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A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is $6 \mathrm{~ms}^{-1}$. The kinetic energy of the other mass is
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486 J
The linear momentum of exploding part will remain conserved.
Applying conservation of linear momentum,
we write,
$m_1 u_1=m_2 u_2$
Here,
$\begin{gathered}m_1=18 \mathrm{~kg}, m_2=12 \mathrm{~kg} \\ u_1=6 \mathrm{~ms}^{-1}, u_2=?\end{gathered}$
$\therefore 18 \times 6=12 u_2$
$\Rightarrow u_2=\frac{18 \times 6}{12} = 9 \mathrm{~ms}^{-1}$
Thus, kinetic energy of 12 kg mass
$\begin{aligned} K_2 & =\frac{1}{2} m_1 u_2^2 \\ & =\frac{1}{2} \times 12 \times(9)^2 \\ & =6 \times 81 \\ & =486 \mathrm{~J}\end{aligned}$
Applying conservation of linear momentum,
we write,
$m_1 u_1=m_2 u_2$
Here,
$\begin{gathered}m_1=18 \mathrm{~kg}, m_2=12 \mathrm{~kg} \\ u_1=6 \mathrm{~ms}^{-1}, u_2=?\end{gathered}$
$\therefore 18 \times 6=12 u_2$
$\Rightarrow u_2=\frac{18 \times 6}{12} = 9 \mathrm{~ms}^{-1}$
Thus, kinetic energy of 12 kg mass
$\begin{aligned} K_2 & =\frac{1}{2} m_1 u_2^2 \\ & =\frac{1}{2} \times 12 \times(9)^2 \\ & =6 \times 81 \\ & =486 \mathrm{~J}\end{aligned}$
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