Question

A bullet of $10g$ strikes sand-bag at a speed of ${10}^{3}m/s$ and comes to stop after travelling 5 cm. Calculate:
(i) The resistive force exerted by the sand on the bullet.
(ii) Time of collision.

Answer 1

(i)
Given:
Mass, ($m)=10g=\frac{10}{1000}kg=0.01kg$
Initial velocity, $u={10}^{3}m/s$
Stopping distance $s=\frac{5}{100}m=0.05m$
Since the bullet finally stops, final velocity, $v=0$

Let the acceleration be $a$.

Using the equation of motion,
$v^2-u^2=2as$
$0-({10}^3{)}^2=2a\times 0.05$
$a=-\frac{{10}^6}{0.1}$
$a=-{10}^7{ms}^{-2}$
Note that the negative value of acceleration shows that the bullet is slowing down.

According to Newton's second law of motion force,
$F=ma$
$F={10}^{-2}\times {10}^7={10}^{5}N$

(ii)
Let the time for stopping be $t$.

Using the equation of motion:
$v=u+at$
$0={10}^3-{10}^{7}t$
$t=\frac{{10}^3}{{10}^7}={10}^{-4}s$