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A bullet fired into a fixed target loses half of its velocity after penetrating $3\, cm$. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion ?.......$cm$
$1.5$
$1$
$3$
$2$
Solution
Let initial velocity of the bullet $=u$
After penetrating $3 \mathrm{cm}$ its velocity becomes $\frac{\mathrm{u}}{2}$
From $v^{2}=u^{2}-2 a s$
$\left(\frac{\mathrm{u}}{2}\right)^{2}=\mathrm{u}^{2}-2 \mathrm{a}(3) \Rightarrow 6 \mathrm{a}=\frac{3 \mathrm{u}^{2}}{4} \Rightarrow \mathrm{a}=\frac{\mathrm{u}^{2}}{8}$
Let further it will penetrate through distance
$x$ and stops at point $C .$
For distance $B C, \mathrm{v}=0, \mathrm{u}=\mathrm{u} / 2, \mathrm{s}=\mathrm{x}, \mathrm{a}=\mathrm{u}^{2} / 8$
From
$v^{2}=u^{2}-2 a s \Rightarrow 0=\left(\frac{u}{2}\right)^{2}-2\left(\frac{u^{2}}{8}\right) \cdot x$
$\Rightarrow x=1 \mathrm{cm}$
