Mastering Density and Pressure for Your Physics Exam

What the examiner is testing

Examiners assess your ability to select and apply the correct formula for density or pressure to calculate an unknown quantity, often requiring unit conversions or rearrangement. Marks are typically awarded for correctly identifying the formula, substituting values, and providing the final answer with the correct unit.

The method

1. Identify the known quantities: Read the problem carefully and list all given values for mass, volume, density, force, area, or pressure.
2. Check and convert units: Ensure all units are consistent. For density, convert mass to kilograms (kg) and volume to cubic metres (\(\text{m}^3\)). For pressure, convert force to Newtons (N) and area to square metres (\(\text{m}^2\)). Remember \(1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3\) and \(1 \text{ cm}^2 = 1 \times 10^{-4} \text{ m}^2\).
3. Select the correct formula:
   • For density: \( \rho = \frac{m}{V} \)
   • For pressure: \( P = \frac{F}{A} \)
4. Rearrange the formula (if necessary): If you need to find mass, volume, force, or area, rearrange the chosen formula before substituting values.
5. Substitute the values: Plug the numerical values (with their converted units) into the rearranged formula.
6. Calculate the answer: Perform the calculation and write down the numerical result.
7. State the final answer with correct units: Include the appropriate unit for density (\(\text{kg/m}^3\)), pressure (\(\text{Pa}\) or \(\text{N/m}^2\)), mass (\(\text{kg}\)), volume (\(\text{m}^3\)), force (\(\text{N}\)), or area (\(\text{m}^2\)).

Worked example

A block of metal has a mass of 1.5 kg and a volume of \(500 \text{ cm}^3\). Calculate its density.

$$ \text{Mass } (m) = 1.5 \text{ kg} $$
$$ \text{Volume } (V) = 500 \text{ cm}^3 $$
Convert volume to \(\text{m}^3\):
$$ V = 500 \times (10^{-2} \text{ m})^3 = 500 \times 10^{-6} \text{ m}^3 = 0.0005 \text{ m}^3 $$
Formula for density:
$$ \rho = \frac{m}{V} $$
Substitute values:
$$ \rho = \frac{1.5 \text{ kg}}{0.0005 \text{ m}^3} $$
Calculate:
$$ \rho = 3000 \text{ kg/m}^3 $$
Sanity check: This is a typical density for a metal, so the answer is reasonable.

Worked example: a harder one

A rectangular block of wood measures \(20 \text{ cm} \times 10 \text{ cm} \times 5 \text{ cm}\) and has a mass of 750 g. When placed on a surface, what is the maximum pressure it can exert?

Initial thought: Calculate density first.
If we calculate density, we get \(\rho = \frac{0.75 \text{ kg}}{(0.2 \times 0.1 \times 0.05) \text{ m}^3} = \frac{0.75 \text{ kg}}{0.001 \text{ m}^3} = 750 \text{ kg/m}^3\). This is correct, but it doesn't help us find pressure. The question asks for pressure, so we need to use \(P = \frac{F}{A}\).

Correct approach:
1. Identify knowns and what's needed for pressure:
* Mass \( (m) = 750 \text{ g} = 0.75 \text{ kg} \)
* Dimensions: \(20 \text{ cm}, 10 \text{ cm}, 5 \text{ cm}\)
* We need Force \( (F) \) and Area \( (A) \).
2. Calculate the force: The force exerted by the block is its weight.
$$ F = m \times g $$
Using \(g = 10 \text{ N/kg}\):
$$ F = 0.75 \text{ kg} \times 10 \text{ N/kg} = 7.5 \text{ N} $$
3. Determine the area for maximum pressure: Maximum pressure occurs when the force is applied over the smallest possible area. The smallest area will be formed by the two smallest dimensions of the block.
Smallest dimensions are \(10 \text{ cm}\) and \(5 \text{ cm}\).
$$ A = 10 \text{ cm} \times 5 \text{ cm} = 50 \text{ cm}^2 $$
4. Convert area to \(\text{m}^2\):
$$ A = 50 \times (10^{-2} \text{ m})^2 = 50 \times 10^{-4} \text{ m}^2 = 0.005 \text{ m}^2 $$
5. Calculate pressure:
$$ P = \frac{F}{A} $$
$$ P = \frac{7.5 \text{ N}}{0.005 \text{ m}^2} $$
$$ P = 1500 \text{ Pa} $$

Practice

1. A liquid has a density of \(800 \text{ kg/m}^3\). What is the mass of \(2.5 \text{ m}^3\) of this liquid?
2. A force of \(250 \text{ N}\) is applied uniformly over an area of \(0.05 \text{ m}^2\). Calculate the pressure exerted.
3. A solid metal cylinder has a radius of \(3 \text{ cm}\) and a height of \(10 \text{ cm}\). Its mass is \(2.1 \text{ kg}\). Calculate the density of the metal in \(\text{kg/m}^3\).
4. A student stands on one foot. Their mass is \(60 \text{ kg}\). The area of the sole of one shoe is \(150 \text{ cm}^2\). Calculate the pressure exerted on the ground. (Take \(g = 10 \text{ N/kg}\)).

Answers:
1. \(2000 \text{ kg}\)
2. \(5000 \text{ Pa}\)
3. \(7426 \text{ kg/m}^3\) (to 4 significant figures)
4. Working for Q4:
Force \( (F) \) exerted is the student's weight:
$$ F = m \times g = 60 \text{ kg} \times 10 \text{ N/kg} = 600 \text{ N} $$
Area \( (A) \) of one shoe sole:
$$ A = 150 \text{ cm}^2 $$
Convert area to \(\text{m}^2\):
$$ A = 150 \times (10^{-2} \text{ m})^2 = 150 \times 10^{-4} \text{ m}^2 = 0.015 \text{ m}^2 $$
Calculate pressure:
$$ P = \frac{F}{A} = \frac{600 \text{ N}}{0.015 \text{ m}^2} = 40000 \text{ Pa} $$

The three mistakes that lose marks

1. Incorrect unit conversion: Forgetting to convert \(\text{cm}^3\) to \(\text{m}^3\) (multiplying by \(10^{-6}\) not \(10^{-2}\) or \(10^{-3}\)) or \(\text{cm}^2\) to \(\text{m}^2\) (multiplying by \(10^{-4}\) not \(10^{-2}\)).
   • Wrong answer: Calculating density with volume in \(\text{cm}^3\) and mass in \(\text{kg}\) gives a value \(10^6\) times too large, e.g., \(3000 \text{ kg/m}^3\) becomes \(3 \times 10^9 \text{ kg/m}^3\).
2. Using mass instead of weight for force in pressure calculations: Pressure is force per unit area, and the force exerted by an object due to gravity is its weight, not its mass.
   • Wrong answer: Using \(60 \text{ kg}\) directly as the force in \(P = F/A\) instead of \(600 \text{ N}\) would give a pressure of \(4000 \text{ Pa}\) instead of \(40000 \text{ Pa}\).
3. Not identifying the correct area for maximum/minimum pressure: For a block, maximum pressure is on the smallest face, minimum pressure on the largest face.
   • Wrong answer: For the harder worked example, using the largest area (\(20 \text{ cm} \times 10 \text{ cm} = 200 \text{ cm}^2\)) would give a pressure of \(375 \text{ Pa}\) instead of \(1500 \text{ Pa}\).

30-second recap

Density is mass per unit volume, \(\rho = m/V\). Pressure is force per unit area, \(P = F/A\). Always ensure units are consistent (kg, \(\text{m}^3\), N, \(\text{m}^2\)) before calculation. Remember that force due to gravity is weight (\(F = m \times g\)).