Question

A ball of mass $m$ falls from rest under gravity in a resisting medium (Exerting drag during motion). The resistance force is given by $f=kv$, which can be considered as a friction force offered by the resisting medium. The correct velocity $\left(v\right)$ versus time $\left(t\right)$ graph for body is:

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given,
Weight of the body$=mg$,
Opposite drag force acting on the body, $f=-kv$

Now, the net force on the body can be given by,
$$F_{net}=mg-f$$If $a$ is acceleration on the body due to this net force, then

$F_{net}=ma=mg-kv$

$\Rightarrow m\frac{dv}{dt}=mg-kv$

$\Rightarrow \frac{m}{k}\left(\frac{dv}{dt}\right)=\frac{mg}{k}-v$

Taking variable $v$ on one side and time on other side and assuming the velocity is $v$ at time $t$.

$\Rightarrow {\int }_0^v\frac{dv}{(\frac{mg}{k}-v)}=\frac{k}{m}{\int }_0^{t}dt$


$\Rightarrow [\ln (\frac{mg}{k}-v)\times (-1\left)\right]{|}_0^v=\frac{k}{m}\left[t\right]{|}_0^t$

$\Rightarrow \ln \left(\frac{mg-kv}{k}\right)-\ln \left(\frac{mg}{k}\right)=\frac{-k}{m}t$

$\Rightarrow \ln \left(\frac{mg-kv}{mg}\right)=\frac{-k}{m}t$

$\Rightarrow 1-\frac{kv}{mg}=\exp \left(\frac{-kt}{m}\right)$

$\Rightarrow v=\frac{mg}{k}[1-\exp \left(\frac{-kt}{m}\right)]$

From the above expression for velocity, we can interprete that $v$ varies with time in exponential manner.

At $t=0$,

$v=\frac{mg}{k}[1-\exp \left(\frac{-k\times 0}{m}\right)]=\frac{mg}{k}[1-1]=0$

And at $t=\infty$,

$v=\frac{mg}{k}[1-\exp (-\infty )]=\frac{mg}{k}$

$(\because \exp (-\infty )=0)$

Hence, option (a) represents correct graph for $\left(v\right)$ vs $\left(t\right)$.