Stop That Bullet

$initial\quad KE=\frac { 1 }{ 2 } m{ v }^{ 2 }\\ but\quad velovity\quad becomes\quad half\quad in\quad 3\quad cm\\ \therefore \quad final\quad KE\quad =\quad \frac { 1 }{ 2 } m\left( \frac { { v }^{ 2 } }{ 2 } \right) \quad =\quad \cfrac { 1 }{ 4 } initial\quad KE\\ thus\quad KE\quad changes\quad by\quad a\quad factor\quad of\quad \cfrac { 3 }{ 4 } \\ \cfrac { 3 }{ 4 } \quad drop\quad in\quad KE\quad \rightarrow \quad 3cm\\ \therefore \quad remaining\quad \cfrac { 1 }{ 4 } \quad drop\quad in\quad KE\quad \rightarrow \quad 1cm$

There is a linear relationship between distance penetrated and energy lost due to friction (since the equation is $W = F \dot |r|$ , and the frictional force is constant).

Then, since the initial velocity is halved, $\frac{3}{4}$ of the energy is lost. There remains a third of that energy left to lose, so it will take another $1$ centimeter.

As an exercise, it may help to show what I said in a more formal, mathematical way.

applying work energy theorem

(1/2)mv^2 - F*3 = 1/2m(v/2)^2

F= (mv^2)/8

(mv^2)/8 * x = mv^/2

this gives x=4

since it has alreadu travelled 3 cm so iit will travel 1cm more

since, v^2 -u^2=2as, v=v,u=v/2, s=3, substitute a=v^2 /8, now use v=v,u=0,a is calculated, s is found to be 4, 3 is already moved,therefore 1 is left

by the given information, let the initial velocity be x and we are given that it loses half of its velocity when it penetrates 3 cm. therefore its velocity after 3cm = x/2. Now applying 3rd equaiton of motion, i.e. v^2 - u^2 = 2as x^2/4 - x^2 = 2 * 3 * a this gives acceleration = (-x^2/8) now when it comes to rest its velocity will be 0. again applying 3rd equation of motion 0^2 - x^2 = 2 * (-x^2/8) *s this gives s = 4cm now since it has covered 3 cm it needs to cover 1 cm more to stop

use newtons equation 3 :- v2 = u2 +2as twice in the problem.