Question

Question 18
Using second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes sand-bag at a speed of ${10}^{3}m{s}^{-1}$ and gets embedded after travelling 5 cm. Calculate

(i) the resistive force exerted by the sand on the bullet
(ii) the time taken by the bullet to come

Answer 1

(i) Mass, $m=10g=\frac{10}{1000}kg$

Initial velocity, $u={10}^{3}m{s}^{-1}$

Final velocity, $v=0$

Distance $s=\frac{5}{100}m$
Let the acceleration be $a$.

Using, the third equation of motion

$v^2-u^2=2as$

$0-({10}^3{)}^2=2a\times \frac{5}{100}$

$\Rightarrow$$a=\frac{-1000\times 1000}{2.5}\times 100$

$=-{10}^{7}m{s}^{-2}$

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

(ii) Let time be $t$.

Using, the first equation of motion

$v=u+at$

$0={10}^3-{10}^{7}t$

${10}^{7}t={10}^3$

$\Rightarrow$ $t=\frac{{10}^3}{{10}^7}$$={10}^{-4}s$