Question

Having gone through a plank of thickness $h$, a bullet changes its speed from $V_0$ to $\frac{V_0}{2}$. Find the time of motion of the bullet in the plank assuming the resistant force to be proportional to the speed.

• A. $\frac{2hln2}{V_0}$
• B. $\frac{hln2}{V_0}$
• C. $\frac{2h}{V_0}$
• D. None of these

Answer 1

The correct option is A $\frac{2hln2}{V_0}$

Let us assume that bullet covers the distance $h$ in time $t$.


It is given that the resistant force is proportional to the speed,
$F_R\propto V$

$\Rightarrow F_R=-k^{{}^{\prime }}V$

Where, $-\text{ve}$ sign indicates that the force is resistive in nature and $k^{{}^{\prime }}$ is any constant

Or, $a=\frac{-k^{{}^{\prime }}}{m}V=-kV$

$a=-kV.....\left(1\right)$

Also as $a=\frac{dV}{dt}$

We have

$a=\frac{dV}{dt}=\frac{dV}{dt}\frac{dx}{dx}=\frac{dV}{dx}\frac{dx}{dt}$

$\Rightarrow a=\frac{dV}{dx}V=-kV.....\left(2\right)$

On solving $\left(1\right)$ and $\left(2\right)$ we get,

$\begin{array}{cc}V\frac{dV}{dx}=-kV& a=-kV\\ dV=-kdx& \frac{dV}{dt}=-kV\\ {\int }_{V_0}^{\frac{V_0}{2}}dV=-k{\int }_0^{h}dx& {\int }_{V_0}^{\frac{V_0}{2}}\frac{dV}{v}=-k{\int }_0^{t}dt\\ -\frac{V_0}{2}=-kh& ln\frac{V_0}{2}-ln{V}_0=-kt\\ k=\frac{V_0}{2h}\dots \dots \left(1\right)& t=\frac{ln2}{k}\dots \dots \left(2\right)\end{array}$

$\text{From (1) and (2)}$
$t=\frac{2hln2}{V_0}$