A student investigates a wooden sphere rolling down a plastic channel and falling to the floor. The channel is set up at the end of a bench. The sphere is initially held in the channel at the position shown in Fig. 1.1. (i) On Fig. 1.1, mark and label the height $h$ of the sphere above the bench before it is released. (ii) Describe how the student ensures that the sphere is released from the same point each time. (b) The sphere is released, rolls down the channel and lands…

(a)(i) Vertical line drawn and labelled h from the bench surface to the sphere · (a)(ii) Make a mark on the channel at the release position · (b)(i) Smooth downward curved path from the end of the channel to the floor · (b)(ii) Horizontal line labelled d from below the channel exit to the landing point on the floor · (c) Place a tray of sand on the floor to record the impact mark · (d) 66 cm · (e)(i) Yes, because if h = 0 the sphere has no initial energy and no horizontal…

Physics 5054/41 — October/November 2013

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

questions

marks

minutes

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

Q1Experimental ContextsObservations and MeasurementsUse of Techniques, Apparatus and MaterialsPlanning Experiments and InvestigationsAnalysis, Conclusions and EvaluationFree sample

A student investigates a wooden sphere rolling down a plastic channel and falling to the floor.

The channel is set up at the end of a bench.
The sphere is initially held in the channel at the position shown in Fig. 1.1.

Fig. 1.1

(a)

(i)

On Fig. 1.1, mark and label the height $h$ of the sphere above the bench before it is released.

DifficultyEasy

Worked solution

Answer

Draw a vertical line or double-headed arrow from the bench surface up to the sphere (or the center/base of the sphere) and label it $h$ .

Final answer

Vertical line drawn and labelled h from the bench surface to the sphere

Detailed explanation

Walkthrough

The height $h$ represents the vertical distance between the level of the bench surface and the sphere in its initial release position on the curved channel.

To mark this correctly on Fig. 1.1:

Identify the horizontal line representing the surface of the bench.
Identify the position of the sphere.
Draw a vertical line or arrow spanning between the bench level and the sphere.
Label this line clearly with the letter $h$ .

Key Takeaways

Height is always measured vertically relative to a defined reference level (here, the top of the bench).

Common Mistakes

Drawing a line along the curve of the track instead of measuring vertically.
Measuring from the floor rather than from the bench.

Things to Be Careful About

Ensure the line is drawn vertically by eye.

Techniques used

identify vertical height on an apparatus diagramdraw and label a dimension line

(ii)

Describe how the student ensures that the sphere is released from the same point each time.

DifficultyEasy

Worked solution

Answer

Make a mark (or pencil line) on the track / channel at the release position.

Final answer

Make a mark on the channel at the release position

Detailed explanation

Walkthrough

To ensure fair testing and accurate repetition of measurements, the release position must remain identical across trials.

A practical and simple technique is to place a physical mark (such as a pencil line, marker dot, or a small piece of tape) directly on the curved channel at the chosen height $h$ . The sphere can then be lined up precisely with this mark before every release.

Key Takeaways

Marking reference points on apparatus ensures high repeatability and minimizes human positioning error.

Common Mistakes

Giving vague answers like 'be careful' or 'look closely' without describing a physical method.

Things to Be Careful About

The suggestion must be a concrete, practical physical action.

Techniques used

suggest a method for ensuring repeatable initial conditions

(b)

The sphere is released, rolls down the channel and lands on the floor.
When the sphere leaves the end of the channel, it is travelling horizontally.

On Fig. 1.1,

(i)

draw a possible path of the sphere after it leaves the channel and until it hits the floor,

DifficultyMedium-Easy

Worked solution

Answer

Draw a smooth curved downward path (a projectile trajectory / parabola) starting horizontally from the exit of the channel at the edge of the bench and curving downwards until it hits the floor.

Final answer

Smooth downward curved path from the end of the channel to the floor

Detailed explanation

Walkthrough

When the sphere leaves the horizontal end of the channel, it possesses an initial horizontal velocity and zero vertical velocity. As it moves through the air, gravity causes it to accelerate vertically downwards while its horizontal velocity remains approximately constant. This produces a characteristic parabolic projectile path (curving downwards towards the floor).

Key Takeaways

Projectiles launched horizontally follow a downward curving parabolic path.

Common Mistakes

Drawing a straight diagonal line from the bench to the floor.
Drawing an upward arc before falling.

Things to Be Careful About

The path must start horizontally right at the channel exit.

Techniques used

draw a projectile trajectory

(ii)

mark and label the horizontal distance $d$ travelled by the sphere after it leaves the channel and until it hits the floor.

DifficultyEasy

Worked solution

Answer

Draw a horizontal line or double-headed arrow extending from the vertical line through the end of the channel / bench edge to the point where the drawn path meets the floor, and label it $d$ .

Final answer

Horizontal line labelled d from below the channel exit to the landing point on the floor

Detailed explanation

Walkthrough

The horizontal distance $d$ is the range of the projectile. It is measured horizontally from the point directly beneath the release edge (the exit of the channel) to the landing point on the floor.

Key Takeaways

Range $d$ is purely horizontal distance measured from the launch position to the landing position.

Common Mistakes

Drawing a diagonal line from the launch point to the impact point.
Measuring $d$ from the leg of the bench rather than the launch point (channel exit).

Things to Be Careful About

The line must be strictly horizontal.

Techniques used

identify horizontal range on a diagramdraw and label horizontal distance

(c)

Suggest a method for finding the point where the sphere hits the floor.

DifficultyMedium-Easy

Worked solution

Answer

Place a tray of sand on the floor (or cover the floor with carbon paper / white paper, or coat the sphere with paint / ink) so that it leaves a mark where it hits.

Final answer

Place a tray of sand on the floor to record the impact mark

Detailed explanation

Walkthrough

Because the sphere bounces and moves quickly upon hitting the floor, locating the exact point of first impact by eye alone is difficult and unreliable.

Acceptable practical methods to record the impact point:

Placing a shallow tray of smooth sand on the floor so the sphere leaves a visible crater.
Laying a sheet of carbon paper over white paper so the impact leaves a distinct dot.
Dipping the sphere in paint, water, or chalk powder so it marks the floor on contact.
Having an observer watch closely at floor level with a marker.

Key Takeaways

Using recording media (sand, carbon paper, wet markers) allows accurate capture of instantaneous landing positions.

Common Mistakes

Saying 'measure it with a ruler' without explaining how the exact impact point is first identified.

Things to Be Careful About

Ensure the suggested method clearly describes how the point is marked or captured.

Techniques used

suggest an experimental method to record impact location

(d)

With $h$ set at $30\text{ cm}$ , the student repeats the experiment and measures $d$ six times.
The student obtains the following values of $d$ in $\text{cm}$ .

68.5 \quad 64.0 \quad 67.0 \quad 66.5 \quad 65.0 \quad 64.5

Calculate the average distance $d_{av}$ .
Give your answer to a suitable number of significant figures.

$d_{av}$ = ______ $\text{cm}$

DifficultyMedium-Easy

Worked solution

Working

\begin{aligned} d_{av} &= \frac{68.5 + 64.0 + 67.0 + 66.5 + 65.0 + 64.5}{6} \\[2ex] &= \frac{395.5}{6} \\[2ex] &= 65.916... \text{ cm} \\[2ex] &\approx 66\text{ cm} \end{aligned}

Answer

Final answer

66 cm

Detailed explanation

Walkthrough

To find the average distance $d_{av}$ :

Sum all six individual distance values:

68.5 + 64.0 + 67.0 + 66.5 + 65.0 + 64.5 = 395.5\text{ cm}

Divide by the total number of readings (6):

\frac{395.5}{6} = 65.916...\text{ cm}

Look at the precision of the table values in Fig. 1.2: all values of $d_{av}$ are given to the nearest whole centimetre (2 significant figures). Rounding $65.916...$ to the nearest whole number gives $66\text{ cm}$ .

Key Takeaways

The mean of repeated trials gives a more reliable estimate of the true value.
The number of significant figures should be consistent with the data provided in the table.

Common Mistakes

Giving too many decimal places (e.g. $65.917\text{ cm}$ ), which is inconsistent with the data in Fig. 1.2.

Things to Be Careful About

Round correctly ( $65.9$ rounds up to $66$ ).

Techniques used

calculate the mean of repeated measurementsround to appropriate significant figures

(e)

The student repeats the experiment with different values of $h$ . The results obtained for $h$ and $d_{av}$ are recorded in Fig. 1.2.

Fig. 1.2

$h\text{ / cm}$	$d_{av}\text{ / cm}$
2	14
5	22
10	33
15	45
20	54
25	60
30

On Fig. 1.2, write your value for $d_{av}$ from (d).

(i)

By considering the experimental arrangement, suggest, with a reason, whether $d_{av} = 0$ when $h = 0$ .

DifficultyMedium-Easy

Worked solution

Answer

Yes, because when $h = 0$ , the sphere has no gravitational potential energy / is not moving, so it has no horizontal velocity when it leaves the bench (or falls straight down / does not roll).

Final answer

Yes, because if h = 0 the sphere has no initial energy and no horizontal velocity

Detailed explanation

Walkthrough

When $h = 0$ :

The sphere is placed at the very bottom horizontal section of the channel.
It has no initial gravitational potential energy relative to the channel exit to convert into kinetic energy.
Therefore, it will not roll forward and will have zero horizontal velocity ( $v = 0$ ).
Without horizontal velocity, it cannot travel horizontally ( $d_{av} = 0$ ).

(Alternatively, candidates can argue 'No' if they explain a systematic effect, e.g. the finite radius of the sphere means its center of mass is still above the bench surface). The standard expected physical answer is Yes because there is no horizontal speed.

Key Takeaways

Extrapolating a physical system to zero often checks if the theoretical model matches physical boundary conditions.

Common Mistakes

Answering 'yes' or 'no' without providing a supporting physical reason.

Things to Be Careful About

Ensure the reasoning clearly links zero height to zero speed / energy.

Techniques used

evaluate physical limits and boundary conditions

(ii)

On Fig. 1.3, plot the graph of $d_{av}\text{ / cm}$ on the y-axis against $h\text{ / cm}$ on the x-axis.
Start your axes from the origin. Draw a smooth curve of best fit.

Fig. 1.3

DifficultyMedium

Worked solution

Working

Axes and labels:
- x-axis: $h\text{ / cm}$
- y-axis: $d_{av}\text{ / cm}$
Scales:
- Starting from $(0, 0)$ :
- x-axis: $2\text{ cm} \equiv 5\text{ cm}$ (or $10\text{ small squares} = 5\text{ cm}$ , extending from $0$ to at least $30\text{ cm}$ )
- y-axis: $2\text{ cm} \equiv 10\text{ cm}$ (or $10\text{ small squares} = 10\text{ cm}$ , extending from $0$ to at least $70\text{ cm}$ )
Points plotted:
- $(0, 0)$
- $(2, 14)$
- $(5, 22)$
- $(10, 33)$
- $(15, 45)$
- $(20, 54)$
- $(25, 60)$
- $(30, 66)$
Best-fit line: A smooth, continuous curve passing through or evenly balancing all plotted points from the origin.

Answer

Final answer

Smooth curve of best fit plotted through the points starting from the origin

Detailed explanation

Walkthrough

To obtain all 4 marks on the graph plotting task:

Axes and Units (1 mark):
- Label the horizontal axis $h\text{ / cm}$ .
- Label the vertical axis $d_{av}\text{ / cm}$ .
Scales (1 mark):
- Both axes must start at $(0,0)$ .
- The data range for $h$ is $0$ to $30\text{ cm}$ . A standard scale of $2\text{ cm} \equiv 5\text{ cm}$ uses 6 large grid blocks (12 cm of grid).
- The data range for $d_{av}$ is $0$ to $66\text{ cm}$ . A standard scale of $2\text{ cm} \equiv 10\text{ cm}$ uses 7 large grid blocks (14 cm of grid).
- Both scales are linear, easy to read, and occupy well over half of the available grid area.
Plotting Points (1 mark):
- Plot all 7 points from the table plus $(0,0)$ accurately to within half a small grid square using small, neat crosses ( $\times$ ) or circled dots ( $\odot$ ).
Line of Best Fit (1 mark):
- Draw a single, smooth curve passing smoothly through the trend of points without sharp bends, kinks, or double lines.

Key Takeaways

Always include quantity and unit on axis labels.
Scales must be linear and use more than half the grid.
Curves of best fit must be drawn with a single smooth sweep.

Common Mistakes

Forcing a straight line with a ruler through non-linear data.
Using non-linear scales or awkward increments (e.g. multiples of 3 or 7).
Drawing thick, 'furry', or multiple sketched lines.

Things to Be Careful About

Ensure the line passes through or near the origin $(0,0)$ as required by the instruction to start axes from the origin.

Techniques used

choose linear scales spanning more than half the gridplot experimental data points accuratelydraw a smooth curve of best fit

(iii)

Another student suggests that $d_{av}$ is directly proportional to $h$ .
Use your graph to explain whether this student is correct.

DifficultyEasy

Worked solution

Answer

The student is not correct because the graph is a curve (not a straight line).

Final answer

Incorrect, because the graph is a curve / not a straight line

Detailed explanation

Walkthrough

For two quantities to be directly proportional:

The graph of $y$ against $x$ must be a straight line.
The straight line must pass through the origin $(0,0)$ .

From the plotted graph, the line is a curve with a decreasing gradient, not a straight line. Therefore, $d_{av}$ is not directly proportional to $h$ .

Key Takeaways

Direct proportionality requires a straight line passing through $(0,0)$ . Any curvature means the relationship is not directly proportional.

Common Mistakes

Stating 'yes because as $h$ increases, $d_{av}$ increases' (this is a positive correlation, not direct proportionality).

Things to Be Careful About

The explanation must state that the graph is curved / not a straight line.

Techniques used

interpret direct proportionality from graph characteristics

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