Question

offee filters behave Ince small parachutes, with a drag force that is proportional to the velocity squared, \(F_{\text {drag }}=K v^{2}\). A single coffee filter, when dropped from a height of \(2.0 \mathrm{~m}\), reaches the ground in a time of \(3.0 \mathrm{~s}\). When a second coffee filter is nestled within the first, the drag force remains the same, but the weight is doubled. Find the time for the combined filters to reach the ground. (Neglect the brief period when the filters are accelerating up to their terminal speed.)

Short answer

Answer: The combined filters take approximately 2.12 seconds to reach the ground.

Step-by-step solution

Analyze the motion of a single coffee filter

We start by analyzing the motion of a single coffee filter. It falls from a height of 2.0 m and takes 3.0 s to reach the ground. We are given that \(F_{\text {drag }}=K v^{2}\), where K is the constant of proportionality. First, we will find the terminal velocity for a single coffee filter. As the motion is almost a free fall with a drag force, we can write Newton's second law in the vertical direction as \(F_{\text {drag }}=m \cdot g\), where m is the mass and g is the acceleration due to gravity. Since the motion is at terminal velocity, \(F_{\text {drag }}\) is balanced by the weight of the coffee filter. Hence, \(K v^{2}=m \cdot g\) or \(v^{2} = \frac{m \cdot g}{K}\). We do not know K yet, but we will find it in step 2.

Finding the drag force constant K

Now, we need to find the drag force constant K. We are given the time it takes for a single coffee filter to drop from a height of 2.0 m which is 3.0 seconds. So we can use the equation, \(y = vt\), where y is the distance fallen, v is the terminal velocity, and t is the time it takes to reach the ground. We will substitute v from step 1 and solve for K. \(2.0 \, \text{m} = \sqrt{\frac{m \cdot g}{K}} \cdot 3.0 \,\text{s}\) Solving for K, \(\displaystyle K = \frac{m \cdot g}{ \left(\frac{2.0 \mathrm{~m}}{3.0 \mathrm{~s}}\right)^2}.\)

Analyze the motion of combined filters

Next, we will analyze the motion of the combined filters that are nestled within each other. The total weight of the combined filters is doubled, so the equation becomes \(F_{\text {drag }}=2m \cdot g\). The drag force remains the same, so we can write \(K v'^{2}=2m \cdot g\), where \(v'\) is the new terminal velocity. Dividing this equation by the equation from step 1, we get \(\displaystyle \frac{v'^{2}}{v^{2}} = 2\). Therefore, the terminal velocity of the combined filters, \(v' = v \sqrt{2}\).

Find the time for the combined filters to reach the ground

Finally, we need to find the time it takes for the combined filters to reach the ground. We can use the equation \(y = v't'\), where y is the distance fallen, and \(t'\) is the time it takes to reach the ground. \(2.0 \, \text{m} = \sqrt{2} \cdot v \cdot t'\) We will substitute v from step 1, and solve for \(t'\). \(\displaystyle t' = \frac{2.0 \mathrm{~m}}{\sqrt{2} \cdot \sqrt{\frac{m \cdot g}{K}}} = \frac{3.0 \, \text{s}}{\sqrt{2}} \approx 2.12 \,\text{s}\). So, the time for the combined filters to reach the ground is about 2.12 seconds.

Key concepts

Terminal Velocity

In the context of physics, *terminal velocity* is a crucial concept to understand when dealing with objects falling through a fluid, like air. Terminal velocity is the constant speed reached by an object when the drag force acting on it is equal to the gravitational force pulling it downward. This means, at terminal velocity, an object no longer accelerates and continues to move at a steady speed.

For the coffee filter in the exercise, reaching terminal velocity implies that the drag force, which depends on the velocity squared, balances out the gravitational pull. Thus, its speed remains the same until it touches the ground. This concept helps in predicting how long the object will take to fall to the ground, given certain conditions, such as height and drag coefficient.

Newton's Second Law

*Newton's Second Law of Motion* provides the foundation for understanding motion dynamics when forces are involved. The law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it is written as \[ F = m imes a \] where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.

In the coffee filter scenario, the forces at play are gravity, pulling it downward, and the drag force acting upwards. As the filter falls, these forces reach equilibrium at terminal velocity, meaning the net force is zero and the acceleration stops. Using Newton's Second Law allows us to derive essential equations describing the motion and forces experienced by the filter.

Proportionality Constant

In this exercise, the *proportionality constant* \( K \) plays a significant role in defining the drag force acting on the falling filters. The relationship between the drag force \( F_{\text{drag}} \) and velocity \( v \) is described by the equation: \[ F_{\text{drag}} = K v^2 \] This constant \( K \) depends on factors like the shape, surface area, and properties of the object moving through the fluid.

To discover \( K \), we use the known height and time for the single filter to reach the ground, and relate it to its terminal velocity. This allows us to compute \( K \), which is vital in determining the behavior of other configurations, such as when two filters are combined while dropping.

Free Fall Motion

*Free fall motion* describes the movement of an object solely under the influence of gravity, with negligible air resistance. In real-world scenarios, air resistance often plays a role by acting as a drag force, which alters the simple nature of free fall.

For the coffee filter, though not in pure free fall due to air resistance, the principles still apply for understanding initial motion until it reaches terminal velocity. The equation \[ y = v \times t \] can be used to describe the distance fallen over time when terminal velocity is achieved. This provides a clear way to calculate falling times in scenarios where drag force is a contributing factor. The assumption of neglecting the short acceleration period simplifies the problem, focusing directly on analyzing uniform motion at terminal speed.