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A \( 12 \mathrm{~kg} \) bomb at rest explodes into two pieces of \( 4 \mathrm{~kg} \) and \( 8 \mathrm{~kg} \). If the momentum of \( 4 \mathrm{~kg} \) piece
is \( 20 \mathrm{Ns} \), the kinetic energy of the \( 8 \mathrm{~kg} \) piece is
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is \( 20 \mathrm{Ns} \), the kinetic energy of the \( 8 \mathrm{~kg} \) piece is
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The correct answer is:
\( 25 \mathrm{~J} \)
Given, mass of bomb, $M=12 \mathrm{~kg} ; m_{1}=4 \mathrm{~kg}, m_{2}=8 \mathrm{~kg}$ momentum of $m_{1}=p_{1}=20 \mathrm{~N}$
Now given, $p_{1}=20 \mathrm{~N}$, momentum of $8 \mathrm{~kg}$ piece $p_{2}=20 \mathrm{~N}$
Thus, kinetic energy of $8 \mathrm{~kg}$ piece is
$\frac{1}{2} \frac{p^{2}}{m_{2}}=\frac{1}{2} \times \frac{20 \times 20}{8}=\frac{100}{4}=25 \mathrm{~J}$
Now given, $p_{1}=20 \mathrm{~N}$, momentum of $8 \mathrm{~kg}$ piece $p_{2}=20 \mathrm{~N}$
Thus, kinetic energy of $8 \mathrm{~kg}$ piece is
$\frac{1}{2} \frac{p^{2}}{m_{2}}=\frac{1}{2} \times \frac{20 \times 20}{8}=\frac{100}{4}=25 \mathrm{~J}$
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