9702/22 Oct/Nov 2025 — Physics Solved Past Paper

Q1MediumPhysical Quantities and UnitsForces, Density and PressureDynamics

Scientists are investigating the variation in air pressure at different locations on a mountain.

(a)

The scientists take measurements of several physical quantities at each location.

Complete Table 1.1 by stating the SI base unit for each quantity and identifying with a tick ( $\checkmark$ ) whether each quantity is a scalar or a vector. Use the space for any working.

Table 1.1

quantity measured	SI base unit	scalar	vector
air temperature
air pressure

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Air temperature: SI unit $\text{K}$ ; scalar $\checkmark$
Air pressure: $\text{Pa} = \text{N m}^{-2} = \text{kg m}^{-1}\text{ s}^{-2}$ ; scalar $\checkmark$

Final answer

Air temperature: K, scalar. Air pressure: kg m^-1 s^-2 (Pa), scalar.

Detailed explanation

Background Concept

An SI base unit is one of the fundamental units (e.g. kilogram, metre, second, kelvin). Many quantities use derived units, built from base units using physical definitions.

A scalar has magnitude only (e.g. temperature, pressure). A vector has magnitude and direction (e.g. force, velocity).

Pressure is defined by

P = \frac{F}{A}

so its unit is force per unit area.

Understanding the Question

You must fill a table for two measured quantities:

give the correct SI unit (and for pressure, it is safest to show it in SI base units),
decide whether each quantity has direction (vector) or not (scalar).

Approach

Recall the SI unit for temperature.
For pressure, start from $P = F/A$ and convert $\text{Pa}$ into base units using $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .
Decide scalar/vector by asking: “does this quantity need a direction to be fully described?”

Step-by-Step Reasoning

Temperature: SI unit is the kelvin ( $\text{K}$ ). Temperature does not have a direction, so it is a scalar.
Pressure:
- Unit of pressure is the pascal: $\text{Pa}$ .
- Convert to base units:

\text{Pa} = \text{N m}^{-2}

and since

\text{N} = \text{kg m s}^{-2},

then

\text{Pa} = (\text{kg m s}^{-2})\,\text{m}^{-2} = \text{kg m}^{-1}\text{ s}^{-2}.

Pressure acts equally in all directions at a point and has no single direction itself, so it is a scalar.

Key Takeaways

Temperature: $\text{K}$ , scalar.
Pressure: $\text{Pa} = \text{kg m}^{-1}\text{ s}^{-2}$ , scalar.
Derived units can be rewritten in SI base units using definitions.

Common Mistakes

Saying pressure is a vector because it “pushes”: pressure is scalar; force is the vector.
Giving $\text{N m}^{-2}$ without converting to base units when the question asks for base units.
Writing Celsius ( $^\circ\text{C}$ ) instead of kelvin.

Things to Be Careful About

The table heading says “SI base unit”: for pressure, show the conversion to base units clearly.
Do not confuse the unit name (pascal) with its base-unit form ( $\text{kg m}^{-1}\text{ s}^{-2}$ ).

Techniques used

identify the SI unit for each physical quantityexpress a derived unit in SI base unitsclassify quantities as scalars or vectors

(b)

7M

(i)

At one location, the density of the air is $1.1\ \text{kg m}^{-3}$ . A spherical weather balloon is filled with a gas and released from rest. The balloon has radius $0.90\ \text{m}$ .

Calculate the upthrust acting on the balloon when it is released.

upthrust = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Volume displaced:

V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.90)^3 = 3.05\ \text{m}^3

Upthrust:

U = \rho g V = (1.1)(9.81)(3.05) = 32.9\ \text{N}

Answer

$U \approx 33\ \text{N}$

Final answer

33 N

Detailed explanation

Background Concept

Upthrust (buoyant force) occurs when an object is in a fluid (liquid or gas). The key result (Archimedes’ principle) is:

U = \text{weight of fluid displaced} = \rho g V

where:

$\rho$ is the fluid density,
$g$ is gravitational field strength,
$V$ is the volume of fluid displaced (for a fully immersed object, this is the object’s volume).

Understanding the Question

A spherical balloon of radius $0.90\ \text{m}$ is released in air of density $1.1\ \text{kg m}^{-3}$ . At the instant it is released, the upthrust depends only on the volume of air displaced (i.e. the balloon’s volume) and the air density.

You are asked to calculate the upthrust in newtons.

Approach

Find the volume of the sphere using $V = \frac{4}{3}\pi r^3$ .
Use $U = \rho g V$ .
Quote the result to sensible significant figures (typically 2 s.f. from given data).

Step-by-Step Reasoning

Volume of the balloon (and hence air displaced):

V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.90)^3

Compute $0.90^3 = 0.729$ , so

V = 4.18879 \times 0.729 \approx 3.05\ \text{m}^3.

Upthrust equals weight of displaced air:

U = \rho g V = (1.1)(9.81)(3.05) \approx 32.9\ \text{N}.

Rounding appropriately:

U \approx 33\ \text{N}.

Key Takeaways

Use $U = \rho g V$ for buoyancy.
For a sphere, $V = \frac{4}{3}\pi r^3$ .
In gases, upthrust is usually smaller than in liquids because $\rho$ is smaller.

Common Mistakes

Using the density of the gas inside the balloon instead of the surrounding air density.
Forgetting that $V$ is in $\text{m}^3$ (not $\text{cm}^3$ ).
Using diameter instead of radius.

Things to Be Careful About

Ensure $r$ is in metres before cubing.
Use a consistent value of $g$ (e.g. $9.81\ \text{m s}^{-2}$ ) throughout.

Techniques used

apply Archimedes' principle to relate upthrust to displaced fluid weightcalculate the volume of a spheresubstitute into a formula with consistent SI units

(ii)

Explain why an upthrust acts on the balloon.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Air pressure increases with depth, so the pressure (and force) on the bottom of the balloon is greater than on the top. This produces a resultant upward force (upthrust), equal to the weight of displaced air.

Final answer

Pressure is greater at the bottom than the top, giving a net upward force equal to the weight of displaced air.

Detailed explanation

Background Concept

In a fluid at rest, pressure increases with depth because the fluid above has weight:

\Delta p = \rho g \Delta h

Pressure acts normally (perpendicularly) on a surface, producing a force

F = pA.

If pressure is larger on the lower surface than the upper surface, the upward force is larger than the downward force, giving a net upward force: upthrust.

Understanding the Question

You are not being asked to calculate anything here. You must explain the physical reason the balloon experiences an upward buoyant force when surrounded by air.

Approach

State that pressure in a fluid increases with depth.
Link pressure difference to different forces on top and bottom surfaces.
Conclude there is a net upward force (upthrust), and optionally connect to Archimedes’ principle.

Step-by-Step Reasoning

The balloon is surrounded by air, which exerts pressure on all sides.
The bottom of the balloon is at a slightly greater depth than the top, so the air pressure at the bottom is slightly larger.
Since force on a surface is $F = pA$ , the upward force on the bottom surface is greater than the downward force on the top surface.
The difference between these forces is the upthrust, acting upward.
This net force is equal to the weight of the displaced air (Archimedes’ principle).

Key Takeaways

Upthrust comes from a pressure gradient in the surrounding fluid.
Pressure differences translate into force differences via $F = pA$ .

Common Mistakes

Saying upthrust is caused by “air pushing up” without mentioning the pressure difference between bottom and top.
Confusing upthrust with the force from the gas inside the balloon; buoyancy depends primarily on the surrounding fluid.

Things to Be Careful About

Make it clear it is the surrounding air pressure that causes upthrust.
Mentioning “greater pressure at the bottom” is usually essential for full credit.

Techniques used

link pressure variation with depth to a net forceexplain buoyancy using force imbalance on opposite sides of an objectrelate upthrust to displaced fluid weight

(iii)

The balloon has weight $19\ \text{N}$ .

Calculate the magnitude of the initial acceleration of the balloon.

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

3M Similar

DifficultyMedium

Worked solution

Working

Resultant upward force:

F_{\text{net}} = U - W = 32.9 - 19 = 13.9\ \text{N}

Mass of balloon:

m = \frac{W}{g} = \frac{19}{9.81} = 1.94\ \text{kg}

Newton’s second law:

a = \frac{F_{\text{net}}}{m} = \frac{13.9}{1.94} = 7.17\ \text{m s}^{-2}

Answer

$a \approx 7.2\ \text{m s}^{-2}$

Final answer

7.2 m s^-2

Detailed explanation

Background Concept

Newton’s second law relates resultant force to acceleration:

F_{\text{net}} = ma.

Weight is a force:

W = mg \quad \Rightarrow \quad m = \frac{W}{g}.

For a balloon released from rest, the initial acceleration depends on the net force at that instant (upthrust upward, weight downward; other forces like drag are zero at the instant of release because speed is zero).

Understanding the Question

You are given the balloon’s weight ( $19\ \text{N}$ ) and you have already found the upthrust from part (i). You must find the magnitude of the initial acceleration just after release.

Approach

Find the net upward force: $F_{\text{net}} = U - W$ .
Convert weight to mass using $m = W/g$ .
Use $a = F_{\text{net}}/m$ .

Step-by-Step Reasoning

Forces:

Upthrust $U$ acts upward.
Weight $W$ acts downward.
At the moment of release, air resistance is negligible because $v = 0$ .

So the resultant force is

F_{\text{net}} = U - W.

Using $U = 32.9\ \text{N}$ and $W = 19\ \text{N}$ :

F_{\text{net}} = 32.9 - 19 = 13.9\ \text{N}.

Find mass from weight:

m = \frac{W}{g} = \frac{19}{9.81} = 1.94\ \text{kg}.

Apply Newton’s second law:

a = \frac{F_{\text{net}}}{m} = \frac{13.9}{1.94} = 7.17\ \text{m s}^{-2} \approx 7.2\ \text{m s}^{-2}.

Key Takeaways

Always use the resultant force to find acceleration.
Convert correctly between weight and mass.
At the instant of release, drag is zero because speed is zero.

Common Mistakes

Using $m = 19\ \text{kg}$ instead of converting from weight.
Adding forces instead of subtracting ( $U + W$ instead of $U - W$ ).
Including drag at release (drag depends on speed).

Things to Be Careful About

The question asks for magnitude, so you can give a positive value even though you might choose up as positive in your working.
Keep consistent $g$ and significant figures from earlier parts (error carried forward is usually allowed).

Techniques used

calculate resultant force from upthrust and weightconvert weight to mass using m = W/gapply Newton's second law to find acceleration

(c)

A quantity $c$ relating to the motion of the balloon is calculated from three measured quantities $k$ , $F$ and $v$ using the formula

c = \frac{2kF}{v^2}

The percentage uncertainties in the measured quantities are given in Table 1.2.

Table 1.2

measured quantity	percentage uncertainty
$k$	$5\%$
$F$	$3\%$
$v$	$4\%$

The calculated value of $c$ is $1.8$ .

Determine the absolute uncertainty in $c$ .

absolute uncertainty = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

c = \frac{2kF}{v^2}

Percentage uncertainty in $c$ :

\frac{\Delta c}{c}(\%) = 5 + 3 + 2(4) = 16\%

Absolute uncertainty:

\Delta c = 0.16 \times 1.8 = 0.288 \approx 0.29

Answer

Absolute uncertainty in $c$ is $0.29$ .

Final answer

0.29

Detailed explanation

Background Concept

When quantities are multiplied/divided, percentage (or fractional) uncertainties add:

If $q = ab$ , then $\frac{\Delta q}{q} \approx \frac{\Delta a}{a} + \frac{\Delta b}{b}$ .
If $q = \frac{a}{b}$ , then $\frac{\Delta q}{q} \approx \frac{\Delta a}{a} + \frac{\Delta b}{b}$ .
If $q = a^n$ , then $\frac{\Delta q}{q} \approx |n|\frac{\Delta a}{a}$ .

Constants (like the factor 2) are taken as exact unless stated otherwise.

To convert percentage uncertainty to absolute uncertainty:

\Delta q = \left(\frac{\Delta q}{q}\right) q.

Understanding the Question

You are given

c = \frac{2kF}{v^2}

and percentage uncertainties: $k: 5\%$ , $F: 3\%$ , $v: 4\%$ . The calculated value is $c = 1.8$ . You must find the absolute uncertainty $\Delta c$ .

Approach

Add the percentage uncertainties for $k$ and $F$ .
Because $v$ is squared in the denominator, double its percentage uncertainty.
Multiply the total percentage uncertainty by $c$ to get absolute uncertainty.

Step-by-Step Reasoning

Combine percentage uncertainties:

From $k$ : $5\%$ .
From $F$ : $3\%$ .
From $v^2$ : since $v$ has $4\%$ , then $v^2$ contributes $2 \times 4\% = 8\%$ .

Total percentage uncertainty:

5\% + 3\% + 8\% = 16\%.

Convert to absolute uncertainty using $c = 1.8$ :

\Delta c = 16\% \text{ of } 1.8 = 0.16 \times 1.8 = 0.288.

Rounded sensibly:

\Delta c \approx 0.29.

Key Takeaways

For products/quotients: add percentage uncertainties.
For a power: multiply the percentage uncertainty by the power.
Absolute uncertainty = (fractional uncertainty) $\times$ value.

Common Mistakes

Forgetting to double the uncertainty because of $v^2$ .
Subtracting uncertainties because $v^2$ is in the denominator (you still add).
Giving $16\%$ as the final answer instead of converting to an absolute uncertainty.

Things to Be Careful About

Treat the constant 2 as exact (no uncertainty contribution).
Round the absolute uncertainty to an appropriate number of significant figures (usually 1–2 s.f.), consistent with exam conventions.

Techniques used

combine percentage uncertainties for products and quotientsaccount for powers when propagating uncertaintiesconvert percentage uncertainty to absolute uncertainty

Q2MediumForces, Density and PressureDynamicsKinematics

A spacecraft in deep space uses jets of hot gas from its thrusters to change its velocity. Fig. 2.1 shows a side view of the spacecraft and some of its thrusters.

Fig. 2.1

Thruster A is a distance of $1.6\ \text{m}$ leftwards from the centre of gravity of the spacecraft. Thruster C is a distance of $0.40\ \text{m}$ upwards from the centre of gravity of the spacecraft.

Thrusters A and B can produce forces on the spacecraft in the upwards direction only. Thruster C can produce a force on the spacecraft in the leftwards direction only. All the thrusters shown produce forces entirely in the same plane as the centre of gravity.

(a)

3M

(i)

Thruster A is activated, producing a force of $60\ \text{N}$ upwards on the spacecraft. Thruster C is also activated, producing a force of $220\ \text{N}$ in the leftwards direction on the spacecraft.

Calculate the resultant moment due to these forces about the centre of gravity.

resultant moment = ______ $\text{N m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Moment from A about C.G.:

\tau_A = Fd = 60 \times 1.6 = 96\ \text{N m} \ (\text{clockwise})

Moment from C about C.G.:

\tau_C = Fd = 220 \times 0.40 = 88\ \text{N m} \ (\text{anticlockwise})

Resultant moment:

\tau = 96 - 88 = 8\ \text{N m} \ (\text{clockwise})

Answer

$8\ \text{N m}$ clockwise

Final answer

8 N m clockwise

Detailed explanation

Background Concept

The moment (torque) of a force about a point is a measure of its turning effect:

\tau = F \times d

where $F$ is the force and $d$ is the perpendicular distance from the point (pivot) to the line of action of the force.

The direction (sense) of the moment is either clockwise (CW) or anticlockwise (ACW). When more than one force acts, the resultant moment is the algebraic sum of the individual moments, taking one sense as positive.

Understanding the Question

Two thrusters produce forces in perpendicular directions:

Thruster A: $60\ \text{N}$ upwards, acting at a point $1.6\ \text{m}$ to the left of the centre of gravity.
Thruster C: $220\ \text{N}$ leftwards, acting at a point $0.40\ \text{m}$ above the centre of gravity.

We must find the net (resultant) moment about the centre of gravity, including the correct sense (CW/ACW).

Approach

For each force, identify the perpendicular distance from the centre of gravity to that force’s line of action.
Calculate each moment using $\tau = Fd$ .
Decide whether each moment is clockwise or anticlockwise.
Subtract because the moments oppose.

Step-by-Step Reasoning

diagram-a-i-1

Moment from thruster A

Line of action is vertical (upwards force).
Perpendicular distance from C.G. is the horizontal separation $d = 1.6\ \text{m}$ .

\tau_A = 60 \times 1.6 = 96\ \text{N m}

Because the force is upward on the left of the C.G., it tends to rotate the spacecraft clockwise.

Moment from thruster C

Line of action is horizontal (leftward force).
Perpendicular distance from C.G. is the vertical separation $d = 0.40\ \text{m}$ .

\tau_C = 220 \times 0.40 = 88\ \text{N m}

A leftward force acting above the C.G. tends to rotate anticlockwise.

Resultant moment
The two moments oppose, so subtract:

\tau_{\text{resultant}} = 96 - 88 = 8\ \text{N m}

Direction is the same as the larger moment (from A): clockwise.

Key Takeaways

Use $\tau = Fd$ with the perpendicular distance to the line of action.
Always state the sense of rotation when moments are involved.
Resultant moment is the algebraic sum (moments can oppose).

Common Mistakes

Using the wrong distance (e.g. using $1.6\ \text{m}$ for thruster C or $0.40\ \text{m}$ for thruster A).
Forgetting that clockwise and anticlockwise moments can cancel.
Not stating the direction (CW/ACW) when required.

Things to Be Careful About

The perpendicular distance is measured to the line of action, not to the point of application in a straight-line sense.
Keep units as $\text{N m}$ (not $\text{m N}$ or missing units).

Techniques used

calculate the moment of a force using perpendicular distancedetermine the sense of rotation about a pivotadd moments algebraically to find a resultant moment

(ii)

State and explain whether the forces from A and C are a couple.

1M Similar

(b)

Thrusters A and C are now switched off and the spacecraft is stationary. Thruster B is activated at time $t_1$ , producing a constant force on the spacecraft until the fuel runs out at time $t_2$ . As the fuel is used, the total mass of the spacecraft decreases.

On Fig. 2.2, sketch the variation of speed of the spacecraft with time from $t_1$ to $t_2$ .

Fig. 2.2

2M Similar

DifficultyMedium

Worked solution

Answer

From $t_1$ the speed increases with time and the gradient increases (curve becomes steeper) because $a = F/m$ increases as $m$ decreases.

So sketch a rising curve from $t_1$ to $t_2$ that is concave upwards (increasing gradient).

Final answer

Speed increases with increasing gradient (concave-up curve) from t1 to t2.

Detailed explanation

Background Concept

Newton’s second law states:

F = ma

so the acceleration is

a = \frac{F}{m}

On a speed–time graph, the gradient at any point equals the acceleration:

\text{gradient} = \frac{\Delta v}{\Delta t} = a

Therefore, if acceleration increases with time, the speed–time graph must become progressively steeper (concave upwards).

Understanding the Question

The spacecraft starts stationary. Thruster B provides a constant force from $t_1$ to $t_2$ . As fuel is burnt, the spacecraft’s mass decreases. We must sketch how speed varies between $t_1$ and $t_2$ .

Approach

Use $a = F/m$ .
Since $F$ is constant and $m$ decreases, $a$ increases.
Increasing $a$ means increasing gradient on the speed–time graph.
Draw a curve starting at $v=0$ at $t_1$ that rises and becomes steeper towards $t_2$ .

Step-by-Step Reasoning

At $t_1$ , speed is zero (stationary). Immediately after $t_1$ there is a forward (upward) thrust from B, so speed begins to increase.

Because the thrust is constant:

a(t) = \frac{F}{m(t)}

As fuel is used, $m(t)$ decreases, so $a(t)$ increases. Since acceleration equals the gradient of the speed–time graph, the gradient must increase with time.

So the correct sketch is an increasing curve that is concave upwards between $t_1$ and $t_2$ .

diagram-b-1

Key Takeaways

Constant force does not necessarily mean constant acceleration if mass changes.
On a $v$ – $t$ graph: increasing acceleration means a curve that gets steeper.

Common Mistakes

Drawing a straight line (constant gradient), which would imply constant acceleration and hence constant mass.
Drawing a curve that flattens (decreasing gradient), which would imply acceleration is decreasing.

Things to Be Careful About

The sketch asked is only from $t_1$ to $t_2$ .
Make sure the curve starts at the correct point: the spacecraft is stationary at $t_1$ , so $v=0$ at $t_1$ .

Techniques used

use Newton's second law to relate force, mass and accelerationinterpret the gradient of a speed-time graph as accelerationdeduce how changing mass affects acceleration for constant force

(c)

The spacecraft now splits apart into a carrier and a payload as shown in Fig. 2.3.

Fig. 2.3

During the split, an average force of $5500\ \text{N}$ acts on the payload for a time of $0.36\ \text{s}$ . The velocity of the payload increases by $8.5\ \text{m s}^{-1}$ in the upwards direction.

The combined mass of the carrier and payload is $2.5 \times 10^3\ \text{kg}$ .

7M

(i)

State the principle of conservation of momentum.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

In an isolated system (no resultant external force), the total momentum remains constant, i.e. total momentum before an interaction equals total momentum after.

Final answer

Total momentum of an isolated system remains constant (before = after) if no resultant external force acts.

Detailed explanation

Background Concept

Linear momentum is

p = mv

The principle of conservation of momentum states that if the resultant external force is zero, then the total momentum of a system does not change.

Equivalently:

\sum p_{\text{before}} = \sum p_{\text{after}}

This is because external force is the rate of change of momentum:

F_{\text{ext}} = \frac{\Delta p_{\text{total}}}{\Delta t}

So if $F_{\text{ext}}=0$ , then $\Delta p_{\text{total}}=0$ .

Understanding the Question

The spacecraft splits into carrier and payload. The forces involved in the split are internal forces between the two parts, so (in deep space) the system can be treated as having negligible external forces. The question asks you to state the principle used to relate their momenta before and after the split.

Approach

Give the standard two-part statement:

Condition: no resultant external force / isolated system.
Conclusion: total momentum remains constant (before = after).

Step-by-Step Reasoning

Identify the system: carrier + payload.
In deep space, external forces are negligible over the short split time, so it is approximately isolated.
Therefore total momentum of carrier + payload is conserved.

Key Takeaways

Momentum conservation is a system rule: apply it to all interacting bodies together.
Always mention the condition: no resultant external force.

Common Mistakes

Saying only “momentum is conserved” without stating the condition for conservation.
Confusing momentum conservation with energy conservation.

Things to Be Careful About

Conservation applies to the total momentum vector, including directions (signs).

Techniques used

state momentum conservation in terms of total momentum before and afterspecify the condition of zero resultant external force

(ii)

Show that the mass of the payload is $230\ \text{kg}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Impulse on payload:

Ft = 5500 \times 0.36 = 1980\ \text{N s}

Ft = \Delta p = m\Delta v

m = \frac{1980}{8.5} = 2.33 \times 10^2\ \text{kg} \approx 230\ \text{kg}

Answer

$230\ \text{kg}$

Final answer

230 kg

Detailed explanation

Background Concept

Impulse is the product of average force and time:

\text{impulse} = Ft

Impulse equals the change in momentum:

Ft = \Delta p

For a body of constant mass,

\Delta p = m\Delta v

Understanding the Question

An average force of $5500\ \text{N}$ acts on the payload for $0.36\ \text{s}$ , and the payload’s velocity increases by $8.5\ \text{m s}^{-1}$ upwards. We are asked to show the payload mass is $230\ \text{kg}$ .

Approach

Compute the impulse $Ft$ .
Set $Ft = m\Delta v$ .
Rearrange for $m$ and substitute values.

Step-by-Step Reasoning

Compute impulse:

Ft = 5500 \times 0.36 = 1980\ \text{N s}

Use impulse–momentum:

1980 = m(8.5)

So

m = \frac{1980}{8.5} = 232.94\ \text{kg}

Rounded to 2 significant figures (consistent with the given data),

m \approx 230\ \text{kg}

Key Takeaways

Impulse links force and time to change in momentum.
For constant mass, $\Delta p = m\Delta v$ is the quickest route to find $m$ .

Common Mistakes

Using $Ft = mv$ instead of $Ft = m\Delta v$ .
Forgetting that $8.5\ \text{m s}^{-1}$ is the change in velocity already.
Rounding too early and losing accuracy.

Things to Be Careful About

Use the average force if the force is not constant (the question explicitly says average).
Quote the final mass to a sensible number of significant figures (here, $230\ \text{kg}$ ).

Techniques used

use impulse as force multiplied by timeequate impulse to change in momentumrearrange to find mass from change in velocity

(iii)

Calculate the magnitude of the change in velocity of the carrier.

change in velocity = ______ $\text{m s}^{-1}$

3M Similar

DifficultyMedium

Worked solution

Working

Payload mass (from (ii)) $m_p = 230\ \text{kg}$ , so carrier mass

m_c = 2.5 \times 10^3 - 230 = 2270\ \text{kg}

Initially total momentum $=0$ , so after split

m_p(8.5) = m_c\,\Delta v_c

\Delta v_c = \frac{230 \times 8.5}{2270} = 0.86\ \text{m s}^{-1}

Answer

$0.86\ \text{m s}^{-1}$

Final answer

0.86 m s^-1

Detailed explanation

Background Concept

For an isolated system, total momentum is conserved:

\sum p_{\text{before}} = \sum p_{\text{after}}

Momentum is a vector, so direction matters. If the system starts at rest, the total momentum is zero; after separation, the two parts must have equal and opposite momenta.

Understanding the Question

The spacecraft (total mass $2.5 \times 10^3\ \text{kg}$ ) splits into:

payload of mass $m_p$ moving upwards with speed increase $8.5\ \text{m s}^{-1}$ ,
carrier of mass $m_c$ moving in the opposite direction.

We must find the magnitude of the carrier’s change in velocity.

Approach

Find the carrier mass from total mass: $m_c = m_{\text{total}} - m_p$ .
Use conservation of momentum with initial momentum zero.
Set payload momentum upward equal to carrier momentum downward in magnitude.
Solve for $\Delta v_c$ .

Step-by-Step Reasoning

diagram-c-iii-1

1. Mass of the carrier
Using $m_p = 230\ \text{kg}$ :

m_c = 2500 - 230 = 2270\ \text{kg}

2. Apply momentum conservation
Initially the spacecraft is stationary, so total momentum is zero.
After the split, take upwards as positive.

Payload momentum:

p_p = m_p v_p = 230 \times 8.5

Carrier must have equal and opposite momentum:

p_c = -p_p

So in magnitudes:

m_p(8.5) = m_c(\Delta v_c)

Solve:

\Delta v_c = \frac{230 \times 8.5}{2270} = 0.86\ \text{m s}^{-1}

This is the magnitude; the direction would be downwards.

Key Takeaways

If the system starts from rest, momenta after separation must be equal and opposite.
The more massive object gets the smaller change in velocity.

Common Mistakes

Forgetting to subtract to find $m_c$ .
Adding velocities instead of using momentum conservation.
Ignoring direction and getting the sign wrong (even though this question asks for magnitude).

Things to Be Careful About

Use consistent significant figures (2–3 s.f. is appropriate here).
If you used a more precise payload mass from (ii), you would get a very similar final value (error carried forward is usually allowed).

Techniques used

apply conservation of momentum to an initially stationary systemuse equal and opposite momenta after separationcompute speed change from momentum and mass ratio

Q3MediumForces, Density and PressureDeformation of Solids

A spring is fixed at one end and attached to the frame of a pulley at the other end. A cable is passed around the wheel of the pulley. The spring is stretched to a fixed length using the cable and pulley.

Fig. 3.1 shows the view from above of the spring, cable and pulley.

Fig. 3.1

The spring obeys Hooke’s law and has a spring constant $k$ of $250\ \text{N m}^{-1}$ . A force $F$ acts on the spring. The tension in the cable is $T$ . The pulley is in equilibrium.

(a)

On Fig. 3.2, draw labelled arrows to show the directions of the forces acting on the pulley.

Fig. 3.2

2M Similar

DifficultyMedium-Easy

Worked solution

diagram-1

Final answer

See diagram

Detailed explanation

Background Concept

For a body in equilibrium, the resultant force on it is zero. To analyse this, we draw a force diagram showing all external forces acting on the body. Each force is drawn as an arrow pointing in the direction the force acts on the chosen body.

In a light, smooth cable, the tension is the same throughout the cable. If the cable touches a pulley and leaves in two straight sections, then the pulley experiences two tension forces from the two cable sections, each acting along the direction of that section.

Understanding the Question

We are looking at the pulley (frame/wheel) as the object. The spring is attached to the pulley frame, so the spring exerts a force $F$ on the pulley. The cable passes around the pulley wheel and leaves as two parallel straight segments to the right; each segment pulls on the pulley with tension $T$ . The pulley is stated to be in equilibrium, so these forces must balance.

Approach

Choose the pulley as the body.
Add forces due to each interaction:
- spring on pulley ( $F$ ), along the spring direction;
- cable on pulley: two forces of magnitude $T$ , each along the direction of a cable segment.
Ensure directions are consistent with the geometry (spring to the left, two cable segments to the right).

Step-by-Step Reasoning

The spring is stretched between the fixed support (left) and the pulley frame (right), so it pulls the pulley back towards the fixed end: force $F$ on the pulley is to the left.
The cable has two straight, parallel sections on the right-hand side of the pulley. Each section exerts a pull on the pulley along the cable direction, so there are two forces of magnitude $T$ acting to the right on the pulley (one for the upper segment and one for the lower segment).
These three forces are the ones to show for the pulley in equilibrium.

diagram-1

Key Takeaways

For a pulley with a cable leaving in two sections, the pulley experiences two tension forces, one from each section.
In equilibrium, a correct force diagram is the essential first step.

Common Mistakes

Drawing only one tension force instead of two.
Drawing the spring force in the wrong direction (it acts to pull the pulley back toward the fixed end).
Labeling cable forces as $F$ instead of $T$ .

Things to Be Careful About

Forces must be drawn on the pulley (not on the cable or on the spring).
Tension forces act along the cable segments (here, horizontal to the right as shown).
If the diagram shows two separate cable strands, you must include two separate $T$ forces.

Techniques used

identify all external forces acting on an objectrepresent forces as labelled vectors on a force diagramuse equilibrium to infer consistent force directions

(b)

The force $F$ is $110\ \text{N}$ .

3M

(i)

Determine $T$ .

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

1M Similar

DifficultyEasy

Worked solution

Working

For equilibrium on the pulley:

F = 2T

T = \frac{F}{2} = \frac{110}{2} = 55\ \text{N}

Answer

$T = 55\ \text{N}$

Final answer

55 N

Detailed explanation

Background Concept

For a body in equilibrium, the vector sum of forces is zero. If all forces act along one line, equilibrium reduces to “total force to the right = total force to the left”.

A pulley with a cable around it experiences a force from each straight section of cable leaving the pulley. If the tension in each section is $T$ and the two sections are parallel, then the resultant pull from the cable on the pulley is $2T$ in that direction.

Understanding the Question

We are told $F = 110\ \text{N}$ acts on the spring and the pulley is in equilibrium. From the diagram context, the spring pulls the pulley one way with force $F$ and the cable pulls the pulley the opposite way with two equal tension forces $T$ (one from each cable segment). We must find $T$ .

Approach

Use horizontal force equilibrium for the pulley:

spring force magnitude $F$ balances the sum of the two cable tensions $T + T$ .
So set $F = 2T$ and solve.

Step-by-Step Reasoning

The cable has two segments pulling the pulley to the right, each with magnitude $T$ , so total rightward force is $2T$ .
The spring pulls the pulley to the left with magnitude $F$ .
Equilibrium gives:

F = 2T

Substitute $F = 110\ \text{N}$ :

T = \frac{110}{2} = 55\ \text{N}

Key Takeaways

A pulley pulled by two equal strands has a total pull of $2T$ .
Equilibrium often reduces to a simple force-balance equation.

Common Mistakes

Using $F = T$ (forgetting there are two strands).
Doubling the wrong force (writing $2F = T$ ).

Things to Be Careful About

The two tensions are separate forces acting on the pulley; only add them if they act in the same direction (they do here: both to the right).

Techniques used

apply equilibrium of forcesrelate multiple tensions to a single balancing forcerearrange a linear equation

(ii)

Calculate the extension of the spring.

extension = ______ $\text{m}$

2M Similar

(c)

A second identical spring with the same spring constant of $250\ \text{N m}^{-1}$ is now also connected to the pulley, as shown in Fig. 3.3.

Fig. 3.3

The tension in the cable is kept the same. The pulley is again in equilibrium.

4M

(i)

Determine the extension of the springs.

extension = ______ $\text{m}$

2M Similar

DifficultyMedium

Worked solution

Working

From (b)(i), $T = 55\ \text{N}$ so the pull from the cable is $2T = 110\ \text{N}$ .

Two identical springs in parallel:

F = (k + k)x = 2kx

x = \frac{F}{2k} = \frac{110}{2 \times 250} = 0.22\ \text{m}

Answer

extension $= 0.22\ \text{m}$

Final answer

0.22 m

Detailed explanation

Background Concept

When springs are connected in parallel (both attached between the same two points), they have the same extension $x$ . The forces they exert add:

F_{\text{total}} = F_1 + F_2 = kx + kx = 2kx

So the equivalent spring constant is:

k_{\text{eq}} = k_1 + k_2

Understanding the Question

A second identical spring is added in parallel with the first between the fixed support and the pulley frame. The cable tension $T$ is kept the same as before, and the pulley remains in equilibrium. So the total force that must be balanced by the springs is unchanged, but it is now shared between two springs.

Approach

Use the same cable effect as before: with tension $T$ in each strand, the cable pulls with $2T$ .
In equilibrium, springs must provide equal and opposite total force $F_{\text{total}} = 2T$ .
Use the parallel-springs relation $F_{\text{total}} = 2kx$ to find the extension $x$ .

Step-by-Step Reasoning

From earlier, the tension in each cable segment is $T = 55\ \text{N}$ . With two segments:

F_{\text{total}} = 2T = 110\ \text{N}

With two identical springs in parallel, both extend by the same amount $x$ , and each exerts force $kx$ , so total restoring force is:

F_{\text{total}} = 2kx

Solve for $x$ :

x = \frac{F_{\text{total}}}{2k} = \frac{110}{2 \times 250} = 0.22\ \text{m}

This is half the original extension because the system is twice as stiff.

Key Takeaways

Parallel springs add their spring constants: stiffer system, smaller extension for the same load.
Always decide whether springs are in series or in parallel by checking whether they share the same extension (parallel) or the same force (series).

Common Mistakes

Treating parallel springs as if they were in series (using $1/k_{\text{eq}} = 1/k + 1/k$ ).
Using $F = kx$ with $k = 250$ instead of $k_{\text{eq}} = 500$ .

Things to Be Careful About

“Tension kept the same” means the external pull from the cable ( $2T$ ) is the same as before, so the total spring force required is unchanged.

Techniques used

use equilibrium to determine the total force applied to the spring systemcombine spring constants for parallel springsapply Hooke’s law to an equivalent spring

(ii)

The elastic potential energy stored in the spring in Fig. 3.1 is $E_1$ . The total elastic potential energy stored in the two springs in Fig. 3.3 is $E_2$ .

Calculate the ratio $\frac{E_1}{E_2}$ .

ratio = ______

2M Similar

DifficultyMedium

Worked solution

Working

Elastic energy in a spring:

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

Fig. 3.1: $x_1 = 0.44\ \text{m}$

E_1 = \frac{1}{2}k x_1^2

Fig. 3.3: $x_2 = 0.22\ \text{m}$ for each spring

E_2 = 2 \left(\frac{1}{2}k x_2^2\right) = kx_2^2

\frac{E_1}{E_2} = \frac{\frac{1}{2}k x_1^2}{k x_2^2} = \frac{1}{2}\left(\frac{x_1}{x_2}\right)^2

\frac{E_1}{E_2} = \frac{1}{2}\left(\frac{0.44}{0.22}\right)^2 = \frac{1}{2}(2^2) = 2

Answer

ratio $\dfrac{E_1}{E_2} = 2$

Final answer

2

Detailed explanation

Background Concept

The elastic potential energy stored in a Hooke’s-law spring is the work done to stretch it:

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

This comes from the area under a force–extension graph (a triangle) because $F$ increases linearly from $0$ to $kx$ .

When there are two springs, the total energy stored is the sum of the energies in each spring.

Understanding the Question

$E_1$ is the energy in the single-spring setup (Fig. 3.1). $E_2$ is the total energy stored when two identical springs are used in parallel (Fig. 3.3). We already found the extensions:

single spring: $x_1 = 0.44\ \text{m}$ ,
each of the two parallel springs: $x_2 = 0.22\ \text{m}$ .
We must find the ratio $E_1/E_2$ .

Approach

Write energy for the one spring: $E_1 = \tfrac{1}{2}k x_1^2$ .
Write energy for the two-spring case: energy per spring is $\tfrac{1}{2}k x_2^2$ , then multiply by 2 to get $E_2$ .
Form the ratio and simplify; $k$ cancels.

Step-by-Step Reasoning

Single spring:

E_1 = \frac{1}{2}k x_1^2

Two parallel springs:

energy in one spring is $\tfrac{1}{2}k x_2^2$ ,
total energy is twice that:

E_2 = 2\left(\frac{1}{2}k x_2^2\right) = kx_2^2

Now take the ratio:

\frac{E_1}{E_2} = \frac{\frac{1}{2}k x_1^2}{k x_2^2} = \frac{1}{2}\left(\frac{x_1}{x_2}\right)^2

With $x_1/x_2 = 0.44/0.22 = 2$ :

\frac{E_1}{E_2} = \frac{1}{2}(2^2) = 2

So the single-spring setup stores twice as much elastic energy as the two-parallel-spring setup, even though the applied force is the same (because extension is larger in the less stiff system).

Key Takeaways

Elastic energy scales with the square of extension: $E \propto x^2$ .
For multiple springs, total energy is the sum across springs.
Doubling stiffness (parallel springs) halves extension for the same force, so energy changes significantly.

Common Mistakes

Forgetting to double the energy for two springs (using $E_2 = \tfrac{1}{2}k x_2^2$ instead of $2 \times$ that).
Using $x_1 = x_2$ (they are not equal here).
Using $E = kx$ or $E = Fx$ instead of $E = \tfrac{1}{2}kx^2$ .

Things to Be Careful About

$E_2$ is the total energy in both springs, not the energy in one spring.
Do the algebra first: cancelling $k$ helps avoid unnecessary arithmetic errors.

Techniques used

use elastic potential energy formula for a springsubstitute extensions for different configurationssimplify a ratio algebraically

Q4MediumSuperpositionWaves

A laser emits visible light of a single frequency in a vacuum. The light is incident normally on a double slit and then forms a pattern of bright and dark fringes on a screen, as shown in Fig. 4.1.

Fig. 4.1

The separation of the slits is $1.0 \times 10^{-3}\ \text{m}$ . The distance from the slits to the screen is $4.8\ \text{m}$ . The distance between the centres of adjacent dark fringes on the screen is $3.3\ \text{mm}$ .

(a)

Explain how the pattern of bright and dark fringes is formed.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Light diffracts at each slit so the two slits act as two coherent sources.
At a point on the screen the waves from the two slits superpose.
Where path difference $= m\lambda$ the waves arrive in phase giving constructive interference (bright fringe).
Where path difference $= \left(m + \frac{1}{2}\right)\lambda$ the waves arrive in antiphase giving destructive interference (dark fringe).

Final answer

Bright fringes where path difference = mλ; dark fringes where path difference = (m + 1/2)λ, due to superposition of coherent waves from the two slits.

Detailed explanation

Background Concept

Interference occurs when two (or more) waves overlap at a point and combine according to the principle of superposition: the resultant displacement is the sum of the individual displacements.

For a stable (unchanging) fringe pattern, the two sources must be coherent: they have a constant phase difference and the same frequency.

For two coherent sources, the type of interference at a point depends on the path difference (difference in distance travelled by the two waves to that point):

Constructive interference (bright): path difference $= m\lambda$ (integer $m$ ), waves arrive in phase.
Destructive interference (dark): path difference $= \left(m + \frac{1}{2}\right)\lambda$ , waves arrive $\pi$ radians out of phase.

Understanding the Question

A laser shines normally onto a double slit and produces alternating bright and dark fringes on a distant screen. The question asks for a physical explanation of how those fringes arise.

So we need to mention:

each slit producing a spreading (diffracted) wave,
coherence (both slits fed by the same laser),
superposition at the screen,
the path-difference conditions for bright and dark.

Approach

Explain the chain:

Laser light reaches both slits from the same source, so the two emerging waves are coherent.
The waves overlap on the screen.
Different points on the screen correspond to different path differences.
Use the path-difference conditions to identify bright (constructive) and dark (destructive) fringes.

Step-by-Step Reasoning

diagram-a-1

Diffraction at each slit: each slit is narrow enough to cause spreading, so light from each slit can reach many points on the screen.
Coherent sources: because both slits are illuminated by the same laser, the waves leaving the slits have the same frequency and a fixed phase relationship.
Superposition on the screen: at any chosen point $P$ on the screen, two waves arrive: one from slit $S_1$ and one from slit $S_2$ .
Path difference controls phase difference:

If $S_2P - S_1P = m\lambda$ , the phase difference is $2\pi m$ , so the waves add to a maximum amplitude → bright fringe.
If $S_2P - S_1P = \left(m + \tfrac{1}{2}\right)\lambda$ , the phase difference is $(2m+1)\pi$ , so the waves cancel → dark fringe.

As you move along the screen, the path difference changes smoothly, so you get alternating bright and dark fringes.

Key Takeaways

Double-slit fringes are an interference effect requiring coherence.
Bright: path difference $m\lambda$ ; dark: path difference $\left(m + \tfrac{1}{2}\right)\lambda$ .
Diffraction at each slit allows overlap of the two waves on the screen.

Common Mistakes

Saying “bright where waves meet” without stating the path-difference / phase condition.
Forgetting to mention that the two slits must act as coherent sources.
Mixing up the conditions for bright and dark fringes.

Things to Be Careful About

“Coherent” means constant phase difference, not merely “same frequency” (though same frequency is necessary).
The pattern is stable only if the phase relationship does not drift randomly with time.

Techniques used

state that the two slits act as coherent sourcesapply the principle of superposition at the screenuse path difference conditions for constructive and destructive interference

(b)

Calculate the frequency of the light emitted by the laser.

frequency = ______ $\text{Hz}$

4M Similar

DifficultyMedium

Worked solution

Working

Fringe spacing (dark to adjacent dark):

\Delta x = \frac{\lambda D}{a} \;\Rightarrow\; \lambda = \frac{a\Delta x}{D}

\lambda = \frac{(1.0\times10^{-3})(3.3\times10^{-3})}{4.8} = 6.9\times10^{-7}\ \text{m}

f = \frac{c}{\lambda} = \frac{3.0\times10^{8}}{6.9\times10^{-7}} = 4.3\times10^{14}\ \text{Hz}

Answer

$4.3\times10^{14}\ \text{Hz}$

Final answer

4.3 × 10^14 Hz

Detailed explanation

Background Concept

For a double-slit pattern (small angles), the separation $\Delta x$ between adjacent fringes is

\Delta x = \frac{\lambda D}{a}

where:

$\lambda$ is the wavelength,
$D$ is the distance from the slits to the screen,
$a$ is the slit separation.

This spacing is the same whether you measure between adjacent bright fringes or adjacent dark fringes (each step corresponds to a path-difference change of one full wavelength).

Once $\lambda$ is known, for light in vacuum:

c = f\lambda \quad \Rightarrow \quad f = \frac{c}{\lambda}

Understanding the Question

Given:

slit separation $a = 1.0 \times 10^{-3}\ \text{m}$
slit-to-screen distance $D = 4.8\ \text{m}$
spacing between centres of adjacent dark fringes $\Delta x = 3.3\ \text{mm} = 3.3\times10^{-3}\ \text{m}$

We must find the frequency $f$ of the laser light.

Approach

Use the measured fringe spacing with $\Delta x = \lambda D/a$ to calculate $\lambda$ .
Use $f = c/\lambda$ with $c = 3.0\times10^{8}\ \text{m s}^{-1}$ .

Step-by-Step Reasoning

Convert the fringe spacing to metres:

3.3\ \text{mm} = 3.3\times10^{-3}\ \text{m}

Rearrange the fringe-spacing formula to get wavelength:

\lambda = \frac{a\Delta x}{D}

Substitute values:

\lambda = \frac{(1.0\times10^{-3})(3.3\times10^{-3})}{4.8} = 6.875\times10^{-7}\ \text{m} \approx 6.9\times10^{-7}\ \text{m}

This is in the visible range, which is a good check.

Use $f = c/\lambda$ :

f = \frac{3.0\times10^{8}}{6.875\times10^{-7}} = 4.36\times10^{14}\ \text{Hz} \approx 4.3\times10^{14}\ \text{Hz}

Key Takeaways

Adjacent bright-to-bright or dark-to-dark spacing equals $\lambda D/a$ .
Always convert mm to m before substituting.
For EM waves in vacuum, use $c = f\lambda$ .

Common Mistakes

Using $a/\Delta x$ instead of $\Delta x/a$ (inverting the formula).
Forgetting to convert $3.3\ \text{mm}$ to metres.
Using $c = 3.0\times10^{8}$ with $\lambda$ left in mm, giving a frequency off by $10^3$ .

Things to Be Careful About

The formula assumes small angles and that $D \gg a$ (true here: $4.8\ \text{m}$ is much larger than $10^{-3}\ \text{m}$ ).
Quote the final frequency to sensible significant figures (limited by given data: typically 2 s.f.).

Techniques used

use the double-slit fringe spacing relationconvert millimetres to metres consistentlyuse c = fλ to determine frequency

(c)

The double slit is removed. A second laser is placed beside the first laser. The second laser produces visible light of a different frequency from that of the first laser. The beams of light from the two lasers overlap on the screen.

Explain why a steady pattern of bright and dark fringes is not formed on the screen.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The two lasers are not coherent (no constant phase relationship, and different frequencies), so the phase difference at any point changes with time and the interference pattern averages out (no steady fringes).

Final answer

No steady fringes because the two lasers are not coherent (phase difference varies with time), so the pattern washes out.

Detailed explanation

Background Concept

A steady interference pattern requires coherent sources: same frequency and a constant phase difference.

If the phase difference at a point changes with time, the intensity at that point fluctuates between bright and dark. The screen (and your eye) effectively averages over time, so you observe a uniform illumination rather than fixed fringes.

Understanding the Question

The original double slit produced fringes because both waves came from the same laser and therefore had a stable phase relationship.

Now the double slit is removed and instead two separate lasers shine onto the same screen, with different frequencies. The question asks why this does not produce a steady bright/dark fringe pattern.

Approach

State the coherence requirement, then explain that two independent lasers (especially with different frequencies) cannot maintain a fixed phase difference at the screen, so the fringe pattern is not stationary.

Step-by-Step Reasoning

With two independent lasers, the relative phase between their electric fields is not locked.
Because their frequencies differ, the phase difference changes continually:

\Delta \phi(t) = 2\pi (f_2 - f_1)t + \text{constant}

Therefore, at any fixed point on the screen, the interference varies rapidly in time from constructive to destructive.
Time averaging removes the contrast, so there is no steady pattern of bright and dark fringes.

Key Takeaways

Interference fringes require coherence.
Different frequencies (and independent sources) cause a varying phase difference, washing out fringes.

Common Mistakes

Saying “different wavelengths so no fringes” without mentioning the real reason: no constant phase difference.
Forgetting that even same-frequency independent lasers are generally not coherent unless phase-locked.

Things to Be Careful About

It is the stability of the phase difference that matters for a steady pattern; instantaneous interference can still occur, but it will not remain fixed in position or contrast.

Techniques used

identify the requirement of coherence for a stable interference patternlink changing phase difference to fringe washing out

Q5MediumD.C. CircuitsElectricity

Fig. 5.1 shows a circuit containing a battery, two fixed resistors X and Y, and a light-dependent resistor (LDR) Z.

Fig. 5.1

The battery has electromotive force (e.m.f.) $5.0\ \text{V}$ and internal resistance $4.7\ \Omega$ . The current in X is $I_1$ and the current in Y is $I_2$ .

The resistance of X is $100\ \Omega$ . The resistance of Z varies with the intensity of light incident on it as shown in Fig. 5.2.

Fig. 5.2

(a)

State Kirchhoff’s first law.

1M Similar

(b)

The intensity of light incident on Z is $130\ \text{W m}^{-2}$ . The current in the battery is $38\ \text{mA}$ .

10M

(i)

Show that the terminal potential difference of the battery is $4.8\ \text{V}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Lost volts:

Ir = (0.038)(4.7) = 0.18\ \text{V}

Terminal p.d.:

V = \mathcal{E} - Ir = 5.0 - 0.18 = 4.82\ \text{V} \approx 4.8\ \text{V}

Answer

$4.8\ \text{V}$

Final answer

4.8 V

Detailed explanation

Background Concept

A real battery can be modelled as an ideal source of e.m.f. $\mathcal{E}$ in series with an internal resistance $r$ . When a current $I$ is drawn, some energy per unit charge is dissipated inside the battery, giving a “lost volts” drop $Ir$ .

The terminal potential difference (p.d.) across the external circuit is then

V = \mathcal{E} - Ir

This applies when the battery is delivering current to the circuit.

Understanding the Question

Given:

$\mathcal{E} = 5.0\ \text{V}$
$r = 4.7\ \Omega$
battery current $I = 38\ \text{mA} = 0.038\ \text{A}$

You must show the terminal p.d. is $4.8\ \text{V}$ .

Approach

Convert the current into amperes, calculate lost volts $Ir$ , then subtract from the e.m.f.

Step-by-Step Reasoning

Convert current:

I = 38\ \text{mA} = 0.038\ \text{A}

Lost volts across internal resistance:

Ir = 0.038 \times 4.7 = 0.1786\ \text{V} \approx 0.18\ \text{V}

Terminal p.d.:

V = \mathcal{E} - Ir = 5.0 - 0.1786 = 4.8214\ \text{V}

To 2 s.f. (matching the data),

V \approx 4.8\ \text{V}

Key Takeaways

Terminal p.d. is less than e.m.f. when current flows because of internal resistance.
Use $V = \mathcal{E} - Ir$ for a delivering cell.

Common Mistakes

Using $V = \mathcal{E} + Ir$ (wrong sign).
Forgetting to convert $\text{mA}$ to $\text{A}$ .

Things to Be Careful About

Quote the terminal p.d. to appropriate significant figures; $4.82\ \text{V}$ or $4.8\ \text{V}$ would typically be accepted depending on marking.
Ensure $r$ is in $\Omega$ and $I$ in $\text{A}$ so $Ir$ is in volts.

Techniques used

use the lost volts equation $V = \mathcal{E} - Ir$substitute values with consistent SI unitsround to an appropriate number of significant figures

(ii)

Calculate the current $I_2$ in Y.

$I_2$ = ______ $\text{A}$

3M Similar

DifficultyMedium

Worked solution

Working

From Fig. 5.2 at $130\ \text{W m}^{-2}$ , take $R_Z \approx 120\ \Omega$ .

Branch $X$ and $Z$ are in series:

R_{XZ} = 100 + 120 = 220\ \Omega

Terminal p.d. across each parallel branch is $V = 4.8\ \text{V}$ , so

I_1 = \frac{4.8}{220} = 2.18 \times 10^{-2}\ \text{A}

Total battery current $I = 3.8 \times 10^{-2}\ \text{A}$ and $I = I_1 + I_2$ :

I_2 = 3.8 \times 10^{-2} - 2.18 \times 10^{-2} = 1.62 \times 10^{-2}\ \text{A}

Answer

I_2 = 1.6 \times 10^{-2}\ \text{A}

Final answer

1.6 × 10^-2 A

Detailed explanation

Background Concept

In a parallel circuit, each branch has the same potential difference across it (equal to the supply terminal p.d.). For any resistor or series combination of resistors in a branch, the branch current is found using Ohm’s law:

I = \frac{V}{R}

At the junction where the current splits, Kirchhoff’s first law applies:

I = I_1 + I_2

where $I$ is the total current from the battery and $I_1$ , $I_2$ are the branch currents.

Understanding the Question

The external circuit consists of two parallel branches:

Top branch: fixed resistor $X = 100\ \Omega$ in series with LDR $Z$ .
Bottom branch: fixed resistor $Y$ .

Given when the light intensity on $Z$ is $130\ \text{W m}^{-2}$ :

battery current $I = 38\ \text{mA} = 0.038\ \text{A}$
terminal p.d. $V \approx 4.8\ \text{V}$ (from part (i))
$R_Z$ must be read from the graph.

You need $I_2$ (current in $Y$ ).

Approach

Read $R_Z$ at $130\ \text{W m}^{-2}$ from Fig. 5.2.
Compute the total resistance of the top branch: $R_{XZ} = 100 + R_Z$ .
Use $I_1 = V/R_{XZ}$ because the branch p.d. is $V$ .
Use Kirchhoff’s first law: $I_2 = I - I_1$ .

Step-by-Step Reasoning

Reading the graph: at intensity $130\ \text{W m}^{-2}$ , the LDR resistance is approximately $R_Z \approx 120\ \Omega$ (your exact value may differ slightly depending on how you read the curve).
Series in the top branch means resistances add:

R_{XZ} = R_X + R_Z = 100 + 120 = 220\ \Omega

The terminal p.d. $V$ is across the whole parallel network, so it is also across the top branch. Hence

I_1 = \frac{V}{R_{XZ}} = \frac{4.8}{220} = 2.18 \times 10^{-2}\ \text{A}

At the junction, total current splits:

I = I_1 + I_2 \;\Rightarrow\; I_2 = I - I_1

So

I_2 = 0.038 - 0.0218 = 0.0162\ \text{A} \approx 1.6 \times 10^{-2}\ \text{A}

Key Takeaways

Parallel branches share the same p.d.; series components share the same current.
For a split, use Kirchhoff’s first law: total current equals the sum of branch currents.
Reading a value from a graph is often the first step in a circuit calculation.

Common Mistakes

Adding currents directly using $I = V/R$ with the wrong resistance (e.g. using $100\ \Omega$ instead of $100 + R_Z$ ).
Using the e.m.f. $5.0\ \text{V}$ instead of the terminal p.d. $4.8\ \text{V}$ .
Forgetting that $38\ \text{mA} = 0.038\ \text{A}$ .

Things to Be Careful About

Your numerical answer depends on how you read $R_Z$ from the curve; keep your reading sensible and consistent with the axes.
Use the terminal p.d. across the external circuit, not the e.m.f., when finding branch currents.
Round at the end to avoid unnecessary rounding error.

Techniques used

read a resistance value from a graphapply $I = V/R$ to a series branchapply Kirchhoff’s first law at a junctionsubtract branch currents from total current

(iii)

Calculate the power dissipated in Y.

power = ______ $\text{W}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For resistor $Y$ , p.d. across it is the terminal p.d. $V = 4.8\ \text{V}$ :

P = VI_2 = 4.8 \times 1.62 \times 10^{-2} = 7.8 \times 10^{-2}\ \text{W}

Answer

P = 7.8 \times 10^{-2}\ \text{W}

Final answer

7.8 × 10^-2 W

Detailed explanation

Background Concept

Electrical power is the rate of transfer of electrical energy. For a component with potential difference $V$ across it and current $I$ through it,

P = VI

For a resistor you can also use

P = I^2R \quad \text{or} \quad P = \frac{V^2}{R}

Understanding the Question

You have already found the current in resistor $Y$ (called $I_2$ ). Resistor $Y$ is connected directly across the battery terminals (as one branch of the parallel network), so the p.d. across $Y$ equals the terminal p.d. of the battery (found in part (i) to be $4.8\ \text{V}$ ).

You are asked for the power dissipated in $Y$ .

Approach

Use $P = VI$ with:

$V$ = terminal p.d. across the parallel branches
$I$ = branch current $I_2$

Step-by-Step Reasoning

Since $Y$ is a single component in its branch, the p.d. across it is the branch p.d., which equals the battery terminal p.d.

V_Y = 4.8\ \text{V}

With $I_2 = 1.62 \times 10^{-2}\ \text{A}$ ,

P_Y = V_Y I_2 = 4.8 \times 1.62 \times 10^{-2} = 7.78 \times 10^{-2}\ \text{W}

So to 2 s.f.,

P_Y \approx 7.8 \times 10^{-2}\ \text{W}

Key Takeaways

In parallel, each branch has the same p.d.
Power in a component can be found quickly using $P = VI$ .

Common Mistakes

Using $5.0\ \text{V}$ instead of the terminal p.d.
Using the total battery current instead of the branch current $I_2$ .

Things to Be Careful About

Keep units consistent: $V$ in volts, $I$ in amperes gives $P$ in watts.
If you use $P = I^2R$ or $P = V^2/R$ , make sure you are using the resistance of $Y$ (not the whole circuit).

Techniques used

use $P = VI$ for power in a componentuse the branch potential difference equal to the terminal potential differencesubstitute the previously found branch current

(iv)

The intensity of the light incident on Z decreases.

State and explain the effect on the terminal potential difference of the battery.

3M Similar

DifficultyMedium

Worked solution

Answer

As light intensity decreases, $R_Z$ increases, so the resistance of the $X$ – $Z$ branch increases and the total external resistance increases. Hence the battery current $I$ decreases. Since

V = \mathcal{E} - Ir

a smaller $I$ gives a smaller $Ir$ , so the terminal p.d. increases (towards $5.0\ \text{V}$ ).

Final answer

Terminal p.d. increases (towards 5.0 V) because lower light makes R_Z larger, reducing current so lost volts Ir decreases.

Detailed explanation

Background Concept

An LDR has a resistance that increases when light intensity decreases. In circuits with an internal resistance $r$ , the terminal p.d. is

V = \mathcal{E} - Ir

So the terminal p.d. depends on the current drawn. If the current decreases, the lost volts $Ir$ decreases and $V$ rises closer to the e.m.f.

Also, for a fixed e.m.f. source, changing a resistance in the external circuit changes the total external resistance and hence the current drawn.

Understanding the Question

The LDR $Z$ is in series with resistor $X$ in the top branch, and this whole top branch is in parallel with resistor $Y$ . You are told the intensity of light incident on $Z$ decreases.

You must state what happens to the battery terminal p.d. and explain using circuit reasoning (including internal resistance).

Approach

Follow the cause-and-effect chain:

Lower light $\Rightarrow$ $R_Z$ increases.
Top branch resistance $R_{XZ}$ increases.
The parallel combination’s equivalent resistance increases (or, at least, the total current drawn decreases).
Battery current $I$ decreases.
Lost volts $Ir$ decreases.
Terminal p.d. $V = \mathcal{E} - Ir$ therefore increases.

Step-by-Step Reasoning

From the given graph behaviour of an LDR, decreasing intensity makes the LDR resistance larger:

\text{intensity decreases} \Rightarrow R_Z \text{ increases}

Since $X$ and $Z$ are in series in the top branch,

R_{XZ} = R_X + R_Z

so $R_{XZ}$ increases.

With a larger resistance in one of the parallel branches, that branch current decreases (because the same branch p.d. produces a smaller current $I_1 = V/R_{XZ}$ ). Therefore the total current drawn from the battery decreases:

I = I_1 + I_2 \quad \text{and } I_1 \downarrow \Rightarrow I \downarrow

(Here $I_2$ through $Y$ may also change because the terminal p.d. changes; the key point is that overall the circuit draws less current when the external resistance increases.)

The internal voltage drop is $Ir$ . If $I$ decreases, $Ir$ decreases.
Using

V = \mathcal{E} - Ir

when $Ir$ becomes smaller, $V$ becomes larger. So the terminal p.d. increases, approaching $\mathcal{E} = 5.0\ \text{V}$ .

Key Takeaways

Lower light on an LDR increases its resistance.
Increasing external resistance reduces the current drawn.
With internal resistance, reduced current means reduced lost volts and increased terminal p.d.

Common Mistakes

Saying the terminal p.d. decreases because “resistance increases” (that would be true for the p.d. across the internal resistance drop $Ir$ , not the terminal p.d.).
Forgetting to mention internal resistance and lost volts in the explanation.

Things to Be Careful About

Be explicit that it is the battery current (total current) that determines the lost volts $Ir$ .
State the direction of change: $I$ decreases, so $Ir$ decreases, so terminal p.d. increases.

Techniques used

link LDR resistance change to branch resistance changededuce effect on total current in a parallel networkuse $V = \mathcal{E} - Ir$ to predict the change in terminal potential difference

Q6Medium-EasyParticle PhysicsWork, Energy and Power

(a)

State what is meant by a fundamental particle.

1M Similar

(b)

5M

(i)

Particle Q is a meson with a charge of 0.

Determine a possible quark composition for Q.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

A meson is a quark–antiquark pair.

For example:

Q = u\bar{u}

Charge $= +\frac{2}{3} + \left(-\frac{2}{3}\right) = 0$ .

Answer

$u\bar{u}$ (or $d\bar{d}$ / $s\bar{s}$ ).

Final answer

u u-bar (e.g. u\bar{u})

Detailed explanation

Background Concept

A meson is a hadron made from one quark and one antiquark.

Quark charges:

$u$ has charge $+\frac{2}{3}e$ so $\bar{u}$ has charge $-\frac{2}{3}e$
$d$ has charge $-\frac{1}{3}e$ so $\bar{d}$ has charge $+\frac{1}{3}e$
$s$ has charge $-\frac{1}{3}e$ so $\bar{s}$ has charge $+\frac{1}{3}e$

The total charge of a particle is the sum of the charges of its constituents.

Understanding the Question

We are told particle $Q$ :

is a meson (so it must be quark–antiquark), and
has charge 0.

We must give one possible quark composition that satisfies both conditions.

Approach

Start with the rule “meson = quark + antiquark”.
Choose a quark flavour and pair it with its own antiquark; their charges cancel to give net charge zero.

Step-by-Step Reasoning

Pick $u$ :

$u$ has charge $+\frac{2}{3}e$ .
$\bar{u}$ has charge $-\frac{2}{3}e$ .

Add them:

+\frac{2}{3}e - \frac{2}{3}e = 0

So $u\bar{u}$ is a valid neutral meson composition.

Similarly:

$d\bar{d}$ gives $-\frac{1}{3}e + \frac{1}{3}e = 0$ .
$s\bar{s}$ gives $-\frac{1}{3}e + \frac{1}{3}e = 0$ .

Key Takeaways

Meson = quark–antiquark.
Neutral meson can be formed by a quark and its corresponding antiquark.

Common Mistakes

Writing three quarks (that would be a baryon, not a meson).
Forgetting that antiquark charges are the opposite sign.
Giving a pair like $ud$ (two quarks) rather than quark–antiquark.

Things to Be Careful About

The question asks for a “possible” composition: you only need one correct example.
Use the bar notation clearly, e.g. $\bar{u}$ , $\bar{d}$ .

Techniques used

use that a meson is a quark–antiquark pairadd quark and antiquark charges to obtain the net chargeselect a valid quark flavour combination

(ii)

Particle Q has a mass of $0.67\ \text{u}$ and a kinetic energy of $2.1 \times 10^{-16}\ \text{J}$ .

Calculate the speed of particle Q.

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

3M Similar

DifficultyMedium-Easy

Worked solution

Working

m = 0.67\,\text{u} = 0.67 \times 1.66 \times 10^{-27} = 1.11 \times 10^{-27}\ \text{kg}

E_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}

v = \sqrt{\frac{2(2.1 \times 10^{-16})}{1.11 \times 10^{-27}}} = 6.2 \times 10^{5}\ \text{m s}^{-1}

Answer

$6.2 \times 10^{5}\ \text{m s}^{-1}$

Final answer

6.2 × 10^5 m s^-1

Detailed explanation

Background Concept

For speeds much less than the speed of light, kinetic energy is given by

E_k = \frac{1}{2}mv^2

where:

$E_k$ is kinetic energy in joules (J),
$m$ is mass in kilograms (kg),
$v$ is speed in $\text{m s}^{-1}$ .

Here the mass is given in atomic mass units, where

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

Understanding the Question

We are given for particle $Q$ :

mass $m = 0.67\ \text{u}$ ,
kinetic energy $E_k = 2.1 \times 10^{-16}\ \text{J}$ .

We must calculate its speed $v$ .

Approach

Convert the mass from $\text{u}$ to kg.
Rearrange $E_k = \frac{1}{2}mv^2$ to get $v = \sqrt{\frac{2E_k}{m}}$ .
Substitute values and evaluate, keeping sensible significant figures.

Step-by-Step Reasoning

1) Convert mass

m = 0.67\ \text{u} = 0.67 \times (1.66 \times 10^{-27})\ \text{kg}

m = 1.11 \times 10^{-27}\ \text{kg} \quad (\text{to 3 s.f.})

2) Rearrange kinetic energy formula

Starting from

E_k = \frac{1}{2}mv^2

Multiply both sides by 2 and divide by $m$ :

v^2 = \frac{2E_k}{m}

So

v = \sqrt{\frac{2E_k}{m}}

3) Substitute

v = \sqrt{\frac{2(2.1 \times 10^{-16})}{1.11 \times 10^{-27}}}

Compute inside the square root:

\frac{4.2 \times 10^{-16}}{1.11 \times 10^{-27}} = 3.78 \times 10^{11}

Then

v = \sqrt{3.78 \times 10^{11}} = 6.15 \times 10^{5}\ \text{m s}^{-1}

Given the data are to 2 s.f., round to:

v \approx 6.2 \times 10^{5}\ \text{m s}^{-1}

(As a quick check, $6.2 \times 10^{5} \ll 3.0 \times 10^{8}$ , so the non-relativistic formula is reasonable.)

Key Takeaways

Always convert $\text{u}$ to kg before using mechanics equations in SI.
Use $E_k = \frac{1}{2}mv^2$ and rearrange carefully.

Common Mistakes

Using $m = 0.67$ as if it were already in kg.
Forgetting the square root when solving for $v$ .
Significant figures: giving many digits when inputs are only 2 s.f.

Things to Be Careful About

Keep powers of ten consistent when dividing numbers in standard form.
Ensure the final unit is $\text{m s}^{-1}$ .
If $v$ came out as a large fraction of $c$ , you would need a relativistic treatment, but here it is clearly not necessary.

Techniques used

convert atomic mass units to kilogramsapply the kinetic energy equation to solve for speedcheck non-relativistic validity by comparing with the speed of light

(c)

Radium-228 ( $^{228}_{88}\text{Ra}$ ) is a radioactive nuclide.

3M

(i)

State the number of electrons in a neutral atom of radium-228.

number of electrons = ______

1M Similar

(ii)

A nucleus of radium-228 undergoes a series of decays to form nucleus X. During the process, 5 $\alpha$ -particles and 4 $\beta^-$ particles are emitted.

Determine the number of protons and the number of neutrons in nucleus X.

number of protons = ______
number of neutrons = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Start: $^{228}_{88}\text{Ra}$ .

After $5\alpha$ :

A = 228 - 5(4) = 208, \quad Z = 88 - 5(2) = 78

After $4\beta^-$ (each increases $Z$ by 1):

Z = 78 + 4 = 82

Neutrons:

N = A - Z = 208 - 82 = 126

Answer

number of protons $= 82$

number of neutrons $= 126$

Final answer

Protons: 82; Neutrons: 126

Detailed explanation

Background Concept

Radioactive decays change the numbers in nuclide notation:

$\alpha$ -decay

An $\alpha$ particle is a helium nucleus $^{4}_{2}\text{He}$ .
So when a nucleus emits an $\alpha$ particle:

nucleon number decreases by 4: $A \to A - 4$
proton number decreases by 2: $Z \to Z - 2$

$\beta^-$ -decay

In $\beta^-$ decay, a neutron turns into a proton + electron + antineutrino.
So:

$A$ stays the same
$Z$ increases by 1: $Z \to Z + 1$

Neutron number is then found from

A = Z + N \Rightarrow N = A - Z

Understanding the Question

We start with $^{228}_{88}\text{Ra}$ .
A series of decays leads to a final nucleus $X$ after emitting:

5 $\alpha$ particles
4 $\beta^-$ particles

We must find the number of protons ( $Z$ ) and neutrons ( $N$ ) in nucleus $X$ .

Approach

Update $A$ and $Z$ for the 5 $\alpha$ decays.
Update $Z$ for the 4 $\beta^-$ decays (no change to $A$ ).
Compute $N = A - Z$ .

Step-by-Step Reasoning

1) Apply 5 alpha decays

Each $\alpha$ reduces $A$ by 4 and $Z$ by 2.

Total change for 5 $\alpha$ :

\Delta A = -5 \times 4 = -20, \quad \Delta Z = -5 \times 2 = -10

So:

A = 228 - 20 = 208

Z = 88 - 10 = 78

2) Apply 4 beta-minus decays

Each $\beta^-$ increases $Z$ by 1; $A$ unchanged.

Total change:

\Delta Z = +4

So:

Z = 78 + 4 = 82

Thus nucleus $X$ is $^{208}_{82}X$ .

3) Find neutrons

N = A - Z = 208 - 82 = 126

Key Takeaways

Track $A$ and $Z$ separately.
$\alpha$ : $A - 4$ , $Z - 2$ .
$\beta^-$ : $A$ unchanged, $Z + 1$ .

Common Mistakes

Changing $A$ during $\beta^-$ decay (it does not change).
Using the wrong sign for $\beta^-$ (it increases $Z$ , not decreases it).
Forgetting to multiply the $\alpha$ changes by 5.

Things to Be Careful About

Do $\alpha$ and $\beta^-$ effects systematically; many students lose marks by mixing steps.
Final neutron number must be calculated from the final $A$ and $Z$ , not from the starting values.

Techniques used

apply alpha decay changes to nucleon and proton numbersapply beta-minus decay change to proton numberuse A = Z + N to find neutron number

Physics 9702/22 — October/November 2025

Answer

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Understanding the Question

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