Question

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?

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

Answer 1

The correct option is D $1.0cm$

Applying the work-energy theorem for the bullet:

The total kinetic energy of bullet = work done by regarding force + Final kinetic energy of the bullet.

$\frac{1}{2}m{v}_0^2=f\left(3cm\right)+\frac{1}{2}m{\left(\frac{v_0}{2}\right)}^2$

$\frac{3}{8}m{v}_0^2=f\left(3cm\right)...\left(i\right)$

Now to find the distance covered to come to rest can be obtained as:

$\frac{1}{2}m{v}_0^2=f(l.cm)....\left(ii\right)$

Now equating the equation (I) and (II):
$\frac{l}{3}=\frac{1}{2}.\frac{8}{3}\Rightarrow l=4cm\Delta l=4-3=1cm$