Explosion of Stationary Bomb – Rankers Physics
Topic: Center of Mass , Momentum and Collision
Subtopic: Collision

Explosion of Stationary Bomb

A stationary bomb explodes into two parts, \(4\text{ kg}\) and \(8\text{ kg}\). The velocity of the \(8\text{ kg}\) mass is \(6\text{ ms}^{-1}\). The KE of the other body is:
\(48\text{ J}\)
\(24\text{ J}\)
\(288\text{ J}\)
\(16\text{ J}\)

Solution:

By conservation of momentum, \(m_1 v_1 = m_2 v_2\), which gives \(4 \times v_1 = 8 \times 6\), so \(v_1 = 12\text{ ms}^{-1}\). The kinetic energy of the \(4\text{ kg}\) body is \(K = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 4 \times 12^2 = 288\text{ J}\).

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