# 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.

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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 $10^{-4}\ s$.

Updated on 10-Oct-2022 13:28:47