Question

A bullet is moving with a velocity v collide a stationary block and losees its 10% energy by travelling 1cm distance find the distance of penitration

Answer 1

$\displaystyle\large\boxed {\sf {Distance\ of\ Penetration=10\ cm}}$

By travelling 1 cm it loses 10% of its kinetic energy actually. Thus, change in kinetic energy,

$\displaystyle\longrightarrow\sf{\Delta K=-\dfrac {1}{10}\cdot\dfrac {1}{2}mu^2}$

[10% = 1/10. Negative sign is because final kinetic energy is less than initial one. Loss of energy is there.]

$\displaystyle\longrightarrow\sf{\Delta K=-\dfrac {1}{20}mu^2}$

By work energy theorem,

$\displaystyle\longrightarrow\sf{Fs=\Delta K}$

$\displaystyle\longrightarrow\sf{Fs=-\dfrac {1}{20}mu^2}$

Here, $\displaystyle\sf {s=1\ cm=10^{-2}\ m.}$ Then,

$\displaystyle\longrightarrow\sf{10^{-2}F=-\dfrac {1}{20}mu^2}$

$\displaystyle\longrightarrow\sf{F=-\dfrac {1}{20\times 10^{-2}}mu^2}$

$\displaystyle\longrightarrow\sf{ma=-5mu^2}$

$\displaystyle\longrightarrow\sf{a=-5u^2}$

Thus, by third kinematic equation, the distance of penetration,

$\displaystyle\longrightarrow\sf{s=\dfrac {v^2-u^2}{2a}}$

Penetration stops when the velocity becomes zero, i.e., $\displaystyle\sf {v=0.}$ Thus,

$\displaystyle\longrightarrow\sf{s=\dfrac {0^2-u^2}{2\times-5u^2}}$

$\displaystyle\longrightarrow\sf{s=\dfrac {-u^2}{-10u^2}}$

$\displaystyle\longrightarrow\sf{\underline {\underline {s=\dfrac {1}{10}\ m}}}$

$\displaystyle\longrightarrow\sf{\underline {\underline {s=0.1\ m}}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {s=10\ cm}}}$