Question

Sand is poured gently at a constant rate of $0.1\text{kg/s}$ on the trolley of initial mass $2\text{kg}$ kept over a smooth surface at rest. The external force acting on the trolley is $10\text{N}$ as shown in the figure. Then, identify the correct statement(s) for the trolley after time $10\text{sec}$:

• A. Mass of trolley is $3\text{kg}$
• B. Net force on trolley is $\frac{20}{3}\text{N}$
• C. Velocity of trolley is $\frac{100}{3}\text{m/s}$
• D. Velocity of trolley is $\frac{200}{3}\text{m/s}$

Answer 1

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

Velocity of trolley is $\frac{100}{3}\text{m/s}$

Given, initial mass of trolley, $m_0=2\text{kg}$
Rate of mass addition for trolley,
$\frac{dm}{dt}=\mu =0.1\text{kg/s}$
Final mass of trolley after $10\text{s}$ is
$m=m_0+\mu t$, where $t=10\text{s}$
$\Rightarrow m=2+0.1\times 10$
$\Rightarrow m=3\text{kg}$

Let the velocity of the trolley be $v$ at $t=10\text{s}$. Velocity of sand is zero as it is poured gently into the trolley.


$\Rightarrow$ Velocity of sand w.r.t trolley is,
$v_{st}=v_s-v_t=0-v=-v$
The velocity of sand w.r.t trolley is in left direction and mass is added, therefore thrust force $\left(F_t\right)$ will act in the direction of $v_{st}$ i.e towards left.

Hence, net force on trolley,
$F_{net}=F-F_t$
$\Rightarrow F_{net}=F-v\frac{dm}{dt}$
$\Rightarrow ma=F-v\frac{dm}{dt}$
Substituting the values of $m\&v$ in the above equation,
$\Rightarrow (m_0+\mu t)\frac{dv}{dt}=F-\mu v...\left(1\right)$
$\Rightarrow \underset{0}{\overset{v}{\int }}\frac{dv}{F-\mu v}=\underset{0}{\overset{t}{\int }}\frac{dt}{m_0+\mu t}$
Applying limits of velocity $v=0\to v$ and time $t=0\to t$ to get the velocity of trolley at any instant of time $t$
$\Rightarrow {\left[\frac{\ln (F-\mu v)}{-\mu }\right]}_0^v={\left[\frac{\ln (m_0+\mu t)}{\mu }\right]}_0^t$
$\Rightarrow [-\ln (F-\mu v)+\ln F]=\left[\ln \right(m_0+\mu t)-\ln m_0]$

Applying property, $\ln a-\ln b=\ln \frac{a}{b}$
$\Rightarrow \ln \frac{F}{F-\mu v}=\ln \frac{m_0+\mu t}{m_0}$
$\Rightarrow \frac{F}{F-\mu v}=\frac{m_0+\mu t}{m_0}$
Substituting values of $m_0=2\text{kg},\mu =0.1\text{kg/s},t=10\text{s},F=10\text{N}$
$\Rightarrow \frac{10}{10-0.1v}=\frac{2+0.1\times 10}{2}$
$\Rightarrow v=\frac{100}{3}\text{m/s}$ will be the velocity of trolley at $t=10\text{s}$.
Hence, net force on trolley will be,
$F_{\text{net}}=F-F_t$
$\Rightarrow F_{\text{net}}=10-(0.1\times \frac{100}{3})=\frac{20}{3}\text{N}$ will be the net force on the trolley at $t=10\text{s}$.
$\therefore$ options (a), (b), (c) are correct.