Question

A trolley of initial mass $m_0$ is kept over a smooth surface, as shown in the figure. A constant force $F$ is applied on it. Sand kept inside the trolley drains out from its floor at a constant rate of $\mu \text{kg/s}$. After time, $t$ find the velocity of trolley.

• A. $\frac{F}{\mu }\ln \left(\frac{m_0-\mu t}{m_0}\right)$
• B. $\frac{F}{\mu }\ln \left(\frac{m_0}{m_0-\mu t}\right)$
• C. $\frac{F}{2\mu }\ln \left(\frac{m_0-\mu t}{m_0}\right)$
• D. $\frac{F}{2\mu }\ln \left(\frac{m_0}{m_0-\mu t}\right)$

Answer 1

The correct option is B $\frac{F}{\mu }\ln \left(\frac{m_0}{m_0-\mu t}\right)$

Mass of the trolley after time $t$,

$m=m_0-\mu t$

Since, the surface is smooth, the speed of trolley and sand will be the same. So, there will be no thrust force on the trolley.

Therefore, $F_{net}=F$

Let speed of trolley is $v$ after time $t$.

Now, $F_{net}=ma$

$F=(m_0-\mu t)\frac{dv}{dt}$

Rearranging and integrating both sides,

${\int }_0^{v}dv={\int }_0^t\frac{Fdt}{m_0-\mu t}$

$\Rightarrow v=\frac{-F}{\mu }\left[\ln \right(m_0-\mu t){]}_0^t$

$=\frac{-F}{\mu }\ln \left(\frac{m_0-\mu t}{m_0}\right)$

$\therefore v=\frac{F}{\mu }\ln \left(\frac{m_0}{m_0-\mu t}\right)$

Hence, $\left(\text{B}\right)$ is the correct answer.