Question

The velocity of a bullet changes from $v_0$ to $v$ after the bullet has passed through a distance $h$ in the plank. Assuming resistance offered is proportional to $v^2$, find the time of motion in the plank.

• A. $t=\frac{v_0-v}{h\log \frac{v_0}{ev}}$
• B. $t=\sqrt{\frac{h^2}{v_{0}v}\log \frac{v_0}{cv}}$
• C. $t=\frac{{(v}_0-v)h}{v_{0}v{\log }_e\frac{v_0}{v}}$
• D. $t=\frac{{(v}_0-v)h}{v_{0}v}$

Answer 1

The correct option is C $t=\frac{{(v}_0-v)h}{v_{0}v{\log }_e\frac{v_0}{v}}$

Here, $F=-k{v}^2\Rightarrow m\frac{dv}{dt}=-k{v}^2$
integrating, ${\int }_{v_0}^v\frac{dv}{v^2}=-\frac{k}{m}{\int }_0^{t}dt$
$(\frac{1}{v}-\frac{1}{v_0})=\frac{k}{m}t.....\left(1\right)$
agian,
$m\frac{dv}{dx}.\frac{dx}{dt}=-k{v}^2$
$m\frac{dv}{dx}.v=-k{v}^2$
${\int }_{v_0}^v\frac{dv}{v}=-\frac{k}{m}{\int }_0^{h}dx$
$lo{g}_e\frac{v_0}{v}=\frac{k}{m}h......\left(2\right)$

Dividing (1) by (2), we get:

$t=\frac{(\frac{1}{v}-\frac{1}{v_0})}{lo{g}_e\frac{v_0}{v}}h=\frac{(v_0-v)h}{v{v}_{0}lo{g}_e\frac{v_0}{v}}$