Question

In the figure shown a cart moves on a smooth horizontal surface due to an external constant force of magnitude $F$. The initial mass of the cart is $M_0$ and velocity is zero. Sand falls on to the cart with negligible velocity at constant rate $\mu \text{kg/s}$ and sticks to the cart. The velocity of the cart at time $t$ is:

• A. $\frac{Ft}{M_0+\mu t}$
• B. $\frac{Ft}{M_0}{e}^{\mu t}$
• C. $\frac{Ft}{M_0}$
• D. $\frac{Ft}{M_0+\mu t}{e}^{\mu t}$

Answer 1

The correct option is A $\frac{Ft}{M_0+\mu t}$


Let, $dm$ is the mass of sand added in time $dt$.

The force acting on the cart at any time $t$ when its velocity is $v$,

$F=m\frac{dv}{dt}+(v-u)\frac{dm}{dt}....\left(1\right)$

Here,
$u=$ initial velocity of sand $=0$,
$m=M_0+\mu t=$ mass at time $t$,
and $\frac{dm}{dt}=\mu$

Substituting the above values, in $\left(1\right)$

$F=(M_0+\mu t)\frac{dv}{dt}+v\mu$

$\Rightarrow (F-\mu v)dt=(M_0+\mu t)dv$

Rearranging and integrating the equation, we get

$\underset{0}{\overset{t}{\int }}\frac{dt}{M_0+\mu t}=\underset{0}{\overset{v}{\int }}\frac{dv}{F-\mu v}$

$\Rightarrow \frac{1}{\mu }\left[\text{ln}\right(M_0+\mu t){]}_0^t=-\frac{1}{\mu }[\text{ln}(F-\mu v){]}_0^v$

$\Rightarrow \text{ln}\frac{(M_0+\mu t)}{M_0}=\text{ln}\left(\frac{F}{F-\mu v}\right)$

$\Rightarrow F-\mu v=\frac{M_{0}F}{M_0+\mu t}$

$\Rightarrow v=\frac{Ft}{M_0+\mu t}$

Hence, option (a) is correct answer.