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A bomb of mass \(9 \mathrm{~kg}\) explodes into two pieces of mass \(3 \mathrm{~kg}\) and \(6 \mathrm{~kg}\). The velocity of mass \(3 \mathrm{~kg}\) is \(16 \mathrm{~m} / \mathrm{s}\). The kinetic energy of mass \(6 \mathrm{~kg}\) (in joule) is
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192
Given, mass of bomb, \(M=9 \mathrm{~kg}\)
According to conservation of linear momentum,
\(\begin{aligned}
& M \times 0 & =m_1 v_1+m_2 v_2 \\
\Rightarrow & 9 \times 0 & =3 \times 16+6 \times v_2 \\
\Rightarrow & v_2 & =\frac{-3 \times 16}{6}=-8 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
\(\therefore\) Kinetic energy of mass, \(m_2=6 \mathrm{~kg}\) (given)
\(\begin{aligned}
K & =\frac{1}{2} m_2 v_2^2=\frac{1}{2} \times 6 \times(-8)^2 \\
& =\frac{1}{2} \times 6 \times 64=192 \mathrm{~J}
\end{aligned}\)
According to conservation of linear momentum,
\(\begin{aligned}
& M \times 0 & =m_1 v_1+m_2 v_2 \\
\Rightarrow & 9 \times 0 & =3 \times 16+6 \times v_2 \\
\Rightarrow & v_2 & =\frac{-3 \times 16}{6}=-8 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
\(\therefore\) Kinetic energy of mass, \(m_2=6 \mathrm{~kg}\) (given)
\(\begin{aligned}
K & =\frac{1}{2} m_2 v_2^2=\frac{1}{2} \times 6 \times(-8)^2 \\
& =\frac{1}{2} \times 6 \times 64=192 \mathrm{~J}
\end{aligned}\)
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