Question

A flatcar having mass $m_0=50\text{kg}$, is moving to the right due to a constant horizontal force $F=20\text{N}$. Raindrops fall perpendicularly on the flatcar at a constant rate equal to $\mu =10\text{kg/s}$. Find the acceleration of the flatcar (at $t=10\text{sec}$). Assume friction is negligibly small.

• A. $0.044{\text{m/s}}^2$
• B. $0.011{\text{m/s}}^2$
• C. $0.022{\text{m/s}}^2$
• D. $0{\text{m/s}}^2$

Answer 1

The correct option is A $0.044{\text{m/s}}^2$

Given, initial velocity of flatcar is zero. Let velocity be $v$ after time $t=10\text{sec}$ and mass be $m$.


At time $t=0$, $v=0\&m=m_0$

$v_r=$ relative speed of loading $=-v$
(since the rain drops are falling perpendicularly, their horizontal component of velocity is zero)
Hence, thrust force $F_t=v_r\frac{dm}{dt}=-v\mu$



Net force on the flatcar at time $t$ is
$F_{net}=F_{ext}+F_t$
$\Rightarrow m\frac{dv}{dt}=F-\mu v$ ...(I)
At time $t$, $m=m_0+\mu t$
$\Rightarrow (m_0+\mu t)\frac{dv}{dt}=F-\mu v$

Integrating both sides,
${\int }_0^v\frac{dv}{F-\mu v}={\int }_0^t\frac{1}{m_0+\mu t}dt$
$\Rightarrow \frac{-1}{\mu }\ln \left[\frac{F-\mu v}{F}\right]=\frac{1}{\mu }\ln \left[\frac{m_0+\mu t}{m_0}\right]$
$\Rightarrow \frac{F}{F-\mu v}=\frac{m_0+\mu t}{m_0}$
or $v=\frac{Ft}{m_0+\mu t}$ ...(II)

From eq (I), $\frac{dv}{dt}=$ acceleration of flatcar at time $t$
or $\frac{dv}{dt}=\frac{F-\mu v}{m}$
$\Rightarrow a=\frac{F-\mu \left(\frac{Ft}{m_0+\mu t}\right)}{m_0+\mu t}=\frac{m_{0}F}{(m_0+\mu t{)}^2}$

At $t=10\text{s}$,
$a=\frac{50\times 20}{(50+10\times 10{)}^2}$
$\therefore a=0.044{\text{m/s}}^2$