A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm . How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

Options

A. ` 1.0 cm`
B. ` 1.5 cm `
C. ` 2.0` cm
D. `3.0 cm`

Answer

A

Explanation

Let initial velocity of the bullet = u
After penetrating 3 cm its velocity becomes ` (u)/(2)`
From ` v^(2) =u^(2) -2as ,` we get
`((u)/(2))^(2) =u^(2) -2a(3) rArr " "6a =(3u^(2))/(4) rArr a= (u^(2))/(8) `
Let further it will penetrate through distance x and stops at point C. For distance ` BC, v=0 ,u =u//2 ,s =x,a =u^(2) //8`
From ` v^(2) =u^(2) -2as , 0=((u)/(2) ) ^(2) -2((u^(2))/(8)) .x rArr x=1 cm `