Background Information: The amount of air resistance an
object encounters is directly proportional to its surface area and velocity.
Terminal velocity, vt, is achieved when the
air resistance equals the object's weight and the object can no longer accelerate.
It reaches a state of dynamic equilibrium.Theory. The air resistance any
group of filters encounters is directly proportional to their cross-sectional area and to
their instantaneous velocity. The formula that represents this relationship is:
where k is a constant that is proportional to the filters'
cross-sectional area. At terminal velocity, vt,
the filters are in dynamic equilibrium,
Solving equation #2 for vt produces the result,
| vt = Ö(mg / k) |
Equation #3 |
Since each group of filters has relatively small mass, this terminal
velocity is reached almost immediately after the filters are released. Thus, for the
most part, they are traveling the entire time at vt. This allows us to use the equation
to determine the time required for each group of filters to reach the
ground.
After solving equation #4 for time, t = s / vt, both times can be set equal to each other, since both groups in each trial
hit the ground simultaneously.
| s1 / vt1 = s2 / vt2 |
Equation #5 |
Rearranging equation #5,
| s1 / s2
= vt1 / vt2 |
Equation #6 |
Substituting equation #3 vt = Ö(mg / k) for each
group's terminal velocity into equation #6 produces,
| s1 / s2
= Ö[(m1g / k) / (m2g / k)] |
Equation #7 |
In this equation (#7), g and k cancel,
| s1 / s2
= Ö[m1 / m2] |
Equation #8 |
Squaring both sides of equation #8 results in the relationship,
| (s1
/ s2) ² = m1 / m2 |
Equation #9 |
This result means that the
distance required to achieve terminal velocity for two objects, both released from rest,
having the same surface area is directly proportional to the square root of their
masses. That is, if one object has 9 times the mass of the other object, it will
require three times the distance to reach its terminal velocity. |
