Question

A freight car is moving on smooth horizontal track without any external force. Rain is falling at the rate of u m/s at an angle $\theta$ with the vertical. Rain drops are collected in the car at the rate of $\mu$ kg/s. If initial mass of the car is $m_0$ and velocity $v_0$ then find its velocity after time it.

• A. $v=2ucos\theta (\frac{m_0}{m_0+\mu t}-1)+\frac{m_0{v}_0}{(m_0+\mu t)}$
• B. $v=usin\theta (\frac{m_0}{m_0+\mu t}-1)+\frac{3m_0{v}_0}{(m_0+\mu t)}$
• C. $v=usin\theta (\frac{m_0}{m_0+\mu t}-1)+\frac{m_0{v}_0}{(m_0+\mu t)}$
• D. $Zero$

Answer 1

The correct option is C

$v=usin\theta (\frac{m_0}{m_0+\mu t}-1)+\frac{m_0{v}_0}{(m_0+\mu t)}$

After time t mass of the car with water is $m=(m_0+\mu t)$ kg. Let at that moment speed of the car be v.
$\Rightarrow v_{relative}=(-usin\theta -v)m\frac{dv}{dt}=F_{ext}+V_{rel}\frac{dm}{dt}As(m_0+\mu t)\frac{dv}{dt}=0-(usin\theta +v)\mu \Rightarrow {\int }_{v_0}^v\frac{dv}{(usin\theta +v)}=-{\int }_0^t\frac{dt}{(\frac{m_0}{\mu }+t)}\Rightarrow ln\left(\frac{usin\theta +v}{usin\theta +v_0}\right)=-ln\left[\frac{(\frac{m_0}{\mu }+t)}{\frac{m_0}{\mu }}\right]\Rightarrow \frac{usin\theta +v}{usin\theta +v_0}=\frac{m_0}{(m_0+\mu t)}\Rightarrow v=usin\theta (\frac{m_0}{m_0+\mu t}-1)+\frac{m_0{v}_0}{(m_0+\mu t)}$