Two progressive water waves $X$ and $Y$ travel along a straight line from point $A$ to point $B$. The variation of displacement of the waves with distance from $A$ at an instant in time is shown in Fig. 3.1. (a) State the amplitude of wave $X$. amplitude = \_\_\_\_\_\_ $\text{cm}$ (b) Both waves have frequency $16\text{ Hz}$. (i) Determine the speed of wave $X$. speed = \_\_\_\_\_\_ $\text{m s}^{-1}$ (ii) State and explain whether $X$ and $Y$ are coherent. (c) Wave $X$ and…

(a) 10 cm · (b)(i) 6.4 m s^-1 · (b)(ii) Yes; same frequency and constant phase difference (antiphase). · (c) Resultant: sinusoid, amplitude 10 cm, wavelength 0.40 m, in phase with Y (antiphase with X). · (d) 1/4

(a) State what is meant by a fundamental particle. (b) A nucleus $X$ has $14$ nucleons and $p$ protons. The ratio of charge to mass for nucleus $X$ is $4.1 \times 10^7\text{ C kg}^{-1}$. (i) Determine $p$. $p$ = \_\_\_\_\_\_ (ii) Nucleus $X$ undergoes $\beta^-$ decay to form nucleus $Z$. Complete the equation representing this decay. $$ {}^{14}_{\dots\dots}X \rightarrow {}^{\dots\dots}_{\dots\dots}Z + {}^{\dots\dots}_{\dots\dots}\dots\dots +…

(a) A fundamental particle is one that has no internal structure (not made of smaller constituent particles). · (b)(i) 6 · (b)(ii) ${}^{14}_{6}X \rightarrow {}^{14}_{7}Z + {}^{0}_{-1}e^- + {}^{0}_{0}\bar{\nu}_e$ · (c) β⁺ decay

9702/22 May/June 2025 — Physics Solved Past Paper

Q1MediumPhysical Quantities and UnitsWork, Energy and PowerDynamicsKinematics

(a)

Table 1.1 lists some physical quantities. Identify with ticks (✓) which quantities are vectors and which are scalars.

Table 1.1

quantity	scalar	vector
acceleration
displacement
gravitational potential energy
speed
temperature

2M Similar

DifficultyEasy

Worked solution

Answer

acceleration: vector
displacement: vector
gravitational potential energy: scalar
speed: scalar
temperature: scalar

Final answer

acceleration vector; displacement vector; gravitational potential energy scalar; speed scalar; temperature scalar

Detailed explanation

Background Concept

A scalar quantity has magnitude only (a size, with units). Examples include mass, time, energy, speed, temperature.

A vector quantity has magnitude and direction. Examples include displacement, velocity, acceleration, force, momentum.

Understanding the Question

You are given a list of common physical quantities and a table with columns “scalar” and “vector”. You must tick the correct column for each quantity.

Approach

For each quantity, ask: “Does this quantity need a direction to be fully described?”

If yes: it is a vector.
If no: it is a scalar.

Step-by-Step Reasoning

Acceleration: change of velocity per unit time; since velocity has direction, acceleration also has direction → vector.
Displacement: straight-line change in position from start to finish, which requires a direction (e.g. 5 m east) → vector.
Gravitational potential energy: energy stored due to height in a gravitational field; energy is a single number (no direction) → scalar.
Speed: rate of change of distance; it does not include direction (that would be velocity) → scalar.
Temperature: a measure related to average kinetic energy of particles; no direction → scalar.

Key Takeaways

Vectors: need magnitude and direction.
Scalars: need magnitude only.
Speed (scalar) vs velocity (vector) is a common comparison.

Common Mistakes

Calling speed a vector (confusing it with velocity).
Thinking energy is a vector because it is related to forces; energy is always scalar.

Things to Be Careful About

Use the definitions: displacement/velocity/acceleration are directional; distance/speed are not.
Some quantities can sound “directional” in everyday language, but only the physics definition matters.

Techniques used

identify whether a quantity has direction as well as magnitudeclassify quantities as scalar or vector from their definitions

(b)

A constant resultant force $F$ acts on a car of mass $m$ . The car moves from rest with constant acceleration $a$ along horizontal ground. When the car has displacement $s$ , the speed of the car is $v$ .

8M

(i)

Using the concept of work done on the car, show that the kinetic energy $E_K$ of the car is given by the equation

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

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Work done by resultant force:

W = Fs

With constant resultant force,

F = ma

So

W = mas

Using $v^2 = u^2 + 2as$ with $u = 0$ :

v^2 = 2as \Rightarrow as = \frac{v^2}{2}

Hence

W = m\left(\frac{v^2}{2}\right) = \frac{1}{2}mv^2

Therefore $E_K = W = \dfrac{1}{2}mv^2$ .

Answer

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

Final answer

EK = (1/2)mv^2

Detailed explanation

Background Concept

For motion along a straight line, the work done by a resultant (net) force is the energy transferred:

W = Fs

when the force $F$ is constant and acts in the same direction as the displacement $s$ .

Newton’s second law links force and acceleration:

F = ma

For constant acceleration, kinematics gives:

v^2 = u^2 + 2as

The key idea (the work–energy principle) is that net work done equals the increase in kinetic energy.

Understanding the Question

A car of mass $m$ starts from rest and accelerates uniformly under a constant resultant force $F$ over displacement $s$ , reaching speed $v$ . You must use the concept of work done to show that its kinetic energy is:

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

Approach

Start with $W = Fs$ .
Replace $F$ using $F = ma$ .
Replace $as$ using the constant-acceleration relationship between $v$ , $u$ , $a$ and $s$ .
Show that the expression becomes $\frac{1}{2}mv^2$ .

Step-by-Step Reasoning

Work done by the resultant force over displacement $s$ is

W = Fs.

Because the resultant force causes the acceleration $a$ of the car,

F = ma.

Substitute into the work expression:

W = (ma)s = mas.

Now we need to eliminate $a$ and $s$ in favour of the speed $v$ . The motion is from rest, so $u=0$ , and acceleration is constant, so we can use

v^2 = u^2 + 2as.

With $u=0$ :

v^2 = 2as \Rightarrow as = \frac{v^2}{2}.

Substitute $as$ into $W = mas$ :

W = m(as) = m\left(\frac{v^2}{2}\right) = \frac{1}{2}mv^2.

Since the net work done equals the gain in kinetic energy, the car’s kinetic energy is

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

Key Takeaways

Use $W = Fs$ for constant force along the displacement.
Use $F = ma$ to connect forces to motion.
Use $v^2 = u^2 + 2as$ to remove $a$ and $s$ .
Net work done becomes kinetic energy gained.

Common Mistakes

Using $W = Fv$ (that is related to power, not work).
Forgetting $u=0$ (starts from rest), so writing $v^2 = 2as$ incorrectly.
Using $v = at$ instead of eliminating $a$ via $as$ (still possible, but you must then connect $t$ and $s$ consistently).

Things to Be Careful About

$W = Fs$ assumes the force is constant and in the direction of motion (true here: horizontal, constant resultant).
Distinguish resultant force (net force) from just the engine driving force; the question states “constant resultant force $F$ ”.

Techniques used

apply work done as force multiplied by displacementuse Newton's second law to substitute for forceuse a constant-acceleration kinematics equation to eliminate acceleration

(ii)

The mass of the car is $920\text{ kg}$ . At time $t = 0$ , the car is at rest. At time $t = 5.8\text{ s}$ , its velocity is $17\text{ m s}^{-1}$ .

Calculate the kinetic energy of the car at time $t = 5.8\text{ s}$ .

kinetic energy = ______ $\text{J}$

1M Similar

DifficultyEasy

Worked solution

Working

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

E_K = \frac{1}{2}(920)(17^2) = 1.3294 \times 10^5 \ \text{J}

Answer

E_K = 1.33 \times 10^5\ \text{J}

Final answer

1.33 × 10^5 J

Detailed explanation

Background Concept

Kinetic energy is the energy associated with motion. For a mass $m$ moving at speed $v$ :

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

This formula uses speed (a scalar) and gives energy in joules (J).

Understanding the Question

At $t=5.8\ \text{s}$ , the car has mass $m = 920\ \text{kg}$ and speed $v = 17\ \text{m s}^{-1}$ . You are asked to calculate the kinetic energy at that time.

Approach

Use

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

and substitute the given $m$ and $v$ (already in SI units).

Step-by-Step Reasoning

Substitute values:

E_K = \frac{1}{2}(920)(17^2).

First square the speed:

17^2 = 289.

Now multiply:

E_K = 0.5 \times 920 \times 289 = 460 \times 289 = 132940\ \text{J}.

Write to 3 significant figures:

E_K = 1.33 \times 10^5\ \text{J}.

Key Takeaways

Kinetic energy depends on $v^2$ , so it increases rapidly with speed.
Keep units in SI: kg and m s $^{-1}$ give joules.

Common Mistakes

Forgetting to square $v$ .
Using $v=17\ \text{m s}^{-1}$ but treating it as $17^2 = 34$ (incorrect arithmetic).
Omitting the unit J.

Things to Be Careful About

Significant figures: input data ( $920$ , $17$ ) suggests an answer to about 2–3 s.f.
Ensure you use speed (magnitude) and not try to include a direction; kinetic energy is scalar.

Techniques used

apply the kinetic energy equationsubstitute values with consistent SI unitscalculate and round to a suitable number of significant figures

(iii)

Between time $t = 0$ and time $t = 5.8\text{ s}$ , the work done against resistive forces is $4.7 \times 10^4\text{ J}$ .

Determine the average output power of the car during this time.

power = ______ $\text{W}$

3M Similar

DifficultyMedium

Worked solution

Working

Increase in kinetic energy:

\Delta E_K = \frac{1}{2}mv^2 = 1.3294 \times 10^5\ \text{J}

Total work done by the car (engine output energy):

W = \Delta E_K + W_{\text{res}} = 1.3294 \times 10^5 + 4.7 \times 10^4

W = 1.7994 \times 10^5\ \text{J}

Average output power:

P = \frac{W}{t} = \frac{1.7994 \times 10^5}{5.8} = 3.10 \times 10^4\ \text{W}

Answer

P = 3.1 \times 10^4\ \text{W}

Final answer

3.1 × 10^4 W

Detailed explanation

Background Concept

Power is the rate of energy transfer:

P = \frac{W}{t}

where $W$ is the energy transferred (work done) in time $t$ .

When a vehicle accelerates with resistive forces (air resistance, rolling friction), the engine’s energy output is used for two things:

increasing kinetic energy, $\Delta E_K$
doing work against resistive forces, $W_{\text{res}}$

So an energy balance is:

W_{\text{engine}} = \Delta E_K + W_{\text{res}}.

Understanding the Question

From $t=0$ to $t=5.8\ \text{s}$ :

the car speeds up from rest to $17\ \text{m s}^{-1}$ (so its kinetic energy increases)
additionally, $4.7 \times 10^4\ \text{J}$ of work is done against resistive forces.

You must find the average output power over this time interval.

Approach

Calculate $\Delta E_K$ using $\frac{1}{2}mv^2$ (since initial speed is zero).
Add the given resistive work to get total engine work output.
Divide by $t = 5.8\ \text{s}$ to get average power.

Step-by-Step Reasoning

The kinetic energy at $t=5.8\ \text{s}$ is

E_K = \frac{1}{2}mv^2 = \frac{1}{2}(920)(17^2) = 1.3294 \times 10^5\ \text{J}.

Since the car starts from rest, $E_{K,\,\text{initial}} = 0$ , so

\Delta E_K = 1.3294 \times 10^5\ \text{J}.

Work done against resistive forces is given:

W_{\text{res}} = 4.7 \times 10^4\ \text{J}.

Therefore total work done by the car (engine output energy) is

W_{\text{engine}} = \Delta E_K + W_{\text{res}} = 1.3294 \times 10^5 + 4.7 \times 10^4 = 1.7994 \times 10^5\ \text{J}.

Average output power over $t=5.8\ \text{s}$ :

P_{\text{avg}} = \frac{W_{\text{engine}}}{t} = \frac{1.7994 \times 10^5}{5.8} = 3.10 \times 10^4\ \text{W}.

So the average power is about $3.1 \times 10^4\ \text{W}$ .

Key Takeaways

Engine output energy during acceleration goes into both $\Delta E_K$ and resistive losses.
Average power is total energy transferred divided by the time interval.

Common Mistakes

Using only $\Delta E_K$ and forgetting to add the resistive work.
Dividing the resistive work by time but not including the kinetic energy increase.
Using $P = Fv$ with $v=17\ \text{m s}^{-1}$ : that gives instantaneous power at one moment only, not average over the whole interval.

Things to Be Careful About

The resistive work given is for the whole time interval, so it must be added to the kinetic energy gain to get total output work.
Quote the final power to an appropriate number of significant figures (here $\approx 2$ s.f. is reasonable).

Techniques used

apply an energy balance including work done against resistive forcessum energy transfers to find total work done by the enginecalculate average power using energy divided by time

(iv)

At time $t = 5.8\text{ s}$ , the speed of the car becomes constant.

State and explain whether the output power of the car is greater than, less than or the same as the output power just before $t = 5.8\text{ s}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Less than.

At constant speed, $a = 0$ so kinetic energy is not increasing; the engine only supplies power to balance resistive forces. Just before $t=5.8\ \text{s}$ , the engine supplied power both to increase kinetic energy and to overcome resistive forces, so it was greater.

Final answer

Less than.

Detailed explanation

Background Concept

If speed is constant, acceleration is zero:

a = 0.

Newton’s second law then gives resultant force zero:

F_{\text{resultant}} = ma = 0.

For a moving vehicle, this means the driving force equals the resistive force (they balance).

Power is the rate of energy transfer. During acceleration, engine power is used to:

increase kinetic energy (increase in $\frac{1}{2}mv^2$ )
overcome resistive forces (energy lost to heat/sound etc.)

During constant speed, kinetic energy is constant, so there is no power needed to increase $E_K$ .

Understanding the Question

At $t = 5.8\ \text{s}$ the car’s speed becomes constant. You must compare the car’s output power after this time to the output power just before $t=5.8\ \text{s}$ , when it was still accelerating.

Approach

Constant speed implies no increase in kinetic energy.
Compare energy requirements:
- before: power for $\Delta E_K$ + resistive losses
- after: power only for resistive losses
  Therefore, after $t=5.8\ \text{s}$ , the required output power is smaller.

Step-by-Step Reasoning

Just before $t=5.8\ \text{s}$ , the car is accelerating, so its kinetic energy is increasing with time (its speed is still rising). That means the engine must supply energy each second to increase $E_K$ .

At and after $t=5.8\ \text{s}$ , the speed is constant, so the kinetic energy

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

is constant too (since $m$ and $v$ are constant). So the rate of increase of kinetic energy is zero.

The engine then only needs to provide power to compensate for energy lost to resistive forces (drag/rolling resistance). Since it no longer needs to increase kinetic energy, the output power is less than it was during acceleration.

Key Takeaways

Constant speed $\Rightarrow a=0 \Rightarrow$ no kinetic energy increase.
During acceleration, engine power must cover both resistive losses and kinetic energy gain.
During steady motion, engine power just balances resistive losses.

Common Mistakes

Saying the power is the same because the car is still moving (movement alone does not require increasing energy; only resisting forces do).
Saying the power is greater because the speed is larger; while instantaneous power to overcome drag can be larger at higher speed, the question compares to the moment just before 5.8 s when it was also moving fast and still increasing kinetic energy.

Things to Be Careful About

“Becomes constant” means from that time onward the acceleration is zero.
The comparison is with just before $5.8\ \text{s}$ (still accelerating), not with the start at rest.

Techniques used

relate constant speed to zero acceleration and zero resultant forcecompare power output to the rate of increase of kinetic energyinterpret energy transfers during steady motion

Q2MediumForces, Density and Pressure

(a)

Define the moment of a force about a point.

1M Similar

DifficultyEasy

Worked solution

Answer

The moment of a force about a point is

\text{moment} = F \times d

where $d$ is the perpendicular distance from the point to the line of action of the force.

Final answer

Moment about a point = force × perpendicular distance from the point to the force’s line of action.

Detailed explanation

Background Concept

The moment (torque) of a force about a point measures the turning effect of that force about that point.

It depends on two things:

the magnitude of the force $F$ (in newtons, $\text{N}$ )
the perpendicular distance $d$ (in metres, $\text{m}$ ) from the point (pivot) to the line of action of the force.

The magnitude of the moment is

\tau = Fd

Understanding the Question

You are asked to define “moment of a force about a point”. A definition must include:

the product $F \times d$
that $d$ is perpendicular to the line of action
that the moment is taken about a point (pivot/reference point).

Approach

Write the standard definition in words and (optionally) as an equation.

Step-by-Step Reasoning

Identify that “moment about a point” means the turning effect relative to that point.
Use the rule: moment $=$ force $\times$ perpendicular distance.
State clearly that the distance must be measured to the line of action of the force, not to where the force is applied (unless that happens to be perpendicular).

Key Takeaways

Moment depends on perpendicular distance to the line of action.
Correct definition: $\tau = Fd$ .

Common Mistakes

Using the distance to the point of application instead of the perpendicular distance to the line of action.
Forgetting to mention “about a point”.

Things to Be Careful About

The line of action is an infinite line in the direction of the force; the perpendicular distance is measured to that line.
In calculations, ensure $d$ is in metres to get moment in $\text{N m}$ .

Techniques used

state the definition of momentidentify the perpendicular distance from a point to a line of action

(b)

A tree of mass $270\text{ kg}$ grows out of sloping ground and is supported by a post, as shown in Fig. 2.1.

Fig. 2.1

The ground applies a total force $R$ on the tree at point $Q$ .
The centre of gravity of the tree is a horizontal distance of $1.2\text{ m}$ from $Q$ .
The post applies a force $F$ of $1800\text{ N}$ perpendicular to the line $PQ$ . The line of action of $F$ passes through point $P$ at an angle $\theta$ to the vertical. $P$ is a horizontal distance of $1.6\text{ m}$ from $Q$ .
The tree is in equilibrium and all forces act on the tree in the same plane.

7M

(i)

By taking moments about point $Q$ , show that $\theta$ is $25^\circ$ .

3M Similar

DifficultyMedium

Worked solution

Working

Taking moments about $Q$ (equilibrium):

Weight $W = mg = 270 \times 9.81 = 2.65 \times 10^3\ \text{N}$ .

Moment of weight about $Q$ :

W \times 1.2

Since $F$ is perpendicular to $PQ$ , moment of $F$ about $Q$ is

F \times PQ

and with $PQ$ making angle $\theta$ to the horizontal, its horizontal projection is $PQ\cos\theta = 1.6$ , so

PQ = \frac{1.6}{\cos\theta}.

Equating moments:

1800 \left(\frac{1.6}{\cos\theta}\right) = (270 \times 9.81)(1.2)

\cos\theta = \frac{1800 \times 1.6}{270 \times 9.81 \times 1.2} = 0.906

\theta \approx 25^\circ.

Answer

$\theta = 25^\circ$

Final answer

25°

Detailed explanation

Background Concept

For a body in rotational equilibrium, the net moment about any point is zero. This is the principle of moments:

\sum \tau_{\text{clockwise}} = \sum \tau_{\text{anticlockwise}}

The moment of a force about a point is

\tau = F \times d_\perp

where $d_\perp$ is the perpendicular distance from the point to the line of action.

A useful alternative form is

\tau = rF\sin\phi

where $r$ is the distance from the point to the point of application, and $\phi$ is the angle between $\vec{r}$ and the force.

Understanding the Question

You have a tree in equilibrium. About point $Q$ :

the weight $W = mg$ acts vertically downward through the centre of gravity, with a given perpendicular distance (horizontal) of $1.2\ \text{m}$ from $Q$ .
the post exerts a force $F = 1800\ \text{N}$ at $P$ .
you are told $F$ is perpendicular to $PQ$ , which is a key clue because it makes the moment calculation simple.
the horizontal distance from $Q$ to $P$ is $1.6\ \text{m}$ .

You must show that the angle $\theta$ (the angle the force makes to the vertical) is $25^\circ$ .

Approach

Take moments about $Q$ to eliminate the unknown reaction force $R$ (because its line of action passes through $Q$ , so its moment about $Q$ is zero).
Write the moment of the weight as $W \times 1.2$ .
Write the moment of $F$ as $F \times PQ$ because $F \perp PQ$ .
Use geometry to express $PQ$ in terms of the given horizontal distance $1.6\ \text{m}$ and $\theta$ .
Solve for $\theta$ .

Step-by-Step Reasoning

diagram-b1-geometry

Moment of the reaction at $Q$ : zero, because it acts at $Q$ .
Moment due to weight:
- $W = mg = 270 \times 9.81 = 2.65 \times 10^3\ \text{N}$ .
- The weight acts vertically; the perpendicular distance from $Q$ to a vertical line is the horizontal distance, given as $1.2\ \text{m}$ .
- So moment magnitude is $W(1.2)$ .
Moment due to $F$ :
- You are told $F \perp PQ$ .
- Using $\tau = rF\sin\phi$ with $r = PQ$ and $\phi = 90^\circ$ ,

\tau_F = PQ \times F

Relate $PQ$ to the horizontal distance:
- The force $F$ is at angle $\theta$ to the vertical, so it is at angle $(90^\circ-\theta)$ to the horizontal.
- Since $PQ$ is perpendicular to $F$ , $PQ$ makes angle $\theta$ to the horizontal.
- The horizontal projection of $PQ$ is then $PQ\cos\theta$ , which is given as $1.6\ \text{m}$ .

PQ\cos\theta = 1.6 \quad \Rightarrow \quad PQ = \frac{1.6}{\cos\theta}

Equate clockwise and anticlockwise moments:

1800 \left(\frac{1.6}{\cos\theta}\right) = (270 \times 9.81)(1.2)

Solve for $\cos\theta$ :

\cos\theta = \frac{1800 \times 1.6}{270 \times 9.81 \times 1.2} \approx 0.906

So

\theta \approx \cos^{-1}(0.906) \approx 25^\circ.

Key Takeaways

Taking moments about a point where an unknown force acts can eliminate that force.
If a force is perpendicular to the line from pivot to point of application, the moment is simply $F \times r$ .
Use projections (here, horizontal projection) to relate an unknown length to a given one.

Common Mistakes

Using $1.6\ \text{m}$ directly as the moment arm for $F$ (it is not generally the perpendicular distance to the line of action).
Forgetting that because $F \perp PQ$ , the angle in $rF\sin\phi$ is $90^\circ$ .
Using $g = 10\ \text{m s}^{-2}$ and then getting a noticeably different angle if rounding too aggressively.

Things to Be Careful About

The $1.2\ \text{m}$ is a perpendicular distance for the weight because the weight is vertical.
Keep angles consistent: $F$ is given at $\theta$ to vertical; converting to angle to horizontal may be needed.
In equilibrium, choose the moment point strategically (here $Q$ is best).

Techniques used

apply the principle of moments about a chosen pointrelate a sloping length to its horizontal projection using trigonometryequate clockwise and anticlockwise moments in equilibrium

(ii)

On Fig. 2.2, draw a labelled scale vector triangle to represent the forces acting on the tree. The weight of the tree has been drawn to scale.

Fig. 2.2

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

In equilibrium, the three forces $W$ , $F$ and $R$ form a closed vector triangle.

From the tail of the given $W$ vector, draw $F$ to scale ( $1800\ \text{N}$ ) at $25^\circ$ to the vertical in the correct direction.

Then draw $R$ to scale to join the head of $F$ to the head of $W$ (closing the triangle), and label $F$ and $R$ .

diagram-b2-triangle

Final answer

Scale vector triangle of W, F and R drawn (closed triangle).

Detailed explanation

Background Concept

If an object is in equilibrium under three coplanar forces, then:

the resultant force is zero, and
the three force vectors can be drawn head-to-tail to form a closed triangle (often called a force triangle or vector triangle).

Mathematically:

\vec{W} + \vec{F} + \vec{R} = \vec{0}

So if you draw $\vec{W}$ , then $\vec{F}$ starting at the head of $\vec{W}$ (or vice versa), the final vector $\vec{R}$ must close the polygon.

Understanding the Question

You are told:

the tree is in equilibrium,
all forces lie in one plane,
the weight vector $W$ has already been drawn to scale on Fig. 2.2.

You must add the other two forces:

$F = 1800\ \text{N}$ at the known direction (angle $\theta = 25^\circ$ to the vertical),
$R$ (the ground reaction at $Q$ ), whose direction and magnitude are found from closing the triangle.

Approach

Use the given weight arrow as one side of the triangle.
Draw $F$ with the correct length (scale) and direction (at $25^\circ$ to vertical).
Draw the final side $R$ to close the triangle and label it.

Step-by-Step Reasoning

diagram-b2-triangle

Start with the weight vector $\vec{W}$ already drawn (downwards).
Choose an end of $\vec{W}$ $W$ (commonly the tail) and from there draw $\vec{F}$ $F$ :
- length proportional to $1800\ \text{N}$ according to the same scale as the weight
- direction at $25^\circ$ to the vertical (matching the force direction on the tree).
The remaining force $\vec{R}$ must connect the head of the last drawn vector back to the starting point to make a closed triangle.
Put arrowheads so each vector direction is clear, and label each vector ( $W$ , $F$ , $R$ ).

Key Takeaways

Three-force equilibrium in a plane implies a closed force triangle.
Scale drawing: direction is as important as magnitude.

Common Mistakes

Drawing vectors not head-to-tail (triangle does not close).
Drawing $F$ at $25^\circ$ to the horizontal instead of to the vertical.
Not labelling $F$ and $R$ or missing arrowheads.

Things to Be Careful About

Use the same scale as the given weight arrow.
The angle must be measured from the vertical line.
$R$ is not necessarily vertical; its direction comes from the triangle closure, not from guessing.

Techniques used

use the condition of equilibrium that forces form a closed triangledraw vectors head-to-tail to scaleuse a given vector scale and direction constraints to complete a force polygon

(iii)

The tree exerts a pressure of $150\text{ kPa}$ on the top of the post.

Determine the surface area of the tree in contact with the post.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Pressure:

p = \frac{F}{A}

A = \frac{F}{p} = \frac{1800}{150 \times 10^3} = 1.2 \times 10^{-2}\ \text{m}^2

Answer

A = 0.012\ \text{m}^2

Final answer

0.012 m^2

Detailed explanation

Background Concept

Pressure is defined as normal force per unit area:

p = \frac{F}{A}

where:

$p$ is pressure in pascals ( $\text{Pa} = \text{N m}^{-2}$ )
$F$ is the normal contact force in newtons ( $\text{N}$ )
$A$ is the contact area in square metres ( $\text{m}^2$ )

Understanding the Question

The tree presses on the top of the post with a pressure of $150\ \text{kPa}$ . The contact force between the tree and the post is the same magnitude as the force the post exerts on the tree (Newton’s third law). From earlier information, the post’s force is $F = 1800\ \text{N}$ .

You are asked to find the surface area of contact between the tree and the post.

Approach

Convert $150\ \text{kPa}$ into pascals.
Rearrange $p = F/A$ to $A = F/p$ .
Substitute values and calculate.

Step-by-Step Reasoning

Convert pressure:

150\ \text{kPa} = 150 \times 10^3\ \text{Pa}

Rearrange the definition:

A = \frac{F}{p}

Substitute:

A = \frac{1800}{150 \times 10^3} = 1.2 \times 10^{-2}\ \text{m}^2

So the contact area is $0.012\ \text{m}^2$ .

Key Takeaways

Use $p = F/A$ and convert $\text{kPa} \rightarrow \text{Pa}$ .
Newton’s third law links the contact forces between two objects.

Common Mistakes

Forgetting to convert $150\ \text{kPa}$ to $150000\ \text{Pa}$ .
Using the weight $mg$ instead of the contact force from the post ( $1800\ \text{N}$ ).
Giving area in $\text{cm}^2$ without converting to $\text{m}^2$ .

Things to Be Careful About

Pressure must be in $\text{Pa}$ for SI consistency.
The force used must be the normal contact force causing the pressure on the post.

Techniques used

use the definition of pressure as force per unit areaconvert kilopascals to pascalsrearrange an equation to solve for area

Q3MediumWavesSuperposition

Two progressive water waves $X$ and $Y$ travel along a straight line from point $A$ to point $B$ . The variation of displacement of the waves with distance from $A$ at an instant in time is shown in Fig. 3.1.

Fig. 3.1

(a)

State the amplitude of wave $X$ .

amplitude = ______ $\text{cm}$

1M Similar

(b)

Both waves have frequency $16\text{ Hz}$ .

3M

(i)

Determine the speed of wave $X$ .

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 3.1, $\lambda = 0.40\ \text{m}$ .

v = f\lambda

v = 16 \times 0.40 = 6.4\ \text{m s}^{-1}

Answer

$6.4\ \text{m s}^{-1}$

Final answer

6.4 m s^-1

Detailed explanation

Background Concept

For a progressive wave,

v = f\lambda

where:

$v$ is wave speed ( $\text{m s}^{-1}$ ),
$f$ is frequency ( $\text{Hz}$ ),
$\lambda$ is wavelength ( $\text{m}$ ), the distance between successive points in phase (e.g. crest-to-crest).

A displacement–distance graph at a fixed time shows the spatial pattern, so you can read $\lambda$ directly along the distance axis.

Understanding the Question

Both waves have frequency $f = 16\ \text{Hz}$ . Using Fig. 3.1, you read the wavelength of wave $X$ and then calculate its speed.

Approach

Read $\lambda$ for wave $X$ from the graph (distance between adjacent crests/troughs).
Apply $v=f\lambda$ .
Substitute and calculate, keeping SI units.

Step-by-Step Reasoning

From Fig. 3.1, wave $X$ repeats its shape every $0.40\ \text{m}$ , so:

\lambda = 0.40\ \text{m}

Use the wave equation:

v = f\lambda

Substitute $f=16\ \text{Hz}$ and $\lambda=0.40\ \text{m}$ :

v = 16 \times 0.40 = 6.4\ \text{m s}^{-1}

Key Takeaways

A displacement–distance graph lets you read wavelength directly.
Wave speed depends on both $f$ and $\lambda$ via $v=f\lambda$ .

Common Mistakes

Reading half a wavelength (crest to trough) instead of a full wavelength.
Using the amplitude values when finding speed (amplitude does not appear in $v=f\lambda$ ).

Things to Be Careful About

Frequency in $\text{Hz}$ is already $\text{s}^{-1}$ , so the units work automatically: $\text{Hz}\times\text{m}=\text{m s}^{-1}$ .
Ensure the wavelength you read is for wave $X$ (though here $X$ and $Y$ have the same $\lambda$ ).

Techniques used

read wavelength from a displacement–distance graphuse the wave equation to relate speed, frequency, and wavelengthsubstitute values with consistent units

(ii)

State and explain whether $X$ and $Y$ are coherent.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$X$ and $Y$ are coherent because they have the same frequency and a constant phase difference (they are in antiphase, i.e. phase difference $\pi$ ).

Final answer

Yes; same frequency and constant phase difference (antiphase).

Detailed explanation

Background Concept

Two waves are coherent if they maintain a constant phase difference (and, in practice for exam questions, they also have the same frequency).

“Same frequency” means the phase difference does not drift with time.
“Constant phase difference” means that if one wave is, for example, at a crest when the other is at a trough, that relationship stays the same.

Being in antiphase means a phase difference of $\pi$ radians (or $180^{\circ}$ ).

Understanding the Question

You are told both waves have frequency $16\ \text{Hz}$ , and Fig. 3.1 shows that when wave $X$ has a crest, wave $Y$ has a trough at the same position (and vice versa). You must decide whether that makes them coherent and explain.

Approach

Recall the definition: coherence requires a constant phase difference.
Use the diagram description (“in antiphase”) and the given equal frequency to justify that the phase difference is constant.

Step-by-Step Reasoning

The question states both waves have $f=16\ \text{Hz}$ , so their phase difference will not change due to different frequencies.
The diagram information states they are in antiphase, meaning:

\Delta \phi = \pi

$\Delta\phi=\pi$ is a fixed (constant) value, so the waves are coherent.

Key Takeaways

Coherence is about a stable phase relationship.
“Antiphase” is still coherent if it stays antiphase (constant $\pi$ phase difference).

Common Mistakes

Saying “not coherent because they are in antiphase” (antiphase is a constant phase difference, so it can be coherent).
Mentioning only “same frequency” but not stating “constant phase difference”.

Things to Be Careful About

Coherence does not require the waves to be in phase; any fixed phase difference (including $\pi$ ) is acceptable.
If the frequencies were different, the phase difference would change with time and they would not be coherent.

Techniques used

use the definition of coherenceidentify constant phase difference from the diagramlink equal frequency to coherence conditions

(c)

Wave $X$ and wave $Y$ superpose to form a resultant wave.

On Fig. 3.2, sketch the variation of displacement of the resultant wave with distance from $A$ at the instant of time shown in Fig. 3.1.

Fig. 3.2

2M Similar

DifficultyMedium

Worked solution

Working

Resultant displacement $= y_X + y_Y$ .

From Fig. 3.1 the waves have the same wavelength and are in antiphase, with amplitudes $A_X=10\ \text{cm}$ and $A_Y=20\ \text{cm}$ .

So the resultant is a sinusoid of wavelength $0.40\ \text{m}$ and amplitude $20-10=10\ \text{cm}$ , in phase with wave $Y$ (opposite phase to $X$ ).

resultant-wave

Answer

Sketch: sinusoidal wave, amplitude $10\ \text{cm}$ , wavelength $0.40\ \text{m}$ , with crests where $Y$ has crests (and troughs where $Y$ has troughs).

Final answer

Resultant: sinusoid, amplitude 10 cm, wavelength 0.40 m, in phase with Y (antiphase with X).

Detailed explanation

Background Concept

The principle of superposition states that when waves overlap, the resultant displacement at any point is the algebraic sum of the individual displacements:

y_R = y_1 + y_2

If two waves have the same wavelength and frequency:

In phase ( $\Delta\phi = 0$ ): amplitudes add (constructive interference).
In antiphase ( $\Delta\phi = \pi$ ): amplitudes subtract (destructive interference).

For antiphase waves of unequal amplitude $A_1$ and $A_2$ , the resultant amplitude is:

A_R = \lvert A_2 - A_1 \rvert

and the resultant oscillates in phase with the larger-amplitude wave.

Understanding the Question

Fig. 3.1 shows waves $X$ and $Y$ along the line from $A$ to $B$ at the same instant. They have:

the same wavelength ( $0.40\ \text{m}$ ),
amplitudes $A_X=10\ \text{cm}$ and $A_Y=20\ \text{cm}$ ,
antiphase relationship (crest of one aligns with trough of the other).

You must sketch, on Fig. 3.2, the resultant displacement vs distance at that same instant.

Approach

Because both waves have the same wavelength and are perfectly antiphase everywhere, one wave’s displacement is the negative of the other (scaled by amplitude).
So at each position you can add them quickly by thinking “the larger one wins by the difference in amplitudes”.
The resultant will still be a sinusoid with the same wavelength, but with reduced amplitude.

Step-by-Step Reasoning

Read amplitudes from Fig. 3.1:

A_X = 10\ \text{cm},\quad A_Y = 20\ \text{cm}

Antiphase means when $X$ is $+10\ \text{cm}$ , $Y$ is $-20\ \text{cm}$ at the same $x$ , so the sum is $-10\ \text{cm}$ . Similarly, when $X$ is $-10\ \text{cm}$ , $Y$ is $+20\ \text{cm}$ , sum is $+10\ \text{cm}$ .
Therefore the resultant is a sinusoid with amplitude:

A_R = 20 - 10 = 10\ \text{cm}

The wavelength stays the same ( $0.40\ \text{m}$ ) because you are adding two sinusoids of the same spatial period.
The sign (phase) follows wave $Y$ because it has the larger amplitude: wherever $Y$ is positive, the resultant is positive (but smaller).

resultant-wave

Key Takeaways

Superposition means add displacements point-by-point.
For equal $\lambda$ and perfect antiphase, the resultant amplitude is the difference of amplitudes.
The resultant is in phase with the wave of larger amplitude.

Common Mistakes

Adding amplitudes ( $10+20=30\ \text{cm}$ ) even though the waves are in antiphase.
Drawing a non-sinusoidal shape even though both waves are sinusoids with the same wavelength.
Keeping the phase of wave $X$ instead of matching the larger wave $Y$ .

Things to Be Careful About

Use algebraic addition: positive + negative can cancel.
Keep the same wavelength ( $0.40\ \text{m}$ ) on the sketch.
Make sure peaks/troughs line up with the correct wave (with $Y$ here).

Techniques used

add displacements algebraically to superpose wavesuse phase relationship to determine constructive or destructive interferencededuce resultant amplitude from the difference of amplitudes in antiphase

(d)

The intensity of wave $X$ is $I_X$ . The intensity of wave $Y$ is $I_Y$ .

Use Fig. 3.1 to determine the ratio $\frac{I_X}{I_Y}$ .

ratio = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 3.1: $A_X = 10\ \text{cm}$ and $A_Y = 20\ \text{cm}$ .

Intensity $I \propto A^2$ , so

\frac{I_X}{I_Y} = \frac{A_X^2}{A_Y^2} = \frac{10^2}{20^2} = \frac{1}{4}

Answer

$\dfrac{I_X}{I_Y} = \dfrac{1}{4}$

Final answer

1/4

Detailed explanation

Background Concept

For many waves (including water waves at this level), the intensity is proportional to the square of the amplitude:

I \propto A^2

So for two waves $X$ and $Y$ ,

\frac{I_X}{I_Y} = \frac{A_X^2}{A_Y^2}

This works because intensity is related to energy transferred per unit time per unit area, and wave energy typically scales with $A^2$ .

Understanding the Question

Fig. 3.1 provides the amplitudes of waves $X$ and $Y$ . You must use those to calculate the ratio $I_X/I_Y$ .

Approach

Read $A_X$ and $A_Y$ from the displacement–distance graph.
Use $I \propto A^2$ .
Form the ratio and simplify.

Step-by-Step Reasoning

From Fig. 3.1:

A_X = 10\ \text{cm},\quad A_Y = 20\ \text{cm}

Apply the proportionality:

\frac{I_X}{I_Y} = \frac{A_X^2}{A_Y^2}

Substitute:

\frac{I_X}{I_Y} = \frac{10^2}{20^2} = \frac{100}{400} = \frac{1}{4}

Key Takeaways

Intensity depends on amplitude squared, so doubling amplitude makes intensity four times larger.
Ratios are convenient because units cancel (cm is fine here).

Common Mistakes

Using $I \propto A$ instead of $I \propto A^2$ .
Using peak-to-peak displacement instead of amplitude.
Inverting the ratio (calculating $I_Y/I_X$ ).

Things to Be Careful About

You do not need to convert cm to m because you are taking a ratio; the unit cancels as long as both amplitudes use the same unit.
Make sure you square both amplitudes before simplifying.

Techniques used

read amplitudes from the graphuse intensity proportional to amplitude squaredform an intensity ratio

Q4MediumDynamicsKinematics

A small ball is dropped from rest from height $h_1$ above the ground and falls vertically downwards. The ball collides with the ground and bounces back vertically upwards, reaching a maximum height $h_2$ . Fig. 4.1 shows the ball just before and just after hitting the ground.

Fig. 4.1

The ball has mass $0.25\text{ kg}$ and is in contact with the ground for a time of $0.18\text{ s}$ .
Just before the ball hits the ground, it has speed $5.2\text{ m s}^{-1}$ . Just after it leaves the ground, it has speed $3.6\text{ m s}^{-1}$ .
Air resistance acting on the ball is negligible.

(a)

State and explain whether the collision is elastic or inelastic.

1M Similar

DifficultyEasy

Worked solution

Answer

The collision is inelastic because the speed (and hence kinetic energy) after the bounce is less than before, so kinetic energy is not conserved.

Final answer

Inelastic (kinetic energy not conserved).

Detailed explanation

Background Concept

An elastic collision is one in which the total kinetic energy of the colliding bodies is conserved (as well as momentum).

An inelastic collision is one in which kinetic energy is not conserved; some kinetic energy is transformed into other forms such as internal energy (heating), sound, and deformation.

In a bounce off the ground, momentum of the ball changes because an impulse acts during the contact time. Kinetic energy may or may not be conserved; that decides elastic vs inelastic.

Understanding the Question

We are told the ball hits the ground with speed $5.2\,\text{m s}^{-1}$ and leaves the ground with speed $3.6\,\text{m s}^{-1}$ . We must state whether the collision is elastic or inelastic and explain using these speeds.

Approach

Compare the kinetic energy just before and just after impact. Since $E_k = \frac{1}{2}mv^2$ and the mass is unchanged, comparing $v^2$ is enough.

Step-by-Step Reasoning

Before collision:

E_{k1} = \frac{1}{2}m(5.2)^2

After collision:

E_{k2} = \frac{1}{2}m(3.6)^2

Because $(3.6)^2 < (5.2)^2$ , we have $E_{k2} < E_{k1}$ . Therefore kinetic energy decreased during the collision, so it was not conserved.

Hence the collision is inelastic.

Key Takeaways

Elastic: kinetic energy conserved.
Inelastic: kinetic energy not conserved (even though momentum changes due to impulse).
If the rebound speed is smaller than the impact speed, the collision must be inelastic.

Common Mistakes

Saying “momentum is not conserved so it is inelastic”: momentum of the ball alone is not conserved because an external force (ground) acts, but that does not define elastic/inelastic.
Confusing “inelastic” with “completely inelastic” (objects stick together). Here the ball rebounds, so it is not completely inelastic.

Things to Be Careful About

Use kinetic energy (or speed) to classify the collision, not momentum.
Speeds are magnitudes; the direction change does not by itself indicate elasticity.

Techniques used

compare kinetic energy before and after a collisionuse the definition of an elastic collision (kinetic energy conserved)

(b)

4M

(i)

Calculate the change in momentum of the ball during the collision with the ground.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Take upward as positive.

Before collision: $u = -5.2\,\text{m s}^{-1}$

After collision: $v = +3.6\,\text{m s}^{-1}$

\Delta p = m(v-u) = 0.25\bigl(3.6-(-5.2)\bigr)=0.25\times 8.8 = 2.2\,\text{kg m s}^{-1}

Answer

$2.2\,\text{kg m s}^{-1}$ (upwards)

Final answer

2.2 kg m s^-1 (upwards)

Detailed explanation

Background Concept

Linear momentum is

p = mv

where $m$ is mass and $v$ is velocity.

The change in momentum is

\Delta p = p_{\text{after}} - p_{\text{before}} = m(v-u)

Because momentum depends on velocity, direction matters: we must use a sign convention.

Understanding the Question

We are given:

$m = 0.25\,\text{kg}$
speed just before impact $= 5.2\,\text{m s}^{-1}$ downward
speed just after leaving ground $= 3.6\,\text{m s}^{-1}$ upward

We need the ball’s change in momentum during the collision.

Approach

Pick upward as positive. Convert the given speeds into signed velocities ( $u$ before, $v$ after), then use $\Delta p = m(v-u)$ .

Step-by-Step Reasoning

Choose sign convention: upward $+$ .
Write velocities:

Before collision the ball is moving downward, so

u = -5.2\,\text{m s}^{-1}

After collision it is moving upward, so

v = +3.6\,\text{m s}^{-1}

Calculate change in momentum:

\Delta p = m(v-u)

\Delta p = 0.25\bigl(3.6-(-5.2)\bigr)

The bracket is the total change in velocity: from $-5.2$ to $+3.6$ is a change of $+8.8\,\text{m s}^{-1}$ .

So

\Delta p = 0.25 \times 8.8 = 2.2\,\text{kg m s}^{-1}

The positive sign means the momentum change is upward.

Key Takeaways

Use velocity (with sign), not speed, for momentum.
A bounce reverses direction, so $v-u$ involves adding magnitudes.

Common Mistakes

Using $5.2$ and $3.6$ both as positive, giving $\Delta p = 0.25(3.6-5.2)$ which is wrong because it ignores direction.
Forgetting units: momentum change is in $\text{kg m s}^{-1}$ .

Things to Be Careful About

State the direction (or ensure your sign convention makes it clear).
Keep the same sign convention for part (ii) so the force direction is consistent.

Techniques used

choose a sign convention for vertical directionuse $\Delta p = m(v-u)$ to find change in momentuminterpret the sign/direction of momentum change

(ii)

Determine the average force on the ball during the collision with the ground.

force = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{2.2}{0.18} = 12.2\,\text{N}

Answer

$12\,\text{N}$ upwards

Final answer

12 N upwards

Detailed explanation

Background Concept

Impulse is the change in momentum:

\text{impulse} = F_{\text{avg}}\Delta t = \Delta p

So the average resultant force over the contact time is

F_{\text{avg}} = \frac{\Delta p}{\Delta t}

Understanding the Question

During the $0.18\,\text{s}$ that the ball is in contact with the ground, its momentum changes by $\Delta p$ (found in part (i) as $2.2\,\text{kg m s}^{-1}$ upward). We are asked for the average force on the ball during this collision.

Approach

Use the impulse relation $F_{\text{avg}} = \Delta p/\Delta t$ with the momentum change from (i) and the given contact time.

Step-by-Step Reasoning

From (i):

\Delta p = 2.2\,\text{kg m s}^{-1}

Contact time:

\Delta t = 0.18\,\text{s}

Average force:

F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{2.2}{0.18} = 12.2\,\text{N}

This is upward (same direction as $\Delta p$ ).

To appropriate significant figures, $F_{\text{avg}} \approx 12\,\text{N}$ upward.

Key Takeaways

Average force during a collision is found from impulse: $F_{\text{avg}} = \Delta p/\Delta t$ .
Direction of force matches direction of change of momentum.

Common Mistakes

Using $F = ma$ with $a$ estimated poorly; the question gives contact time specifically to use impulse.
Dividing the wrong way round (using $\Delta t/\Delta p$ ).

Things to Be Careful About

This value is the average resultant force on the ball during the collision, based on momentum change.
If a question specifically asked for the average contact force from the ground, you would need to consider weight as well (since $N - mg = \Delta p/\Delta t$ ). Here the standard impulse calculation gives the required average force.

Techniques used

use impulse relation $F_{\text{avg}} = \Delta p / \Delta t$substitute values consistently with chosen directionquote the magnitude and direction of the average force

(c)

Calculate the ratio $\frac{h_2}{h_1}$ .

ratio = ______

3M Similar

DifficultyMedium

Worked solution

Working

For the fall from rest:

(5.2)^2 = 2gh_1

For the rise to rest:

(3.6)^2 = 2gh_2

Divide:

\frac{h_2}{h_1} = \frac{(3.6)^2}{(5.2)^2} = \frac{12.96}{27.04} = 0.479

Answer

$\dfrac{h_2}{h_1} = 0.48$

Final answer

0.48

Detailed explanation

Background Concept

When air resistance is negligible, a falling or rising object moves under constant acceleration due to gravity $g$ .

A useful constant-acceleration equation is

v^2 = u^2 + 2as

For vertical motion:

take upward as positive
acceleration is $a = -g$

For a drop from rest through height $h$ , the speed at the bottom satisfies

v^2 = 2gh

(similarly for an object projected upward with initial speed $u$ that rises a height $h$ until $v=0$ ).

Understanding the Question

We have two separate journeys:

The ball is dropped from rest from height $h_1$ and reaches speed $5.2\,\text{m s}^{-1}$ just before hitting the ground.
After bouncing, it leaves the ground at speed $3.6\,\text{m s}^{-1}$ upward and rises to maximum height $h_2$ where its speed becomes zero.

We must find the ratio $h_2/h_1$ .

Approach

Use $v^2 = u^2 + 2as$ for each part of the motion:

For the drop: $u=0$ , $a=g$ downward, distance $=h_1$ .
For the rise: final speed $v=0$ , acceleration $=-g$ , distance $=h_2$ .

Then divide the two equations so that $2g$ cancels, leaving a ratio in terms of the given speeds only.

Step-by-Step Reasoning

1) Falling from rest

Taking downward as the direction of acceleration for the fall (or using magnitudes):

v^2 = 2gh_1

with $v = 5.2\,\text{m s}^{-1}$ , so

(5.2)^2 = 2gh_1

2) Rising to a maximum height

At maximum height, final speed is zero. Using magnitudes again:

u^2 = 2gh_2

with upward launch speed $u = 3.6\,\text{m s}^{-1}$ , so

(3.6)^2 = 2gh_2

3) Form the ratio

Divide the second equation by the first:

\frac{(3.6)^2}{(5.2)^2} = \frac{2gh_2}{2gh_1} = \frac{h_2}{h_1}

Compute:

\frac{h_2}{h_1} = \frac{12.96}{27.04} = 0.479 \approx 0.48

Key Takeaways

For motion under gravity with negligible air resistance: $v^2 = 2gh$ links speed and vertical height change.
Forming ratios can cancel constants like $g$ .
Because $h \propto v^2$ , halving the speed would quarter the height.

Common Mistakes

Using $h_2/h_1 = 3.6/5.2$ instead of squaring the speeds.
Mixing up which speed corresponds to which height.
Forgetting that at the top of the rebound the speed is $0$ .

Things to Be Careful About

You can use either a consistent sign convention or use the magnitude forms carefully; both must reflect constant acceleration.
The ratio is dimensionless, so no units are included in the final answer.
Round appropriately (typically 2 s.f. here): $0.48$ .

Techniques used

apply $v^2 = u^2 + 2as$ for vertical motion under gravityrelate impact and rebound speeds to drop and rise heightsform a ratio to eliminate $g$

Q5Medium-EasyDeformation of SolidsWork, Energy and Power

(a)

Define the Young modulus.

1M Similar

DifficultyEasy

Worked solution

Answer

Young modulus is the ratio of tensile stress to tensile strain (within the limit of proportionality / elastic limit):

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

Final answer

Young modulus is the ratio of tensile stress to tensile strain (within the elastic region).

Detailed explanation

Background Concept

Young modulus (Young’s modulus), usually written as $E$ , quantifies how stiff a material is when it is stretched.

It is defined using:

Tensile stress:

\sigma = \frac{F}{A}

where $F$ is tensile force and $A$ is cross-sectional area.

Tensile strain:

\varepsilon = \frac{\Delta L}{L}

where $\Delta L$ is extension and $L$ is original length.

Young modulus is then:

E = \frac{\sigma}{\varepsilon}

This definition is only valid in the region where stress is proportional to strain (Hooke’s law region), i.e. before plastic deformation.

Understanding the Question

The question asks you to define Young modulus, so you need a clear statement involving stress and strain, and ideally the condition that it applies in the elastic/proportional region.

Approach

Recall the definition: Young modulus is stress divided by strain. Mention the “elastic / proportional” condition because outside that region the ratio is not constant.

Step-by-Step Reasoning

Start from the basic definition of Young modulus as a material property in extension.
Use the standard relationship:

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

State the condition: it is taken for elastic deformation (up to limit of proportionality).

Key Takeaways

Young modulus measures stiffness in tension/compression.
Definition: $E = \sigma/\varepsilon$ .
Only constant in the Hooke’s law (linear elastic) region.

Common Mistakes

Writing $E = \frac{\text{strain}}{\text{stress}}$ (inverted).
Forgetting to mention it applies within the elastic / proportional region.
Confusing Young modulus with spring constant $k$ (spring constant depends on the object’s dimensions; $E$ is a material property).

Things to Be Careful About

Stress uses original cross-sectional area $A$ .
Strain uses original length $L$ .
Units: stress is Pa, strain is dimensionless, so $E$ is in Pa.

Techniques used

state the definition of Young modulusrelate stress and strain using their definitions

(b)

A wire of unstretched length $0.81\text{ m}$ is made of a metal with Young modulus $95\text{ GPa}$ . The wire obeys Hooke’s law and has a constant cross-sectional area. Fig. 5.1 shows the force–extension graph for the wire.

Fig. 5.1

6M

(i)

Determine the cross-sectional area of the wire.

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

3M Similar

DifficultyMedium-Easy

Worked solution

Working

From the graph, gradient

\frac{F}{x} = \frac{500}{4.0 \times 10^{-3}} = 1.25 \times 10^{5}\ \text{N m}^{-1}

For a wire obeying Hooke’s law:

F = \frac{EA}{L}x \;\;\Rightarrow\;\; \frac{F}{x} = \frac{EA}{L}

So

A = \frac{L}{E}\left(\frac{F}{x}\right)= \frac{0.81}{95 \times 10^{9}}\left(1.25 \times 10^{5}\right) = 1.07 \times 10^{-6}\ \text{m}^2

Answer

A \approx 1.1 \times 10^{-6}\ \text{m}^2

Final answer

1.1 × 10⁻⁶ m²

Detailed explanation

Background Concept

For a uniform wire of original length $L$ and cross-sectional area $A$ , the Young modulus $E$ relates force and extension (in the elastic, Hooke’s law region) by:

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

Rearranging gives the useful working form:

F = \frac{EA}{L}x

So if you plot $F$ against $x$ , the gradient is:

\text{gradient} = \frac{F}{x} = \frac{EA}{L}

Understanding the Question

You are given:

$L = 0.81\ \text{m}$
$E = 95\ \text{GPa} = 95 \times 10^{9}\ \text{Pa}$
A straight-line force–extension graph passing through the origin and the point $(x, F) = (4.0 \times 10^{-3}\ \text{m},\ 500\ \text{N})$ .

You are asked to find the cross-sectional area $A$ of the wire.

Approach

Use the straight-line graph to find the gradient $F/x$ .
Use the relationship $\frac{F}{x} = \frac{EA}{L}$ .
Rearrange to solve for $A$ .

Step-by-Step Reasoning

Read one convenient point on the line (since it passes through the origin, one point is enough):
- $F = 500\ \text{N}$ at $x = 4.0 \times 10^{-3}\ \text{m}$ .
Compute the gradient:

\frac{F}{x} = \frac{500}{4.0 \times 10^{-3}} = 1.25 \times 10^{5}\ \text{N m}^{-1}

Use the wire equation:

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

Rearrange:

A = \frac{L}{E}\left(\frac{F}{x}\right)

Substitute values:

A = \frac{0.81}{95 \times 10^{9}}\left(1.25 \times 10^{5}\right) = 1.07 \times 10^{-6}\ \text{m}^2

Quote to a sensible number of significant figures (limited by 95 GPa and 4.0):

A \approx 1.1 \times 10^{-6}\ \text{m}^2

Key Takeaways

For a Hooke’s law wire, the $F$ – $x$ graph is linear through the origin.
Gradient of $F$ vs $x$ equals $EA/L$ .
Convert $\text{GPa}$ to $\text{Pa}$ before substituting.

Common Mistakes

Forgetting to convert $95\ \text{GPa}$ to $95 \times 10^{9}\ \text{Pa}$ .
Using $x$ in “ $10^{-3}\ \text{m}$ units” incorrectly (the axis label means you must multiply by $10^{-3}$ ).
Rearranging wrongly (e.g. $A = \frac{E}{L}(F/x)$ instead of $A = \frac{L}{E}(F/x)$ ).

Things to Be Careful About

The gradient is $\Delta F / \Delta x$ ; because the line goes through the origin, $F/x$ at any point is fine.
Unit check: $\frac{F}{x}$ has units $\text{N m}^{-1}$ and $\frac{EA}{L}$ also gives $\text{N m}^{-1}$ .
Significant figures: do not overstate precision (e.g. $1.065789...$ is unnecessary).

Techniques used

find the gradient of a force-extension graphuse the wire extension relation $F = \frac{EA}{L}x$rearrange to determine cross-sectional area

(ii)

The extension of the wire is initially $2.0 \times 10^{-3}\text{ m}$ .

Determine the work done to increase the extension of the wire to $3.0 \times 10^{-3}\text{ m}$ .

work done = ______ $\text{J}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

From the straight line, $F \propto x$ so

F = \left(\frac{500}{4.0 \times 10^{-3}}\right)x = (1.25 \times 10^{5})x

At $x_1 = 2.0 \times 10^{-3}\ \text{m}$ :

F_1 = (1.25 \times 10^{5})(2.0 \times 10^{-3}) = 250\ \text{N}

At $x_2 = 3.0 \times 10^{-3}\ \text{m}$ :

F_2 = (1.25 \times 10^{5})(3.0 \times 10^{-3}) = 375\ \text{N}

Work done = area under graph between $x_1$ and $x_2$ (trapezium):

W = \frac{1}{2}(F_1 + F_2)(x_2 - x_1) = \frac{1}{2}(250+375)(1.0 \times 10^{-3}) = 0.3125\ \text{J}

Answer

W \approx 0.31\ \text{J}

Final answer

0.31 J

Detailed explanation

Background Concept

The work done by a force when it causes an extension is:

W = \int F\,\mathrm{d}x

On a force–extension graph, this integral is the area under the graph between the two extensions.

For a Hooke’s law wire, $F = kx$ (a straight line through the origin), so the area can be found using simple shapes (triangle or trapezium).

Understanding the Question

The wire is initially extended to $x_1 = 2.0 \times 10^{-3}\ \text{m}$ and then further extended to $x_2 = 3.0 \times 10^{-3}\ \text{m}$ .

You are asked for the additional work done in increasing the extension from $2.0\ \text{mm}$ to $3.0\ \text{mm}$ , i.e. the area under the $F$ – $x$ graph between those two $x$ values.

Approach

Use the straight-line graph to find the force at each extension ( $F_1$ at $x_1$ and $F_2$ at $x_2$ ).
The work between $x_1$ and $x_2$ is the area under the line segment: a trapezium of width $(x_2-x_1)$ and parallel sides $F_1$ and $F_2$ .
Compute:

W = \frac{1}{2}(F_1 + F_2)(x_2 - x_1)

Step-by-Step Reasoning

First find the constant gradient (spring constant for the whole wire):

\frac{F}{x} = \frac{500}{4.0 \times 10^{-3}} = 1.25 \times 10^{5}\ \text{N m}^{-1}

So $F = (1.25 \times 10^{5})x$ .

Force at $x_1 = 2.0 \times 10^{-3}\ \text{m}$ :

F_1 = (1.25 \times 10^{5})(2.0 \times 10^{-3}) = 250\ \text{N}

Force at $x_2 = 3.0 \times 10^{-3}\ \text{m}$ :

F_2 = (1.25 \times 10^{5})(3.0 \times 10^{-3}) = 375\ \text{N}

Work done from $x_1$ to $x_2$ is trapezium area:

W = \frac{1}{2}(F_1 + F_2)(x_2 - x_1)

Here, $x_2-x_1 = 1.0 \times 10^{-3}\ \text{m}$ , so:

W = \frac{1}{2}(250+375)(1.0 \times 10^{-3}) = 0.3125\ \text{J}

Rounded:

W \approx 0.31\ \text{J}

(Equivalent method: use $W = \frac{1}{2}k(x_2^2-x_1^2)$ .)

Key Takeaways

Work done in stretching equals area under the $F$ – $x$ graph.
Between two non-zero extensions, use a trapezium (or use $\frac{1}{2}k(x_2^2-x_1^2)$ ).
For Hooke’s law, force scales linearly with extension.

Common Mistakes

Finding the total energy at $x_2$ (area from $0$ to $x_2$ ) instead of the additional energy from $x_1$ to $x_2$ .
Using a triangle area with height $F_2$ and base $(x_2-x_1)$ (this would ignore the non-zero starting force $F_1$ ).
Forgetting that the extension axis is in $10^{-3}\ \text{m}$ .

Things to Be Careful About

Units: $\text{N} \times \text{m} = \text{J}$ .
Don’t round intermediate forces too aggressively if it affects the final rounding.
The straight-line assumption relies on “wire obeys Hooke’s law”; if it were curved, you would need a different area estimate.

Techniques used

use the area under a force-extension graph to find work doneuse proportionality of force and extension for a Hooke's law wirecalculate the area of a trapezium between two extensions

Q6MediumElectricityD.C. Circuits

(a)

Define electric potential difference across a component.

1M Similar

(b)

A circuit contains four resistors and a battery of electromotive force (e.m.f.) $8.0\text{ V}$ with negligible internal resistance. When the variable resistor has resistance $R$ , the currents in the circuit are $0.030\text{ A}$ , $I_1$ and $I_2$ , as shown in Fig. 6.1.

Fig. 6.1

7M

(i)

Determine the charge passing through the battery in a time of $4.0\text{ minutes}$ .

charge = ______ $\text{C}$

2M Similar

(ii)

Calculate $I_1$ .

$I_1$ = ______ $\text{A}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Resistance in $I_1$ branch:

R_{\text{branch}} = 430 + 240 = 670\ \Omega

p.d. across branch $= 8.0\ \text{V}$ :

I_1 = \frac{V}{R} = \frac{8.0}{670} = 1.19 \times 10^{-2}\ \text{A}

Answer

$1.2 \times 10^{-2}\ \text{A}$

Final answer

1.2 × 10^-2 A

Detailed explanation

Background Concept

Resistors in series add: $R_{\text{series}} = R_1 + R_2$ .
In a parallel circuit, each branch has the same potential difference as the supply (here the battery has negligible internal resistance, so the full $8.0\ \text{V}$ is across each branch).
Ohm’s law:

V = IR

Understanding the Question

You need the current $I_1$ in the branch that contains the $430\ \Omega$ and $240\ \Omega$ resistors in series. That whole branch is connected directly across the $8.0\ \text{V}$ battery.

Approach

Add the series resistances in the $I_1$ branch.
Use $I = V/R$ with $V = 8.0\ \text{V}$ for that branch.

Step-by-Step Reasoning

Combine the series resistors:

R_{I_1} = 430 + 240 = 670\ \Omega

Apply Ohm’s law to the branch:

I_1 = \frac{V}{R_{I_1}} = \frac{8.0}{670} = 0.01194\ \text{A}

Rounded appropriately:

I_1 \approx 1.2 \times 10^{-2}\ \text{A}

Key Takeaways

Series resistors: add them.
Parallel branches: same p.d. across each branch.

Common Mistakes

Treating the two branches as series with each other and adding all four resistors.
Using the total current $0.030\ \text{A}$ in $V = IR$ for this branch.

Things to Be Careful About

The $8.0\ \text{V}$ is across each branch (because the branches are in parallel).
Keep enough significant figures during intermediate steps to avoid rounding errors later.

Techniques used

combine series resistorsapply V = IR to a parallel branch

(iii)

Calculate $I_2$ .

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

1M Similar

(iv)

Determine $R$ .

$R$ = ______ $\Omega$

2M Similar

DifficultyMedium

Worked solution

Working

p.d. across $I_2$ branch $= 8.0\ \text{V}$ , so

R_{I_2} = \frac{V}{I_2} = \frac{8.0}{0.01806} = 4.43 \times 10^{2}\ \Omega

But

R_{I_2} = 210 + R

So

R = 4.43 \times 10^{2} - 210 = 2.33 \times 10^{2}\ \Omega

Answer

$2.3 \times 10^{2}\ \Omega$

Final answer

2.3 × 10^2 Ω

Detailed explanation

Background Concept

In a parallel circuit, each branch has the same potential difference as the supply. For any branch:

R_{\text{branch}} = \frac{V}{I_{\text{branch}}}

If the branch contains series resistors, their resistances add.

Understanding the Question

The $I_2$ branch contains a $210\ \Omega$ resistor in series with the variable resistor of resistance $R$ . You already found the current in that branch $I_2$ in part (iii). Since the battery has negligible internal resistance, the full $8.0\ \text{V}$ is across that branch.

Approach

Use $R_{I_2} = V/I_2$ to find the total resistance of the $I_2$ branch.
Subtract $210\ \Omega$ to obtain the variable resistor resistance $R$ .

Step-by-Step Reasoning

Total resistance of the $I_2$ branch:

R_{I_2} = \frac{8.0}{I_2}

Using a more accurate $I_2$ (to reduce rounding error), e.g. $I_2 = 0.01806\ \text{A}$ :

R_{I_2} = \frac{8.0}{0.01806} = 443\ \Omega

Since the branch resistors are in series:

R_{I_2} = 210 + R

So

R = 443 - 210 = 233\ \Omega \approx 2.3 \times 10^{2}\ \Omega

Key Takeaways

Same p.d. across parallel branches lets you find branch resistance from $V/I$ .
Series resistors add; then isolate the unknown by subtraction.

Common Mistakes

Using the total circuit current $0.030\ \text{A}$ instead of the branch current $I_2$ .
Adding $210\ \Omega$ instead of subtracting it.

Things to Be Careful About

Keep extra significant figures for $I_1$ and $I_2$ before the final rounding, because $R$ depends sensitively on $I_2$ .
Final answer should be sensible: it must be positive and of the same order as the other resistors in the circuit.

Techniques used

apply V = IR to find branch resistancecombine series resistors by additionrearrange an equation to isolate an unknown resistance

(c)

The variable resistor in (b) is fitted with a scale so that its resistance can be accurately determined.
The resistor of resistance $240\ \Omega$ is now replaced by a new resistor $X$ of unknown resistance. A galvanometer is connected as shown in Fig. 6.2.

Fig. 6.2

With reference to ratios of resistances, explain how this circuit can be used to determine the resistance of $X$ .

2M Similar

DifficultyMedium

Worked solution

Answer

Adjust the variable resistor until the galvanometer shows zero (no current).
At balance the midpoints are at the same potential, so the potential-divider ratios are equal:

\frac{210}{R} = \frac{430}{X}

Hence

X = \frac{430}{210}R

Read $R$ from the variable-resistor scale and calculate $X$ .

Final answer

Balance bridge (galvanometer zero), then use 210/R = 430/X so X = (430/210)R.

Detailed explanation

Background Concept

This is a bridge (Wheatstone-bridge type) null method. Each side branch acts as a potential divider. If the galvanometer connects between the midpoints, it measures the potential difference between those midpoints.

If the midpoint potentials are equal, the galvanometer p.d. is zero and no current flows through it (null condition).
For a series pair $R_1$ (top) and $R_2$ (bottom) across a supply, the p.d. across $R_1$ is a fraction of the supply:

V_{R_1} = V_{\text{s}}\frac{R_1}{R_1+R_2}

At balance, the fractions on both branches match.

diagram-c-bridge-balance

Understanding the Question

The $240\ \Omega$ resistor is replaced by an unknown resistor $X$ , and a galvanometer is connected between the junctions of the two branches. The variable resistor has a scale, so once you set it, you know its resistance accurately.

You must explain (using ratios of resistances) how to find $X$ .

Approach

Vary the variable resistor until the galvanometer reads zero.
At this null point, the midpoint potentials are equal, so the voltage-divider ratios on each side are the same.
Convert that condition into a resistance ratio and rearrange to calculate $X$ .

Step-by-Step Reasoning

Let the variable resistor be $R$ .

At balance, no current flows through the galvanometer, so the two junctions it connects are at the same potential.

Each branch is a series pair across the same supply, so the fraction of the supply voltage dropped across the top resistor in each branch must be equal:

\frac{210}{210+R} = \frac{430}{430+X}

Cross-multiplying:

210(430+X) = 430(210+R)

The $210\times 430$ terms cancel, leaving:

210X = 430R

So:

X = \frac{430}{210}R

Experimentally: adjust $R$ until the galvanometer reads zero, read $R$ from its scale, and calculate $X$ from the above ratio.

Key Takeaways

A null method uses a zero reading (galvanometer zero) to avoid needing to know the galvanometer resistance or its calibration.
Bridge balance gives a simple ratio relationship between resistances.

Common Mistakes

Thinking the galvanometer current is used directly; at balance it should be zero.
Writing an incorrect ratio (e.g. swapping top and bottom resistors in one branch without consistency).

Things to Be Careful About

The balance condition compares ratios of the two resistors in each branch in the same order (top-to-bottom).
State clearly that you adjust $R$ to get zero deflection, then read $R$ and compute $X$ ; this is what makes it a practical method.

Techniques used

use a null (zero deflection) method with a galvanometerapply the balance condition for a bridge circuit using resistance ratiosrearrange the ratio to calculate an unknown resistance

Q7Medium-EasyParticle Physics

(a)

State what is meant by a fundamental particle.

1M Similar

(b)

A nucleus $X$ has $14$ nucleons and $p$ protons. The ratio of charge to mass for nucleus $X$ is $4.1 \times 10^7\text{ C kg}^{-1}$ .

6M

(i)

Determine $p$ .

$p$ = ______

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Charge on nucleus $q = pe$ .

Mass of nucleus $m \approx 14m_n$ .

\frac{q}{m} = \frac{pe}{14m_n} = 4.1 \times 10^7\ \text{C kg}^{-1}

Using $e = 1.60 \times 10^{-19}\ \text{C}$ and $m_n = 1.67 \times 10^{-27}\ \text{kg}$ ,

p = \frac{\left(4.1 \times 10^7\right)\left(14\right)\left(1.67 \times 10^{-27}\right)}{1.60 \times 10^{-19}} \approx 6.0

Answer

$p = 6$

Final answer

6

Detailed explanation

Background Concept

For a nucleus with proton number $p$ (often written $Z$ ):

The nuclear charge is

q = pe

where $e = 1.60 \times 10^{-19}\ \text{C}$ is the elementary charge.

The nuclear mass is approximately the number of nucleons $A$ times the nucleon mass $m_n$ :

m \approx Am_n

This approximation is used because the mass of protons and neutrons are very similar and binding energy corrections are ignored at this level.

So the charge-to-mass ratio is

\frac{q}{m} \approx \frac{pe}{Am_n}

Understanding the Question

You are told nucleus $X$ has:

nucleon number $A = 14$ ,
proton number $p$ (unknown),
charge-to-mass ratio $q/m = 4.1 \times 10^7\ \text{C kg}^{-1}$ .

You must determine the integer $p$ .

Approach

Write $q = pe$ .
Approximate $m \approx 14m_n$ .
Substitute into $q/m$ and rearrange to solve for $p$ .
Your final value should be close to an integer because $p$ counts protons.

Step-by-Step Reasoning

Start from the given ratio:

\frac{q}{m} = 4.1 \times 10^7\ \text{C kg}^{-1}

Write expressions for $q$ and $m$ :

q = pe

m \approx 14m_n

Substitute:

\frac{pe}{14m_n} = 4.1 \times 10^7

Rearrange for $p$ :

p = \frac{\left(4.1 \times 10^7\right)\left(14m_n\right)}{e}

Insert constants ( $m_n = 1.67 \times 10^{-27}\ \text{kg}$ , $e = 1.60 \times 10^{-19}\ \text{C}$ ):

p = \frac{\left(4.1 \times 10^7\right)\left(14\right)\left(1.67 \times 10^{-27}\right)}{1.60 \times 10^{-19}} \approx 6.0

So $p = 6$ .

Key Takeaways

Nucleus charge $= pe$ .
Nuclear mass $\approx Am_n$ .
A calculated proton number should be an integer (within rounding).

Common Mistakes

Using $A = 14$ as the number of protons (it is total nucleons).
Forgetting to multiply by 14 for the mass.
Using the electron mass instead of nucleon mass.
Rounding to the wrong integer (always check the value is very close to an integer).

Things to Be Careful About

Keep powers of ten consistent; this calculation involves $10^{7}$ , $10^{-27}$ and $10^{-19}$ .
Use standard constants with appropriate significant figures; the final $p$ must be an integer.

Techniques used

use charge of nucleus as integer multiples of the elementary chargeapproximate nuclear mass using nucleon number times nucleon massrearrange a ratio equation to solve for an unknown integer

(ii)

Nucleus $X$ undergoes $\beta^-$ decay to form nucleus $Z$ .

Complete the equation representing this decay.

{}^{14}_{\dots\dots}X \rightarrow {}^{\dots\dots}_{\dots\dots}Z + {}^{\dots\dots}_{\dots\dots}\dots\dots + {}^{\dots\dots}_{\dots\dots}\dots\dots

3M Similar

DifficultyMedium-Easy

Worked solution

Working

In $\beta^-$ decay: nucleon number unchanged, proton number increases by $1$ .

{}^{14}_{6}X \rightarrow {}^{14}_{7}Z + {}^{0}_{-1}e^- + {}^{0}_{0}\bar{\nu}_e

Answer

{}^{14}_{6}X \rightarrow {}^{14}_{7}Z + {}^{0}_{-1}e^- + {}^{0}_{0}\bar{\nu}_e

Final answer

${}^{14}_{6}X \rightarrow {}^{14}_{7}Z + {}^{0}_{-1}e^- + {}^{0}_{0}\bar{\nu}_e$

Detailed explanation

Background Concept

In nuclear equations, two key quantities are conserved:

Nucleon number $A$ (total protons + neutrons)
Charge / proton number $Z$

For $\beta^-$ decay:

A neutron in the nucleus changes into a proton.
The nucleus emits an electron (the $\beta^-$ particle) and an electron antineutrino.

Symbolically:

n \rightarrow p + e^- + \bar{\nu}_e

Consequences for the nucleus:

$A$ stays the same (still the same total number of nucleons)
$Z$ increases by 1 (one more proton)

Understanding the Question

You found in part (b)(i) that nucleus $X$ has $A = 14$ and $Z = p = 6$ .

The question says nucleus $X$ undergoes $\beta^-$ decay to form nucleus $Z$ (new nucleus, labelled $Z$ ). You must complete the nuclear equation by filling in:

the missing proton numbers and nucleon numbers,
the identities of the emitted particles.

Approach

Write $X$ in nuclide notation using $A = 14$ and proton number $6$ .
For $\beta^-$ decay, increase proton number by 1 and keep nucleon number unchanged for the daughter nucleus.
Add the emitted electron and antineutrino with correct nuclide notation.
Check conservation of $A$ and $Z$ on both sides.

Step-by-Step Reasoning

Start with $X$ :

{}^{14}_{6}X

After $\beta^-$ decay:

nucleon number stays $14$ ,
proton number becomes $6 + 1 = 7$ ,
so daughter nucleus is

{}^{14}_{7}Z

The emitted $\beta^-$ particle is an electron:

{}^{0}_{-1}e^-

and the other emitted particle is an electron antineutrino:

{}^{0}_{0}\bar{\nu}_e

So the completed equation is:

{}^{14}_{6}X \rightarrow {}^{14}_{7}Z + {}^{0}_{-1}e^- + {}^{0}_{0}\bar{\nu}_e

Check conservation:

Nucleon number: $14 = 14 + 0 + 0$
Proton number: $6 = 7 + (-1) + 0$

Both are conserved, so the equation is correct.

Key Takeaways

In $\beta^-$ decay: $A$ unchanged, $Z$ increases by 1.
Always include the (anti)neutrino in $\beta$ decay equations.
Verify by checking conservation of nucleon number and charge.

Common Mistakes

Writing $\beta^-$ as ${}^{0}_{+1}e$ (wrong sign).
Omitting the antineutrino.
Changing the nucleon number (it does not change in $\beta$ decay).
Increasing $Z$ in $\beta^+$ decay (it actually decreases by 1 for $\beta^+$ ).

Things to Be Careful About

The nuclide notation for an electron is ${}^{0}_{-1}e^-$ .
The antineutrino has ${}^{0}_{0}\bar{\nu}_e$ (zero nucleon number and zero charge).
Use the proton number you found in part (b)(i) for $X$ .

Techniques used

apply conservation of nucleon number in nuclear equationsapply conservation of charge (proton number) in nuclear equationsidentify the particles emitted in beta-minus decay

(c)

A sample of a radioactive substance emits particles that are positively charged and have a continuous range of kinetic energies.

State and explain whether the nuclei in the sample are undergoing $\alpha$ -decay, $\beta^+$ decay or $\beta^-$ decay.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\beta^+$ decay.

The emitted particles are positively charged, so they are not $\beta^-$ particles. $\alpha$ particles are positively charged but have discrete energies, whereas a continuous range of kinetic energies indicates $\beta$ decay (energy shared with a neutrino). Hence $\beta^+$ .

Final answer

β⁺ decay

Detailed explanation

Background Concept

Three common nuclear decays at AS level are:

$\alpha$ decay: emits an $\alpha$ particle (a helium nucleus ${}^{4}_{2}\text{He}$ ), charge $+2e$ . Because it is a two-body decay (daughter nucleus + $\alpha$ ), the kinetic energy of the $\alpha$ particle is typically discrete (one main value).
$\beta^-$ decay: emits an electron ${}^{0}_{-1}e^-$ . The electron’s kinetic energy is continuous because the decay also emits an antineutrino, so energy is shared between particles.
$\beta^+$ decay: emits a positron ${}^{0}_{+1}e^+$ . The positron’s kinetic energy is also continuous because a neutrino is emitted and shares the energy.

So:

charge tells you the sign of the emitted particle,
continuous energy spectrum strongly suggests a $\beta$ decay (because of the neutrino).

Understanding the Question

The sample emits particles that are:

positively charged,
have a continuous range of kinetic energies.

You must decide which of $\alpha$ , $\beta^+$ , or $\beta^-$ decay is occurring, and explain using these observations.

Approach

Use the charge information to rule out $\beta^-$ (negative).
Then use the energy spectrum to decide between $\alpha$ (discrete energy) and $\beta^+$ (continuous energy).
Conclude the decay and explain both clues.

Step-by-Step Reasoning

Charge clue:
- $\beta^-$ particles are electrons with charge $-e$ , so they cannot match “positively charged”.
- That leaves either $\alpha$ (charge $+2e$ ) or $\beta^+$ (positrons, charge $+e$ ).
Energy spectrum clue:
- $\alpha$ decay produces $\alpha$ particles with (approximately) one kinetic energy because the decay is effectively two-body, so the energy is fixed by conservation laws.
- $\beta^+$ decay produces positrons with a continuous range of kinetic energies because the decay also emits a neutrino; the available decay energy is shared in varying proportions between the positron and the neutrino.

Since the particles are positively charged and have a continuous range of kinetic energies, the decay must be $\beta^+$ decay.

Key Takeaways

Positive charge eliminates $\beta^-$ .
Continuous kinetic energy spectrum is the signature of $\beta$ decay (neutrino involved).
$\alpha$ emissions are typically monoenergetic (discrete energy).

Common Mistakes

Choosing $\alpha$ decay just because the particles are positive, ignoring the continuous energy spectrum.
Saying “continuous energy means multiple decays happening” (the key reason is energy sharing with a neutrino).
Mixing up the charges of $\beta^-$ and $\beta^+$ .

Things to Be Careful About

The question says “particles ... positively charged” (not “nuclei”), so think about the emitted radiation.
Always use both pieces of evidence (charge + energy distribution) to justify the choice for full marks.

Techniques used

identify the decay type from the charge of emitted radiationuse the kinetic energy spectrum to distinguish alpha from beta decaylink continuous energy spectra to emission with a neutrino

Physics 9702/22 — May/June 2025

Answer

Background Concept

Understanding the Question

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Step-by-Step Reasoning

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