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2.Motion in Straight Line
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A bullet fired into a fixed target loses half of its velocity after penetrating $1\,cm.$ How much further it will penetrate before coming to rest, assuming that it faces constant resistance to motion
A$1.5\,cm$
B$1.0\,cm$
C$3.0\,cm$
D$\frac {1}{3}\,cm$
Solution
$A$ to $B$$:$ $\left(\frac{\mathrm{u}}{2}\right)^{2}=\mathrm{u}^{2}-2 \mathrm{a} \times 1$ $…(i)$
$\mathrm{B}$ to $\mathrm{C}: 0=\left(\frac{\mathrm{u}}{2}\right)^{2}-2 \mathrm{a} \times \mathrm{d}$ $…(ii)$
From $(\text { i }), 2 a=\frac{3 u^{2}}{4}$ in $(\text { ii })$
$0=\frac{\mathrm{u}^{2}}{4}-\frac{3 \mathrm{u}^{2}}{4} \mathrm{d} \quad \Rightarrow \mathrm{d}=\frac{1}{3} \mathrm{cm}$
$\mathrm{B}$ to $\mathrm{C}: 0=\left(\frac{\mathrm{u}}{2}\right)^{2}-2 \mathrm{a} \times \mathrm{d}$ $…(ii)$
From $(\text { i }), 2 a=\frac{3 u^{2}}{4}$ in $(\text { ii })$
$0=\frac{\mathrm{u}^{2}}{4}-\frac{3 \mathrm{u}^{2}}{4} \mathrm{d} \quad \Rightarrow \mathrm{d}=\frac{1}{3} \mathrm{cm}$
Standard 11
Physics
