Question

Question 14
A bullet of mass 10 g travelling horizontally with a velocity of $150m{s}^{-1}$ 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.

Answer 1

Given, initial velocity, u = 150 $m{s}^{-1}$, final velocity, $v=0$ (since the bullet finally comes to rest), mass of the bullet, $m=10g=0.01kg$ and time taken to come to rest, $t=0.03s$.
Let the acceleration of a bullet be $a$.
We use first equation of motion, $v=u+at$,
$\Rightarrow$$0=150+a\times 0.03$
$\Rightarrow$a = $\frac{-150}{0.03}=-5000m{s}^{-2}$
(Negative sign indicates that the velocity of the bullet is decreasing.)
Now we use third equation of motion:
$v^2=u^2+2as$,
$\Rightarrow$$0={150}^2+2\times (-5000)s$
$\Rightarrow$$s=\frac{22500}{10000}=2.25m$
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton's second law of motion:
Force, $F=m\times a=0.01\times 5000=50N$
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.