Question

A flat cart of mass $m_0$ starts moving to the right due to a constant horizontal force F at $t=0$. Sand spills on the flat cart from a stationary hopper. The velocity of loading is constant and is equal to $\mu$ kg/s. Find the time dependence of the velocity and the acceleration of the flatcar in the process of loading. The friction is negligibly small.

• A. Initial acceleration is equal to $F/m_0$
• B. Acceleration at time t is $F/(m_0+\mu t)$
• C. Kinetic energy of loaded cart at an instant is equal to work done by force F up to that instant
• D. Momentum of loaded cart at an instant is equal to impulse of force F up to that instant

Answer 1

The correct option is B Acceleration at time t is $F/(m_0+\mu t)$

At $t=0,v=0$ and $m=m_0$

At $t=t,v=v$ and $m_0+\mu t$

Here $v\left(rel\right)=u-v=0-v=-v$

Here $u$ is the velocity of mass being added in the horizontal direction which is zero ;

$\frac{dm}{dt}=\mu$

applying, $F+v\left(rel\right)=\frac{dm}{dt}=\frac{mdv}{dt}$ we get

$F+(-v)\mu =(m_0+\mu t)\frac{dv}{dt}$

or $\int \frac{dv}{(f-\mu v)}=\int \frac{dt}{(m_0+\mu t)}$

Integrating $v$ from $0$ to $v$ and $t$ from $0$ to $t$ we get

$ln\left[\frac{F}{(F-\mu v)}\right]=ln\left[\frac{(m_0+\mu t)}{m_0}\right]$

or, $v=\frac{Ft}{(m_0+\mu t}....\left(i\right)$ This is the time dependence of velocity;

Hence the acceleration of the flat car is,

$a=\frac{(F-\mu v)}{m}$

Putting value of $v$ from equation $\left(i\right)$

$a=\frac{F{m}_0}{(m_0+\mu t{)}^2}$