Physics 5054/32 — May/June 2012

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 · Planning Experiments and Investigations

Q15MObservations and MeasurementsExperimental ContextsUse of Techniques, Apparatus and MaterialsAnalysis, Conclusions and EvaluationFree sample

In this experiment, you will determine the mass of a piece of modelling clay.

You have been provided with

a piece of modelling clay,
a metre rule,
a knife-edge,
a 50 g mass.

(a)

Balance the metre rule on the knife-edge. The knife-edge shows the position of the centre of mass of the rule. Record the position of the knife-edge.

position of knife-edge = ______

DifficultyEasy

Worked solution

Answer

50.0 cm

(Note: This is a practical experiment. The candidate records their own reading from the balanced rule. The mark scheme accepts any value in the range 48.0 cm to 52.0 cm, measured to the nearest mm or 0.1 mm. The value 50.0 cm is used here as a representative example.)

Final answer

50.0 cm (candidate's reading, expected in range 48.0 to 52.0 cm)

Detailed explanation

Walkthrough

When a uniform (or near-uniform) object like a metre rule is balanced on a knife-edge, the pivot must be directly below its centre of mass. The candidate balances the rule, reads the position on the rule's scale where the knife-edge sits, and records it. The reading must be taken to the precision the scale allows, which is typically the nearest millimetre (0.1 cm), and must include the unit.

Key Takeaways

The centre of mass of a freely balanced object is directly above the pivot.
Readings must be taken to the precision of the instrument and always include units.

Common Mistakes

Forgetting to write the unit (cm).
Reading to the wrong precision (e.g., only to the nearest cm when the scale has mm divisions).
Not allowing for a zero error if the rule is not perfectly aligned with the scale's zero (though less likely here, it is a general technique).

Things to Be Careful About

The mark scheme accepts any value between 48.0 cm and 52.0 cm. A perfectly balanced uniform rule would read exactly 50.0 cm, but slight variations in the rule's mass distribution or the bench surface are expected.
Ensure the reading is to the nearest 0.1 cm (or 0.1 mm if using a finer scale, though a standard metre rule is to the nearest mm).

Techniques used

find the centre of mass of a metre rule by balancing itread a position on a scale to the appropriate precision

(b)

Place the modelling clay at one end of the metre rule and adjust the position of the 50 g mass at the other end until the rule balances. The knife-edge must remain below the centre of mass of the rule. The arrangement is shown in Fig. 1.1.

Fig. 1.1

(i)

Determine the horizontal distance $x$ from the centre of the modelling clay to the knife-edge and the horizontal distance $y$ from the knife-edge to the centre of the 50 g mass.

$x$ = ______
$y$ = ______

DifficultyMedium-Easy

Worked solution

Answer

$x = 40.0 \text{ cm}$
$y = 32.0 \text{ cm}$

(Note: These are example values that satisfy the constraints $x < 50.0 \text{ cm}$ and $y < x$ . The candidate should record their own measured values from the experiment, read to the nearest mm or 0.1 mm with units.)

Final answer

x = 40.0 cm, y = 32.0 cm (example values; candidate's own readings to nearest mm with units)

Detailed explanation

Walkthrough

The candidate places the clay at one end and the 50 g mass at the other, adjusting the 50 g mass until the rule balances again. The knife-edge remains below the centre of mass of the rule (so the rule's own weight produces no moment). The candidate then measures:

$x$ : the horizontal distance from the centre of the clay to the knife-edge.
$y$ : the horizontal distance from the knife-edge to the centre of the 50 g mass.

These distances must be measured using the metre rule to the nearest millimetre (0.1 cm). The mark scheme requires $x < 50.0 \text{ cm}$ (the clay is on the left half) and $y < x$ (the 50 g mass must be closer to the pivot than the clay, since it is lighter).

Key Takeaways

Moments are calculated using perpendicular distances from the pivot to the line of action of the force.
Measurements must be precise and include units.

Common Mistakes

Forgetting the units (cm).
Measuring to the wrong end of the mass or clay (not the centre).
Recording $y > x$ , which is physically impossible if the 50 g mass is balancing a heavier piece of clay at a greater distance.

Things to Be Careful About

Always measure to the nearest millimetre and include a trailing zero if the scale allows (e.g., 40.0 cm, not just 40 cm).
Ensure the distances are measured horizontally from the pivot to the centre of the objects.

Techniques used

measure distances from a pivot using a metre ruleensure measurements are taken to the nearest millimetre

(ii)

Explain how you ensured that your measurement of $y$ was taken to the centre of the 50 g mass.

DifficultyMedium-Easy

Worked solution

Answer

Take readings from both sides of the 50 g mass and calculate the average.

(Alternatively: Use the slot in the mass to locate the centre, or measure the diameter of the mass and halve it, adding this to the left-hand reading or subtracting from the right-hand reading.)

Final answer

Take readings from both sides of the mass and average them (or use a slot/diameter to locate the centre).

Detailed explanation

Walkthrough

The 50 g mass has a physical width, so the distance $y$ must be measured to its centre, not its edge. To find the centre accurately:

Read the position on the metre rule at the left edge of the mass.
Read the position at the right edge of the mass.
Calculate the average of these two readings to find the position of the centre.

Other acceptable techniques include using a slot in the mass (if provided) as a direct guide to the centre, or measuring the width of the mass, halving it, and adding/subtracting from an edge reading.

Key Takeaways

When measuring distances to objects with width, always measure to their centre of mass.
Averaging readings from both sides reduces the error in locating the centre.

Common Mistakes

Measuring only to one edge of the mass.
Saying "estimate the centre" without specifying a method.

Things to Be Careful About

The mark scheme accepts any valid method to locate the centre: averaging edge readings, using a slot, or using the diameter/width. Ensure the explanation is clear and physically sound.

Techniques used

locate the centre of a mass by averaging readings from both sidesuse physical features of the mass to guide the measurement

(iii)

Calculate the mass $M$ of the modelling clay using the relationship

M = \frac{50y}{x}\text{ gram}.

$M$ = ______

DifficultyMedium-Easy

Worked solution

Working

M = \frac{50y}{x}

M = \frac{50 \times 32.0}{40.0}

M = 40.0 \text{ g}

(Note: Using the example values $x = 40.0 \text{ cm}$ and $y = 32.0 \text{ cm}$ . The candidate must substitute their own measured values for $x$ and $y$ .)

Final answer

40.0 g (example calculation; candidate's own value based on their measurements of x and y)

Detailed explanation

Walkthrough

The principle of moments states that for a balanced object, the sum of clockwise moments equals the sum of anticlockwise moments about the pivot. Since the rule is balanced at its centre of mass, its own weight produces no moment. The moments are:

Anticlockwise moment: $M \times x$ (weight of clay $imes$ distance $x$ )
Clockwise moment: $50 \times y$ (weight of 50 g mass $imes$ distance $y$ )

Equating them:

M \times x = 50 \times y

M = \frac{50y}{x}

Substitute the measured values of $x$ and $y$ (in cm, since the units cancel) into the formula. The result is the mass $M$ in grams.

Using the example values $x = 40.0 \text{ cm}$ and $y = 32.0 \text{ cm}$ :

M = \frac{50 \times 32.0}{40.0} = 40.0 \text{ g}

Key Takeaways

The principle of moments can be used to find an unknown mass when distances and a known mass are measured.
Units of distance cancel out in the ratio $y/x$ , so the mass is directly in grams if the 50 g mass is used.

Common Mistakes

Forgetting the unit (g) in the final answer.
Using incorrect values for $x$ and $y$ (e.g., swapping them).
Not giving the answer to 2 or 3 significant figures.

Things to Be Careful About

The mark scheme accepts values around 40.0 g (specifically $40.0 \pm 3.0 \text{ g}$ ) depending on the candidate's measurements.
Ensure the final answer is given to 2 or 3 significant figures with the correct unit (g).
The formula is already provided, so no derivation is needed; just substitute and calculate.

Techniques used

apply the principle of moments to calculate an unknown masssubstitute measured values into the given formula

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