A bullet is fired into a fixed target looses one third of its velocity after travelling $4 \mathrm{~cm}$. It penetrates further $\mathrm{D} \times 10^{-3} \mathrm{~m}$ before coming to rest. The value of $\mathrm{D}$ is :
Solution
<p>$$\begin{aligned}
& v^2-u^2=2 a S \\
& \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \\
& \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \\
& -\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \ldots(1) \\
& 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \\
& -\frac{4 u^2}{9}=-2 a x \ldots(2)
\end{aligned}$$</p>
<p>$(1)/(2)$</p>
<p>$$\begin{aligned}
& \frac{5}{4}=\frac{4 \times 10^{-2}}{\mathrm{x}} \\
& \mathrm{x}=\frac{16}{5} \times 10^{-2} \\
& \mathrm{x}=3 \cdot 2 \times 10^{-2} \mathrm{~m} \\
& \mathrm{x}=32 \times 10^{-3} \mathrm{~m}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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