Answer all the questions in the spaces provided. (a) Define density. (b) Explain how the difference in the densities of solids, liquids and gases may be related to the spacing of their molecules. (c) A paving slab has a mass of $68\ \text{kg}$ and dimensions $50\ \text{mm} \times 600\ \text{mm} \times 900\ \text{mm}$. (i) Calculate the density, in $\text{kg m}^{-3}$, of the material from which the paving slab is made. density = \_\_\_\_\_\_ $\text{kg m}^{-3}$ (ii) Calculate…

Physics 9702/21 — October/November 2011

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

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Topics Forces, Density and Pressure · Dynamics · Physical Quantities and Units · Work, Energy and Power · Kinematics · Electric Fields · +4 more

Q1Forces, Density and PressureDynamicsFree sample

Answer all the questions in the spaces provided.

(a)

Define density.

DifficultyEasy

Worked solution

Answer

Density is mass per unit volume.

\rho = \frac{m}{V}

Final answer

Density is mass per unit volume (\rho = m/V).

Detailed explanation

Background Concept

Density, symbol $\rho$ , describes how much mass is contained in a given volume.

It is defined by

\rho = \frac{m}{V}

where $m$ is mass (in $\text{kg}$ ) and $V$ is volume (in $\text{m}^3$ ). The SI unit of density is $\text{kg m}^{-3}$ .

Understanding the Question

You are asked to define density, so you should give a clear statement (and optionally the equation) linking density to mass and volume.

Approach

State “mass per unit volume” and, to be precise, write the defining equation $\rho = m/V$ .

Step-by-Step Reasoning

Density compares mass to volume.
“Per unit volume” means “divide mass by volume”, giving $\rho = m/V$ .

Key Takeaways

Density is defined, not derived: $\rho = m/V$ .
SI unit: $\text{kg m}^{-3}$ .

Common Mistakes

Writing $\rho = V/m$ (inverting the fraction).
Giving units incorrectly (e.g. $\text{kg m}^{3}$ instead of $\text{kg m}^{-3}$ ).

Things to Be Careful About

Use the correct symbol $\rho$ .
Ensure the definition involves mass and volume, not weight.

Techniques used

recall the definition of densityexpress a definition using an equation and symbols

(b)

Explain how the difference in the densities of solids, liquids and gases may be related to the spacing of their molecules.

DifficultyMedium-Easy

Worked solution

Answer

For the same amount of substance, mass is (approximately) fixed but the volume depends on particle spacing.

In solids, molecules are very close together so volume is small, so density is high.
In liquids, molecules are slightly further apart so volume is larger, so density is lower than a solid.
In gases, molecules are very far apart so volume is much larger, so density is very low.

Final answer

Solids have closely packed molecules (small volume) so high density; liquids are slightly more spaced so lower density; gases have large spacing (large volume) so very low density.

Detailed explanation

Background Concept

Density is

\rho = \frac{m}{V}

For a given sample containing a certain number of molecules, the mass $m$ depends mainly on how many molecules there are and their molecular mass. The main difference between solids, liquids and gases is the volume $V$ , which is strongly affected by the average spacing between molecules.

If particles are farther apart, the same number of particles occupies a larger volume, so the density decreases.

Understanding the Question

You must explain why solids, liquids and gases have different densities by referring to how closely their molecules are spaced.

The key idea is: spacing affects volume, and density depends inversely on volume.

Approach

Use $\rho = m/V$ and argue that for comparable samples the mass is similar, but:

solids: smallest spacing → smallest volume → largest density
liquids: intermediate spacing → intermediate volume → intermediate density
gases: largest spacing → largest volume → smallest density

Step-by-Step Reasoning

Start from the definition:

\rho = \frac{m}{V}

Consider a sample containing a fixed number of molecules. Its mass $m$ is fixed.
In a solid, molecules are packed closely in a regular arrangement, so the sample occupies a small volume $V$ . Therefore $\rho$ is large.
In a liquid, molecules are still close but not fixed in a lattice; average spacing is a little larger than in a solid, so $V$ is a bit larger and $\rho$ is a bit smaller.
In a gas, molecules are much farther apart. The same number of molecules occupies a very large volume, so $V$ is very large and $\rho$ is very small.

Key Takeaways

Differences in density between states are mainly due to differences in volume, not mass.
Greater molecular spacing → greater volume → lower density.

Common Mistakes

Saying gases are less dense because their molecules are “lighter” (molecular mass doesn’t change with state).
Not explicitly linking spacing to volume and then to density.

Things to Be Careful About

Use comparative language: “closer together” vs “further apart”.
Make the chain clear: spacing → volume → density.
Avoid absolute claims like “liquids always less dense than solids” (water/ice is an exception), but the general trend with spacing is what is being tested.

Techniques used

link macroscopic density to microscopic particle spacingcompare particle separation in solids, liquids and gasesexplain how volume changes with spacing for a fixed amount of substance

(c)

A paving slab has a mass of $68\ \text{kg}$ and dimensions $50\ \text{mm} \times 600\ \text{mm} \times 900\ \text{mm}$ .

(i)

Calculate the density, in $\text{kg m}^{-3}$ , of the material from which the paving slab is made.

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

DifficultyMedium-Easy

Worked solution

Working

Dimensions in metres: $0.050\ \text{m} \times 0.600\ \text{m} \times 0.900\ \text{m}$ .

V = 0.050 \times 0.600 \times 0.900 = 0.0270\ \text{m}^3

\rho = \frac{m}{V} = \frac{68}{0.0270} = 2.52 \times 10^3\ \text{kg m}^{-3}

Answer

$2.52 \times 10^3\ \text{kg m}^{-3}$

Final answer

2.52 × 10^3 kg m^-3

Detailed explanation

Background Concept

For a uniform material,

\rho = \frac{m}{V}

where $\rho$ is density, $m$ is mass, and $V$ is volume.

For a rectangular block (cuboid),

V = lwh

You must use SI units: metres for length so that volume is in $\text{m}^3$ and density ends up in $\text{kg m}^{-3}$ .

Understanding the Question

A paving slab has mass $68\ \text{kg}$ and dimensions $50\ \text{mm} \times 600\ \text{mm} \times 900\ \text{mm}$ . You need its density in $\text{kg m}^{-3}$ .

So you must:

Convert each dimension from $\text{mm}$ to $\text{m}$ .
Compute its volume.
Divide mass by volume.

Approach

Convert: $1\ \text{mm} = 1\times 10^{-3}\ \text{m}$ .
Find $V = lwh$ .
Use $\rho = m/V$ .

Step-by-Step Reasoning

Convert dimensions:

50\ \text{mm} = 0.050\ \text{m},\quad 600\ \text{mm} = 0.600\ \text{m},\quad 900\ \text{mm} = 0.900\ \text{m}

Calculate volume of the cuboid:

V = 0.050 \times 0.600 \times 0.900 = 0.0270\ \text{m}^3

Use the density equation:

\rho = \frac{m}{V} = \frac{68\ \text{kg}}{0.0270\ \text{m}^3} = 2.52 \times 10^3\ \text{kg m}^{-3}

The value $\approx 2.5\times 10^3\ \text{kg m}^{-3}$ is typical of stone/concrete, which is a good reasonableness check.

Key Takeaways

Always convert lengths to metres before finding volume.
Density is mass divided by volume.

Common Mistakes

Forgetting to convert $\text{mm}$ to $\text{m}$ (this changes the volume by a factor of $10^9$ ).
Using $50, 600, 900$ directly as if they were in metres.
Writing units as $\text{kg m}^{-2}$ or $\text{kg m}^{3}$ .

Things to Be Careful About

Volume must be in $\text{m}^3$ for density in $\text{kg m}^{-3}$ .
Significant figures: $68$ (2 s.f.) and dimensions (often 2–3 s.f.), so $2$ – $3$ s.f. is appropriate.

Techniques used

convert dimensions into SI unitscalculate volume from cuboid dimensionsapply \rho = m/Vquote an answer with correct unit and significant figures

(ii)

Calculate the maximum pressure a slab could exert on the ground when resting on one of its surfaces.

pressure = ______ $\text{Pa}$

DifficultyMedium

Worked solution

Working

Maximum pressure occurs for minimum contact area.

Smallest face: $50\ \text{mm} \times 600\ \text{mm} = 0.050\ \text{m} \times 0.600\ \text{m}$ .

A = 0.050 \times 0.600 = 0.0300\ \text{m}^2

Weight:

W = mg = 68 \times 9.81 = 6.67 \times 10^2\ \text{N}

Pressure:

p = \frac{W}{A} = \frac{6.67 \times 10^2}{0.0300} = 2.22 \times 10^4\ \text{Pa}

Answer

$2.22 \times 10^4\ \text{Pa}$

Final answer

2.22 × 10^4 Pa

Detailed explanation

Background Concept

Pressure is defined as normal force per unit area:

p = \frac{F}{A}

For an object resting on the ground, the force on the ground is its weight (assuming it is at rest and the ground provides an equal and opposite normal reaction):

W = mg

To get the maximum pressure for a fixed weight, you need the smallest contact area because pressure is inversely proportional to area.

Understanding the Question

You are asked for the maximum pressure the slab could exert when resting on one of its surfaces.

Given the slab dimensions, it can rest on three different faces with different areas. Since the weight is constant, the maximum pressure happens when it stands on the face with the smallest area.

Approach

List the three possible face areas and pick the smallest.
Convert the chosen dimensions to metres and calculate the area in $\text{m}^2$ .
Calculate the weight $W = mg$ .
Use $p = W/A$ .

Step-by-Step Reasoning

Possible contact face areas (in $\text{mm}^2$ ):

$50 \times 600$
$50 \times 900$
$600 \times 900$

The smallest is $50 \times 600$ .

Convert to metres and find area:

A = 0.050\ \text{m} \times 0.600\ \text{m} = 0.0300\ \text{m}^2

Weight of the slab:

W = mg = 68\ \text{kg} \times 9.81\ \text{m s}^{-2} = 6.67 \times 10^2\ \text{N}

Pressure on the ground:

p = \frac{W}{A} = \frac{6.67 \times 10^2}{0.0300} = 2.22 \times 10^4\ \text{Pa}

Key Takeaways

Maximum pressure occurs with minimum contact area.
Use $W = mg$ for the force and $p = F/A$ for pressure.
Convert areas into $\text{m}^2$ to get pressure in Pa.

Common Mistakes

Using the largest face area (this would give minimum pressure, not maximum).
Using mass $m$ directly in $p = F/A$ instead of weight $W$ .
Forgetting to convert $\text{mm}^2$ to $\text{m}^2$ .

Things to Be Careful About

Pressure uses the force perpendicular to the surface; here it is the weight.
Ensure the area is in $\text{m}^2$ (not $\text{mm}^2$ ).
Use a sensible value for $g$ (typically $9.81$ or $9.8\ \text{m s}^{-2}$ ) and keep consistent significant figures.

Techniques used

identify the orientation that gives minimum contact areacalculate weight using W = mgapply p = F/Aconvert area into SI units

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