = E_{\alpha} + E_{\gamma}

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

E_{\gamma} = 0.074\times 10^6 \ \text{eV} \times 1.60\times 10^{-19}\ \text{J eV}^{-1} = 1.18\times 10^{-14}\ \text{J}

\lambda = \frac{hc}{E_{\gamma}} = \frac{6.63\times 10^{-34}\times 3.00\times 10^8}{1.18\times 10^{-14}}

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

Answer

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

Final answer

1.68 × 10^-11 m

Detailed explanation

Background Concept

In a nuclear decay, the total energy released $Q$ is shared between the kinetic energy of the decay products and any photon energy emitted.

If a gamma photon is emitted, its energy is related to its wavelength by

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

So once you know $E_{\gamma}$ you can find

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

You must use $E_{\gamma}$ in joules if you use SI values of $h$ and $c$ .

Understanding the Question

Given:

total energy released $Q = 4.274\ \text{MeV}$ ,
kinetic energy of alpha particle $E_{\alpha} = 4.200\ \text{MeV}$ ,
assume thorium recoil KE is negligible.

Asked: wavelength of the gamma radiation.

The key clue is “rebound kinetic energy ... negligible”, which means we can take the energy split as just alpha KE plus gamma photon energy.

Approach

Use energy conservation: $Q = E_{\alpha} + E_{\gamma}$ .
Find $E_{\gamma}$ by subtraction.
Convert MeV to J.
Use $\lambda = hc/E_{\gamma}$ .

Step-by-Step Reasoning

With negligible thorium KE:

Q = E_{\alpha} + E_{\gamma}

so

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

Convert $0.074\ \text{MeV}$ to joules:

0.074\ \text{MeV} = 0.074\times 10^6\ \text{eV} = 7.4\times 10^4\ \text{eV}

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

E_{\gamma} = 7.4\times 10^4 \times 1.60\times 10^{-19} = 1.18\times 10^{-14}\ \text{J}

Apply the photon relation:

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

with $h = 6.63\times 10^{-34}\ \text{J s}$ and $c = 3.00\times 10^8\ \text{m s}^{-1}$ :

\lambda = \frac{6.63\times 10^{-34}\times 3.00\times 10^8}{1.18\times 10^{-14}} = 1.68\times 10^{-11}\ \text{m}

This is in the gamma-ray region (very short wavelength).

Key Takeaways

Use $Q = \text{(total kinetic energies)} + \text{(photon energies)}$ .
Gamma photon energy gives wavelength via $\lambda = hc/E$ .
Unit conversion (MeV (\to) J) is essential when using SI constants.

Common Mistakes

Using $E_{\gamma} = 4.274\ \text{MeV}$ (forgetting the alpha already has 4.200 MeV).
Mixing eV/MeV with SI values of $h$ and $c$ without conversion to joules.
Writing $\lambda = E/(hc)$ instead of $hc/E$ .

Things to Be Careful About

The photon energy here is the difference between the total release and the alpha kinetic energy, so it is much smaller than either number.
Quote wavelength to sensible significant figures (limited by given energies).

Techniques used

use energy conservation to find gamma photon energy from the decay 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 rebound kinetic energy, some of $Q$ goes into thorium KE, so $E_{\gamma}$ is smaller than in (b)(ii).

Since $\lambda = hc/E_{\gamma}$ , the true gamma wavelength is greater than the value calculated in (b)(ii).

Final answer

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

Detailed explanation

Background Concept

Energy released $Q$ in a decay must be shared between all forms of energy of the products:

kinetic energy of the alpha particle,
kinetic energy of the recoiling daughter nucleus,
energy of any emitted photons.

For a photon,

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

So decreasing photon energy increases wavelength.

Understanding the Question

In (b)(ii) you assumed the thorium nucleus recoil kinetic energy is negligible, so you effectively set $Q = E_{\alpha} + E_{\gamma}$ .

Now you are told this is not true and you must say how your computed wavelength compares to the true one, without doing any extra arithmetic.

Approach

Recognise that if thorium recoil KE is not negligible, energy conservation becomes $Q = E_{\alpha} + E_{\text{Th}} + E_{\gamma}$ .
Because $E_{\text{Th}} > 0$ , the remaining $E_{\gamma}$ must be smaller.
Use $\lambda = hc/E_{\gamma}$ to infer the direction of change in wavelength.

Step-by-Step Reasoning

If recoil is not negligible, then

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

Compared with the earlier assumption $E_{\text{Th}} \approx 0$ , adding a positive term $E_{\text{Th}}$ means the photon energy must reduce to keep the same total $Q$ .

So $E_{\gamma}$ is smaller than the value used in (b)(ii). Since

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

a smaller $E_{\gamma}$ implies a larger $\lambda$ .

Therefore the wavelength you calculated in (b)(ii) is an underestimate.

Key Takeaways

Recoil of the daughter nucleus always takes some energy.
Less photon energy means a longer photon wavelength.

Common Mistakes

Saying the wavelength is smaller (mixing up the inverse relation between $E$ and $\lambda$ ).
Thinking the total energy release $Q$ changes; $Q$ is fixed by the mass difference.

Things to Be Careful About

This is a comparison question: you must state direction (greater/less) and link it correctly to $\lambda = hc/E$ .

Techniques used

apply energy conservation qualitativelyrelate photon energy change to wavelength change using lambda = hc/E

(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

In $\beta$ decay of $^{60}_{27}\text{Co}$ , the daughter nucleus can be left in excited state(s) and then de-excites by gamma emission.

There are (at least) different possible energy-level transitions / a cascade of transitions, so different photon energies are produced and hence different wavelengths.

Final answer

Because Co-60 beta decay can produce de-excitation via different (often cascaded) nuclear energy-level transitions, giving gamma photons of more than one energy and hence more than one wavelength.

Detailed explanation

Background Concept

Gamma rays are emitted when a nucleus drops from a higher (excited) nuclear energy level to a lower level. Because energy levels are discrete, each transition produces a photon with a specific energy difference:

E_{\gamma} = \Delta E

and therefore a specific wavelength

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

If there is only one possible transition (one fixed $\Delta E$ ), you get a single gamma wavelength. If there are multiple transitions (different $\Delta E$ values), you get multiple wavelengths.

Understanding the Question

You are told uranium-238 gamma radiation has a single wavelength (i.e. one dominant photon energy).

You are asked why cobalt-60 does not give a single wavelength of gamma radiation in a sample.

The key idea is that cobalt-60 decay involves beta emission, which can leave the daughter nucleus in excited states that may de-excite in more than one step.

Approach

Explain qualitatively:

Beta decay often leaves the daughter nucleus excited.
De-excitation can occur through more than one transition (e.g. a cascade via an intermediate level, or different routes).
Different transitions mean different gamma photon energies, hence different wavelengths.

Step-by-Step Reasoning

In $\beta^-$ decay, a neutron converts to a proton and an electron is emitted. The daughter nucleus is frequently produced in an excited state.
The excited nucleus can drop to the ground state in stages (for example, excited (\to) intermediate (\to) ground), emitting a gamma photon at each step.
Each step has its own energy difference $\Delta E$ , so each gamma photon has its own $E_{\gamma}$ .
Because $\lambda = hc/E_{\gamma}$ , different $E_{\gamma}$ values correspond to different wavelengths.

So the gamma radiation from a cobalt-60 sample is not a single wavelength; it contains multiple discrete wavelengths (gamma lines).

Key Takeaways

Gamma rays come from transitions between discrete nuclear energy levels.
More than one transition (or a cascade) produces more than one gamma photon energy and wavelength.

Common Mistakes

Saying “beta particles have a range of energies so gamma does too” without mentioning nuclear energy-level transitions (gamma energies depend on level differences).
Claiming gamma is continuous like the beta spectrum.

Things to Be Careful About

The reason for multiple wavelengths should be stated in terms of different nuclear transitions / multiple energy differences, not vague statements about “random energies”.

Techniques used

link gamma emission to transitions between discrete energy levelsinfer multiple photon energies from multiple possible transitions

Q9MediumAstronomy and Cosmology

(a)

State Wien’s displacement law.

2M Similar

DifficultyEasy

Worked solution

Answer

For a black body,

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

where $b \approx 2.9 \times 10^{-3}\ \text{m K}$ .

Final answer

$\lambda_{\text{max}} T = 2.9 \times 10^{-3}\ \text{m K}$

Detailed explanation

Background Concept

A black body spectrum has a single peak when plotted as intensity (or power emitted per unit wavelength) against wavelength. The wavelength at which the emission is greatest is called the peak wavelength $\lambda_{\text{max}}$ .

Wien’s displacement law links the position of this peak to the absolute temperature $T$ of the emitting surface:

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

where $b$ is Wien’s constant (approximately $2.9 \times 10^{-3}\ \text{m K}$ ). Hotter objects peak at shorter wavelengths.

Understanding the Question

The question asks for the statement of Wien’s displacement law (2 marks). That means you need both:

the proportionality/equation linking $\lambda_{\text{max}}$ and $T$ ;
the constant (or an equivalent statement such as “ $\lambda_{\text{max}} \propto 1/T$ ”).

Approach

Write the standard equation, identify $\lambda_{\text{max}}$ as the wavelength of maximum emission, and give the numerical value of the constant with units.

Step-by-Step Reasoning

For a black body, the peak wavelength shifts with temperature.
The relationship is inverse: as $T$ increases, $\lambda_{\text{max}}$ decreases.
The quantitative law is:

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

State Wien’s constant:

b \approx 2.9 \times 10^{-3}\ \text{m K}

Key Takeaways

$\lambda_{\text{max}}$ is the wavelength where the spectrum peaks.
Wien’s law: $\lambda_{\text{max}} \propto 1/T$ .
The constant is about $2.9 \times 10^{-3}\ \text{m K}$ .

Common Mistakes

Forgetting to use kelvin for $T$ .
Writing the law as $\lambda_{\text{max}} \propto T$ (wrong direction).
Omitting the constant or its units.

Things to Be Careful About

Use the symbol $\lambda_{\text{max}}$ (not just any wavelength).
Quote the constant to an acceptable significant figure (typically 2 s.f. is fine) and include units $\text{m K}$ .

Techniques used

state the mathematical form of Wien’s displacement lawquote the Wien constant with correct 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

Answer

From Fig. 9.1, $F \propto d^{-2}$ , consistent with

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

so star X radiates approximately isotropically with constant luminosity $L$ .

From Fig. 9.2, $\lambda_{\text{max}}$ for star X is smaller than for the Sun, so $T_X > 5770\ \text{K}$ . Using Wien’s law with $\lambda_{\text{max,X}} \approx 3.5 \times 10^{-7}\ \text{m}$ ,

T_X = \frac{2.9 \times 10^{-3}}{3.5 \times 10^{-7}} \approx 8.3 \times 10^{3}\ \text{K}

The curve for star X in Fig. 9.2 has a larger peak / larger area under the curve than the Sun, so star X has a greater total power output (greater luminosity) than the Sun.

Final answer

Three conclusions: (i) $F \propto 1/d^2$ so isotropic emission with constant luminosity, (ii) $T_X > 5770\ \text{K}$ (e.g. $\approx 8\times10^3\ \text{K}$ from Wien’s law), (iii) star X is more luminous than the Sun (greater total emitted power).

Detailed explanation

Background Concept

There are two key ideas here:

Inverse-square spreading of radiation
If a source has luminosity $L$ (total power output), then at distance $d$ this power is spread over a sphere of area $4\pi d^2$ . The radiant flux intensity (power per unit area) is

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

So if $L$ is constant, then $F \propto d^{-2}$ .

Black-body spectra and temperature
For a (roughly) black-body emitter:

Wien’s law: $\lambda_{\text{max}}T = b$ tells you the temperature from the peak wavelength.
The total emitted power (luminosity) relates to surface area and temperature via Stefan–Boltzmann:

L = 4\pi r^2 \sigma T^4.

Also, if you compare two emission curves $P$ vs $\lambda$ (rate of emission against wavelength), the area under the curve is proportional to the total emitted power.

Understanding the Question

You are given two graphs:

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

You must state three conclusions about star X that are supported by these graphs. These can be qualitative (e.g. “hotter than the Sun”) or quantitative (e.g. “ $T \approx 8\times10^3\ \text{K}$ ”).

Approach

Pick conclusions that come directly from:

The straight-line $F$ vs $d^{-2}$ trend (inverse-square law, constant luminosity, isotropic emission).
The peak wavelength shift (use Wien to compare/compute temperature).
The overall size of the emission curve (compare total emitted power/luminosity).

Step-by-Step Reasoning

Conclusion from Fig. 9.1 (inverse square)

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

F \propto d^{-2}.

This matches the theoretical law

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

So it is reasonable to conclude star X behaves like an isotropic radiator with constant luminosity (at least over the distances used).

Conclusion from Fig. 9.2 (temperature from peak wavelength)

Star X peaks at a shorter wavelength than the Sun.
By Wien’s displacement law,

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

shorter $\lambda_{\text{max}}$ implies higher $T$ .

If you estimate $\lambda_{\text{max,X}}$ from the graph (for example, about $3.5\times 10^{-7}\ \text{m}$ ), then

T_X = \frac{2.9 \times 10^{-3}}{3.5 \times 10^{-7}} \approx 8.3 \times 10^3\ \text{K}.

Even without calculation, the direction is clear: $T_X > 5770\ \text{K}$ .

Conclusion from Fig. 9.2 (luminosity/total power)

The curve for star X is higher (larger peak) and overall encloses a larger area than the Sun’s curve.
Since the graph is rate of emission $P$ against $\lambda$ , the total emitted power is proportional to the integral (area):

L \propto \int P(\lambda)\, d\lambda.

So star X has greater total power output (greater luminosity) than the Sun.

(If you want to go further: with Stefan–Boltzmann, a larger luminosity combined with a larger $T$ tells you something about radius $r$ , but without reliable numerical ratios from the graphs, the safest credited conclusion is simply “more luminous”.)

Key Takeaways

A straight line of $F$ vs $d^{-2}$ through the origin is strong evidence for an inverse-square relationship $F \propto 1/d^2$ .
Shorter peak wavelength $\Rightarrow$ higher surface temperature (Wien’s law).
Larger area under the emission curve $\Rightarrow$ larger luminosity (total power output).

Common Mistakes

Confusing $F$ (flux received) with $L$ (power emitted): $F$ depends on distance; $L$ does not.
Using $\lambda_{\text{max}}$ incorrectly (choosing a non-peak point on the curve).
Claiming a definite radius for star X without enough numerical information.

Things to Be Careful About

Read $\lambda_{\text{max}}$ from the peak of the curve, and convert the axis scale correctly (here $10^{-7}\ \text{m}$ ).
If you give a quantitative temperature, it should be consistent with your read-off of $\lambda_{\text{max}}$ .
For “three conclusions”, make them distinct (e.g. inverse-square behaviour, higher temperature, higher luminosity) rather than repeating the same idea in different words.

Techniques used

use an inverse-square graph to infer proportionality between flux and distanceapply Wien’s displacement law to estimate surface temperature from peak wavelengthcompare spectra to infer relative luminosity from relative total emitted power

(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 would be shifted to longer wavelengths (peak at larger $\lambda$ ), i.e. redshifted, because the galaxy is moving away from Earth (Doppler / cosmological redshift).

Final answer

Shifted to longer wavelengths (redshifted peak), due to recession (Doppler/cosmological redshift).

Detailed explanation

Background Concept

When a light source is moving away from an observer, the observed wavelength increases. This is redshift. For relatively small speeds, the Doppler idea is that recession stretches the wave, giving a larger wavelength:

\lambda_{\text{observed}} > \lambda_{\text{emitted}}.

In cosmology, an overall redshift also occurs due to the expansion of space, but the key observable effect is the same: spectral features shift to longer wavelengths.

Understanding the Question

Fig. 9.2 shows $P$ vs $\lambda$ for star X as it would be at the source (or equivalently, without considering recession effects). The question asks how the line (curve) for star X would look if it were instead obtained from measurements made on Earth, given the galaxy is moving away.

So we must describe the appearance of the curve (its position on the wavelength axis) and give the reason.

Approach

Use the definition of redshift: recession increases observed wavelength. For a continuous spectrum, that means the whole curve shifts right (towards larger $\lambda$ ), including the position of the peak.

Step-by-Step Reasoning

diagram-c-redshift

Star X emits a spectrum with a peak at some emitted wavelength $\lambda_{\text{max,emit}}$ .
Because the galaxy is receding, every wavelength is observed as a larger wavelength:

\lambda_{\text{obs}} = (1+z)\,\lambda_{\text{emit}}

(where $z$ is the redshift).
3. Therefore the entire $P$ vs $\lambda$ curve is displaced to the right: the peak appears at

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

That is exactly what “redshifted” means.

Key Takeaways

Recession causes redshift: observed wavelengths are longer.
For a whole spectrum, the peak and all features shift to the right on a $\lambda$ axis.

Common Mistakes

Saying the curve shifts to shorter wavelength (that would be blueshift, for an approaching source).
Only mentioning “spectral lines” without linking it to the black-body curve shown.

Things to Be Careful About

The question asks how the curve would appear differently: the key mark is the direction of shift on the wavelength axis.
Keep the reason explicit: “moving away” $\Rightarrow$ redshift (Doppler/cosmological).

Techniques used

apply the Doppler/cosmological redshift idea to infer a shift in observed wavelengthrelate recession to an increase in observed wavelengthdescribe how a spectrum changes under wavelength scaling

Q10Medium-EasyMedical Physics

(a)

Define specific acoustic impedance.

2M Similar

DifficultyEasy

Worked solution

Answer

Specific acoustic impedance $Z$ is the product of the density $\rho$ and the speed of sound $c$ in the medium:

Z = \rho c.

Final answer

Specific acoustic impedance is Z = ρc (product of density and speed of sound in the medium).

Detailed explanation

Background Concept

When a sound (or ultrasound) wave travels through a medium, it involves oscillations of pressure and particle motion. A useful material property that characterises how “resistant” a medium is to these oscillations is the specific acoustic impedance.

For a plane progressive wave in a material,

Z = \rho c,

where:

$\rho$ is the density of the medium (in $\text{kg m}^{-3}$ ),
$c$ is the speed of sound in the medium (in $\text{m s}^{-1}$ ),
$Z$ has units $\text{kg m}^{-2}\,\text{s}^{-1}$ .

(Equivalent statement: $Z$ is the ratio of acoustic pressure to particle velocity in the wave.)

Understanding the Question

You are asked to define specific acoustic impedance. For full credit, you should state what it is and give the standard defining relation.

Approach

Provide the accepted definition used in ultrasound physics: state that it depends on the medium, and write $Z = \rho c$ (and optionally mention the ratio form).

Step-by-Step Reasoning

Identify the key quantities that set impedance for sound waves in a medium: density $\rho$ and speed of sound $c$ .
Write the defining equation:

Z = \rho c.

This matches the standard A Level definition and the units given in the question table.

Key Takeaways

Specific acoustic impedance is a material property.
It is given by $Z = \rho c$ (units $\text{kg m}^{-2}\,\text{s}^{-1}$ ).

Common Mistakes

Giving only units or only words without the equation.
Confusing impedance with intensity or attenuation.
Writing $Z = \rho / c$ (incorrect relationship).

Things to Be Careful About

Ensure the symbol is clear: $Z$ is impedance, $\rho$ is density, $c$ is speed of sound.
Use the correct SI-derived units (as in the table): $\text{kg m}^{-2}\,\text{s}^{-1}$ .

Techniques used

recall and state a precise definitionquote the defining equation for a material property

(b)

Explain how ultrasound waves are detected by a piezoelectric crystal.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Ultrasound incident on the crystal causes it to compress and expand (vibrate) at the wave frequency. By the piezoelectric effect this produces an alternating p.d. (charge separation) across the crystal faces, which is detected as an electrical signal.

Final answer

Ultrasound makes the crystal vibrate (compress/expand), and the piezoelectric effect produces an alternating p.d./charge across it that is detected as an electrical signal.

Detailed explanation

Background Concept

A piezoelectric crystal converts between mechanical deformation and electrical signals:

Direct piezoelectric effect (detection): applying stress/strain to the crystal causes charge separation, producing a p.d.
Inverse piezoelectric effect (generation): applying an alternating p.d. makes the crystal vibrate and emit ultrasound.

In ultrasound detection, the incoming wave is a sequence of compressions and rarefactions (pressure variations). These exert a rapidly varying force on the crystal.

Understanding the Question

The question asks how ultrasound waves are detected by the crystal. So you must describe the chain:

ultrasound pressure variations → 2) crystal deformation/vibration → 3) electrical output (p.d./charge) → 4) signal measured.

Approach

State that the ultrasound causes alternating compression/expansion of the crystal, and then invoke the direct piezoelectric effect to say an alternating p.d. is produced that can be amplified/recorded.

Step-by-Step Reasoning

An ultrasound wave arriving at the crystal produces alternating high and low pressures at its surface.
These pressure changes apply an alternating stress, so the crystal is repeatedly compressed and allowed to expand (it vibrates at the ultrasound frequency).
In a piezoelectric material, mechanical strain causes separation of charge within the crystal lattice.
Therefore an alternating potential difference appears across the electrodes on the crystal faces.
This alternating voltage is taken as the detected signal (often sent to an amplifier and processing electronics).

Key Takeaways

Detection uses the direct piezoelectric effect.
Ultrasound pressure variations cause crystal deformation, producing an a.c. voltage at the same frequency.

Common Mistakes

Describing only ultrasound generation (applying an a.c. voltage) instead of detection.
Saying the crystal “detects intensity” without mentioning conversion to an electrical signal.
Forgetting that the output is alternating (not steady d.c.).

Things to Be Careful About

Use correct causality: wave → deformation → charge/p.d.
Mention a measurable electrical quantity (p.d. / voltage / charge) to secure the mark.

Techniques used

link mechanical wave pressure variations to crystal deformationapply the piezoelectric effect to infer an electrical output signalstate how the signal is detected as a p.d. or charge

(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.

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 0.864.

Answer

Intensity reflection coefficient $\approx 0.86$ .

Final answer

0.86

Detailed explanation

Background Concept

When a wave reaches 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 mismatch of specific acoustic impedances.

The intensity reflection coefficient $R$ is

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

where $Z_1$ and $Z_2$ are the specific acoustic impedances of the two media. The square appears because intensity is proportional to (wave amplitude) $^2$ .

Understanding the Question

You are given $Z$ values for water and steel and asked for the intensity reflection coefficient for ultrasound incident on a water–steel boundary. That means:

take $Z_1 = Z_{\text{water}} = 1.48 \times 10^6\ \text{kg m}^{-2}\text{s}^{-1}$ ,
take $Z_2 = Z_{\text{steel}} = 4.04 \times 10^7\ \text{kg m}^{-2}\text{s}^{-1}$ ,
substitute into the formula for $R$ .

Approach

Use the standard boundary formula for $R$ at normal incidence. Compute the ratio of difference to sum, then square it, and finally round sensibly (typically to 2 s.f.).

Step-by-Step Reasoning

Write the formula:

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

Substitute the given impedances:

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.

Evaluate difference and sum:

Z_2 - Z_1 = 3.892\times 10^7, \qquad Z_2 + Z_1 = 4.188\times 10^7.

Form the ratio and square:

\frac{Z_2 - Z_1}{Z_2 + Z_1} \approx \frac{3.892}{4.188} \approx 0.929, \qquad R \approx (0.929)^2 \approx 0.864.

So about $86\%$ of the incident intensity is reflected at a water–steel boundary (very large mismatch).

Key Takeaways

Reflection is large when there is a large impedance mismatch.
Use

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

for normal incidence.

Common Mistakes

Forgetting to square the ratio (giving the amplitude reflection coefficient instead of intensity).
Swapping in the wrong $Z$ values (e.g. using tissue instead of water).
Arithmetic slips with powers of ten (keeping everything in standard form helps).

Things to Be Careful About

$R$ is a ratio, so it has no unit and must be between $0$ and $1$ .
Rounding: keep enough digits during calculation, round at the end (e.g. $0.86$ to 2 s.f.).

Techniques used

select and apply the intensity reflection coefficient formulasubstitute specific acoustic impedances into a boundary equationsquare an amplitude coefficient to obtain an intensity coefficient

(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

The impedances of body tissue and water are very similar, so only a small fraction of the ultrasound intensity is reflected at the boundary and most of the ultrasound is transmitted into the water.

Final answer

Very little reflection; most ultrasound is transmitted.

Detailed explanation

Background Concept

At a boundary, reflection depends mainly on the mismatch between the two specific acoustic impedances:

If $Z_1$ and $Z_2$ are very different, reflection is large.
If $Z_1$ and $Z_2$ are close, reflection is small and transmission is large.

This is why coupling gel (with impedance similar to tissue) is used in ultrasound scanning: it reduces reflections at the skin boundary and lets more ultrasound enter the body.

Understanding the Question

You are told (in the table) that:

$Z_{\text{tissue}} = 1.38 \times 10^6\ \text{kg m}^{-2}\text{s}^{-1}$
$Z_{\text{water}} = 1.48 \times 10^6\ \text{kg m}^{-2}\text{s}^{-1}$

The question asks what is likely to happen when ultrasound hits a tissue–water boundary, without calculation. So you should compare the impedances qualitatively.

Approach

Compare $Z$ values: since they are close, conclude that the reflection coefficient is small and transmission is large. State the physical outcome in words.

Step-by-Step Reasoning

Look at the impedances: $1.38 \times 10^6$ and $1.48 \times 10^6$ differ by only a small fraction.
Small impedance mismatch means the ratio

\frac{Z_2 - Z_1}{Z_2 + Z_1}

is small.
3. Since intensity reflection is proportional to the square of that ratio, the reflected intensity fraction is even smaller.
4. Therefore, only a small amount of the ultrasound is reflected; most continues through into the other medium.

Key Takeaways

Similar impedances (\Rightarrow) good transmission, little reflection.
Large mismatches (e.g. tissue–air, water–steel) cause strong reflections.

Common Mistakes

Claiming “most is reflected” just because there is a boundary (reflection depends on impedance mismatch, not just the existence of a boundary).
Confusing reflection with absorption/attenuation inside a material.

Things to Be Careful About

The question says “without calculation”: you should not attempt to compute $R$ ; a clear comparison of the sizes of $Z$ is enough.
Use language that matches what $R$ represents: “small fraction of intensity reflected” and “most transmitted.”

Techniques used

compare specific acoustic impedances to infer reflection magnitudeuse qualitative reasoning about impedance matching at a boundarystate likely outcomes in terms of reflection and transmission

Physics 9702/41 — October/November 2025

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