Question

a bullet loses 25% of its initial kinetic energy after penetrating through a distance of 2 cm in a target the for the distance it travel before coming to rest is

Answer 1

Answer:

6cm

Explanation:

Given initially it loses 25% of its energy to penetrate 2cm , and now in order to come to rest it penetrates some x cm and now it loses its remaining energy completely !

25% energy - 2cm

(100-25)% energy - x cm

On solving we get x = 6cm !!

Answer 2

Answer:

The distance traveled by the bullet before coming to rest is 6 cm.

Explanation:

Now given, Initially the bullet losses 25% of its energy and it covers 2cm distance.

so,from work-energy theorem,we can write:

work-done(W) =change of kinetic energy( ΔE)

→ W = FS = 25% of initial K.E.

       → 2 F = (1/4) $\frac{1}{2} m u^{2}$ =  $\frac{1}{8 } m u^{2}$   [F = ma ,resistance force]

       → 2 ma =  $\frac{1}{8 } m u^{2}$

       → 16 a = $u^{2}$          ................................(i)

The bullet finally comes to a rest.

so,final velocity(v) = 0

now, $v^{2} = u^{2} -2as$

         → $u^{2} =2as$

         → 16a = 2 * a * $s$       [using equation (i)]

         → $s$ = 8

Therefore,distance travelled by the bullet before coming to rest is (8-2) cm =  6 cm.