(a) Explain what is meant by the accuracy of a measured value. (b) Two solid cubes, A and B, are measured to determine the density of their materials. Table 1.1 shows the measurements for cube A. Table 1.1 quantity measurement --- --- length of side $(1.53 \pm 0.01)\ \text{cm}$ mass $(31.3 \pm 0.5)\ \text{g}$ (i) Show that the calculated density of the material of cube A is $8.7 \times 10^{3}\ \text{kg m}^{-3}$. (ii) Calculate the percentage uncertainty in the density of the…

(a) Accuracy is how close a measured value is to the true (accepted) value. · (b)(i) 8.7 × 10^3 kg m^-3 · (b)(ii) 3.6% · (b)(iii) Yes; their density ranges overlap within uncertainties, so they could be the same material.

9702/22 Feb/Mar 2025 — Physics Solved Past Paper

Q1MediumPhysical Quantities and UnitsForces, Density and Pressure

(a)

Explain what is meant by the accuracy of a measured value.

1M Similar

(b)

Two solid cubes, A and B, are measured to determine the density of their materials.

Table 1.1 shows the measurements for cube A.

Table 1.1

quantity	measurement
length of side	$(1.53 \pm 0.01)\ \text{cm}$
mass	$(31.3 \pm 0.5)\ \text{g}$

6M

(i)

Show that the calculated density of the material of cube A is $8.7 \times 10^{3}\ \text{kg m}^{-3}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Side length:

L = 1.53\ \text{cm} = 1.53 \times 10^{-2}\ \text{m}

Mass:

m = 31.3\ \text{g} = 3.13 \times 10^{-2}\ \text{kg}

Volume of cube:

V = L^3 = \left(1.53 \times 10^{-2}\right)^3 = 3.58 \times 10^{-6}\ \text{m}^3

Density:

\rho = \frac{m}{V} = \frac{3.13 \times 10^{-2}}{3.58 \times 10^{-6}} = 8.74 \times 10^{3}\ \text{kg m}^{-3}

Answer

\rho \approx 8.7 \times 10^{3}\ \text{kg m}^{-3}

Final answer

8.7 × 10^3 kg m^-3

Detailed explanation

Background Concept

Density $\rho$ is defined as mass per unit volume:

\rho = \frac{m}{V}

For a cube of side length $L$ :

V = L^3

To get $\rho$ in $\text{kg m}^{-3}$ , use SI units: mass in kg and length in m.

Understanding the Question

Cube A has:

side length $L = (1.53 \pm 0.01)\ \text{cm}$
mass $m = (31.3 \pm 0.5)\ \text{g}$

You must calculate the density and show it rounds to $8.7 \times 10^{3}\ \text{kg m}^{-3}$ .

Approach

Convert $L$ to metres and $m$ to kilograms.
Calculate $V = L^3$ .
Use $\rho = m/V$ .
Round to match the given form (2 s.f. in $8.7$ ).

Step-by-Step Reasoning

Convert units:
- $1\ \text{cm} = 10^{-2}\ \text{m}$ so

L = 1.53\ \text{cm} = 1.53 \times 10^{-2}\ \text{m}

$1\ \text{g} = 10^{-3}\ \text{kg}$ so

m = 31.3\ \text{g} = 31.3 \times 10^{-3}\ \text{kg} = 3.13 \times 10^{-2}\ \text{kg}

Volume of a cube:

V = L^3 = \left(1.53 \times 10^{-2}\right)^3

Compute the power carefully:

$1.53^3 = 3.58$ (approximately)
$(10^{-2})^3 = 10^{-6}$

So

V = 3.58 \times 10^{-6}\ \text{m}^3

Density:

\rho = \frac{m}{V} = \frac{3.13 \times 10^{-2}}{3.58 \times 10^{-6}}

Divide numbers and subtract powers:

\rho = \left(\frac{3.13}{3.58}\right) \times 10^{4} = 0.874 \times 10^{4} = 8.74 \times 10^{3}\ \text{kg m}^{-3}

Rounding to 2 s.f. gives:

\rho = 8.7 \times 10^{3}\ \text{kg m}^{-3}

Key Takeaways

Always convert to SI before calculating derived units.
For cubes, $V = L^3$ , so small changes in $L$ can significantly affect $V$ .

Common Mistakes

Forgetting to convert cm to m (would give density off by a factor of $10^{6}$ ).
Using $V = 6L^2$ (surface area) instead of $L^3$ .
Incorrect power handling: $(10^{-2})^3 = 10^{-6}$ .

Things to Be Careful About

Use consistent units to get $\text{kg m}^{-3}$ .
Keep sufficient significant figures during working, then round only at the end.

Techniques used

convert measured quantities into SI unitscalculate the volume of a cube from its side lengthapply the definition of density as mass divided by volumeround to an appropriate number of significant figures

(ii)

Calculate the percentage uncertainty in the density of the material of cube A.

percentage uncertainty = ______ %

2M Similar

DifficultyMedium

Worked solution

Working

\rho = \frac{m}{L^3}

Fractional (percentage) uncertainty:

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta L}{L}

\frac{\Delta m}{m} = \frac{0.5}{31.3} = 0.0160 = 1.6\%

\frac{\Delta L}{L} = \frac{0.01}{1.53} = 0.00654 = 0.654\%

3\frac{\Delta L}{L} = 3 \times 0.654\% = 1.96\%

\text{percentage uncertainty in }\rho = 1.6\% + 1.96\% = 3.56\% \approx 3.6\%

Answer

\boxed{3.6\%}

Final answer

3.6%

Detailed explanation

Background Concept

When a quantity is calculated from measured values, its uncertainty depends on how those measurements combine.

For products and quotients, fractional uncertainties add:

Q = \frac{ab}{c} \Rightarrow \frac{\Delta Q}{Q} = \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c}

For powers, multiply the fractional uncertainty by the power:

Q = a^n \Rightarrow \frac{\Delta Q}{Q} = n\frac{\Delta a}{a}

Here,

\rho = \frac{m}{L^3}

So $m$ contributes once, and $L$ contributes three times.

Understanding the Question

You are given uncertainties in:

mass $m = (31.3 \pm 0.5)\ \text{g}$
length $L = (1.53 \pm 0.01)\ \text{cm}$

You must find the percentage uncertainty in $\rho$ .

Approach

Write $\rho$ in terms of measured quantities: $\rho = m/L^3$ .
Use fractional uncertainty rules:
- add the fractional uncertainty of $m$
- add $3$ times the fractional uncertainty of $L$
Convert to a percentage.

Step-by-Step Reasoning

Start with the relationship:

\rho = \frac{m}{L^3}

Apply uncertainty propagation:

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta L}{L}

Mass fractional uncertainty:

\frac{\Delta m}{m} = \frac{0.5}{31.3} = 0.0160

As a percentage:

0.0160 \times 100\% = 1.6\%

Length fractional uncertainty:

\frac{\Delta L}{L} = \frac{0.01}{1.53} = 0.00654

As a percentage:

0.00654 \times 100\% = 0.654\%

Because $L$ is cubed:

3 \times 0.654\% = 1.96\%

Total percentage uncertainty:

1.6\% + 1.96\% = 3.56\% \approx 3.6\%

Key Takeaways

For $\rho = m/L^3$ , the length uncertainty is tripled because of the cube.
Add fractional uncertainties for multiplied/divided quantities.

Common Mistakes

Forgetting the factor of 3 for $L^3$ .
Subtracting uncertainties because it is a division (you still add fractional uncertainties).
Mixing units (cm vs m): fractional uncertainties are unit-independent, but values must be consistent when doing other calculations.

Things to Be Careful About

Keep extra figures during intermediate steps, then round the final percentage.
Quote the final uncertainty to a sensible number of significant figures (here $3.6\%$ ).

Techniques used

use fractional uncertainty rules for multiplication and powersadd percentage uncertainties for quantities multiplied/dividedmultiply fractional uncertainty by the power for a repeated factorconvert fractional uncertainty into a percentage

(iii)

The density of the material of cube B is determined to be $9.2 \times 10^{3}\ \text{kg m}^{-3} \pm 6\%$ .

State and explain whether cube A and cube B could be made from the same material.

2M Similar

DifficultyMedium

Worked solution

Working

For cube A:

\rho_A = 8.7 \times 10^{3}\ \text{kg m}^{-3} \pm 3.6\%

Absolute uncertainty:

\Delta \rho_A = 0.036 \times 8.7 \times 10^{3} = 0.31 \times 10^{3}\ \text{kg m}^{-3}

So range for A:

(8.7 \pm 0.31) \times 10^{3} \Rightarrow 8.39 \times 10^{3} \text{ to } 9.01 \times 10^{3}\ \text{kg m}^{-3}

For cube B:

\rho_B = 9.2 \times 10^{3}\ \text{kg m}^{-3} \pm 6\%

Range for B:

(9.2 \pm 0.06 \times 9.2)\times 10^{3} = (9.2 \pm 0.55)\times 10^{3}

\Rightarrow 8.65 \times 10^{3} \text{ to } 9.75 \times 10^{3}\ \text{kg m}^{-3}

Answer

Yes. The uncertainty ranges overlap (about $8.65$ to $9.01 \times 10^{3}\ \text{kg m}^{-3}$ ), so the two densities are consistent with the same material.

Final answer

Yes; their density ranges overlap within uncertainties, so they could be the same material.

Detailed explanation

Background Concept

When two measured (or calculated) quantities have uncertainties, you decide whether they could represent the same true value by checking consistency.

A common exam method is to compare the ranges:

\text{range} = \text{best estimate} \pm \text{uncertainty}

If the ranges overlap, the results are consistent (they could be the same). If there is no overlap, they are inconsistent (unlikely to be the same).

Understanding the Question

Cube A has a calculated density $\rho_A = 8.7 \times 10^{3}\ \text{kg m}^{-3}$ and from part (ii) a percentage uncertainty (about $3.6\%$ ).

Cube B has:

\rho_B = 9.2 \times 10^{3}\ \text{kg m}^{-3} \pm 6\%

You must state and explain whether A and B could be the same material.

Approach

Convert each percentage uncertainty into an absolute uncertainty.
Write the min and max possible densities for A and B.
Check whether the intervals overlap.
Make a conclusion that matches the overlap test.

Step-by-Step Reasoning

Cube A absolute uncertainty (using $3.6\%$ ):

\Delta \rho_A = 0.036 \times 8.7 \times 10^{3} = 0.31 \times 10^{3}\ \text{kg m}^{-3}

So A could be anywhere between:

(8.7 - 0.31)\times 10^{3} = 8.39 \times 10^{3}

and

(8.7 + 0.31)\times 10^{3} = 9.01 \times 10^{3}\ \text{kg m}^{-3}

Cube B absolute uncertainty:

\Delta \rho_B = 0.06 \times 9.2 \times 10^{3} = 0.55 \times 10^{3}\ \text{kg m}^{-3}

So B could be between:

(9.2 - 0.55)\times 10^{3} = 8.65 \times 10^{3}

and

(9.2 + 0.55)\times 10^{3} = 9.75 \times 10^{3}\ \text{kg m}^{-3}

Compare the ranges:

A: $8.39$ to $9.01 \times 10^{3}$
B: $8.65$ to $9.75 \times 10^{3}$

There is overlap from $8.65$ to $9.01 \times 10^{3}\ \text{kg m}^{-3}$ . Therefore the two results are consistent.

Key Takeaways

Use uncertainty ranges to judge if two results could agree.
Overlap of ranges implies “could be the same” (consistent within uncertainty).

Common Mistakes

Comparing only the central values ( $8.7$ vs $9.2$ ) and concluding “different” without using uncertainties.
Using percentage uncertainty of one cube only and ignoring the other.
Treating overlap as proof they are the same; the correct wording is “could be made from the same material”.

Things to Be Careful About

Convert percentages to decimals correctly (e.g. $6\% = 0.06$ ).
Keep powers of ten consistent when forming the ranges.
Use cautious language: consistent with / could be rather than “are definitely”.

Techniques used

convert percentage uncertainty into an absolute uncertaintycalculate the range of possible values using uncertaintiescompare two ranges to decide whether results are consistentstate a conclusion consistent with overlapping uncertainties

Q2MediumForces, Density and Pressure

(a)

State the principle of moments.

2M Similar

DifficultyEasy

Worked solution

Answer

For a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point (so the resultant moment about the point is zero).

Final answer

In equilibrium, total clockwise moment about a point equals total anticlockwise moment (net moment is zero).

Detailed explanation

Background Concept

A moment of a force about a point (pivot) measures the turning effect of the force.

\text{moment} = F \times d

where $F$ is the force and $d$ is the perpendicular distance from the pivot to the line of action of the force.

For a rigid body to be in rotational equilibrium, it must have no angular acceleration, so the resultant (net) moment about any point must be zero.

Understanding the Question

You are asked to state (not calculate) the condition involving moments that must be true when an object is in equilibrium.

The expected statement must refer to clockwise and anticlockwise turning effects about the same point.

Approach

Recall the standard equilibrium condition:

either “sum of clockwise moments = sum of anticlockwise moments”,
or “algebraic sum of moments about a point is zero”.

Write it clearly and include the idea of a chosen pivot point.

Step-by-Step Reasoning

Choose a point (pivot) about which moments are considered.
In equilibrium, clockwise turning effects must balance anticlockwise turning effects.
Therefore the total clockwise moment equals the total anticlockwise moment (equivalently net moment is zero).

Key Takeaways

Moment $= F \times$ perpendicular distance.
Equilibrium requires no resultant moment.
A correct equivalent statement is acceptable (clockwise = anticlockwise, or sum of moments = 0).

Common Mistakes

Saying “forces are equal” rather than “moments are equal”.
Missing the word perpendicular when defining moment (if asked).
Stating the principle without mentioning about a point.

Things to Be Careful About

It must be about the same point.
Moments depend on perpendicular distance, not the distance along the beam unless the force is perpendicular to the beam (as it is in many beam problems).

Techniques used

state the equilibrium condition for moments about a pointdistinguish clockwise and anticlockwise moments

(b)

A solid plastic cylinder floats in water. It is used to support one end of a horizontal uniform beam AB as shown in Fig. 2.1.

Fig. 2.1

The beam has length $6.0\ \text{m}$ and weight $1700\ \text{N}$ . The beam is attached to solid ground with a hinge at end A.

The cylinder is floating vertically in the water. The top of the cylinder is attached at its centre to the beam at a horizontal distance of $5.0\ \text{m}$ from end A. The cylinder applies a vertical force of $1300\ \text{N}$ to the beam.

A person of weight $660\ \text{N}$ stands on the beam at point P.

The beam AB is in equilibrium.

8M

(i)

By taking moments about end A, determine the distance $x$ from A to P.

distance = ______ $\text{m}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Taking moments about $A$ :

1300 \times 5.0 = 1700 \times 3.0 + 660 \times x

660x = 6500 - 5100 = 1400

x = \frac{1400}{660} = 2.12\ \text{m}

Answer

$2.1\ \text{m}$

Final answer

2.1 m

Detailed explanation

Background Concept

For a horizontal beam in equilibrium about a hinge, we often take moments about the hinge because the unknown hinge reaction forces then have zero moment (their perpendicular distance is zero).

For equilibrium:

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

All forces here are vertical, so the perpendicular distance from $A$ is just the horizontal distance along the beam.

Understanding the Question

The beam of length $6.0\ \text{m}$ is hinged at $A$ .

Beam weight $1700\ \text{N}$ acts at its centre, $3.0\ \text{m}$ from $A$ (downward).
Cylinder pushes on the beam with $1300\ \text{N}$ at $5.0\ \text{m}$ from $A$ (upward).
Person of weight $660\ \text{N}$ acts downward at point $P$ , distance $x$ from $A$ .

The beam is in equilibrium, so moments about $A$ balance. We must find $x$ .

diagram-bi

Approach

Choose pivot at $A$ .
Identify which forces tend to rotate the beam clockwise and anticlockwise about $A$ .
Write the moments equation and solve for $x$ .

Step-by-Step Reasoning

About pivot $A$ :

The upward force at $5.0\ \text{m}$ tends to lift the beam on the right, producing an anticlockwise moment:

M_{\text{ACW}} = 1300 \times 5.0

The beam’s weight acts downward at $3.0\ \text{m}$ , producing a clockwise moment:

M_{\text{CW, beam}} = 1700 \times 3.0

The person’s weight acts downward at $x$ , also producing a clockwise moment:

M_{\text{CW, person}} = 660 \times x

Equilibrium:

1300 \times 5.0 = 1700 \times 3.0 + 660x

Compute:

6500 = 5100 + 660x

660x = 1400

x = 2.12\ \text{m} \approx 2.1\ \text{m}

Key Takeaways

Taking moments about the hinge removes hinge forces from the equation.
For vertical forces on a horizontal beam, moment arm is the horizontal distance from pivot.
Write a clear clockwise = anticlockwise equation.

Common Mistakes

Putting the beam weight at $6.0\ \text{m}$ instead of at $3.0\ \text{m}$ .
Using $6.0 - x$ for the person’s distance when $x$ is defined from $A$ .
Mixing clockwise/anticlockwise signs inconsistently.

Things to Be Careful About

Check which forces produce clockwise vs anticlockwise turning about $A$ .
Quote $x$ to sensible significant figures (here data are mostly 2 s.f., so $2.1\ \text{m}$ is appropriate).

Techniques used

take moments about a pivot to eliminate unknown hinge forcesequate clockwise and anticlockwise moments for equilibriumsolve a linear equation for the unknown distance

(ii)

The bottom of the cylinder is submerged in the water to depth $y$ as shown in Fig. 2.2. The beam is still attached to the cylinder but not shown.

Fig. 2.2

The cylinder has mass $11\ \text{kg}$ and diameter $0.78\ \text{m}$ . The beam exerts a vertical force of $1300\ \text{N}$ on the cylinder. The cylinder is in equilibrium.

Show that the upthrust acting on the cylinder is $1400\ \text{N}$ .

1M Similar

DifficultyEasy

Worked solution

Working

For the cylinder in equilibrium:

U = 1300 + mg

U = 1300 + 11 \times 9.81 = 1.41 \times 10^3\ \text{N} \approx 1.4 \times 10^3\ \text{N}

Answer

Upthrust $= 1400\ \text{N}$

Final answer

1400 N

Detailed explanation

Background Concept

For an object floating at rest, the vertical forces balance:

\sum F_y = 0

So the upthrust (buoyant force) equals the total downward forces.

Also, if two objects push on each other, Newton’s third law says the forces are equal in magnitude and opposite in direction.

Understanding the Question

The beam exerts a downward vertical force of $1300\ \text{N}$ on the cylinder. The cylinder also has its own weight $mg$ downward. The water provides an upthrust $U$ upward.

We must show that $U$ is $1400\ \text{N}$ (to appropriate s.f.).

diagram-bii

Approach

Write the equilibrium condition for vertical forces on the cylinder:

\text{upward force} = \text{total downward force}

Then substitute $m = 11\ \text{kg}$ and $g$ .

Step-by-Step Reasoning

Forces on the cylinder:

Upward: upthrust $U$ .
Downward: force from beam $1300\ \text{N}$ and weight $mg$ .

Equilibrium gives:

U = 1300 + mg

Compute weight:

mg = 11 \times 9.81 = 108\ \text{N} \ (\text{to 3 s.f.})

So:

U = 1300 + 108 = 1408\ \text{N} \approx 1.4 \times 10^3\ \text{N}

which is $1400\ \text{N}$ to 2 s.f.

Key Takeaways

For equilibrium, net force is zero.
Upthrust balances all downward forces, not just the object’s own weight.
Action–reaction explains why the beam’s force on the cylinder is downward if the cylinder’s force on the beam is upward.

Common Mistakes

Using $U = 1300 - mg$ instead of adding.
Forgetting the cylinder’s own weight.
Using $g = 10\ \text{m s}^{-2}$ without consistency (may shift rounding).

Things to Be Careful About

Ensure you are balancing forces on the cylinder, not on the beam.
State the equilibrium equation before substituting numbers (this is usually what earns the method mark).

Techniques used

draw up a vertical force balance for equilibriumuse Newton's third law to relate interaction forcescalculate weight using W = mg

(iii)

The water has density $990\ \text{kg m}^{-3}$ .

Calculate the depth $y$ .

$y$ = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Upthrust:

U = \rho gV = \rho g (Ay)

A = \pi r^2 = \pi (0.39)^2 = 0.478\ \text{m}^2

y = \frac{U}{\rho gA} = \frac{1400}{990 \times 9.81 \times 0.478} = 0.302\ \text{m}

Answer

$y = 0.30\ \text{m}$

Final answer

0.30 m

Detailed explanation

Background Concept

Archimedes’ principle states that the upthrust on an object in a fluid equals the weight of fluid displaced:

U = \rho g V_{\text{displaced}}

For a vertical cylinder of constant cross-sectional area $A$ submerged to depth $y$ , the displaced volume is:

V_{\text{displaced}} = Ay

So:

U = \rho gAy quad \Rightarrow \quad y = \frac{U}{\rho gA}

Understanding the Question

You are given:

upthrust from (ii): $U = 1400\ \text{N}$ ,
water density $\rho = 990\ \text{kg m}^{-3}$ ,
cylinder diameter $0.78\ \text{m}$ so radius $r = 0.39\ \text{m}$ .

You must calculate the submerged depth $y$ .

diagram-biii

Approach

Find the cylinder cross-sectional area $A = \pi r^2$ .
Use $U = \rho gAy$ .
Rearrange to get $y$ and substitute values.

Step-by-Step Reasoning

Radius:

r = \frac{0.78}{2} = 0.39\ \text{m}

Area:

A = \pi r^2 = \pi (0.39)^2 = 0.478\ \text{m}^2

Apply Archimedes:

U = \rho gAy

Rearrange:

y = \frac{U}{\rho gA}

Substitute:

y = \frac{1400}{990 \times 9.81 \times 0.478} = 0.302\ \text{m}

So $y \approx 0.30\ \text{m}$ to 2 s.f.

Key Takeaways

Upthrust depends on displaced volume, not the object’s mass directly.
For a cylinder, displaced volume is area $\times$ submerged depth.
Convert diameter to radius correctly before using $\pi r^2$ .

Common Mistakes

Using diameter as the radius in $\pi r^2$ .
Using $U = \rho g h$ (hydrostatic pressure) instead of $U = \rho gV$ (upthrust).
Forgetting units and mixing cm with m.

Things to Be Careful About

Use $r = 0.39\ \text{m}$ , not $0.78\ \text{m}$ .
Keep consistent significant figures (final $y$ to about 2 s.f. is appropriate here).

Techniques used

apply Archimedes' principle U = rho g Vcalculate cross-sectional area of a cylinderrearrange an equation to find the submerged depth

(iv)

The person can stand anywhere between A and B.

On Fig. 2.3, sketch the variation of the depth of the bottom of the cylinder with the distance of the person from A, for distances between $0$ and $6.0\ \text{m}$ . Numerical values are not required.

Fig. 2.3

2M Similar

DifficultyMedium

Worked solution

Answer

As the person moves further from $A$ , the clockwise moment about $A$ increases, so the upward force from the cylinder must increase.

Since upthrust (and hence submerged depth $y$ ) increases with this force, $y$ increases linearly with distance from $A$ .

Sketch: a straight line with positive gradient (non-zero value at distance $0$ ).

Final answer

Depth increases linearly with distance from A (straight line with positive gradient).

Detailed explanation

Background Concept

For the beam:

Rotational equilibrium gives a moments equation relating the support force from the cylinder to the position of the person.

For the cylinder:

Upthrust is proportional to displaced volume:

U = \rho gAy

With $\rho$ , $g$ , and $A$ constant, we have:

U \propto y

So if the required upthrust increases, the depth $y$ must increase in direct proportion.

Understanding the Question

The person can stand anywhere between $A$ and $B$ (distance from $A$ between $0$ and $6.0\ \text{m}$ ). You must sketch how the depth of the bottom of the cylinder varies with the person’s distance from $A$ .

No numerical scale is needed—only the correct shape/trend.

diagram-biv

Approach

Let the person’s distance from $A$ be $d$ .
Use moments about $A$ to express the cylinder’s upward force on the beam, $F$ , in terms of $d$ .
Relate $F$ to the cylinder’s upthrust $U$ (they differ only by the constant cylinder weight).
Since $U \propto y$ , conclude the form of $y$ vs $d$ and sketch it.

Step-by-Step Reasoning

1) Beam equilibrium (moments about A)

Let the cylinder force on the beam be $F$ (upward) at $5.0\ \text{m}$ .

Equilibrium about $A$ :

F(5.0) = 1700(3.0) + 660(d)

Rearrange:

F = \frac{1700\times 3.0}{5.0} + \frac{660}{5.0}d

This has the form:

F = \text{constant} + (\text{constant})\, d

So $F$ increases linearly with $d$ .

2) Cylinder equilibrium

For the cylinder:

U = F + mg

Since $mg$ is constant, $U$ also increases linearly with $d$ .

3) Link to depth

Because $U = \rho gAy$ and $\rho gA$ is constant:

y \propto U

Therefore $y$ increases linearly with distance $d$ .

At $d=0$ , $F$ is still positive (it must support the beam’s own weight), so $y$ has a non-zero intercept.

Key Takeaways

Person further from hinge $\Rightarrow$ larger clockwise moment.
Support force must rise to balance moments.
Floating cylinder: upthrust $\propto$ submerged depth, so depth rises in the same way.
Linear moments equation $\Rightarrow$ straight-line sketch.

Common Mistakes

Drawing a curve (e.g. quadratic) when the moments relationship is linear.
Drawing the line through the origin; even with the person at $A$ , the cylinder still supports part of the beam’s weight, so depth is not zero.
Sketching decreasing depth with distance (wrong direction of effect).

Things to Be Careful About

The independent variable is the person’s distance from $A$ (0 to 6.0 m).
You are sketching depth of the bottom of the cylinder below the water surface; greater upthrust means greater submerged length, hence greater depth.

Techniques used

express the support force using a moments equationlink upthrust to displaced volume to infer proportionalitysketch a qualitative graph from a linear relationship

Q3MediumKinematicsWork, Energy and PowerDynamics

(a)

A truck R of mass $9400\ \text{kg}$ moves with constant acceleration in a straight line down a slope, as illustrated in Fig. 3.1.

Fig. 3.1

At point A the speed of the truck is $13\ \text{m s}^{-1}$ and at point B the speed of the truck is $22\ \text{m s}^{-1}$ . A and B are a distance of $180\ \text{m}$ apart.

5M

(i)

Calculate the acceleration of the truck between A and B.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For constant acceleration,

v^2 = u^2 + 2as

a = \frac{v^2-u^2}{2s} = \frac{22^2-13^2}{2\times 180} = \frac{315}{360} = 0.875

Answer

a = 0.88\ \text{m s}^{-2}

Final answer

0.88 m s^-2

Detailed explanation

Background Concept

For motion in a straight line with constant acceleration, the kinematic (SUVAT) equations apply. One useful equation that links speeds and distance without time is

v^2 = u^2 + 2as

where:

$u$ is the initial speed,
$v$ is the final speed,
$a$ is the (constant) acceleration,
$s$ is the displacement along the line of motion.

This equation is valid only when acceleration is constant and motion is one-dimensional.

Understanding the Question

The truck moves down a slope in a straight line with constant acceleration. At point A: $u=13\ \text{m s}^{-1}$ . At point B: $v=22\ \text{m s}^{-1}$ . The distance along the slope between A and B is $s=180\ \text{m}$ . We are asked to calculate the acceleration $a$ .

Approach

Use the SUVAT equation that contains $u$ , $v$ , $s$ , and $a$ but does not require time, then rearrange to make $a$ the subject.

Step-by-Step Reasoning

Start from

v^2 = u^2 + 2as

Rearrange:

a = \frac{v^2-u^2}{2s}

Substitute values:

a = \frac{22^2-13^2}{2\times 180}

Compute:

$22^2 = 484$
$13^2 = 169$
$484-169 = 315$
$2\times 180 = 360$

So

a = \frac{315}{360} = 0.875\ \text{m s}^{-2} \approx 0.88\ \text{m s}^{-2}

Key Takeaways

For constant acceleration, pick the SUVAT equation that uses the variables given.
$v^2 = u^2 + 2as$ is ideal when time is not given.

Common Mistakes

Using $v = u + at$ even though $t$ is not given.
Forgetting the factor of $2$ in $2as$ .
Using $s=180$ with the wrong sign convention (here, taking motion down the slope as positive makes $a$ positive).

Things to Be Careful About

Check the condition “constant acceleration” before using SUVAT.
Quote acceleration with correct unit $\text{m s}^{-2}$ and sensible significant figures.

Techniques used

select the appropriate constant-acceleration SUVAT equationsubstitute given speeds and displacement into the equationrearrange to solve for acceleration

(ii)

Determine the gain in kinetic energy of the truck between A and B.

gain in kinetic energy = ______ $\text{J}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Gain in kinetic energy

\Delta E_k = \frac{1}{2}m\left(v^2-u^2\right)

\Delta E_k = \frac{1}{2}(9400)\left(22^2-13^2\right)=4700\times 315 = 1.4805\times 10^6

Answer

\Delta E_k \approx 1.48\times 10^6\ \text{J}

Final answer

1.48 × 10^6 J

Detailed explanation

Background Concept

The kinetic energy of an object of mass $m$ moving at speed $v$ is

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

If its speed changes from $u$ to $v$ , the gain in kinetic energy is the difference:

\Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2}m\left(v^2-u^2\right)

Understanding the Question

The truck has mass $m=9400\ \text{kg}$ . Its speed increases from $13\ \text{m s}^{-1}$ at A to $22\ \text{m s}^{-1}$ at B. We must find how much its kinetic energy increases between these points.

Approach

Compute kinetic energy at B and at A (or use the combined change formula), then subtract to get the gain.

Step-by-Step Reasoning

Use the change formula:

\Delta E_k = \frac{1}{2}m\left(v^2-u^2\right)

Substitute:

\Delta E_k = \frac{1}{2}(9400)\left(22^2-13^2\right)

Calculate the bracket first:

22^2-13^2 = 484-169 = 315

Now multiply:

\Delta E_k = 4700\times 315 = 1\,480\,500\ \text{J} = 1.4805\times 10^6\ \text{J}

So the gain in kinetic energy is about $1.48\times 10^6\ \text{J}$ (often rounded to $1.5\times 10^6\ \text{J}$ depending on sig figs).

Key Takeaways

Kinetic energy depends on $v^2$ , so a moderate speed increase can cause a large energy increase.
Change in kinetic energy can be computed efficiently using $\frac{1}{2}m(v^2-u^2)$ .

Common Mistakes

Using $\Delta E_k = \frac{1}{2}m(v-u)^2$ (this is incorrect).
Forgetting to square the speeds.
Writing the unit as $\text{kg m}^2\text{s}^{-2}$ instead of the accepted $\text{J}$ (joule).

Things to Be Careful About

Keep the mass in $\text{kg}$ and speeds in $\text{m s}^{-1}$ so the energy comes out in joules.
Sensible rounding: the speeds are given to 2 s.f., so an energy to 2–3 s.f. is appropriate.

Techniques used

use the kinetic energy expressioncalculate change in kinetic energy between two speedshandle substitution and arithmetic with correct units and significant figures

(b)

A short time after passing point B truck R moves in a straight line on horizontal ground. The driver of the truck applies the brakes. Fig. 3.2 shows the variation with time of the momentum of the truck.

Fig. 3.2

5M

(i)

Define force.

1M Similar

(ii)

Show that the average resultant force $F$ acting on truck R between time $t = 0$ and $t = 15\ \text{s}$ is $-1.2 \times 10^{4}\ \text{N}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Average force:

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

From the graph:

p(0) = 21\times 10^4,\quad p(15) = 3\times 10^4\ \text{kg m s}^{-1}

\Delta p = 3\times 10^4 - 21\times 10^4 = -18\times 10^4 = -1.8\times 10^5

F_{\text{avg}} = \frac{-1.8\times 10^5}{15} = -1.2\times 10^4\ \text{N}

Answer

F_{\text{avg}} = -1.2\times 10^4\ \text{N}

Final answer

-1.2 × 10^4 N

Detailed explanation

Background Concept

Force is related to momentum by

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

On a graph of momentum $p$ against time $t$ :

the gradient at a point gives the instantaneous force,
the average gradient between two times gives the average force:

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

A negative value means the force acts opposite to the direction of the original momentum (a braking force).

Understanding the Question

You are given a momentum–time graph for truck R after braking begins. You must show that the average resultant force between $t=0$ and $t=15\ \text{s}$ is $-1.2\times 10^4\ \text{N}$ . This means:

read momentum at $t=0$ and $t=15\ \text{s}$ ,
calculate $\Delta p$ ,
divide by $15\ \text{s}$ .

Note: the $y$ -axis is labelled “momentum / $10^4\ \text{kg m s}^{-1}$ ”, so a reading of 21 corresponds to $21\times 10^4\ \text{kg m s}^{-1}$ .

Approach

Use average force formula $F_{\text{avg}} = \Delta p/\Delta t$ with the two momentum values taken from the graph.

Step-by-Step Reasoning

Read off from the graph:

p(0) = 21\times 10^4\ \text{kg m s}^{-1}

p(15) = 3\times 10^4\ \text{kg m s}^{-1}

Compute the change:

\Delta p = p(15) - p(0) = 3\times 10^4 - 21\times 10^4 = -18\times 10^4 = -1.8\times 10^5

Divide by the time interval $\Delta t = 15\ \text{s}$ :

F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{-1.8\times 10^5}{15} = -1.2\times 10^4\ \text{N}

The negative sign indicates the resultant force is opposite the truck’s initial direction of motion (as expected for braking).

Key Takeaways

The force is the slope of the momentum–time graph.
Average force over an interval is the change in momentum divided by the time taken.

Common Mistakes

Forgetting the $\times 10^4$ scale factor on the momentum axis.
Using $p(0)-p(15)$ and then forgetting to include the negative sign.
Dividing by $25\ \text{s}$ (time to reach rest) instead of the required $15\ \text{s}$ .

Things to Be Careful About

Always include units: momentum in $\text{kg m s}^{-1}$ and force in $\text{N}$ .
Keep sign conventions consistent: if initial momentum is positive, braking reduces it so $\Delta p$ is negative.

Techniques used

read values from a momentum–time graphuse F = Δp/Δt to find average resultant forceapply sign conventions to indicate direction

(iii)

An identical truck S has the same initial momentum as truck R. Truck S experiences a constant force equal to the force $F$ in (b)(ii).

State and explain whether truck S will take more, less or the same amount of time to come to rest as truck R.

3M Similar

DifficultyMedium

Worked solution

Working

For truck S, with constant force $F$ ,

F = \frac{\Delta p}{\Delta t} \Rightarrow t = \frac{p_0}{|F|}

p_0 = 21\times 10^4 = 2.1\times 10^5\ \text{kg m s}^{-1}

t_S = \frac{2.1\times 10^5}{1.2\times 10^4} = 17.5\ \text{s}

Truck R comes to rest at about $25\ \text{s}$ (from the graph).

Answer

Truck S takes less time to come to rest (about $18\ \text{s}$ ).

The force on R decreases in magnitude with time (momentum–time graph becomes less steep), so its average force over the whole stop is smaller and it takes longer.

Final answer

Less time (≈ 18 s); truck R’s braking force decreases so it stops later (≈ 25 s).

Detailed explanation

Background Concept

Using momentum form of Newton’s second law:

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

If $F$ is constant, then $\mathrm{d}p/\mathrm{d}t$ is constant, so $p$ decreases linearly with time.
The time to reduce momentum from $p_0$ to zero under a constant opposing force of magnitude $|F|$ is

0 = p_0 - |F|t \Rightarrow t = \frac{p_0}{|F|}

For a variable force, the slope (gradient) of the $p$ – $t$ graph changes with time; a smaller magnitude slope means momentum is being removed more slowly.

Understanding the Question

Two trucks start with the same initial momentum.

Truck R: braking force is not constant (shown by a curved $p$ – $t$ graph). It reaches $p=0$ at about $25\ \text{s}$ .
Truck S: experiences a constant braking force equal to the average force found for R between $0$ and $15\ \text{s}$ , i.e. $F=-1.2\times 10^4\ \text{N}$ .

We must decide whether S takes more/less/same time to stop compared to R, and explain using momentum and force ideas.

Approach

For truck S, use constant-force momentum relation to calculate stopping time.
For truck R, use the graph to identify its stopping time (where momentum reaches zero).
Explain the difference by discussing how the gradient (hence force) changes for truck R.

Step-by-Step Reasoning

1. Stopping time for truck S (constant force):
Initial momentum (from the same starting point as R) is

p_0 = 21\times 10^4 = 2.1\times 10^5\ \text{kg m s}^{-1}

Constant braking force magnitude is $|F|=1.2\times 10^4\ \text{N}$ . For constant $F$ ,

t_S = \frac{p_0}{|F|} = \frac{2.1\times 10^5}{1.2\times 10^4} = 17.5\ \text{s}

So truck S would stop in about $18\ \text{s}$ .

2. Stopping time for truck R (from graph):
From the graph, momentum approaches zero at about $t \approx 25\ \text{s}$ . So truck R takes about $25\ \text{s}$ to stop.

3. Comparison and explanation:
Truck S stops sooner ( $\approx 18\ \text{s}$ ) than truck R ( $\approx 25\ \text{s}$ ).

The momentum–time curve for truck R is steep initially and becomes progressively less steep. Since the gradient is $\mathrm{d}p/\mathrm{d}t$ and equals force, this means the braking force on R is large at first but then decreases in magnitude. When the braking force becomes smaller, momentum is removed more slowly, increasing the total time to reach zero momentum.

In contrast, truck S maintains a constant force equal to $-1.2\times 10^4\ \text{N}$ throughout, so it continues removing momentum at a steady rate and comes to rest earlier.

Key Takeaways

Constant force $\Rightarrow$ linear change of momentum with time.
The gradient of a $p$ – $t$ graph gives force; a curve means force is changing.
For a given initial momentum, a larger (or sustained) braking force gives a shorter stopping time.

Common Mistakes

Saying “same time because same average force” without noticing that the given average force is only for the first $15\ \text{s}$ .
Confusing the momentum–time graph with a velocity–time graph.
Ignoring the sign of the force and treating $-1.2\times 10^4\ \text{N}$ as if it were accelerating the truck forwards.

Things to Be Careful About

Use the correct initial momentum value including the $\times 10^4$ factor.
When comparing times, read the stopping time for truck R at the point where $p=0$ .
The key physics explanation must refer to gradient = force and the fact that the gradient decreases in magnitude for truck R (force becomes less effective with time).

Techniques used

use F = dp/dt to relate stopping time to forcecompare constant-force braking with variable-force braking from graph shapeestimate stopping time from initial momentum and constant average force

Q4MediumSuperpositionWaves

A device containing a microwave emitter and receiver is placed in front of a large metal sheet in a vacuum as shown in Fig. 4.1.

Fig. 4.1

The line XY is perpendicular to the metal sheet. The device emits microwaves of frequency $6.3\ \text{GHz}$ .

(a)

When the device is at position P, a stationary wave is formed between the device and the sheet.

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

4M Similar

DifficultyMedium

Worked solution

Answer

Microwaves from the emitter travel towards the metal sheet and are reflected.
The reflected wave has the same frequency (and wavelength) and travels back along XY in the opposite direction.
The incident and reflected waves superpose to give interference.
Where the waves are always in antiphase, destructive interference gives a node (zero amplitude); where they are always in phase, constructive interference gives an antinode (maximum amplitude).
Nodes and antinodes are fixed; adjacent nodes are separated by $\lambda/2$ (node to nearest antinode is $\lambda/4$ ).

Final answer

A stationary wave forms by superposition of the incident and reflected microwaves, giving fixed nodes (zero amplitude, antiphase) and antinodes (maximum amplitude, in phase), spaced by (\lambda/2) between nodes.

Detailed explanation

Background Concept

A stationary (standing) wave is produced when two waves of the same frequency and wavelength travel in opposite directions along the same line and overlap.

The resultant displacement (or electric field for microwaves) at any point is found by the principle of superposition: the instantaneous displacements add.

Node: a point where the resultant amplitude is always zero (complete destructive interference at all times).
Antinode: a point where the resultant amplitude is maximum (complete constructive interference).
Spacing: adjacent nodes are separated by $\lambda/2$ , and a node to its nearest antinode is $\lambda/4$ .

Understanding the Question

The device emits microwaves towards a large metal sheet in a vacuum. The sheet reflects the microwaves back along the same line XY. You are asked to explain how this reflection leads to a stationary wave pattern and what nodes and antinodes are.

Approach

State that the metal sheet reflects the incident microwaves.
Emphasise that the reflected wave travels back in the opposite direction with the same frequency.
Use superposition to explain how interference produces fixed points of zero and maximum amplitude.
Identify which phase relationships produce nodes and antinodes, and state the standard spacing.

Step-by-Step Reasoning

The emitter produces a travelling wave that moves from the device towards the metal sheet.
A metal sheet is a strong reflector for microwaves, so an additional travelling wave is produced that moves back from the sheet towards the device.
Both waves have the same frequency $f$ (reflection does not change frequency) and therefore the same wavelength $\lambda$ in the same medium (vacuum).
Because they are travelling in opposite directions, at each position the two waves combine:
- If at a particular position the two waves are always equal and opposite (phase difference $\pi$ ), they cancel at all times. The resultant amplitude is zero: a node.
- If at a particular position the two waves are always in step (phase difference $0$ ), they reinforce. The resultant amplitude is maximum: an antinode.
These positions do not move (they depend only on position along XY), so the pattern is “stationary”.

diagram-a-1

Key Takeaways

Stationary waves come from two counter-propagating waves with the same $f$ and $\lambda$ .
Nodes are permanent zero-amplitude points (destructive interference); antinodes are permanent maximum-amplitude points (constructive interference).
Key spacings: node-node $= \lambda/2$ , node-antinode $= \lambda/4$ .

Common Mistakes

Saying a stationary wave is “one wave that stops moving” (it is a superposition of two travelling waves).
Mixing up the spacing: writing node-node as $\lambda/4$ .
Forgetting to mention that reflection produces a wave travelling in the opposite direction.

Things to Be Careful About

“Amplitude detected” corresponds to the wave’s field (or intensity) at the receiver position; nodes mean the detected amplitude is minimum (ideally zero).
Frequency stays the same on reflection; the medium determines the wave speed and hence wavelength.

Techniques used

describe reflection of a wave at a boundaryapply the principle of superposition to two counter-propagating wavesuse phase relationships to identify nodes and antinodesstate the spacing between adjacent nodes and antinodes

(b)

4M

(i)

Calculate the wavelength of the microwaves.

wavelength = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

\lambda = \frac{c}{f} = \frac{3.0 \times 10^{8}}{6.3 \times 10^{9}}

\lambda = 4.76 \times 10^{-2}\ \text{m} \approx 4.8 \times 10^{-2}\ \text{m}

Answer

$\boxed{4.8 \times 10^{-2}\ \text{m}}$

Final answer

4.8 × 10^-2 m

Detailed explanation

Background Concept

For any wave,

v = f\lambda

where $v$ is wave speed, $f$ is frequency, and $\lambda$ is wavelength.

Microwaves are electromagnetic waves, and in a vacuum they travel at the speed of light:

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

Understanding the Question

You are given the microwave frequency $f = 6.3\ \text{GHz}$ and told the waves travel in a vacuum. You must find $\lambda$ .

Approach

Convert $6.3\ \text{GHz}$ to Hz, then rearrange $v=f\lambda$ to $\lambda = v/f$ using $v=c$ .

Step-by-Step Reasoning

Convert frequency:

6.3\ \text{GHz} = 6.3 \times 10^{9}\ \text{Hz}.

Rearrange and substitute:

\lambda = \frac{c}{f} = \frac{3.0 \times 10^{8}}{6.3 \times 10^{9}}\ \text{m}.

Evaluate:

\lambda = 4.76 \times 10^{-2}\ \text{m} \approx 4.8 \times 10^{-2}\ \text{m}.

Key Takeaways

In vacuum, electromagnetic waves have $v=c$ .
Always convert GHz to Hz before substituting.

Common Mistakes

Using $3.0 \times 10^{8}$ with $6.3$ instead of $6.3 \times 10^{9}$ .
Writing the wavelength in cm without converting to metres when the answer line requires metres.

Things to Be Careful About

Powers of ten: dividing by $10^{9}$ gives a factor of $10^{-9}$ , so the result must be much smaller than 1 m.
Significant figures: frequency is given to 2 s.f., so $\lambda$ should be quoted to 2 s.f.

Techniques used

use the wave equation to relate speed, frequency and wavelengthconvert between gigahertz and hertzsubstitute values and round to appropriate significant figures

(ii)

At point P the receiver detects a maximum amplitude of the stationary wave.

The device is moved slowly from point P along the line XY and the receiver detects a series of minimum and maximum amplitudes. The first time a minimum amplitude is detected by the receiver is when the device is at point Q.

Determine the distance between P and Q.

distance = ______ $\text{m}$

1M Similar

DifficultyEasy

Worked solution

Working

From a maximum (antinode) to the nearest minimum (node),

PQ = \frac{\lambda}{4} = \frac{4.76 \times 10^{-2}}{4}

PQ = 1.19 \times 10^{-2}\ \text{m} \approx 1.2 \times 10^{-2}\ \text{m}

Answer

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

Final answer

1.2 × 10^-2 m

Detailed explanation

Background Concept

In a stationary wave, nodes (minimum amplitude) and antinodes (maximum amplitude) occur at fixed positions.

Key spacing rules:

node to adjacent node: $\lambda/2$
node to adjacent antinode (or antinode to adjacent node): $\lambda/4$

Understanding the Question

At $P$ the receiver is at a maximum amplitude position (an antinode). As the device is moved, the first time it detects a minimum amplitude is at $Q$ (a node). The question asks for the distance from an antinode to the nearest node.

Approach

Use the stationary-wave spacing: antinode to nearest node equals $\lambda/4$ . Use the wavelength found in (i).

Step-by-Step Reasoning

$P$ is an antinode, $Q$ is the nearest node encountered.
The nearest node is a quarter wavelength away:

PQ = \frac{\lambda}{4}.

With $\lambda = 4.76 \times 10^{-2}\ \text{m}$ :

PQ = \frac{4.76 \times 10^{-2}}{4} = 1.19 \times 10^{-2}\ \text{m} \approx 1.2 \times 10^{-2}\ \text{m}.

Key Takeaways

Moving from a maximum to the first minimum in a standing wave corresponds to $\lambda/4$ .

Common Mistakes

Using $\lambda/2$ instead of $\lambda/4$ .
Measuring between successive maxima (which would be $\lambda/2$ ) instead of maximum to minimum.

Things to Be Careful About

“First minimum” means the nearest node, not the next one further away.
Keep your answer in metres, matching the answer line.

Techniques used

identify maxima and minima as antinodes and nodes in a stationary waveuse the node-antinode spacing in a stationary wavesubstitute the wavelength to find a distance

(iii)

The intensity of the microwaves emitted by the device is increased. The frequency of the microwaves is unchanged. The device is moved slowly along the line XY from point Q until the next maximum amplitude is detected at point R.

State and explain whether the distance QR is greater than, less than or the same as distance PQ.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$QR$ is the same as $PQ$ .

Increasing intensity increases amplitude but does not change frequency, so $\lambda$ is unchanged. The separation of a node and adjacent antinode remains $\lambda/4$ , so $QR = PQ$ .

Final answer

Same as PQ

Detailed explanation

Background Concept

For waves in a fixed medium (vacuum here), the wavelength is set by

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

If $f$ is unchanged and $v$ is unchanged, then $\lambda$ is unchanged.

Intensity is related to amplitude (for electromagnetic waves, intensity $I \propto A^2$ ). Changing intensity changes amplitude, not the spacing of nodes/antinodes.

Understanding the Question

At $Q$ the receiver detects a minimum (a node). The device is then moved until the next maximum is detected at $R$ (an antinode). You are asked how $QR$ compares with $PQ$ after the emitted intensity is increased but frequency stays the same.

Approach

Recognise that node/antinode positions depend on wavelength only. Since frequency is unchanged in vacuum, wavelength is unchanged. Therefore the node-to-adjacent-antinode distance stays $\lambda/4$ , so $QR$ equals $PQ$ .

Step-by-Step Reasoning

At $Q$ the receiver is at a node (minimum amplitude).
The next maximum occurs at the adjacent antinode, a distance $\lambda/4$ away.
Increasing the emitted intensity increases the amplitude of both incident and reflected waves, so the maxima become larger and minima remain minima, but the positions of nodes/antinodes do not shift.
Because $f$ is unchanged and the waves travel in vacuum ( $v=c$ constant),

\lambda = \frac{c}{f}

is unchanged.

Therefore,

QR = \frac{\lambda}{4} = PQ.

Key Takeaways

Changing intensity changes amplitude, not wavelength.
Node/antinode spacing in a stationary wave depends on $\lambda$ only.

Common Mistakes

Saying $QR$ is larger because the wave is “stronger” (strength affects amplitude, not spacing).
Confusing intensity with frequency.

Things to Be Careful About

The question explicitly says frequency is unchanged; that is the key clue that spacing is unchanged.
“Next maximum” from a node means the adjacent antinode, not skipping one (which would add another $\lambda/2$ ).

Techniques used

relate intensity change to amplitude changelink node/antinode positions to wavelengthcompare distances using stationary-wave spacing

Q5Medium-EasyWaves

A stationary loudspeaker emits sound of constant frequency. A microphone is placed near to the loudspeaker and connected to a cathode-ray oscilloscope (CRO). The trace on the screen of the CRO is shown in Fig. 5.1.

Fig. 5.1

The time-base of the CRO is set to $5.0 \times 10^{-4}\ \text{s cm}^{-1}$ .

(a)

The speed of the sound emitted by the loudspeaker is $330\ \text{m s}^{-1}$ .

Determine the wavelength of the sound.

wavelength = ______ $\text{m}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

From the CRO trace, one period occupies $4.0\ \text{cm}$ .

Time-base $= 5.0 \times 10^{-4}\ \text{s cm}^{-1}$ , so

T = 4.0 \times 5.0 \times 10^{-4} = 2.0 \times 10^{-3}\ \text{s}

Using $\lambda = vT$ :

\lambda = 330 \times 2.0 \times 10^{-3} = 0.66\ \text{m}

Answer

$\lambda = 0.66\ \text{m}$

Final answer

0.66 m

Detailed explanation

Background Concept

For a progressive wave,

v = f\lambda

where:

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

Since frequency is the reciprocal of period,

f = \frac{1}{T}

we can also write

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

A CRO time-base tells you how much time corresponds to $1\ \text{cm}$ horizontally, so the horizontal length of one cycle gives the period $T$ .

Understanding the Question

You are given:

speed of sound $v = 330\ \text{m s}^{-1}$ ,
CRO time-base $5.0 \times 10^{-4}\ \text{s cm}^{-1}$ ,
the trace shows one complete cycle spanning $4$ major divisions, and each major division is $1\ \text{cm}$ .

You must find the wavelength $\lambda$ of the sound.

Approach

Read the horizontal distance for one full cycle from the CRO trace (in cm).
Convert this to a time using the time-base to get the period $T$ .
Use $\lambda = vT$ to calculate wavelength.

Step-by-Step Reasoning

Period from the trace: one full cycle (e.g. crest to next crest) spans $4.0\ \text{cm}$ .
Convert cm to seconds:

T = (4.0\ \text{cm})\times \left(5.0 \times 10^{-4}\ \frac{\text{s}}{\text{cm}}\right) = 2.0 \times 10^{-3}\ \text{s}.

Calculate wavelength:

\lambda = vT = 330\ \text{m s}^{-1} \times 2.0 \times 10^{-3}\ \text{s} = 0.66\ \text{m}.

Key Takeaways

CRO time-base converts horizontal distance ( $\text{cm}$ ) into time ( $\text{s}$ ).
The wave relations $f=1/T$ and $v=f\lambda$ combine to give $\lambda=vT$ .

Common Mistakes

Using peak-to-peak vertical height to find period (period is horizontal).
Forgetting the time-base is per cm and not per division (here they are the same because $1$ division $= 1\ \text{cm}$ ).
Using $\lambda = v/f$ but calculating $f$ incorrectly (e.g. using $f=T$ instead of $f=1/T$ ).

Things to Be Careful About

Make sure you measure one complete cycle (same point on the wave to the next identical point).
Keep track of powers of ten: $5.0 \times 10^{-4}$ multiplied by $4$ gives $2.0 \times 10^{-3}$ .
Give wavelength in metres and to appropriate significant figures (here $2$ s.f. matches the data).

Techniques used

read the period from an oscilloscope time-baseconvert horizontal divisions to time using the time-base settinguse $\lambda = vT$ (or $v=f\lambda$) to calculate wavelength

(b)

The loudspeaker now moves in a straight line while emitting the same sound of constant frequency. The period of the trace on the CRO increases continuously.

Describe the motion of the loudspeaker.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Since $T$ increases, the observed frequency decreases ( $f = 1/T$ ).

For a moving source, a decrease in observed frequency means the loudspeaker is moving away from the microphone.

Because the period increases continuously, the loudspeaker’s speed away from the microphone is increasing (it is accelerating away).

Final answer

It moves away from the microphone with increasing speed (accelerating away).

Detailed explanation

Background Concept

The Doppler effect is the change in observed frequency due to relative motion between a source and an observer.

For a moving source and a stationary observer in still air:

If the source moves towards the observer, wavefronts are compressed in front: smaller $\lambda'$ , so larger observed frequency $f'$ .
If the source moves away, wavefronts are spread out behind: larger $\lambda'$ , so smaller observed frequency $f'$ .

The CRO displays the microphone signal versus time, so the horizontal period $T$ is linked to observed frequency by

f' = \frac{1}{T}.

So an increasing period means the observed frequency is decreasing.

Understanding the Question

Initially the loudspeaker is stationary and emits constant frequency. Then it moves in a straight line while emitting the same sound frequency at the source.

The key observation is: the period shown on the CRO increases continuously.

The CRO is connected to the microphone, so it shows the frequency arriving at the microphone.
Therefore, the frequency arriving at the microphone is continuously decreasing.

The question asks you to describe the loudspeaker’s motion (direction and whether speed is changing).

Approach

Convert “period increases” into “observed frequency decreases” using $f'=1/T$ .
Use Doppler-effect direction rule: decreasing observed frequency means the source is moving away.
“Continuously increases” indicates the Doppler shift is getting larger with time, so the speed of recession is increasing (acceleration away).

Step-by-Step Reasoning

From CRO period to frequency:
If $T$ increases, then

f' = \frac{1}{T}

decreases.

Relate frequency change to motion direction (Doppler):
For sound, a moving source that is receding from the observer produces a lower observed frequency. Therefore the loudspeaker must be moving away from the microphone.
Use “increases continuously” to infer acceleration:
If the loudspeaker moved away at constant speed, the observed frequency would be constant (so the CRO period would be constant).

Because the period keeps increasing, the observed frequency keeps decreasing more and more, which implies the speed away from the microphone is increasing: the loudspeaker is accelerating away.

diagram-5b-1

Key Takeaways

CRO period $T$ is the reciprocal of observed frequency $f'$ .
For a moving source: moving away (\Rightarrow) lower observed frequency (larger period).
A continuously changing period indicates changing speed (acceleration).

Common Mistakes

Saying the loudspeaker is moving towards the microphone (that would make $T$ decrease).
Confusing “source frequency constant” with “observed frequency constant” (Doppler effect changes the observed frequency).
Not commenting on acceleration: “period increases continuously” implies the speed is not constant.

Things to Be Careful About

The loudspeaker emits constant frequency in its own rest frame; the microphone measures the frequency it receives.
“Straight line” does not automatically mean towards/away; you must infer direction from whether $f'$ increases or decreases.
“Continuously” is the clue that the speed is changing rather than constant.

Techniques used

relate period change to frequency change using $f = 1/T$use Doppler-effect reasoning to connect observed frequency to source directioninfer acceleration from a continuous change in observed frequency

Q6MediumElectricity

A cylindrical copper wire P of length $0.24\ \text{m}$ is shown in Fig. 6.1.

Fig. 6.1

The current in the wire is $0.85\ \text{A}$ .
The resistance of the wire is $3.3\ \text{m}\Omega$ .
The total number of charge carriers $N$ in the wire is $2.6 \times 10^{22}$ .
The resistivity of copper is $1.8 \times 10^{-8}\ \Omega\ \text{m}$ .

(a)

Calculate the potential difference between the two ends of the wire.

potential difference = ______ $\text{V}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

V = IR

R = 3.3\ \text{m}\Omega = 3.3 \times 10^{-3}\ \Omega

V = 0.85 \times 3.3 \times 10^{-3} = 2.8 \times 10^{-3}\ \text{V}

Answer

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

Final answer

2.8 × 10⁻³ V

Detailed explanation

Background Concept

For a conductor at constant temperature, the potential difference $V$ across it is related to the current $I$ through it by Ohm’s law:

V = IR

where $R$ is the resistance in ohms ($\Omega$). A common pitfall is unit conversion: milliohms (m$\Omega$) must be converted to ohms.

Understanding the Question

You are given the current in wire P ( $I = 0.85\ \text{A}$ ) and its resistance ( $R = 3.3\ \text{m}\Omega$ ). The task is to find the potential difference between the ends of the wire.

Approach

Convert $R$ from m$\Omega $to \$ \Omega$.
Substitute into $V = IR$ .

Step-by-Step Reasoning

Convert resistance:

3.3\ \text{m}\Omega = 3.3 \times 10^{-3}\ \Omega

Apply Ohm’s law:

V = IR = 0.85 \times 3.3 \times 10^{-3} = 2.805 \times 10^{-3}\ \text{V}

Round appropriately:

V \approx 2.8 \times 10^{-3}\ \text{V}

Key Takeaways

Use $V = IR$ for a simple potential difference calculation.
Always convert prefixes correctly (m$\Omega\rightarrow\Omega $means \$ \times 10^{-3}$).

Common Mistakes

Using $R = 3.3\ \Omega$ instead of $3.3 \times 10^{-3}\ \Omega$ (gives an answer 1000 times too large).
Missing the unit of volts in the final answer.

Things to Be Careful About

Significant figures: inputs are typically 2 s.f., so quote $V$ to 2 s.f.
Keep powers of ten explicit to avoid calculator entry errors.

Techniques used

apply Ohm's lawconvert milliohms to ohmssubstitute values with correct powers of ten

(b)

5M

(i)

Show that the cross-sectional area of the wire is $1.3 \times 10^{-6}\ \text{m}^{2}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

A = \frac{(1.8 \times 10^{-8}) \times 0.24}{3.3 \times 10^{-3}} = 1.3 \times 10^{-6}\ \text{m}^2

Answer

$A = 1.3 \times 10^{-6}\ \text{m}^2$

Final answer

1.3 × 10⁻⁶ m²

Detailed explanation

Background Concept

The resistance of a uniform cylindrical wire is related to its resistivity by

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

where:

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

Understanding the Question

You are given $R$ , $\rho$ , and $L$ for wire P and asked to show that its cross-sectional area is $1.3 \times 10^{-6}\ \text{m}^2$ .

Approach

Rearrange $R = \rho L/A$ to make $A$ the subject, then substitute the given values (converting $R$ from m$\Omega $to \$ \Omega$).

Step-by-Step Reasoning

Rearrange:

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

Substitute:

A = \frac{(1.8 \times 10^{-8}\ \Omega\,\text{m})\times (0.24\ \text{m})}{3.3\times 10^{-3}\ \Omega}

Compute:

1.8\times 0.24 = 0.432

Powers of ten:

\frac{10^{-8}}{10^{-3}} = 10^{-5}

So:

A = \frac{0.432}{3.3} \times 10^{-5} \approx 0.131 \times 10^{-5} = 1.31 \times 10^{-6}\ \text{m}^2

Hence $A = 1.3 \times 10^{-6}\ \text{m}^2$ .

Key Takeaways

Know and use $R = \rho L/A$ .
Converting m$\Omega $to \$ \Omega$ is essential.

Common Mistakes

Forgetting to convert $3.3\ \text{m}\Omega$ to $3.3\times 10^{-3}\ \Omega$ .
Rearranging incorrectly (e.g. $A = \rho/(RL)$ ).

Things to Be Careful About

Units: using $\rho$ in $\Omega,\text{m} $forces$ L $in m and$ A $in m$ ^2$.
Significant figures: final shown value matches the stated target (2 s.f.).

Techniques used

use the resistivity relationrearrange an equation to make the required quantity the subjectsubstitute values with correct powers of ten

(ii)

Show that the number density of charge carriers in the wire is $8.3 \times 10^{28}\ \text{m}^{-3}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\text{Volume} = AL = (1.3 \times 10^{-6})\times 0.24 = 3.12 \times 10^{-7}\ \text{m}^3

n = \frac{N}{\text{Volume}} = \frac{2.6 \times 10^{22}}{3.12 \times 10^{-7}} = 8.3 \times 10^{28}\ \text{m}^{-3}

Answer

$8.3 \times 10^{28}\ \text{m}^{-3}$

Final answer

8.3 × 10²⁸ m⁻³

Detailed explanation

Background Concept

Number density $n$ is the number of particles per unit volume:

n = \frac{N}{V}

For a uniform wire, the volume is

V = AL

where $A$ is cross-sectional area and $L$ is length.

Understanding the Question

You are told the total number of charge carriers in the wire is $N = 2.6 \times 10^{22}$ . Using the wire’s volume (found from $A$ and $L$ ), you must show that the number density is $8.3 \times 10^{28}\ \text{m}^{-3}$ .

Approach

Find the wire’s volume $V = AL$ .
Use $n = N/V$ .

Step-by-Step Reasoning

Using $A = 1.3 \times 10^{-6}\ \text{m}^2$ and $L = 0.24\ \text{m}$ :

V = AL = (1.3 \times 10^{-6})(0.24) = 3.12 \times 10^{-7}\ \text{m}^3

Then

n = \frac{2.6 \times 10^{22}}{3.12 \times 10^{-7}} = \left(\frac{2.6}{3.12}\right)\times 10^{29} \approx 0.833\times 10^{29} = 8.33\times 10^{28}\ \text{m}^{-3}

So $n \approx 8.3 \times 10^{28}\ \text{m}^{-3}$ .

Key Takeaways

Number density is always “total number divided by volume”.
Keep track of standard form carefully when dividing.

Common Mistakes

Using $n = N/(A+L)$ or other incorrect combinations.
Forgetting that volume is in $\text{m}^3$ .

Things to Be Careful About

Use consistent SI units: $A$ in $\text{m}^2$ and $L$ in m.
When dividing by $10^{-7}$ , the power of ten increases by 7.

Techniques used

calculate volume from cross-sectional area and lengthuse number density as number per unit volumehandle powers of ten in division

(iii)

Calculate the average drift speed of the charge carriers (electrons) in the wire.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

I = nAqv \Rightarrow v = \frac{I}{nAq}

v = \frac{0.85}{(8.3 \times 10^{28})(1.3 \times 10^{-6})(1.6 \times 10^{-19})}

v = 4.9 \times 10^{-5}\ \text{m s}^{-1}

Answer

$4.9 \times 10^{-5}\ \text{m s}^{-1}$

Final answer

4.9 × 10⁻⁵ m s⁻¹

Detailed explanation

Background Concept

In a metal wire, the current is due to charge carriers (electrons) drifting through the conductor. The current is related to drift speed by

I = nAqv

where:

$I$ is current (A),
$n$ is number density of charge carriers (m $^{-3}$ ),
$A$ is cross-sectional area (m $^2$ ),
$q$ is the magnitude of charge of one carrier ( $1.6\times 10^{-19}\ \text{C}$ for an electron),
$v$ is average drift speed (m s $^{-1}$ ).

Understanding the Question

You have already found (or been asked to show) that $A = 1.3\times 10^{-6}\ \text{m}^2$ and $n = 8.3\times 10^{28}\ \text{m}^{-3}$ . With $I = 0.85\ \text{A}$ , you must calculate the drift speed $v$ .

Approach

Rearrange $I = nAqv$ to make $v$ the subject, then substitute $I$ , $n$ , $A$ , and $q$ .

Step-by-Step Reasoning

Rearrange:

v = \frac{I}{nAq}

Substitute:

v = \frac{0.85}{(8.3 \times 10^{28})(1.3 \times 10^{-6})(1.6 \times 10^{-19})}

Work through the denominator in parts:

Numerical part: $8.3\times 1.3\times 1.6 = 17.264$ .
Powers of ten: $10^{28} \times 10^{-6} \times 10^{-19} = 10^{3}$ .

So

nAq = 17.264 \times 10^{3} = 1.7264 \times 10^{4}

Hence

v = \frac{0.85}{1.7264 \times 10^{4}} \approx 4.9 \times 10^{-5}\ \text{m s}^{-1}

This small value is expected: drift speeds in metals are typically very slow.

Key Takeaways

Use $I = nAqv$ for drift speed problems.
Drift speed is usually tiny compared with random thermal speeds.

Common Mistakes

Using $q = -1.6\times 10^{-19}\ \text{C}$ and carrying a negative sign into the speed (speed should be positive; direction is handled separately).
Mixing up $n$ (number density) with $N$ (total number).

Things to Be Careful About

Ensure $A$ is in $\text{m}^2$ and $n$ in $\text{m}^{-3}$ so the units give $\text{m s}^{-1}$ .
Power-of-ten errors are common: combine indices systematically.

Techniques used

use the drift current equation I = nAqvrearrange to find drift speedsubstitute and compute with standard form

(c)

A different copper wire Q has the same volume as wire P, but non-uniform radius, as shown in Fig. 6.2.

Fig. 6.2

The radius $r_1$ at end X of wire Q is the same as the radius of wire P. Radius $r_2$ is less than $r_1$ .

6M

(i)

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

4M Similar

DifficultyMedium

Worked solution

Answer

Wire Q has a greater resistance than wire P.

Treat Q as many short lengths in series:

R = \sum \frac{\rho\,\Delta x}{A(x)} \ (\text{or } R = \int \frac{\rho\,dx}{A(x)})

Towards the end where radius is smaller, $A(x)$ is smaller so $\rho\Delta x/A(x)$ is larger. Although Q has the same total volume as P, the thin sections contribute a large resistance and this outweighs the reduced resistance of thicker sections, so overall $R_Q > R_P$ .

Final answer

Resistance of Q is greater than resistance of P.

Detailed explanation

Background Concept

For a uniform wire,

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

If the cross-sectional area is not uniform, you cannot use one single $A$ . Instead, split the wire into many tiny lengths $dx$ . Each tiny piece has resistance

dR = \frac{\rho\,dx}{A(x)}

and the total is the sum (integral) of all pieces:

R = \int \frac{\rho\,dx}{A(x)}

This shows directly that smaller area causes larger resistance per unit length.

Understanding the Question

Wire Q is copper (same $\rho$ ), has the same total volume as wire P, but its radius decreases from $r_1$ at end X to a smaller $r_2$ at the other end. You must compare its total resistance with that of uniform wire P.

Approach

Think of the wire as a series combination of short sections.
Compare how resistance depends on area: $dR \propto 1/A(x)$ .
Use the fact that Q has sections with area smaller than the uniform area of P (at least near the thin end), which increases $R$ significantly.

Step-by-Step Reasoning

diagram-6c-i

Consider a short element of wire at position $x$ of length $dx$ . Its resistance is

dR = \frac{\rho\,dx}{A(x)}

For wire P, $A$ is constant, so every equal length contributes the same resistance per unit length.
For wire Q, $A(x)$ decreases towards the thin end, so for those regions $A(x)$ is smaller and therefore $dR$ is larger.
The wire has the same total volume as P, so if some parts are thinner (smaller $A$ ), other parts must be thicker to keep the total volume the same. However, because resistance depends on $1/A$ , the increase in resistance from a small-area section is larger than the decrease in resistance from an equally long large-area section.
Therefore the total resistance is greater for Q:

R_Q > R_P.

Key Takeaways

A non-uniform conductor must be treated as series elements: $R = \int \rho\,dx/A(x)$ .
Thin regions dominate the resistance because of the $1/A$ dependence.

Common Mistakes

Saying “same volume so same resistance”: resistance depends on how the area is distributed along the length, not just the total volume.
Using $R = \rho L/A$ with a single “average area” without justification.

Things to Be Careful About

The current through a wire in steady state is the same through every cross-section, but resistance depends on geometry and material.
If you mention an integral, you do not need to evaluate it; the key is the dependence on $1/A(x)$ and the qualitative comparison.

Techniques used

model a non-uniform wire as series elementsuse R = rho L / A locally and sum contributionsreason qualitatively using how resistance depends on cross-sectional area

(ii)

On Fig. 6.3, sketch a graph of the variation of the average drift speed of the charge carriers with distance from end X of wire Q.

Fig. 6.3

2M Similar

DifficultyMedium-Easy

Worked solution

diagram-6c-ii

Answer

The drift speed increases with distance from X (smallest at X, largest at the thinner end), with the curve rising more steeply towards the end where the radius is smallest.

Final answer

Drift speed increases from X towards the thinner end (upward-curving).

Detailed explanation

Background Concept

The drift speed $v$ of charge carriers in a wire is related to current by

I = nAqv

For a given wire material, $n$ and $q$ are constant. In steady state, the same current $I$ flows through every cross-section. Therefore

v = \frac{I}{nAq} \propto \frac{1}{A}.

So whenever the cross-sectional area decreases, the drift speed must increase to maintain the same current.

Understanding the Question

Wire Q is tapered: at end X the radius is $r_1$ (same as wire P), and further along the wire the radius decreases to $r_2 < r_1$ . The question asks for a sketch of how average drift speed changes with distance from X.

Approach

Use $v \propto 1/A$ .
Recognise $A(x) = \pi r(x)^2$ decreases as you move away from X.
Therefore $v(x)$ must increase. Because area depends on $r^2$ , a steady decrease in radius generally makes $1/A$ increase more rapidly as the wire gets thinner, giving an upward-curving increase.

Step-by-Step Reasoning

Current is constant along the wire: $I$ is the same through every cross-section.
Since $n$ and $q$ are constant, from $I = nAqv$ :

v(x) = \frac{I}{nqA(x)}.

At $x=0$ (end X), $A$ is largest, so $v$ is smallest.
Moving away from X, $r$ decreases, so $A(x)$ decreases, so $v(x)$ increases.
The sketch should therefore start at some positive drift speed at X and rise as $x$ increases. A reasonable sketch shows the slope increasing as the wire becomes thinner (upward curvature).

diagram-6c-ii

Key Takeaways

With constant current, drift speed is inversely proportional to cross-sectional area.
Tapered (narrowing) wires force carriers to drift faster in the narrower region.

Common Mistakes

Drawing drift speed decreasing (it would only decrease if area increased).
Starting the graph at zero drift speed at X: current is non-zero so drift speed is non-zero.
Drawing a constant drift speed despite changing radius.

Things to Be Careful About

The question asks for a sketch: the key marking points are correct start (non-zero), correct trend (increasing), and sensible curvature (increasing more rapidly as area gets smaller).
Do not confuse drift speed with electron random thermal speed (which is much larger and not what changes here).

Techniques used

use proportionality v \propto 1/A from I = nAqvinterpret how cross-sectional area varies with positionsketch a qualitative graph with correct trend and curvature

Q7MediumParticle PhysicsDynamics

An isolated stationary nucleus Q decays into nucleus R and an $\alpha$ -particle. The $\alpha$ -particle has speed $1.5 \times 10^{7}\ \text{m s}^{-1}$ .

(a)

Complete the equation for this decay.

{}_{88}^{\dots}Q \rightarrow {}_{\dots}^{222}R + {}_{2}^{4}\alpha

1M Similar

DifficultyEasy

Worked solution

Answer

{}_{88}^{226}Q \rightarrow {}_{86}^{222}R + {}_{2}^{4}\alpha

Final answer

{}{88}^{226}Q \rightarrow {}{86}^{222}R + {}_{2}^{4}\alpha

Detailed explanation

Background Concept

In any nuclear decay equation, two key quantities must be conserved:

Nucleon number (mass number) $A$ : total number of protons + neutrons.
Proton number (atomic number) $Z$ : total number of protons (this is also the nuclear charge in units of $e$ ).

In alpha decay, an $\alpha$ -particle is a helium nucleus:

{}_{2}^{4}\alpha

So when a nucleus emits an $\alpha$ -particle, the parent nucleus loses $4$ nucleons and $2$ protons.

Understanding the Question

You are given the decay:

{}_{88}^{\dots}Q \rightarrow {}_{\dots}^{222}R + {}_{2}^{4}\alpha

You must fill in the missing mass number of $Q$ and the missing proton number of $R$ .

Approach

Use conservation on each side of the equation:

Conserve mass number $A$ to find $A$ for $Q$ .
Conserve atomic number $Z$ to find $Z$ for $R$ .

Step-by-Step Reasoning

Mass numbers:

A(Q) = A(R) + A(\alpha) = 222 + 4 = 226

So $Q$ must be ${}_{88}^{226}Q$ .

Atomic numbers:

88 = Z(R) + 2 \Rightarrow Z(R) = 86

So $R$ must be ${}_{86}^{222}R$ .

Hence:

{}_{88}^{226}Q \rightarrow {}_{86}^{222}R + {}_{2}^{4}\alpha

Key Takeaways

Balance nuclear equations by conserving mass number and atomic number.
In alpha decay: $A$ decreases by $4$ and $Z$ decreases by $2$ for the daughter nucleus.

Common Mistakes

Subtracting $4$ from the atomic number (it should be $2$ ).
Mixing up which nucleus is the parent and which is the daughter.
Forgetting that $A$ and $Z$ must balance separately.

Things to Be Careful About

Always check both conservation laws: $A$ and $Z$ .
Write nuclide notation in the correct form ${}_{Z}^{A}X$ .

Techniques used

apply conservation of nucleon number to balance mass numbersapply conservation of proton number to balance atomic numberscomplete nuclide notation for a decay equation

(b)

By considering momentum, calculate the speed of nucleus R after the decay.

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

3M Similar

DifficultyMedium

Worked solution

Working

Initially nucleus $Q$ is stationary, so total momentum is zero.

m_{\alpha}v_{\alpha} = m_R v_R

Using $m \propto A$ :

v_R = \frac{4}{222} \times 1.5 \times 10^{7}

v_R = 2.7 \times 10^{5}\ \text{m s}^{-1}

Answer

$2.7 \times 10^{5}\ \text{m s}^{-1}$

Final answer

2.7 \times 10^5 m s^-1

Detailed explanation

Background Concept

Momentum is defined as

p = mv

It is a vector quantity (it has direction). If a system is isolated (no external forces), then total momentum is conserved.

For a decay where the initial nucleus is at rest, the initial total momentum is zero. After decay into two particles, their momenta must be equal in magnitude and opposite in direction so that the vector sum remains zero.

For nuclear particles, a good approximation at this level is that nuclear mass is proportional to mass number $A$ :

m \approx A u

where $u$ is the atomic mass unit. In ratios (like $m_{\alpha}/m_R$ ), the factor $u$ cancels.

Understanding the Question

A stationary nucleus $Q$ decays into a nucleus $R$ and an $\alpha$ -particle.
The $\alpha$ -particle speed is $1.5 \times 10^{7}\ \text{m s}^{-1}$ .
You must calculate the speed of nucleus $R$ .

From part (a), $R$ has mass number $222$ , and the $\alpha$ has mass number $4$ .

Approach

Use conservation of momentum: initial momentum $=0$ , so final momenta must balance.
Write $m_{\alpha}v_{\alpha} = m_R v_R$ (magnitudes).
Replace masses by mass numbers: $m_{\alpha}/m_R = 4/222$ .
Solve for $v_R$ .

Step-by-Step Reasoning

Initial total momentum:

p_{\text{initial}} = 0

Final total momentum (taking the direction of the $\alpha$ as positive):

m_{\alpha}v_{\alpha} - m_R v_R = 0

So (in magnitudes):

m_{\alpha}v_{\alpha} = m_R v_R

Now use $m \propto A$ :

\frac{m_{\alpha}}{m_R} = \frac{4}{222}

Hence

v_R = \frac{m_{\alpha}}{m_R} v_{\alpha} = \frac{4}{222} \times 1.5 \times 10^{7}

Calculate:

\frac{4}{222} \approx 1.80 \times 10^{-2}

So

v_R \approx 1.80 \times 10^{-2} \times 1.5 \times 10^{7} = 2.7 \times 10^{5}\ \text{m s}^{-1}

(And it moves in the opposite direction to the $\alpha$ -particle, though only the speed was asked.)

Key Takeaways

For an isolated decay from rest into two products, the two products have equal and opposite momentum.
Using mass numbers as proportional to masses makes the calculation quick: $v_R = (4/222)v_{\alpha}$ .

Common Mistakes

Setting $m_{\alpha}v_{\alpha} + m_R v_R = 0$ but then forgetting the sign indicates opposite directions.
Using $226$ (the parent mass number) instead of $222$ for $R$ .
Treating momentum as a scalar and not recognising the recoil is opposite to the emitted $\alpha$ .

Things to Be Careful About

The system must be isolated for momentum conservation (the question states this).
Keep units as $\text{m s}^{-1}$ throughout.
Quote the final speed to an appropriate number of significant figures (here $2$ s.f. is reasonable).

Techniques used

apply conservation of momentum to a two-body separationuse nuclide mass numbers as proportional to nuclear massesrearrange a momentum equation to solve for an unknown speedmaintain consistent significant figures and units

(c)

State three quantities that are conserved during the decay.

1 ________________________________________________

2 ________________________________________________

3 ________________________________________________

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Nucleon number (mass number)
Proton number / charge
Momentum

Final answer

Nucleon number; charge (proton number); momentum

Detailed explanation

Background Concept

In nuclear and particle processes, several quantities are conserved. At AS level, the most commonly examined are:

Nucleon number (mass number $A$ ) in nuclear reactions/decays.
Charge (equivalently proton number $Z$ for nuclei).
Momentum (vector) for an isolated system.
Energy (including rest mass energy) for an isolated system.

Different syllabuses/emphases may list slightly different sets, but for alpha decay, these are the standard conservation statements used to balance equations and describe recoil.

Understanding the Question

You are asked to state three quantities that are conserved during the decay of a stationary nucleus into a daughter nucleus and an $\alpha$ -particle.

The question is not asking for calculations, just correct physics statements.

Approach

Pick three distinct conservation laws that always apply to this decay and that are commonly credited:

nucleon number,
charge (proton number),
momentum.

(Energy is also conserved, but only three are needed.)

Step-by-Step Reasoning

In alpha decay, the nucleus emits ${}_{2}^{4}\alpha$ . To allow the decay equation to balance, the total mass number before equals the total mass number after. So nucleon number is conserved.
The total charge before must equal total charge after. For nuclei, this is equivalent to conserving proton number. So charge/proton number is conserved.
The system is isolated and initially at rest, so total momentum is initially zero and must remain zero: the products recoil with equal and opposite momentum. So momentum is conserved.

Key Takeaways

Use nucleon number and proton number conservation to write correct decay equations.
Use momentum conservation to explain recoil in decays.

Common Mistakes

Stating “mass is conserved” (rest mass is not conserved; mass can convert to kinetic energy).
Giving the same idea twice (e.g. listing “proton number” and “charge” as two separate quantities).
Stating “velocity is conserved” (it is not).

Things to Be Careful About

If you choose “energy”, it should be clearly stated as total energy (including rest energy), not just “kinetic energy”.
Ensure the three quantities you list are genuinely distinct and are conservation laws that apply to nuclear decay.

Techniques used

recall conservation laws relevant to nuclear decaystate conserved quantities using standard physics terminologyselect three distinct conserved quantities

Physics 9702/22 — February/March 2025

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

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

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

1) Beam equilibrium (moments about A)

2) Cylinder equilibrium

3) Link to depth

Key Takeaways