Question

A bullet fired towards a wall reduces its half kinetic energy after the penetration of 6cm

Answer 1

Given -bullet fired into target loses half of its velocity after penetrating 3 cm.

So let us take, initial velocity, u = 2w and after s = 3cm travel, final velocity, v = w ( from given)

now from the equation,

(v^2) = (u^2) + 2as (Kinematic equation)

(w^2) = (4*(w^2)) + (2*a*3)

(w^2) - (4*(w^2)) = 2*a*3

-(3*(w^2)) = 2*a*3

-(w^2) = 2*a

a = -(w^2)/2 (object get decelerated)

Asked to find, further peneration distance before the object comes to rest.

let us take

u = w, v =0 and a = -(w^2)/2

(v^2) = (u^2) + 2as

0 = (w^2) + (2*(-(w^2)/2)*s)

(w^2) = (w^2)*s

s = 1 cm

so, the objects penerates further 1 cm with above found deceleration before it comes to rest.

:) Stay happy always :)