Question

A bullet of mass $10g$ travelling horizontally with a velocity of $150{ms}^{-1}$ strikes a stationary wooden block and comes to rest in $0.03s$. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer 1

Initial velocity of the bullet, $u=150$ $m{s}^{-1}$

Final velocity of the bullet, $v=0$

Mass of the bullet, $m=10g=0.01kg$

Time taken by the bullet to come to rest, $t=0.03s$

Let distance of penetration be $s$

$v=u+at$

$0=150+a\left(0.03\right)$

$⟹a=-5000{ms}^{-2}$ (- sign shows retardation)

$v^2=u^2+2as$

$0={150}^2+2\times (-5000)\times s$

$s=2.25m$

Magnitude of the force exerted by the wooden block, $|F|=m|a|$

$|F|=0.01\times 5000=50N$