Question

A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.1 m/s. The drag force is of the form bv2 What is the value of b?

Answer 1

`2.72\cdot 10^(-4) kg/m`

Step-by-step explanation:

The terminal speed is reached by an object in free fall when the force of gravity becomes equal to the air resistance:

`W=R`

This can be rewritten as

`mg=bv^2`

where

m is the mass of the object

g is the gravitational acceleration

b is the coefficient of the air resistance

v is the speed of the object

In this problem, we have m = 2.3 g = 0.0023 kg, g=9.8 m/s^2 and v=9.1 m/s. Substituting and re-arranging the equation we find

`b=(mg)/(v^2)=((0.0023 kg)(9.8 m/s^2))/((9.1 m/s)^2)=2.72\cdot 10^(-4) kg/m`

Answer 2

At terminal velocity, drag force becomes equal to weight. Therefore:
weight = bv²
0.0023 x 9.81 = b x 9.1²
b = 2.72 x 10⁻⁴