Question

A bullet of mass $10g$ travelling horizontally with a velocity of $150m{s}^{-1}$ 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

Given,

Mass = $m=10g=10\times {10}^{-3}kg$,

Final velocity = $v=0$,

Initial velocity = $u=150m/s$,

Time = $t=0.03s$.

Let the uniform deceleration of bullet be 'a'.

Using the first equation of motion we get,

$\Rightarrow v=u+at$

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

$\Rightarrow a=\frac{150}{0.03}=-5000m/{\sec }^2$

Using the third equation of motion we get,

$s=\frac{v^2-u^2}{2a}=\frac{0^2-(150{)}^2}{2(-5000)}=\frac{225}{100}=2.25m$.

Therefore, the distance of penetration of the bullet into the block is 2.25 m.

We know that,

$F=ma$

$F=10\times {10}^{-3}\times 5\times {10}^3$

$F=50N$

Therefore, the magnitude of the force exerted by the wooden block on the bullet is 50 N.