A child kicks a ball so that it leaves horizontal ground with a velocity of $28\ \text{m s}^{-1}$ at an angle of $34^{\circ}$ to the horizontal, as shown in Fig. 1.1. Air resistance is negligible. The ball leaves the ground at time $t = 0$. (i) Calculate the horizontal component $v_H$ and the vertical component $v_V$ of the velocity of the ball immediately after it has left the ground. $v_H$ = \_\_\_\_\_\_ $\text{m s}^{-1}$ $v_V$ = \_\_\_\_\_\_ $\text{m s}^{-1}$ (ii) Show…

(a)(i) v_H = 23.2 m s^-1, v_V = 15.7 m s^-1 · (a)(ii) 1.6 s · (a)(iii) Horizontal line at v_H = 23.2 m s^-1 from t = 0 to 3.2 s (label H). · (a)(iv) Straight line with gradient −9.81 from +15.7 m s^-1 at t=0 to 0 at t=1.6 s to −15.7 m s^-1 at t=3.2 s (label V). · (b)(i) p = mv · (b)(ii) 4.1 N downward · (b)(iii) 0.41 kg

A bungee jumper of mass $64\ \text{kg}$ secures one end of an elastic rope to a bridge. The other end is attached to the bungee jumper. The jumper falls from rest from the bridge and descends into the valley below, as shown in Fig. 3.1. Fig. 3.2 shows the variation of the tension $T$ in the rope with the vertical distance $h$ of the jumper below the level of the bridge. (a) The rope obeys Hooke’s law. State Hooke’s law. (i) Determine the unstretched length of the rope.…

(a) Extension is proportional to applied force, provided the elastic limit is not exceeded. · (b)(i) 40 m · (b)(ii) 24 N m^-1 · (c)(i) 75 kJ · (c)(ii) 75 kJ · (d) 0 m s^-1 (momentarily at rest)

9702/24 Oct/Nov 2025 — Physics Solved Past Paper

Q1MediumPhysical Quantities and UnitsKinematicsDynamics

A child kicks a ball so that it leaves horizontal ground with a velocity of $28\ \text{m s}^{-1}$ at an angle of $34^{\circ}$ to the horizontal, as shown in Fig. 1.1.

Fig. 1.1

Air resistance is negligible. The ball leaves the ground at time $t = 0$ .

(a)

7M

(i)

Calculate the horizontal component $v_H$ and the vertical component $v_V$ of the velocity of the ball immediately after it has left the ground.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

v_H = 28\cos 34^\circ

v_H = 23.2\ \text{m s}^{-1}

v_V = 28\sin 34^\circ

v_V = 15.7\ \text{m s}^{-1}

Answer

$v_H = 23.2\ \text{m s}^{-1}$

$v_V = 15.7\ \text{m s}^{-1}$

Final answer

v_H = 23.2 m s^-1, v_V = 15.7 m s^-1

Detailed explanation

Background Concept

A velocity is a vector, so it has both magnitude and direction. When a vector makes an angle $\theta$ to the horizontal, it can be resolved into perpendicular components:

horizontal component: $v_H = v\cos\theta$
vertical component: $v_V = v\sin\theta$

These come from right-triangle trigonometry, where the velocity vector is the hypotenuse.

Understanding the Question

The ball leaves the ground at speed $28\ \text{m s}^{-1}$ at an angle of $34^\circ$ above the horizontal. You are asked for:

the horizontal component $v_H$ immediately after launch
the vertical component $v_V$ immediately after launch

Approach

Treat the velocity vector as the hypotenuse of a right triangle.

Use cosine for the adjacent (horizontal) side.
Use sine for the opposite (vertical) side.
Then substitute the given values.

Step-by-Step Reasoning

Horizontal component:

v_H = v\cos\theta = 28\cos 34^\circ

v_H = 23.2\ \text{m s}^{-1}

Vertical component:

v_V = v\sin\theta = 28\sin 34^\circ

v_V = 15.7\ \text{m s}^{-1}

Key Takeaways

Resolve vectors using $\cos$ for the component next to the angle and $\sin$ for the component opposite the angle.
Components must be in the correct directions (horizontal and vertical here).

Common Mistakes

Swapping sine and cosine (giving $v_H = v\sin\theta$ ).
Using the wrong angle (e.g. measuring from the vertical instead of the horizontal).
Rounding too early so later parts (like time to maximum height) do not match.

Things to Be Careful About

The given angle is to the horizontal, so $v_H$ uses $\cos$ and $v_V$ uses $\sin$ .
Keep enough significant figures so that subsequent calculations (e.g. showing $t=1.6\ \text{s}$ ) work cleanly.

Techniques used

resolve a velocity vector into perpendicular componentsuse trigonometric relationships for componentssubstitute values and evaluate with appropriate significant figures

(ii)

Show that the ball reaches its maximum height at time $t = 1.6\ \text{s}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

At maximum height, $v_V = 0$ .

v_V = u_V + at

0 = 15.7 - 9.81t

t = \frac{15.7}{9.81} = 1.60\ \text{s}

Answer

$t = 1.6\ \text{s}$

Final answer

1.6 s

Detailed explanation

Background Concept

For vertical projectile motion (air resistance negligible), acceleration is constant and downward:

a = -g \approx -9.81\ \text{m s}^{-2}

Using upward as positive, the kinematics equation

v = u + at

applies, where $u$ is initial velocity component, $v$ is velocity at time $t$ .

At the maximum height of a projectile, the vertical velocity becomes zero momentarily before reversing direction.

Understanding the Question

You are told the ball is launched and air resistance is negligible. The question asks you to show that the time when it reaches maximum height is $1.6\ \text{s}$ . From part (a)(i), the initial vertical velocity is $u_V = 15.7\ \text{m s}^{-1}$ .

Approach

Consider only the vertical motion:

Initial vertical velocity $u_V$ is known.
Vertical acceleration is $a = -g$ .
At maximum height, the vertical velocity $v_V = 0$ .
Use $v = u + at$ and solve for $t$ .

Step-by-Step Reasoning

Take upward as positive.
At maximum height:

v_V = 0

Use the kinematics equation:

v_V = u_V + at

Substitute $u_V = 15.7\ \text{m s}^{-1}$ and $a = -9.81\ \text{m s}^{-2}$ :

0 = 15.7 - 9.81t

Rearrange:

9.81t = 15.7

t = \frac{15.7}{9.81} = 1.60\ \text{s}

So to 2 s.f., $t = 1.6\ \text{s}$ .

Key Takeaways

Maximum height occurs when vertical velocity is zero.
Vertical motion is uniformly accelerated with acceleration $-g$ .

Common Mistakes

Using the total speed $28\ \text{m s}^{-1}$ instead of the vertical component.
Taking $a = +g$ when upward is defined as positive.
Forgetting that only the vertical component changes; horizontal motion is separate.

Things to Be Careful About

Sign convention: if upward is positive, then $a$ must be negative.
Use a consistent value of $g$ ; small rounding differences can change the final time slightly.

Techniques used

apply constant-acceleration kinematics to vertical motionuse the condition v = 0 at maximum heightsubstitute into v = u + at

(iii)

On Fig. 1.2, sketch the variation of $v_H$ with time $t$ between $t = 0$ and $t = 3.2\ \text{s}$ . Label your line H.

Fig. 1.2

1M Similar

DifficultyMedium-Easy

Worked solution

Since air resistance is negligible, horizontal acceleration is $0$ , so $v_H$ is constant at $23.2\ \text{m s}^{-1}$ .

diagram-a-iii

Final answer

Horizontal line at v_H = 23.2 m s^-1 from t = 0 to 3.2 s (label H).

Detailed explanation

Background Concept

In projectile motion with negligible air resistance:

There is no horizontal force.
Therefore horizontal acceleration $a_H = 0$ .
So the horizontal velocity component remains constant throughout the flight.

On a velocity–time graph, constant velocity is shown by a horizontal straight line.

Understanding the Question

You must sketch $v_H$ against time $t$ from $0$ to $3.2\ \text{s}$ . From part (a)(i), $v_H = 23.2\ \text{m s}^{-1}$ immediately after launch. With no air resistance, nothing changes this horizontal component.

Approach

Start the graph at $t=0$ with $v_H = 23.2\ \text{m s}^{-1}$ .
Keep it constant up to $t=3.2\ \text{s}$ .
Label the line as H.

Step-by-Step Reasoning

Because the only significant force is weight, and weight acts vertically, the horizontal resultant force is zero:

F_H = 0 \Rightarrow a_H = 0

Hence $v_H$ does not change with time:

v_H(t) = 23.2\ \text{m s}^{-1}\ \text{for}\ 0 \le t \le 3.2\ \text{s}

So the sketch is a horizontal line at $+23.2\ \text{m s}^{-1}$ between $t=0$ and $t=3.2\ \text{s}$ .

diagram-a-iii

Key Takeaways

With no air resistance, horizontal velocity stays constant in projectile motion.
On a $v$ – $t$ graph, constant velocity is a horizontal straight line.

Common Mistakes

Drawing a decreasing $v_H$ (that would require air resistance).
Starting at $28\ \text{m s}^{-1}$ instead of the horizontal component.
Drawing the line through the origin (incorrect intercept).

Things to Be Careful About

The value is positive and constant for the whole time interval.
Make sure the line is clearly labelled H and spans the full time to $3.2\ \text{s}$ .

Techniques used

identify constant horizontal velocity when horizontal acceleration is zerosketch a velocity–time graph with correct intercept and gradientlabel the required line on the graph

(iv)

On Fig. 1.2, sketch the variation of $v_V$ with time $t$ between $t = 0$ and $t = 3.2\ \text{s}$ . Assume that velocity in the upward direction is positive. Label your line V.

3M Similar

DifficultyMedium

Worked solution

Vertical acceleration $a = -9.81\ \text{m s}^{-2}$ .
Initial $v_V = +15.7\ \text{m s}^{-1}$ at $t=0$ .
At maximum height $v_V = 0$ at $t=1.6\ \text{s}$ .
At $t=3.2\ \text{s}$ , $v_V = 15.7 - 9.81(3.2) = -15.7\ \text{m s}^{-1}$ .

diagram-a-iv

Final answer

Straight line with gradient −9.81 from +15.7 m s^-1 at t=0 to 0 at t=1.6 s to −15.7 m s^-1 at t=3.2 s (label V).

Detailed explanation

Background Concept

In projectile motion without air resistance, the only acceleration is due to gravity, which is constant and downward:

a_V = -g

If upward is defined as positive, then $g$ is negative in the kinematics equations.

The vertical velocity changes uniformly with time:

v_V = u_V + a_V t = u_V - gt

On a velocity–time graph, uniform acceleration corresponds to a straight line whose gradient equals the acceleration.

Understanding the Question

You must sketch $v_V$ against time from $0$ to $3.2\ \text{s}$ , taking upward velocity as positive. From earlier parts:

initial vertical velocity $u_V = +15.7\ \text{m s}^{-1}$ at $t=0$
at maximum height (at $t=1.6\ \text{s}$ ), $v_V = 0$
Because the ball lands at $t=3.2\ \text{s}$ (same height as launch), the vertical velocity at landing should be the negative of the initial vertical velocity.

Approach

Use the fact that acceleration is constant:

The graph must be a straight line.
Its gradient is $-g$ .
Mark key points: $(0, +15.7)$ , $(1.6, 0)$ , and $(3.2, -15.7)$ , then draw the straight line through them and label it V.

Step-by-Step Reasoning

Start with the vertical kinematics equation:

v_V = u_V + a_V t

Here $u_V = +15.7\ \text{m s}^{-1}$ and $a_V = -9.81\ \text{m s}^{-2}$ , so:

v_V = 15.7 - 9.81t

Key points:

At $t=0$ :

v_V = 15.7\ \text{m s}^{-1}

At maximum height, $v_V = 0$ occurs when $t = 1.6\ \text{s}$ (already shown).
At $t=3.2\ \text{s}$ :

v_V = 15.7 - 9.81(3.2) = 15.7 - 31.4 = -15.7\ \text{m s}^{-1}

So the sketch is a straight line decreasing from $+15.7$ at $t=0$ , crossing $0$ at $t=1.6$ , and reaching $-15.7$ at $t=3.2$ .

diagram-a-iv

Key Takeaways

Vertical velocity changes uniformly with time because acceleration is constant.
The gradient of a $v$ – $t$ graph equals acceleration.
At the same launch and landing height (no air resistance), the landing vertical velocity is equal in magnitude and opposite in direction to the initial vertical velocity.

Common Mistakes

Drawing a curve instead of a straight line (would imply changing acceleration).
Keeping $v_V$ positive for the whole motion (forgetting it becomes downward after the peak).
Using $+g$ as the gradient even though upward is defined as positive.

Things to Be Careful About

Sign convention: upward positive means the line must slope downwards (negative gradient).
Make sure the line is correctly labelled V and extends exactly from $t=0$ to $t=3.2\ \text{s}$ .

Techniques used

use v = u + at for vertical motion with constant accelerationdetermine the gradient of a velocity–time graph from accelerationidentify key points (initial value, zero crossing, final value) to sketch a straight line

(b)

The total change in momentum of the ball between leaving the ground at $t = 0$ and landing on the ground at $t = 3.2\ \text{s}$ is $13\ \text{kg m s}^{-1}$ .

4M

(i)

Define momentum.

1M Similar

(ii)

Calculate the force that acts on the ball while it is in the air.

force = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Average force:

F = \frac{\Delta p}{\Delta t} = \frac{13}{3.2}

F = 4.06\ \text{N}

(downward)

Answer

$4.1\ \text{N}$ downward

Final answer

4.1 N downward

Detailed explanation

Background Concept

Newton’s second law can be written in momentum form:

F = \frac{\Delta p}{\Delta t}

This gives the resultant (average) force over a time interval if the change in momentum over that interval is known.

If air resistance is negligible for a projectile, the only force while in the air is the weight $mg$ , acting vertically downward. So the change in momentum is consistent with a downward force.

Understanding the Question

You are told the total change in momentum of the ball from $t=0$ (leaving the ground) to $t=3.2\ \text{s}$ (landing) is $13\ \text{kg m s}^{-1}$ . You are asked to calculate the force acting on the ball while in the air.

Because air resistance is negligible, this force is essentially the weight, and it acts downward.

Approach

Use the momentum form of Newton’s second law to find the average force:

Take $\Delta p = 13\ \text{kg m s}^{-1}$ .
Take $\Delta t = 3.2\ \text{s}$ .
Compute $F = \Delta p/\Delta t$ and state the direction.

Step-by-Step Reasoning

Apply:

F = \frac{\Delta p}{\Delta t}

Substitute:

F = \frac{13}{3.2} = 4.06\ \text{N}

The momentum change is due to gravity, so the resultant force is downward.
Rounding suitably gives $4.1\ \text{N}$ downward.

Key Takeaways

Resultant force equals rate of change of momentum.
With negligible air resistance, the force on the projectile in flight is its weight.

Common Mistakes

Using $F = ma$ directly without having $m$ (this part is designed to avoid needing $m$ yet).
Forgetting units: $\text{kg m s}^{-1}$ divided by $\text{s}$ gives $\text{N}$ .
Not stating direction when the force is clearly vertical.

Things to Be Careful About

This calculation gives the average resultant force over the whole flight time.
Ensure you divide by $3.2\ \text{s}$ (not $1.6\ \text{s}$ ).

Techniques used

use average force as rate of change of momentumsubstitute change in momentum and time intervalstate the direction of the resultant force from the momentum change

(iii)

Determine the mass of the ball.

mass = ______ $\text{kg}$

1M Similar

Q2MediumForces, Density and Pressure

Fig. 2.1 shows a square metal sheet of non-uniform density, with a thin wooden rod fixed at its centre. One of the corners of the sheet is labelled X.

Fig. 2.1

The rod has negligible mass. The mass of the metal sheet is $2.8\ \text{kg}$ .
The rod is supported so that the rod is horizontal and the metal sheet is vertical.

(a)

Define the torque of a couple.

2M Similar

DifficultyEasy

Worked solution

Answer

A couple consists of two equal and opposite parallel forces separated by a perpendicular distance.

The torque (moment) of the couple is

\tau = Fd

where $d$ is the perpendicular separation of the forces (independent of the point about which moments are taken).

Final answer

$\tau = Fd$ for two equal and opposite parallel forces separated by perpendicular distance $d$ .

Detailed explanation

Background Concept

A couple is a special kind of force system: two forces of equal magnitude, opposite direction, and parallel lines of action, separated by a perpendicular distance. The net force is zero, so a couple produces rotation without translation.

The torque (moment) of a couple is the turning effect produced, and its magnitude is given by

\tau = Fd

where $F$ is the magnitude of either force in the couple and $d$ is the perpendicular separation between their lines of action.

A key property is that the torque of a couple is the same about any point (because shifting the origin changes both moments equally and oppositely).

Understanding the Question

You are asked to define torque of a couple, so the answer must include:

what a couple is (equal/opposite/parallel forces, separation), and
the expression for its torque in terms of force and separation.

Approach

State the defining features of a couple, then write the standard expression for its torque using the perpendicular separation.

Step-by-Step Reasoning

Identify what makes a couple: two forces, same size, opposite directions, parallel, separated.
The turning effect is the moment of one force about the line of action of the other, giving magnitude $F \times d$ .
Emphasise that $d$ must be the perpendicular separation.

Key Takeaways

A couple produces rotation only (no resultant force).
Torque of a couple: $\tau = Fd$ , with $d$ the perpendicular separation.

Common Mistakes

Using $\tau = Fr$ without specifying $r$ is the perpendicular distance.
Forgetting to mention the forces are equal and opposite.
Giving $2Fd$ (counting both forces) instead of the correct $Fd$ .

Things to Be Careful About

Always say perpendicular separation.
Torque direction can be clockwise/anticlockwise, but for a definition the magnitude and the correct force system are essential.

Techniques used

define a couple as two equal and opposite parallel forcesstate torque of a couple as force multiplied by perpendicular separation

(b)

When the rod is supported in such a way that it can rotate freely within its support, the sheet hangs in equilibrium with point X vertically above the rod, as shown in Fig. 2.2.

Fig. 2.2

On Fig. 2.2, draw a line to indicate the range of possible positions for the centre of gravity of the metal sheet.

1M Similar

DifficultyMedium-Easy

Worked solution

q2b

Answer

Draw a straight line along the diagonal through $X$ and the rod (centre), extending below the rod towards the opposite corner.

Final answer

Line along the diagonal through X and the rod (towards the opposite corner).

Detailed explanation

Background Concept

For an object suspended and free to rotate, it settles in equilibrium when its centre of gravity (CoG) is vertically below the point of support (so the line of action of weight passes through the pivot). If it were not, the weight would produce a moment about the support and the object would rotate.

Understanding the Question

The sheet can rotate freely about the rod. In the shown equilibrium position (Fig. 2.2), corner $X$ is vertically above the rod. The question asks you to indicate the range of possible positions of the CoG consistent with that orientation.

Because the rod is the effective pivot, the weight must act through the rod; therefore the CoG must lie on the vertical line through the rod. Since $X$ is vertically above the rod in this equilibrium, that vertical line coincides with the diagonal of the square passing through $X$ and the centre.

Approach

Use the condition for rotational equilibrium of a hanging object:

line of action of $W=mg$ passes through the support.
Then map that onto the diagram: draw the corresponding line on the sheet.

Step-by-Step Reasoning

Pivot is at the rod through the sheet’s centre.
For no turning effect about the rod, the moment of the weight about the rod must be zero.
Moment is zero when the perpendicular distance from the rod to the line of action of weight is zero, i.e. the line of action passes through the rod.
The line of action of weight is vertical through the CoG.
Therefore the CoG must be on the vertical line through the rod.
In Fig. 2.2, the vertical line through the rod also passes through $X$ (since $X$ is vertically above the rod), so this line is exactly the diagonal from $X$ through the centre to the opposite corner.

q2b

Key Takeaways

A freely suspended object rotates until its CoG is vertically below the support.
Therefore the CoG lies somewhere on the vertical line through the support (pivot).

Common Mistakes

Drawing a horizontal line through the rod (wrong direction for the weight’s line of action).
Marking a single point instead of a line (the question asks for a range/locus).
Drawing the other diagonal (the one not passing through $X$ ).

Things to Be Careful About

The CoG must lie on the correct diagonal because the figure tells you which corner is vertically above the rod.
Even though physically stable equilibrium implies the CoG is below the rod, mark schemes for “range of possible positions” often credit the full correct line through the rod and $X$ .

Techniques used

apply the condition for rotational equilibrium about a supportuse the fact that the line of action of weight passes through the pivotidentify the locus of the centre of gravity from the hanging orientation

(c)

When a torque of $3.3\ \text{N m}$ is applied to the rod, the sheet is held in equilibrium with two of its edges horizontal, as shown in Fig. 2.3. Point X is at the top-left corner.

Fig. 2.3

2M

(i)

Explain whether the torque applied to the rod to hold the sheet in equilibrium is clockwise or anticlockwise.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The centre of gravity is to the right of the rod, so the weight produces a clockwise moment about the rod.

Therefore the applied torque must be anticlockwise to hold the sheet in equilibrium.

Final answer

Anticlockwise.

Detailed explanation

Background Concept

A force causes a turning effect (moment/torque) about a pivot if its line of action does not pass through the pivot. For a force $F$ acting at perpendicular distance $d$ from the pivot,

\tau = Fd

The direction of the torque (clockwise/anticlockwise in the plane of the page) depends on where the force acts relative to the pivot.

In equilibrium, the net torque about the pivot is zero, so any applied torque must balance the torque due to weight.

Understanding the Question

In Fig. 2.3 the sheet is held with edges horizontal. Because the sheet’s density is non-uniform, its CoG is not at the rod. From part (b), the CoG lies along a diagonal; in this orientation that diagonal places the CoG down-and-right of the rod. The question asks whether the external torque applied to the rod is clockwise or anticlockwise.

Approach

Decide which side of the rod the CoG lies on in Fig. 2.3.
Use the direction of the weight’s turning effect about the rod.
The applied torque must be opposite to balance it.

Step-by-Step Reasoning

In Fig. 2.3, the CoG is horizontally displaced to the right of the rod.
The weight acts vertically downward through the CoG.
A downward force acting to the right of a pivot tends to pull the right side down and lift the left side up, which is a clockwise rotation.
To keep the sheet stationary, the applied torque must oppose this clockwise moment.

Therefore, the applied torque is anticlockwise.

Key Takeaways

Determine torque direction by imagining which side would go down/up.
In equilibrium, applied torque must oppose the torque caused by weight.

Common Mistakes

Guessing clockwise/anticlockwise without linking it to where the CoG lies.
Reversing the direction because of viewing the sheet from the wrong side.

Things to Be Careful About

Always state torque direction about the pivot (the rod).
Be consistent about the viewing direction (front view of Fig. 2.3).

Techniques used

identify the direction of the moment produced by weight about a pivotuse equilibrium to relate applied torque direction to the weight moment

(ii)

Show that the centre of gravity of the sheet has a horizontal displacement of $0.12\ \text{m}$ from the rod.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For equilibrium about the rod,

\tau = mgx

x = \frac{\tau}{mg} = \frac{3.3}{2.8 \times 9.81} = 0.120\ \text{m}

Answer

Horizontal displacement $= 0.12\ \text{m}$ .

Final answer

0.12 m

Detailed explanation

Background Concept

If a rigid body is in rotational equilibrium about a pivot, the net torque about that pivot is zero. Here, the sheet experiences:

an applied torque $\tau$ from the rod, and
a torque due to its weight $W = mg$ acting at the centre of gravity.

If the horizontal (perpendicular) distance from the pivot to the line of action of the weight is $x$ , then the magnitude of the gravitational torque is

\tau_W = mgx

Understanding the Question

You are told that an applied torque of $3.3\ \text{N m}$ holds the sheet in equilibrium in the orientation of Fig. 2.3. Because the sheet is not uniform, its weight acts at a point horizontally displaced from the rod. The question asks you to show that this horizontal displacement is $0.12\ \text{m}$ .

Given:

$m = 2.8\ \text{kg}$
$\tau = 3.3\ \text{N m}$
take $g = 9.81\ \text{m s}^{-2}$

Unknown: $x$ .

Approach

Equate the applied torque to the gravitational moment about the rod:

\tau = mgx

Then rearrange for $x$ and substitute values.

Step-by-Step Reasoning

Write equilibrium of moments about the rod:

\tau = mgx

Rearrange:

x = \frac{\tau}{mg}

Substitute:

x = \frac{3.3}{2.8 \times 9.81}

Calculate:

2.8 \times 9.81 = 27.468

x = \frac{3.3}{27.468} = 0.120\ \text{m} \approx 0.12\ \text{m}

Key Takeaways

Torque from weight about a pivot is $mg$ times the perpendicular distance to its line of action.
In equilibrium, applied torque balances gravitational torque.

Common Mistakes

Using the diagonal distance to the CoG instead of the perpendicular (horizontal) distance $x$ .
Using $g = 10\ \text{m s}^{-2}$ and then not matching the stated value.
Dropping units for torque or distance.

Things to Be Careful About

$x$ here is the horizontal displacement because the weight is vertical.
Quote the final value to sensible significant figures (here $0.12\ \text{m}$ ).

Techniques used

apply the moment equation for equilibrium about a pivotset applied torque equal to the moment of weightrearrange to find perpendicular displacement

(d)

The square metal sheet has an average density of $3000\ \text{kg m}^{-3}$ and a uniform thickness of $4.0\ \text{mm}$ .

Show that the side length of the sheet is $0.48\ \text{m}$ .

3M Similar

DifficultyMedium

Worked solution

Working

V = \frac{m}{\rho} = \frac{2.8}{3000} = 9.33 \times 10^{-4}\ \text{m}^3

Thickness

t = 4.0\ \text{mm} = 4.0 \times 10^{-3}\ \text{m}

Area

A = \frac{V}{t} = \frac{9.33 \times 10^{-4}}{4.0 \times 10^{-3}} = 0.233\ \text{m}^2

For a square, $A = L^2$ :

L = \sqrt{A} = \sqrt{0.233} = 0.483\ \text{m} \approx 0.48\ \text{m}

Answer

Side length $= 0.48\ \text{m}$ .

Final answer

0.48 m

Detailed explanation

Background Concept

Density is defined as

\rho = \frac{m}{V}

so the volume is

V = \frac{m}{\rho}

For a sheet of uniform thickness $t$ , the volume is also

V = At

where $A$ is the area of the face. For a square of side length $L$ ,

A = L^2

so

L = \sqrt{A}

Understanding the Question

You are given:

mass $m = 2.8\ \text{kg}$
average density $\rho = 3000\ \text{kg m}^{-3}$
thickness $t = 4.0\ \text{mm}$ (must convert to metres)

You must show the side length is $0.48\ \text{m}$ .

Approach

Use $V = m/\rho$ to find the sheet volume.
Use $A = V/t$ to find the face area.
Use $L = \sqrt{A}$ for a square.

Step-by-Step Reasoning

Find volume:

V = \frac{2.8}{3000} = 9.33 \times 10^{-4}\ \text{m}^3

Convert thickness to SI:

4.0\ \text{mm} = 4.0 \times 10^{-3}\ \text{m}

Find area:

A = \frac{V}{t} = \frac{9.33 \times 10^{-4}}{4.0 \times 10^{-3}} = 0.233\ \text{m}^2

Side length of square:

L = \sqrt{0.233} = 0.483\ \text{m}

Rounded appropriately:

L \approx 0.48\ \text{m}

Key Takeaways

Density links mass and volume.
For a uniform sheet: $V = At$ .
Square side length comes from $L = \sqrt{A}$ .

Common Mistakes

Forgetting to convert $4.0\ \text{mm}$ into metres.
Using $A = V \times t$ instead of $A = V/t$ .
Rounding too early (can distort the final square root).

Things to Be Careful About

Keep SI units throughout so that $\rho$ in $\text{kg m}^{-3}$ is consistent.
Check plausibility: half-side would be $\sim 0.24\ \text{m}$ , reasonable for a 2.8 kg metal sheet of this thickness and density.

Techniques used

use density to calculate volume from massconvert thickness to SI unitsrelate volume to area and thicknessuse geometry of a square to find side length from area

(e)

Use the answer in (b) and the information in (c) and (d) to determine the position of the centre of gravity of the sheet. Indicate this position on Fig. 2.3 with a point labelled Y.

2M Similar

DifficultyMedium

Worked solution

Working

From (b), the centre of gravity lies on the diagonal through $X$ and the rod (towards the opposite corner).

In Fig. 2.3, this diagonal is at $45^\circ$ to the horizontal, so horizontal and vertical displacements from the rod are equal.

Given the horizontal displacement is $0.12\ \text{m}$ , the centre of gravity is $0.12\ \text{m}$ to the right and $0.12\ \text{m}$ below the rod.

(With $L=0.48\ \text{m}$ , this is $L/4$ in each direction, so it lies within the sheet.)

q2e

Answer

Point $Y$ is on the diagonal from $X$ to the opposite corner, at $(+0.12\ \text{m},\,-0.12\ \text{m})$ from the rod (right and down).

Final answer

On the X-to-opposite-corner diagonal, 0.12 m right and 0.12 m down from the rod.

Detailed explanation

Background Concept

Two ideas combine here:

Hanging equilibrium: when an object is free to rotate about a pivot, it settles so that its CoG lies somewhere on the vertical line through the pivot.
Torque from weight: if, in some orientation, the object is held by an applied torque, the weight produces a moment $mgx$ where $x$ is the perpendicular distance from the pivot to the weight’s line of action.

Also, in a square, a diagonal makes $45^\circ$ to the horizontal/vertical, so moving along a diagonal changes horizontal and vertical coordinates equally in magnitude.

Understanding the Question

You must locate the CoG on Fig. 2.3 using:

(b) the CoG lies along a particular diagonal (because in the free-hanging position, that diagonal is vertical),
(c) the horizontal displacement from the rod is $0.12\ \text{m}$ in the held orientation,
(d) the side length is $0.48\ \text{m}$ , letting you judge whether the point is inside the sheet and how far it is relative to the square.

Approach

Transfer the result of (b): the CoG is on the diagonal through $X$ and the centre.
Use (c): the CoG must be $0.12\ \text{m}$ horizontally from the centre (to the right in Fig. 2.3).
Because the CoG is on a $45^\circ$ diagonal, set vertical displacement magnitude equal to horizontal displacement.
Mark the point on Fig. 2.3.

Step-by-Step Reasoning

From (b), the CoG is somewhere on the diagonal joining corner $X$ to the opposite corner, passing through the rod at the centre.
In Fig. 2.3, that diagonal slopes downwards to the right (from $X$ at top-left towards bottom-right).
From (c)(ii), the horizontal displacement of the CoG from the rod is

x = 0.12\ \text{m}

and since the torque direction showed the CoG is to the right, take $+0.12\ \text{m}$ horizontally.
4. Any point on a $45^\circ$ diagonal has equal horizontal and vertical offsets from the centre in magnitude. Therefore the vertical displacement must be $0.12\ \text{m}$ downward.

So relative to the rod (centre), the CoG is at

(\Delta x,\Delta y) = (+0.12,\,-0.12)\ \text{m}

Use (d) to check scale: side length $L=0.48\ \text{m}$ , so from the centre to an edge is $L/2 = 0.24\ \text{m}$ . Since $0.12 < 0.24$ , the point is inside the sheet and roughly halfway from centre to the right-hand edge and halfway from centre to the bottom edge.

q2e

Key Takeaways

The CoG must lie on a line fixed by the hanging equilibrium orientation.
The moment from weight gives a perpendicular (here horizontal) displacement.
On a square diagonal, horizontal and vertical displacements are equal.

Common Mistakes

Placing the CoG on the wrong diagonal (not the one through $X$ ).
Using the diagonal distance from the centre instead of the horizontal displacement.
Putting $Y$ above the centre: the CoG must be below the pivot for stable hanging.

Things to Be Careful About

The $0.12\ \text{m}$ from (c)(ii) is a horizontal displacement because the force is vertical.
In Fig. 2.3, “right and down” corresponds to moving along the diagonal from $X$ towards the opposite corner.
Use $L=0.48\ \text{m}$ to keep the plotted point realistic (not outside the square).

Techniques used

combine a geometric locus for the centre of gravity with a measured displacementuse 45° geometry of a square diagonal to relate horizontal and vertical componentsplace the centre of gravity relative to the rod using the sheet’s dimensions

Q3MediumDeformation of SolidsWork, Energy and Power

A bungee jumper of mass $64\ \text{kg}$ secures one end of an elastic rope to a bridge. The other end is attached to the bungee jumper. The jumper falls from rest from the bridge and descends into the valley below, as shown in Fig. 3.1.

Fig. 3.1

Fig. 3.2 shows the variation of the tension $T$ in the rope with the vertical distance $h$ of the jumper below the level of the bridge.

Fig. 3.2

(a)

The rope obeys Hooke’s law.

State Hooke’s law.

1M Similar

DifficultyEasy

Worked solution

Answer

Hooke’s law: the extension $x$ is proportional to the applied force (tension) $T$ provided the elastic limit is not exceeded.

Final answer

Extension is proportional to applied force, provided the elastic limit is not exceeded.

Detailed explanation

Background Concept

For an ideal spring (or elastic material) in its elastic region, the restoring force increases linearly with extension. This is written as

T = kx

where:

$T$ is the tension (force) in the spring/rope,
$x$ is the extension beyond the natural (unstretched) length,
$k$ is the spring constant.

The relationship only holds while the material behaves elastically (i.e. below the limit of proportionality / elastic limit). Beyond that, $T$ is no longer proportional to $x$ .

Understanding the Question

You are told the bungee rope obeys Hooke’s law, and you are asked simply to state what Hooke’s law is. To gain the mark you must include both (i) the proportionality and (ii) the condition.

Approach

Write the standard statement: “extension is proportional to force” and add the condition “provided the elastic limit is not exceeded”.

Step-by-Step Reasoning

Identify the two relevant quantities: tension $T$ and extension $x$ .
State the proportionality: $T \propto x$ .
Add the condition: only valid within the elastic region (elastic limit not exceeded).

Key Takeaways

Hooke’s law is a linear force-extension relationship.
Always state the condition (elastic limit / limit of proportionality).

Common Mistakes

Omitting the condition (elastic limit not exceeded).
Saying “length is proportional to force” instead of “extension is proportional to force”.
Mixing up proportionality and equality without defining $k$ .

Things to Be Careful About

Use “extension” (change in length), not “length”.
“Elastic limit not exceeded” (or equivalent) must be present for full credit.

Techniques used

state a proportionality law with its condition of validitylink force and extension for an elastic material

(b)

3M

(i)

Determine the unstretched length of the rope.

length = ______ $\text{m}$

1M Similar

(ii)

Determine the spring constant $k$ of the rope.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For $h > 40\ \text{m}$ , extension $x = h - 40$ .

From the graph at $h = 120\ \text{m}$ , $T \approx 1.9 \times 10^{3}\ \text{N}$ , so $x = 120 - 40 = 80\ \text{m}$ .

k = \frac{T}{x} = \frac{1.9 \times 10^{3}}{80} = 23.75\ \text{N m}^{-1}

Answer

$k \approx 24\ \text{N m}^{-1}$

Final answer

24 N m^-1

Detailed explanation

Background Concept

In the Hooke’s law region,

T = kx

So a graph of $T$ against $x$ is a straight line through the origin with gradient $k$ .

Here the graph is $T$ against $h$ (distance below the bridge), not directly against extension. The extension starts only after the rope becomes taut.

Understanding the Question

You must find the spring constant $k$ from the straight-line part of the tension-distance graph. The rope is slack up to the unstretched length; after that, any additional distance fallen produces extension.

Approach

Use part (b)(i) to identify the natural length $L_0$ .
Convert a chosen $h$ value to extension using $x = h - L_0$ .
Use $k = T/x$ (equivalently, the gradient of the straight-line section).

Step-by-Step Reasoning

From the graph, tension becomes non-zero at $h=40\ \text{m}$ . Hence $L_0 = 40\ \text{m}$ .
Pick a clear point on the straight-line region: at $h = 120\ \text{m}$ , read $T \approx 1.9 \times 10^{3}\ \text{N}$ .
Compute extension at that point:

x = h - L_0 = 120 - 40 = 80\ \text{m}

Apply Hooke’s law:

k = \frac{T}{x} = \frac{1.9 \times 10^{3}}{80} \approx 24\ \text{N m}^{-1}

(Using two points and calculating gradient gives the same result.)

Key Takeaways

The spring constant is the gradient of a $T$ vs $x$ graph.
When given $T$ vs position, you must first convert position into extension using the natural length.

Common Mistakes

Using $h$ directly as the extension (forgetting that $x = h - 40$ ).
Using the wrong part of the graph (including the flat region where $T=0$ ).
Forgetting units: $k$ is in $\text{N m}^{-1}$ .

Things to Be Careful About

Read the tension value from the graph as accurately as possible.
Use the straight-line region only (where Hooke’s law applies).

Techniques used

use the gradient of a force-extension graph to find a spring constantconvert the horizontal variable into extension beyond natural lengthuse Hooke’s law in the linear region

(c)

For the position of the bungee jumper at a distance of $120\ \text{m}$ below the bridge:

4M

(i)

show that the loss of gravitational potential energy since leaving the bridge is $75\ \text{kJ}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\Delta E_p = mgh = 64 \times 9.8 \times 120 = 7.53 \times 10^{4}\ \text{J}

7.53 \times 10^{4}\ \text{J} = 75\ \text{kJ}

Answer

$75\ \text{kJ}$

Final answer

75 kJ

Detailed explanation

Background Concept

The change in gravitational potential energy (GPE) for a mass $m$ moving vertically through height $h$ in a uniform gravitational field is

\Delta E_p = mgh

A downward movement reduces GPE; the “loss of GPE” is the magnitude $mgh$ .

Understanding the Question

At $120\ \text{m}$ below the bridge, the jumper has descended vertically by $h=120\ \text{m}$ . You are asked to show the energy lost from the gravitational store is $75\ \text{kJ}$ .

Approach

Use $\Delta E_p = mgh$ with $m=64\ \text{kg}$ , $g \approx 9.8\ \text{m s}^{-2}$ , $h=120\ \text{m}$ , then convert J to kJ.

Step-by-Step Reasoning

Substitute values:

\Delta E_p = 64 \times 9.8 \times 120

Calculate:

\Delta E_p = 7.53 \times 10^{4}\ \text{J}

Convert to kJ using $1\ \text{kJ} = 10^{3}\ \text{J}$ :

7.53 \times 10^{4}\ \text{J} = 75.3\ \text{kJ} \approx 75\ \text{kJ}

Key Takeaways

GPE change depends only on vertical displacement: $mgh$ .
“Loss” usually means the positive magnitude of the decrease.

Common Mistakes

Using $h=80\ \text{m}$ (confusing with extension) instead of the full $120\ \text{m}$ drop.
Forgetting the $10^{3}$ factor when converting J to kJ.
Using the wrong unit for $g$ .

Things to Be Careful About

The height change here is the full distance below the bridge, not the rope extension.
Quote the final result to a sensible number of significant figures consistent with the data (here, $75\ \text{kJ}$ ).

Techniques used

apply the gravitational potential energy change formulasubstitute values with consistent unitsconvert joules to kilojoules

(ii)

show that the elastic potential energy in the rope is $75\ \text{kJ}$ .

2M Similar

DifficultyMedium

Worked solution

Working

At $h=120\ \text{m}$ , extension $x = 120 - 40 = 80\ \text{m}$ .

Elastic potential energy $E_e$ = area under $T$ – $x$ graph:

E_e = \frac{1}{2}Tx = \frac{1}{2} \left(1.9 \times 10^{3}\right)(80) = 7.6 \times 10^{4}\ \text{J} \approx 75\ \text{kJ}

Answer

$75\ \text{kJ}$

Final answer

75 kJ

Detailed explanation

Background Concept

Elastic potential energy stored in a spring/elastic rope equals the work done to stretch it. Work done by a variable force is the area under a force-extension graph:

E_e = \int_0^x T\,\mathrm{d}x

If Hooke’s law applies ( $T=kx$ ), the force increases linearly from $0$ to $T$ , so the area is a triangle:

E_e = \frac{1}{2}Tx = \frac{1}{2}kx^2

Understanding the Question

You must find the elastic energy stored in the rope when the jumper is $120\ \text{m}$ below the bridge. The graph gives $T$ against $h$ , and the rope only starts stretching after $h=40\ \text{m}$ .

Approach

Find the extension at $h=120\ \text{m}$ using $x = h - 40$ .
Read the corresponding tension $T$ from the graph.
Since the line is straight (Hooke’s law), use the triangle area under the $T$ vs $x$ graph: $E_e = \tfrac{1}{2}Tx$ .

Step-by-Step Reasoning

Natural length from the flat part is $40\ \text{m}$ , so at $h=120\ \text{m}$ :

x = 120 - 40 = 80\ \text{m}

From the graph at $h=120\ \text{m}$ , $T \approx 1.9 \times 10^{3}\ \text{N}$ .
Because $T$ increases linearly with $x$ (Hooke’s law), the $T$ – $x$ graph is a straight line from the origin to $(x,T)$ , so the elastic energy is the triangular area:

E_e = \frac{1}{2}Tx = \frac{1}{2} \left(1.9 \times 10^{3}\right)(80)

E_e = 7.6 \times 10^{4}\ \text{J} = 76\ \text{kJ} \approx 75\ \text{kJ}

diagram-c-ii-area

Key Takeaways

Elastic energy is the area under a force-extension graph.
When a force varies linearly, use the triangle area $\tfrac{1}{2}Fx$ .
Extension starts only after the rope becomes taut.

Common Mistakes

Using $h=120\ \text{m}$ as the extension instead of $x=80\ \text{m}$ .
Using $E_e = Tx$ (rectangle) instead of $\tfrac{1}{2}Tx$ (triangle).
Taking the area under the entire $T$ vs $h$ graph including the slack region incorrectly; the slack region contributes zero anyway because $T=0$ .

Things to Be Careful About

Make sure the base of the triangle is the extension, not the total drop.
Read $T$ from the graph at the correct $h$ value and use an appropriate number of significant figures.
Keep units consistent: $\text{N} \times \text{m} = \text{J}$ .

Techniques used

convert position into extension beyond the natural lengthcalculate elastic potential energy as area under a force-extension graphuse the triangle area formula for a linear force-extension relation

(d)

Explain what can be deduced from the information in (c) about the speed of the bungee jumper when at a distance of $120\ \text{m}$ below the bridge.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

From (c), loss of GPE $= 75\ \text{kJ}$ and elastic energy gained $= 75\ \text{kJ}$ , so no energy remains as kinetic energy.

Hence the bungee jumper’s speed at $h=120\ \text{m}$ is $0\ \text{m s}^{-1}$ (momentarily at rest).

Final answer

0 m s^-1 (momentarily at rest)

Detailed explanation

Background Concept

Energy conservation for this situation (ignoring any energy losses such as air resistance) is:

\text{loss of GPE} = \text{gain in elastic energy} + \text{gain in KE}

Kinetic energy is

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

So if the kinetic energy is zero, the speed must be zero.

Understanding the Question

In part (c), you found that by the time the jumper is $120\ \text{m}$ below the bridge, the gravitational potential energy lost is $75\ \text{kJ}$ and the elastic potential energy stored in the rope is also $75\ \text{kJ}$ . You are asked what this implies about the jumper’s speed at that point.

Approach

Use the energy balance. If all the lost GPE has been transferred into elastic energy, then there is none left to be kinetic energy. Then use $E_k = \tfrac{1}{2}mv^2$ to deduce the speed.

Step-by-Step Reasoning

Energy transferred from gravity: $75\ \text{kJ}$ .
Energy stored elastically: $75\ \text{kJ}$ .
Therefore the kinetic energy at that instant is

E_k = 75\ \text{kJ} - 75\ \text{kJ} = 0

With $E_k = \tfrac{1}{2}mv^2$ , if $E_k=0$ then $v=0$ .

Physically, this means $h=120\ \text{m}$ is the lowest point of the motion (turning point): the jumper is momentarily at rest before moving upward.

Key Takeaways

Compare energy changes to determine whether an object is moving.
Equal loss of GPE and gain of elastic energy implies zero kinetic energy at that moment.
A turning point corresponds to $v=0$ .

Common Mistakes

Saying “speed is maximum” because the drop is largest; actually at the lowest point the speed is zero.
Forgetting kinetic energy exists and concluding incorrectly without an energy balance.
Assuming air resistance must be included when the question’s deduction is clearly based on the equality in (c).

Things to Be Careful About

The conclusion $v=0$ follows if we treat the energy changes in (c) as exact (i.e. negligible losses). In a real jump, some energy would be dissipated, but the exam deduction uses the given results.
Make it clear this is an instantaneous statement: “momentarily at rest” at $120\ \text{m}$ .

Techniques used

apply conservation of mechanical energycompare energy transfers between gravitational, elastic and kinetic storesdeduce speed from kinetic energy

Q4Medium-EasySuperpositionWaves

(a)

State the principle of superposition.

2M Similar

DifficultyEasy

Worked solution

Answer

When two or more waves overlap, the resultant displacement at any point is the algebraic (vector) sum of the displacements due to the individual waves at that point (at that instant).

Final answer

When two or more waves overlap, the resultant displacement at any point is the algebraic sum of the individual displacements at that point.

Detailed explanation

Background Concept

The principle of superposition is the key rule for adding waves. For a wave quantity such as displacement (or electric field for an electromagnetic wave), the medium responds linearly, so overlapping disturbances simply add.

In symbols, if two waves produce displacements (or field values) $y_1$ and $y_2$ at the same place and time, then the resultant is

y = y_1 + y_2.

Understanding the Question

You are asked to state the superposition principle. This is a definition-mark question, so you need a precise statement using the words “resultant”, “at a point” and “algebraic sum”.

Approach

Recall the standard definition used for interference and stationary waves: the total displacement (or field) is the sum of the individual displacements (or fields) at the same point and at the same instant.

Step-by-Step Reasoning

Identify the relevant quantity being added: displacement (for mechanical waves) or field (for EM waves).
State that when waves overlap, you add their instantaneous values.
Emphasise that it is the algebraic sum (so signs / directions matter).

Key Takeaways

Superposition is the rule that allows interference patterns and stationary waves.
You add instantaneous displacements/fields at the same point.

Common Mistakes

Saying “amplitudes add” (it is the instantaneous displacement/field that adds, not always the amplitudes).
Missing “at that point” and “at that instant”.

Things to Be Careful About

Use “algebraic sum” (or “vector sum” where appropriate).
Keep it general: it applies to two or more waves.

Techniques used

state a physical principle as a precise definitionuse correct terminology for wave displacement and superpositiondescribe what happens at a point when waves overlap

(b)

An electromagnetic wave of wavelength $0.026\ \text{m}$ in free space is incident normally on an aluminium sheet, as shown in Fig. 4.1.

Fig. 4.1

The wave reflects at the aluminium sheet and a stationary wave is formed in the region between the transmitter and the sheet.

7M

(i)

Explain how the stationary wave, including its nodes and antinodes, is formed.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

The incident wave reflects from the aluminium sheet, producing a reflected wave of the same frequency (and wavelength) travelling in the opposite direction.

The incident and reflected waves superpose. At positions where they are always in antiphase, destructive interference occurs giving nodes of zero amplitude. At positions where they are always in phase, constructive interference occurs giving antinodes of maximum amplitude. These node and antinode positions are fixed, so a stationary wave is formed.

Final answer

Incident and reflected waves of the same frequency travel in opposite directions and superpose; fixed points of destructive interference give nodes (zero amplitude) and fixed points of constructive interference give antinodes (maximum amplitude), forming a stationary wave.

Detailed explanation

Background Concept

A stationary (standing) wave is formed when two progressive waves of the same frequency and wavelength (and usually similar amplitudes) travel in opposite directions along the same line and interfere.

Using superposition, the resultant displacement/field at each point is the sum of the two wave contributions. Because the phase difference between the two waves depends on position, some positions always cancel (nodes) and some always reinforce (antinodes).

Node: resultant amplitude is always zero (complete destructive interference at all times).
Antinode: resultant amplitude is maximum (complete constructive interference at some instants).

Understanding the Question

An electromagnetic wave is sent normally towards an aluminium sheet. Aluminium is a good conductor, so it reflects the electromagnetic wave strongly. You are asked to explain how the reflected wave combines with the incident wave to produce a stationary wave pattern between the transmitter and the sheet, and to describe nodes and antinodes.

Approach

State that reflection produces a second wave travelling back towards the transmitter.
Note that the two waves have the same $f$ and $\lambda$ but travel in opposite directions.
Apply superposition: interference produces fixed nodes (destructive) and antinodes (constructive), creating a stationary pattern.

Step-by-Step Reasoning

The transmitter produces an incident electromagnetic wave travelling towards the sheet.
At the aluminium sheet, the wave is reflected, so a second wave travels back in the opposite direction.
Both waves occupy the same region between transmitter and sheet at the same time.
By superposition, the resultant field/displacement at a point is the sum of the two contributions.
At certain positions, the waves are always $\pi$ out of phase (path difference corresponds to an odd multiple of $\frac{\lambda}{2}$ ), so the sum is always zero: these are nodes.
Halfway between nodes, the waves are in phase (path difference an integer multiple of $\lambda$ ), so the resultant amplitude is largest: these are antinodes.
Because the phase difference depends only on position, the nodes and antinodes stay at fixed positions: that fixed pattern is the stationary wave.

diagram-4b-i-1

Key Takeaways

A standing wave requires two waves of same $f$ and $\lambda$ travelling in opposite directions.
Nodes are permanent points of zero amplitude; antinodes are permanent points of maximum amplitude.

Common Mistakes

Saying “reflection changes the frequency” (it does not; the reflected wave has the same frequency).
Confusing nodes with points that are “momentarily zero” (nodes are zero at all times).
Forgetting to mention superposition/interference.

Things to Be Careful About

Use the terms constructive/destructive interference correctly.
Make it clear the pattern is fixed in space (unlike a progressive wave).

Techniques used

describe reflection producing a wave travelling in the opposite directionapply the principle of superposition to two counter-propagating wavesidentify conditions for constructive and destructive interferencerelate interference to nodes and antinodes at fixed positions

(ii)

Calculate the frequency of the electromagnetic wave.

frequency = ______ $\text{Hz}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For an electromagnetic wave in free space, $v = c = 3.00 \times 10^8\ \text{m s}^{-1}$ .

f = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{0.026} = 1.15 \times 10^{10}\ \text{Hz}

Answer

$1.15 \times 10^{10}\ \text{Hz}$

Final answer

1.15 × 10^10 Hz

Detailed explanation

Background Concept

For any wave,

v = f\lambda

where $v$ is wave speed, $f$ is frequency, and $\lambda$ is wavelength. For electromagnetic waves in free space (vacuum/air to a very good approximation),

v = c = 3.00 \times 10^8\ \text{m s}^{-1}.

Understanding the Question

You are given the wavelength in free space: $\lambda = 0.026\ \text{m}$ . You are asked to find the frequency. Since it is in free space, use speed $c$ .

Approach

Rearrange $v = f\lambda$ to $f = \frac{v}{\lambda}$ , then substitute $v=c$ and the given $\lambda$ .

Step-by-Step Reasoning

Start with

v = f\lambda.

Rearrange for frequency:

f = \frac{v}{\lambda}.

Use $v=c$ for an EM wave in free space:

f = \frac{3.00 \times 10^8}{0.026}.

Calculate:

\frac{3.00}{0.026} = 115.38\ \Rightarrow\ f = 1.1538 \times 10^{10}\ \text{Hz}.

Quote to 3 s.f.:

f = 1.15 \times 10^{10}\ \text{Hz}.

Key Takeaways

For EM waves in free space, always use $c = 3.00 \times 10^8\ \text{m s}^{-1}$ .
Frequency is found from $f = c/\lambda$ .

Common Mistakes

Using $v=340\ \text{m s}^{-1}$ (that is for sound, not EM waves).
Converting $0.026\ \text{m}$ incorrectly (it is already in metres).
Forgetting standard form or missing the unit Hz.

Things to Be Careful About

Check powers of ten: dividing by $2.6 \times 10^{-2}$ increases the result by $10^2$ .
Significant figures: follow the data (here typically 2–3 s.f.).

Techniques used

use the wave equation to relate speed, frequency and wavelengthsubstitute the speed of electromagnetic waves in free spacepresent the result in standard form with correct units

(iii)

State the principal region of the electromagnetic spectrum to which the wave belongs.

1M Similar

(iv)

Determine the distance between a node and an adjacent antinode.

distance = ______ $\text{m}$

1M Similar

Q5MediumElectricityD.C. CircuitsPhysical Quantities and Units

A student uses a circuit containing an ammeter, a voltmeter and a cell to take measurements to determine the resistance of a length of nichrome wire.

(a)

3M

(i)

Define resistance.

1M Similar

(ii)

Draw a circuit diagram to show how the components should be connected. Use the symbol for a resistor to represent the nichrome wire.

2M Similar

DifficultyMedium-Easy

Worked solution

diagram-5-a-ii-1

Final answer

Ammeter in series with the nichrome wire; voltmeter in parallel across the wire.

Detailed explanation

Background Concept

To measure resistance using $R = V/I$ , you must measure:

the current $I$ through the wire (ammeter),
the potential difference $V$ across the wire (voltmeter).

Meter connection rules:

An ammeter has very low resistance, so it must be placed in series so that the same current as the wire passes through it.
A voltmeter has very high resistance, so it must be placed in parallel across the wire so it measures the p.d. across the wire without significantly changing the current.

Understanding the Question

You have a cell, an ammeter, a voltmeter, and a length of nichrome wire (to be drawn as a resistor symbol). The task is to draw the correct circuit diagram to measure $V$ across and $I$ through the wire.

Approach

Draw a simple series loop containing the cell, ammeter, and the resistor (wire). Then connect the voltmeter across the resistor only.

Step-by-Step Reasoning

diagram-5-a-ii-1

Put the ammeter in the main loop so current must pass through it and then through the nichrome wire.
Connect the voltmeter across the nichrome wire (one lead to each end of the resistor symbol).

This arrangement allows you to use $R = V/I$ with the measured values.

Key Takeaways

Ammeter: series.
Voltmeter: parallel (across the component of interest).
Use the resistor symbol to represent the nichrome wire.

Common Mistakes

Putting the voltmeter in series (would give very small current / disrupt circuit).
Putting the ammeter in parallel (could short-circuit and damage the meter).
Connecting the voltmeter across the cell instead of across the wire (measures supply voltage, not the p.d. across the wire).

Things to Be Careful About

The voltmeter must be across only the nichrome wire (resistor symbol), not across the ammeter as well.
Use correct circuit symbols and clear junctions (dots) where wires join.

Techniques used

place an ammeter in series with a componentplace a voltmeter in parallel with a componentuse standard circuit symbols to represent components

(b)

The student also measures the length and the diameter of the wire. Table 5.1 shows the measurements recorded for each quantity.

Table 5.1

quantity	measurement
length	$(0.864 \pm 0.001)\ \text{m}$
diameter	$(0.496 \pm 0.002)\ \text{mm}$
voltmeter reading	$(1.38 \pm 0.02)\ \text{V}$
ammeter reading	$(0.276 \pm 0.001)\ \text{A}$

7M

(i)

Show that the resistance of the wire is $5.00\ \Omega$ .

1M Similar

(ii)

Calculate, to three significant figures, the resistivity $\rho$ of the nichrome.

$\rho$ = ______ $\Omega\text{m}$

3M Similar

DifficultyMedium

Worked solution

Working

Diameter $d = 0.496\ \text{mm} = 4.96 \times 10^{-4}\ \text{m}$

A = \pi\left(\frac{d}{2}\right)^2 = \pi\left(2.48 \times 10^{-4}\right)^2 = 1.93 \times 10^{-7}\ \text{m}^2

Using $R = \rho L/A$ ,

\rho = \frac{RA}{L} = \frac{(5.00)(1.93 \times 10^{-7})}{0.864} = 1.12 \times 10^{-6}\ \Omega\text{m}

Answer

$\rho = 1.12 \times 10^{-6}\ \Omega\text{m}$

Final answer

1.12 × 10^-6 Ωm

Detailed explanation

Background Concept

Resistivity $\rho$ is a material property relating resistance to the dimensions of a uniform wire:

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

where:

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

Rearranging gives:

\rho = \frac{RA}{L}

For a circular wire, area is

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

Understanding the Question

Given (from the table):

$L = 0.864\ \text{m}$ ,
diameter $d = 0.496\ \text{mm}$ (must convert to m),
and from part (i) $R = 5.00\ \Omega$ .

You must calculate $\rho$ to three significant figures.

Approach

Convert $d$ from mm to m.
Use $A = \pi(d/2)^2$ .
Substitute into $\rho = RA/L$ .

Step-by-Step Reasoning

Unit conversion:

d = 0.496\ \text{mm} = 0.496 \times 10^{-3}\ \text{m} = 4.96 \times 10^{-4}\ \text{m}

Radius and area:

r = \frac{d}{2} = 2.48 \times 10^{-4}\ \text{m}

A = \pi r^2 = \pi(2.48 \times 10^{-4})^2 = 1.93 \times 10^{-7}\ \text{m}^2

Resistivity:

\rho = \frac{RA}{L} = \frac{(5.00)(1.93 \times 10^{-7})}{0.864}

\rho = 1.12 \times 10^{-6}\ \Omega\text{m}

The unit is $\Omega\text{m}$ because $R$ is $\Omega$ , multiplied by $\text{m}^2$ , divided by m.

Key Takeaways

Resistivity is found using $\rho = RA/L$ .
Always convert mm to m before using area in SI.
Diameter must be squared (through the area), which matters later for uncertainty.

Common Mistakes

Forgetting to convert $\text{mm}$ to $\text{m}$ (gives $\rho$ wrong by a factor of $10^{6}$ ).
Using $A = \pi d^2$ instead of $A = \pi(d/2)^2$ (factor of 4 error).
Using $\rho = RL/A$ instead of $\rho = RA/L$ .

Things to Be Careful About

Keep at least 3 s.f. in intermediate steps (especially area) so the final value rounds correctly to three significant figures.
Ensure the final power of ten is sensible for a metal/alloy (nichrome is typically around $10^{-6}\ \Omega\text{m}$ ).

Techniques used

rearrange $R = \rho L/A$ to find resistivityconvert units to SI before substitutioncalculate cross-sectional area from diameter

(iii)

Calculate the percentage uncertainty in $\rho$ .

percentage uncertainty = ______ $\%$

2M Similar

DifficultyMedium

Worked solution

Working

\rho = \frac{V}{I}\frac{\pi d^2}{4L}\ \Rightarrow\ \%\Delta\rho = \%\Delta V + \%\Delta I + 2\%\Delta d + \%\Delta L

\%\Delta V = \frac{0.02}{1.38}\times 100 = 1.45\%

\%\Delta I = \frac{0.001}{0.276}\times 100 = 0.362\%

\%\Delta d = \frac{0.002}{0.496}\times 100 = 0.403\%\ \Rightarrow\ 2\%\Delta d = 0.806\%

\%\Delta L = \frac{0.001}{0.864}\times 100 = 0.116\%

\%\Delta\rho = 1.45 + 0.362 + 0.806 + 0.116 = 2.73\%

Answer

$2.73\%$

Final answer

2.73 %

Detailed explanation

Background Concept

When a quantity is found by multiplying/dividing measured values, the percentage uncertainties add.

If

Q = \frac{ab}{c}

then approximately

\frac{\Delta Q}{Q} \approx \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c}

In percentage form:

\%\Delta Q \approx \%\Delta a + \%\Delta b + \%\Delta c

If a measured quantity is raised to a power, e.g. $Q \propto d^2$ , then its percentage uncertainty is multiplied by 2.

Understanding the Question

You have measured $L$ , $d$ , $V$ , and $I$ each with an uncertainty. You calculated

\rho = \frac{RA}{L}

and since $R = V/I$ and $A \propto d^2$ , the uncertainty in $\rho$ depends on all four measurements. The question asks for the percentage uncertainty in $\rho$ .

Approach

Express $\rho$ in terms of the measured quantities $V, I, d, L$ .
Identify whether each is multiplied/divided and whether any are squared.
Calculate each percentage uncertainty and add them.

Step-by-Step Reasoning

Write $\rho$ in terms of $V, I, d, L$ :

\rho = \frac{RA}{L} = \frac{V}{I} \cdot \frac{\pi(d/2)^2}{L} = \frac{V}{I}\frac{\pi d^2}{4L}

So $\rho \propto V \times d^2 \div (I \times L)$ . Therefore

\%\Delta\rho = \%\Delta V + 2\%\Delta d + \%\Delta I + \%\Delta L

Now compute each:

Voltmeter:

\%\Delta V = \frac{0.02}{1.38}\times 100 = 1.45\%

Ammeter:

\%\Delta I = \frac{0.001}{0.276}\times 100 = 0.362\%

Diameter (then multiply by 2 because of $d^2$ ):

\%\Delta d = \frac{0.002}{0.496}\times 100 = 0.403\%

2\%\Delta d = 0.806\%

Length:

\%\Delta L = \frac{0.001}{0.864}\times 100 = 0.116\%

Add them:

\%\Delta\rho = 1.45 + 0.362 + 0.806 + 0.116 = 2.73\%

Key Takeaways

For products/quotients, percentage uncertainties add.
If $\rho \propto d^2$ , the diameter’s percentage uncertainty is doubled.

Common Mistakes

Forgetting the factor of 2 from $d^2$ .
Subtracting uncertainties because a quantity is in the denominator (you still add percentage uncertainties).
Using absolute uncertainties directly without converting to percentage first.

Things to Be Careful About

Be consistent: use the measured value in the denominator for each percentage calculation.
Keep enough significant figures during the uncertainty calculation, then round the final percentage sensibly (often to 2–3 s.f.).

Techniques used

identify how each measured quantity contributes to a derived quantityadd percentage uncertainties for products and quotientsmultiply the percentage uncertainty by the power in a relationship

(iv)

Determine the absolute uncertainty in $\rho$ .

absolute uncertainty = ______ $\Omega\text{m}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\Delta \rho = \frac{2.73}{100}(1.12 \times 10^{-6}) = 3.06 \times 10^{-8}\ \Omega\text{m}

Answer

$\Delta \rho = 3.1 \times 10^{-8}\ \Omega\text{m}$

Final answer

3.1 × 10^-8 Ωm

Detailed explanation

Background Concept

Percentage uncertainty tells you the size of the uncertainty relative to the value:

\%\Delta Q = \frac{\Delta Q}{Q} \times 100

So

\Delta Q = \frac{\%\Delta Q}{100} \times Q

This converts a percentage uncertainty into an absolute uncertainty (in the units of $Q$ ).

Understanding the Question

From part (ii),

\rho = 1.12 \times 10^{-6}\ \Omega\text{m}

From part (iii), the percentage uncertainty is $2.73\%$ . You must find the absolute uncertainty $\Delta \rho$ in $\Omega\text{m}$ .

Approach

Convert $2.73\%$ into a fraction by dividing by 100, then multiply by $\rho$ .

Step-by-Step Reasoning

Fractional uncertainty:

\frac{2.73}{100} = 0.0273

Absolute uncertainty:

\Delta\rho = 0.0273 \times (1.12 \times 10^{-6}) = 3.06 \times 10^{-8}\ \Omega\text{m}

Round the uncertainty sensibly (typically to 2 s.f.):

\Delta\rho \approx 3.1 \times 10^{-8}\ \Omega\text{m}

Key Takeaways

Absolute uncertainty $= (\text{percentage}/100) \times \text{value}$ .
Absolute uncertainty has the same unit as the quantity.

Common Mistakes

Forgetting to divide by 100 (giving an answer 100 times too large).
Quoting the absolute uncertainty with mismatched units.

Things to Be Careful About

Round uncertainties appropriately (often 1 or 2 significant figures), then quote the final value of the quantity to match the decimal place if required.

Techniques used

convert percentage uncertainty to a fractional uncertaintycalculate absolute uncertainty from fractional uncertainty and the calculated valueround an uncertainty to a sensible number of significant figures

Q6Medium-EasyParticle Physics

Fig. 6.1 shows four alpha particles W, X, Y and Z moving towards a gold nucleus that is in thin gold foil.

Fig. 6.1

(a)

5M

(i)

The paths of particles X and Z are shown.

Complete Fig. 6.1 to show possible paths for particles W and Y.

3M Similar

DifficultyMedium-Easy

Worked solution

diagram-6a-i

Answer

$W$ passes essentially straight (at most a very small deflection).

$Y$ is deflected through a very large angle (may be back-scattered, i.e. turned back).

Final answer

W: approximately straight; Y: large-angle deflection/back-scatter.

Detailed explanation

Background Concept

In Rutherford (alpha-particle) scattering, an alpha particle ( $+2e$ ) experiences an electrostatic repulsive force from the positively charged nucleus. The force increases very strongly as separation decreases.

If the alpha particle passes far from the nucleus, the force is small, so the path is almost a straight line.
If it passes close to the nucleus, the force becomes large, so the direction changes significantly and it can even be scattered backwards.

Understanding the Question

The diagram shows four alpha particles aimed at a gold nucleus. Two trajectories are already drawn:

$Z$ goes far from the nucleus and is not deflected (straight line).
$X$ passes nearer and is deflected away.

You must complete the diagram with reasonable paths for:

$W$ (far from the nucleus)
$Y$ (aimed very close to the nucleus)

Approach

Use the idea that the closer the alpha particle goes to the nucleus (smaller impact parameter), the larger the repulsive force and hence the larger the change in direction (scattering angle).

So:

$W$ should resemble $Z$ (little or no deflection).
$Y$ should show the most dramatic deflection (large angle, possibly back-scattering).

Step-by-Step Reasoning

diagram-6a-i

Identify that $W$ is far from the nucleus (like $Z$ ). Because the electric field is weaker far away, the sideways force component is tiny, so the path remains almost straight.
Identify that $Y$ would pass very close to the nucleus. Here the electric field is much stronger, producing a large repulsive force for a short time, so the particle’s momentum direction changes by a large angle.
Therefore draw $Y$ bending sharply away from the nucleus; for an extremely close approach it can turn through nearly $180^\circ$ (back-scatter).

Key Takeaways

Large scattering angles occur only for close approaches to the nucleus.
Most distant passes give negligible deflection.
A sketch should show a smooth curve (not a kink), because the force acts continuously.

Common Mistakes

Drawing $W$ deflected more than $X$ even though $W$ is further away.
Drawing $Y$ as a straight line despite being aimed close to the nucleus.
Drawing a trajectory that curves towards the nucleus (would imply attraction, not repulsion for alpha particles and a positive nucleus).

Things to Be Careful About

The nucleus is positive, so alpha particles are repelled.
Paths should be smooth curves with the deflection greatest near the nucleus.
A “possible path” for $Y$ includes very large deflections, including back-scattering.

Techniques used

infer the effect of electrostatic repulsion on an alpha-particle trajectoryrelate impact parameter to scattering anglesketch qualitatively correct particle paths around a nucleus

(ii)

Describe what may be inferred about the structure of an atom from the path of particle X.

1M Similar

(iii)

When a beam containing many alpha particles is incident on thin gold foil, nearly all of the alpha particles follow paths that are similar to the path of particle Z.

Describe what may be inferred from this about the structure of an atom.

1M Similar

DifficultyEasy

Worked solution

Answer

Nearly all follow path $Z$ (no deflection), so most of the atom is empty space and the nucleus occupies a very small volume compared with the atom.

Final answer

Atom is mostly empty space; nucleus is very small.

Detailed explanation

Background Concept

In scattering experiments, significant deflections occur only when the projectile passes close enough for a strong interaction. If most projectiles are not deflected, then most of the target’s volume must not produce a strong force.

Understanding the Question

A beam of many alpha particles hits thin gold foil. Nearly all follow paths like $Z$ , i.e. straight through with negligible deflection. You are asked what this implies about atomic structure.

Approach

Interpret “nearly all undeflected” as “most alpha particles do not come close to the concentrated charged region”. That leads to the conclusion that the charged/massive part of the atom is tiny, so the rest is mostly empty space.

Step-by-Step Reasoning

If the atom were filled with spread-out positive charge, many alpha particles would feel a force and be deflected.
The observation that almost all go straight means that, for almost all trajectories, the alpha particle experiences negligible force.
Therefore the region that produces strong repulsion (the nucleus) must occupy only a tiny fraction of the atomic volume.
Hence most of the atom is empty space.

Key Takeaways

“Most undeflected” (\Rightarrow) interaction region is very small.
Supports the nuclear model: tiny nucleus, large empty space.

Common Mistakes

Claiming “there is no nucleus” because most go straight (incorrect; rare large deflections show the nucleus exists).
Saying “electrons cause the scattering” (electrons are too light to cause large deflections of alpha particles).

Things to Be Careful About

The inference is about volume/size: the nucleus is tiny relative to the atom.
Thin foil matters: alpha particles typically meet at most one nucleus, making the interpretation clearer.

Techniques used

use the predominance of undeflected alpha particles to infer empty space in the atomrelate low probability of scattering to a small nuclear size

(b)

State the mass and the charge, in terms of the atomic mass unit $u$ and the elementary charge $e$ , of an alpha particle.

mass = ______ $u$
charge = ______ $e$

2M Similar

(c)

There are two types of hadron.
The hadrons that are in alpha particles are each composed of three quarks.

3M

(i)

State the name of this type of hadron.

1M Similar

(ii)

Show, by reference to their constituent quarks, that the hadrons in an alpha particle have charges of either zero or $+1e$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Up quark charge $= +\frac{2}{3}e$ , down quark charge $= -\frac{1}{3}e$ .

Proton $= uud$ :

q_p = \left(\frac{2}{3}+\frac{2}{3}-\frac{1}{3}\right)e = +1e

Neutron $= udd$ :

q_n = \left(\frac{2}{3}-\frac{1}{3}-\frac{1}{3}\right)e = 0

Answer

Charges are $0$ (neutron) or $+1e$ (proton).

Final answer

Proton: +1e; neutron: 0.

Detailed explanation

Background Concept

Quarks have fractional charges:

up quark $u$ : $+\frac{2}{3}e$
down quark $d$ : $-\frac{1}{3}e$

Baryons are made of three quarks. The proton and neutron (the nucleons found in atomic nuclei, including alpha particles) have quark compositions:

proton: $uud$
neutron: $udd$

The total charge of a baryon is the sum of the charges of its constituent quarks.

Understanding the Question

An alpha particle contains nucleons (protons and neutrons). You must show, using quark compositions, that these hadrons have charges either $0$ or $+1e$ .

Approach

Write the quark charge values.
Write the quark compositions of proton and neutron.
Add the three quark charges for each and simplify in units of $e$ .

Step-by-Step Reasoning

First list quark charges:

$u$ : $+\frac{2}{3}e$
$d$ : $-\frac{1}{3}e$

Proton has quarks $uud$ :

q_p = \left(+\frac{2}{3}e\right)+\left(+\frac{2}{3}e\right)+\left(-\frac{1}{3}e\right)

Factor out $e$ and add fractions:

q_p = \left(\frac{2}{3}+\frac{2}{3}-\frac{1}{3}\right)e = \left(\frac{3}{3}\right)e = +1e

Neutron has quarks $udd$ :

q_n = \left(+\frac{2}{3}e\right)+\left(-\frac{1}{3}e\right)+\left(-\frac{1}{3}e\right)

So

q_n = \left(\frac{2}{3}-\frac{1}{3}-\frac{1}{3}\right)e = 0e = 0

Therefore, the hadrons present (protons and neutrons) have charges of either $+1e$ or $0$ .

Key Takeaways

Remember $u = +\frac{2}{3}e$ , $d = -\frac{1}{3}e$ .
Proton $uud \Rightarrow +1e$ .
Neutron $udd \Rightarrow 0$ .

Common Mistakes

Swapping quark charges (e.g. using $u=-\frac{2}{3}e$ ).
Using the wrong compositions (e.g. proton as $udd$ ).
Arithmetic error with fractions, especially signs.

Things to Be Careful About

Always keep the factor $e$ outside the bracket until the end; it reduces errors.
The question asks for charges of the hadrons in an alpha particle: those are protons and neutrons, not the alpha particle’s overall charge.

Techniques used

use standard quark compositions of proton and neutronadd fractional quark charges to obtain hadron chargeexpress the final charge in units of the elementary charge

Physics 9702/24 — October/November 2025

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