Question

Coffee filters behave like 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 \(2.00 \mathrm{~m},\) reaches the ground in a time of \(3.00 \mathrm{~s}\). When a second coffee filter is nestled within the first, the drag coefficient 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: Approximately 0.45 seconds.

Step-by-step solution

Determine the speed at which the filters reach the ground

First, let's find the speed at which the single coffee filter reaches the ground when dropped from a height of 2.00 m. We can use the following equation of motion to find the final velocity, \(v_f\): \(v_f^2 = v_i^2 + 2as\) Where: \(v_f\) is the final velocity \(v_i\) is the initial velocity (0 for a dropped object) \(a\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)) \(s\) is the height from which the object is dropped (2.00 m) \(v_f^2 = 0 + 2(9.81)(2.00)\) \(v_f = \sqrt{39.24}\) \(v_f \approx 6.26 \mathrm{~m/s}\) The single filter reaches the ground with a velocity of approximately 6.26 m/s.

Calculate the drag force acting on the single filter

As given in the problem statement, the drag force, \(F_\text{drag}\), is proportional to the velocity squared, \(Kv^2\). Using the velocity we found in Step 1, we can calculate the drag force acting on the single filter: \(F_\text{drag} = Kv^2\) \(F_\text{drag} = K(6.26)^2\)

Determine the terminal velocity of combined filters

Now it's time to find the terminal velocity of the combined filters. Since the drag force is the same for both filters, and the weight is doubled, the terminal velocity of the combined filters will be smaller. The terminal velocity of the combined filters, \(v_\text{combined}\), can be found using this relationship: \(F_\text{drag} = K v_\text{combined}^2\) We can use the equation we found in Step 2 for the drag force acting on the single filter: \(K(6.26)^2 = K v_\text{combined}^2\) To find the terminal velocity of combined filters, we divide by \(K\) and take the square root of both sides: \(v_\text{combined} = \sqrt{(6.26)^2/2}\) \(v_\text{combined} \approx 4.43 \mathrm{~m/s}\) The terminal velocity of the combined filters is approximately 4.43 m/s.

Calculate the time for the combined filters to reach the ground

Now that we have the terminal velocity of the combined filters, we can use the following equation of motion to find the time it takes for them to reach the ground: \(s = v_\text{combined}t\) \(t = \frac{s}{v_\text{combined}}\) \(t = \frac{2.00}{4.43}\) \(t \approx 0.45 \mathrm{~s}\) The combined filters take approximately 0.45 seconds to reach the ground.

Key concepts

Drag Force

When an object moves through a fluid such as air or water, it experiences a resistance known as drag force. This force opposes the object's motion and increases with the velocity of the object. In our exercise, the drag force on a coffee filter falling through the air is given by the equation
\( F_{\text{drag}} = K v^2 \).
Here, \( K \) is a constant that incorporates factors like air density and the cross-sectional area of the filter, and \( v \) is the velocity of the falling coffee filter.
At terminal velocity, the drag force exactly balances the weight of the object, preventing further acceleration and resulting in a constant velocity. For the single coffee filter, we can determine the drag force at terminal velocity using the velocity it reaches just before hitting the ground. This force, once known, can also be applied to understand the impact when the weight is doubled, as is the case with two nested filters.

Equations of Motion

The equations of motion describe the relationship between an object's velocity, acceleration, and displacement over time. One equation often used is
\( v_f^2 = v_i^2 + 2as \),
where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement.
For a freely falling object, assuming initial velocity is zero and acceleration due to gravity, we can use this equation to determine the velocity of the coffee filter at the point of impact. Understanding the equations of motion is crucial for solving problems involving moving objects, as it allows us to predict future movement based on current conditions.

Acceleration due to Gravity

Acceleration due to gravity is the acceleration experienced by an object solely due to the gravitational pull of the Earth. Near the surface of the Earth, this value is approximately \( 9.81 \text{m/s}^2 \) and acts towards the center of the Earth.
In our exercise, the coffee filters are dropped from a certain height, and gravity is the only force causing acceleration down until drag force equals the force due to gravity, reaching terminal velocity. This acceleration due to gravity is a pivotal concept in our calculations, providing the basis to determine the terminal velocity and the time it takes for the coffee filters to reach the ground.
Since the acceleration due to gravity is constant near Earth's surface, it simplifies our calculations for motion and is an essential factor in predicting the behavior of falling objects.