Question

A body of mass m is placed on a smooth horizontal surface. The mass of body is decreasing exponential with disintegration constant lambda. Assuming that the mass is ejected backward with a relative velocity u, and initially the body was at rest, find the velocity of body after time t

Answer 1

Given:

A body of mass m is placed on a smooth horizontal surface. The mass of body is decreasing exponential with disintegration constant lambda. Assuming that the mass is ejected backward with a relative velocity u, and initially the body was at rest.

To find:

Velocity after time t .

Calculation:

As per the relation , the mass of object after time t will be :

$\boxed{ \rm{m = m_{0} \: {e}^{ - \lambda t} }}$

So , change in mass w.r.t time :

$\rm{ \therefore \: \dfrac{dm}{dt} = ( - m_{0} \lambda) \times {e}^{ - \lambda t} }$

Now , thrust force will be F :

$\rm{ \therefore \: F = u \times ( - \dfrac{dm}{dt} )}$

$\rm{ = > \: F = u \times ( m_{0} \lambda {e}^{ - \lambda t} )}$

$\rm{ = > \: m \times ( \dfrac{dv}{dt} ) = u \times ( m_{0} \lambda {e}^{ - \lambda t} )}$

$\rm{ = > \: (m_{0} {e}^{ - \lambda t} )\times ( \dfrac{dv}{dt} ) = u \times ( m_{0} \lambda {e}^{ - \lambda t} )}$

$\rm{ = > \: \cancel{(m_{0} {e}^{ - \lambda t} )}\times ( \dfrac{dv}{dt} ) = u \times ( \cancel{m_{0}} \: \lambda \: \cancel{ {e}^{ - \lambda t}} )}$

$\rm{ = > \dfrac{dv}{dt} = u \lambda}$

$\rm{ = > dv = (u \lambda) \: dt}$

$\displaystyle{ = > \rm{ \int dv = (u \lambda) \int\: dt}}$

$\rm{ = > v = u \lambda t}$

So , final answer is :

$\blue{ \boxed{ \huge{ \bold{ \red{v = u \lambda t}}}}}$