# A bullet of mass 10 g travelling horizontally with a velocity

Question.
A bullet of mass 10 g travelling horizontally with a velocity of 150 ms–1 strikes a stationary wooden block and comes to rest in 0.03 s. 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.

Solution:

Mass of bullet, $\mathrm{m}=10 \mathrm{~g}=0.01 \mathrm{~kg} ;$ initial velocity of bullet, $\mathrm{u}=150 mathrm{~ms}^{-1} ;$ final velocity of
bullet, $\mathrm{v}=0 ;$ time, $\mathrm{t}=0.03 \mathrm{~s} ;$ acceleration of bullet, $\mathrm{a}=? ;$ force exerted by wooden block,

$\mathrm{F}=? ;$ distance penetrated by bullet, $\mathrm{s}=?$

We know, $v=u+$ at

or $\quad 0=150+\mathrm{a} \times 0.03$

or $\quad-\mathrm{a} \times 0.03=150$

or $\quad a=\frac{-150}{0.03}=-5000 \mathrm{~ms}^{-2}$

We know, $s=u t+\frac{1}{2} a t^{2}$

$=150 \times 0.03+\frac{1}{2} \times(-5000) \times(0.03)^{2}$

$=4.5-2.25=2.25 \mathrm{~m}$

We know, $\mathrm{F}=\mathrm{ma}$

Force acting on bullet,

$\mathrm{F}=0.01 \times(-5000)=-50 \mathrm{~N}$

Negative sign denotes that wooden block exerts force in a direction opposite to the direction of motion of the bullet.