Question

Sand kept inside the trolley drains out from its floor at a constant rate of $0.2\text{kg/s}$. A force $15\text{N}$ is acting on the trolley as shown in figure. After time $5\text{sec}$, which of the following option(s) is/are true?

• A. Mass of trolley is $1\text{kg}$
• B. Net force on trolley is $15\text{N}$
• C. Velocity of trolley is $75\ln 2\text{m/s}$
• D. Velocity of trolley is $75\ln 3\text{m/s}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Mass of trolley is $1\text{kg}$
B Net force on trolley is $15\text{N}$
C Velocity of trolley is $75\ln 2\text{m/s}$

Initial mass of trolley, $m_0=2\text{kg}$
Rate of mass reduction, $\mu =\frac{dm}{dt}=0.2\text{kg/s}$
Final mass of trolley after $5\text{s}$ is
$m=m_0-\mu t$
$\Rightarrow m=2-(0.2\times 5)$
$\Rightarrow m=1\text{kg}$
Let, at $t=5\text{sec}$, velocity of trolley be $v$.
Since sand is drained out from the trolley, its exit velocity will be the same as the velocity of trolley.


Relative velocity with which sand leaves the trolley
$v_{st}=v_s-v_t=v-v=0$
When mass is reducing with time, then $F_t$ acts opposite to the direction of $v_{st}$
As velocity of sand w.r.t trolley is zero,
$F_t=v_{st}\frac{dm}{dt}=0$
Therefore, no thrust force will act,
i.e $F_{net}=F$
$\Rightarrow ma=F$
$\Rightarrow (m_0-\mu t)\frac{dv}{dt}=F$
$\Rightarrow \underset{0}{\overset{v}{\int }}dv=\underset{0}{\overset{t}{\int }}\frac{Fdt}{m_0-\mu t}$
$\Rightarrow v=F{\left[\frac{\ln (m_0-\mu t)}{-\mu }\right]}_0^t$
$\Rightarrow v=\frac{-F}{\mu }\left[\ln \right(m_0-\mu t)-\ln (m_0\left)\right]$
$\Rightarrow v=\frac{-F}{\mu }\ln \frac{m_0-\mu t}{m_0}=\frac{F}{\mu }\ln \frac{m_0}{m_0-\mu t}$
Substituting values of $F=15\text{N},m_0=2\text{kg},\mu =0.2\text{kg/s},t=5\text{s}$
$\Rightarrow v=\frac{15}{0.2}\ln \frac{2}{2-(0.2\times 5)}$
$\therefore v=75\ln 2\text{m/s}$
Hence, net force on trolley is,
$F_{\text{net}}=F=15\text{N}$

$\Rightarrow \left(a\right),\left(b\right),\left(c\right)$ are correct choices.