(a) A source oscillates with frequency $f$ to produce a progressive wave of wavelength $\lambda$. The source takes time $t$ to produce $n$ complete oscillations. (i) State what is meant by a progressive wave. (ii) State expressions, in terms of some or all of $f$, $\lambda$ and $n$, for: - the distance moved by a wavefront in time $t$ distance = \_\_\_\_\_\_ - time $t$. time $t$ = \_\_\_\_\_\_ (iii) Use your answers in (ii) to determine an expression for the speed $v$ of the…

(a)(i) A travelling disturbance that transfers energy from one point to another, with no net transfer of matter. · (a)(ii) distance = n\lambda, t = n/f · (a)(iii) v = f\lambda · (b)(i) 7.5 GHz · (b)(ii) 6 cm · (b)(iii) Minimum intensity (destructive interference) at Q. · (b)(iv) Amplitude is maximum at P, then alternates through maxima/minima, ending at a minimum at Q.

9702/24 May/June 2025 — Physics Solved Past Paper

Q1MediumForces, Density and Pressure

(a)

Define the moment of a force.

1M Similar

(b)

A trapdoor has a hinge at end A, as shown in Fig. 1.1.

Fig. 1.1

The trapdoor has length $80\text{ cm}$ and weight $75\text{ N}$ . The mass of the trapdoor is uniformly distributed along its length.

A force $F$ acts at right angles to the trapdoor at end B so that the trapdoor is held in equilibrium at an angle of $42^{\circ}$ to the horizontal.

5M

(i)

State the principle of moments.

2M Similar

(ii)

Calculate the component of the weight that is perpendicular to the trapdoor.

component of weight = ______ $\text{N}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Angle between trapdoor and horizontal $= 42^\circ$ .

Component of weight perpendicular to trapdoor:

W_\perp = W\cos 42^\circ = 75\cos 42^\circ = 55.7\ \text{N}

Answer

$56\ \text{N}$

Final answer

56 N

Detailed explanation

Background Concept

To find the turning effect or to analyse forces on an inclined object, it is often useful to resolve a force into components:

perpendicular to the object (normal to the surface/rod)
parallel to the object

If a force $W$ makes an angle $\theta$ with the perpendicular direction, then the perpendicular component is $W\cos\theta$ . Equivalently, if $\theta$ is the angle between the force and the object itself, then the perpendicular component is $W\sin\theta$ .

Understanding the Question

The trapdoor is at $42^\circ$ to the horizontal. The weight $W = 75\ \text{N}$ acts vertically downward. The question asks for the component of this weight that is perpendicular to the trapdoor.

This is the component that would act “into” the door, at right angles to its surface.

Approach

Use the given angle of the trapdoor to work out the relevant angle for resolving $W$ .
Resolve $W$ into a component perpendicular to the trapdoor using sine/cosine.

A fast recognition here is that the perpendicular component comes out as $W\cos 42^\circ$ .

Step-by-Step Reasoning

diagram-bii-1

The trapdoor is tilted $42^\circ$ above the horizontal.
The vertical direction is $90^\circ$ to the horizontal.
The angle between the trapdoor and the vertical is:

90^\circ - 42^\circ = 48^\circ

The component of $W$ perpendicular to the trapdoor is:

W_\perp = W\sin 48^\circ

and since $\sin 48^\circ = \cos 42^\circ$ :

W_\perp = 75\cos 42^\circ = 55.7\ \text{N} \approx 56\ \text{N}

Key Takeaways

Resolve forces relative to the object (parallel/perpendicular) using the correct angle.
Use complementary angles: $\sin(90^\circ-\theta)=\cos\theta$ .

Common Mistakes

Using $75\sin 42^\circ$ instead of $75\cos 42^\circ$ (wrong component).
Treating $42^\circ$ as the angle between the weight and the trapdoor (it is not).

Things to Be Careful About

Always sketch or imagine a right-angled triangle of components to decide whether to use sine or cosine.
Give the answer with a sensible number of significant figures and include the unit $\text{N}$ .

Techniques used

resolve a force into components parallel and perpendicular to a surfaceuse trigonometry to relate angles between forces and the object

(iii)

Calculate the magnitude of the force $F$ .

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

2M Similar

DifficultyMedium

Worked solution

Working

Take moments about hinge at $A$ .

Weight acts at centre: distance from $A$ along trapdoor $= 0.40\ \text{m}$ .

Using $W_\perp = 75\cos 42^\circ = 55.7\ \text{N}$ ,

F(0.80) = W_\perp(0.40)

F = \frac{55.7 \times 0.40}{0.80} = 27.9\ \text{N}

Answer

$28\ \text{N}$

Final answer

28 N

Detailed explanation

Background Concept

For a rigid body in equilibrium:

\sum \tau = 0

A moment is found from:

\tau = F \times d_\perp

where $d_\perp$ is the perpendicular distance from the pivot to the force’s line of action.

A very useful exam technique is to take moments about a pivot where unknown reaction forces act (here, the hinge at $A$ ). Any force whose line of action passes through $A$ then produces zero moment, simplifying the equation.

Understanding the Question

The trapdoor (length $0.80\ \text{m}$ ) is hinged at $A$ and held in equilibrium at $42^\circ$ to the horizontal by a force $F$ applied at end $B$ .

Given:

weight $W = 75\ \text{N}$ acting at the centre (uniform mass distribution) so at $0.40\ \text{m}$ from $A$ along the door
$F$ acts at right angles to the trapdoor at $B$

Find $F$ .

Approach

Choose pivot at the hinge $A$ .
Write a moment balance: clockwise moments = anticlockwise moments.
Use perpendicular distances:
- For $F$ : since it is perpendicular to the door at $B$ , its moment arm is simply $AB = 0.80\ \text{m}$ .
- For the weight: either use the perpendicular distance to the vertical line of action, or (often simpler) use the component of the weight perpendicular to the door $W_\perp$ acting at $0.40\ \text{m}$ .

Step-by-Step Reasoning

diagram-biii-1

Perpendicular component of weight to the trapdoor:

W_\perp = 75\cos 42^\circ = 55.7\ \text{N}

Moments about $A$ :

Anticlockwise moment from $F$ at $B$ :

\tau_F = F \times 0.80

Clockwise moment from the weight component perpendicular to the door, acting at the centre:

\tau_W = W_\perp \times 0.40

Equilibrium condition:

F(0.80) = W_\perp(0.40)

Solve for $F$ :

F = \frac{55.7 \times 0.40}{0.80} = 27.9\ \text{N} \approx 28\ \text{N}

(Notice the geometry makes $F$ half of $W_\perp$ because the lever arm is twice as long.)

Key Takeaways

In rotational equilibrium: clockwise moments = anticlockwise moments.
Taking moments about the hinge avoids needing to consider hinge reaction forces.
If a force is not perpendicular to the lever arm, use the perpendicular component (or perpendicular distance).

Common Mistakes

Using the full weight $75\ \text{N}$ instead of the perpendicular component to the door.
Using $0.80\ \text{m}$ for the weight’s lever arm instead of $0.40\ \text{m}$ (weight acts at the centre).
Mixing centimetres and metres (e.g. using 80 instead of 0.80).

Things to Be Careful About

Always use SI units in calculations of moments: distances in metres.
Be consistent about which forces cause clockwise vs anticlockwise rotation.
Because $F$ is stated to act at right angles to the door, you do not need to multiply by any sine/cosine for its moment arm.

Techniques used

take moments about the hinge to eliminate unknown reaction forcesuse the perpendicular component of a force when calculating a momentequate clockwise and anticlockwise moments for equilibrium

Q2MediumDynamicsKinematics

An object of constant mass moves in a straight line. The variation with time $t$ of the momentum $p$ of the object is shown in Fig. 2.1.

Fig. 2.1

(a)

Define momentum.

1M Similar

(b)

Calculate the change in momentum of the object from time $t = 0$ to $t = 12\text{ s}$ .

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

1M Similar

DifficultyEasy

Worked solution

Working

From the graph: $p(0)=3.0\ \text{kg m s}^{-1}$ and $p(12)=-1.5\ \text{kg m s}^{-1}$ .

\Delta p = p(12)-p(0)=-1.5-3.0=-4.5\ \text{kg m s}^{-1}

Answer

$-4.5\ \text{kg m s}^{-1}$

Final answer

-4.5 kg m s^-1

Detailed explanation

Background Concept

The change in any quantity is calculated as

\Delta p = p_{\text{final}} - p_{\text{initial}}

A negative change means the momentum has decreased (or changed direction, depending on signs).

Understanding the Question

The momentum–time graph shows $p$ at different times. You must find how much momentum changes between $t=0$ and $t=12\ \text{s}$ .

Given from the graph:

at $t=0$ , $p=3.0\ \text{kg m s}^{-1}$
at $t=12\ \text{s}$ , $p=-1.5\ \text{kg m s}^{-1}$

Approach

Read $p$ at the two times, then compute $p(12)-p(0)$ .

Step-by-Step Reasoning

Read initial momentum: $p(0)=3.0\ \text{kg m s}^{-1}$ .
Read final momentum: $p(12)=-1.5\ \text{kg m s}^{-1}$ .
Subtract:

\Delta p = -1.5 - 3.0 = -4.5\ \text{kg m s}^{-1}

The negative sign is expected because the graph falls with time.

Key Takeaways

Always use “final minus initial”.
Signs on a momentum graph matter: negative momentum corresponds to motion in the opposite direction.

Common Mistakes

Doing $3.0-(-1.5)=+4.5$ (wrong order).
Ignoring the negative sign at $t=12\ \text{s}$ .

Things to Be Careful About

Quote the unit $\text{kg m s}^{-1}$ .
Don’t confuse “change” with “magnitude of change” unless asked for magnitude.

Techniques used

read values from a graph at specified timescalculate change in a quantity using final minus initial

(c)

Calculate the magnitude of the resultant force acting on the object.

force = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Resultant force:

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

Using points $(0,3.0)$ and $(8.0,0)$ :

F = \frac{0-3.0}{8.0-0} = -0.375\ \text{N}

Magnitude $=0.375\ \text{N}$ .

Answer

$0.38\ \text{N}$

Final answer

0.38 N

Detailed explanation

Background Concept

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

F = \frac{\mathrm{d}p}{\mathrm{d}t}

So on a graph of momentum $p$ against time $t$ , the gradient is the resultant force. The unit works out because

\frac{\text{kg m s}^{-1}}{\text{s}} = \text{kg m s}^{-2} = \text{N}

Understanding the Question

You are given a straight-line $p$ – $t$ graph. You must find the magnitude of the resultant force, so you find the gradient and then take its absolute value.

Approach

Pick two clear points on the straight line, compute $\Delta p/\Delta t$ , then take the magnitude.

Step-by-Step Reasoning

Choose convenient points on the line, e.g. $(t,p)=(0,3.0)$ and $(8.0,0)$ .
Calculate the gradient:

\frac{\Delta p}{\Delta t} = \frac{0-3.0}{8.0-0} = -0.375

Interpret this as the force:

F = -0.375\ \text{N}

The negative sign means the force is opposite to the positive direction of momentum.
4. The question asks for magnitude, so

|F| = 0.375\ \text{N} \approx 0.38\ \text{N}

Key Takeaways

The gradient of a $p$ – $t$ graph is the resultant force.
“Magnitude” means you report a positive value.

Common Mistakes

Using $p/t$ instead of $\Delta p/\Delta t$ .
Calculating $\Delta t/\Delta p$ (inverting the gradient).
Forgetting to take the magnitude when asked.

Things to Be Careful About

Use two points that lie exactly on the line to reduce reading error.
Include the unit $\text{N}$ and round sensibly (typically 2 s.f. here).

Techniques used

use the gradient of a momentum–time graph to find forcecalculate a gradient using two points on a straight linetake the magnitude of a signed quantity

(d)

Describe the variation of the speed of the object from time $t = 0$ to $t = 8.0\text{ s}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Since $p=mv$ and $m$ is constant, $v\propto p$ .
From $t=0$ to $8.0\ \text{s}$ the momentum decreases linearly to zero, so the speed decreases uniformly to zero (object comes to rest at $t=8.0\ \text{s}$ ).

Final answer

Speed decreases uniformly to zero, reaching zero at t = 8.0 s.

Detailed explanation

Background Concept

For constant mass,

p = mv

So velocity $v$ is directly proportional to momentum $p$ . Speed is the magnitude of velocity.

If $p$ is positive, the object moves in the positive direction.
If $p$ becomes zero, velocity is zero (the object is momentarily at rest).

Understanding the Question

You must describe how the speed changes between $t=0$ and $t=8.0\ \text{s}$ . The $p$ – $t$ graph shows $p$ falling from $+3.0$ to $0$ in a straight line over this interval.

Approach

Use $p=mv$ to map the momentum graph into a velocity (and therefore speed) description. A straight-line change in $p$ means a straight-line change in $v$ .

Step-by-Step Reasoning

Because $m$ is constant, dividing the entire momentum graph by $m$ would give the velocity–time graph.
From $0$ to $8\ \text{s}$ , the momentum decreases linearly from $+3.0$ to $0$ .
Therefore velocity decreases linearly from some positive value to $0$ .
Since speed is the magnitude and the velocity stays positive until it reaches zero, the speed also decreases steadily (uniformly) to zero.

Key Takeaways

With constant mass, $p$ and $v$ have the same “shape” as functions of time.
When $p$ reaches zero, the object is at rest.

Common Mistakes

Saying the speed is constant because the force is constant (constant force implies constant acceleration, not constant speed).
Saying the speed becomes negative (speed cannot be negative; velocity can).

Things to Be Careful About

The question only asks up to $8.0\ \text{s}$ , where the momentum reaches zero; after that the direction changes (negative momentum).

Techniques used

use p = mv to relate momentum to velocity for constant massinfer how speed changes from the trend of momentum

(e)

By reference to Fig. 2.1, explain why the resultant force acting on the object during the first $8.0\text{ s}$ of its motion cannot be due to air resistance.

2M Similar

DifficultyMedium

Worked solution

Answer

From Fig. 2.1 the gradient of the $p$ – $t$ graph is constant for $0$ to $8.0\ \text{s}$ , so the resultant force is constant.
Air resistance depends on speed, so as the object slows (speed decreasing to zero at $8.0\ \text{s}$ ) the air resistance would decrease to zero and would not remain constant. Hence the resultant force cannot be due to air resistance.

Final answer

Because the p–t graph has constant gradient (constant force) while air resistance would decrease with speed and be zero at t = 8 s when v = 0.

Detailed explanation

Background Concept

Air resistance (drag) is a resistive force that acts opposite to the direction of motion and depends on speed. Common models are:

low speeds: $F_D \propto v$
higher speeds: $F_D \propto v^2$

In all cases, if $v$ decreases, the drag force decreases; and if $v=0$ , the drag force is $0$ .

Also, from momentum form of Newton’s second law:

F = \frac{\mathrm{d}p}{\mathrm{d}t}

So a constant gradient on a $p$ – $t$ graph means constant resultant force.

Understanding the Question

You are asked to use the graph to argue that the resultant force during the first $8.0\ \text{s}$ cannot be air resistance.
From the graph:

The $p$ – $t$ line is straight, so its gradient is constant $→$ force constant.
Momentum falls to $0$ at $t=8.0\ \text{s}$ → velocity (and speed) is $0$ at $t=8.0\ \text{s}$ .

Approach

Use the straight-line $p$ – $t$ graph to deduce the force is constant.
Use the fact that speed decreases to zero to deduce that any air resistance would decrease to zero.
Show these two conclusions are incompatible.

Step-by-Step Reasoning

On a $p$ – $t$ graph, gradient $= \mathrm{d}p/\mathrm{d}t$ .
The graph is a straight line from $t=0$ to $t=8\ \text{s}$ , so $\mathrm{d}p/\mathrm{d}t$ is constant → resultant force is constant.
But $p=mv$ and $m$ is constant, so as $p$ decreases to $0$ , the speed decreases to $0$ .
Air resistance depends on speed, so it would decrease as the object slows, and it would be exactly $0$ at the instant the object is at rest.
Therefore the resultant force being constant (and non-zero even approaching $t=8\ \text{s}$ ) cannot be explained by air resistance.

Key Takeaways

Straight line on $p$ – $t$ → constant resultant force.
Drag force is not constant when speed changes; it falls to zero at $v=0$ .

Common Mistakes

Saying “air resistance is always constant” (it is not; it varies with speed).
Forgetting that $p=0$ implies $v=0$ for constant mass.

Things to Be Careful About

The key evidence from the graph is the constant gradient and the momentum reaching zero at $8\ \text{s}$ .
Use wording like “depends on speed” and “would be zero at rest” to secure the explanation marks.

Techniques used

use the gradient of the p–t graph to infer the force behaviourapply the speed-dependence of air resistancecompare physical expectations with graph evidence

(f)

At time $t = 0$ the displacement $d$ of the object is zero.

On Fig. 2.2, sketch the variation of $d$ with time $t$ from $t = 0$ to $t = 12\text{ s}$ .

Numerical values of $d$ are not required.

Fig. 2.2

3M Similar

DifficultyMedium

Worked solution

Answer

diagram-f-1

Final answer

Sketch: displacement increases concave down to a maximum at t = 8 s, then decreases (still concave down), remaining above zero at t = 12 s.

Detailed explanation

Background Concept

Displacement $d$ , velocity $v$ , and acceleration $a$ are related by:

gradient of a displacement–time graph is velocity:

v = \frac{\mathrm{d}d}{\mathrm{d}t}

if acceleration is constant, then velocity changes linearly with time and displacement is a quadratic (parabolic) function of time.

Here we are given momentum $p$ rather than velocity. With constant mass,

p = mv \ \Rightarrow \ v = \frac{p}{m}

So the shape of the velocity–time graph is the same as the shape of the momentum–time graph (just scaled by $1/m$ ).

Understanding the Question

You must sketch $d$ against $t$ from $0$ to $12\ \text{s}$ , starting at $d=0$ when $t=0$ . Numerical values are not required, but the correct shape is.

From the momentum graph:

$p$ decreases linearly from $+3$ at $t=0$ to $0$ at $t=8\ \text{s}$ .
then continues linearly to $-1.5$ at $t=12\ \text{s}$ .

So velocity is positive then becomes zero at $8\ \text{s}$ , then becomes negative after $8\ \text{s}$ .

Approach

Convert momentum behaviour to velocity behaviour using $v=p/m$ .
Use “gradient of $d$ $d$ – $t$ $t$ equals $v$ $v$ ”:
- positive $v$ means $d$ increases;
- zero $v$ means a horizontal tangent;
- negative $v$ means $d$ decreases.
Use the fact that $v$ changes linearly (straight line in $p$ – $t$ ) to set the curvature: $d$ should be a smooth concave-down parabola.

Step-by-Step Reasoning

For $0 < t < 8\ \text{s}$ $0 < t < 8 s$ : $p>0 \Rightarrow v>0$ $p > 0 \Rightarrow v > 0$ .
- Therefore $d$ is increasing.
- But $p$ (hence $v$ ) is decreasing, so the gradient of the $d$ – $t$ graph decreases with time.
- This means the curve rises but flattens out: concave down.
At $t=8\ \text{s}$ $t = 8 s$ : $p=0 \Rightarrow v=0$ $p = 0 \Rightarrow v = 0$ .
- The displacement curve has zero gradient here: a turning point (maximum displacement).
For $8 < t \le 12\ \text{s}$ $8 < t \leq 12 s$ : $p<0 \Rightarrow v<0$ $p < 0 \Rightarrow v < 0$ .
- Therefore $d$ decreases after $8\ \text{s}$ (object moves back in the negative direction).
- The momentum continues to become more negative linearly, so the speed in the negative direction increases; the slope of the $d$ – $t$ graph becomes more negative.
- With constant negative acceleration, the curve remains concave down.
Since the negative momentum area from $8$ to $12\ \text{s}$ is smaller than the positive area from $0$ to $8\ \text{s}$ , the object has not returned to $d=0$ by $t=12\ \text{s}$ , so the curve at $t=12\ \text{s}$ is still above the axis.

A correct sketch is therefore: start at origin, rise concave down to a maximum at $t=8\ \text{s}$ , then fall (still concave down) but stay above $d=0$ at $t=12\ \text{s}$ .

diagram-f-1

Key Takeaways

Use $p=mv$ (constant mass) to translate a $p$ – $t$ graph into a $v$ – $t$ description.
Use “gradient of $d$ – $t$ is $v$ ” to sketch displacement.
Linear $v(t)$ implies parabolic $d(t)$ .

Common Mistakes

Drawing $d$ as a straight line (would imply constant velocity).
Making a sharp corner at $t=8\ \text{s}$ (displacement should be smooth; velocity changes continuously).
Making the curve change concavity at $t=8\ \text{s}$ (acceleration is constant, so concavity stays the same).
Bringing the curve back to $d=0$ at $t=12\ \text{s}$ (not consistent with areas under the velocity graph).

Things to Be Careful About

The turning point must be exactly at $t=8\ \text{s}$ because $p=0 \Rightarrow v=0$ there.
The displacement at $t=12\ \text{s}$ should still be positive (above the axis).
Label axes correctly: $t$ horizontal, $d$ vertical, and the curve must start at $(0,0)$ as stated.

Techniques used

relate momentum to velocity using p = mvuse the gradient of a displacement–time graph as velocityinfer curvature from constant acceleration (linearly changing velocity)

Q3MediumWork, Energy and PowerDeformation of SolidsDynamics

The lower end of a vertical spring is fixed to a horizontal surface, as shown in Fig. 3.1.

Fig. 3.1

The mass of the spring is negligible. A block of mass $5.5\text{ kg}$ drops vertically onto the spring and is brought to rest as the spring is compressed.

(a)

The block has kinetic energy $110\text{ J}$ as it makes contact with the spring.

Calculate the speed of the block as it makes contact with the spring.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2(110)}{5.5}} = \sqrt{40}

v = 6.32\ \text{m s}^{-1}

Answer

$6.3\ \text{m s}^{-1}$

Final answer

6.3 m s^-1

Detailed explanation

Background Concept

Kinetic energy is the energy an object has because it is moving. For a mass $m$ moving with speed $v$ ,

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

This comes from the work done to accelerate the object from rest. It applies in normal (non-relativistic) situations such as this one.

Understanding the Question

You are told that the block (mass $5.5\ \text{kg}$ ) has kinetic energy $110\ \text{J}$ at the instant it first touches the spring. The question asks for the speed at that instant.

So:

known: $E_k = 110\ \text{J}$ , $m = 5.5\ \text{kg}$
unknown: $v$ .

Approach

Use the kinetic energy formula and rearrange it to solve for $v$ . Then substitute the given values.

Step-by-Step Reasoning

Start with

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

Rearrange:

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

Substitute $E_k = 110\ \text{J}$ and $m = 5.5\ \text{kg}$ :

v = \sqrt{\frac{2(110)}{5.5}} = \sqrt{40} = 6.32\ \text{m s}^{-1}

Rounding to 2 s.f. gives $6.3\ \text{m s}^{-1}$ .

Key Takeaways

Use $E_k = \tfrac{1}{2}mv^2$ to link energy and speed.
Rearranging for $v$ introduces a square root.

Common Mistakes

Forgetting the square root and giving $v^2$ as the final answer.
Using the wrong mass (e.g. confusing the spring as having mass when it is stated negligible).
Dropping units or giving $\text{m s}^{-2}$ instead of $\text{m s}^{-1}$ .

Things to Be Careful About

Significant figures: here a 2 s.f. answer is appropriate ( $6.3\ \text{m s}^{-1}$ ).
Keep $m$ in $\text{kg}$ and $E_k$ in $\text{J}$ so that $v$ comes out in $\text{m s}^{-1}$ .

Techniques used

apply the kinetic energy formularearrange an equation to make speed the subjectsubstitute values and evaluate with correct units

(b)

The gravitational potential energy of the block decreases by $20\text{ J}$ as the spring is compressed to its maximum compression $x_0$ .

Show that $x_0$ is $0.37\text{ m}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Decrease in GPE:

\Delta E_p = mgx_0

20 = 5.5 \times 9.81 \times x_0

x_0 = \frac{20}{5.5 \times 9.81} = 0.370\ \text{m}

Answer

$0.37\ \text{m}$

Final answer

0.37 m

Detailed explanation

Background Concept

Gravitational potential energy (GPE) changes when an object moves vertically in a uniform gravitational field. Near Earth’s surface,

\Delta E_p = mg\Delta h

where $m$ is mass, $g$ is gravitational field strength, and $\Delta h$ is the vertical displacement. If the object moves downward by $x_0$ , its GPE decreases by $mgx_0$ .

Understanding the Question

As the spring compresses to its maximum compression $x_0$ , the block moves downward by the same distance $x_0$ . You are told the block’s GPE decreases by $20\ \text{J}$ during this compression. You must use this to show that $x_0 = 0.37\ \text{m}$ .

Known:

$m = 5.5\ \text{kg}$
decrease in GPE $= 20\ \text{J}$
$g \approx 9.81\ \text{m s}^{-2}$

Unknown: $x_0$ .

Approach

Use $\Delta E_p = mgx_0$ (magnitude of the decrease). Rearrange to find $x_0$ .

Step-by-Step Reasoning

The magnitude of the GPE decrease is

mgx_0 = 20

So

x_0 = \frac{20}{mg} = \frac{20}{5.5 \times 9.81}

Calculate:

x_0 = 0.370\ \text{m} \approx 0.37\ \text{m}

This matches what you were asked to show.

Key Takeaways

For vertical motion near Earth, GPE change is $mg\Delta h$ .
When an object moves down, GPE decreases; the magnitude is still $mg\Delta h$ .

Common Mistakes

Using $20 = \tfrac{1}{2}kx^2$ (that would be spring energy, not GPE).
Using $g = 10\ \text{m s}^{-2}$ and then not matching the required $0.37\ \text{m}$ .
Confusing $x_0$ (compression distance) with speed or acceleration.

Things to Be Careful About

The vertical distance moved during compression is the compression $x_0$ .
Use consistent units: $\text{J} = \text{N m}$ , so dividing by $mg$ (in N) gives m.
Keep enough significant figures in intermediate steps to obtain $0.37\ \text{m}$ .

Techniques used

apply the gravitational potential energy change formulaequate energy change to work done by weightrearrange to determine a distance

(c)

Assume that, as the spring compresses, all of the energy lost by the block is converted into the elastic potential energy of the spring.

Use the data from (a) and (b) to determine the maximum elastic potential energy of the spring.

Show your working.

maximum elastic potential energy = ______ $\text{J}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Energy lost by block during compression:

E = E_k + \text{decrease in GPE} = 110 + 20 = 130\ \text{J}

Answer

$130\ \text{J}$

Final answer

130 J

Detailed explanation

Background Concept

When a block interacts with a spring, energy can be transferred between:

kinetic energy of the block,
gravitational potential energy of the block (if it moves vertically),
elastic potential energy stored in the spring.

If we assume no energy is dissipated (no sound, heat, deformation losses), then the total mechanical energy is conserved: energy lost by the block becomes energy stored in the spring.

Understanding the Question

From the instant the block first contacts the spring until it reaches maximum compression:

the block’s kinetic energy falls from $110\ \text{J}$ to $0$ (it comes to rest), so it loses $110\ \text{J}$ of kinetic energy.
the block moves downward and its GPE decreases by $20\ \text{J}$ .

The question states to assume all energy lost by the block becomes elastic potential energy of the spring. You must find that maximum elastic potential energy.

Approach

Add the two energy losses (loss of kinetic energy + decrease in GPE). That total must equal the spring’s maximum elastic potential energy.

Step-by-Step Reasoning

Initial kinetic energy at contact: $110\ \text{J}$ .
Final kinetic energy at maximum compression: $0\ \text{J}$ .
So the decrease in kinetic energy is $110\ \text{J}$ .

The GPE decreases by $20\ \text{J}$ during the same interval.

Total energy transferred to the spring is therefore

E_{\text{spring,max}} = 110 + 20 = 130\ \text{J}

Key Takeaways

During compression, both the loss of kinetic energy and the decrease in GPE can contribute to energy stored in the spring.
With no dissipation, energy lost by the block equals energy gained by the spring.

Common Mistakes

Using only $110\ \text{J}$ and forgetting the extra $20\ \text{J}$ from the drop during compression.
Subtracting $20\ \text{J}$ instead of adding it (a decrease in GPE means additional energy is released).
Thinking the spring energy must be less because the block stops; actually all the mechanical energy must be stored somewhere at that instant.

Things to Be Careful About

Sign convention: “GPE decreases by $20\ \text{J}$ ” means the block loses $20\ \text{J}$ of GPE, which becomes available to store in the spring.
The maximum elastic potential energy occurs at maximum compression, when the block’s speed is zero.

Techniques used

apply conservation of mechanical energyadd kinetic energy and gravitational potential energy changesidentify energy transferred to elastic potential energy

(d)

The variation of the force $F$ acting on the spring with the compression $x$ of the spring is shown in Fig. 3.2.

Fig. 3.2

Use the information in (b) and your answer in (c) to show that the maximum force $F_0$ exerted on the spring by the block is $700\text{ N}$ .

2M Similar

DifficultyMedium

Worked solution

Working

Maximum elastic potential energy is area under $F$ - $x$ graph:

E = \frac{1}{2}F_0 x_0

Using $E = 130\ \text{J}$ and $x_0 = 0.37\ \text{m}$ ,

F_0 = \frac{2E}{x_0} = \frac{2(130)}{0.37} = 7.0 \times 10^2\ \text{N}

Answer

$F_0 = 700\ \text{N}$

Final answer

700 N

Detailed explanation

Background Concept

For a spring that obeys Hooke’s law, force is proportional to extension/compression:

F = kx

A force–displacement graph for such a spring is a straight line through the origin. The work done in compressing the spring from $0$ to $x_0$ equals the area under the $F$ – $x$ graph:

W = \int_0^{x_0} F\,dx

That work done is stored as elastic potential energy in the spring.

Understanding the Question

You are given:

maximum compression $x_0 = 0.37\ \text{m}$ (from part (b)),
maximum elastic potential energy $E = 130\ \text{J}$ (from part (c)),
a linear $F$ vs $x$ graph from $(0,0)$ to $(x_0, F_0)$ .

You must use these to show the maximum force $F_0$ is $700\ \text{N}$ .

Approach

Because the graph is a straight line from the origin, the area under it is a triangle. So:

E = \text{area} = \frac{1}{2} \times (\text{base}) \times (\text{height}) = \frac{1}{2} x_0 F_0

Rearrange for $F_0$ and substitute the values.

Step-by-Step Reasoning

Since $F$ increases linearly with $x$ , the $F$ – $x$ graph is triangular. The base is $x_0$ and the height is $F_0$ .

So elastic energy stored is

E = \frac{1}{2}F_0 x_0

Rearrange:

F_0 = \frac{2E}{x_0}

Substitute $E = 130\ \text{J}$ and $x_0 = 0.37\ \text{m}$ :

F_0 = \frac{2(130)}{0.37} = \frac{260}{0.37} \approx 7.03 \times 10^2\ \text{N}

To appropriate significant figures:

F_0 \approx 700\ \text{N}

Key Takeaways

Energy stored in a spring equals the work done compressing it.
Work done is the area under an $F$ – $x$ graph.
For a Hooke’s law spring, that area is a triangle: $E = \tfrac{1}{2}F_0 x_0$ .

Common Mistakes

Using $E = F_0 x_0$ (rectangle area) instead of half that for a triangle.
Mixing up $x_0$ (compression) with the vertical drop from earlier parts.
Forgetting that $\text{J} = \text{N m}$ ; units help confirm the formula.

Things to Be Careful About

The graph must pass through the origin for the triangular area argument; the question shows it does.
Keep enough precision in intermediate calculation (e.g. $260/0.37$ ) before rounding to $700\ \text{N}$ .

Techniques used

use area under a force-displacement graph to find work done/energy storedmodel the graph as a triangle and calculate its arearearrange to obtain the maximum force

(e)

Use the information in (d) to determine, for the instant that the block is first brought to rest by the spring, the magnitude of:

4M

(i)

the resultant force acting on the block

resultant force = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At maximum compression, spring force on block $= F_0 = 700\ \text{N}$ (upward).

Weight:

mg = 5.5 \times 9.81 = 54\ \text{N}

Resultant force on block:

F_R = 700 - 54 = 6.46 \times 10^2\ \text{N}

Answer

$6.5 \times 10^2\ \text{N}$

Final answer

6.5 × 10^2 N

Detailed explanation

Background Concept

Forces add vectorially. In vertical 1D problems, choose an upward or downward positive direction and add forces with signs.

At the instant the block is at maximum compression, it is momentarily at rest but not in equilibrium (unless forces balance). The resultant force is the net force:

F_R = \sum F

Weight acts downward with magnitude $mg$ . The spring exerts an upward force on the block.

Understanding the Question

At the instant the block is first brought to rest by the spring (i.e. at maximum compression):

the spring force is at its maximum $F_0 = 700\ \text{N}$ (from part (d)), upward on the block.
the weight of the block is $mg$ , downward.

You are asked for the magnitude of the resultant force on the block at that instant.

Approach

Draw/visualise the free-body diagram, take upward as positive, then calculate

F_R = F_0 - mg

Step-by-Step Reasoning

e-i-fbd

Calculate the weight:

mg = 5.5 \times 9.81 = 53.955\ \text{N} \approx 54\ \text{N}

Take upward as positive:

spring force: $+700\ \text{N}$
weight: $-54\ \text{N}$

So the resultant force is

F_R = 700 - 54 = 646\ \text{N}

Magnitude to 2 s.f.:

|F_R| \approx 6.5 \times 10^2\ \text{N}

(Direction would be upward, but the question asks only for magnitude.)

Key Takeaways

At maximum compression, the spring force is largest.
Resultant force is not zero just because the speed is momentarily zero.
Net force is found by subtracting weight from the upward spring force.

Common Mistakes

Setting resultant force to zero because the block is at rest at that instant.
Adding $mg$ and $F_0$ instead of subtracting (directions are opposite).
Using $g = 9.81$ in earlier parts but switching to a very rough value here and losing consistency.

Things to Be Careful About

Always include directions when forming the resultant; only at the end take magnitude if asked.
Keep $mg$ to a sensible precision; it is much smaller than $700\ \text{N}$ but still affects the result by nearly $8\%$ .

Techniques used

draw or visualise a free-body diagramidentify forces acting and their directionscalculate the resultant force as a vector sum in one dimension

(ii)

the acceleration of the block.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $F_R = 6.46 \times 10^2\ \text{N}$ ,

a = \frac{F_R}{m} = \frac{6.46 \times 10^2}{5.5} = 1.18 \times 10^2\ \text{m s}^{-2}

Answer

$1.2 \times 10^2\ \text{m s}^{-2}$

Final answer

1.2 × 10^2 m s^-2

Detailed explanation

Background Concept

Newton’s second law links resultant force and acceleration:

F_R = ma

where $F_R$ is the net (resultant) force on the object, $m$ is its mass, and $a$ is its acceleration. The acceleration is in the direction of the resultant force.

Understanding the Question

At maximum compression the block is momentarily at rest, but it experiences an upward resultant force (spring force larger than weight). You are asked for the magnitude of the acceleration at that instant.

Known:

$m = 5.5\ \text{kg}$
spring force at that instant $F_0 = 700\ \text{N}$
weight $mg \approx 54\ \text{N}$
resultant force magnitude $\approx 6.46 \times 10^2\ \text{N}$ .

Unknown: $a$ .

Approach

Use the same force situation as part (e)(i) to get the resultant force.
Apply $a = F_R/m$ .

Step-by-Step Reasoning

e-ii-fbd

Resultant force (upward) is

F_R = F_0 - mg = 700 - (5.5 \times 9.81) \approx 646\ \text{N}

Apply Newton’s second law:

a = \frac{F_R}{m} = \frac{646}{5.5} = 117.5\ \text{m s}^{-2}

Magnitude to 2 s.f.:

a \approx 1.2 \times 10^2\ \text{m s}^{-2}

(The acceleration direction would be upward, but only magnitude is requested.)

Key Takeaways

Zero speed at an instant does not imply zero acceleration.
Use $F_R = ma$ with the net force, not any single force.

Common Mistakes

Using $a = g$ (this is not free fall; the spring exerts a large additional force).
Dividing by the wrong mass or accidentally using weight instead of mass.
Using $700\ \text{N}$ directly as the resultant force and forgetting to subtract $mg$ .

Things to Be Careful About

Ensure $F_R$ is in newtons and $m$ in kilograms so $a$ comes out in $\text{m s}^{-2}$ .
Quote acceleration to sensible significant figures consistent with earlier parts.

Techniques used

apply Newton's second lawuse the resultant force from force balance to find accelerationcalculate magnitude of acceleration from F/m

Q4MediumWavesSuperposition

(a)

A source oscillates with frequency $f$ to produce a progressive wave of wavelength $\lambda$ . The source takes time $t$ to produce $n$ complete oscillations.

4M

(i)

State what is meant by a progressive wave.

1M Similar

(ii)

State expressions, in terms of some or all of $f$ , $\lambda$ and $n$ , for:

the distance moved by a wavefront in time $t$

distance = ______

time $t$ .

time $t$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

In $t$ , the source makes $n$ oscillations, so the wavefront advances $n$ wavelengths:

\text{distance} = n\lambda

Also

f = \frac{n}{t} \;\Rightarrow\; t = \frac{n}{f}

Answer

\text{distance} = n\lambda, \qquad t = \frac{n}{f}

Final answer

distance = n\lambda, t = n/f

Detailed explanation

Background Concept

Frequency $f$ is the number of complete oscillations per second:

f = \frac{\text{number of oscillations}}{\text{time taken}}

Wavelength $\lambda$ is the distance between adjacent points in phase (e.g. crest-to-crest). A key interpretation is:

in one period $T$ , a wave travels forward by one wavelength $\lambda$ .
since $T = 1/f$ , in time corresponding to $n$ oscillations (i.e. $n$ periods), the wave travels $n\lambda$ .

Understanding the Question

The stem says: a source oscillates with frequency $f$ , wavelength $\lambda$ , and takes time $t$ to produce $n$ oscillations.

You must write expressions for:

the distance moved by a wavefront in time $t$ ,
the time $t$ itself,
using some or all of $f$ , $\lambda$ , and $n$ .

Approach

Convert “ $n$ complete oscillations” into “ $n$ periods”.
Use the fact that each period corresponds to one wavelength of forward travel.
Use the definition of frequency to link $n$ and $t$ .

Step-by-Step Reasoning

Distance travelled by a wavefront in time $t$

In one oscillation (one period), the wave pattern moves forward by $\lambda$ .
In $n$ oscillations, it moves forward by $n\lambda$ .

So:

\text{distance} = n\lambda

Time $t$
Using the definition of frequency:

f = \frac{n}{t}

Rearrange for $t$ :

t = \frac{n}{f}

Key Takeaways

$f = n/t$ (frequency definition).
Each oscillation corresponds to an advance of one wavelength for the travelling wave pattern.

Common Mistakes

Writing $\text{distance} = \lambda / n$ or $n/\lambda$ (mixing up proportionality).
Writing $t = f/n$ (algebra error).

Things to Be Careful About

The wavefront distance is about the wave pattern, not about a particle in the medium.
Keep the meanings clear: $n$ is a pure number (no units).

Techniques used

relate frequency to oscillations per unit timeuse the definition of wavelength as distance moved in one periodexpress a distance travelled in terms of number of wavelengths

(iii)

Use your answers in (ii) to determine an expression for the speed $v$ of the wave in terms of $f$ and $\lambda$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

v = \frac{\text{distance}}{t} = \frac{n\lambda}{n/f}

v = f\lambda

Answer

v = f\lambda

Final answer

v = f\lambda

Detailed explanation

Background Concept

Wave speed $v$ is the rate at which the wave pattern (e.g. a crest) moves:

v = \frac{\text{distance travelled}}{\text{time taken}}

For a progressive wave, a wave crest moves one wavelength $\lambda$ in one period $T$ . Since $f = 1/T$ , the standard wave equation is:

v = f\lambda

Understanding the Question

You previously found:

distance moved by a wavefront in time $t$ is $n\lambda$ ,
$t = n/f$ .

Now you must combine these to obtain an expression for wave speed $v$ in terms of $f$ and $\lambda$ only.

Approach

Use $v = \text{distance}/t$ and substitute the expressions from (ii). The number of oscillations $n$ should cancel.

Step-by-Step Reasoning

Start with speed definition:

v = \frac{\text{distance}}{t}

Substitute from (ii):

v = \frac{n\lambda}{n/f}

Dividing by $n/f$ is multiplying by $f/n$ :

v = n\lambda \times \frac{f}{n}

Cancel $n$ :

v = f\lambda

Key Takeaways

Deriving $v = f\lambda$ comes directly from speed = distance/time and the meaning of wavelength and frequency.
The result is independent of how many oscillations $n$ you consider.

Common Mistakes

Forgetting to invert when dividing by a fraction: writing $v = n\lambda \times n/f$ .
Leaving $n$ in the final expression (it should cancel).

Things to Be Careful About

Units check: $f$ is $\text{s}^{-1}$ and $\lambda$ is $\text{m}$ , so $f\lambda$ has units $\text{m s}^{-1}$ , consistent with speed.

Techniques used

use speed = distance / timesubstitute expressions for distance and timesimplify algebraically to obtain the wave equation

(b)

Two identical microwave sources X and Y emit waves in phase. The sources are separated by a distance of $30\text{ cm}$ , as shown in Fig. 4.1.

Fig. 4.1

The intensity of the microwaves is to be investigated at points P and Q.

Line PQ is parallel to line XY. Distance XP is equal to distance YP. Distance YQ is $72\text{ cm}$ and angle XYQ is $90^{\circ}$ .

The wavelength of the microwaves is $4.0\text{ cm}$ .

8M

(i)

Calculate the frequency, in $\text{GHz}$ , of the microwaves.

frequency = ______ $\text{GHz}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For microwaves, $v \approx c = 3.0 \times 10^{8}\ \text{m s}^{-1}$ .

\lambda = 4.0\ \text{cm} = 4.0 \times 10^{-2}\ \text{m}

f = \frac{c}{\lambda} = \frac{3.0 \times 10^{8}}{4.0 \times 10^{-2}} = 7.5 \times 10^{9}\ \text{Hz}

7.5 \times 10^{9}\ \text{Hz} = 7.5\ \text{GHz}

Answer

f = 7.5\ \text{GHz}

Final answer

7.5 GHz

Detailed explanation

Background Concept

All electromagnetic (e.m.) waves in air/vacuum travel at speed

c = 3.0 \times 10^{8}\ \text{m s}^{-1}

They satisfy the wave equation

v = f\lambda

For e.m. waves, $v \approx c$ , so:

f = \frac{c}{\lambda}

Also, $1\ \text{GHz} = 10^{9}\ \text{Hz}$ .

Understanding the Question

You are given microwave wavelength $\lambda = 4.0\ \text{cm}$ and asked for frequency in GHz. The only subtlety is converting cm to m and then Hz to GHz.

Approach

Convert $\lambda$ to metres.
Use $f = c/\lambda$ .
Convert Hz to GHz.

Step-by-Step Reasoning

Convert wavelength:

4.0\ \text{cm} = 4.0 \times 10^{-2}\ \text{m}

Compute frequency:

f = \frac{3.0 \times 10^{8}}{4.0 \times 10^{-2}} = 7.5 \times 10^{9}\ \text{Hz}

Convert to GHz:

7.5 \times 10^{9}\ \text{Hz} = 7.5\ \text{GHz}

Key Takeaways

Use $f = c/\lambda$ for e.m. waves.
Always convert wavelength to SI (metres) before calculating.

Common Mistakes

Using $\lambda = 4.0\ \text{m}$ instead of $4.0\ \text{cm}$ .
Forgetting that GHz is $10^{9}$ Hz.
Using $3.0 \times 10^{8}$ with $\lambda$ still in cm.

Things to Be Careful About

Significant figures: $c$ is usually taken as $3.0 \times 10^{8}$ (2 s.f.) and $\lambda$ is 2 s.f., so $f$ to 2 s.f. is appropriate (7.5 GHz).

Techniques used

use v = f\lambdaconvert centimetres to metresuse c for the speed of electromagnetic waves in air

(ii)

Show that the difference between the path lengths XQ and YQ is $6\text{ cm}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Triangle $XYQ$ is right-angled at $Y$ , with $XY = 30\ \text{cm}$ and $YQ = 72\ \text{cm}$ .

XQ = \sqrt{(72)^2 + (30)^2} = \sqrt{6084} = 78\ \text{cm}

XQ - YQ = 78 - 72 = 6\ \text{cm}

Answer

Difference in path lengths $= 6\ \text{cm}$ .

Final answer

6 cm

Detailed explanation

Background Concept

For interference questions, we often need the path difference between two waves reaching the same point. If the sources are at $X$ and $Y$ , and the point is $Q$ , then the path difference is:

\Delta = XQ - YQ

(You can take the magnitude; the important thing is how many wavelengths it corresponds to.)

Here, the geometry provides a right-angled triangle, so Pythagoras’ theorem applies:

\text{hypotenuse}^2 = \text{adjacent}^2 + \text{opposite}^2

Understanding the Question

Given:

the sources are separated by $XY = 30\ \text{cm}$ ,
$YQ = 72\ \text{cm}$ ,
angle $XYQ = 90^\circ$ (so $XY \perp YQ$ ).

You must show that $XQ$ is longer than $YQ$ by $6\ \text{cm}$ .

Approach

Recognise $\triangle XYQ$ is right-angled at $Y$ .
Use Pythagoras to find $XQ$ .
Subtract $YQ$ from $XQ$ .

Step-by-Step Reasoning

diagram-bii-1

Use Pythagoras:

XQ^2 = XY^2 + YQ^2 = 30^2 + 72^2 = 900 + 5184 = 6084

So:

XQ = \sqrt{6084} = 78\ \text{cm}

Now path difference:

XQ - YQ = 78 - 72 = 6\ \text{cm}

Key Takeaways

Right angle given implies Pythagoras.
Path difference is found by subtracting the two path lengths.

Common Mistakes

Using $30 + 72$ instead of Pythagoras.
Squaring incorrectly or forgetting the square root.
Finding $YQ - XQ$ and then giving a negative value without stating magnitude.

Things to Be Careful About

Keep units consistent (both given in cm here, so no conversion needed in this part).
The “show that” mark is earned by demonstrating the key calculation clearly.

Techniques used

apply Pythagoras’ theorem to a right-angled trianglecalculate two path lengths from given geometryfind a path difference by subtraction

(iii)

State and explain what may be deduced about the intensity of the microwaves at point Q.

3M Similar

DifficultyMedium

Worked solution

Working

\Delta = XQ - YQ = 6\ \text{cm}, \qquad \lambda = 4.0\ \text{cm}

\frac{\Delta}{\lambda} = \frac{6}{4.0} = 1.5 = \frac{3}{2}

Path difference $= \left(m + \tfrac{1}{2}\right)\lambda$ so the waves arrive in antiphase.

Answer

The intensity at $Q$ is a minimum (destructive interference; ideally zero) because $\Delta = 1.5\lambda$ gives a phase difference of an odd multiple of $\pi$ .

Final answer

Minimum intensity (destructive interference) at Q.

Detailed explanation

Background Concept

Two coherent sources emitting in phase produce interference.

At a point, the path difference $\Delta$ determines the phase difference $\phi$ :

\phi = 2\pi \frac{\Delta}{\lambda}

Interference conditions:

Constructive interference (maximum intensity):

\Delta = m\lambda \quad (m = 0,1,2,\dots)

Destructive interference (minimum intensity):

\Delta = \left(m + \tfrac{1}{2}\right)\lambda

For perfect equal amplitudes, destructive interference would give zero resultant amplitude.

Understanding the Question

You have already shown that at point $Q$ the path difference is $6\ \text{cm}$ . The wavelength is $4.0\ \text{cm}$ and the sources are in phase.

You must state and explain what can be deduced about the intensity at $Q$ .

Approach

Express the path difference as a multiple of the wavelength.
Compare with the conditions for constructive/destructive interference.
State the resulting intensity (maximum or minimum) and explain via phase difference.

Step-by-Step Reasoning

Given:

\Delta = 6\ \text{cm}, \quad \lambda = 4.0\ \text{cm}

Find the number of wavelengths:

\frac{\Delta}{\lambda} = \frac{6}{4.0} = 1.5 = \frac{3}{2}

So:

\Delta = 1.5\lambda = \left(1 + \tfrac{1}{2}\right)\lambda

This matches the destructive condition $\left(m + \tfrac{1}{2}\right)\lambda$ , so the waves arrive at $Q$ in antiphase.

You can also show the phase difference explicitly:

\phi = 2\pi \frac{\Delta}{\lambda} = 2\pi (1.5) = 3\pi

Since $3\pi$ differs from $\pi$ by $2\pi$ , it is still effectively an odd multiple of $\pi$ , meaning the two wave contributions oppose each other.

Therefore the resultant amplitude (and hence intensity) is minimal at $Q$ .

Key Takeaways

Convert path difference to a fraction/multiple of $\lambda$ .
Half-integer multiples of $\lambda$ correspond to destructive interference.

Common Mistakes

Saying “ $1.5\lambda$ is constructive because it’s a multiple of $\lambda$ ” (it is not an integer multiple).
Mixing up amplitude and intensity: intensity $\propto A^2$ .
Forgetting the sources are in phase (the interference conditions depend on that starting phase).

Things to Be Careful About

If amplitudes from X and Y are not equal at Q, the minimum may not be exactly zero, but it is still a minimum. Cambridge often accepts “minimum (may be zero)” wording.

Techniques used

convert path difference into a phase difference using wavelengthapply conditions for constructive and destructive interferencededuce relative intensity from interference condition

(iv)

A microwave detector is positioned at P and connected to a cathode-ray oscilloscope (CRO). The controls of the CRO are adjusted so that a waveform is shown on the screen.

Describe the changes to the amplitude of the waveform as the detector is moved from P to Q.

2M Similar

DifficultyMedium

Worked solution

Answer

At $P$ , $XP = YP$ so path difference $=0$ and the CRO waveform has maximum amplitude.

As the detector moves from $P$ to $Q$ , the path difference increases, so the amplitude alternates between maxima and minima (constructive and destructive interference).

Since at $Q$ the path difference is $6\ \text{cm} = 1.5\lambda$ , the amplitude at $Q$ is a minimum.

Final answer

Amplitude is maximum at P, then alternates through maxima/minima, ending at a minimum at Q.

Detailed explanation

Background Concept

The CRO displays a voltage signal from the detector that is proportional to the resultant wave amplitude at the detector. For two coherent sources, the resultant amplitude depends on their phase difference.

As you move the detector through space, the path lengths from the two sources change, so the path difference $\Delta$ changes, causing alternating:

maxima when $\Delta = m\lambda$ ,
minima when $\Delta = \left(m + \tfrac{1}{2}\right)\lambda$ .

Understanding the Question

The detector starts at $P$ , where $XP = YP$ , so $\Delta = 0$ .
It moves along the line from $P$ to $Q$ .
Earlier parts establish that at $Q$ , the path difference is $6\ \text{cm}$ , and $\lambda = 4.0\ \text{cm}$ , so $\Delta = 1.5\lambda$ at $Q$ .

You must describe how the CRO waveform amplitude changes as you move from $P$ to $Q$ .

Approach

Identify whether the starting point is a maximum or minimum (use path difference).
Determine whether the end point is a maximum or minimum.
State that as you move, you pass through alternating maxima and minima.

Step-by-Step Reasoning

At $P$ :

XP = YP \Rightarrow \Delta = 0 = 0\lambda

So $P$ is a constructive interference point → CRO amplitude is maximum.

Moving from $P$ towards $Q$ increases the asymmetry in path lengths, so $\Delta$ increases from $0$ up to $1.5\lambda$ .

The sequence of interference as $\Delta$ increases from $0$ to $1.5\lambda$ is:

at $\Delta = 0$ : maximum,
at $\Delta = 0.5\lambda$ : minimum,
at $\Delta = 1.0\lambda$ : maximum,
at $\Delta = 1.5\lambda$ : minimum (this is $Q$ ).

So the CRO amplitude falls from a maximum to a minimum, rises back to a maximum, then falls again to a minimum at $Q$ .

diagram-biv-1

Key Takeaways

CRO amplitude tracks resultant wave amplitude.
Moving the detector changes path difference, producing alternating constructive/destructive interference.
Identify the endpoints using $\Delta$ in units of $\lambda$ .

Common Mistakes

Saying amplitude “steadily decreases” (ignores alternating maxima/minima).
Confusing intensity and amplitude: intensity changes more sharply because $I \propto A^2$ .
Assuming Q is a maximum because it is “further away”.

Things to Be Careful About

In real experiments, minima may not reach exactly zero due to unequal amplitudes or reflections, but the pattern is still alternating maxima/minima.
Use the given wavelength to decide whether the endpoint is constructive or destructive.

Techniques used

relate CRO amplitude to received wave amplitude and intensitytrack how path difference changes as position changesapply alternating constructive and destructive interference conditions

Q5MediumElectricityD.C. Circuits

(a)

3M

(i)

State and explain the effect, if any, on the resistance of a filament wire in a lamp as the current in the wire decreases.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

As the current decreases, the filament becomes cooler, so its resistance decreases (lower temperature (\Rightarrow) lower resistivity).

Final answer

Resistance decreases (filament cools so resistivity falls).

Detailed explanation

Background Concept

A filament lamp is non-ohmic because its resistance depends on its temperature. The resistivity (\rho) of a metal increases with temperature, so the resistance

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

increases when the filament gets hotter.

The filament is heated mainly by electrical power

P = IV = I^2R.

So changing current changes heating, which changes temperature, which changes resistance.

Understanding the Question

You are asked what happens to the resistance of the filament wire when the current decreases, and you must explain the cause.

Approach

State whether (R) increases or decreases.
Explain using: lower current (\Rightarrow) less heating power (\Rightarrow) lower temperature (\Rightarrow) lower resistivity (\Rightarrow) lower resistance.

Step-by-Step Reasoning

When current decreases, the rate of electrical energy transfer to the filament decreases (from (P = I^2R) or (P=IV)).
The filament temperature therefore falls.
For a metal filament, a lower temperature means fewer lattice vibrations and less scattering of electrons, so resistivity (\rho) decreases.
Since (R \propto \rho), the resistance decreases.

Key Takeaways

Filament lamps are non-ohmic because (R) depends strongly on temperature.
Decreasing current cools the filament, reducing (\rho) and hence (R).

Common Mistakes

Saying resistance increases when current decreases (reverses the temperature argument).
Using (R = V/I) without considering that (V) may also change; the key physics is the temperature dependence.

Things to Be Careful About

The explanation must mention temperature (or heating) and the metal’s resistivity changing with temperature to secure the mark.

Techniques used

link current to filament temperature via heating effectuse the temperature dependence of resistivity to infer resistance change

(ii)

On Fig. 5.1, sketch the $I-V$ characteristic of a filament lamp.

Fig. 5.1

2M Similar

DifficultyMedium-Easy

Worked solution

diagram-5a-ii

Answer

Curve through the origin that becomes less steep as (V) increases (decreasing gradient).

Final answer

Non-linear curve through origin with decreasing gradient as V increases.

Detailed explanation

Background Concept

For an (I)-(V) graph, the gradient is

\text{gradient} = \frac{\Delta I}{\Delta V} \approx \frac{I}{V} = \frac{1}{R}.

So:

a steeper graph means smaller resistance,
a shallower graph means larger resistance.

A filament lamp heats up as the voltage/current increase. Hotter filament (\Rightarrow) larger resistivity (\Rightarrow) larger resistance. Therefore the resistance increases at higher currents.

Understanding the Question

You must sketch the (I)-(V) characteristic for a filament lamp on given axes ((I) vertical, (V) horizontal). The key feature is that the lamp is non-ohmic, so the graph is not a straight line.

Approach

Start at the origin (when (V=0), (I=0)).
At small (V), filament is cool (\Rightarrow) low (R) (\Rightarrow) large gradient.
As (V) increases, filament heats (\Rightarrow) (R) increases (\Rightarrow) gradient decreases.

Step-by-Step Reasoning

diagram-5a-ii

Draw a curve that initially rises quite steeply from the origin.
As (V) increases further, make the curve bend over so that it becomes less steep (concave down in the first quadrant).
This represents the increasing resistance as the filament temperature increases.

Key Takeaways

Filament lamp: non-linear (I)-(V) curve.
Increasing (V) heats filament, increasing (R), so (I) rises more slowly: decreasing gradient.

Common Mistakes

Drawing a straight line through the origin (that would be an ohmic conductor).
Drawing a curve that becomes steeper with (V) (wrong direction of temperature effect for metals).

Things to Be Careful About

The curve must pass through ((0,0)).
If you draw only the first quadrant, ensure the shape clearly shows decreasing gradient with increasing (V).

Techniques used

use the changing gradient of an I-V graph to represent changing resistancesketch a non-linear characteristic passing through the origin

(b)

A battery of electromotive force (e.m.f.) $E$ and negligible internal resistance is connected in parallel with two filament lamps A and B, as shown in Fig. 5.2.

Fig. 5.2

The current in the battery is $3.3\text{ A}$ and the current in lamp A is $1.5\text{ A}$ . The power dissipated in lamp A is $18\text{ W}$ .

5M

(i)

Calculate the e.m.f. $E$ of the battery.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For lamp A,

P = IV \Rightarrow V = \frac{P}{I} = \frac{18}{1.5} = 12\ \text{V}.

In parallel with negligible internal resistance, supply e.m.f. (E) equals the p.d. across lamp A.

Answer

E = 12\ \text{V}

Final answer

12 V

Detailed explanation

Background Concept

Electrical power transferred in a component is

P = IV,

where (I) is the current through the component and (V) is the potential difference across it.

In a parallel circuit, each branch is connected across the same two nodes, so the potential difference across each branch is the same as the supply potential difference. If the battery has negligible internal resistance, the terminal p.d. is essentially the e.m.f. (E).

Understanding the Question

Lamp A and lamp B are connected in parallel to a battery of e.m.f. (E) (internal resistance negligible). You are given for lamp A:

current (I_A = 1.5\ \text{A})
power (P_A = 18\ \text{W})

You must find (E).

Approach

Use (P = IV) for lamp A to find the p.d. across it.
Because the lamps are in parallel and internal resistance is negligible, set (E) equal to this p.d.

Step-by-Step Reasoning

Apply (P = IV) to lamp A:

V_A = \frac{P_A}{I_A} = \frac{18\ \text{W}}{1.5\ \text{A}} = 12\ \text{V}.

In parallel, (V_A = E). Therefore (E = 12\ \text{V}).

Key Takeaways

Use (P=IV) to find p.d. when power and current are known.
In parallel circuits, all branches share the same p.d. as the supply.

Common Mistakes

Using the total battery current (3.3 A) in (P=IV) for lamp A instead of (I_A).
Forgetting that in parallel, (V_A) equals the battery voltage.

Things to Be Careful About

The statement “negligible internal resistance” is the clue that terminal p.d. (\approx E).
Keep units consistent: (\text{W} / \text{A} = \text{V}).

Techniques used

use power equation P = IV to find the potential differenceuse the parallel-circuit rule that p.d. across each branch equals the supply e.m.f.

(ii)

The filament wire of lamp B has a cross-sectional area of $1.4 \times 10^{-9}\text{ m}^2$ . The number of free (conduction) electrons per unit volume in the metal of the filament wire is $3.4 \times 10^{28}\text{ m}^{-3}$ .

Calculate the average drift speed of the free electrons in the filament wire of lamp B.

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

3M Similar

DifficultyMedium

Worked solution

Working

Current in lamp B:

I_B = 3.3 - 1.5 = 1.8\ \text{A}.

Using (I = Anvq):

v = \frac{I}{Anq} = \frac{1.8}{\left(1.4\times 10^{-9}\right)\left(3.4\times 10^{28}\right)\left(1.6\times 10^{-19}\right)} = 2.36\times 10^{-1}\ \text{m s}^{-1}.

Answer

v = 0.24\ \text{m s}^{-1}

Final answer

0.24 m s^-1

Detailed explanation

Background Concept

In a metal, the electric current is due to drifting conduction electrons. The current is related to drift speed by

I = Anvq,

where:

(I) is current (A)
(A) is cross-sectional area of the wire (m(^2))
(n) is number density of free electrons (m(^{-3}))
(v) is drift speed (m s(^{-1}))
(q) is charge on each carrier (for electrons, (q = 1.6\times 10^{-19}\ \text{C})).

In a parallel circuit, currents add at a junction (Kirchhoff’s first law):

I_{\text{total}} = I_A + I_B.

Understanding the Question

Two lamps A and B are in parallel. You are given:

total current from the battery: (I_{\text{total}} = 3.3\ \text{A})
current in lamp A: (I_A = 1.5\ \text{A})
for lamp B filament: (A = 1.4\times 10^{-9}\ \text{m}^2), (n = 3.4\times 10^{28}\ \text{m}^{-3})

You must calculate the drift speed of electrons in lamp B, so you need (I_B) first, then use (I = Anvq).

Approach

Use current conservation to find (I_B) from the total current.
Rearrange (I = Anvq) to (v = I/(Anq)).
Substitute values in SI units, including (q = 1.6\times 10^{-19}\ \text{C}), and calculate.

Step-by-Step Reasoning

Find the current in branch B:

I_B = I_{\text{total}} - I_A = 3.3 - 1.5 = 1.8\ \text{A}.

Rearrange drift-current equation:

v = \frac{I}{Anq}.

Substitute:

\begin{aligned} Anq &= \left(1.4\times 10^{-9}\right)\left(3.4\times 10^{28}\right)\left(1.6\times 10^{-19}\right)\\ &= (1.4\times 3.4\times 1.6)\times 10^{-9+28-19}\\ &= 7.616\times 10^0\\ &= 7.616. \end{aligned}

So

\begin{aligned} v &= \frac{1.8}{7.616}\\ &= 0.236\ \text{m s}^{-1}\\ &\approx 0.24\ \text{m s}^{-1}. \end{aligned}

Key Takeaways

In parallel: currents split, and (I_{\text{total}} = I_A + I_B).
Drift speed comes from (v = I/(Anq)); it is usually small because (n) is very large.

Common Mistakes

Using (I = 3.3\ \text{A}) in (I = Anvq) instead of the branch current (I_B).
Forgetting the electron charge (q = 1.6\times 10^{-19}\ \text{C}).
Errors with powers of ten when multiplying (10^{-9}), (10^{28}), and (10^{-19}).

Things to Be Careful About

(A) must be in (\text{m}^2) and (n) in (\text{m}^{-3}) (already given in SI here).
Quote the final drift speed to 2–3 s.f. with unit (\text{m s}^{-1}).
The negative charge of electrons does not affect the speed (a magnitude); direction would be opposite to conventional current if asked.

Techniques used

use Kirchhoff's first law at a junction to find branch currentapply I = Anvq to determine drift speedsubstitute physical constants with correct SI unitshandle powers of ten in standard form

Q6MediumD.C. CircuitsElectricity

A battery of electromotive force (e.m.f.) $6.0\text{ V}$ and negligible internal resistance is connected in series with a variable resistor and a uniform resistance wire XY, as shown in Fig. 6.1.

Fig. 6.1

Wire XY has length $2.00\text{ m}$ and resistance $8.0\ \Omega$ . The resistance $R$ of the variable resistor is adjusted so that the potential difference across wire XY is $2.4\text{ V}$ .

(a)

Determine $R$ .

$R$ = ______ $\Omega$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Resistance of wire XY $= 8.0\ \Omega$ and p.d. across it $= 2.4\ \text{V}$ .

I = \frac{V}{R} = \frac{2.4}{8.0} = 0.30\ \text{A}

Total supply $= 6.0\ \text{V}$ , so p.d. across variable resistor $= 6.0 - 2.4 = 3.6\ \text{V}$ .

R = \frac{V}{I} = \frac{3.6}{0.30} = 12\ \Omega

Answer

$R = 12\ \Omega$

Final answer

12 Ω

Detailed explanation

Background Concept

In a series circuit, the same current flows through every component. The potential differences (p.d.s) across components add to the supply e.m.f. (here internal resistance is negligible, so the terminal p.d. is $6.0\ \text{V}$ ).

Ohm’s law for a component of resistance $R$ is

V = IR

So if we know the p.d. across a component and its resistance, we can find the circuit current.

Understanding the Question

A $6.0\ \text{V}$ ideal battery is connected in series with:

a variable resistor of resistance $R$ (unknown), and
a uniform resistance wire XY of resistance $8.0\ \Omega$ .

The p.d. across XY is adjusted to be $2.4\ \text{V}$ . We must find $R$ .

Approach

Use $V = IR$ on wire XY to find the series current $I$ .
Subtract the p.d. across XY from the supply to get the p.d. across $R$ .
Use $R = V/I$ for the variable resistor.

Step-by-Step Reasoning

For wire XY:

I = \frac{V_{XY}}{R_{XY}} = \frac{2.4}{8.0} = 0.30\ \text{A}

Since it is a series circuit, $0.30\ \text{A}$ also flows through the variable resistor.

The supply is $6.0\ \text{V}$ , so

V_R = 6.0 - 2.4 = 3.6\ \text{V}

Then

R = \frac{V_R}{I} = \frac{3.6}{0.30} = 12\ \Omega

Key Takeaways

Series circuit: same current through each component.
P.d.s add in series: $V_{\text{supply}} = V_1 + V_2 + \dots$ .
Use $V=IR$ to link p.d., current, and resistance.

Common Mistakes

Using $6.0\ \text{V}$ as the p.d. across the variable resistor (forgetting $2.4\ \text{V}$ is already across the wire).
Treating the wire and resistor as if they were in parallel (they are in series).
Arithmetic slip: $2.4/8.0$ must give $0.30\ \text{A}$ .

Things to Be Careful About

Internal resistance is stated negligible, so the full $6.0\ \text{V}$ is available across the series combination.
Quote the final resistance with appropriate significant figures (here $12\ \Omega$ is fine).

Techniques used

apply Ohm's law to a circuit componentuse series-circuit current equivalenceuse potential difference sharing in series circuits

(b)

Explain why the potential difference $V$ between any two points on wire XY is proportional to the distance $L$ between those points.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

For a uniform wire, cross-sectional area and resistivity are constant, so

R \propto L \quad (R = \rho L/A).

In the wire, the current $I$ is the same through all sections, so

V = IR.

Hence for any section of the wire,

V \propto R \propto L,

so the p.d. between two points is proportional to their separation along the wire.

Final answer

Because the wire is uniform so R ∝ L and, with constant current, V = IR ⇒ V ∝ L.

Detailed explanation

Background Concept

A uniform resistance wire has constant resistivity $\rho$ and constant cross-sectional area $A$ . Its resistance depends on its length $L$ by

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

So for a uniform wire, $R \propto L$ .

If a steady current $I$ flows through the wire, the potential difference across any section of resistance $R_{\text{section}}$ is given by Ohm’s law:

V_{\text{section}} = I R_{\text{section}}.

If $I$ is the same throughout the wire (series path, no branches), then $V_{\text{section}} \propto R_{\text{section}}$ .

Understanding the Question

You are asked to justify the key potentiometer idea: along the uniform wire XY, the p.d. between two points depends linearly on how far apart those points are.

So we need to connect “distance along the wire” to “resistance of that piece” and then to “potential difference across that piece”.

Approach

Consider a short section of the wire of length $L$ .
Use $R = \rho L/A$ to show that the section’s resistance is proportional to its length.
Use $V = IR$ with the same current $I$ through all sections to show $V \propto L$ .

Step-by-Step Reasoning

Take any two points on the wire, separated by distance $L$ .

Because the wire is uniform:

R_{\text{between points}} = \frac{\rho L}{A} \implies R_{\text{between points}} \propto L.

Since the wire is part of a series circuit, the current in the wire is the same through every section (no current is “used up” in earlier parts of the wire).

Applying Ohm’s law to just that section:

V_{\text{between points}} = I R_{\text{between points}}.

With $I$ constant for all sections, this gives

V_{\text{between points}} \propto R_{\text{between points}} \propto L.

So the p.d. is proportional to the distance between the points.

diagram-b1

Key Takeaways

Uniform wire: $R \propto L$ .
Same current through each section of a series conductor.
Therefore $V = IR$ makes p.d. proportional to length (constant potential gradient).

Common Mistakes

Saying “voltage is shared equally” without the condition of equal lengths (it is shared in proportion to resistance, and resistance is proportional to length only for a uniform wire).
Forgetting to mention constant cross-sectional area / resistivity when using $R \propto L$ .
Confusing current and voltage: it is current that is the same in series, not voltage.

Things to Be Careful About

The statement $V \propto L$ only holds if the wire is uniform (constant $\rho$ and $A$ ) and if the current is steady.
If significant current is taken from a point on the wire (a “load” connected), the current in the wire would not be constant along its length and the linear relationship would no longer be exact.

Techniques used

use proportionality of resistance with length for a uniform wireapply Ohm's law to relate potential difference to resistancecombine proportionalities to deduce a linear relationship

(c)

A cell of e.m.f. $E$ and internal resistance $r$ is connected to the circuit, as shown in Fig. 6.2.

Fig. 6.2

Resistance $R$ is unchanged.

The movable connection P is positioned on wire XY so that the galvanometer reading is zero. Distance XP is $1.24\text{ m}$ .

4M

(i)

Calculate $E$ .

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

2M Similar

DifficultyMedium

Worked solution

Working

Potential gradient along XY:

k = \frac{V_{XY}}{L_{XY}} = \frac{2.4}{2.00} = 1.20\ \text{V m}^{-1}

At balance (galvanometer zero), $E = V_{XP}$ :

E = kL_{XP} = 1.20 \times 1.24 = 1.49\ \text{V}

Answer

$E = 1.49\ \text{V}$

Final answer

1.49 V

Detailed explanation

Background Concept

A potentiometer uses a uniform wire carrying a steady current to create a uniform potential gradient (constant p.d. per unit length).

If the p.d. across the entire wire of length $L_{XY}$ is $V_{XY}$ , then the potential gradient is

k = \frac{V_{XY}}{L_{XY}} \quad (\text{V m}^{-1}).

For any point P at distance $L_{XP}$ from X, the p.d. between X and P is

V_{XP} = kL_{XP}.

At the null point (galvanometer reading zero), no current flows in the secondary circuit, so the cell’s terminal p.d. equals its e.m.f. (because there is no voltage drop across its internal resistance). Therefore,

E = V_{XP}.

Understanding the Question

The main circuit is unchanged from earlier: wire XY still has a known p.d. across it. A test cell (e.m.f. $E$ , internal resistance $r$ ) is connected between X and a sliding contact P via a galvanometer.

When the galvanometer reads zero, P is at a balance position. You are told $XP = 1.24\ \text{m}$ and must calculate $E$ .

Approach

Find the potential gradient along XY using the given p.d. across the full wire and its length.
Multiply by the balance length $XP$ to get $V_{XP}$ .
Use the null condition to set $E = V_{XP}$ .

Step-by-Step Reasoning

Across the full wire XY:

$V_{XY} = 2.4\ \text{V}$
$L_{XY} = 2.00\ \text{m}$

So the potential gradient is

k = \frac{2.4}{2.00} = 1.20\ \text{V m}^{-1}.

At balance, with $XP = 1.24\ \text{m}$ :

V_{XP} = kL_{XP} = 1.20 \times 1.24 = 1.488\ \text{V}.

Because the galvanometer reads zero, no current flows through the test cell, so there is no $Ir$ drop inside it and its terminal p.d. equals its e.m.f. Hence

E = 1.49\ \text{V} \; (\text{to 3 s.f.}).

diagram-ci1

Key Takeaways

Potentiometer null point: galvanometer zero means $E$ equals the p.d. across the corresponding wire length.
Use a potential gradient $k = V/L$ for uniform wires.
Internal resistance does not affect the result at null because current in the test circuit is zero.

Common Mistakes

Using the full $6.0\ \text{V}$ instead of the $2.4\ \text{V}$ across the potentiometer wire.
Forgetting that $L_{XY} = 2.00\ \text{m}$ is needed to find the gradient.
Thinking $E$ depends on $r$ here; at balance it does not.

Things to Be Careful About

Ensure lengths are in metres if you want $k$ in $\text{V m}^{-1}$ .
Rounding: $1.488\ \text{V}$ can be quoted as $1.49\ \text{V}$ (3 s.f.) or $1.5\ \text{V}$ (2 s.f.); keep consistent with expected exam precision.

Techniques used

use the null condition of a potentiometer to equate emfsdetermine potential gradient along a uniform wirescale potential difference by length ratio

(ii)

The value of $R$ is now decreased.

State and explain the change that must be made to the position of P on wire XY so that the galvanometer reads zero again.

2M Similar

DifficultyMedium

Worked solution

Answer

Decreasing $R$ increases the current in the main circuit, so the p.d. across XY (and the potential gradient along the wire) increases.

Since at balance $E = V_{XP} = kL_{XP}$ and $E$ is unchanged, a larger $k$ means $L_{XP}$ must be smaller.

So P must be moved towards X (reduce $XP$ ) to obtain zero galvanometer reading again.

Final answer

Move P towards X (smaller XP) because the potential gradient increases when R is decreased.

Detailed explanation

Background Concept

For the potentiometer wire, the p.d. per unit length (potential gradient) is

k = \frac{V_{XY}}{L_{XY}}.

The p.d. across the wire is also

V_{XY} = I R_{XY}

where $I$ is the main-circuit current and $R_{XY}$ is fixed (wire resistance doesn’t change).

At the balance (null) point,

E = V_{XP} = kL_{XP}.

So if $k$ changes but the cell e.m.f. $E$ does not, the balance length $L_{XP}$ must adjust inversely.

Understanding the Question

You reduce the variable resistor value $R$ in the main circuit while keeping the same $6.0\ \text{V}$ supply and the same potentiometer wire XY.

You need to state what change to the slider position P is required to get the galvanometer back to zero, and explain why.

Approach

Decide how reducing $R$ affects the main-circuit current $I$ .
Use $V_{XY} = IR_{XY}$ to decide how the p.d. across XY changes.
Convert that into how the potential gradient $k$ changes.
Use the balance condition $E = kL_{XP}$ to infer the required change in $L_{XP}$ , hence the direction to move P.

Step-by-Step Reasoning

The main circuit is a series circuit: supply, variable resistor $R$ , and wire XY.
If $R$ is decreased, the total series resistance decreases.

Therefore the main current increases:

I = \frac{V}{R_{\text{total}}} \quad \Rightarrow \quad R_{\text{total}} \downarrow \; \Rightarrow \; I \uparrow.

The resistance of the potentiometer wire is fixed ( $R_{XY}$ constant), so the p.d. across the wire becomes larger:

V_{XY} = IR_{XY} \quad \Rightarrow \quad I \uparrow \; \Rightarrow \; V_{XY} \uparrow.

Since $L_{XY}$ is fixed, the potential gradient increases:

k = \frac{V_{XY}}{L_{XY}} \uparrow.

At balance,

E = kL_{XP}.

$E$ is the property of the test cell and is unchanged. If $k$ increases, then to keep $kL_{XP}$ the same, $L_{XP}$ must decrease.

So the contact P must move closer to X, reducing $XP$ , until the p.d. across XP matches $E$ again.

diagram-cii1

Key Takeaways

Reducing a series resistor increases current.
Higher current through the potentiometer wire increases its potential gradient.
Balance condition $E = kL$ implies larger $k$ requires smaller balance length $L$ .

Common Mistakes

Saying P must move towards Y (this would increase $L_{XP}$ and make $V_{XP}$ even bigger when $k$ has increased).
Thinking the galvanometer reads zero when $V_{XP} = 0$ ; actually it reads zero when $V_{XP}$ equals the cell e.m.f. in magnitude (with correct polarity).
Forgetting that $E$ is unchanged by changing $R$ in the main circuit.

Things to Be Careful About

The reasoning depends on the wire’s resistance being constant; heating could slightly change it in real life, but this is usually neglected at this level.
Direction statement must be clear: “towards X” or “reduce $XP$ ”.

Techniques used

predict the effect of changing series resistance on circuit currentrelate potential gradient to current in a uniform wireuse the balance condition to infer how the balance length changes

Q7Medium-EasyParticle Physics

(a)

State the names of two different leptons.

1 ______

2 ______

2M Similar

(b)

In the following list, underline all the particles that are hadrons.

antineutrino beta-plus meson neutron

1M Similar

(c)

By reference to quark composition, show that the charge of a proton is $+1.6 \times 10^{-19}\text{ C}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Proton composition is $uud$ .

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

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

With $e = 1.6 \times 10^{-19}\ \text{C}$ ,

Q = +1.6 \times 10^{-19}\ \text{C}

Answer

$+1.6 \times 10^{-19}\ \text{C}$

Final answer

+1.6 × 10^-19 C

Detailed explanation

Background Concept

Quarks have fractional charges in multiples of the elementary charge $e$ :

up-type quarks ( $u$ , $c$ , $t$ ) have charge $+\frac{2}{3}e$
down-type quarks ( $d$ , $s$ , $b$ ) have charge $-\frac{1}{3}e$

A proton is a baryon made of three quarks, with composition $uud$ .

The elementary charge is:

e = 1.6 \times 10^{-19}\ \text{C}

Understanding the Question

You must use the quark model of the proton to calculate its total charge and show that it equals $+1.6 \times 10^{-19}\ \text{C}$ . This is effectively showing that the proton has charge $+e$ .

Approach

Write the proton quark content ( $uud$ ).
Replace each quark by its charge in units of $e$ .
Add the three charges to get the total in units of $e$ .
Multiply by $e = 1.6 \times 10^{-19}\ \text{C}$ .

Step-by-Step Reasoning

Proton quark composition: $p = uud$
Charges of each quark: $Q_u = +\frac{2}{3}e, \qquad Q_d = -\frac{1}{3}e$
Total proton charge: $Q_p = Q_u + Q_u + Q_d$ Substitute: $Q_p = \left(+\frac{2}{3}e\right) + \left(+\frac{2}{3}e\right) + \left(-\frac{1}{3}e\right)$
Add the fractions: $Q_p = \left(\frac{2}{3}+\frac{2}{3}-\frac{1}{3}\right)e = \left(\frac{3}{3}\right)e = 1e$
Convert to coulombs using $e = 1.6 \times 10^{-19}\ \text{C}$ : $Q_p = 1 \times 1.6 \times 10^{-19}\ \text{C} = +1.6 \times 10^{-19}\ \text{C}$

Key Takeaways

Proton is $uud$ .
Add quark charges (in units of $e$ ) to get the hadron’s total charge.
Proton total is $+1e = +1.6 \times 10^{-19}\ \text{C}$ .

Common Mistakes

Using the wrong quark composition (e.g. writing $udd$ , which is a neutron).
Swapping the quark charges (writing $u$ as $-\frac{1}{3}e$ ).
Forgetting to multiply by $e$ to convert from “in units of $e$ ” to coulombs.
Missing the sign of the charge.

Things to Be Careful About

Keep the elementary charge symbol $e$ separate from scientific notation (do not confuse it with “ $\times 10^$ ”).
Show the fractional-charge addition clearly; that is usually where the marks are.
Final answer must include the unit $\text{C}$ and the positive sign is implied but may be written.

Techniques used

use quark composition of a protonsum quark charges in units of the elementary chargeconvert from multiples of $e$ to coulombs

Physics 9702/24 — May/June 2025

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question