KCET · Physics · Center of Mass Momentum and Collision
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
- A \( 25 \mathrm{~J} \)
- B \( 20 \mathrm{~J} \)
- C \( 50 \mathrm{~J} \)
- D (40 J
Answer & Solution
Correct Answer
(A) \( 25 \mathrm{~J} \)
Step-by-step Solution
Detailed explanation
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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