Question

directions: answer all three questions. the suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. the parts within a question may not have equal weight. show all your work in the pink booklet in the spaces provided after each part, not in this green insert. mech. 1. students are to conduct an experiment to investigate the relationship between the terminal speed of a stack of falling paper coffee filters and its mass. their procedure involves stacking a number of coffee filters, like the one shown in the figure above, and dropping the stack from rest. the students change the number of filters in the stack to vary the mass m while keeping the shape of the stack the same. as a stack of coffee filters falls, there is an air resistance (drag) force acting on the filters. (a) the students suspect that the drag force (f_d) is proportional to the square of the speed (v): (f_d = cv^{2}), where (c) is a constant. using this relationship, derive an expression relating the terminal speed (v_t) to the mass (m). the students conduct the experiment and obtain the following data.

mass of the stack of filters, (m) (kg)	(1.12\times10^{-3})	(2.04\times10^{-3})	(2.96\times10^{-3})	(4.18\times10^{-3})	(5.10\times10^{-3})
terminal speed, (v_t) (m/s)	(0.51)	(0.62)	(0.82)	(0.92)	(1.06)

(b) (i) assuming the functional relationship for the drag force above, use the grid below to plot a linear graph as a function of (m) to verify the relationship. use the empty boxes in the data table, as appropriate, to record any calculated values you are graphing. label the vertical axis as appropriate, and place numbers on both axes.

Explanation:

Step1: Analyze forces at terminal speed

At terminal speed \(v_T\), the net - force on the stack of coffee filters is zero. The gravitational force \(F_g = mg\) and the drag force \(F_D = Cv_T^{2}\) are equal, so \(mg = Cv_T^{2}\).

Step2: Solve for \(v_T\) in terms of \(m\)

Rearrange the equation \(mg = Cv_T^{2}\) to get \(v_T=\sqrt{\frac{mg}{C}}\).

Step3: For part (b)(i)

We know from \(mg = Cv_T^{2}\) that \(m=\frac{C}{g}v_T^{2}\). To get a linear graph of \(m\) as a function of some variable related to \(v_T\), we can plot \(m\) versus \(v_T^{2}\). Calculate \(v_T^{2}\) for each data - point:
For \(m_1 = 1.12\times10^{-3}\text{ kg}\), \(v_{T1}=0.51\text{ m/s}\), \(v_{T1}^{2}=(0.51)^2 = 0.2601\text{ m}^2/\text{s}^2\)
For \(m_2 = 2.04\times10^{-3}\text{ kg}\), \(v_{T2}=0.62\text{ m/s}\), \(v_{T2}^{2}=(0.62)^2 = 0.3844\text{ m}^2/\text{s}^2\)
For \(m_3 = 2.96\times10^{-3}\text{ kg}\), \(v_{T3}=0.82\text{ m/s}\), \(v_{T3}^{2}=(0.82)^2 = 0.6724\text{ m}^2/\text{s}^2\)
For \(m_4 = 4.18\times10^{-3}\text{ kg}\), \(v_{T4}=0.92\text{ m/s}\), \(v_{T4}^{2}=(0.92)^2 = 0.8464\text{ m}^2/\text{s}^2\)
For \(m_5 = 5.10\times10^{-3}\text{ kg}\), \(v_{T5}=1.06\text{ m/s}\), \(v_{T5}^{2}=(1.06)^2 = 1.1236\text{ m}^2/\text{s}^2\)

On the graph, the horizontal axis is \(m\) (in kg) and the vertical axis is \(v_T^{2}\) (in \(\text{m}^2/\text{s}^2\)). Plot the points \((1.12\times 10^{-3},0.2601)\), \((2.04\times 10^{-3},0.3844)\), \((2.96\times 10^{-3},0.6724)\), \((4.18\times 10^{-3},0.8464)\), \((5.10\times 10^{-3},1.1236)\) and draw a best - fit line. If the relationship \(F_D = Cv_T^{2}\) is correct, the graph should be a straight line passing through the origin (since when \(m = 0\), \(v_T = 0\)).

Answer:

(a) \(v_T=\sqrt{\frac{mg}{C}}\)
(b)(i) Calculate \(v_T^{2}\) for each data - point as shown above and plot \(m\) (horizontal axis) versus \(v_T^{2}\) (vertical axis) on the provided grid, then draw a best - fit line.