Physics 5054/41 — May/June 2017

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

questions

marks

minutes

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

Q1Experimental ContextsAnalysis, Conclusions and EvaluationObservations and MeasurementsFree sample

A strip of paper is attached to a small toy car. As the toy car moves, it pulls the strip of paper through a timer. The timer marks a dot on the paper every $0.020\text{ s}$ .

Fig. 1.1 shows a section of the paper strip with the first four dots marked. The first of these dots to be marked on the paper is labelled A.

Fig. 1.1

(a)

(i)

Tick the box that describes the motion of the car.

[ ] acceleration

[ ] constant speed

[ ] deceleration

DifficultyMedium-Easy

Worked solution

Answer

[x] deceleration

Final answer

deceleration

Detailed explanation

Walkthrough

As the toy car moves, it pulls the paper strip through the ticker timer. The first dot marked is dot A at the left. Subsequent dots are printed every $0.020\text{ s}$ as the tape moves. Looking at the successive intervals from left to right (from dot A onwards):

The spacing between consecutive dots decreases as time goes on.
Since each dot is made at equal time intervals of $0.020\text{ s}$ , a smaller distance between consecutive dots means the car covers less distance per unit time, which indicates that the car is slowing down (decelerating).

Key Takeaways

For ticker timer tape, the direction of motion is away from the timer. Dot A was the first dot made, so subsequent intervals show the motion over subsequent equal time steps.
Decreasing spacing between consecutive dots in the direction of time indicates deceleration.

Common Mistakes

Confusing the direction of motion: thinking that the car was accelerating because the rightmost dots look further from dot A than the first interval is from zero, rather than looking at the distance between consecutive dots (the spacing between successive dots decreases).

Things to Be Careful About

Always examine the spacing between successive consecutive dots, not the cumulative distance from the start.

Techniques used

interpret ticker tape dot spacingdeduce deceleration from decreasing distance intervals

(ii)

Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

Less distance is travelled in each successive equal time interval of $0.020\text{ s}$ (the distance between consecutive dots decreases).

Final answer

Less distance travelled in equal time intervals of 0.020 s

Detailed explanation

Walkthrough

To earn the mark, state clearly that the distance between consecutive dots decreases, which means less distance is travelled in each fixed time interval of $0.020\text{ s}$ . Speed is $\frac{\text{distance}}{\text{time}}$ ; because the time interval is constant, a decreasing distance interval means decreasing speed (deceleration).

Key Takeaways

Constant time between dots ( $0.020\text{ s}$ ) means distance per interval is directly proportional to average speed during that interval.
Decreasing gap size means decreasing speed.

Common Mistakes

Stating simply "the dots get closer together" without mentioning that this happens in equal time intervals or relates to distance per unit time.

Things to Be Careful About

Explicitly mention that the time interval between consecutive dots is constant / equal ( $0.020\text{ s}$ ).

Techniques used

link distance between dots to speedrelate changing distance in equal time intervals to deceleration

(b)

The distance of each dot from A is $d$ . Dot A was marked on the strip at time $t = 1.000\text{ s}$ .

Take measurements from Fig. 1.1 and, in the space below, draw a table of results for $d$ against $t$ .

DifficultyMedium

Worked solution

Answer

$t\text{ / s}$	$d\text{ / cm}$
$1.000$	$0.0$
$1.020$	$5.0$
$1.040$	$9.0$
$1.060$	$12.0$

Final answer

Table with columns t / s (1.000, 1.020, 1.040, 1.060) and d / cm (0.0, 5.0, 9.0, 12.0)

Detailed explanation

Walkthrough

Determine the time values $t$ :
- Dot A is at $t = 1.000\text{ s}$ .
- Each subsequent dot is made $0.020\text{ s}$ $0.020 s$ later:
  - 2nd dot: $t = 1.000 + 0.020 = 1.020\text{ s}$
  - 3rd dot: $t = 1.020 + 0.020 = 1.040\text{ s}$
  - 4th dot: $t = 1.040 + 0.020 = 1.060\text{ s}$
Measure the distances $d$ from dot A using a ruler on the full-scale printed strip:
- Dot 1 (A): $d = 0.0\text{ cm}$
- Dot 2: $d = 5.0\text{ cm}$
- Dot 3: $d = 9.0\text{ cm}$
- Dot 4: $d = 12.0\text{ cm}$
Draw and label the table:
- Include a column for time with heading time / s or $t\text{ / s}$ .
- Include a column for distance with heading distance / cm or $d\text{ / cm}$ .
- Record all 4 sets of corresponding values.

Key Takeaways

Column headings must have both quantity and unit separated by a slash (e.g., $t\text{ / s}$ , $d\text{ / cm}$ ).
Distances must be measured cumulatively from reference point A.

Common Mistakes

Forgetting units in table headings.
Recording interval distances between adjacent dots (e.g., 5.0, 4.0, 3.0) instead of total distance from dot A ( $d$ ).
Starting time at $0\text{ s}$ instead of the given $t = 1.000\text{ s}$ .

Things to Be Careful About

Ensure ruler measurements are taken to the centre of each dot.
Record measured values to a consistent precision (to the nearest millimetre / $0.1\text{ cm}$ ).

Techniques used

measure distances from a full-scale diagramcalculate cumulative times from a ticker-timer frequencyconstruct a results table with proper column headings and units

(c)

Use your data to calculate the average speed of the car between $t = 1.000\text{ s}$ and $t = 1.060\text{ s}$ .

Use the equation

\text{average speed} = \frac{\text{distance travelled}}{\text{time taken}}

average speed = ______

DifficultyMedium-Easy

Worked solution

Working

\begin{aligned} \text{distance travelled} &= 12.0\text{ cm} = 0.12\text{ m} \\ \text{time taken} &= 1.060 - 1.000 = 0.060\text{ s} \\ \text{average speed} &= \frac{\text{distance travelled}}{\text{time taken}} = \frac{12.0}{0.060} = 200\text{ cm/s} \quad (\text{or } 2.0\text{ m/s}) \end{aligned}

Answer

$200\text{ cm/s}$ (or $2.0\text{ m/s}$ )

Final answer

200 cm/s

Detailed explanation

Walkthrough

From the table in part (b):
- Total distance travelled between $t = 1.000\text{ s}$ and $t = 1.060\text{ s}$ is $d = 12.0\text{ cm}$ (or $0.12\text{ m}$ ).
- Time taken $\Delta t = 1.060\text{ s} - 1.000\text{ s} = 0.060\text{ s}$ .
Use the given equation: $\text{average speed} = \frac{\text{distance travelled}}{\text{time taken}} = \frac{12\text{ cm}}{0.060\text{ s}} = 200\text{ cm/s}$ In SI units: $\text{average speed} = \frac{0.12\text{ m}}{0.060\text{ s}} = 2.0\text{ m/s}$

Key Takeaways

$\text{Average speed} = \frac{\text{total distance}}{\text{total time}}$ .
Either $\text{cm/s}$ or $\text{m/s}$ is acceptable, provided the unit matches the numerical value.

Common Mistakes

Dividing by $1.060\text{ s}$ instead of the time interval $\Delta t = 0.060\text{ s}$ .
Incorrect unit conversion (e.g. giving $200\text{ m/s}$ instead of $200\text{ cm/s}$ ).

Things to Be Careful About

Always check that the numerical value and the stated unit correspond properly.

Techniques used

calculate average speed from total distance and total time

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