Question

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 1

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`