Question

A bullet fired into a trunk of a tree loses $1/4$ of its kinetic energy in traveling a distance of $5cm$. Assuming constant retardation before stopping, it travels a further distance of:

• A. $150cm$
• B. $1.5cm$
• C. $1.25cm$
• D. $15cm$

Answer 1

Answer: (D)

$15cm$

$2as=v^2-u^2$
$2\times a\times 5=v^2-u^2$
$\Rightarrow 2\times a\times 5=\frac{3}{4}{u}^2-u^2$

$\Rightarrow a=\frac{-u^2}{40}$

After $5cm$ it loses ${\frac{1}{4}}^{th}$ of its initial $KE$.

The acceleration will be same throughout.

Again using $2as=v^2-u^2$ when $v$ becomes $0$.

$2\times \frac{-1}{40}{u}^2\times s=-u^2$

$\Rightarrow s=20$

So bullet further travels $15m$.