Question

A bullet after getting fired from a gun goes through a plank of thickness h and changes its velocity from u to v and resisting force is proportional to cube of velocity. The time of motion of the bullet in the plank is proportional to

• A. $(u^2-v^2)u^{-2}$
• B. $(u-v)(uv{)}^{-1}$
• C. $(u^2-v^2)u^{-1}$
• D. $(u^2-v^2)(uv{)}^{-2}$

Answer 1

The correct option is D $(u^2-v^2)(uv{)}^{-2}$

$F\alpha v^3$

$F=k{v}^3$

$a=\frac{F}{m}=\frac{k{v}^3}{m}$

We know that,

$a=\frac{dv}{dt}$

$\frac{dv}{dt}=\frac{k{v}^3}{m}$

$\frac{1}{v^3}dv=\left(\frac{k}{m}\right)dt$

${\left[\frac{v^{-2}}{-2}\right]}_u^v=\left(\frac{k}{m}\right){\int }_0^{t}dt$

$-\frac{1}{2}{\left[\frac{1}{v^2}\right]}_u^v=\left(\frac{k}{m}\right)[t{]}_0^t$

$-\frac{1}{2}\left[\frac{u^2-v^2}{v^2{u}^2}\right]=\left(\frac{k}{m}\right)t$

$t\alpha \left(\frac{u^2-v^2}{v^2{u}^2}\right)$

$t\alpha (u^2-v^2)(uv{)}^{-2}$