Question

A cart filled with sand of mass $m_0=50\text{kg}$ is moving with constant velocity because of a constant force $F=10\text{N}$ acting in the direction of the cart's velocity vector. Sand starts spilling through a hole in the bottom with a constant velocity at the rate of $\mu =10\text{kg/s}$.. Find the velocity of the cart at the time $t=4\text{s}$. Friction is to be neglected. [Take $\ln \left(5\right)=1.61$]

• A. $2.7\text{m/s}$
• B. $1.61\text{m/s}$
• C. $3.2\text{m/s}$
• D. $16.1\text{m/s}$

Answer 1

The correct option is B $1.61\text{m/s}$


For a variable mass system with constant rate of ejection $\frac{dm}{dt}$,
$\overrightarrow{F_{ext}}-\overrightarrow{u}\frac{dm}{dt}=m\frac{\overrightarrow{dv}}{dt}$
$\overrightarrow{F_{ext}}=m\frac{\overrightarrow{dv}}{dt}+\overrightarrow{u}\frac{dm}{dt}$

Here, relative velocity of ejection $\overrightarrow{u}=0$ (since sand has same horizontal velocity component as the cart)

Initially at $t=0,m=m_0$
After time $t$, $m=m_0-\mu t$

$\Rightarrow F=m\frac{dv}{dt}$
$\Rightarrow m\left(dv\right)=Fdt$

Integrating both sides and putting the value of $m$
${\int }_0^{v}dv={\int }_0^t\frac{F}{m_0-\mu t}\cdot dt$
$\Rightarrow v=\frac{F}{\mu }\left[\ln \left(\frac{m_0}{m_0-\mu t}\right)\right]$
At $t=4\text{s}$,
$\overrightarrow{v}=\frac{10}{10}\left[\ln \left(\frac{50}{50-10\times 4}\right)\right]=1.61\text{m/s}$