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.

Using second law of motion, derive the relation between force and acce

1.	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.
Answer» The required relation is `F=ma` Mass of bullet, `m=10g=10^(-2)kg` speed of bullet, `v=10^(3)ms^(-1)` distance travelled, `s=5cm=5xx10^(-2)m` As work done= `K.E` bullet `:. Fxxs=1/2mv^(2)` `F=(mv^(2))/(2s)=(10^(-2)(10^(3))^(2))/(2xx5xx10^(-2))=10^(5)N` From `v=u+at` `0=10^(3)+((-10^(5))/(10^(-2)))xxt, t=10^(3)/(10^(7))=10^(-4)s`