Physics 5054/42 — May/June 2016

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

questions

marks

minutes

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

Q1Use of Techniques, Apparatus and MaterialsObservations and MeasurementsAnalysis, Conclusions and EvaluationFree sample

A student investigates the oscillations of a metre rule.

(a)

The student starts by supporting one end of a heavy rod using a stand, boss and clamp.
Fig. 1.1 shows how he assembled the apparatus.

Fig. 1.1

(i)

Explain why this is not the best way to assemble one stand, boss and clamp to support a heavy rod.

DifficultyMedium-Easy

Worked solution

Answer

The clamp is at the end of the rod, so the heavy rod extending to the right creates a large turning effect (moment) about the boss. This makes the stand likely to topple, tilt or become unbalanced to the right.

Final answer

The heavy rod creates a large moment about the boss, making the stand likely to topple or tilt to the right.

Detailed explanation

Walkthrough

The question asks why the assembly in Fig. 1.1 is not the best way to support a heavy rod. In Fig. 1.1, the clamp is attached near the top of the boss, and the heavy rod extends horizontally to the right. This arrangement acts like a lever: the weight of the heavy rod acts at a large distance from the pivot point (the boss), creating a large clockwise moment. This large moment will cause the stand to tilt, bend, or topple to the right. A safer assembly would have the rod extending in the opposite direction, or the clamp positioned closer to the stand to reduce the moment arm.

Key Takeaways

When setting up apparatus with stands and clamps, consider the centre of mass and the turning effects (moments) of any heavy loads. Loads extending away from the support create instability.

Common Mistakes

Saying "it is heavy" without explaining the mechanism (the large moment/turning effect).
Saying "it will break" instead of "topple" or "tilt".

Things to Be Careful About

Use precise language: "topple", "tilt", "become unbalanced", or mention the "turning effect" or "moment". Avoid vague terms like "it will fall" without explanation.

Techniques used

assess the stability of a stand and clamp assemblyrecognise the turning effect of a heavy load on a clamp

(ii)

In the space to the right of Fig. 1.1, sketch a better way of assembling this apparatus.

DifficultyEasy

Worked solution

Answer

Final answer

Sketch showing the rod turned around so it extends to the left, over the base of the stand.

Detailed explanation

Walkthrough

To make the assembly stable, the heavy rod should extend over the base of the stand rather than away from it. This can be achieved by turning the rod around the other way, so the clamp is still at the top but the rod now extends to the left, over the heavy base. This places the centre of mass of the rod over the base, preventing it from toppling.

Key Takeaways

A stable setup places the load over the base of support. Turning the rod around achieves this.

Common Mistakes

Drawing the rod extending to the right still.
Changing the position of the boss to the bottom of the stand (the mark scheme specifically accepts turning the rod around).

Things to Be Careful About

Ensure the sketch is clear and shows the rod extending over the base. The mark scheme specifically looks for "rod turned around the other way".

Techniques used

sketch a stable stand and clamp assemblyreorient the rod to reduce the moment arm

(b)

Two clamps are used to support the rod horizontally, about $60\text{ cm}$ above the bench, as shown in Fig. 1.2.

Fig. 1.2

The student is given a metre rule and two set-squares to check that the rod is horizontal.

(i)

Draw on Fig. 1.2 to show how the apparatus is used.

DifficultyMedium-Easy

Worked solution

Answer

Final answer

Draw a vertical rule (or set-square edge) standing on the bench at two different points along the rod, with the top of the rule touching the underside of the rod.

Detailed explanation

Walkthrough

To check if the rod is horizontal, the student can measure the height of the rod above the bench at two different points. This is done by placing a rule or the edge of a set-square vertically on the bench, with its top edge touching the underside of the rod. This is repeated at two different positions along the rod. If the rod is horizontal, the height measured at both points will be the same.

Key Takeaways

Set-squares and rules can be used as vertical references to check horizontal alignment by measuring the gap or height at multiple points.

Common Mistakes

Drawing the set-squares on top of the rod instead of underneath.
Not showing the set-squares or rule standing vertically on the bench.

Things to Be Careful About

The mark scheme requires the rule/set-square to be vertical and correctly used (touching the bench and the rod). Ensure the drawing clearly shows this.

Techniques used

use set-squares to check horizontal alignmentmeasure height above bench at two points

(ii)

Explain how he can tell that the rod is horizontal.

DifficultyEasy

Worked solution

Answer

The height (or distance) above the bench is the same at both places where the set-squares (or rule) are placed.

Final answer

The height above the bench is the same at two places.

Detailed explanation

Walkthrough

If the rod is horizontal, its distance from the horizontal bench must be constant. Therefore, when the student measures the height from the bench to the underside of the rod at two different points using the set-squares, the readings will be identical. If the readings differ, the rod is not horizontal.

Key Takeaways

A horizontal line is at a constant height above a horizontal reference surface.

Common Mistakes

Saying "the rule is parallel to the bench" without referencing the measurement.
Not mentioning that the heights must be the same at two places.

Things to Be Careful About

Use the word "height" or "distance" above the bench. Ensure it is clear that two measurements are being compared.

Techniques used

compare heights at two points to verify horizontal alignment

(c)

The student suspends the metre rule from the rod using two pieces of thread of equal length.
A half-metre rule is placed on the bench, under the metre rule, so that the end A of the metre rule is above the $0\text{ cm}$ end of the half-metre rule, as shown in Fig. 1.3.

Fig. 1.3

(i)

Describe how the student can check that the end A of the metre rule is exactly above the $0\text{ cm}$ end of the half-metre rule.

DifficultyMedium-Easy

Worked solution

Answer

Use a vertical ruler, set-square, or plumb line hanging from end A of the metre rule down to the half-metre rule to check that end A is directly above the 0 cm mark.

Final answer

Use a vertical ruler, set-square, or plumb line to check alignment.

Detailed explanation

Walkthrough

To ensure end A of the metre rule is exactly above the 0 cm end of the half-metre rule, a vertical reference line is needed. The student can hang a plumb line from end A, or place a set-square or ruler vertically from end A down to the half-metre rule. If end A is directly above the 0 cm mark, the vertical tool will align with the 0 cm mark on the half-metre rule.

Key Takeaways

Vertical alignment can be checked using a plumb line, set-square, or ruler placed vertically from the upper point to the lower scale.

Common Mistakes

Suggesting to look from the side (parallax error).
Not specifying a vertical tool.

Things to Be Careful About

The mark scheme accepts "vertical ruler", "set square", or "plumb line". Ensure the method establishes a vertical line.

Techniques used

use a plumb line or vertical ruler to align pointscheck vertical alignment between two scales

(ii)

The metre rule is moved to the left until the end A is above the $10\text{ cm}$ mark on the half-metre rule. It is then released. As the metre rule swings, the amplitude of the swing decreases.

The student counts the number $N$ of swings until end A no longer passes the $5\text{ cm}$ mark on the half-metre rule. He repeats this several times and his results are shown below.

53 \quad 55 \quad 52 \quad 51 \quad 53

Calculate $N_{\text{av}}$ , the average value of $N$ .
Give your answer to 2 significant figures.

$N_{\text{av}}$ = ______

DifficultyEasy

Worked solution

Working

N_{\text{av}} = \frac{53 + 55 + 52 + 51 + 53}{5} = \frac{264}{5} = 52.8

Rounding to 2 significant figures:

N_{\text{av}} = 53

Answer

Final answer

Detailed explanation

Walkthrough

The student counts the number of swings $N$ until the amplitude decreases enough that end A no longer passes the 5 cm mark. Five repeated measurements are given: 53, 55, 52, 51, 53. To find the average $N_{\text{av}}$ , sum these values and divide by the number of measurements (5).
Sum = 53 + 55 + 52 + 51 + 53 = 264.
Average = 264 / 5 = 52.8.
The question asks for the answer to 2 significant figures. 52.8 rounded to 2 significant figures is 53.

Key Takeaways

Always calculate the average of repeated measurements to reduce random errors. Pay attention to the required number of significant figures.

Common Mistakes

Forgetting to divide by the number of measurements.
Rounding 52.8 to 52 or 530.
Not rounding to 2 significant figures as requested.

Things to Be Careful About

The mark scheme gives "53 cao" (correct answer only), so 52.8 would not score if 2 sig figs are required. Ensure the final answer is exactly 53.

Techniques used

calculate the average of repeated measurementsround to the correct number of significant figures

(d)

The student is given one square piece of card of side $l$ . He attaches the centre of the card to the end A of the metre rule with a small piece of Blu-tack, as shown in Fig. 1.4.

Fig. 1.4

The value of $l$ is $19.0\text{ cm}$ .
The student repeats the experiment in (c) to obtain a value for $N_{\text{av}}$ .
He then cuts the card to make the square smaller and repeats the experiment with smaller values of $l$ . Fig. 1.5 shows the student's results.

Fig. 1.5

$l\text{ / cm}$	$N_{\text{av}}$
19.0	3
15.0	6
11.0	12
7.0	22
3.0	35
0

(i)

Complete Fig. 1.5 by recording your value for $N_{\text{av}}$ from (c)(ii) for $l = 0$ .

DifficultyEasy

Worked solution

Answer

$l\text{ / cm}$	$N_{\text{av}}$
19.0	3
15.0	6
11.0	12
7.0	22
3.0	35
0	53

Final answer

Detailed explanation

Walkthrough

The value of $N_{\text{av}}$ calculated in part (c)(ii) for $l = 0$ (no card attached) is 53. This value is recorded in the last row of the table in Fig. 1.5.

Key Takeaways

Results from calculations should be correctly transcribed into results tables.

Common Mistakes

Writing 52.8 instead of 53.
Leaving the cell blank.

Things to Be Careful About

The table already has the unit in the column header ( $N_{\text{av}}$ ), so just write the number 53.

Techniques used

record calculated average value in a results table

(ii)

On Fig. 1.6, plot the graph of $l\text{ / cm}$ on the $y$ -axis against $N_{\text{av}}$ on the $x$ -axis.
Start your axes from (0,0). Draw the smooth curve of best fit.

Fig. 1.6

DifficultyMedium

Worked solution

Answer

Final answer

Graph plotted with $N_{\text{av}}$ on x-axis (0 to 40), $l$ on y-axis (0 to 20), points plotted accurately, smooth curve of best fit drawn.

Detailed explanation

Walkthrough

The graph plots $l$ (y-axis) against $N_{\text{av}}$ (x-axis). The data points are:
(3, 19.0), (6, 15.0), (12, 11.0), (22, 7.0), (35, 3.0), (53, 0).

Axes: x-axis is $N_{\text{av}}$ (no unit, or unitless count), y-axis is $l$ / cm. Start both axes at (0,0).
Scales: x-axis needs to go up to at least 53 (e.g., 0 to 60, with 10 small squares = 10 units). y-axis needs to go up to at least 19 (e.g., 0 to 20, with 10 small squares = 2 cm).
Plotting: Plot all 6 points accurately. Points should be within half a small square.
Curve: Draw a smooth curve of best fit. The curve should start at (53, 0) on the x-axis and curve upwards to the left, passing near or through all other points. It is not a straight line.

Key Takeaways

When plotting graphs, ensure axes are correctly labelled with quantity and unit, scales are linear and use at least half the grid, points are plotted accurately, and a smooth curve of best fit is drawn (not a line of best fit for non-linear data).

Common Mistakes

Swapping the x and y axes ( $l$ on x-axis, $N_{\text{av}}$ on y-axis).
Using awkward scales (e.g., 0 to 50 with 10 small squares = 5 units).
Drawing a straight line through the points instead of a smooth curve.
Plotting points inaccurately.

Things to Be Careful About

The mark scheme awards 1 mark for axes labelled correctly (quantity on both, unit on y-axis only, correct way round).
1 mark for scales linear, not awkward, starting from (0,0).
1 mark for points plotted accurately.
1 mark for smooth best fit curve.
Ensure the curve is smooth and does not necessarily pass through every point, but represents the trend.

Techniques used

plot a graph with correct axis labels and scalesdraw a smooth curve of best fit through plotted points

(iii)

The graph shows that as $l$ decreases, $N_{\text{av}}$ increases.

Two quantities $x$ and $y$ are inversely proportional if they obey the equation

x = \frac{k}{y}

where $k$ is a constant.

By taking two pairs of values from the graph, show that $N_{\text{av}}$ is not inversely proportional to $l$ .

DifficultyMedium

Worked solution

Working

If $N_{\text{av}}$ is inversely proportional to $l$ , then $l \times N_{\text{av}} = k$ (a constant) for all pairs of values.

Using values from the table/graph:
For $l = 19.0$ , $N_{\text{av}} = 3$ : $19.0 \times 3 = 57$
For $l = 3.0$ , $N_{\text{av}} = 35$ : $3.0 \times 35 = 105$

Since $57 \neq 105$ , the product $l \times N_{\text{av}}$ is not constant.

Answer

$l \times N_{\text{av}}$ is not constant (e.g., $19.0 \times 3 = 57$ and $3.0 \times 35 = 105$ ), so $N_{\text{av}}$ is not inversely proportional to $l$ .

Final answer

The product $l \times N_{\text{av}}$ is not constant (e.g., 57 and 105), so they are not inversely proportional.

Detailed explanation

Walkthrough

Two quantities $x$ and $y$ are inversely proportional if $x = k/y$ , which means $x \times y = k$ (a constant). To show that $N_{\text{av}}$ is not inversely proportional to $l$ , we calculate the product $l \times N_{\text{av}}$ for at least two pairs of values and show that they are not equal.

Using values from the table:

Pair 1: $l = 19.0$ cm, $N_{\text{av}} = 3$ . Product = $19.0 \times 3 = 57$ .
Pair 2: $l = 3.0$ cm, $N_{\text{av}} = 35$ . Product = $3.0 \times 35 = 105$ .

Since $57 \neq 105$ , the product is not constant, and therefore $N_{\text{av}}$ is not inversely proportional to $l$ .

Key Takeaways

To test for inverse proportionality, check if the product of the two variables is constant across different data points.

Common Mistakes

Calculating the ratio $l / N_{\text{av}}$ instead of the product.
Using only one pair of values (need at least two to compare).
Making arithmetic errors in the multiplication.

Things to Be Careful About

The mark scheme awards 1 mark for seeing $l \times N_{\text{av}}$ for one pair, and 1 mark for two correct values calculated and not equal. Ensure you show the calculation for at least two pairs.

Techniques used

test for inverse proportionality by checking if the product of two variables is constantcalculate $l \times N_{\text{av}}$ for multiple data pairs

(iv)

Suggest why the student starts with $l = 19\text{ cm}$ and then reduces $l$ , rather than starts with $l = 3\text{ cm}$ and then increases $l$ .

DifficultyMedium-Easy

Worked solution

Answer

Starting with $l = 19$ cm and reducing $l$ means less card is used overall. The student can use just one piece of card for the entire experiment, cutting it smaller each time. If they started with $l = 3$ cm and increased $l$ , they would need multiple larger pieces of card, which may not be available or would require cutting from a larger sheet, wasting material.

Answer

Less card is used; the student can use just one piece of card and cut it smaller each time.

Final answer

Less card is used; can use just one piece of card.

Detailed explanation

Walkthrough

The student attaches a square card of side $l$ to the metre rule. If they start with $l = 19$ cm and cut it down to 15 cm, 11 cm, etc., they are simply cutting off pieces from a single large card. This minimises waste and means they only need one piece of card initially. If they started with $l = 3$ cm and wanted to increase to 19 cm, they would need to add more card or use multiple different sized cards, which is impractical and wastes material.

Key Takeaways

Experimental procedures should be designed to minimise waste and maximise the use of available apparatus. Starting with the largest required size and reducing it is more material-efficient.

Common Mistakes

Saying "it is easier to cut" without mentioning material usage.
Not mentioning that only one piece of card is needed.

Things to Be Careful About

The mark scheme specifically looks for "less card used" or "can use just one piece of card". Ensure the answer focuses on material efficiency.

Techniques used

justify experimental procedure based on material usagerecognise the advantage of reducing card size over increasing it

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