A lead bullet of mass 0.05 kg is fired with a velocity of 200 m/s into a lead block of mass 0.95 kg. Given that the lead block can move freely, the final kinetic energy after impact is
A. 100J B. 150J C. 50J D. 200J
Correct Answer: C
Explanation
From the law of conservation of linear momentum, \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) Since the collision is inelastic, we have \((0.05 \times 200) - (0.95 \times 0) = (0.05 + 0.95)V\) \(10 = V\) \(V = 10ms^{-1}\) Hence the Kinetic energy = \(\frac{1}{2} (0.05 + 0.95) \times 10^2\) = \(\frac{1}{2} \times 100\) = 50J
