Question

Having gone through a plank of thickness $h$, a bullet changed its velocity from $v_0$ to $v$. The time of motion of the bullet in the plank is , assuming the resistance force to be proportional to the square of the velocity. Find $x$.

Answer 1

According to the problem
$m\frac{dv}{dt}=-k{v}^2$ or, $m\frac{dv}{v^2}=-kdt$
Integrating within the limits,
${\int }_{v_0}^v\frac{dv}{v^2}=-\frac{k}{m}{\int }_0^{t}dt$ or, $t=\frac{m}{k}\frac{(v_0-v)}{v_{0}v}$ (1)
To find the value of $k$, rewrite
$m\frac{dv}{ds}=-k{v}^2$ or, $\frac{dv}{v}=-\frac{k}{m}ds$
On integrating
${\int }_{v_0}^v\frac{dv}{v}=-\frac{k}{m}{\int }_0^{k}ds$
So, $k=\frac{m}{h}\ln \frac{v_0}{v}$ (2)
Putting the value of $k$ from (2) in (1), we get
$t=\frac{h(v_0-v)}{v_{0}v\ln \frac{v_0}{v}}$

$\therefore x=1$