Using second law of motion, derive the relation between force and acceleration. A bullet of `10g` strikes a sand-bag at a speed of `10^(3)ms^(-1)` and gets embedded after travelling `5cm`.Calculate
(i) the resistive force exerted by the sand on the bullet (ii) the time taken by the bullet to come to rest.

Explanation

If a body of mass (m), moving at velocity (v) accelerates uniformity at (a) for time 't'. So that its velocity changes to v, then
Initial momentum `p_(1)=mu`
and final mementum `p_(2)=mu`
Change in momentum=`p_(2)-p_(1)=mv-mu=m(u-v)`
According to the second law of motion, force `F alpha ("Change in momentum")/("Time")`
`Rightarrow F alpha (p_(2)-p_(1))/(t)`
`Rightarrow F alpha (m(v-u))/(t)`
`Rightarrow F alpha ma ("where", (v-u)/(t)=a)`
`"Here" F-kma`
`therefore F=ma`
Given, m=10g=`(10)/(1000)kg=0.01kg.u=10^(3)m//s, v=0 [therefore 1kg=1000g]`
and , `s=5cm=(5)/(100)m=0.05m [therefore 1m=100cm]`
`"(i) From the third equation of motion" v^(2)=u^(2)+2as`
`Rightarrow v^(2)-u^(2)=2as`
`Rightarrow a=(v^(2)-u^(2))/(2s)=((0)^(2)-(10^(3))^(2))/(2xx0.05)`
`a=(-10^(6))/(-.1)=10^(7)m//s`
Now, force applied by the bullet. F=ms =`0.01xx10^(-7)=-10^(5)N` [negative sign shown against the direction of motion]
The resistive force exerted by the sand on the bullet=`10^(5)N`
(ii) From first equation of motion. `v=u+at`
`Rightarrow at=v-u`
Now, time taken by bullet come to rest. `t=(v-u)/(a)`
`Rightarrow t=(0-10^(3))/(-10^(7))`
`Rightarrow t=(-10^(3))/(-10^(7))=10^(3)xx10^(-7)`
`Rightarrow t=10^(-4)s`
Hence, force (F) =`10^(5)N` and time=`(t)=10^(-4)s`