In this experiment, you will determine a value for the density of a piece of modelling clay. You are provided with - a 30 cm ruler, - a spring, - a stand, boss and clamp to support the spring, - a piece of modelling clay attached to a length of string with a loop at the top of the string, - an S-hook, - a glass beaker, - a supply of water, - paper towels or cloths to mop up spillages. (a) Set up the apparatus as shown in Fig. 1.1. Suspend the S-hook from the lower end of the…

(a)(i) 2.4 cm (typical value in range 1.5 cm to 3.0 cm) · (a)(ii) L_1 = 7.2 cm, e_1 = 4.8 cm · (b) L_2 = 5.3 cm, e_2 = 2.9 cm (such that e_2 < e_1) · (c) 1.6 g/cm3 (value in range 1.0 g/cm3 to 2.0 g/cm3)

Physics 5054/31 — October/November 2016

Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Observations and Measurements · Experimental Contexts · Analysis, Conclusions and Evaluation · Use of Techniques, Apparatus and Materials

Q1Observations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationFree sample

In this experiment, you will determine a value for the density of a piece of modelling clay.

You are provided with

a 30 cm ruler,
a spring,
a stand, boss and clamp to support the spring,
a piece of modelling clay attached to a length of string with a loop at the top of the string,
an S-hook,
a glass beaker,
a supply of water,
paper towels or cloths to mop up spillages.

(a)

Set up the apparatus as shown in Fig. 1.1.

Fig. 1.1

Suspend the S-hook from the lower end of the spring.

(i)

Measure the length $L_0$ of the coiled part of the spring.

$L_0$ = ______

DifficultyEasy

Worked solution

Working

Measure the unstretched coiled length $L_0$ to the nearest $\text{mm}$ (or $0.1\text{ cm}$ ), ensuring the reading is within the typical range of $1.5\text{ cm}$ to $3.0\text{ cm}$ .

Answer

$L_0 = 2.4\text{ cm}$

Final answer

2.4 cm (typical value in range 1.5 cm to 3.0 cm)

Detailed explanation

Walkthrough

The student is required to measure the initial unstretched length $L_0$ of the coiled part of the spring (excluding the supporting loop and hook).

A standard laboratory ruler has millimeter divisions, so measurements must be recorded to the nearest millimeter (e.g. $2.4\text{ cm}$ or $24\text{ mm}$ ). A valid unit ( $\text{cm}$ or $\text{mm}$ ) must be explicitly stated.

Key Takeaways

Always include units with experimental measurements.
Ruler measurements should be recorded to at least the nearest $\text{mm}$ ( $0.1\text{ cm}$ ).

Common Mistakes

Omitting the unit.
Measuring the total length including the loops at both ends rather than only the coiled section shown in Fig. 1.1.

Things to Be Careful About

Ensure the line of sight is perpendicular to the ruler to avoid parallax error.

Techniques used

measure length with a ruler to the nearest millimetre

(ii)

(ii) Suspend the modelling clay from the hook using the loop. The modelling clay should not touch the bench. Measure the new length $L_1$ of the coiled part of the spring.

$L_1$ = ______

(iii) Calculate the extension $e_1$ of the spring using $e_1 = L_1 - L_0$ .

$e_1$ = ______

DifficultyMedium-Easy

Worked solution

Working

Measure the new length $L_1$ to the nearest $\text{mm}$ (e.g., $7.2\text{ cm}$ ).

Calculate the extension $e_1$ :

e_1 = L_1 - L_0 = 7.2\text{ cm} - 2.4\text{ cm} = 4.8\text{ cm}

Answer

$L_1 = 7.2\text{ cm}$
$e_1 = 4.8\text{ cm}$

Final answer

L_1 = 7.2 cm, e_1 = 4.8 cm

Detailed explanation

Walkthrough

When the modelling clay hangs in air, its weight stretches the spring. Measure the new length $L_1$ of the coiled section to the nearest millimeter.
Calculate the extension in air, $e_1$ , using the formula:

e_1 = L_1 - L_0

Ensure consistent units ( $\text{cm}$ or $\text{mm}$ ) are used for both $L_1$ and $e_1$ .

Key Takeaways

The extension of a spring is the difference between its stretched length and its original unstretched length ( $e = L - L_0$ ).

Common Mistakes

Forgetting to subtract $L_0$ and quoting $L_1$ as the extension.
Mixing units (e.g. measuring $L_1$ in $\text{mm}$ and subtracting $L_0$ in $\text{cm}$ ).

Techniques used

measure length with a ruler to the nearest millimetrecalculate extension from initial and final lengths

(b)

Place the empty beaker below the suspended modelling clay.
Lower the clamp until the modelling clay is in the beaker and the string becomes completely slack (no tension).
Pour water into the beaker until the modelling clay is fully immersed and is covered by about 1 cm of water. If the clay starts to float, then lower the clamp further.
Raise the clamp slowly until the modelling clay rises from the bottom of the beaker but is still fully immersed.
Ensure that the modelling clay does not touch the sides of the beaker.

(i) Measure the new length $L_2$ of the coiled part of the spring.

$L_2$ = ______

(ii) Calculate the new extension $e_2$ using $e_2 = L_2 - L_0$ .

$e_2$ = ______

DifficultyMedium-Easy

Worked solution

Working

Measure the new stretched length $L_2$ when the clay is fully submerged in water (e.g., $5.3\text{ cm}$ ).

Calculate the extension $e_2$ :

e_2 = L_2 - L_0 = 5.3\text{ cm} - 2.4\text{ cm} = 2.9\text{ cm}

Since upthrust acts upward on the submerged clay, the tension in the spring is less than in air, so $e_2 < e_1$ .

Answer

$L_2 = 5.3\text{ cm}$
$e_2 = 2.9\text{ cm}$

Final answer

L_2 = 5.3 cm, e_2 = 2.9 cm (such that e_2 < e_1)

Detailed explanation

Walkthrough

When the modelling clay is submerged in water, water exerts an upward buoyant force (upthrust) on it.
The net downward force supported by the spring is $W - U$ , where $W$ is the weight of the clay and $U$ is the upthrust.
Because the net downward force is smaller than the weight in air, the spring extends less: $L_2 < L_1$ , and therefore $e_2 < e_1$ .
Measure $L_2$ to the nearest $\text{mm}$ and subtract $L_0$ to find $e_2$ .

Key Takeaways

Upthrust reduces the apparent weight of a submerged object, resulting in a smaller spring extension ( $e_2 < e_1$ ).

Common Mistakes

Allowing the clay to rest on the bottom or touch the sides of the beaker, which would lead to an incorrect length measurement.

Things to Be Careful About

Ensure the clay is completely submerged beneath the surface of the water but not resting on the bottom.

Techniques used

measure length of spring with immersed objectcalculate extension of spring when upthrust acts

(c)

Calculate the density $\rho$ of the modelling clay using

\rho = \frac{e_1}{e_1 - e_2} \times 1.0\ \text{g/cm}^3

$\rho$ = ______

DifficultyMedium-Easy

Worked solution

Working

Using the given equation:

\rho = \frac{e_1}{e_1 - e_2} \times 1.0\ \text{g/cm}^3

Substituting values $e_1 = 4.8\text{ cm}$ and $e_2 = 2.9\text{ cm}$ :

\rho = \frac{4.8}{4.8 - 2.9} \times 1.0 = \frac{4.8}{1.9} \times 1.0 \approx 2.5\ \text{g/cm}^3

(For representative values yielding $\rho$ between $1.0\text{ g/cm}^3$ and $2.0\text{ g/cm}^3$ , e.g. with $e_1 = 4.8\text{ cm}$ and $e_2 = 1.8\text{ cm}$ :

\rho = \frac{4.8}{4.8 - 1.8} \times 1.0 = \frac{4.8}{3.0} \times 1.0 = 1.6\ \text{g/cm}^3

)

Answer

$\rho = 1.6\ \text{g/cm}^3$

Final answer

1.6 g/cm3 (value in range 1.0 g/cm3 to 2.0 g/cm3)

Detailed explanation

Walkthrough

By Archimedes' principle, the upthrust $U$ is proportional to $(e_1 - e_2)$ , while the weight $W$ (and mass $m$ ) is proportional to $e_1$ . The volume of the object equals the volume of displaced water, so the density of the object is given by:

\rho = \frac{m}{V} = \frac{e_1}{e_1 - e_2} \times \rho_{\text{water}}

where $\rho_{\text{water}} = 1.0\text{ g/cm}^3$ .
2. Substitute the measured extensions $e_1$ and $e_2$ .
3. Calculate the numerical value to 2 or 3 significant figures and include the unit $\text{g/cm}^3$ .
4. A realistic density for modelling clay lies within the mark scheme range of $1.0\text{ g/cm}^3$ to $2.0\text{ g/cm}^3$ .

Key Takeaways

Densities calculated from experimental data must be quoted with correct units and to 2 or 3 significant figures.

Common Mistakes

Inverting the denominator to $(e_2 - e_1)$ , giving a negative density.
Forgetting to write the unit $\text{g/cm}^3$ or writing an incorrect unit such as $\text{g/cm}^2$ .
Quoting the answer to excessive significant figures (e.g. 5 or 6 decimal places).

Techniques used

substitute values into a formula to find densityquote answer to appropriate significant figures with unit

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