In this experiment, you will measure your reaction time using a falling ruler. Your partner (student A) holds a metre ruler vertically so that the zero mark...
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In this experiment, you will measure your reaction time using a falling ruler. Your partner (student A) holds a metre ruler vertically so that the zero mark is level with the top of you's (student B's) thumb and finger. Student B holds your thumb and finger open on either side of the ruler without touching it. Student A releases the ruler without warning and student B catches it as quickly as possible. The distance d from the zero mark to the point where student B catches the ruler is recorded. The experiment is repeated five times. Fig. 5.1 shows the arrangement. The five ruler readings at the catch position are shown in Fig. 5.2. The equation d = ½gt² relates the distance fallen d to the reaction time t, where g = 9.8 m/s².
(a) Record the five distances fallen from the ruler readings in Fig. 5.2. [2]
(b) Measure the average of your five distance values. [1]
(c) Use d = ½gt² to calculate the reaction time for the largest and smallest distances. Show your working. [2]
(d) Calculate the mean reaction time from all five trials. [1]
(e) State one variable that must be kept the same for a fair test. [1]
(f) Give one reason why the first trial might be less reliable than the others. [1]
(g) Describe one way to improve the reliability of the reaction time value obtained. [1]
(h) Explain why a larger distance fallen corresponds to a longer reaction time. [1]
Answer Details
(a) Reading the five catch positions from the ruler diagrams in Fig. 5.2: [2]
| Trial | Distance d / cm |
|---|---|
| 1 | 18.5 |
| 2 | 15.2 |
| 3 | 16.8 |
| 4 | 14.5 |
| 5 | 17.0 |
Each arrow in the diagram points to where the ruler was caught. Read the scale carefully to the nearest millimetre.
(b) [1]
\[ d_{\text{avg}} = \frac{18.5 + 15.2 + 16.8 + 14.5 + 17.0}{5} = \frac{82.0}{5} = 16.4 \text{ cm} \]
(c) Rearranging \( d = \tfrac{1}{2}gt^{2} \) gives \( t = \sqrt{\dfrac{2d}{g}} \). [2]
For the largest distance, \( d = 18.5 \text{ cm} = 0.185 \text{ m} \):
\[ t = \sqrt{\frac{2 \times 0.185}{9.8}} = \sqrt{0.03776} = 0.194 \text{ s} \]
For the smallest distance, \( d = 14.5 \text{ cm} = 0.145 \text{ m} \):
\[ t = \sqrt{\frac{2 \times 0.145}{9.8}} = \sqrt{0.02959} = 0.172 \text{ s} \]
Always convert centimetres to metres before substituting into the equation.
(d) Calculate all five reaction times, then take the mean: [1]
| Trial | d / m | t / s |
|---|---|---|
| 1 | 0.185 | 0.194 |
| 2 | 0.152 | 0.176 |
| 3 | 0.168 | 0.185 |
| 4 | 0.145 | 0.172 |
| 5 | 0.170 | 0.186 |
\[ t_{\text{mean}} = \frac{0.194 + 0.176 + 0.185 + 0.172 + 0.186}{5} = \frac{0.913}{5} = 0.183 \text{ s} \]
(e) The same person must catch the ruler every time. Other acceptable answers: the same hand is used, the same ruler, or the fingers start at the same position (level with the zero mark). Keeping these constant ensures only reaction time varies between trials. [1]
(f) The catcher may not yet be familiar with the procedure or may be anxious, leading to a slower or less consistent response. Practice effects mean later trials are usually more reliable. [1]
(g) Take more than five trials and calculate a new mean, or remove the highest and lowest values before averaging. More trials reduce the impact of any single anomalous result on the calculated mean. [1]
(h) The ruler falls under gravity with acceleration \( g = 9.8 \text{ m/s}^{2} \). From \( d = \tfrac{1}{2}gt^{2} \), a longer reaction time \( t \) means the ruler falls for a longer duration before being caught, producing a greater distance \( d \). The relationship is quadratic, so even a small increase in reaction time gives a noticeably larger distance. [1]
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