Question

A flatcar of mass $m_0=20\text{kg}$ is moving to the right due to a constant horizontal force $F=10\text{N}$. Sand spills on the flatcar from a stationary hopper. The velocity of loading is constant and equal to $\mu =1\text{kg/s}$. Find the time dependence of the velocity of the flatcar in the process of loading. The friction is negligibly small.

• A. $\frac{10t}{(20+t{)}^2}$
• B. $0$
• C. $\frac{20t}{20+t}$
• D. $\frac{10t}{20+t}$

Answer 1

The correct option is D $\frac{10t}{20+t}$

Given, $F=10\text{N}$
For a variable mass system where mass is added at a rate $\mu$,
$m\cdot \frac{d\overrightarrow{v}}{dt}$ =$\overrightarrow{F_{ext}}+\mu u$
where $v$ = velocity of car, $m_0$ = mass of car $=20\text{kg}$ and $\mu =1\text{kg/s}$
Here, $m=m_0+\mu t$ after $t$ seconds
and relative velocity of mass added $u=-v$
(in horizontal direction, sand has zero velocity)
$\Rightarrow (m_0+\mu t)\frac{dv}{dt}=\overrightarrow{F}-\mu v$

Rearranging the terms,
$\Rightarrow m_{0}dv+\mu tdv=\overrightarrow{F}dt-\mu vdt$
$\Rightarrow m_{0}dv+\mu (tdv+vdt)=\overrightarrow{F}dt$
Intergrating $v$ from $0$ to $v$ and $t$ from $0$ to $t$:
$\Rightarrow \int m_{0}dv+\int \mu (tdv+vdt)=\int \overrightarrow{F}dt$
$\Rightarrow m_{0}v+\int \mu d\left(vt\right)=\overrightarrow{F}t$
$\Rightarrow m_{0}v+\mu vt=\overrightarrow{F}t$
or $\overrightarrow{v}=\frac{\overrightarrow{F}t}{(m_o+\mu t)}=\frac{10t}{20+t}$