Question

A bullet fired towards a wall reduces its half kinetic energy after the penetration of 6 cm. The further penetration of bullet in the block is

• A. 2 cm
• B. 1 cm
• C. 6 cm
• D. 3 cm

Answer 1

Answer: (C)

6 cm

Kinetic energy , $K=\frac{1}{2}m{v}^2$ so $K\propto v^2$

If initial velocity is $u$, the final velocity will be $v=\frac{u}{\sqrt{2}}$ (as the kinetic energy is half on its initial value)

Given , penetration distance, $d=6cm$

Using, $v^2-u^2=2aS$,

or $\frac{u^2}{2}-u^2=2a\left(6\right)$

or $a=-\frac{u^2}{24}$

Now, the bullet penetrate further distance $d^{\prime }$ with initial velocity $v=\frac{u}{\sqrt{2}}$ before coming to rest.

Thus, final velocity is zero.

So, $0^2-(u/\sqrt{2}{)}^2=2a{d}^{\prime }$

or $-u^2/2=2(-u^2/24)d^{\prime }$

or $d^{\prime }=6cm$