Question

Two hail stones with radii in the ratio of $1 : 2$ fall from a great height through the atmosphere. Then the ratio of their momenta after they have attained terminal velocity is :

• A. 1 : 1
• B. 1 : 4
• C. 1 : 16
• D. 1 : 32

Answer 1

Answer: (D)

Terminal velocity is attained when gravitational force is equal and opposite to atmospheric drag.
It is given by $v=\frac{2}{9} \frac{r^{2}(\rho-\sigma)}{\eta} g$
where $\eta$ is coefficient of viscosity, $r$ the radius, $\rho$ and $\sigma$ the density.
$\therefore \frac{v_{1}}{v_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$
Ratio of masses of the two hail stones is
$\frac{m_{1}}{m_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3} \cdot \rho}{\frac{4}{3} \pi r_{2}^{3} \cdot \rho}=\frac{1}{8}$
The ratio of momenta $=\frac{p_{1}}{p_{2}}=\frac{m_{1} v_{1}}{m_{2} v_{2}}$
$=\frac{1}{8} \times \frac{1}{4}=\frac{1}{32} $
$\Rightarrow $ $\frac{d^{2} y}{d t^{2}}+\frac{k}{m} y=0$