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Using second law of motion, derive the relation between force and acceleration. A bullet of $10\ g$ strikes a sand-bag at a speed of $10^3\ ms^{-1}$ 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 to rest.
Let an object of mass $m$ starts moving with an initial speed $u$ and in time $t$ its velocity becomes $v$,
Then, initial momentum $P_1=mu$
Final momentum $P_2=mv$
Change in momentum $=P_2-P_1=mv-mu$
Rate of change of momentum $=\frac{change\ in\ momentum(P_2-P_1)}{time(t)}$
$=\frac{mv-mu}{t}$
$=\frac{m(v-u)}{t}$ .................. $(i)$
By using, the equation of motion $v=u+at$
$a=\frac{v-u}{t}$, on putting this value in $(i)$
Rate of momentum change $=ma$
According to Newton's second law of motion, the force applied is directly proportional to the rate of change of momentum.
$F\propto ma$
Or $F=Kma$ [$K$ is a constant, its value considered to be $1$]
Or $F=ma$
$(i)$. Here, mass of the bullet $m=10\ g=\frac{10}{1000}\ kg=10^{-2}\ kg$
Initial speed $u=10^3\ m/s$
Final speed $v=0$
Distance travelled $s=5\ cm=\frac{5}{100}\ m=5\times 10^{-2}\ m$
On using third equation of motion $v^2=u^2+2as$
$2as=v^2-u^2$
Or $2\times a\times 5\times10^{-2}=0^2-(10^3)^2$
Or $a\times10^{-1}=0-10^6=-10^6$
Or $a=\frac{-10^6}{10^{-1}}$
Or $a=-10^{-7}\ m/s^2$ [$-ve$ sign indicates retadation]
So, force exerted by the bullet on sand $=ma=10^{-2}\times(-10^{7}$
$=-10^5\ N$
Therefore, resistive force exerted by sand on bullet $=-$force exrted by bullet on sand [Newton's third law of motion]
$=-(-10^5\ N)$
$=10^5\ N$
$(ii)$. Let the bullet takes time $t$ to come into rest
On using first equation of motion, $v=u+at$
$0=10^3+(-10^7)\times t$
Or $t=\frac{-10^3}{10^7}$
Or $t=10^{-4}\ s$
Therefore, the bullet will come into rest after $10^{-4}\ s$.
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