Question

A bomb of mass 9 kg explodes into the fragments of masses 3 kg and 6 kg. The velocity of mass 3 kg is 16 m/sec. The energy of explosion is equal to:

• A. 384 J
• B. 576 J
• C. 192 J
• D. 1152 J

Answer 1

Answer: (B)

576 J

Answer 2

Conservation of momentum just before and after the impact yields

Pi = P_f

⇒ 0 = $\left| m_{1}{\overrightarrow{v}}_{1} + m_{2}{\overrightarrow{v}}_{2} \right|$ ⇒ m₁ v₁ = m₂ v₂ = p (say)

⇒ Energy of explosion = ∆ KE_system

= $\frac{p^{2}}{2m_{1}} + \frac{p^{2}}{2m_{2}}$= $\frac{p^{2}}{2m_{1}}\left\lbrack 1 + \frac{m_{1}}{m_{2}} \right\rbrack$

Putting P = m₁ v₁, where m₁ = 3 kg, M₂ = 6 kg , v = 16 m/sec.

We obtain, E = $\frac{(3 \times 16)^{2}}{2 \times 3}$ $\left\lbrack 1 + \frac{3}{6} \right\rbrack$

⇒ E = 576 J . Hence correct choice is