Physics 9702 / 23 — October/November 2025

(b)

A rocket is launched vertically from the surface of the Earth.

Fig. 1.1 shows the variation of the velocity of the rocket with time for the first $20\text{ s}$ after its launch.

Fig. 1.1

3M

(i)

Determine the acceleration of the rocket.

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

DifficultyMedium-Easy

Worked solution

Working

From the graph, the straight line passes through $(0\ \text{s},0\ \text{m s}^{-1})$ and $(20\ \text{s},320\ \text{m s}^{-1})$ .

a = \frac{\Delta v}{\Delta t} = \frac{320-0}{20-0} = 16\ \text{m s}^{-2}

Answer

$16\ \text{m s}^{-2}$

Final answer

16 m s^-2

Detailed explanation

Background Concept

For a velocity–time graph, the gradient gives the acceleration:

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

If the graph is a straight line, the acceleration is constant, so any two points on the line can be used.

Understanding the Question

The graph shows velocity increasing linearly from launch to $t=20\ \text{s}$ . You are asked to determine the rocket’s acceleration, i.e. the gradient of the $v$ – $t$ graph.

From the given graph description, the line goes from $(0,0)$ to approximately $(20\ \text{s}, 320\ \text{m s}^{-1})$ .

Approach

Pick two clear points on the straight line (best to use endpoints for accuracy).
Compute the gradient using $\Delta v/\Delta t$ .
Quote the value with unit $\text{m s}^{-2}$ .

Step-by-Step Reasoning

Take the endpoints:

At $t=0\ \text{s}$ , $v=0\ \text{m s}^{-1}$ .
At $t=20\ \text{s}$ , $v=320\ \text{m s}^{-1}$ .

Then

a = \frac{\Delta v}{\Delta t} = \frac{320 - 0}{20 - 0} = 16

So

a = 16\ \text{m s}^{-2}

Key Takeaways

Gradient of a $v$ – $t$ graph is acceleration.
Straight-line $v$ – $t$ graph implies constant acceleration.

Common Mistakes

Using $\Delta t/\Delta v$ instead of $\Delta v/\Delta t$ .
Reading the velocity at $t=20\ \text{s}$ incorrectly from the axis scale.
Forgetting the unit $\text{m s}^{-2}$ .

Things to Be Careful About

Use two widely separated points to reduce percentage reading error.
Ensure you are using velocity (vertical axis) and time (horizontal axis) in the correct order for the gradient.

Techniques used

use the gradient of a velocity–time graph to find accelerationselect two well-separated points on a straight-line graphcalculate a gradient as change in y over change in x

(ii)

Show that the height of the rocket above the surface of the Earth at a time of $20\text{ s}$ after launch is $3.2\text{ km}$ .

DifficultyMedium-Easy

Worked solution

Working

Height $=$ area under the $v$ – $t$ graph from $0$ to $20\ \text{s}$ .

Triangle with base $20\ \text{s}$ and height $320\ \text{m s}^{-1}$ :

h = \frac{1}{2} \times 20 \times 320 = 3200\ \text{m} = 3.2\ \text{km}

Answer

$3.2\ \text{km}$

Final answer

3.2 km

Detailed explanation

Background Concept

Displacement (and hence height for vertical motion) is found from velocity by integrating over time. On a velocity–time graph, this is represented by the area under the graph:

s = \int v\,\mathrm{d}t

For straight-line segments, the area can be found using simple shapes (rectangles, triangles, trapezia).

Understanding the Question

You must show that at $t=20\ \text{s}$ the rocket is $3.2\ \text{km}$ above the launch point, using the velocity–time graph given (a straight line from $v=0$ at $t=0$ to $v=320\ \text{m s}^{-1}$ at $t=20\ \text{s}$ ).

Approach

Recognise that height gained equals the area under the $v$ – $t$ graph from $0$ to $20\ \text{s}$ .
The line starts at the origin, so the region under the line is a triangle.
Find its area and convert metres to kilometres.

Step-by-Step Reasoning

diagram-bii-1

The graph is a straight line from $(0,0)$ to $(20,320)$ , so the area under it is a right triangle.

Base (time) $= 20\ \text{s}$ .
Height (velocity) $= 320\ \text{m s}^{-1}$ .

Area:

h = \frac{1}{2} \times (20) \times (320) = 3200\ \text{m}

Convert to kilometres:

3200\ \text{m} = 3.2\ \text{km}

So the height at $20\ \text{s}$ is $3.2\ \text{km}$ .

Key Takeaways

Area under a velocity–time graph gives displacement.
With a straight-line increase from zero, displacement is the area of a triangle.

Common Mistakes

Using the gradient (acceleration) instead of the area to find height.
Calculating $20 \times 320$ without the $\tfrac{1}{2}$ (that would be the area of the bounding rectangle, not the triangle).
Forgetting to convert $3200\ \text{m}$ to $3.2\ \text{km}$ .

Things to Be Careful About

Keep track of units: $\text{s} \times \text{m s}^{-1} = \text{m}$ .
The question asks for height above the surface, so take displacement from $t=0$ to $20\ \text{s}$ (not from some other time interval).

Techniques used

use the area under a velocity–time graph to determine displacementcalculate the area of a triangle from its base and heightconvert between metres and kilometres

(c)

The mass of the rocket in (b) is $2.9 \times 10^{6}\ \text{kg}$ . Assume that this mass remains constant.

For this rocket, from launch to its height at a time of $20\text{ s}$ after launch:

6M

(i)

calculate the gain in gravitational potential energy $\Delta E_P$

$\Delta E_P$ = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

Height gained in $20\ \text{s}$ : $h = 3.2\ \text{km} = 3.2 \times 10^{3}\ \text{m}$ .

\Delta E_P = mgh = (2.9 \times 10^{6})(9.81)(3.2 \times 10^{3})

\Delta E_P = 9.1 \times 10^{10}\ \text{J}

Answer

$9.1 \times 10^{10}\ \text{J}$

Final answer

9.1 × 10^10 J

Detailed explanation

Background Concept

A gain in gravitational potential energy near Earth’s surface is given by

\Delta E_P = mgh

where:

$m$ is mass,
$g$ is gravitational field strength (about $9.81\ \text{m s}^{-2}$ near Earth’s surface),
$h$ is vertical height gained.

Understanding the Question

The rocket has mass $2.9 \times 10^{6}\ \text{kg}$ (assumed constant). From launch to $t=20\ \text{s}$ it rises by $3.2\ \text{km}$ (found in part (b)(ii)). You must calculate the increase in gravitational potential energy over this rise.

Approach

Convert $h$ to metres.
Substitute into $\Delta E_P = mgh$ .
Give the answer in joules and standard form.

Step-by-Step Reasoning

Convert height:

h = 3.2\ \text{km} = 3.2 \times 10^{3}\ \text{m}

Now substitute:

\Delta E_P = (2.9 \times 10^{6})(9.81)(3.2 \times 10^{3})

Multiply the powers of ten: $10^{6} \times 10^{3} = 10^{9}$ , and multiply the numerical factors:

2.9 \times 9.81 \times 3.2 \approx 91

So

\Delta E_P \approx 91 \times 10^{9} = 9.1 \times 10^{10}\ \text{J}

Key Takeaways

Use $\Delta E_P = mgh$ for vertical motion near Earth.
Always convert km to m before substitution.

Common Mistakes

Using $h = 3.2$ instead of $3.2 \times 10^{3}$ (forgetting km to m conversion).
Omitting $g$ or using incorrect units.
Not giving the answer in joules or not using standard form for a large number.

Things to Be Careful About

The question says mass remains constant, so you do not need to account for fuel burn.
Using $g = 9.8$ instead of $9.81$ may give a slightly different final value; either is typically acceptable if used consistently.

Techniques used

apply the gravitational potential energy change equationsubstitute values with correct SI unitsuse standard form and appropriate significant figures

(ii)

calculate the gain in kinetic energy $\Delta E_K$

$\Delta E_K$ = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

From the graph at $t=20\ \text{s}$ , $v = 320\ \text{m s}^{-1}$ (and $u=0$ ).

\Delta E_K = \frac{1}{2}mv^2 = \frac{1}{2}(2.9 \times 10^{6})(320)^2

\Delta E_K = 1.5 \times 10^{11}\ \text{J}

Answer

$1.5 \times 10^{11}\ \text{J}$

Final answer

1.5 × 10^11 J

Detailed explanation

Background Concept

Kinetic energy of a mass $m$ moving at speed $v$ is

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

If an object starts from rest, the gain in kinetic energy is just its final kinetic energy.

Understanding the Question

The rocket starts from rest at launch ( $u=0$ ). From the velocity–time graph, at $t=20\ \text{s}$ the velocity is $320\ \text{m s}^{-1}$ . With constant mass $m = 2.9 \times 10^{6}\ \text{kg}$ , you must calculate the increase in kinetic energy from $t=0$ to $t=20\ \text{s}$ .

Approach

Read the final speed from the graph.
Use $\Delta E_K = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$ .
Since $u=0$ , simplify to $\Delta E_K = \tfrac{1}{2}mv^2$ .

Step-by-Step Reasoning

Final speed:

v = 320\ \text{m s}^{-1}

Compute gain in kinetic energy:

\Delta E_K = \frac{1}{2}(2.9 \times 10^{6})(320)^2

Square the speed:

(320)^2 = 102400

Then

\Delta E_K = \frac{1}{2}(2.9 \times 10^{6})(102400) = 1.4848 \times 10^{11}\ \text{J}

To appropriate significant figures:

\Delta E_K \approx 1.5 \times 10^{11}\ \text{J}

Key Takeaways

Use $\Delta E_K = \tfrac{1}{2}mv^2$ for a start from rest.
Speed must be in $\text{m s}^{-1}$ and mass in kg.

Common Mistakes

Forgetting to square the speed.
Using $v=400\ \text{m s}^{-1}$ by misreading the axis.
Using $\Delta E_K = mv^2$ (missing the factor $\tfrac{1}{2}$ ).

Things to Be Careful About

This is a very large energy, so standard form is expected.
Ensure you use the velocity at $20\ \text{s}$ from the graph, not the average velocity.

Techniques used

read final speed from a velocity–time graphapply the kinetic energy equationsubstitute values with correct SI units

(iii)

determine the average power output of the rocket engines. Assume that resistive forces are negligible.

power = ______ $\text{W}$

DifficultyMedium

Worked solution

Working

With negligible resistive forces, energy output by engines

E = \Delta E_P + \Delta E_K

E = (9.1 \times 10^{10}) + (1.5 \times 10^{11}) = 2.4 \times 10^{11}\ \text{J}

Average power over $20\ \text{s}$ :

P = \frac{E}{t} = \frac{2.4 \times 10^{11}}{20} = 1.2 \times 10^{10}\ \text{W}

Answer

$1.2 \times 10^{10}\ \text{W}$

Final answer

1.2 × 10^10 W

Detailed explanation

Background Concept

Power is the rate at which energy is transferred:

P = \frac{\Delta E}{\Delta t}

If resistive forces are negligible, the energy supplied by the engines is converted into mechanical energy of the rocket (increase in gravitational potential energy and kinetic energy):

E_{\text{engines}} = \Delta E_P + \Delta E_K

Understanding the Question

From launch to $t=20\ \text{s}$ , the rocket:

rises by $3.2\ \text{km}$ (so gains GPE),
speeds up to $320\ \text{m s}^{-1}$ (so gains KE).

You are asked for the average power output over the first $20\ \text{s}$ , assuming no energy is lost to air resistance etc.

Approach

Take the energy supplied by engines as the sum $\Delta E_P + \Delta E_K$ .
Divide by the time interval $20\ \text{s}$ .

Step-by-Step Reasoning

From earlier parts:

\Delta E_P \approx 9.1 \times 10^{10}\ \text{J}

\Delta E_K \approx 1.5 \times 10^{11}\ \text{J}

Total energy transferred by the engines:

E = \Delta E_P + \Delta E_K \approx (9.1 \times 10^{10}) + (1.5 \times 10^{11})

E \approx 2.4 \times 10^{11}\ \text{J}

Average power over $t=20\ \text{s}$ :

P = \frac{E}{t} = \frac{2.4 \times 10^{11}}{20} = 1.2 \times 10^{10}\ \text{W}

Key Takeaways

With negligible resistive forces, engine energy becomes $\Delta E_P + \Delta E_K$ .
Average power is total energy transferred divided by total time.

Common Mistakes

Using only $\Delta E_K$ or only $\Delta E_P$ instead of both.
Dividing by the wrong time (e.g. using $t=10\ \text{s}$ ).
Confusing average power with instantaneous power (which would vary during the launch).

Things to Be Careful About

The assumption “resistive forces negligible” is what allows you to ignore energy lost to heating the air.
Keep consistent significant figures when adding energies; then round the final power appropriately.

Techniques used

apply energy conservation to relate engine work to energy gainssum changes in gravitational potential energy and kinetic energycalculate average power as energy transferred per unit time

Q2MediumForces, Density and PressureDynamicsPhysical Quantities and Units

(a)

3M

(i)

Define pressure.

(ii)

Explain how hydrostatic pressure results in an upthrust force acting on a solid object immersed in a liquid.

DifficultyMedium-Easy

Worked solution

Answer

Hydrostatic pressure increases with depth, so the pressure on the bottom of the object is greater than on the top.

This gives a larger downward force on the bottom surface than the top surface, producing a resultant upward force (upthrust) on the object (side forces cancel).

Final answer

Pressure increases with depth so the bottom experiences greater pressure than the top; forces from the liquid therefore give a net upward resultant (upthrust), with horizontal components cancelling.

Detailed explanation

Background Concept

In a stationary liquid, hydrostatic pressure depends on depth:

p = \rho g h

where $\rho$ is liquid density, $g$ is gravitational field strength, and $h$ is depth below the surface. Pressure acts equally in all directions at a point, and a pressure on a surface produces a force:

F = pA

Understanding the Question

You must explain why an immersed solid experiences an upward force. The key idea is that pressure is higher deeper down, so the forces from the liquid on different faces are not equal.

Approach

Use $p = \rho g h$ to argue pressure increases with depth.
Compare the pressure (and hence force) on the top and bottom surfaces.
Note horizontal forces cancel, leaving a net upward resultant.

Step-by-Step Reasoning

diagram-a-ii

Consider an object in a liquid. The liquid exerts pressure on every part of the surface.
The top of the object is at a smaller depth $h_\text{top}$ than the bottom $h_\text{bottom}$ .
Since $p = \rho g h$ , we have $p_\text{bottom} > p_\text{top}$ .
Force due to pressure on a surface is $F = pA$ . So the upward force on the bottom face is

F_\text{bottom} = p_\text{bottom}A

and the downward force on the top face is

F_\text{top} = p_\text{top}A

Because $p_\text{bottom} > p_\text{top}$ , $F_\text{bottom} > F_\text{top}$ , giving a net upward force.
Forces on the sides act horizontally in opposite directions and cancel, so the resultant from the liquid is upward: the upthrust.

(At a higher level, this net upthrust is equal to the weight of liquid displaced, i.e. Archimedes’ principle.)

Key Takeaways

Pressure increases with depth.
Net upthrust comes from a pressure difference between bottom and top surfaces.
Side forces cancel.

Common Mistakes

Saying “pressure pushes up” without mentioning the difference in pressure with depth.
Forgetting that pressure acts in all directions, so side forces exist but cancel.

Things to Be Careful About

Use clear language: bottom has greater pressure because it is deeper.
Distinguish pressure ( $p$ ) from force ( $F = pA$ ).

Techniques used

use the increase of pressure with depthcompare forces on top and bottom surfacesdeduce a resultant force from pressure difference

(b)

A small steel ball of radius $r$ and mass $m$ falls vertically at terminal speed $v$ through oil.

The viscous drag force $D$ that acts on the ball is given by

D = 6\pi\eta rv

where $\eta$ is a property of the oil called its viscosity.

5M

(i)

On Fig. 2.1, draw labelled arrows from the ball to show the directions of the three forces that act on the ball as it falls.

Fig. 2.1

DifficultyMedium-Easy

Worked solution

diagram-b-i

Answer

Weight $mg$ acts vertically downward.

Upthrust acts vertically upward.

Viscous drag $D$ acts vertically upward (opposes the downward motion).

Final answer

Forces: weight mg downward; upthrust upward; drag D upward.

Detailed explanation

Background Concept

For an object moving through a fluid:

Weight $W = mg$ acts downward.
Upthrust (buoyant force) acts upward due to pressure differences in the fluid.
Drag (resistive force) acts opposite to the direction of motion; here it is viscous drag.

At terminal speed, acceleration is zero, so the resultant force is zero.

Understanding the Question

The ball is falling down through oil at terminal speed. You must draw arrows from the ball showing the directions of the three forces and label them.

Approach

List the forces on a falling sphere in a fluid and set their directions by physical meaning:

gravity always downward,
buoyancy always upward,
drag opposes motion, so upward for downward motion.

Step-by-Step Reasoning

diagram-b-i

Weight acts toward the centre of Earth, so draw a downward arrow labelled $mg$ .
Upthrust is the net force from fluid pressure, which is upward, so draw an upward arrow labelled “upthrust” (or $U$ ).
Drag opposes the motion. Since the ball moves downward, drag is upward, so draw an upward arrow labelled $D$ .

(If you also use the fact that speed is terminal: the two upward forces together balance weight.)

Key Takeaways

Drag acts opposite to motion.
Upthrust is always upward in a fluid.
Weight is always downward.

Common Mistakes

Drawing drag downward (same direction as motion).
Missing one of the three forces.
Labelling upthrust as “pressure” without indicating it is a force.

Things to Be Careful About

Arrows should start at the ball and point in the correct direction.
Use clear labels: $mg$ , $D$ , and upthrust ( $U$ ).

Techniques used

identify forces acting on an object moving through a fluiduse relative motion to set the direction of dragrepresent forces with a labelled force diagram

(ii)

Determine the SI base units of $\eta$ .

base units ______

DifficultyMedium-Easy

Worked solution

Working

From

D = 6\pi \eta r v

\eta = \frac{D}{6\pi r v}

Units:

[\eta] = \frac{\text{N}}{\text{m}\,(\text{m s}^{-1})} = \frac{\text{kg m s}^{-2}}{\text{m}^2\,\text{s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1}

Answer

\boxed{\text{kg m}^{-1}\text{s}^{-1}}

Final answer

kg m^-1 s^-1

Detailed explanation

Background Concept

To find SI base units of a quantity, use a given physical equation and ensure dimensional consistency (homogeneity). Constants like $6\pi$ have no units.

Useful base-unit facts:

$\text{N} = \text{kg m s}^{-2}$
$v$ has units $\text{m s}^{-1}$
$r$ has units $\text{m}$

Understanding the Question

You are given the drag law for a sphere:

D = 6\pi \eta r v

You must determine the SI base units of viscosity $\eta$ .

Approach

Rearrange to make $\eta$ the subject, then replace each quantity by its SI units and simplify to base units ( $\text{kg}$ , $\text{m}$ , $\text{s}$ , etc.).

Step-by-Step Reasoning

Rearrange:

\eta = \frac{D}{6\pi r v}

Ignore $6\pi$ (dimensionless).
Substitute units:

$D$ in newtons: $\text{N} = \text{kg m s}^{-2}$
$r$ in metres: $\text{m}$
$v$ in $\text{m s}^{-1}$

So

[\eta] = \frac{\text{kg m s}^{-2}}{\text{m}\,(\text{m s}^{-1})}

Simplify denominator: $\text{m}\,(\text{m s}^{-1}) = \text{m}^2\,\text{s}^{-1}$ .
Divide:

[\eta] = \text{kg m s}^{-2} \times \frac{\text{s}}{\text{m}^2} = \text{kg m}^{-1}\text{s}^{-1}

Key Takeaways

Rearrange first, then substitute SI units.
Check that dimensionless constants do not affect units.

Common Mistakes

Treating $6\pi$ as having units.
Using $\text{N m}^{-2}$ for pressure-like quantities (viscosity is not pressure).
Forgetting that $v$ includes $\text{s}^{-1}$ .

Things to Be Careful About

Write the final answer in SI base units (not in derived units like N).
Keep track of negative indices carefully when simplifying.

Techniques used

rearrange an equation to isolate a quantitysubstitute SI units for each variablesimplify to SI base units

(c)

The oil in (b) has a density of $920\ \text{kg m}^{-3}$ and a viscosity of $4.7$ in SI units.

The steel ball has a mass of $2.4 \times 10^{-3}\ \text{kg}$ and a radius of $4.2 \times 10^{-3}\ \text{m}$ .

4M

(i)

Show that the upthrust force acting on the ball is $2.8 \times 10^{-3}\ \text{N}$ .

DifficultyMedium-Easy

Worked solution

Working

Upthrust $U$ equals weight of displaced oil:

U = \rho g V = \rho g \left(\frac{4}{3}\pi r^3\right)

U = 920 \times 9.81 \times \frac{4}{3}\pi (4.2\times 10^{-3})^3

U = 2.8 \times 10^{-3}\ \text{N}

Answer

$2.8 \times 10^{-3}\ \text{N}$

Final answer

2.8 × 10^-3 N

Detailed explanation

Background Concept

Archimedes’ principle: the upthrust on an object equals the weight of fluid displaced.

So for a fully immersed object of volume $V$ in a fluid of density $\rho$ :

U = \rho g V

For a sphere of radius $r$ :

V = \frac{4}{3}\pi r^3

Understanding the Question

You are given the oil density $\rho = 920\ \text{kg m}^{-3}$ and ball radius $r = 4.2\times 10^{-3}\ \text{m}$ . You must show that the buoyant force (upthrust) on the ball is $2.8\times 10^{-3}\ \text{N}$ .

Approach

Compute the sphere’s volume, then multiply by $\rho g$ to get the weight of displaced oil.

Step-by-Step Reasoning

Volume of the ball:

V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (4.2\times 10^{-3})^3

Evaluate $r^3$ :

(4.2\times 10^{-3})^3 = 4.2^3 \times 10^{-9} = 74.088 \times 10^{-9} = 7.4088\times 10^{-8}

So

V \approx 4.18879 \times 7.4088\times 10^{-8} \approx 3.10\times 10^{-7}\ \text{m}^3

Upthrust:

U = \rho g V = 920 \times 9.81 \times 3.10\times 10^{-7}

U \approx 2.8\times 10^{-3}\ \text{N}

Key Takeaways

Upthrust depends on fluid density and displaced volume, not on the object’s mass.
For a sphere, use $V = \frac{4}{3}\pi r^3$ .

Common Mistakes

Using the density of the steel instead of the oil.
Using diameter instead of radius in $r^3$ .
Dropping powers of ten when cubing $10^{-3}$ .

Things to Be Careful About

Keep $r$ in metres before calculating volume.
$10^{-3}$ cubed is $10^{-9}$ (a very common slip).

Techniques used

calculate volume of a sphereapply Archimedes' principle to find upthrustsubstitute numerical values with correct powers of ten

(ii)

Determine the terminal speed $v$ of the ball.

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

DifficultyMedium

Worked solution

Working

At terminal speed, resultant force is zero:

mg = U + D \quad\Rightarrow\quad D = mg - U

D = (2.4\times 10^{-3}\times 9.81) - (2.8\times 10^{-3}) = 2.07\times 10^{-2}\ \text{N}

Using $D = 6\pi\eta r v$ :

v = \frac{D}{6\pi\eta r} = \frac{2.07\times 10^{-2}}{6\pi\times 4.7\times 4.2\times 10^{-3}}

v = 5.6\times 10^{-2}\ \text{m s}^{-1}

Answer

\boxed{v = 5.6\times 10^{-2}\ \text{m s}^{-1}}

Final answer

5.6 × 10^-2 m s^-1

Detailed explanation

Background Concept

When an object falls through a fluid, three vertical forces act:

weight $mg$ downward,
upthrust $U$ upward,
drag $D$ upward (opposes downward motion).

At terminal speed, acceleration is zero, so the resultant force is zero:

\sum F = 0

The drag is given by

D = 6\pi \eta r v

Understanding the Question

You are given $m$ , $r$ , oil viscosity $\eta$ , and the upthrust from part (i). You must calculate the terminal speed $v$ .

Approach

Use terminal-speed condition to form the force balance.
Find the drag force needed to balance weight minus upthrust.
Substitute into $D = 6\pi\eta r v$ and solve for $v$ .

Step-by-Step Reasoning

At terminal speed:

mg = U + D

so

D = mg - U

Calculate weight:

mg = 2.4\times 10^{-3} \times 9.81 = 2.3544\times 10^{-2}\ \text{N}

Subtract upthrust $U = 2.8\times 10^{-3}\ \text{N}$ :

D = 2.3544\times 10^{-2} - 2.8\times 10^{-3} = 2.0744\times 10^{-2}\ \text{N}

Use Stokes’ drag law and rearrange:

v = \frac{D}{6\pi\eta r}

Substitute $\eta = 4.7$ (SI units), $r = 4.2\times 10^{-3}\ \text{m}$ :

v = \frac{2.07\times 10^{-2}}{6\pi\times 4.7\times 4.2\times 10^{-3}} \approx 5.6\times 10^{-2}\ \text{m s}^{-1}

Key Takeaways

Terminal speed means forces balance (not that forces are zero).
For a falling object: $mg$ is balanced by (upthrust + drag).
Rearranging a given formula cleanly is essential for full marks.

Common Mistakes

Setting $mg = D$ and forgetting upthrust.
Using $mg + U = D$ (wrong sign for upthrust).
Substituting $r$ in mm instead of m.

Things to Be Careful About

Use consistent SI units: $r$ in m, $D$ in N, $v$ in $\text{m s}^{-1}$ .
Terminal speed is a constant speed: acceleration $=0$ , so resultant force $=0$ .
Keep sufficient significant figures during intermediate steps to avoid rounding errors.

Techniques used

apply zero resultant force at terminal speedform an equilibrium equation with weight, upthrust and dragrearrange Stokes' law to solve for the terminal speed

Q3MediumDeformation of SolidsElectricityPhysical Quantities and Units

A wire has length $L$ and cross-sectional area $A$ . The wire is made from a metal that has Young modulus $E$ and resistivity $\rho$ .

(a)

Define the Young modulus of a material.

DifficultyEasy

Worked solution

Answer

Young modulus $E$ is the ratio of tensile stress to tensile strain (within the limit of proportionality), i.e.

E = \frac{\text{stress}}{\text{strain}}.

Final answer

Young modulus is stress/strain (within limit of proportionality).

Detailed explanation

Background Concept

Young modulus $E$ is a measure of how stiff a material is when stretched (or compressed) elastically.

Tensile stress is force per unit cross-sectional area:

\text{stress} = \frac{F}{A}

Tensile strain is fractional extension:

\text{strain} = \frac{\Delta L}{L}

Within the limit of proportionality (the straight-line part of a force–extension graph), stress is proportional to strain, so their ratio is constant and equal to $E$ :

E = \frac{\text{stress}}{\text{strain}}

Understanding the Question

You are asked to define Young modulus. So you must give the key ratio (stress divided by strain) and state the condition that it applies in the linear elastic region (limit of proportionality).

Approach

Write the standard definition in words or as an equation. For full credit, include “within the limit of proportionality” (or equivalent wording such as “in the elastic region where Hooke’s law applies”).

Step-by-Step Reasoning

Identify the quantity being defined: Young modulus $E$ .
Recall the definition:

E = \frac{\text{tensile stress}}{\text{tensile strain}}

Add the condition: this is valid where the stress–strain relationship is linear (limit of proportionality).

Key Takeaways

Stress is $F/A$ and strain is $\Delta L/L$ .
Young modulus is their ratio in the linear elastic region.

Common Mistakes

Writing $E = \text{strain}/\text{stress}$ (inverted).
Missing the condition “within the limit of proportionality”.
Confusing Young modulus with spring constant $k$ (property of an object, not the material).

Things to Be Careful About

Use “stress” and “strain” explicitly; “force/extension” is not the same as $E$ .
State the region of validity (limit of proportionality / Hooke’s law region).

Techniques used

state the definition of Young modulusrelate stress and strain to force, area, extension and original length

(b)

3M

(i)

State an expression, in terms of some or all of $L$ , $A$ , $E$ and $\rho$ , for the resistance $R_0$ of the wire.

$R_0$ = ______

(ii)

Show that the spring constant $k_0$ of the wire is given by

k_0 = \frac{EA}{L}

DifficultyMedium-Easy

Worked solution

Working

E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}

F = \frac{EA}{L}x

So

k_0 = \frac{F}{x} = \frac{EA}{L}.

Answer

k_0 = \frac{EA}{L}

Final answer

k0 = EA/L

Detailed explanation

Background Concept

There are two connected descriptions of stretching:

Material description (Young modulus):

E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}

Object description (Hooke’s law / spring constant): for a wire behaving elastically,

F = kx

where $x$ is the extension.

The spring constant $k$ depends on the object’s dimensions as well as the material.

Understanding the Question

You must show that a wire (length $L$ , area $A$ ) made of a material with Young modulus $E$ behaves like a spring with spring constant

k_0 = \frac{EA}{L}.

So you need to connect $E$ to $F$ and extension $x$ and then identify $k_0 = F/x$ .

Approach

Write stress as $F/A$ .
Write strain as $x/L$ .
Substitute into $E = \text{stress}/\text{strain}$ .
Rearrange into the Hooke’s law form $F = kx$ .

Step-by-Step Reasoning

Start from Young modulus:

E = \frac{\text{stress}}{\text{strain}}.

Substitute stress and strain:

\text{stress} = \frac{F}{A}, \qquad \text{strain} = \frac{x}{L}.

So

E = \frac{F/A}{x/L}.

Simplify the complex fraction by multiplying top and bottom:

E = \frac{F}{A} \cdot \frac{L}{x} = \frac{FL}{Ax}.

Rearrange to make $F$ the subject:

F = \frac{EA}{L}x.

Compare with Hooke’s law $F = kx$ . Therefore

k_0 = \frac{EA}{L}.

Key Takeaways

$E$ links stress and strain; $k$ links force and extension.
A longer wire is less stiff ( $k \propto 1/L$ ).
A thicker wire is stiffer ( $k \propto A$ ).

Common Mistakes

Using strain as $L/x$ instead of $x/L$ .
Forgetting the cross-sectional area $A$ in stress.
Writing $k = E/L$ (missing $A$ ).

Things to Be Careful About

This result assumes extension is small and within the limit of proportionality.
Use $x$ (or $\Delta L$ ) consistently as the extension, not the total length.

Techniques used

use definitions of stress and strainrearrange Young modulus equation to relate force and extensionidentify spring constant as force per extension

(c)

The wire is stretched, within the limit of proportionality, by a tensile force $F$ . Assume that any changes in the cross-sectional area of the wire are negligible.

2M

(i)

On Fig. 3.1, sketch the variation with $F$ of the resistance $R$ of the wire.

Fig. 3.1

DifficultyMedium-Easy

Worked solution

Answer

As $F$ increases (within the limit of proportionality), the length increases so $R$ increases.

Sketch: straight line with positive gradient passing through $R=R_0$ at $F=0$ .

diagram-c-i

Final answer

Straight line increasing from R0 at F = 0.

Detailed explanation

Background Concept

The resistance of a uniform wire is

R = \frac{\rho L}{A}.

Here $\rho$ is the resistivity (assumed constant) and the question says changes in cross-sectional area are negligible, so $A$ is constant.

Within the limit of proportionality, the extension $x$ is proportional to the applied force $F$ (Hooke’s law behaviour):

x \propto F.

So the new length is $L' = L + x$ , and therefore resistance changes because $L$ changes.

Understanding the Question

You are asked to sketch how the wire’s resistance $R$ varies as the tensile force $F$ increases, starting from the original resistance $R_0$ at $F=0$ .

Key clues:

“within the limit of proportionality” (\Rightarrow x \propto F)
“changes in cross-sectional area negligible” (\Rightarrow A \text{ constant})

Approach

Use $R = \rho L/A$ .
As the wire stretches, replace $L$ by $L' = L + x$ .
Since $x \propto F$ , the resistance should increase linearly with $F$ .

Step-by-Step Reasoning

Initially, at $F=0$ , the wire has length $L$ and resistance

R_0 = \frac{\rho L}{A}.

Apply force $F$ . The wire extends by $x$ (small and elastic), so the length becomes

L' = L + x.

New resistance:

R = \frac{\rho L'}{A} = \frac{\rho (L+x)}{A} = \frac{\rho L}{A} + \frac{\rho x}{A}.

So

R = R_0 + \frac{\rho}{A}x.

Within the limit of proportionality, $x \propto F$ , so $R$ increases in direct proportion to $F$ :

R = R_0 + (\text{constant})\,F.

So the graph of $R$ against $F$ is a straight line starting at $(F=0, R=R_0)$ and rising.

diagram-c-i

Key Takeaways

If $A$ is constant, stretching increases $L$ and therefore increases $R$ .
In the Hooke’s law region, extension is proportional to force, giving a linear graph.

Common Mistakes

Drawing the line through the origin (it must pass through $R_0$ when $F=0$ ).
Drawing a curve (with the given assumptions, it is linear).
Making $R$ decrease (resistance increases with length).

Things to Be Careful About

The question tells you to neglect changes in $A$ . If $A$ were to decrease significantly, $R$ would increase more strongly, but that is not required here.
Resistivity can change with temperature, but there is no mention of heating; assume $\rho$ constant.

Techniques used

relate resistance to length using the resistivity equationuse proportionality of extension to applied force within Hooke’s law regionsketch a graph using the expected functional dependence

(ii)

On Fig. 3.2, sketch the variation with $F$ of the spring constant $k$ of the wire.

Fig. 3.2

DifficultyMedium

Worked solution

Answer

As $F$ increases, the wire length increases, so

k = \frac{EA}{L'}

decreases from $k_0$ .

Sketch: a decreasing curve starting at $k=k_0$ when $F=0$ .

diagram-c-ii

Final answer

k decreases from k0 as F increases (decreasing curve).

Detailed explanation

Background Concept

For a wire behaving elastically,

k = \frac{F}{x}

and for a uniform wire (material Young modulus $E$ , cross-sectional area $A$ , length $L$ ):

k = \frac{EA}{L}.

This shows stiffness depends on current length: a longer wire is less stiff.

Understanding the Question

You are asked to sketch how the spring constant $k$ changes as the tensile force $F$ increases, given that the wire is stretched within the limit of proportionality and area changes are negligible.

The key point is that as you increase $F$ , the wire extends, so its length is no longer the original $L$ ; it becomes $L' = L + x$ .

Approach

Use the result $k = EA/L$ but apply it to the stretched length $L'$ .
Since $L'$ increases with $F$ , $k$ must decrease with $F$ .
Optionally, use Hooke’s law to show the dependence is not linear.

Step-by-Step Reasoning

Initially, at $F=0$ , the wire length is $L$ and the spring constant is

k_0 = \frac{EA}{L}.

When a force $F$ is applied (still in the proportional region), the wire extends by $x$ , so

L' = L + x.

The spring constant corresponding to length $L'$ is

k = \frac{EA}{L'} = \frac{EA}{L+x}.

As $F$ increases, $x$ increases, so the denominator increases and therefore $k$ decreases.
If you want the shape: within the proportional region,

x = \frac{FL}{EA}.

So

L' = L\left(1 + \frac{F}{EA}\right)

and hence

k = \frac{EA}{L\left(1 + \frac{F}{EA}\right)} = \frac{k_0}{1 + \frac{F}{EA}}.

That is a decreasing curve (it flattens as $F$ increases), starting at $k_0$ when $F=0$ .

diagram-c-ii

Key Takeaways

For a wire, stiffness $k$ is inversely proportional to its length.
Stretching increases length, so the wire becomes less stiff as it is pulled.

Common Mistakes

Drawing $k$ as constant at $k_0$ (only true if the length remained exactly $L$ ).
Drawing $k$ increasing with $F$ .
Drawing a straight-line decrease through the origin (it must start at $k_0$ at $F=0$ ).

Things to Be Careful About

The wire is within the limit of proportionality, so the material is not permanently deformed; however, the geometry still changes while the force is applied.
You are told to neglect changes in $A$ , so only $L$ changes in $k = EA/L$ .

Techniques used

relate spring constant to geometry using k = EA/Lexpress the new length as original length plus extensionsketch a graph showing a decreasing trend from an initial value

(d)

Copper has a resistivity of $1.8 \times 10^{-8}\ \Omega\text{m}$ and a Young modulus of $1.3 \times 10^{11}\ \text{Pa}$ .

A copper wire of diameter $1.6\text{ mm}$ has a resistance of $0.034\ \Omega$ .

3M

(i)

Show that the length of the wire is $3.8\text{ m}$ .

DifficultyMedium-Easy

Worked solution

Working

Diameter $d = 1.6\ \text{mm}$ so radius $r = 0.80\ \text{mm} = 8.0 \times 10^{-4}\ \text{m}$ .

A = \pi r^2 = \pi (8.0 \times 10^{-4})^2 = 2.0 \times 10^{-6}\ \text{m}^2

Using

R = \frac{\rho L}{A} \Rightarrow L = \frac{RA}{\rho}

L = \frac{0.034 \times 2.0 \times 10^{-6}}{1.8 \times 10^{-8}} = 3.8\ \text{m}

Answer

$L = 3.8\ \text{m}$

Final answer

3.8 m

Detailed explanation

Background Concept

For a uniform wire,

R = \frac{\rho L}{A}

where:

$R$ is resistance in $\Omega$ ,
$\rho$ is resistivity in $\Omega\,\text{m}$ ,
$L$ is length in $\text{m}$ ,
$A$ is cross-sectional area in $\text{m}^2$ .

For a circular cross-section,

A = \pi r^2 = \pi\left(\frac{d}{2}\right)^2.

Understanding the Question

You are given:

$\rho = 1.8 \times 10^{-8}\ \Omega\,\text{m}$ ,
diameter $d = 1.6\ \text{mm}$ ,
resistance $R = 0.034\ \Omega$ .

You must show the wire length is $3.8\ \text{m}$ .

Approach

Convert diameter to metres and find radius.
Compute area $A = \pi r^2$ .
Rearrange $R = \rho L/A$ to $L = RA/\rho$ .
Substitute and evaluate carefully with powers of ten.

Step-by-Step Reasoning

Convert and find radius:

r = \frac{d}{2} = \frac{1.6\ \text{mm}}{2} = 0.80\ \text{mm} = 8.0 \times 10^{-4}\ \text{m}.

Area:

A = \pi r^2 = \pi (8.0 \times 10^{-4})^2 = \pi (6.4 \times 10^{-7}) \approx 2.0 \times 10^{-6}\ \text{m}^2.

Rearrange resistivity equation:

R = \frac{\rho L}{A} \Rightarrow L = \frac{RA}{\rho}.

Substitute:

L = \frac{0.034 \times 2.0 \times 10^{-6}}{1.8 \times 10^{-8}}.

Compute the powers of ten and numbers separately:

\frac{10^{-6}}{10^{-8}} = 10^{2}, \qquad \frac{0.034 \times 2.0}{1.8} \approx 0.0378.

So

L \approx 0.0378 \times 10^{2} = 3.78\ \text{m} \approx 3.8\ \text{m}.

Key Takeaways

Always convert mm to m before using SI equations.
For circular wires, area comes from radius squared.
Rearranging $R = \rho L/A$ is standard for these problems.

Common Mistakes

Using diameter instead of radius in $A = \pi r^2$ .
Forgetting to convert $\text{mm}$ to $\text{m}$ (gives an answer off by $10^6$ ).
Rearranging to $L = \rho A/R$ (incorrect).

Things to Be Careful About

Keep enough significant figures during intermediate steps to reach $3.8\ \text{m}$ .
Track powers of ten explicitly when dividing by $10^{-8}$ .

Techniques used

compute cross-sectional area from diameterrearrange the resistivity equation to find lengthhandle unit conversions and powers of ten

(ii)

Use the equation in (b)(ii) to determine the spring constant of the wire.

spring constant = ______ $\text{N m}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

A = \pi\left(8.0 \times 10^{-4}\right)^2 = 2.0 \times 10^{-6}\ \text{m}^2

k = \frac{EA}{L} = \frac{(1.3 \times 10^{11})(2.0 \times 10^{-6})}{3.8} = 6.9 \times 10^{4}\ \text{N m}^{-1}

Answer

$6.9 \times 10^{4}\ \text{N m}^{-1}$

Final answer

6.9 × 10^4 N m^-1

Detailed explanation

Background Concept

A wire in elastic tension behaves like a spring. Its spring constant is

k = \frac{EA}{L}

where:

$E$ is Young modulus ( $\text{Pa} = \text{N m}^{-2}$ ),
$A$ is cross-sectional area ( $\text{m}^2$ ),
$L$ is wire length ( $\text{m}$ ).

This result comes from combining $E = (F/A)/(x/L)$ with $k = F/x$ .

Understanding the Question

You are told to use the equation from (b)(ii) and the wire data:

$E = 1.3 \times 10^{11}\ \text{Pa}$ ,
diameter $1.6\ \text{mm}$ (so you must find $A$ ),
length $L = 3.8\ \text{m}$ from (d)(i).

You must calculate the spring constant in $\text{N m}^{-1}$ .

Approach

Find $A$ from the diameter.
Substitute into $k = EA/L$ .
Check units: $E(\text{N m}^{-2}) \times A(\text{m}^2) = \text{N}$ ; dividing by $L(\text{m})$ gives $\text{N m}^{-1}$ .

Step-by-Step Reasoning

Radius:

r = 0.80\ \text{mm} = 8.0 \times 10^{-4}\ \text{m}.

Area:

A = \pi r^2 = \pi(8.0 \times 10^{-4})^2 \approx 2.0 \times 10^{-6}\ \text{m}^2.

Substitute into $k = EA/L$ :

k = \frac{(1.3 \times 10^{11})(2.0 \times 10^{-6})}{3.8}.

Compute $EA$ first:

EA = 1.3 \times 2.0 \times 10^{11-6} = 2.6 \times 10^{5}\ \text{N}.

Then divide by $3.8$ :

k \approx \frac{2.6 \times 10^{5}}{3.8} = 6.9 \times 10^{4}\ \text{N m}^{-1}.

Key Takeaways

Spring constant of a wire increases with $E$ and $A$ , decreases with $L$ .
Unit-checking helps confirm the result is sensible.

Common Mistakes

Using diameter as radius when finding $A$ .
Using $L$ in cm or mm instead of m.
Writing units as $\text{N m}$ instead of $\text{N m}^{-1}$ .

Things to Be Careful About

Keep consistent SI units throughout.
Use the length found in (d)(i); do not re-calculate $L$ with a rounded area in a different way that changes the final significant figures.

Techniques used

use k = EA/L for a wirecalculate cross-sectional area from diametersubstitute numerical values with correct SI units

Q4Medium-EasySuperpositionWaves

(a)

State what is meant by diffraction of a wave.

(b)

A beam of vertically polarised light of wavelength $540\text{ nm}$ is incident normally on a diffraction grating, as shown in Fig. 4.1.

Fig. 4.1

The diffraction grating has a line spacing of $5.0 \times 10^{-6}\ \text{m}$ .

The light transmitted by the diffraction grating illuminates a circular screen. The diffraction grating is at the centre X of the circle.

The central bright fringe is formed at point O on the screen and has intensity $I_0$ .

P is a point on the screen where the line XP is at a variable angle $\theta$ to the line XO. The intensity $I$ of light on the screen at P varies with $\theta$ .

6M

(i)

Show that the angle $\theta$ at which the first-order bright fringe is formed is $6.2^\circ$ .

DifficultyMedium-Easy

Worked solution

Working

For a diffraction grating,

d\sin\theta = n\lambda

First order: $n=1$ , $d=5.0\times10^{-6}\ \text{m}$ , $\lambda=540\ \text{nm}=5.40\times10^{-7}\ \text{m}$ .

\sin\theta = \frac{\lambda}{d}=\frac{5.40\times10^{-7}}{5.0\times10^{-6}}=0.108

\theta=\sin^{-1}(0.108)=6.2^\circ

Answer

$\theta = 6.2^\circ$

Final answer

6.2°

Detailed explanation

Background Concept

A diffraction grating has many equally spaced slits. Bright fringes (principal maxima) occur when waves from adjacent slits interfere constructively. The condition for a principal maximum is that the path difference between adjacent slits equals an integer number of wavelengths:

d\sin\theta = n\lambda

where:

$d$ is the grating line spacing (distance between adjacent slits),
$\theta$ is the angle from the straight-through direction (the normal),
$n$ is the order number ( $0,1,2,\dots$ ),
$\lambda$ is the wavelength.

Understanding the Question

Light of wavelength $540\ \text{nm}$ is incident normally on a grating with spacing $5.0\times10^{-6}\ \text{m}$ . You must show that the first-order ( $n=1$ ) bright fringe occurs at $6.2^\circ$ .

Approach

Use the grating equation with $n=1$ . Convert the wavelength from nm to m, calculate $\sin\theta$ , then take $\sin^{-1}$ to find $\theta$ .

Step-by-Step Reasoning

Convert wavelength:

540\ \text{nm} = 540\times10^{-9}\ \text{m} = 5.40\times10^{-7}\ \text{m}

Apply the grating equation for first order ( $n=1$ ):

\sin\theta=\frac{n\lambda}{d}=\frac{5.40\times10^{-7}}{5.0\times10^{-6}}=0.108

Convert from sine to angle:

\theta=\sin^{-1}(0.108)\approx 6.2^\circ

This matches the value requested.

Key Takeaways

Principal maxima angles come from $d\sin\theta=n\lambda$ .
Always convert nm to m in calculations.

Common Mistakes

Forgetting to convert $540\ \text{nm}$ into metres.
Using $d=5.0\times10^{-6}$ as if it were lines per metre (it is spacing, not line density).
Using $\theta\approx\sin\theta$ without checking the size of the angle (here it would still be close, but the question expects the proper calculation).

Things to Be Careful About

Ensure $\theta$ is measured from the central direction (the normal), as in the grating equation.
Keep significant figures sensible: $d$ is given to 2 s.f., so $\theta$ to about 2 s.f. is appropriate.

Techniques used

apply the diffraction grating equationconvert wavelength into SI unitssolve a trigonometric equation for the diffraction angle

(ii)

Determine the value of $\theta$ at which the second-order bright fringe is formed.

$\theta$ = ______ $^\circ$

DifficultyEasy

Worked solution

Working

Second order: $n=2$ .

\sin\theta = \frac{2\lambda}{d}=\frac{2(5.40\times10^{-7})}{5.0\times10^{-6}}=0.216

\theta=\sin^{-1}(0.216)=12.5^\circ

Answer

$\theta = 12.5^\circ$

Final answer

12.5°

Detailed explanation

Background Concept

For a diffraction grating, the $n$ th order maximum occurs where the path difference between adjacent slits is $n\lambda$ , giving:

d\sin\theta = n\lambda

Higher order ( $n=2,3,\dots$ ) fringes occur at larger angles, as long as $\sin\theta \le 1$ .

Understanding the Question

You are asked for the angle of the second-order bright fringe for the same grating and wavelength as in part (b)(i).

Given:

$d=5.0\times10^{-6}\ \text{m}$
$\lambda=5.40\times10^{-7}\ \text{m}$
$n=2$

Approach

Substitute $n=2$ into $\sin\theta=n\lambda/d$ and then take the inverse sine.

Step-by-Step Reasoning

Compute $\sin\theta$ :

\sin\theta = \frac{2\lambda}{d} = \frac{2(5.40\times10^{-7})}{5.0\times10^{-6}}=0.216

Find the angle:

\theta=\sin^{-1}(0.216)\approx 12.5^\circ

Key Takeaways

Doubling the order roughly doubles $\sin\theta$ (not $\theta$ itself).
Check $n\lambda/d \le 1$ to ensure the order exists.

Common Mistakes

Writing $\theta_2 = 2\theta_1$ (not generally correct because of the sine relationship).
Leaving the answer as $\sin\theta$ rather than calculating $\theta$ .

Things to Be Careful About

Use degrees (as the axis and question specify).
Don’t round too early; keep enough figures in $\sin\theta$ before applying $\sin^{-1}$ .

Techniques used

apply the diffraction grating equation for a higher ordersolve a trigonometric equation for the diffraction angle

(iii)

On Fig. 4.2, sketch the variation of the intensity $I$ with $\theta$ for values of $\theta$ from $-15^\circ$ to $+15^\circ$ .

Fig. 4.2

DifficultyMedium-Easy

Worked solution

Answer

Sketch sharp principal maxima at

$\theta=0^\circ$ with height $I_0$ ,
$\theta=\pm 6.2^\circ$ (first order),
$\theta=\pm 12.5^\circ$ (second order),
with the pattern symmetric about $\theta=0^\circ$ and intensity close to zero between maxima. The first- and second-order peaks should be shown smaller than the central peak.

Final answer

Sharp symmetric maxima at 0°, ±6.2°, ±12.5°; central peak I0; others smaller; near-zero between.

Detailed explanation

Background Concept

A diffraction grating produces a set of very narrow bright fringes (principal maxima) at angles that satisfy:

d\sin\theta = n\lambda

Because there are many slits, the maxima are sharp and the intensity between them is very low. The pattern is symmetric: for every maximum at $+\theta$ there is a corresponding maximum at $-\theta$ .

In real gratings, higher orders often have lower intensity (due to slit diffraction envelope and grating efficiency), so a sketch typically shows the central maximum largest, then decreasing heights for $\pm 1$ , then $\pm 2$ .

Understanding the Question

You must sketch $I$ against $\theta$ from $-15^\circ$ to $+15^\circ$ .

From earlier parts:

$n=0$ maximum at $\theta=0^\circ$ with intensity $I_0$ (given).
$n=1$ maxima at $\theta=\pm 6.2^\circ$ .
$n=2$ maxima at $\theta=\pm 12.5^\circ$ .

No higher orders fit within $\pm 15^\circ$ .

Approach

Mark the angular positions of the allowed orders on the $\theta$ axis, then draw narrow peaks at those positions, making the graph symmetric. Use the given reference level $I_0$ for the central peak and show reduced heights for higher orders.

Step-by-Step Reasoning

Place the central maximum at $\theta=0^\circ$ and label its height as $I_0$ .
Add first-order maxima at $\theta=+6.2^\circ$ and $\theta=-6.2^\circ$ .
Add second-order maxima at $\theta=+12.5^\circ$ and $\theta=-12.5^\circ$ .
Draw these as sharp (narrow) peaks, because a grating produces sharp principal maxima.
Make the intensity between peaks close to zero.
Show decreasing peak heights for increasing order (central largest; first order smaller; second order smaller again).

Key Takeaways

Grating maxima occur at discrete angles given by the grating equation.
The pattern must be symmetric about $\theta=0^\circ$ .
Many slits produce narrow maxima and low intensity between them.

Common Mistakes

Putting peaks at the wrong angles (e.g. using $6.2^\circ$ and $12.4^\circ$ but not symmetric, or forgetting negative angles).
Drawing broad sinusoidal oscillations rather than sharp grating maxima.
Making the first/second order peaks larger than the central maximum without any justification.

Things to Be Careful About

The sketch is qualitative, but peak positions must be correct.
Use the provided $I_0$ marking correctly: the central peak should reach $I_0$ .
Stay within the stated range $-15^\circ$ to $+15^\circ$ (so do not add third order).

Techniques used

identify principal maxima positions for different ordersuse symmetry of the diffraction pattern about the central maximumsketch a qualitative intensity distribution with narrow bright fringes

(c)

A polarising filter is placed in the path of the light beam that is incident on the diffraction grating in Fig. 4.1. The transmission axis of the filter is at $45^\circ$ to the vertical.

Suggest how the variation of intensity with $\theta$ for the light on the screen compares with the answer in (b)(iii).

DifficultyMedium-Easy

Worked solution

Answer

The polariser at $45^\circ$ transmits intensity

I = I_{\text{in}}\cos^2 45^\circ = \tfrac{1}{2}I_{\text{in}}.

So the diffraction pattern has maxima at the same angles as in (b)(iii), but all intensities (including the central maximum) are reduced by a factor of $\tfrac{1}{2}$ (central becomes $\tfrac{1}{2}I_0$ ).

Final answer

Same fringe angles as (b)(iii), but all intensities are halved (central becomes I0/2).

Detailed explanation

Background Concept

A polarising filter transmits only the component of the electric field along its transmission axis. If plane-polarised light of intensity $I_{\text{in}}$ is incident on a polariser whose axis makes an angle $\phi$ with the light’s polarisation direction, Malus’s law gives the transmitted intensity:

I = I_{\text{in}}\cos^2\phi.

A diffraction grating determines the angles of the bright fringes through the interference condition $d\sin\theta=n\lambda$ , which depends on $d$ , $\lambda$ and $n$ , not on the intensity.

Understanding the Question

Originally the light is vertically polarised. A polariser with axis at $45^\circ$ to the vertical is inserted before the grating. You must compare the new $I$ vs $\theta$ graph with the sketch in (b)(iii).

Approach

Use Malus’s law with $\phi=45^\circ$ to find the factor by which the incident intensity is reduced.
State that diffraction angles (peak positions) are unchanged because $d\sin\theta=n\lambda$ is unchanged.
Therefore the whole intensity pattern is scaled down by the same factor.

Step-by-Step Reasoning

The incident light is vertically polarised; the filter axis is at $45^\circ$ to vertical, so the angle between them is $\phi=45^\circ$ .
Apply Malus’s law:

I_{\text{after filter}} = I_{\text{before filter}}\cos^2 45^\circ = I_{\text{before filter}}\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}I_{\text{before filter}}.

The grating still has the same $d$ , and the light still has the same wavelength $\lambda$ , so the maxima angles from

d\sin\theta=n\lambda

are unchanged.

Because the light entering the grating is half as intense, every bright fringe on the screen is correspondingly reduced in intensity by a factor of $1/2$ . In particular, the central maximum changes from $I_0$ to $I_0/2$ .

Key Takeaways

Malus’s law changes intensity, not wavelength.
Diffraction/interference geometry sets positions of fringes; intensity sets heights.

Common Mistakes

Saying the fringe angles change (they do not, since $d$ and $\lambda$ are unchanged).
Using $\cos 45^\circ$ instead of $\cos^2 45^\circ$ for intensity.
Claiming the light becomes unpolarised after the filter (it remains plane-polarised along the filter axis).

Things to Be Careful About

The reference $I_0$ in part (b) refers to the original central maximum; after inserting the polariser the central maximum is $I_0/2$ .
The overall shape (peak positions and relative distribution) stays the same; it is a uniform vertical scaling of the graph.

Techniques used

apply Malus’s law to compare transmitted intensityinfer that angular positions of diffraction maxima are unchangedscale an intensity pattern without changing its shape

Q5MediumD.C. CircuitsElectricity

(a)

State Kirchhoff’s first law.

(b)

Fig. 5.1 shows a circuit containing a thermistor T that has a negative temperature coefficient.

Fig. 5.1

5M

(i)

The thermistor has resistance $R_0$ at a temperature of $0^\circ\text{C}$ .

On Fig. 5.2, sketch a possible variation of the resistance of the thermistor with temperature between $0^\circ\text{C}$ and $100^\circ\text{C}$ .

Fig. 5.2

DifficultyMedium-Easy

Worked solution

diagram-b-i

Answer

A decreasing (non-linear) curve starting at $R_0$ when $T = 0\ ^\circ\text{C}$ and falling to a smaller resistance at $100\ ^\circ\text{C}$ .

Final answer

A decreasing (non-linear) curve from R0 at 0°C to a smaller resistance at 100°C.

Detailed explanation

Background Concept

A thermistor with a negative temperature coefficient (NTC) has resistance that decreases as temperature increases.

This happens because, in a semiconductor thermistor, increasing temperature increases the number of charge carriers, so conductivity increases and resistance falls. The fall is typically non-linear (often roughly exponential).

Understanding the Question

You are told the thermistor has resistance $R_0$ at $0\ ^\circ\text{C}$ , and you must sketch how its resistance changes up to $100\ ^\circ\text{C}$ . The axes already show the point corresponding to $R_0$ on the resistance axis.

Approach

Start the graph at $(0\ ^\circ\text{C}, R_0)$ .
Because it is NTC, the curve must go downwards as temperature increases.
Make it curved (not a straight line) to show a “possible variation”.

Step-by-Step Reasoning

Mark the point at $T=0\ ^\circ\text{C}$ with resistance $R_0$ .
For higher temperature, resistance must be less than $R_0$ , so the graph must slope downwards to the right.
Draw a smooth non-linear curve decreasing towards $T=100\ ^\circ\text{C}$ (for many NTC thermistors it falls steeply at first and then becomes less steep).

diagram-b-i

Key Takeaways

NTC thermistor: higher $T$ (\Rightarrow) lower resistance.
Sketch should pass through $(0, R_0)$ and decrease to the right.

Common Mistakes

Drawing resistance increasing with temperature (that would be PTC, not NTC).
Not starting the curve at the marked $R_0$ value at $0\ ^\circ\text{C}$ .
Drawing a horizontal line (implies resistance constant).

Things to Be Careful About

The question says “possible variation”, so exact shape is not required, but it must clearly be decreasing and should be smooth.
Keep axes labels consistent: resistance on $y$ -axis, temperature on $x$ -axis.

Techniques used

use the negative temperature coefficient property to determine the trendsketch a non-linear relationship on given axes with correct intercept

(ii)

With reference to the current in the cell, explain why the current in resistor R decreases with increasing temperature of the thermistor.

DifficultyMedium

Worked solution

Answer

As temperature increases, the thermistor resistance decreases, so the equivalent resistance of the parallel combination decreases and the current in the cell increases.

With internal resistance $r$ , the lost p.d. $Ir$ increases, so the terminal p.d. across the parallel network (and hence across $R$ ) decreases.

Since $I_R = V_R / R$ , a smaller p.d. across $R$ means the current in $R$ decreases.

Final answer

Thermistor resistance falls → total cell current rises → larger Ir drop across internal resistance → smaller terminal p.d. across R → IR decreases.

Detailed explanation

Background Concept

There are two key ideas:

Parallel resistance: if two resistors are in parallel, the equivalent resistance $R_{\text{eq}}$ satisfies

\frac{1}{R_{\text{eq}}} = \frac{1}{R} + \frac{1}{R_T}

So if the thermistor resistance $R_T$ decreases, $R_{\text{eq}}$ also decreases.

Internal resistance: a cell with e.m.f. $E$ and internal resistance $r$ behaves like an ideal source $E$ in series with a resistor $r$ . If the cell current is $I$ , then terminal p.d. $V$ is

V = E - Ir

So as $I$ increases, the “lost volts” $Ir$ increases and $V$ decreases.

Understanding the Question

The circuit has the cell (with internal resistance) feeding a parallel network of a fixed resistor $R$ and an NTC thermistor $T$ .

You must explain why the current in resistor $R$ decreases when the thermistor temperature increases, and you must refer to the current in the cell in your explanation.

Approach

Chain the cause-and-effect:

Increase thermistor temperature (\Rightarrow) thermistor resistance decreases (NTC).
Lower thermistor resistance (\Rightarrow) lower equivalent resistance of the parallel part.
Lower external resistance (\Rightarrow) larger current drawn from the cell.
Larger cell current (\Rightarrow) bigger voltage drop across internal resistance $Ir$ .
Bigger internal drop (\Rightarrow) smaller terminal p.d. across the parallel part, so smaller p.d. across $R$ .
With $R$ fixed, smaller p.d. (\Rightarrow) smaller current in $R$ .

Step-by-Step Reasoning

The thermistor is NTC, so when temperature increases, its resistance $R_T$ decreases.
The thermistor is in parallel with $R$ , so decreasing $R_T$ decreases the equivalent resistance $R_{\text{eq}}$ of the parallel pair.
The cell sees a smaller external resistance, so the current drawn from the cell $I$ increases.
Because the cell has internal resistance $r$ , there is a lost p.d. $Ir$ inside the cell. When $I$ increases, $Ir$ increases.
Therefore the terminal p.d. across the parallel network is

V = E - Ir

and this terminal p.d. decreases as $I$ increases.
6. The resistor $R$ is connected directly across this terminal p.d., so $V_R = V$ .
7. Since $R$ is constant, the current through it is

I_R = \frac{V_R}{R}

so when $V_R$ decreases, $I_R$ decreases.

(You may also note that in parallel, a smaller thermistor resistance “steals” more current; however, the key reason $I_R$ decreases is that the voltage across $R$ falls due to the increased internal drop.)

Key Takeaways

NTC thermistor: higher $T$ gives lower $R_T$ .
Lower external resistance increases cell current.
With internal resistance, higher cell current reduces terminal p.d., reducing branch currents.

Common Mistakes

Claiming the p.d. across $R$ stays constant; it does not when the cell has internal resistance.
Saying “current splits so $I_R$ decreases” without mentioning the required link to cell current and lost volts.
Forgetting that $R$ is in parallel, so the p.d. across $R$ equals the terminal p.d.

Things to Be Careful About

Use correct language: “terminal p.d.” is across the external circuit; “lost p.d.” is across internal resistance.
Do not confuse $I$ (cell current) with $I_R$ (current in resistor $R$ ).

Techniques used

relate a change in thermistor resistance to a change in equivalent parallel resistanceapply the internal resistance model to link cell current to terminal p.d.use Ohm's law to connect terminal p.d. to current in a branch resistor

(c)

The electromotive force (e.m.f.) $E$ of the cell in Fig. 5.1 is $1.50\text{ V}$ . The internal resistance $r$ of the cell is $0.12\ \Omega$ .

Resistor R has a resistance of $6.00\ \Omega$ .

At a particular temperature of the thermistor, the current in R is $0.200\text{ A}$ .

For this temperature of the thermistor, determine:

4M

(i)

the current in the cell

current = ______ $\text{A}$

DifficultyMedium-Easy

Worked solution

Working

Voltage across $R$ :

V = I_R R = 0.200 \times 6.00 = 1.20\ \text{V}

Lost p.d. across $r$ :

Ir = E - V = 1.50 - 1.20 = 0.30\ \text{V}

Cell current:

I = \frac{0.30}{0.12} = 2.50\ \text{A}

Answer

$2.50\ \text{A}$

Final answer

2.50 A

Detailed explanation

Background Concept

For a cell of e.m.f. $E$ and internal resistance $r$ , the terminal p.d. $V$ when a current $I$ flows is

V = E - Ir

The quantity $Ir$ is the “lost volts” across the internal resistance.

In a parallel network, the p.d. across each branch is the same. So the p.d. across resistor $R$ equals the terminal p.d. across the parallel combination.

Understanding the Question

You are given:

$E = 1.50\ \text{V}$
$r = 0.12\ \Omega$
$R = 6.00\ \Omega$
current in $R$ is $I_R = 0.200\ \text{A}$

You must find the current in the cell (the total current supplied to the parallel combination).

Approach

Use $V = I_R R$ to find the p.d. across $R$ .
That same p.d. is the terminal p.d. $V$ of the cell.
Use $E - V = Ir$ to find $I$ .

Step-by-Step Reasoning

Find the p.d. across resistor $R$ using Ohm’s law:

V_R = I_R R = 0.200 \times 6.00 = 1.20\ \text{V}

Because $R$ is in parallel with the thermistor, the p.d. across the parallel network is also $1.20\ \text{V}$ . This is the terminal p.d. of the cell:

V = 1.20\ \text{V}

The lost volts across internal resistance are:

Ir = E - V = 1.50 - 1.20 = 0.30\ \text{V}

Solve for the cell current:

I = \frac{Ir}{r} = \frac{0.30}{0.12} = 2.50\ \text{A}

Key Takeaways

Use a known branch current to get the terminal p.d. in a parallel circuit.
Then use $V = E - Ir$ to link terminal p.d. to the cell current.

Common Mistakes

Treating $1.50\ \text{V}$ as the p.d. across $R$ (it is the e.m.f., not the terminal p.d.).
Forgetting that the terminal p.d. is reduced by the internal resistance.

Things to Be Careful About

Keep clear which current is which: $I_R$ is not the cell current.
Use consistent significant figures: here $2.50\ \text{A}$ is appropriate.

Techniques used

use Ohm's law to find the terminal p.d. from a branch current and resistanceapply the internal resistance equation E = V + Irsolve for the cell current using the lost p.d.

(ii)

the resistance of the thermistor.

resistance = ______ $\Omega$

DifficultyMedium-Easy

Worked solution

Working

Terminal p.d. across the parallel network:

V = I_R R = 0.200 \times 6.00 = 1.20\ \text{V}

From (i), $I = 2.50\ \text{A}$ . Hence thermistor current:

I_T = I - I_R = 2.50 - 0.200 = 2.30\ \text{A}

Thermistor resistance:

R_T = \frac{V}{I_T} = \frac{1.20}{2.30} = 0.522\ \Omega

Answer

$0.522\ \Omega$

Final answer

0.522 Ω

Detailed explanation

Background Concept

In a parallel circuit:

The p.d. across each branch is the same.
The total current is the sum of the branch currents (Kirchhoff’s first law):

I = I_R + I_T

Once you know the p.d. across a component and the current through it, you can find its resistance using Ohm’s law:

R = \frac{V}{I}

Understanding the Question

At the stated thermistor temperature you know:

current in resistor $R$ is $0.200\ \text{A}$
from part (i) the current in the cell is $2.50\ \text{A}$
$R = 6.00\ \Omega$

You must find the thermistor resistance $R_T$ at that temperature.

Approach

Find the terminal p.d. across the parallel network using $V = I_R R$ .
Use Kirchhoff’s first law to get thermistor current $I_T = I - I_R$ .
Use $R_T = V/I_T$ .

Step-by-Step Reasoning

The p.d. across $R$ is

V = I_R R = 0.200 \times 6.00 = 1.20\ \text{V}

This is also the p.d. across the thermistor because they are in parallel.

The total current from the cell is the sum of branch currents:

I = I_R + I_T

So

I_T = I - I_R = 2.50 - 0.200 = 2.30\ \text{A}

Apply Ohm’s law to the thermistor:

R_T = \frac{V}{I_T} = \frac{1.20}{2.30} = 0.5217...\ \Omega \approx 0.522\ \Omega

Key Takeaways

In parallel: same p.d., currents add.
Use branch current + p.d. to find branch resistance.

Common Mistakes

Using $1.50\ \text{V}$ as the p.d. across the thermistor (ignoring internal resistance).
Subtracting currents the wrong way round (getting a negative or tiny thermistor current).
Using $R_T = I_T / V$ instead of $R_T = V / I_T$ .

Things to Be Careful About

Ensure you use the terminal p.d. ( $1.20\ \text{V}$ ), not the e.m.f.
Keep enough significant figures in intermediate steps; round at the end to $0.522\ \Omega$ (3 s.f.).

Techniques used

apply Kirchhoff’s first law at a junction to relate total and branch currentsuse Ohm’s law to compute an unknown branch resistance from branch current and p.d.use parallel-circuit p.d. equality to use the same terminal voltage for both branches

Q6Medium-EasyParticle Physics

The nuclide ${}^3_1\text{H}$ is an isotope of hydrogen that is called tritium.

(a)

4M

(i)

Determine the numbers of protons, neutrons and electrons in a neutral atom of tritium.

number of protons = ______
number of neutrons = ______
number of electrons = ______

DifficultyEasy

Worked solution

Answer

For ${}^3_1\text{H}$ :

number of protons $= 1$
number of neutrons $= 3 - 1 = 2$
number of electrons (neutral atom) $= 1$

Final answer

protons = 1, neutrons = 2, electrons = 1

Detailed explanation

Background Concept

A nuclide written as ${}^A_Z\text{X}$ has:

proton number (atomic number) $Z$ = number of protons in the nucleus.
nucleon number (mass number) $A$ = total number of nucleons in the nucleus = protons + neutrons.
So:

\text{neutrons} = A - Z.

For a neutral atom, total positive charge from protons is balanced by electrons, so:

\text{electrons} = \text{protons} = Z.

Understanding the Question

You are told the nuclide is tritium, ${}^3_1\text{H}$ . The question asks for the number of protons, neutrons and electrons in a neutral atom (so it includes the electrons outside the nucleus).

Approach

Read $Z$ from the subscript to get protons.
Use $A - Z$ to get neutrons.
Use neutrality (electrons = protons) to get electrons.

Step-by-Step Reasoning

From ${}^3_1\text{H}$ :

$Z = 1 \Rightarrow$ protons $= 1$ .
$A = 3$ , so neutrons $= A - Z = 3 - 1 = 2$ .
Neutral atom means electrons $= 1$ .

Key Takeaways

In ${}^A_Z\text{X}$ , $Z$ gives protons and $A - Z$ gives neutrons.
Neutral atoms have equal numbers of protons and electrons.

Common Mistakes

Using $A$ as the number of neutrons (it is total nucleons).
Forgetting the word “neutral” and giving electrons $= 0$ (that would be the nucleus, not the atom).

Things to Be Careful About

Protons and neutrons are in the nucleus; electrons are outside.
Always subtract in the correct order: neutrons $= A - Z$ , not $Z - A$ .

Techniques used

interpret nuclide notation to identify proton (atomic) number and nucleon numberuse charge neutrality to determine the number of electronscalculate neutrons from nucleon number minus proton number

(ii)

Draw a labelled diagram to represent a simple model of the arrangement of the protons, neutrons and electrons in a tritium atom.

DifficultyMedium-Easy

Worked solution

diagram-6a-ii

Answer

Nucleus contains $1\,p$ and $2\,n$ ; $1$ electron shown outside the nucleus (in a shell/orbit), with all particles labelled.

Final answer

Labelled model showing nucleus (1p, 2n) and one electron outside.

Detailed explanation

Background Concept

A simple atomic model for A Level typically shows:

a tiny central nucleus containing protons ( $p$ , charge $+e$ ) and neutrons ( $n$ , charge $0$ ),
electrons ( $e^-$ , charge $-e$ ) arranged outside the nucleus (often drawn as shells/orbits for a simple model).
This is not a scale drawing; it is a schematic that shows arrangement and types of particles.

Understanding the Question

You must draw a labelled diagram for tritium. From ${}^3_1\text{H}$ you know the nucleus has 1 proton and 2 neutrons, and (neutral atom) there is 1 electron outside the nucleus.

Approach

Draw a central nucleus (circle).
Put symbols for the correct number of nucleons inside: one $p$ and two $n$ .
Draw one electron on an orbit/shell outside.
Label each particle type clearly.

Step-by-Step Reasoning

diagram-6a-ii

The nucleus is drawn at the centre.
Inside it, include three nucleons: one labelled $p$ (or $p^+$ ) and two labelled $n$ .
Outside, show one electron labelled $e^-$ on a circular shell/orbit.
Ensure labels identify particle types (and ideally charges).

Key Takeaways

Nucleons (protons and neutrons) are in the nucleus.
A neutral atom has the same number of electrons as protons.
A diagram mark is usually for correct placement and clear labels.

Common Mistakes

Drawing three electrons because $A=3$ .
Putting electrons inside the nucleus.
Not labelling the particles (loses the “labelled diagram” mark).

Things to Be Careful About

The diagram is a model: not to scale, but the nucleus must be shown as central with electrons outside.
Count particles correctly: tritium has $1$ electron total (not 2 or 3).

Techniques used

translate nuclide composition into a simple atomic modeluse a labelled schematic to show nucleus and electron arrangementlabel particles with correct charge and location

(b)

Tritium is radioactive and undergoes $\beta^-$ decay to form an isotope of helium (He). Gamma radiation is not emitted during this decay.

3M

(i)

Complete the equation to represent the radioactive decay of tritium.

{}^3_1\text{H} \rightarrow {}^{\dots\dots}_{\dots\dots}\text{He} + {}^{\dots\dots}_{\dots\dots}\beta + {}^0_0\text{X}

DifficultyMedium-Easy

Worked solution

Working

Conserve nucleon number $A$ and proton number $Z$ .

{}^3_1\text{H} \rightarrow {}^3_2\text{He} + {}^0_{-1}\beta + {}^0_0\text{X}

Answer

${}^3_2\text{He}$ and ${}^0_{-1}\beta$ .

Final answer

${}^3_2\text{He} + {}^0_{-1}\beta

Detailed explanation

Background Concept

In nuclear decay equations, two conservation rules are essential:

Nucleon number $A$ is conserved.
Proton number $Z$ (equivalently nuclear charge) is conserved.

In $\beta^-$ decay, inside the nucleus a neutron changes into a proton and emits:

an electron (the $\beta^-$ particle), and
an electron antineutrino to conserve lepton number and energy-momentum.
So $Z$ increases by 1 while $A$ stays the same.

Understanding the Question

Tritium is ${}^3_1\text{H}$ . It undergoes $\beta^-$ decay to become an isotope of helium. You must fill in the missing nuclide numbers for He and for the $\beta$ particle in the equation (gamma not emitted). The equation already includes a particle X with ${}^0_0\text{X}$ .

Approach

Use the rule for $\beta^-$ decay: $A$ unchanged, $Z$ increases by 1 for the daughter nucleus.
Write the daughter nucleus as helium with the found $A$ and $Z$ .
Use standard notation for a beta-minus particle: ${}^0_{-1}\beta$ .
Check conservation of $A$ and $Z$ across the full equation.

Step-by-Step Reasoning

Start with parent:

$A = 3$ , $Z = 1$ .

After $\beta^-$ decay:

$A$ stays $3$ .
$Z$ increases to $2$ .
So the daughter nucleus is:

{}^3_2\text{He}.

A $\beta^-$ particle is an electron, written:

{}^0_{-1}\beta.

Now check balancing:

Nucleon number: LHS $3$ ; RHS $3 + 0 + 0 = 3$ .
Proton number: LHS $1$ ; RHS $2 + (-1) + 0 = 1$ .
So the completed equation is consistent.

Key Takeaways

In $\beta^-$ decay: $A$ unchanged, $Z$ increases by 1.
A $\beta^-$ particle is written as ${}^0_{-1}\beta$ .

Common Mistakes

Writing helium as ${}^3_1\text{He}$ (forgetting $Z$ must increase).
Using ${}^0_{+1}\beta$ (that would be $\beta^+$ / positron emission).
Changing $A$ in $\beta$ decay (it does not change).

Things to Be Careful About

Use the correct nuclear notation with superscript/subscript.
Remember that the antineutrino has ${}^0_0$ (so it does not affect $A$ or $Z$ balancing).

Techniques used

apply conservation of nucleon number in nuclear equationsapply conservation of proton number (charge) in nuclear equationsidentify beta-minus emission as an electron with specific nuclide notation

(ii)

State the name of particle X.

(c)

Determine the quark composition of a tritium nucleus.