Bomb explosion at rest – Rankers Physics

Kinetic Energy and Momentum: Practice Problem & Solution

A bomb of mass \(30\text{ kg}\) at rest explodes into two pieces of masses \(18\text{ kg}\) and \(12\text{ kg}\). The velocity of \(18\text{ kg}\) mass is \(6\text{ ms}^{-1}\). The kinetic energy of the other mass is: (2005)
\(243\text{ J}\)
\(486\text{ J}\)
\(564\text{ J}\)
\(388\text{ J}\)

Solution Explained:

To solve this problem, we apply the core principles of Kinetic Energy and Momentum. Understanding the underlying formula is key to arriving at the correct answer below:

By conservation of momentum, \(m_1v_1 + m_2v_2 = 0\) (since initial momentum is zero). Given \(m_1 = 18\text{ kg}\), \(v_1 = 6\text{ m/s}\), and \(m_2 = 12\text{ kg}\). So, \(18 \times 6 + 12v_2 = 0 \Rightarrow 108 + 12v_2 = 0 \Rightarrow v_2 = -9\text{ m/s}\). The kinetic energy of the other mass is \(KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(12)(-9)^2 = 6 \times 81 = 486\text{ J}\).

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