Physics 9702 / 21 — October/November 2024

(b)

Fig. 1.1 shows a cuboidal glass block.

Fig. 1.1

A student measures the mass $m$ of the block and the side lengths $x$ , $y$ and $z$ . The measurements are shown in Table 1.1.

Table 1.1

quantity	measurement
$m$	$(0.243 \pm 0.001)\text{ kg}$
$x$	$(5.41 \pm 0.01)\text{ cm}$
$y$	$(11.09 \pm 0.01)\text{ cm}$
$z$	$(1.62 \pm 0.01)\text{ cm}$

5M

(i)

Determine the density of the glass.

density = ______ $\text{kg m}^{-3}$

DifficultyMedium-Easy

Worked solution

Working

Convert to metres:

x = 5.41\,\text{cm} = 0.0541\,\text{m},\quad y = 11.09\,\text{cm} = 0.1109\,\text{m},\quad z = 1.62\,\text{cm} = 0.0162\,\text{m}

Volume:

V = xyz = (0.0541)(0.1109)(0.0162) = 9.72 \times 10^{-5}\,\text{m}^3

Density:

\rho = \frac{m}{V} = \frac{0.243}{9.72 \times 10^{-5}} = 2.50 \times 10^{3}\,\text{kg m}^{-3}

Answer

$2.50 \times 10^{3}\ \text{kg m}^{-3}$

Final answer

2.50 × 10^3 kg m^-3

Detailed explanation

Background Concept

For a uniform solid,

\rho = \frac{m}{V}

For a cuboid (rectangular block), the volume is the product of its three perpendicular side lengths:

V = xyz

To obtain density in SI units ( $\text{kg m}^{-3}$ ), mass must be in $\text{kg}$ and volume in $\text{m}^3$ . Since the lengths are given in $\text{cm}$ , they must be converted to $\text{m}$ before calculating $V$ .

Understanding the Question

You are given:

mass $m = (0.243 \pm 0.001)\,\text{kg}$
side lengths $x = (5.41 \pm 0.01)\,\text{cm}$ , $y = (11.09 \pm 0.01)\,\text{cm}$ , $z = (1.62 \pm 0.01)\,\text{cm}$

You must find the density of the glass block, using the central values of the measurements.

Approach

Convert $x$ , $y$ , $z$ from $\text{cm}$ to $\text{m}$ .
Compute the volume $V = xyz$ .
Use $\rho = m/V$ .
Quote the result with appropriate significant figures and the correct SI unit $\text{kg m}^{-3}$ .

Step-by-Step Reasoning

Convert lengths:
- $5.41\,\text{cm} = 5.41 \times 10^{-2}\,\text{m} = 0.0541\,\text{m}$
- $11.09\,\text{cm} = 0.1109\,\text{m}$
- $1.62\,\text{cm} = 0.0162\,\text{m}$
Calculate volume:

V = (0.0541)(0.1109)(0.0162) = 9.72 \times 10^{-5}\,\text{m}^3

Calculate density:

\rho = \frac{0.243\,\text{kg}}{9.72 \times 10^{-5}\,\text{m}^3} = 2.50 \times 10^3\,\text{kg m}^{-3}

The value is sensible for glass (typically around $2.5 \times 10^3\,\text{kg m}^{-3}$ ).

Key Takeaways

Use $\rho = m/V$ and for a cuboid $V = xyz$ .
Convert $\text{cm}$ to $\text{m}$ before calculating $V$ in $\text{m}^3$ .
Final unit must be $\text{kg m}^{-3}$ .

Common Mistakes

Forgetting to convert from $\text{cm}$ to $\text{m}$ (gives a density too small by a factor of $10^6$ because volume involves $\text{cm}^3$ ).
Using $V = x + y + z$ instead of multiplying.
Writing the unit as $\text{kg m}^3$ or $\text{kg}/\text{m}^3$ (CIE expects $\text{kg m}^{-3}$ ).

Things to Be Careful About

Converting all three lengths consistently: $1\,\text{cm} = 10^{-2}\,\text{m}$ .
Significant figures: measurements are typically 3 s.f., so quoting $\rho$ to 3 s.f. is appropriate here.
Keep track of powers of ten when dividing by a small volume.

Techniques used

convert measurements to SI unitscalculate volume of a cuboidapply the density equationuse correct significant figures and units

(ii)

Calculate the percentage uncertainty in the density.

percentage uncertainty = ______ $\%$

DifficultyMedium

Worked solution

Working

For $\rho = \dfrac{m}{xyz}$ ,

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta x}{x} + \frac{\Delta y}{y} + \frac{\Delta z}{z}

\frac{\Delta m}{m} = \frac{0.001}{0.243} = 0.00412

\frac{\Delta x}{x} = \frac{0.01}{5.41} = 0.00185,\quad \frac{\Delta y}{y} = \frac{0.01}{11.09} = 0.000902,\quad \frac{\Delta z}{z} = \frac{0.01}{1.62} = 0.00617

\frac{\Delta \rho}{\rho} = 0.00412 + 0.00185 + 0.000902 + 0.00617 = 0.0130

Percentage uncertainty:

0.0130 \times 100\% = 1.30\%

Answer

$1.3\%$

Final answer

1.3 %

Detailed explanation

Background Concept

When quantities are multiplied or divided, their fractional (or percentage) uncertainties add.

If

Q = \frac{a b}{c d}

then

\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}

This works for powers too: for $Q = a^n$ , the fractional uncertainty is $n\,\Delta a/a$ . Here, the density is

\rho = \frac{m}{xyz}

so we add the fractional uncertainties in $m$ , $x$ , $y$ , and $z$ .

Understanding the Question

You have already calculated $\rho$ in (b)(i). Now you must determine how uncertain that value is, using the uncertainties given with each measured quantity.

The question asks specifically for percentage uncertainty, so your final result should be in $\%$ .

Approach

Write down the rule for fractional uncertainty for $\rho = m/(xyz)$ .
Calculate each fractional uncertainty: $\Delta m/m$ , $\Delta x/x$ , $\Delta y/y$ , $\Delta z/z$ .
Add them to get $\Delta \rho/\rho$ .
Multiply by $100$ to convert to a percentage.

Step-by-Step Reasoning

Start with the relationship:

\rho = \frac{m}{xyz}

So,

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta x}{x} + \frac{\Delta y}{y} + \frac{\Delta z}{z}

Compute each term from the table:

Mass term:

\frac{\Delta m}{m} = \frac{0.001}{0.243} \approx 0.00412 \quad (0.412\%)

Length terms:

\frac{\Delta x}{x} = \frac{0.01}{5.41} \approx 0.00185 \quad (0.185\%)

\frac{\Delta y}{y} = \frac{0.01}{11.09} \approx 0.000902 \quad (0.0902\%)

\frac{\Delta z}{z} = \frac{0.01}{1.62} \approx 0.00617 \quad (0.617\%)

Notice the $z$ measurement contributes the biggest fractional uncertainty because $z$ is the smallest length but has the same absolute uncertainty ( $\pm 0.01\,\text{cm}$ ).

Add them:

\frac{\Delta \rho}{\rho} \approx 0.00412 + 0.00185 + 0.000902 + 0.00617 = 0.0130

Convert to percentage:

\text{percentage uncertainty} = 0.0130 \times 100\% = 1.30\% \approx 1.3\%

Key Takeaways

For products/quotients, add fractional (percentage) uncertainties.
The smallest dimension often dominates the uncertainty in volume because its fractional uncertainty is largest.
Always finish by converting to a percentage if asked.

Common Mistakes

Subtracting uncertainties because one quantity is in the denominator (you still add fractional uncertainties).
Forgetting to multiply by $100$ to get a percentage.
Using absolute uncertainties directly (e.g. adding $0.001 + 0.01 + 0.01 + 0.01$ ), which is invalid.
Converting to metres before calculating uncertainties: not needed, because $\Delta x/x$ is the same in any consistent unit.

Things to Be Careful About

Use the same units for $\Delta x$ and $x$ (here both in $\text{cm}$ is fine).
Keep enough significant figures during intermediate steps so rounding does not distort the final percentage.
Check which measurement dominates: this helps spot arithmetic errors (here $z$ should contribute most).

Techniques used

propagate uncertainties through multiplication and division by adding fractional uncertaintiesconvert fractional uncertainty to percentage uncertaintyidentify dominant sources of uncertainty

(c)

The true value of the density of the glass is different from the answer in (b)(i) because of a systematic error in the measurements.

Suggest one possible cause of this systematic error.

DifficultyMedium-Easy

Worked solution

Answer

Zero error in the measuring instrument (e.g. balance not zeroed / vernier calipers have a zero error), so all readings are shifted.

Final answer

Zero error in an instrument (e.g. balance not zeroed or calipers zero error).

Detailed explanation

Background Concept

A systematic error causes measurements to be consistently too large or too small by a similar amount (or proportion) each time. This shifts the final calculated result away from the true value in a consistent direction.

Common causes include:

zero error (instrument does not read zero when it should)
calibration error (scale spacing incorrect)
consistent parallax due to a fixed viewing angle

This differs from random error, which produces scatter about a mean value and can be reduced by repeats.

Understanding the Question

The density you calculated from the measured $m$ , $x$ , $y$ , and $z$ differs from the true density because there is a systematic error in the measurements. You must suggest one plausible cause.

Any one valid cause is sufficient for 1 mark, but it must be systematic (biasing all measurements the same way), not just “human error”.

Approach

Identify a measurement that could be biased in the same direction every time and state the reason. Typical answers refer to:

the balance (mass)
the ruler/vernier calipers/micrometer (length)

Step-by-Step Reasoning

If the balance is not zeroed before measuring, every mass reading has the same offset. That makes $m$ consistently too large or too small, so $\rho = m/V$ is also consistently too large or too small.
Similarly, if vernier calipers have a zero error, every length is shifted. Because $V = xyz$ , this bias affects the volume and therefore density systematically.

Either of these is a valid single cause.

Key Takeaways

Systematic error = consistent bias, not scatter.
Zero error and calibration error are the most common systematic errors in basic measurements.

Common Mistakes

Writing “reaction time” (relevant to timing, and usually random).
Writing “parallax error” without indicating it is consistent (parallax can be random if viewing angle changes each time).
Saying “instrument not precise” (precision refers to random uncertainty).

Things to Be Careful About

Your suggestion must clearly be systematic (same direction each time).
Link the cause to a specific measurement (mass or one of the lengths) for maximum clarity.

Techniques used

distinguish systematic error from random erroridentify how instrument calibration affects all readings in the same direction

Q2MediumDynamicsWork, Energy and PowerKinematics

(a)

Define linear momentum.

(b)

A car of mass $1800\text{ kg}$ is moving in a straight line. Fig. 2.1 shows the variation with time $t$ of the momentum $p$ of the car.

Fig. 2.1

7M

(i)

Calculate the maximum speed reached by the car.

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

DifficultyMedium-Easy

Worked solution

Working

From graph, $p_{\max} = 7.2 \times 10^{4}\ \text{N s}$ .

v_{\max} = \frac{p_{\max}}{m} = \frac{7.2 \times 10^{4}}{1800} = 40\ \text{m s}^{-1}.

Answer

$40\ \text{m s}^{-1}$

Final answer

40 m s^-1

Detailed explanation

Background Concept

For motion in a straight line, linear momentum and velocity are related by

\vec{p} = m\vec{v}.

If the mass is constant, the speed is found from

v = \frac{p}{m}.

The graph gives momentum as a function of time, so the maximum speed occurs when the momentum is at its maximum.

Understanding the Question

Mass of car: $m = 1800\ \text{kg}$ .
From the momentum–time graph, the largest momentum is at the peak (at $t = 8.0\ \text{s}$ ): $p_{\max} = 7.2 \times 10^{4}\ \text{N s}$ .
Required: maximum speed $v_{\max}$ .

Approach

Read the maximum momentum from the graph.
Use $v = p/m$ .

Step-by-Step Reasoning

Identify the peak momentum on the graph: $p_{\max} = 7.2 \times 10^{4}\ \text{N s}$ .
Substitute into $v = p/m$ :

v_{\max} = \frac{7.2 \times 10^{4}}{1800}.

Calculate:

v_{\max} = 40\ \text{m s}^{-1}.

(Using $\text{N s} = \text{kg m s}^{-1}$ ensures the units reduce to $\text{m s}^{-1}$ .)

Key Takeaways

The maximum speed corresponds to the maximum momentum (for constant mass).
Use $v = p/m$ and keep track of units.

Common Mistakes

Forgetting that the vertical axis is in $10^{4}\ \text{N s}$ (i.e. misreading $7.2$ as $7.2\ \text{N s}$ ).
Using $p = mv^2$ (incorrect).

Things to Be Careful About

Ensure the factor $10^{4}$ from the axis label is included.
Quote speed in $\text{m s}^{-1}$ .

Techniques used

read the maximum value from a graphuse the relationship p = mvrearrange an equation to isolate the required quantity

(ii)

Calculate the maximum kinetic energy of the car.

maximum kinetic energy = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

From (i), $v_{\max} = 40\ \text{m s}^{-1}$ .

E_{k,\max} = \frac{1}{2}mv_{\max}^2 = \frac{1}{2}(1800)(40^2) = 1.44 \times 10^{6}\ \text{J} \approx 1.4 \times 10^{6}\ \text{J}.

Answer

$1.4 \times 10^{6}\ \text{J}$

Final answer

1.4 × 10^6 J

Detailed explanation

Background Concept

Kinetic energy is the energy associated with motion:

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

If you are given momentum instead of speed, you can find speed using $p = mv$ (for constant mass).

Understanding the Question

The graph shows that the car reaches a maximum momentum of $7.2 \times 10^{4}\ \text{N s}$ , and the car’s mass is $1800\ \text{kg}$ . The maximum kinetic energy occurs at the maximum speed (same time as maximum momentum).

Approach

Find $v_{\max}$ using $v = p/m$ (or use your answer from part (i)).
Substitute $v_{\max}$ into $E_k = \tfrac{1}{2}mv^2$ .

Step-by-Step Reasoning

From the peak momentum:

v_{\max} = \frac{7.2 \times 10^{4}}{1800} = 40\ \text{m s}^{-1}.

Kinetic energy at that speed:

E_{k,\max} = \frac{1}{2}(1800)(40^2).

Compute $40^2 = 1600$ , so

E_{k,\max} = 900 \times 1600 = 1.44 \times 10^{6}\ \text{J}.

To 2 s.f. (limited by the graph reading $7.2$ ), this is $1.4 \times 10^{6}\ \text{J}$ .

Key Takeaways

Maximum $E_k$ occurs at maximum $v$ .
Combine $p = mv$ with $E_k = \tfrac{1}{2}mv^2$ when momentum is given.

Common Mistakes

Using the momentum value directly in the kinetic energy formula without converting to speed.
Squaring the mass by mistake.

Things to Be Careful About

Significant figures: the momentum read from the graph is typically 2 s.f., so the final energy should be quoted accordingly.
Keep units consistent: $m$ in $\text{kg}$ and $v$ in $\text{m s}^{-1}$ gives $E_k$ in joules.

Techniques used

convert from momentum to speed using p = mvapply the kinetic energy formulasubstitute values and calculate with correct units

(iii)

Show that the acceleration of the car at time $t = 4.0\text{ s}$ is $5.0\text{ m s}^{-2}$ .

DifficultyMedium

Worked solution

Working

From $t = 0$ to $8.0\ \text{s}$ ,

\frac{\Delta p}{\Delta t} = \frac{7.2 \times 10^{4} - 0}{8.0 - 0} = 9.0 \times 10^{3}\ \text{N}.

F = \frac{\mathrm{d}p}{\mathrm{d}t} = 9.0 \times 10^{3}\ \text{N}, \qquad a = \frac{F}{m} = \frac{9.0 \times 10^{3}}{1800} = 5.0\ \text{m s}^{-2}.

Answer

$a = 5.0\ \text{m s}^{-2}$ at $t = 4.0\ \text{s}$ .

Final answer

5.0 m s^-2

Detailed explanation

Background Concept

A key link between dynamics and momentum is

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

where $F$ is the resultant force in the direction of motion.

If the mass is constant, $p = mv$ and differentiating gives

\frac{\mathrm{d}p}{\mathrm{d}t} = m\frac{\mathrm{d}v}{\mathrm{d}t} = ma,

so you can also use Newton’s second law:

F = ma.

On a momentum–time graph, the gradient equals $\mathrm{d}p/\mathrm{d}t$ , i.e. the resultant force.

Understanding the Question

At $t = 4.0\ \text{s}$ the car is on the rising straight-line section of the momentum–time graph (from $0$ to $8.0\ \text{s}$ ). Because this section is a straight line, the gradient (and hence force and acceleration) is constant throughout that interval. You are asked to show the acceleration is $5.0\ \text{m s}^{-2}$ .

Approach

Find the gradient of the straight-line part from $0$ to $8.0\ \text{s}$ .
Interpret that gradient as $F = \mathrm{d}p/\mathrm{d}t$ .
Use $a = F/m$ .

Step-by-Step Reasoning

Read endpoints of the first straight segment:

At $t = 0$ , $p = 0$ .
At $t = 8.0\ \text{s}$ , $p = 7.2 \times 10^{4}\ \text{N s}$ .

Compute gradient:

\frac{\Delta p}{\Delta t} = \frac{7.2 \times 10^{4}}{8.0} = 9.0 \times 10^{3}.

Units:

\frac{\text{N s}}{\text{s}} = \text{N},

so this gradient is a force. Therefore

F = 9.0 \times 10^{3}\ \text{N}.

Apply Newton’s second law to find acceleration:

a = \frac{F}{m} = \frac{9.0 \times 10^{3}}{1800} = 5.0\ \text{m s}^{-2}.

Because $t = 4.0\ \text{s}$ lies in the same constant-gradient region, $a = 5.0\ \text{m s}^{-2}$ at that time.

Key Takeaways

Gradient of a $p$ – $t$ graph equals resultant force.
If mass is constant, $a = (1/m)(\mathrm{d}p/\mathrm{d}t)$ .
Straight-line segment means constant acceleration.

Common Mistakes

Treating the gradient as acceleration directly (it is force, not acceleration).
Using $p/t$ instead of $\Delta p/\Delta t$ from two points on the straight line.

Things to Be Careful About

Use two well-separated points on the straight-line section to reduce reading error.
Include the $10^{4}$ factor from the graph’s momentum scale when calculating $\Delta p$ .

Techniques used

find a gradient from a straight-line graphuse F = dp/dt to determine resultant forceapply Newton’s second law to find acceleration

(iv)

Determine the distance travelled by the car between times $t = 0$ and $t = 12.0\text{ s}$ .

distance = ______ $\text{m}$

DifficultyMedium

Worked solution

Working

Distance

s = \int_0^{12} v\,\mathrm{d}t = \frac{1}{m}\int_0^{12} p\,\mathrm{d}t.

Area under $p$ – $t$ graph $=$ area of two triangles:

A_1 = \frac{1}{2}(8.0)(7.2 \times 10^{4}) = 2.88 \times 10^{5}

A_2 = \frac{1}{2}(4.0)(7.2 \times 10^{4}) = 1.44 \times 10^{5}

A = A_1 + A_2 = 4.32 \times 10^{5}.

s = \frac{A}{m} = \frac{4.32 \times 10^{5}}{1800} = 240\ \text{m}.

Answer

$240\ \text{m}$

Final answer

240 m

Detailed explanation

Background Concept

Distance travelled is the area under a velocity–time graph:

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

Here you are given momentum rather than velocity, but for constant mass:

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

So

s = \int_0^{12} \frac{p}{m}\,\mathrm{d}t = \frac{1}{m}\int_0^{12} p\,\mathrm{d}t.

That means: distance = (area under the momentum–time graph) divided by mass.

Understanding the Question

You need the distance travelled from $t = 0$ to $t = 12.0\ \text{s}$ . The $p$ – $t$ graph is piecewise linear:

From $0$ to $8.0\ \text{s}$ it rises linearly from $0$ to $7.2 \times 10^{4}\ \text{N s}$ .
From $8.0$ to $12.0\ \text{s}$ it falls linearly back to $0$ .

So the area under the graph is the sum of two triangles.

Approach

Find the total area under the $p$ – $t$ graph by geometry.
Divide by $m$ to convert $\int p\,\mathrm{d}t$ into $\int v\,\mathrm{d}t$ .

Step-by-Step Reasoning

First triangle ( $0$ to $8.0\ \text{s}$ ):

A_1 = \frac{1}{2} \times 8.0 \times 7.2\times 10^{4} = 2.88 \times 10^{5}.

Second triangle ( $8.0$ to $12.0\ \text{s}$ has base $4.0\ \text{s}$ ):

A_2 = \frac{1}{2} \times 4.0 \times 7.2\times 10^{4} = 1.44 \times 10^{5}.

Total area:

A = A_1 + A_2 = 4.32 \times 10^{5}.

Convert to distance using $s = A/m$ :

s = \frac{4.32 \times 10^{5}}{1800} = 240\ \text{m}.

Unit check: $A$ has units $(\text{N s})\cdot \text{s} = \text{N s}^2 = \text{kg m}$ , and dividing by $\text{kg}$ gives $\text{m}$ .

Key Takeaways

Distance is area under the $v$ – $t$ graph.
If given a $p$ – $t$ graph and $m$ is constant, use $v = p/m$ so distance is area under $p$ – $t$ divided by $m$ .
Break piecewise-linear graphs into simple shapes.

Common Mistakes

Treating the area under a momentum–time graph as momentum change (it is not).
Forgetting to divide the area by the mass.
Forgetting the $10^{4}$ factor from the momentum axis.

Things to Be Careful About

Use the correct base lengths: $8.0\ \text{s}$ for the first triangle and $4.0\ \text{s}$ for the second.
Keep consistent powers of ten when multiplying $7.2 \times 10^{4}$ .

Techniques used

convert a momentum-time graph to a velocity-time relationshipuse area under a graph to find a physical quantitysplit an area into simple geometric shapes and sum them

(c)

On Fig. 2.2, sketch the variation with time $t$ of the acceleration $a$ of the car in (b) from $t = 0$ to $t = 12.0\text{ s}$ .

Fig. 2.2

DifficultyMedium

Worked solution

diagram-2c-1

Answer

From $t=0$ to $8.0\ \text{s}$ : $a = +5.0\ \text{m s}^{-2}$ (constant).

From $t=8.0$ to $12.0\ \text{s}$ :

\frac{\Delta p}{\Delta t} = \frac{0-7.2\times 10^{4}}{12.0-8.0} = -1.8\times 10^{4}\ \text{N}

a = \frac{1}{m}\frac{\Delta p}{\Delta t} = \frac{-1.8\times 10^{4}}{1800} = -10\ \text{m s}^{-2}

So sketch a horizontal line at $+5.0$ from $0$ to $8.0\ \text{s}$ , then a horizontal line at $-10$ from $8.0$ to $12.0\ \text{s}$ .

Final answer

a = +5.0 m s^-2 (0–8 s), then a = −10 m s^-2 (8–12 s)

Detailed explanation

Background Concept

Acceleration is the rate of change of velocity:

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

With constant mass $m$ , momentum is $p = mv$ . Differentiating gives

\frac{\mathrm{d}p}{\mathrm{d}t} = m\frac{\mathrm{d}v}{\mathrm{d}t} = ma.

So acceleration can be found directly from a momentum–time graph using

a = \frac{1}{m}\frac{\mathrm{d}p}{\mathrm{d}t}.

That means: acceleration is proportional to the gradient of the $p$ – $t$ graph.

Understanding the Question

You are given the momentum–time graph from part (b). It has two straight-line sections:

$0$ to $8.0\ \text{s}$ : $p$ increases linearly.
$8.0$ to $12.0\ \text{s}$ : $p$ decreases linearly back to zero.

A straight-line section has constant gradient, so acceleration will be constant on each time interval. The task is to sketch $a$ against $t$ from $0$ to $12.0\ \text{s}$ .

Approach

Find the gradient $\Delta p/\Delta t$ for each straight-line section.
Convert each gradient to acceleration using $a = (1/m)(\Delta p/\Delta t)$ .
Sketch a step-like (piecewise constant) acceleration graph with the correct values and time intervals.

Step-by-Step Reasoning

For $0$ to $8.0\ \text{s}$ :

\frac{\Delta p}{\Delta t} = \frac{7.2\times 10^{4}-0}{8.0-0} = 9.0\times 10^{3}\ \text{N}.

Then

a = \frac{9.0\times 10^{3}}{1800} = 5.0\ \text{m s}^{-2}.

So $a$ is a horizontal line at $+5.0$ from $t=0$ to $t=8.0\ \text{s}$ .

For $8.0$ to $12.0\ \text{s}$ :

\frac{\Delta p}{\Delta t} = \frac{0-7.2\times 10^{4}}{12.0-8.0} = -1.8\times 10^{4}\ \text{N}.

Then

a = \frac{-1.8\times 10^{4}}{1800} = -10\ \text{m s}^{-2}.

So $a$ is a horizontal line at $-10$ from $t=8.0$ to $t=12.0\ \text{s}$ .

The sketch should show an abrupt change at $t=8.0\ \text{s}$ (because the slope of the momentum graph changes abruptly there).

diagram-2c-1

Key Takeaways

Gradient of $p$ – $t$ graph gives force ( $F = \mathrm{d}p/\mathrm{d}t$ ).
With constant mass, $a = (1/m)(\mathrm{d}p/\mathrm{d}t)$ .
Straight-line momentum segments produce constant acceleration segments.

Common Mistakes

Sketching acceleration increasing linearly (confusing with momentum): acceleration should be constant on each interval because the momentum graph is straight on each interval.
Getting the sign wrong on the deceleration section (it must be negative because momentum is decreasing).
Using $a = \Delta p/\Delta t$ directly without dividing by $m$ .

Things to Be Careful About

The value $-10\ \text{m s}^{-2}$ is at the lower limit of the provided axis range; place it correctly on the grid.
The change occurs at $t=8.0\ \text{s}$ exactly; ensure the horizontal segments span the correct time intervals.

Techniques used

interpret the gradient of a momentum-time graph as forceuse a = (1/m) dp/dtsketch a piecewise-constant graph from piecewise-linear data

Q3MediumWork, Energy and PowerElectricity

(a)

State what is meant by the work done by a force.

(b)

A block of mass $m$ is raised vertically at constant speed. The vertical height gained by the block is $\Delta h$ , as shown in Fig. 3.1.

Fig. 3.1

Derive an expression, in terms of $m$ and $\Delta h$ , for the change in gravitational potential energy $\Delta E_P$ of the block. State the meaning of any other symbols you use.

DifficultyMedium-Easy

Worked solution

Working

Weight of block $= mg$ .

Work done against weight in raising through height $\Delta h$ :

W = mg\,\Delta h

This work done equals the increase in gravitational potential energy, so

\Delta E_P = mg\,\Delta h

where $g$ is the acceleration of free fall.

Answer

\Delta E_P = mg\,\Delta h

Final answer

(\Delta E_P = mg,\Delta h) (where (g) is gravitational field strength / acceleration of free fall).

Detailed explanation

Background Concept

Near the Earth’s surface, the weight of a mass $m$ is approximately constant and equal to $mg$ , where $g$ is the acceleration of free fall (gravitational field strength).

The change in gravitational potential energy is the energy gained when an object is raised. For a constant weight, it equals the work done against gravity:

\Delta E_P = \text{work done against weight}

Work done by a constant force is

W = Fs

when the displacement is in the direction of the force.

Understanding the Question

A block of mass $m$ is lifted vertically through a height $\Delta h$ at constant speed. The question asks you to derive an expression for the change in gravitational potential energy $\Delta E_P$ in terms of $m$ and $\Delta h$ , and to define any other symbol used (here, $g$ ).

Approach

Identify the force opposing the lift: the weight $mg$ .
Since the block moves up at constant speed, the energy transferred by the lifting force is not increasing kinetic energy; it goes into gravitational potential energy.
Use $W = Fs$ with $F = mg$ and $s = \Delta h$ .

Step-by-Step Reasoning

The weight of the block is

W_t = mg

The block is raised vertically by $\Delta h$ , so the displacement in the direction of weight (downward) is opposite; equivalently, the work done against weight uses the magnitude $mg$ and distance $\Delta h$ :

\text{work done against gravity} = mg\,\Delta h

At constant speed, kinetic energy does not change, so the work done against gravity appears as an increase in gravitational potential energy:

\Delta E_P = mg\,\Delta h

Key Takeaways

$\Delta E_P$ equals work done against weight for vertical lifting near Earth.
Constant speed implies no change in kinetic energy.

Common Mistakes

Using $\Delta E_P = \tfrac{1}{2}mv^2$ (confusing with kinetic energy).
Forgetting to define $g$ .
Using $h$ instead of $\Delta h$ when the question specifies a height gain.

Things to Be Careful About

$g$ should be stated as acceleration of free fall / gravitational field strength (units $\text{m s}^{-2}$ or $\text{N kg}^{-1}$ ).
The derivation assumes $g$ is constant over the height change (valid for ordinary lab-scale heights).

Techniques used

apply work done against weight as force multiplied by vertical displacementuse constant speed to infer no change in kinetic energyexpress change in gravitational potential energy in terms of weight and height gain

(c)

An electric motor has an input power of $900\text{ W}$ . The motor takes $1.0\text{ minute}$ to lift a load of weight $240\text{ N}$ at constant speed through a vertical height of $150\text{ m}$ . Resistive forces are negligible.

8M

(i)

Show that the work done by the motor on the load in $1.0\text{ minute}$ is $36\text{ kJ}$ .

(ii)

Determine the useful output power of the motor.

power = ______ $\text{W}$

(iii)

Use your answer in (c)(ii) to determine the efficiency of the motor.

efficiency = ______

(iv)

Some of the power wasted in the motor is dissipated by the resistance of its coil. This dissipated power is $280\text{ W}$ .

The coil of the motor is made from wire of total length $23\text{ m}$ . The wire has a cross-sectional area of $2.6 \times 10^{-8}\text{ m}^2$ and is made from metal of resistivity $1.7 \times 10^{-8}\ \Omega\text{m}$ .

Calculate the current in the coil.

current = ______ $\text{A}$

DifficultyMedium

Worked solution

Working

Resistance of coil:

R = \rho\frac{L}{A} = 1.7\times10^{-8}\times\frac{23}{2.6\times10^{-8}} = 15\ \Omega

Dissipated power:

P = I^2R \Rightarrow I = \sqrt{\frac{P}{R}} = \sqrt{\frac{280}{15}} = 4.3\ \text{A}

Answer

$4.3\ \text{A}$

Final answer

4.3 A

Detailed explanation

Background Concept

The resistance of a uniform wire is related to its material and dimensions by

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

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

Electrical power dissipated as heating in a resistor can be written as

P = I^2R

This is appropriate when the given power is specifically the power lost in the resistance of the coil.

Understanding the Question

We are told that the wasted (dissipated) power in the motor coil is $280\ \text{W}$ . The coil is made from wire of length $23\ \text{m}$ , area $2.6\times10^{-8}\ \text{m}^2$ , resistivity $1.7\times10^{-8}\ \Omega\text{ m}$ . We must find the current in the coil.

So we need:

the coil resistance from $R = \rho L/A$ ;
then use $P = I^2R$ to solve for $I$ .

Approach

Compute $R$ using the resistivity formula (all values are already in SI).
Rearrange $P = I^2R$ to $I = \sqrt{P/R}$ .
Substitute and round sensibly.

Step-by-Step Reasoning

Resistance of the coil:

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

Substitute:

R = 1.7\times10^{-8}\times\frac{23}{2.6\times10^{-8}}

Notice the $10^{-8}$ cancels, leaving a value of order $10^1$ :

R \approx \frac{1.7\times 23}{2.6} \approx 15\ \Omega

Use the resistive power equation:

P = I^2R

Rearrange:

I = \sqrt{\frac{P}{R}}

Substitute $P=280\ \text{W}$ and $R \approx 15\ \Omega$ :

I = \sqrt{\frac{280}{15}} \approx \sqrt{18.7} \approx 4.3\ \text{A}

Key Takeaways

Use $R = \rho L/A$ when resistance is determined by geometry/material.
When power loss in a resistor is given, $P = I^2R$ is often the quickest route to current.

Common Mistakes

Using $P = VI$ without knowing $V$ across the coil.
Forgetting the square root when rearranging $P = I^2R$ .
Mis-handling powers of ten in $R = \rho L/A$ (the $10^{-8}$ terms cancel here).

Things to Be Careful About

Ensure $\rho$ is in $\Omega\text{ m}$ , $L$ in m, $A$ in $\text{m}^2$ so that $R$ is in $\Omega$ .
Quote current to an appropriate number of significant figures (here, $4.3\ \text{A}$ is reasonable).

Techniques used

calculate resistance from resistivity, length and cross-sectional areause the resistive power equation to relate power, current and resistancerearrange and evaluate using consistent SI units

Q4MediumDeformation of Solids

(a)

Define the Young modulus of a material.

DifficultyEasy

Worked solution

Answer

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

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

Final answer

Young modulus is the ratio of tensile stress to tensile strain (within the limit of proportionality).

Detailed explanation

Background Concept

When a wire is stretched by a tensile force, it experiences:

Tensile stress: force per unit cross-sectional area,

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

Tensile strain: fractional change in length,

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

In the linear-elastic region (up to the limit of proportionality), stress is proportional to strain. The constant of proportionality is the Young modulus $E$ :

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

Understanding the Question

You are asked to define Young modulus. That means you must give the ratio and indicate it applies where the material obeys Hooke’s law (linear relationship).

Approach

Give the standard definition:

write stress and strain (can be in words),
state $E =$ stress/strain, and (ideally) that it is within the limit of proportionality.

Step-by-Step Reasoning

Young modulus compares how much stress is needed to produce a given strain.
In the region where the graph of stress against strain is a straight line, the ratio is constant.
Therefore,

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

Key Takeaways

$E$ measures stiffness of a material in extension.
Definition must include stress/strain and the linear (proportional) region.

Common Mistakes

Writing $E = \frac{F}{x}$ (that is stiffness/spring constant for a particular sample, not a material property).
Forgetting strain is fractional change in length.
Not indicating the condition “within limit of proportionality”.

Things to Be Careful About

Young modulus depends on the material, not the dimensions, whereas $\frac{F}{x}$ depends on wire geometry.
Use correct terms: tensile stress and tensile strain.

Techniques used

state the definition of Young modulus as a ratioidentify the condition of proportional (linear-elastic) behaviour

(b)

A metal wire P that obeys Hooke’s law is stretched within its limit of proportionality.

3M

(i)

On Fig. 4.1, sketch the variation of tensile force $F$ in the wire with its extension $x$ .

Fig. 4.1

(ii)

State the name of the quantity represented by the gradient of the line in Fig. 4.1.

(iii)

State the name of the quantity represented by the area under the line in Fig. 4.1.

(c)

Another wire Q is made from a metal that has twice the Young modulus of the metal of wire P in (b). Wire Q has the same volume as wire P but has double the cross-sectional area of wire P.

The two wires are extended by equal tensile forces within their limits of proportionality.

State and explain how the extension of wire Q compares with the extension of wire P.

DifficultyMedium

Worked solution

Working

For a wire in the proportional region,

x = \frac{FL}{AE}

Same volume: $AL = \text{constant}$ , and $A_Q = 2A_P$ so

L_Q = \frac{A_PL_P}{A_Q} = \frac{L_P}{2}

Also $E_Q = 2E_P$ . Hence

\frac{x_Q}{x_P} = \frac{\frac{FL_Q}{A_QE_Q}}{\frac{FL_P}{A_PE_P}} = \left(\frac{L_Q}{L_P}\right)\left(\frac{A_P}{A_Q}\right)\left(\frac{E_P}{E_Q}\right) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{8}

Answer

Wire Q extends to one-eighth of the extension of wire P: $x_Q = \frac{1}{8}x_P$ .

Final answer

x_Q = (1/8) x_P (wire Q extends one-eighth as much as wire P).

Detailed explanation

Background Concept

In the linear-elastic (Hooke’s law) region for a wire, combining definitions of stress and strain gives the standard extension formula:

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

Rearranging,

x = \frac{FL}{AE}

So extension increases with force $F$ and original length $L$ , and decreases with cross-sectional area $A$ and Young modulus $E$ .

Also, wire volume is

V = AL

If volume is fixed and the area increases, the length must decrease.

Understanding the Question

You have two wires:

Wire P: area $A_P$ , length $L_P$ , Young modulus $E_P$ .
Wire Q: Young modulus is twice: $E_Q = 2E_P$ .
Same volume as P, but double cross-sectional area: $A_Q = 2A_P$ .
Both stretched by the same tensile force $F$ within proportional limits.

You must compare extensions $x_Q$ and $x_P$ and explain.

Approach

Use same-volume condition $AL = \text{constant}$ to find how $L_Q$ compares with $L_P$ .
Use $x = \frac{FL}{AE}$ for each wire.
Form a ratio $\frac{x_Q}{x_P}$ so $F$ cancels and only geometry/material factors remain.

Step-by-Step Reasoning

1) Use equal volumes

Same volume means:

A_PL_P = A_QL_Q

Given $A_Q = 2A_P$ :

A_PL_P = 2A_PL_Q \Rightarrow L_Q = \frac{L_P}{2}

So Q is half as long.

2) Use the extension formula

For P:

x_P = \frac{FL_P}{A_PE_P}

For Q:

x_Q = \frac{FL_Q}{A_QE_Q}

3) Compare using a ratio

\frac{x_Q}{x_P} = \frac{FL_Q/(A_QE_Q)}{FL_P/(A_PE_P)} = \left(\frac{L_Q}{L_P}\right)\left(\frac{A_P}{A_Q}\right)\left(\frac{E_P}{E_Q}\right)

Now substitute the given comparisons:

$\frac{L_Q}{L_P} = \frac{1}{2}$
$\frac{A_P}{A_Q} = \frac{1}{2}$
$\frac{E_P}{E_Q} = \frac{1}{2}$

So

\frac{x_Q}{x_P} = \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} = \frac{1}{8}

Therefore wire Q extends one-eighth as much as wire P under the same force.

Key Takeaways

For a wire, $x = \frac{FL}{AE}$ .
Same volume gives $AL$ constant, so changing $A$ changes $L$ .
Increasing $A$ and $E$ both reduce extension.

Common Mistakes

Assuming $L_Q = L_P$ even though the volume is the same.
Only accounting for the change in $E$ (getting $x_Q = x_P/2$ ) and forgetting $A$ and $L$ changes.
Using $x = \frac{FA}{LE}$ (incorrect rearrangement).

Things to Be Careful About

The wire sample stiffness depends on both material property ( $E$ ) and geometry ( $A$ and $L$ ).
When forming ratios, cancel the common force $F$ to avoid unnecessary algebra errors.
Work strictly within “limits of proportionality” so the linear relation applies.

Techniques used

use the extension formula for a stretched wireuse volume conservation to relate lengths and areascompare extensions by forming a ratio

Q5Medium-EasyParticle Physics

(a)

Potassium-40 ( $^{40}_{19}\text{K}$ ) undergoes $\beta^-$ decay to form a nuclide of element X. Particle Z is emitted during the decay. The equation for the decay is shown.

^{40}_{19}\text{K} \rightarrow \, ^P_Q\text{X} + \, ^R_S\beta^- + \text{Z}

4M

(i)

State the values of $P$ , $Q$ , $R$ and $S$ .

$P$ = ______ $R$ = ______
$Q$ = ______ $S$ = ______

DifficultyMedium-Easy

Worked solution

Working

In $\beta^-$ decay, nucleon number is conserved, so $P = 40$ .

Charge (proton number) is conserved:

19 = Q + (-1)

so $Q = 20$ .

For a $\beta^-$ particle:

R = 0,\quad S = -1

Answer

$P = 40,\ Q = 20,\ R = 0,\ S = -1$

Final answer

P = 40, Q = 20, R = 0, S = -1

Detailed explanation

Background Concept

In any nuclear decay equation, two key quantities are conserved:

Nucleon (mass) number, $A$ : total number of protons + neutrons in the nucleus.
Proton (atomic) number, $Z$ : number of protons (and hence the nuclear charge in units of $+e$ ).

For $\beta^-$ decay, a neutron in the nucleus changes into a proton and emits an electron (the $\beta^-$ particle) and an electron antineutrino:

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

So, for the nucleus as a whole:

$A$ stays the same (a neutron becomes a proton: still one nucleon).
$Z$ increases by 1 (because there is one extra proton).

A $\beta^-$ particle is an electron, written in nuclear notation as:

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

Understanding the Question

You are given:

^{40}_{19}\text{K} \rightarrow \, ^P_Q\text{X} + \, ^R_S\beta^- + \text{Z}

and asked to state $P, Q, R, S$ .

So you must use conservation of nucleon number and proton number to fill the unknowns.

Approach

Use conservation of nucleon number ( $A$ ) to find $P$ (and $R$ for the beta particle).
Use conservation of proton number ( $Z$ ) to find $Q$ (and $S$ for the beta particle).
Recall the standard nuclear notation for $\beta^-$ .

Step-by-Step Reasoning

1) Nucleon number conservation

Potassium-40 has $A = 40$ .
A $\beta^-$ particle has $A = 0$ .
Neutrinos/antineutrinos also have $A = 0$ .

So the daughter nuclide must keep $A = 40$ :

P = 40

Also, for the $\beta^-$ particle:

R = 0

2) Proton number (charge) conservation

Potassium-40 has $Z = 19$ .
A $\beta^-$ particle has $Z = -1$ .
Neutrinos/antineutrinos have $Z = 0$ .

So:

19 = Q + (-1)

Q = 20

and for the beta particle:

S = -1

Therefore:

P = 40,\ Q = 20,\ R = 0,\ S = -1

(You could also note that $Z$ increasing from 19 to 20 means the element becomes calcium, but the question only asks for $P,Q,R,S$ .)

Key Takeaways

Balance mass number $A$ and proton number $Z$ on both sides of any nuclear equation.
In $\beta^-$ decay: $A$ unchanged, $Z$ increases by 1.
$\beta^-$ has nuclear notation $^{0}_{-1}\beta^-$ .

Common Mistakes

Writing $\beta^-$ as $^{1}_{0}\beta$ (confusing it with a neutron/proton).
Changing the mass number $A$ in $\beta^-$ decay (it should stay the same).
Forgetting the negative proton number ( $-1$ ) for the emitted electron.

Things to Be Careful About

Always treat $Z$ as a signed quantity for emitted particles: electron has $Z=-1$ .
Neutrinos/antineutrinos carry no nucleon number and no charge, so they do not affect $A$ or $Z$ balancing.

Techniques used

apply conservation of nucleon (mass) number in a nuclear equationapply conservation of charge / proton number in a nuclear equationuse standard nuclide notation for a beta-minus particle

(ii)

State the name of particle Z.

(iii)

State the name of the class of fundamental particle to which both the $\beta^-$ particle and particle Z belong.

(b)

Determine the quark composition of an alpha-particle.

quark composition ______

DifficultyMedium-Easy

Worked solution

Working

An $\alpha$ -particle has $2$ protons and $2$ neutrons.

Proton: $uud$ .
Neutron: $udd$ .

Total quarks:

2(uud) + 2(udd) = (4u+2d) + (2u+4d) = 6u + 6d

Answer

Quark composition: $6u + 6d$ .

Final answer

6u + 6d

Detailed explanation

Background Concept

Quarks combine to form hadrons:

A proton has quark composition $uud$ (two up quarks and one down quark).
A neutron has quark composition $udd$ (one up quark and two down quarks).

An alpha particle ( $\alpha$ ) is the nucleus of helium-4, containing:

2 protons
2 neutrons

To find the quark composition of a nucleus, you add together the quarks in all its protons and neutrons.

Understanding the Question

You are asked for the quark composition of an alpha particle. Since an alpha particle is made of 2 protons and 2 neutrons, you must express it in terms of the numbers of up ( $u$ ) and down ( $d$ ) quarks.

Approach

Write the quark content for one proton and one neutron.
Multiply by how many protons/neutrons are in the alpha particle.
Add the totals of $u$ and $d$ quarks.

Step-by-Step Reasoning

1) Identify nucleons in an alpha particle

$\alpha$ : $2p + 2n$ .

2) Write quark compositions

$p = uud$
$n = udd$

3) Count quarks from the two protons
Two protons give:

up quarks: $2 \times 2 = 4u$
down quarks: $2 \times 1 = 2d$
So: $4u + 2d$ .

4) Count quarks from the two neutrons
Two neutrons give:

up quarks: $2 \times 1 = 2u$
down quarks: $2 \times 2 = 4d$
So: $2u + 4d$ .

5) Total quarks in the alpha particle
Add them:

(4u+2d) + (2u+4d) = 6u + 6d

So the alpha particle contains six up quarks and six down quarks.

Key Takeaways

Use $p = uud$ and $n = udd$ .
For composite particles like nuclei, multiply and add quark counts across all nucleons.

Common Mistakes

Forgetting that an alpha particle has two neutrons as well as two protons.
Writing $3u+3d$ (counting only one proton + one neutron).
Mixing in electrons (the alpha particle is the nucleus only, not the whole atom).

Things to Be Careful About

State your answer clearly as numbers of quarks, e.g. $6u + 6d$ (either order is fine).
Do not include antiquarks here: protons and neutrons are baryons made of three quarks (not antiquarks).

Techniques used

use standard quark compositions of protons and neutronscount and sum quarks for composite particlesscale quark content according to the number of nucleons present

Q6MediumSuperpositionWaves

Two coherent sources X and Y of microwaves of frequency $2.5 \times 10^{10}\text{ Hz}$ are a distance of $0.18\text{ m}$ apart in a vacuum, as shown in Fig. 6.1.

Fig. 6.1

There is a phase difference of $90^\circ$ between the waves emitted at the two sources.

A microwave detector moves along the line PQ, which is parallel to the line joining the two sources and $2.3\text{ m}$ away from it.

Point O is on the line PQ at a position that is equidistant from the two sources.
Point A is the position on line PQ where the intensity of the microwaves is the greatest.

(a)

4M

(i)

Explain why the position of greatest intensity is not at point O.

DifficultyMedium-Easy

Worked solution

Answer

At $O$ the path lengths from $X$ and $Y$ are equal, so the path difference is $0$ .
Therefore the phase difference at $O$ is still $90^\circ$ , not $0^\circ$ (or $360^\circ$ ), so the waves do not arrive in phase and the intensity at $O$ is not a maximum.

Final answer

At O the path difference is zero so the phase difference remains 90°, hence not in phase and not a maximum.

Detailed explanation

Background Concept

For two coherent sources, the intensity at a point depends on the phase difference (\phi) between the two waves when they arrive at that point.

Constructive interference (maximum intensity): waves arrive in phase

\phi = 0^\circ,\ 360^\circ,\ 720^\circ,\dots

Destructive interference (minimum intensity): waves arrive in antiphase

\phi = 180^\circ,\ 540^\circ,\dots

The total phase difference at the detector comes from two contributions:

Any initial phase difference at the sources.
Additional phase difference caused by a path difference (\Delta x) between the two routes to the detector.

A path difference of one wavelength (\lambda) corresponds to (360^\circ) of phase.
So a path difference (\Delta x) produces phase difference

\Delta \phi_{\text{path}} = 360^\circ \times \frac{\Delta x}{\lambda}

Understanding the Question

Point (O) on line (PQ) is equidistant from sources (X) and (Y). That means the two microwaves travel the same distance to reach (O). However, the question states that the sources emit with a phase difference of (90^\circ). The task is to explain why, despite equal path lengths, (O) is not where the intensity is greatest.

Approach

Use the interference condition for a maximum: the waves must arrive in phase. Check what the phase difference is at (O):

compute/identify path difference at (O)
add it to the given source phase difference
compare the result with (0^\circ) (maximum condition)

Step-by-Step Reasoning

At (O), the distances from (X) and (Y) to (O) are the same (by definition of equidistant).

Hence path difference

\Delta x = 0

If (\Delta x = 0), then the phase difference due to the path is

\Delta \phi_{\text{path}} = 360^\circ \times \frac{0}{\lambda} = 0^\circ

Therefore, the phase difference at (O) is just the original source phase difference:

\phi_O = 90^\circ + 0^\circ = 90^\circ

(90^\circ) is neither (0^\circ) (in phase) nor (180^\circ) (antiphase). The waves do not fully reinforce each other, so (O) cannot be a point of greatest intensity.

Key Takeaways

Equal distances (zero path difference) do not guarantee a maximum if the sources are not emitting in phase.
Total phase difference = (source phase difference) + (path-induced phase difference).

Common Mistakes

Assuming “equidistant from both sources” automatically means maximum intensity.
Forgetting to include the given (90^\circ) phase difference at emission.

Things to Be Careful About

Always distinguish between path difference (geometry) and phase difference at the source (given property of emission).
The condition for maximum is (0^\circ) (mod (360^\circ)), not “small phase difference”.

Techniques used

use phase difference as the condition for constructive or destructive interferencecombine source phase difference with phase difference due to path differencerelate path difference to phase difference using fractions of a wavelength

(ii)

On Fig. 6.1, draw a cross ( $\times$ ) to show the position of the point on line PQ where the intensity minimum that is the closest to point O occurs. Label this point B.

DifficultyMedium

Worked solution

Working

At $O$ , phase difference $=90^\circ$ .
For the nearest minimum, total phase difference $=180^\circ$ , so extra phase needed is $90^\circ$ .
Hence path difference

\Delta x = \frac{90^\circ}{360^\circ}\lambda = \frac{\lambda}{4}

So point $B$ is the closest point on $PQ$ where the path to one source is longer than to the other by $\lambda/4$ .
Since the nearest maximum ( $A$ ) is on the opposite side of $O$ , $B$ is on $PQ$ on the opposite side of $O$ from $A$ .

diagram-1

Answer

$B$ is on line $PQ$ closest to $O$ on the opposite side of $O$ from $A$ (so that the path difference is $+\lambda/4$ ).

Final answer

Point B is the nearest point on PQ to O on the opposite side of O from A (path difference = λ/4).

Detailed explanation

Background Concept

Interference minima and maxima depend on the total phase difference at the observation point:

\phi_{\text{total}} = \phi_{\text{source}} + \Delta \phi_{\text{path}}

with

\Delta \phi_{\text{path}} = 360^\circ\frac{\Delta x}{\lambda}

A minimum occurs when

\phi_{\text{total}} = 180^\circ\ (\text{mod }360^\circ)

Here (\phi_{\text{source}} = 90^\circ), so the “usual” positions of maxima/minima are shifted compared with in-phase sources.

Understanding the Question

You must mark the position (B) on the detector line (PQ) that is the closest minimum to (O). At (O), the detector is equidistant from (X) and (Y), so the path difference is zero. Because the sources are already (90^\circ) out of phase, (O) is not a maximum, and nearby points will give the first minimum and first maximum.

The diagram also shows a point (A) above (O) which is the location of greatest intensity (a maximum). The closest minimum will be on the opposite side of (O) from that maximum.

Approach

Start from the phase difference at (O): (90^\circ).
For a minimum, set total phase difference to (180^\circ).
Find the extra phase needed and convert it to path difference (\Delta x).
Decide where along (PQ) this condition is met relative to (O) and (A).

Step-by-Step Reasoning

At (O): (\Delta x = 0), so (\phi_{\text{total}} = 90^\circ).
Nearest minimum requires:

\phi_{\text{total}} = 180^\circ

So the additional phase needed from the path difference is

\Delta \phi_{\text{path}} = 180^\circ - 90^\circ = 90^\circ

Convert (90^\circ) into a fraction of a wavelength:

90^\circ = 360^\circ\frac{\Delta x}{\lambda} \Rightarrow \Delta x = \frac{90^\circ}{360^\circ}\lambda = \frac{\lambda}{4}

This means (B) is the closest point to (O) where one path is longer than the other by (\lambda/4).
Since point (A) is the maximum on one side of (O), the nearest minimum occurs on the other side (because a maximum and a minimum correspond to opposite signs of the required path difference when there is a fixed source phase).

diagram-1

Key Takeaways

With a non-zero source phase difference, the central equidistant point is not necessarily a maximum.
To find minima/maxima, use total phase difference, not just geometry.
A phase change of (90^\circ) corresponds to a path difference of (\lambda/4).

Common Mistakes

Putting the minimum at (O) because “equal distances means cancellation” (it doesn’t here).
Using the usual in-phase conditions (max at (\Delta x = 0)) without adjusting for the (90^\circ) source phase difference.
Confusing the spacing of fringes (which stays the same) with the shift of the whole pattern (which changes).

Things to Be Careful About

The question asks for the closest minimum: choose the smallest magnitude path difference solution (here (\lambda/4), not (5\lambda/4), etc.).
On a diagram, you cannot show exact distances numerically, but you must place (B) correctly relative to (O) and (A).

Techniques used

set up the condition for destructive interference using total phase differenceconvert a required phase change to a path difference as a fraction of a wavelengthuse symmetry to decide which side of the central point the minimum occurs

(b)

5M

(i)

Show that the wavelength of the microwaves is $0.012\text{ m}$ .

DifficultyMedium-Easy

Worked solution

Working

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

\lambda = \frac{c}{f} = \frac{3.0\times 10^8}{2.5\times 10^{10}}

\lambda = 1.2\times 10^{-2}\ \text{m} = 0.012\ \text{m}

Answer

$\lambda = 0.012\ \text{m}$

Final answer

0.012 m

Detailed explanation

Background Concept

All waves satisfy the wave equation

v = f\lambda

where:

(v) is wave speed ((\text{m s}^{-1}))
(f) is frequency ((\text{Hz}))
(\lambda) is wavelength ((\text{m}))

Electromagnetic waves (including microwaves) travel in a vacuum at the speed of light

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

So in vacuum:

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

Understanding the Question

You are given the microwave frequency (f = 2.5 \times 10^{10}\ \text{Hz}) and told the microwaves travel in a vacuum, so their speed is (c). The task is to show the wavelength is (0.012\ \text{m}).

Approach

Use (\lambda = c/f). Substitute (c = 3.0 \times 10^8) and the given (f), then simplify the powers of ten and the numerical factor.

Step-by-Step Reasoning

Write the equation:

\lambda = \frac{c}{f}

Substitute:

\lambda = \frac{3.0\times 10^8}{2.5\times 10^{10}}

Separate numbers and powers of ten:

\lambda = \left(\frac{3.0}{2.5}\right)\times 10^{8-10} = 1.2\times 10^{-2}\ \text{m}

Convert to decimal:

1.2\times 10^{-2}\ \text{m} = 0.012\ \text{m}.

Key Takeaways

In vacuum, microwaves travel at (c).
Wavelength is found from (\lambda = c/f).

Common Mistakes

Using (v = 340\ \text{m s}^{-1}) (speed of sound) instead of (c).
Power-of-ten errors, e.g. using (10^{10}/10^8) instead of (10^8/10^{10}).

Things to Be Careful About

Keep units consistent: (c) in (\text{m s}^{-1}) and (f) in (\text{s}^{-1}) gives (\lambda) in metres.
When dividing powers of ten: (10^a/10^b = 10^{a-b}).

Techniques used

use the wave equation to relate speed, frequency and wavelengthsubstitute the speed of electromagnetic waves in vacuumhandle powers of ten in a calculation

(ii)

For point A on line PQ, determine the difference in the distances $\Delta x$ travelled by the microwaves from X and the microwaves from Y.

$\Delta x$ = ______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Working

For a maximum at $A$ , total phase difference must be $0^\circ$ (mod $360^\circ$ ).
Source phase difference is $90^\circ$ , so the path must provide $-90^\circ$ (equivalently $270^\circ$ ).
Smallest magnitude path difference is

\Delta x = \frac{90^\circ}{360^\circ}\lambda = \frac{\lambda}{4}

With $\lambda = 0.012\ \text{m}$ ,

\Delta x = \frac{0.012}{4} = 3.0\times 10^{-3}\ \text{m}

Answer

$\Delta x = 0.0030\ \text{m}$

Final answer

0.0030 m

Detailed explanation

Background Concept

For two-source interference, the brightness/intensity depends on the total phase difference between waves at the observation point.

If the sources are not in phase initially, then

\phi_{\text{total}} = \phi_{\text{source}} + \Delta \phi_{\text{path}}.

A maximum occurs when

\phi_{\text{total}} = 0^\circ\ (\text{mod } 360^\circ).

The path phase difference comes from path difference (\Delta x):

\Delta \phi_{\text{path}} = 360^\circ \frac{\Delta x}{\lambda}.

Understanding the Question

Point (A) is stated to be where the intensity is greatest (a maximum) on the detector line (PQ). The sources emit with a fixed phase difference (90^\circ). You are asked to find the path difference (\Delta x) between the distances travelled from (X) and from (Y) to reach (A).

Because (A) is the closest maximum to (O) on the diagram, the corresponding path difference should be the smallest one that satisfies the maximum condition.

Approach

Set the maximum condition for total phase difference. Since the sources already differ by (90^\circ), the path difference must supply an additional phase shift that brings the total to (0^\circ) (or (360^\circ)). Convert that required phase shift into (\Delta x) using fractions of (\lambda).

Step-by-Step Reasoning

Given source phase difference:

\phi_{\text{source}} = 90^\circ.

For a maximum:

\phi_{\text{total}} = 0^\circ \ (\text{mod } 360^\circ).

So the path must contribute either (-90^\circ) or equivalently (270^\circ). The smallest magnitude phase shift is (90^\circ), corresponding to a quarter wavelength path difference.

Convert (90^\circ) to a path difference:

90^\circ = 360^\circ \frac{\Delta x}{\lambda} \Rightarrow \Delta x = \frac{\lambda}{4}.

Using (\lambda = 0.012\ \text{m}):

\Delta x = \frac{0.012}{4} = 0.0030\ \text{m}.

(If you track sign, (A) would correspond to (\Delta x = -\lambda/4) or (+3\lambda/4) depending on which source’s wave arrives ahead. The question asks for the difference in distances; typically the magnitude (\lambda/4) is sufficient.)

Key Takeaways

Maxima occur when total phase difference is (0^\circ) (mod (360^\circ)), even if sources are initially out of phase.
A (90^\circ) phase adjustment corresponds to a path difference of (\lambda/4).

Common Mistakes

Using (\Delta x = 0) for a maximum (only true if sources start in phase).
Using the destructive condition ((180^\circ)) instead of constructive.

Things to Be Careful About

The fringe spacing formula in the next part is unaffected by the initial phase, but the positions of maxima/minima shift.
Be consistent about whether (\Delta x) is a signed difference or an absolute difference; many mark schemes accept the magnitude for this style of question.

Techniques used

apply the phase condition for constructive interference including a source phase offsetconvert a phase difference into a path difference using $\Delta x/\lambda$select the smallest path difference that gives a maximum

(iii)

Use the formula for the double-slit interference of light to calculate the distance between adjacent intensity maxima on line PQ.

distance = ______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Working

Using double-slit fringe spacing

\text{spacing} = \frac{\lambda D}{a}

with $\lambda = 0.012\ \text{m}$ , $D = 2.3\ \text{m}$ , $a = 0.18\ \text{m}$ :

\text{spacing} = \frac{0.012 \times 2.3}{0.18} = 0.153\ \text{m}

Answer

distance $= 0.15\ \text{m}$

Final answer

0.15 m

Detailed explanation

Background Concept

For two coherent sources (or a double slit) separated by distance (a), with a screen (or observation line) a perpendicular distance (D) away, the separation (s) between adjacent bright fringes is

s = \frac{\lambda D}{a}.

This comes from the small-angle approximation where the path difference changes approximately linearly with position along the screen.

A key point: an initial phase difference between sources shifts the entire pattern, but it does not change the spacing (s) between adjacent maxima.

Understanding the Question

You are told:

source separation (a = 0.18\ \text{m})
detector line (PQ) is parallel to the sources and (D = 2.3\ \text{m}) away
wavelength from earlier part (\lambda = 0.012\ \text{m})

You must calculate the distance along (PQ) between adjacent intensity maxima.

Approach

Use the standard fringe spacing equation (s = \lambda D/a) and substitute the given values in metres.

Step-by-Step Reasoning

Write the formula:

s = \frac{\lambda D}{a}

Substitute:

s = \frac{0.012\times 2.3}{0.18}

Calculate:

0.012\times 2.3 = 0.0276

s = \frac{0.0276}{0.18} = 0.153\ \text{m}

Round appropriately (typically 2 s.f. here):

s \approx 0.15\ \text{m}.

Key Takeaways

Fringe spacing for two-source/double-slit interference is (\lambda D/a).
A fixed source phase difference shifts where the maxima occur, but not the distance between them.

Common Mistakes

Using (a) and (D) the wrong way round (writing (aD/\lambda)).
Forgetting all distances must be in metres.
Thinking the (90^\circ) phase difference changes the spacing; it only shifts the pattern.

Things to Be Careful About

The formula is derived using small angles; it is appropriate here because (D) (2.3 m) is much larger than (a) (0.18 m), so angles to fringes are small.
Quote the final answer to a sensible number of significant figures consistent with the given data.

Techniques used

apply the double-slit fringe spacing equationsubstitute wavelength, slit separation and screen distancecalculate spacing between adjacent maxima

Q7Medium-EasyD.C. Circuits

(a)

Fig. 7.1 shows two resistors connected in series with a cell of electromotive force (e.m.f.) $1.50\text{ V}$ and internal resistance $0.28\ \Omega$ .

Fig. 7.1

One of the resistors has resistance $1.0\ \Omega$ . The other resistor has resistance $R$ . The terminal potential difference (p.d.) across the cell is $1.36\text{ V}$ .

5M

(i)

Show that the current $I$ in the circuit is $0.50\text{ A}$ .

DifficultyMedium-Easy

Worked solution

Working

Lost volts across internal resistance:

E - V = 1.50 - 1.36 = 0.14\text{ V}

E - V = Ir \Rightarrow I = \frac{0.14}{0.28} = 0.50\text{ A}

Answer

$I = 0.50\text{ A}$

Final answer

0.50 A

Detailed explanation

Background Concept

A real cell can be modelled as an ideal source of e.m.f. $E$ in series with an internal resistance $r$ . When a current $I$ flows, there is a potential drop inside the cell equal to $Ir$ (often called the “lost volts”).

The terminal potential difference $V$ across the cell’s terminals is then

V = E - Ir

This is smaller than $E$ when the cell is delivering current.

Understanding the Question

You are told the e.m.f. of the cell ( $1.50\text{ V}$ ), its internal resistance ( $0.28\ \Omega$ ), and the terminal p.d. while the external circuit is connected ( $1.36\text{ V}$ ). The circuit current $I$ is the same everywhere because the external resistors are in series.

The question asks you to show that this current is $0.50\text{ A}$ .

Approach

Use the internal resistance model: first find the “lost volts” $E - V$ , then use $E - V = Ir$ to calculate $I$ .

Step-by-Step Reasoning

Find the drop across the internal resistance:

E - V = 1.50 - 1.36 = 0.14\text{ V}

Apply Ohm’s law to the internal resistance $r$ (since the same current $I$ flows through it):

E - V = Ir

Rearrange and substitute:

I = \frac{E - V}{r} = \frac{0.14}{0.28} = 0.50\text{ A}

Key Takeaways

Terminal p.d. is reduced from e.m.f. by the internal drop $Ir$ .
The “lost volts” method is often the quickest way to find $I$ when $E$ , $V$ , and $r$ are given.

Common Mistakes

Using $V = IR$ with $R$ as the external resistance before finding $I$ (you are not given $R$ yet).
Writing $V = E + Ir$ (wrong sign when the cell supplies current).

Things to Be Careful About

Keep track of which voltage is terminal p.d. ( $V$ ) and which is e.m.f. ( $E$ ).
Ensure the internal resistance is in $\Omega$ and current comes out in $\text{A}$ .

Techniques used

use the relation between e.m.f., terminal p.d. and internal resistancecalculate current from the lost volts across the internal resistance

(ii)

Calculate the combined resistance of the two resistors.

resistance = ______ $\Omega$

DifficultyMedium-Easy

Worked solution

Working

Terminal p.d. across external resistors:

V = I R_{\text{ext}} \Rightarrow R_{\text{ext}} = \frac{V}{I} = \frac{1.36}{0.50} = 2.72\ \Omega

Answer

Combined resistance $= 2.72\ \Omega$

Final answer

2.72 Ω

Detailed explanation

Background Concept

For components in series, the same current flows through each component. The total resistance of the external circuit can be treated as a single equivalent resistance $R_{\text{ext}}$ .

If the terminal potential difference across the external circuit is $V$ and the series current is $I$ , then by Ohm’s law:

V = IR_{\text{ext}}

Understanding the Question

In Fig. 7.1, the terminal p.d. across the cell is $1.36\text{ V}$ . This is also the p.d. across the entire external circuit (the two resistors in series), because that external circuit is connected directly across the cell terminals.

From (a)(i), $I = 0.50\text{ A}$ . The question asks for the combined (equivalent) resistance of the two series resistors.

Approach

Treat the two external resistors as a single equivalent resistance $R_{\text{ext}}$ and use $R_{\text{ext}} = V/I$ .

Step-by-Step Reasoning

Identify that $V = 1.36\text{ V}$ is across the external resistors together.
Apply Ohm’s law:

R_{\text{ext}} = \frac{V}{I}

Substitute values:

R_{\text{ext}} = \frac{1.36}{0.50} = 2.72\ \Omega

Key Takeaways

Terminal p.d. is the p.d. across the external circuit connected to the cell.
Equivalent resistance for the external circuit can be found directly from $V/I$ when $V$ is the terminal p.d.

Common Mistakes

Using the e.m.f. $1.50\text{ V}$ instead of the terminal p.d. $1.36\text{ V}$ when calculating the external resistance.
Adding internal resistance to the external resistance here (the question asks only for the two resistors).

Things to Be Careful About

Use the current found in (a)(i); do not recalculate it from other assumptions.
Quote resistance with unit $\Omega$ and to a sensible number of significant figures consistent with the data (here $2.72\ \Omega$ is fine).

Techniques used

use terminal p.d. across external circuitapply Ohm's law to the external resistance

(iii)

Use your answer in (a)(ii) to determine resistance $R$ .

$R$ = ______ $\Omega$

(b)

The circuit in Fig. 7.1 is disconnected and the two resistors are reconnected to the cell, now in parallel with each other.

3M

(i)

On Fig. 7.2, complete the circuit diagram to show this arrangement.

Fig. 7.2

DifficultyEasy

Worked solution

diagram-1

Answer

Two resistors connected in parallel across the cell terminals (one branch $1.0\ \Omega$ , other branch $R$ ).

Final answer

See diagram

Detailed explanation

Background Concept

Two components are in parallel if they are connected between the same pair of nodes, so they share the same potential difference. In a circuit diagram, this means the branches split from one node and rejoin at the other node.

Understanding the Question

You are asked to modify the circuit so that the two resistors ( $1.0\ \Omega$ and $R$ ) are no longer end-to-end (series), but instead each forms its own branch connected directly across the cell terminals.

Approach

Identify the two terminals of the cell (the two nodes provided on Fig. 7.2). Draw two separate branches between these same two nodes: one containing the $1.0\ \Omega$ resistor and one containing the $R$ resistor.

Step-by-Step Reasoning

diagram-1

Keep the given cell symbol and its internal resistance as already drawn.
Use the two wires provided from the cell terminals as the two nodes for the parallel network.
Draw two distinct paths between these nodes:
- Branch 1: a resistor labelled $1.0\ \Omega$ .
- Branch 2: a resistor labelled $R$ .
Ensure both branches start and end at the same pair of nodes (this is what makes them parallel).

Key Takeaways

Parallel components share the same p.d. because they connect across the same two nodes.
A correct circuit diagram must clearly show a split and rejoin (two branches).

Common Mistakes

Drawing the resistors still in series (one after the other) rather than in separate branches.
Connecting one resistor to only one terminal of the cell (an open circuit branch).

Things to Be Careful About

Both resistors must be connected directly across the cell terminals, not across the internal resistance symbol alone.
Label both resistor values clearly ( $1.0\ \Omega$ and $R$ ).

Techniques used

draw a parallel connection sharing the same two nodeslabel each branch with the correct resistor value

(ii)

Explain, without calculation, whether the terminal p.d. across the cell is now less than, equal to or greater than $1.36\text{ V}$ .