Question

A cart of mass $M_0$ is moving with velocity $v_0$. At $t=0$, water starts pouring into the cart from a container above the cart at the rate of $\lambda \text{kg/sec}$. Find the velocity of the cart as a function of time.

• A. $\frac{M_0{v}_0}{M_0-\lambda t}$
• B. $\frac{M_0{v}_0}{M_0+\lambda t}$
• C. $\frac{M_0{v}_0}{M_0+2\lambda t}$
• D. None of these

Answer 1

The correct option is B $\frac{M_0{v}_0}{M_0+\lambda t}$

Mass of water poured per second = rate of change of mass of cart $=\frac{dm}{dt}=\lambda$
Let the velocity of cart be $v$ at any time $t$. Its initial velocity is $v_0$.

The thrust force on the cart is given by
$F=-\left(\frac{dm}{dt}\right)\left(u_{\text{rel}}\right)$
where $u_{\text{rel}}=v$ is the velocity of cart relative to the container in the horizontal direction.
$\therefore F=-\lambda v$

Now the net mass of the cart at any time $t$ is $(M_0+\lambda t)\text{kg}$
Applying Newton's second law for cart in horizontal direction, considering thrust force $\left(F\right)$ on it, as the net external force:
$F=m\frac{dv}{dt}$
$\Rightarrow -\lambda v=(M_0+\lambda t)\frac{dv}{dt}$
Integrating both sides,
$\underset{0}{\overset{t}{\int }}\frac{-\lambda dt}{(M_0+\lambda t)}=\underset{v_0}{\overset{v}{\int }}\frac{dv}{v}$
$\Rightarrow {\begin{array}{c}\frac{-\lambda }{\lambda }\ln (M_0+\lambda t)\end{array}|}_0^t={\begin{array}{c}\ln v\end{array}|}_{v_0}^v$
$\Rightarrow -\left[\ln \right(M_0+\lambda t)-\ln M_0]=\ln \frac{v}{v_0}$
$\Rightarrow -\ln \left(\frac{M_0+\lambda t}{M_0}\right)=\ln \left(\frac{v}{v_0}\right)$
Or, $\ln \left(\frac{M_0}{M_0+\lambda t}\right)=\ln \left(\frac{v}{v_0}\right)$
$\Rightarrow \frac{M_0}{M_0+\lambda t}=\frac{v}{v_0}$
$\therefore v=\frac{M_0{v}_0}{M_0+\lambda t}$