Question

A 6.5-cm-diameter ball has a terminal speed of 18 m/s. what is the ball's mass? use 1.2 kg/m3 for the density of air at room temperature.

Answer 1

The terminal velocity is the highest velocity that an object  can reach when falling through a medium opposing a drag force to the motion, and it is given by
$v_t = \sqrt{ \frac{2 mg}{\rho A C_D} }$
where 
m is the mass
g is the gravitational acceleration
$\rho$ is the density of the medium
$A$ is the projected area of the object
$C_D$ is the drag coefficient

We can consider the ball as a perfect sphere, so its drag coefficient is $C_D=0.47$. The projected area is the cross-sectional area of the ball, so
$A=\pi r^2$
the radius of the ball is half the diameter:
$r= \frac{d}{2}=3.13 cm=0.0313 m$
so the area is
$A=\pi r^2 = \pi (0.0313 m)=3.1 \cdot 10^{-3} m^2$

And so by using the data of the problem and re-arranging the equation, we can calculate m (the mass) from the terminal velocity:
$m= \frac{v_t^2 \rho A C_D}{2 g}= \frac{(18 m/s)^2(1.2 kg/m^3)(3.1 \cdot 10^{-3} m^2)(0.47)}{2 \cdot 9.81 m/s^2}=0.029 kg$

Answer 2

Ball's mass is 0.30 kg.

Given :

Diameter = 6.5 cm

Terminal Speed = 18 m/sec

Density of air at room temperature, $\rm \rho = 1.2\;Kg/m^3$

Solution :

We know that terminal velocity is given by

$\rm v_t = \sqrt{\dfrac{2mg}{\rho A C_D}}$

So,

$\rm m = \dfrac{\rho A C_Dv_t^2}{2 g}$  ------ (1)

Where, m is the mass, A is the area and $\rm C_D$ is the drag coefficient.

Now, consider the ball as a perfect sphere therefore,

$\rm C_D = 0.47$

Now area is,

$\rm A = \pi r^2$

$\rm A = \pi \times (\dfrac{6.5}{2})^2$

$\rm A = 33.18\;cm^2 = 3.3\times 10^-^3\; m^2$

Now putting the values in equation (1) we get,

$\rm m = \dfrac{1.2\times 3.3\times10^-^3\times0.47\times 18^2}{2\times 9.81}$

m = 0.30 kg

Ball's mass is 0.30 kg.

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