A Bullet On Penetrating 30cm Into Its Target A Loses Its Velocity By 50%. What Additional Distance Will Its Penetrate Into The Target Before It Comes To Rest?

A bullet on penetrating 30cm into its target A loses its velocity by 50%. What additional distance will its penetrate into the target before it comes to rest?

10 cm
30 cm
20 cm
None of these

Explanation

Step 1: Relation between velocity and penetration depth
Assume the bullet experiences a constant deceleration $a$ .
From kinematics:

$v^{2} = u^{2} - 2 a s$

Let initial velocity = $u$ .

Step 2: First penetration
Penetrates $s_{1} = 30 cm$ , velocity becomes $u / 2$ :

${(\frac{u}{2})}^{2} = u^{2} - 2 a (30)$ $\frac{u^{2}}{4} = u^{2} - 60 a$ $60 a = u^{2} - \frac{u^{2}}{4} = \frac{3 u^{2}}{4}$ $a = \frac{u^{2}}{80}$

Step 2: First penetration
Penetrates $s_{1} = 30 cm$ , velocity becomes $u / 2$ :

${(\frac{u}{2})}^{2} = u^{2} - 2 a (30)$ $\frac{u^{2}}{4} = u^{2} - 60 a$ $60 a = u^{2} - \frac{u^{2}}{4} = \frac{3 u^{2}}{4}$ $a = \frac{u^{2}}{80}$

Step 3: Additional penetration to stop
From velocity $u / 2$ to 0, distance $s_{2}$ :

$0^{2} = {(\frac{u}{2})}^{2} - 2 a s_{2}$ $0 = \frac{u^{2}}{4} - 2 \cdot \frac{u^{2}}{80} \cdot s_{2}$ $\frac{u^{2}}{4} = \frac{u^{2}}{40} s_{2}$ $s_{2} = 10 cm$