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

Options

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

Answer

B

Explanation

Let initial velocity of the bullet `= u`
After penetrating `3 cm` its velocity becomes `(u)/(2)`.
From `v^2 = u^2 - 2as`
`((u)/(2))^2 = u^2 - 2a(3)`
`rArr 6a = (3 u^2)/(4) rArr a = (u^2)/(8)`
Let it further 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 rArr 0 = ((u)/(2))^2 - 2 ((u^2)/(8)) x`
`rArr x = 1 cm`.