In this experiment, you will investigate how the time taken for a paper cone to fall a fixed distance depends on the radius of the cone. You makes five cone...
Question 1 Report
In this experiment, you will investigate how the time taken for a paper cone to fall a fixed distance depends on the radius of the cone. You makes five cones from identical sheets of card, each with a different base radius. You drops each cone from the same height of 2.00 m and times how long it takes to reach the floor using a stopwatch. You makes three attempts for each cone and calculates the mean. Fig. 19.1 shows the arrangement. Your results are shown in Table 19.1.
| Radius r / cm | Mean fall time t / s |
|---|---|
| 2.0 | 1.28 |
| 3.0 | 1.65 |
| 4.0 | 2.10 |
| 5.0 | 2.58 |
| 6.0 | 3.12 |
(a) Record the mean fall time for each cone in Table 19.1. [1]
(b) Plot a graph of t (y-axis) against r (x-axis) on a grid. [3]
(c) State the relationship between fall time and radius shown by the graph. [1]
(d) Explain, in terms of air resistance, why a larger cone takes longer to fall the same distance. [2]
(e) Describe how you should release the cone to ensure a fair test. [1]
(f) Measure the difference in fall time between the smallest and largest cones. [1]
(g) State one variable you must keep the same, other than the drop height. [1]
Answer Details
(a) The five mean fall times are read from Table 19.1 and recorded: 1.28 s, 1.65 s, 2.10 s, 2.58 s, and 3.12 s for radii 2.0, 3.0, 4.0, 5.0, and 6.0 cm respectively. [1]
(b) The graph of fall time t (y-axis) against radius r (x-axis) is plotted below. Axes are labelled with quantity and unit [1]. All five data points are plotted at the correct positions [1]. A best-fit straight line is drawn through the points [1].
(c) As the radius increases, the fall time increases. The relationship shown by the graph is approximately linear (directly proportional): a larger radius leads to a proportionally longer fall time. [1]
(d) A larger cone has a greater base area. [1] This means it pushes against more air as it falls, producing a larger air resistance (drag) force. The greater drag force opposes the weight more effectively, so the cone reaches a lower terminal velocity and takes longer to cover the 2.00 m distance. [1]
The underlying physics is that drag force increases with the cross-sectional area of the falling object. Since the area of the cone base is \( A = \pi r^2 \), a larger radius gives a much larger area and therefore a proportionally larger resistive force at any given speed.
(e) Hold the cone point-down with the tip at the 2.00 m release height. Release it from rest without pushing, throwing, or spinning it, so that the only initial force acting is gravity. This ensures every cone starts with the same initial velocity (zero), making the test fair. [1]
(f) The difference in fall time between the smallest cone (\( r = 2.0 \) cm, \( t = 1.28 \) s) and the largest cone (\( r = 6.0 \) cm, \( t = 3.12 \) s) is:
\[ \Delta t = 3.12 - 1.28 = 1.84 \text{ s} \] [1]
(g) One variable that must be kept the same (other than drop height) is the type and weight of card used for making each cone. Other acceptable answers include: the mass of each cone, or the shape of the cone (same apex angle for all cones). [1]
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