Question

A large open top container of negligible mass and uniform cross-sectional area $A$ has a small hole of cross-sectional area $A/100$ in its side wall near the bottom. The container is kept on a smooth horizontal floor ad contains a liquid of density $\rho$ and mass $m_0$. Assuming that the liquid starts flowing out horizontally through the hole at $t=0$. The acceleration of the container is $\frac{x}{10}m/s^2$ Find $x:$

Answer 1

Mass of the liquid inside container at $t=0$ is given by
$m_0=\text{density}\times \text{Volume}=\rho \times hA=\rho hA$
Lets say container starts moving with acceleration $a_o$, then force acting on container $F=m_o{a}_o=\rho hA{a}_o......\left(I\right)$
Volume of the liquid flowing out of the hole per second $=\frac{dV}{dt}=v{A}_h=\sqrt{2gh}\times \frac{A}{100}$
Mass of the liquid flowing out of the hole per second $=\frac{dm}{dt}=\rho \frac{dV}{dt}=\rho \times \sqrt{2gh}\times \frac{A}{100}$
Rate of momentum transferred to the container i.e. force acting on container $F=\frac{dp}{dt}=v\frac{dm}{dt}=\rho \times \left(2gh\right)\times \frac{A}{100}=0.02\rho ghA.......\left(II\right)$
Equating $\left(I\right)$ and $\left(II\right)$, we have
$\rho hA{a}_o=0.02\rho ghA\Rightarrow a_o=0.02g=0.2=\frac{2}{10}m/s^2\Rightarrow x=2$