A student measures the volume of two identical straws by different methods. (a) method 1 (i) The student takes one straw and: - cuts it into 5 pieces that are equal in length - lines up the pieces of straw as shown in Fig. 1.1 - ensures the pieces of straw are touching. Fig. 1.1 is drawn actual size. Measure and record lengths $x$ and $D$ to the nearest 0.1 cm. $x$ = \_\_\_\_\_\_ $\text{cm}$ $D$ = \_\_\_\_\_\_ $\text{cm}$ (ii) Use your value of $x$ in (a)(i) to calculate the…

(a)(i) x = 4.3 cm, D = 2.8 cm · (a)(ii) 21.5 cm · (a)(iii) 0.56 cm · (a)(iv) Use set squares at each end lined up with the ruler to ensure accurate measurement of the width D. · (a)(v) 5.3 cm^3 · (b)(i) 24 cm^3 · (b)(ii) 4.8 cm^3 · (c) Internal volume is smaller than external volume; air bubbles in straw; water remains in straw; water sticks to finger; straw damaged.

A student does an experiment to find the mass of a metre rule. (a) A fixed mass is placed on the metre rule at the 5.0 cm mark as shown in Fig. 3.1. The student: - places the pivot below the 25.0 cm mark on the metre rule - places a mass of mass $m = 20\ \text{g}$ on the metre rule - adjusts the position of the 20 g mass until the rule is as close to balance as possible. (i) The position of the 20 g mass when the rule is as close to balance as possible is shown in Fig. 3.2.…

(a)(i) 57 cm · (a)(ii) Move the 20 g mass slowly backwards and forwards along the rule until balance is achieved · (a)(iii) 1/d values: 0.018, 0.027, 0.036, 0.045, 0.056 (or 2 to 3 s.f.) · (a)(iv) d would be greater than 75 cm / off the end of the rule · (b)(i) Graph plotted with correctly labelled axes from (0,0), accurate points, and a straight line of best fit · (b)(ii) 9.6 x 10^-4 (allow 0.00086 to 0.00106) · (c) 120 g (allow 110 g to 130 g) · (d) 28.0 g (allow 25.0 g to…

Physics 5054/41 — October/November 2024

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

questions

marks

minutes

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

Q110MMediumExperimental ContextsObservations and MeasurementsUse of Techniques, Apparatus and MaterialsAnalysis, Conclusions and Evaluation

A student measures the volume of two identical straws by different methods.

(a)

method 1

(i)

The student takes one straw and:

cuts it into 5 pieces that are equal in length
lines up the pieces of straw as shown in Fig. 1.1
ensures the pieces of straw are touching.

Fig. 1.1

Fig. 1.1 is drawn actual size.

Measure and record lengths $x$ and $D$ to the nearest 0.1 cm.

$x$ = ______ $\text{cm}$
$D$ = ______ $\text{cm}$

1M Similar

(ii)

Use your value of $x$ in (a)(i) to calculate the length $l$ of the straw before it was cut into pieces.

$l$ = ______ $\text{cm}$

1M Similar

(iii)

Use your value for $D$ in (a)(i) to calculate the diameter $d$ of one straw. Give your answer to the nearest 0.01 cm.

$d$ = ______ $\text{cm}$

1M Similar

(iv)

Explain how the student uses a ruler and two set squares to make sure that the measurement of $D$ is as accurate as possible.

You may draw a diagram to help your explanation.

1M Similar

(v)

The volume $V_1$ of the straw is given by the equation:

V_1 = \frac{3.14 d^2 l}{4}

Use your answers from (a)(ii) and (a)(iii) to calculate $V_1$ . Give your answer to two significant figures.

$V_1$ = ______ $\text{cm}^3$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Given:

V_1 = \frac{3.14 d^2 l}{4}

Substitute $d = 0.56\text{ cm}$ and $l = 21.5\text{ cm}$ :

V_1 = \frac{3.14 \times (0.56)^2 \times 21.5}{4}

V_1 = \frac{3.14 \times 0.3136 \times 21.5}{4}

V_1 = \frac{21.171136}{4} = 5.292784\text{ cm}^3

Rounding to 2 significant figures:

V_1 = 5.3\text{ cm}^3

Answer

$V_1 = 5.3\text{ cm}^3$

Final answer

5.3 cm^3

Detailed explanation

Walkthrough

The volume of a cylinder is given by $V = \pi r^2 l = \frac{\pi d^2 l}{4}$ . The formula uses 3.14 for $\pi$ .
Substitute the values calculated in previous parts:
$d = 0.56\text{ cm}$ , $l = 21.5\text{ cm}$ .

V_1 = \frac{3.14 \times (0.56)^2 \times 21.5}{4}

Calculate the numerator: $3.14 \times 0.3136 = 0.984704$ . Then $0.984704 \times 21.5 = 21.171136$ . Divide by 4: $21.171136 / 4 = 5.292784$ .
The question asks for the answer to 2 significant figures. $5.292784$ rounded to 2 s.f. is $5.3$ .

Key Takeaways

Always show the substitution into the formula before calculating the final answer. Pay attention to the required number of significant figures.

Common Mistakes

Using the radius instead of the diameter (or forgetting to square the diameter and divide by 4).
Rounding to the wrong number of significant figures (e.g., 5.29 instead of 5.3).
Forgetting the unit $\text{cm}^3$ .

Things to Be Careful About

The formula uses $d^2$ , not $r^2$ . If you used $r = d/2 = 0.28$ , you would calculate $\pi r^2 l = 3.14 \times 0.28^2 \times 21.5 = 5.29$ , which is the same result. Ensure you use the values from (a)(ii) and (a)(iii) exactly as calculated.

Techniques used

substitute values into the volume formula for a cylinder

(b)

method 2

The student:

takes the second straw
immerses the straw fully in a container of water as shown in Fig. 1.2

Fig. 1.2

moves the straw backwards and forwards in the water several times so that the water enters the straw
puts a finger firmly over one end of the straw, and removes the straw from the water
puts the straw above the open end of a 50 $\text{cm}^3$ measuring cylinder, and then removes the finger so that the water is transferred into the measuring cylinder.

The student repeats the process 4 more times for a total of 5 transfers.

(i)

Fig. 1.3 shows the total volume of water $V_T$ in the measuring cylinder.

Fig. 1.3

Write down the reading $V_T$ .

$V_T$ = ______ $\text{cm}^3$

1M Similar

(ii)

Calculate the average volume of water $V_2$ in one straw.

$V_2$ = ______ $\text{cm}^3$

1M Similar

(c)

Suggest two reasons why $V_1$ and $V_2$ are different.

______
______

2M Similar

DifficultyMedium

Worked solution

Answer

The internal volume of the straw is smaller than the external volume calculated in method 1 (method 1 measures outer dimensions, method 2 measures internal capacity).
Air bubbles may remain in the straw during method 2, reducing the volume of water collected.
(Other acceptable reasons: straw became squashed/damaged during cutting; small amount of water remains in straw after transfer; water sticks to the finger; some water falls out during transfer; straw not completely filled with water.)

Final answer

Internal volume is smaller than external volume; air bubbles in straw; water remains in straw; water sticks to finger; straw damaged.

Detailed explanation

Walkthrough

Method 1 calculates the volume based on external dimensions (diameter and length), assuming the straw is a solid cylinder or measuring the outer volume. Method 2 measures the actual internal capacity by filling it with water and transferring it. The difference arises because:

Internal vs. External Volume: Method 1 uses the outer diameter $d$ , which includes the thickness of the straw walls. The actual volume of water the straw can hold (internal volume) is less than the outer volume calculated.
Incomplete Filling/Transfer Losses: In method 2, it is difficult to ensure the straw is completely filled with water. Air bubbles may get trapped. When transferring, some water may stick to the finger, remain in the straw, or spill out. These losses mean $V_2$ is likely less than the true internal volume, and both $V_1$ and $V_2$ differ due to these practical limitations.

Key Takeaways

When comparing two methods of measurement, consider what each method actually measures (outer vs. inner dimensions) and practical errors (losses, bubbles, damage).

Common Mistakes

Saying "human error" or "be more careful" (too vague).
Only giving one reason when two are required.
Not understanding that method 1 measures outer volume while method 2 measures inner capacity.

Things to Be Careful About

The question asks for two reasons. Any two from the list of valid practical errors or the geometric difference (internal vs external) will score. Be specific about the physics or practical aspect.

Techniques used

identify systematic and random errors in volume measurement methods

Q210MMediumExperimental ContextsObservations and MeasurementsAnalysis, Conclusions and EvaluationUse of Techniques, Apparatus and Materials

A student investigates series and parallel combinations of resistors.

The student is provided with two resistors, X and Y, connected in the circuit shown in Fig. 2.1. The resistors are not identical.

The electromotive force (e.m.f.) of the power supply is 3.0 V.

Fig. 2.1

(a)

(i)

The student closes the switch.

Fig. 2.2 shows the readings $V_X$ on the voltmeter and $I_S$ on the ammeter.

Record the readings of $V_X$ and $I_S$ .

Fig. 2.2

$V_X$ = ______ $\text{V}$
$I_S$ = ______ $\text{A}$

The student opens the switch.

2M Similar

(ii)

The resistance of a resistor can be found using the equation:

\text{resistance} = \frac{\text{voltage across the resistor}}{\text{current in the resistor}}

Calculate $R_X$ , the resistance of resistor X.

$R_X$ = ______ $\Omega$

1M Similar

(iii)

Suggest why the switch is opened after the readings of potential difference and current are recorded.

1M Similar

(iv)

The student disconnects the voltmeter and reconnects it across resistor Y.

The potential difference $V_Y$ across Y is given by the equation:

V_Y = 3.0 - V_X

Using this equation and your value of $V_X$ from (a)(i), calculate $V_Y$ .

Calculate the resistance $R_Y$ of Y.

$V_Y$ = ______ $\text{V}$
$R_Y$ = ______ $\Omega$

1M Similar

(b)

(i)

The student now connects X and Y in parallel.

Complete the circuit diagram in Fig. 2.3 to show the two resistors connected in parallel between points W and Z.

Draw the voltmeter connected to measure the potential difference $V_P$ across both resistors.

Fig. 2.3

2M Similar

(ii)

Theory suggests that, if the two resistors are arranged in parallel, the combined resistance $R_T$ is given by:

R_T = \frac{R_X R_Y}{R_X + R_Y}

Use the equation and your values of $R_X$ and $R_Y$ from (a)(ii) and (a)(iv) to calculate $R_T$ .

$R_T$ = ______ $\Omega$

1M Similar

(iii)

The manufacturer suggests that the combined resistance of resistors X and Y when placed in parallel is 2.5 $\Omega$ .

Two quantities can be considered to be equal within the limits of experimental accuracy if their values are within 10% of each other.

State whether your value of $R_T$ calculated in (b)(ii) has the same value as that suggested by the manufacturer. Support your statement with a calculation.

calculation

statement ______

2M Similar

DifficultyMedium

Worked solution

Working

10% of $2.5 \ \Omega = 0.25 \ \Omega$ .
Acceptable range: $2.5 - 0.25 = 2.25 \ \Omega$ to $2.5 + 0.25 = 2.75 \ \Omega$ .
Calculated $R_T = 2.1 \ \Omega$ .
$2.1 \ \Omega$ is not within the range $2.25 - 2.75 \ \Omega$ .

Answer

calculation: $10% \text{ of } 2.5 = 0.25; \text{ range is } 2.25 \text{ to } 2.75 \ \Omega; \ 2.1 \ \Omega \text{ is outside this range}
statement: No

Final answer

No, 2.1 is not within 10% of 2.5 (range 2.25-2.75)

Detailed explanation

Walkthrough

The manufacturer's value is $2.5 \ \Omega$ . The question states that values are considered equal if they are within 10% of each other.
Calculate 10% of $2.5 \ \Omega$ : $0.10 \times 2.5 = 0.25 \ \Omega$ .
The acceptable range is $2.5 - 0.25 = 2.25 \ \Omega$ to $2.5 + 0.25 = 2.75 \ \Omega$ .
The calculated value from part (b)(ii) is $2.1 \ \Omega$ .
Since $2.1 \ \Omega < 2.25 \ \Omega$ , it falls outside the acceptable range.
Therefore, the calculated value does not have the same value as the manufacturer's suggestion within the limits of experimental accuracy.

Key Takeaways

When comparing experimental values to theoretical or manufacturer values, use the given tolerance percentage to define an acceptable range. If the experimental value falls outside this range, the values are not considered equal within experimental accuracy.

Common Mistakes

Calculating the percentage difference incorrectly (e.g., $(2.5 - 2.1) / 2.1 \times 100$ instead of comparing to the manufacturer's value). Forgetting to state a clear 'Yes' or 'No' conclusion. Not showing the calculation to support the statement.

Things to Be Careful About

The mark scheme requires both the calculation (showing the 10% range or percentage difference) and the statement ('No'). Ensure the calculation uses the manufacturer's value (2.5) as the basis for the 10% tolerance. $2.1$ is $16\%$ less than $2.5$ , which is greater than the $10\%$ tolerance.

Techniques used

calculate percentage difference to compare experimental and theoretical values

Q314MMediumExperimental ContextsObservations and MeasurementsUse of Techniques, Apparatus and MaterialsAnalysis, Conclusions and Evaluation

A student does an experiment to find the mass of a metre rule.

(a)

A fixed mass is placed on the metre rule at the 5.0 cm mark as shown in Fig. 3.1.

Fig. 3.1

The student:

places the pivot below the 25.0 cm mark on the metre rule
places a mass of mass $m = 20\ \text{g}$ on the metre rule
adjusts the position of the 20 g mass until the rule is as close to balance as possible.

(i)

The position of the 20 g mass when the rule is as close to balance as possible is shown in Fig. 3.2.

Fig. 3.2

Using Fig. 3.2, determine the distance $d$ of the centre of the 20 g mass from the pivot when the metre rule is as close to balance as possible.

$d$ = ______ $\text{cm}$

1M Similar

(ii)

Describe the technique the student uses to balance the metre rule with the 20 g mass.

1M Similar

(iii)

The student repeats the procedure for values of mass $m = 30\ \text{g}$ , 40 g, 50 g and 60 g.

Table 3.1 shows the results.

Add your value of $d$ for mass $m = 20\ \text{g}$ in (a)(i) to Table 3.1.

Calculate $1/d$ for each mass $m$ , and record all values in Table 3.1.

Give your answers to an appropriate number of significant figures.

Table 3.1

$m / \text{g}$	$d / \text{cm}$	$\frac{1}{d} / \frac{1}{\text{cm}}$
20
30	37
40	28
50	22
60	18

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For $m = 20\text{ g}$ , $d = 57\text{ cm}$ :
$\frac{1}{d} = \frac{1}{57} = 0.0175\text{ cm}^{-1} \approx 0.018\text{ cm}^{-1}$
For $m = 30\text{ g}$ , $d = 37\text{ cm}$ :
$\frac{1}{d} = \frac{1}{37} = 0.0270\text{ cm}^{-1} \approx 0.027\text{ cm}^{-1}$
For $m = 40\text{ g}$ , $d = 28\text{ cm}$ :
$\frac{1}{d} = \frac{1}{28} = 0.0357\text{ cm}^{-1} \approx 0.036\text{ cm}^{-1}$
For $m = 50\text{ g}$ , $d = 22\text{ cm}$ :
$\frac{1}{d} = \frac{1}{22} = 0.0455\text{ cm}^{-1} \approx 0.045\text{ cm}^{-1}$
For $m = 60\text{ g}$ , $d = 18\text{ cm}$ :
$\frac{1}{d} = \frac{1}{18} = 0.0556\text{ cm}^{-1} \approx 0.056\text{ cm}^{-1}$

Answer

$m / \text{g}$	$d / \text{cm}$	$\frac{1}{d} / \frac{1}{\text{cm}}$
20	57	0.018
30	37	0.027
40	28	0.036
50	22	0.045
60	18	0.056

Final answer

1/d values: 0.018, 0.027, 0.036, 0.045, 0.056 (or 2 to 3 s.f.)

Detailed explanation

Walkthrough

Insert the value $d = 57\text{ cm}$ into the table for $m = 20\text{ g}$ .
For each row, calculate $1/d$ $1/ d$ :
- $1/57 = 0.0175$ (to 2 s.f., $0.018$ or to 3 s.f., $0.0175$ )
- $1/37 = 0.0270$ (to 2 s.f., $0.027$ or to 3 s.f., $0.0270$ )
- $1/28 = 0.0357$ (to 2 s.f., $0.036$ or to 3 s.f., $0.0357$ )
- $1/22 = 0.0455$ (to 2 s.f., $0.045$ or $0.0455$ )
- $1/18 = 0.0556$ (to 2 s.f., $0.056$ or $0.0556$ )
Check that all values are recorded consistently to 2 or 3 significant figures.

Key Takeaways

Always maintain consistency in significant figures (usually matching the 2 or 3 s.f. of the raw data).

Common Mistakes

Rounding to 1 significant figure (e.g., $0.02, 0.03, 0.04, 0.05, 0.06$ ).
Writing fractions instead of decimals.

Things to Be Careful About

Do not mix significant figures arbitrarily; keeping 2 or 3 significant figures across all rows is required by the mark scheme.

Techniques used

calculate reciprocal valuesrecord values to appropriate significant figures

(iv)

Suggest why a value of $d$ cannot be found for mass $m = 10\ \text{g}$ .

1M Similar

(b)

(i)

Using the grid provided in Fig. 3.3 on page 13, plot a graph of $1/d$ on the y-axis against $m$ on the x-axis.

Start your axes from the origin (0, 0).

Draw the straight line of best fit.

Fig. 3.3

4M Similar

DifficultyMedium

Worked solution

Answer

Plot the graph following these requirements:

Axes: Horizontal axis labelled $m / \text{g}$ and vertical axis labelled $\frac{1}{d} / \frac{1}{\text{cm}}$ (or $\text{cm}^{-1}$ ).
Scales: Both axes start from $(0, 0)$ . Linear scales occupying more than half the grid in both directions (e.g. x-axis: $10\text{ g}$ per $2\text{ cm}$ or $20\text{ mm}$ ; y-axis: $0.010\text{ cm}^{-1}$ per $2\text{ cm}$ or $20\text{ mm}$ ).
Plotting: Points plotted accurately to within $\pm \frac{1}{2}$ $\pm \frac{1}{2}$ small square:
- $(20, 0.018)$
- $(30, 0.027)$
- $(40, 0.036)$
- $(50, 0.045)$
- $(60, 0.056)$
Best-fit line: Draw a single, thin, straight line of best fit passing through the plotted points.

Final answer

Graph plotted with correctly labelled axes from (0,0), accurate points, and a straight line of best fit

Detailed explanation

Walkthrough

Axes & Labels: Label the y-axis with ' $\frac{1}{d} / \text{cm}^{-1}$ ' or ' $\frac{1}{d} / \frac{1}{\text{cm}}$ ' and the x-axis with ' $m / \text{g}$ '.
Scale Selection: Both axes must start at $(0, 0)$ $(0, 0)$ . Choose a sensible, non-awkward scale:
- x-axis: $0$ to $70\text{ g}$ , where each large grid division ( $2\text{ cm}$ ) represents $10\text{ g}$ .
- y-axis: $0$ to $0.060\text{ cm}^{-1}$ (or $0.070\text{ cm}^{-1}$ ), where each large division ( $2\text{ cm}$ ) represents $0.010\text{ cm}^{-1}$ .
Plot Points: Plot each of the five points from Table 3.1 carefully using neat crosses (x) or circled dots ( $\odot$ ), ensuring accuracy to within half a small square.
Line of Best Fit: Using a clear ruler, draw a single straight line that best balances the points on either side with an even distribution.

Key Takeaways

Graph scales must be easy to read (multiples of 1, 2, 5, or 10) and must span more than half the available grid.
Best-fit lines should be thin, straight, continuous, and not 'dot-to-dot'.

Common Mistakes

Omitting units on axis labels.
Choosing awkward scales like multiples of 3 or 7.
Forgetting to start axes from $(0, 0)$ as requested.

Things to Be Careful About

Ensure the line is drawn cleanly with a single stroke of a sharp pencil.

Techniques used

choose sensible graph scalesplot experimental data pointsdraw a line of best fit

(ii)

Calculate the gradient $G$ of your line.

Show all your working, and indicate on the graph the values you use.

$G$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

Select two points on the line of best fit separated by more than half the line (e.g. $(0, 0)$ and $(60, 0.0575)$ ):

\Delta y = 0.0575 - 0.0000 = 0.0575\text{ cm}^{-1}

\Delta x = 60 - 0 = 60\text{ g}

G = \frac{\Delta y}{\Delta x} = \frac{0.0575}{60} = 9.58 \times 10^{-4}\text{ g}^{-1}\text{cm}^{-1} \approx 9.6 \times 10^{-4}

Answer

$9.6 \times 10^{-4}$ (or $0.00096$ )

Final answer

9.6 x 10^-4 (allow 0.00086 to 0.00106)

Detailed explanation

Walkthrough

Indicate coordinates on the line: Choose two points on the drawn straight line (not necessarily data points) that are far apart, spanning at least half the length of the line (i.e. $\Delta m \geqslant 25\text{ g}$ ). Indicate these points clearly with a triangle or markings on the graph.
Calculate $\Delta y$ and $\Delta x$ : $\Delta y = y_2 - y_1$ $\Delta x = x_2 - x_1$
Compute gradient $G$ : $G = \frac{\Delta y}{\Delta x}$ For example, $\frac{0.0575 - 0}{60 - 0} = 9.58 \times 10^{-4}$ .
The acceptable range according to the mark scheme is $9.6 (\pm 1.0) \times 10^{-4}$ (i.e. $0.00086$ to $0.00106$ ).

Key Takeaways

Always use a large triangle ( $\\Delta x \ge \text{half the axis span}$ ) to minimize reading uncertainty.
Gradient read-offs must come from the best-fit line, not raw table values.

Common Mistakes

Using data points from the table that do not lie exactly on the line.
Using a triangle that is too small (covering less than half the drawn line).
Misplacing decimal points when working with values like $0.00096$ .

Things to Be Careful About

Keep track of significant figures and decimal places carefully when computing reciprocals and gradients.

Techniques used

determine gradient from a graph using a large triangle

(c)

The mass $M$ of the metre rule can be calculated using the equation:

M = 160 - \frac{0.040}{G}

Use your value of $G$ in (b)(ii) to calculate $M$ .

$M$ = ______ $\text{g}$

1M Similar

(d)

Fig. 3.4

The student is given a piece of modelling clay. He places it on the metre rule as shown in Fig. 3.4. He finds that the metre rule is balanced when the modelling clay is a distance of 40.0 cm from the pivot.

Using your graph in Fig. 3.3 on page 13, find the mass of the piece of modelling clay. Show your working.

mass of piece of modelling clay = ______ $\text{g}$

2M Similar

Q46MMediumPlanning Experiments and InvestigationsExperimental ContextsObservations and Measurements

A student uses ice cubes to investigate the time taken for different masses of ice to melt when the ice cubes are placed in water.

Plan an experiment using ice cubes to investigate how the mass of ice affects the time taken for the ice to melt.

The following apparatus is available:

top pan balance
supply of ice cubes
250 $\text{cm}^3$ beaker
supply of cold water
stopwatch

You may also use other apparatus and materials that are usually available in a school laboratory.

In your plan, you should:

explain briefly how to do the investigation
state the key variables to keep constant
draw a table, with column headings, to show how to display readings (you are not required to enter any readings in the table)
explain how to use these readings to reach a conclusion.

You do not have to include a diagram of the apparatus you use but you may do so if it helps your plan.

DifficultyMedium

Worked solution

Method

Fill the beaker with a fixed volume of cold water.
Measure the mass of the ice cubes using the top pan balance and record it as $m$ .
Add the ice to the water and start the stopwatch at the same moment.
Stop the stopwatch when all the ice has completely melted; record the time $t$ .
Repeat steps 2–4 with different masses of ice (for example, 1, 2, 3, 4 and 5 cubes), measuring the mass each time.

Key variables to keep constant

Initial temperature of the water
Initial volume (or mass) of the water
Room temperature
Use the same beaker, with no added insulation

Results table

Mass of ice $m$ / g	Time for ice to melt $t$ / s

Conclusion

Compare the times for the different masses, or plot a graph of time taken against mass of ice. If the time increases as the mass increases (the graph is not horizontal), then the mass of ice affects the time taken to melt.

Final answer

See working — full plan covering the method, controlled variables, a results table and how the readings lead to a conclusion.

Detailed explanation

Walkthrough

This is a planning question: you are not told what happens, you must decide what to do, what to keep the same, what to record, and how to judge the result. Identify the three types of variable first. The independent variable is the mass of ice (or the number of cubes) — it is what you deliberately change. The dependent variable is the time taken for the ice to melt — it is what you measure. Everything that could affect the melting rate other than the mass is a controlled variable.

The mark scheme awards one mark for each of six features, so your plan must contain all of them:

Method (MP1, MP2, MP3): measure the mass of the ice with the top pan balance, add the ice to the water, time how long it takes to disappear completely, and repeat with a range of different masses. The repeat step is what converts a single reading into a reliable investigation.
Controlled variables (MP4): the mark scheme accepts any one of several, but a strong plan lists several. The initial temperature of the water matters because warmer water melts ice faster. The initial volume (or mass) of water matters because more water holds more thermal energy, so it can melt more ice. The room temperature and the insulation of the beaker also change the rate at which thermal energy reaches the ice.
Table (MP5): two columns — mass (or number of cubes) with its unit, and time with its unit. Putting the unit in the column heading is the 5054 convention; it keeps the body of the table as numbers only.
Conclusion (MP6): either compare the times directly, or — more convincingly — plot time against mass. If larger masses take longer (the graph trends upwards, not horizontally), the mass of ice does affect the melting time. The conclusion must say which way the relationship goes and refer to the data or graph.

Key Takeaways

A plan must name the independent, dependent and controlled variables before describing the procedure.
The method must include measuring the independent variable (mass), timing the dependent variable, and repeating over a range of values.
Controlled variables are the things that would otherwise change the melting rate — water temperature, water volume, room temperature, insulation.
A results table shows column headings with units, not readings in the headings.
The conclusion links the pattern in the data (or the slope of the graph) to the physics: more ice needs more thermal energy, so it takes longer.

Common Mistakes

Missing the controlled variables entirely — MP4 is lost.
Giving a table without units in the headings — MP5 is lost.
Ending with "the mass affects the time" without saying how the readings or graph show it — MP6 is lost.
Not saying that the experiment is repeated for different masses — MP3 is lost.
Using hot or boiling water — the question specifies cold water, and a consistent initial temperature is part of the control.
Forgetting to state how the end point is judged (all ice fully melted).

Things to Be Careful About

The time is the dependent variable — it is read off the stopwatch, not calculated.
The independent variable can be recorded as a mass (using the balance) or as the number of cubes; the mark scheme accepts both, but measuring the mass gives a true mass–time relationship.
Keep the controlled variables genuinely constant — for example, the same volume of water and the same beaker for every run.
Write the table headings as "quantity / unit" (e.g. "Mass of ice $m$ / g"), which is the convention expected on Papers 3 and 4.
Room temperature is a controlled variable, not something to be varied — changing it would change the melting rate between runs and spoil the comparison.

Techniques used

design a controlled experiment varying the mass of icemeasure mass with a top pan balancemeasure a time interval with a stopwatchset out a results table with quantity and unit headingsreach a conclusion by comparing readings or plotting a graph