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Worksheet · IGCSE 0625 · 1.1 Measurement · Core

Measuring time: practice

Six original Core questions on the multiple-measurement method and reaction time. Show your working and give a unit with every answer.

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Q13 marks

The multiple-measurement method.

A pendulum swing: count complete swings, starting and stopping at the centre.
Timing a pendulum's swing
(a) Explain why timing a single swing of a pendulum with a stopwatch is not accurate. (b) Describe a better method for finding the time of one swing.
Q22 marks

A pendulum is timed over 20 complete swings. The total time is 29.0 s. Calculate the period (the time for one swing).

Q32 marks

A stack of 250 sheets of paper has a height of 25 mm. Calculate the thickness of one sheet, in mm.

Q43 marks

A student times 10 complete swings four times and records: 14.8 s, 15.0 s, 14.9 s, 15.3 s.

(a) Find the average total time for 10 swings. (b) Calculate the period. (c) One value sits further from the others. Suggest what may have caused it.
Q52 marks

Explain why timing 20 swings and dividing by 20 gives a more accurate period than timing one swing. Refer to reaction time.

The reaction-time error is a large share of one swing but a small share of 20 swings.
The reaction-time error as a share of the timing
Q63 marks

A tap drips steadily. A student wants the time for one drip.

(a) Describe how to measure this accurately. (b) The student counts 30 drips in 48 s. Calculate the time for one drip.

Total: 15 marks. Original work by the TheLucidSTEM team. Written in the style of the papers; no past paper question is reproduced.

Answer key · full worked solutionsclick to reveal
Q1. The multiple-measurement method.

(a) the time for one swing is short, and human reaction time (about 0.2 s) is a large fraction of it, so a single reading is unreliable.
(b) time a number of complete swings, for example 20, then divide the total time by the number of swings to find the period.

Q2. Period from 20 swings.

T = total ÷ number of swings = 29.0 ÷ 20 = 1.45 s.

Q3. Thickness of one sheet.

thickness of one sheet = 25 ÷ 250 = 0.10 mm.

Q4. Repeated timings.

(a) average total = (14.8 + 15.0 + 14.9 + 15.3) ÷ 4 = 60.0 ÷ 4 = 15.0 s.
(b) period = 15.0 ÷ 10 = 1.5 s.
(c) a counting or timing slip, or a slow start or stop (reaction time). It is an anomalous result.

Q5. Reaction time.

the reaction-time error happens once on starting and once on stopping, so it is about the same size whether one swing or twenty is timed. Dividing the total by 20 divides that error by 20 as well, so each swing's share of the error is much smaller.

Q6. Timing a small interval.

(a) count a number of drips over a measured time with a stopwatch, then divide the total time by the number of drips.
(b) time for one drip = 48 ÷ 30 = 1.6 s.

Marking note: award the unit mark only when a correct unit is given. In Q2, Q3, Q4(b) and Q6(b), the division (total ÷ count) must be shown.
Original work by the TheLucidSTEM team. Questions are written in the style of the papers; no past paper question is reproduced. Supplied in editable formats so you can adapt them freely.
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