Question

8.
A bullet moving with a speed of 150 m/s strikes a tree and penetrates 3.5
cm before stopping. The magnitude of acceleration and time takes to stop​

Answer 1

Given,

$\sf{u=150\ m\ s^{-1}}$

$\sf{s=3.5\ cm=3.5\times10^{-2}\ m}$

$\sf{v=0\ m\ s^{-1}}$

The magnitude of the retardation (not acceleration) acting on the bullet is,

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

$\longrightarrow\sf{a=-\dfrac{0^2-150^2}{2\times3.5\times10^{-2}}\ m\ s^{-2}}$

$\longrightarrow\sf{\underline{\underline{a=321428.57\ m\ s^{-2}}}}$

Negative sign is only to obtain the magnitude. Actual retardation is,

$\longrightarrow\sf{a=-321428.57\ m\ s^{-2}}$

And the time taken for the bullet being stopped,

$\longrightarrow\sf{t=\dfrac{v-u}{a}}$

$\longrightarrow\sf{t=\dfrac{0-150}{-321428.57}\ s}$

$\longrightarrow\sf{\underline{\underline{t=4.67\times10^{-4}\ s}}}$