Question

A bullet fired into a fixed target loses half of its velocity after penetrating $4\, cm$. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

• A. $0.2\, cm$
• B. $5\, cm$
• C. $2\, cm$
• D. $1.33\, cm$

Answer 1

Answer: (D)

Let the initial velocity of the bullet $= u$
After penetrating its final velocity $=\frac{ u }{2}$
From $v^{2}-u^{2}=2$ as
$\left(\frac{u}{2}\right)^{2}-u^{2}=2 \times a \times 4$
$\frac{u^{2}}{4}-u^{2}=2 \times a \times 4$
$\frac{-3 u^{2}}{4}=8 \times a \Rightarrow a=\frac{-3 u^{2}}{32}$
Bullet will further penetrate after penetrate $4 cm$.
Initial velocity $=\frac{ u }{2}$
Final velocity $=0$
From $v^{2}-u^{2}=2$ as
$0^{2}-\left(\frac{u}{2}\right)^{2}=2 \times \frac{-3 u^{2}}{32} \times s$
$\frac{-u^{2}}{4}=\frac{-3 u^{2}}{16} \times s$
$\Rightarrow s=\frac{4}{3}=1.33\, cm$