A lead bullet of mass 0.05kg is fired with a velocity of 200ms\(^{-1}\) into a block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is?
A. 50J B. 100J C. 150J D. 200J
Correct Answer: A
Explanation
Given m\(_{1}\) = 0.05kg, u\(_{1}\) = 200ms\(^{-1}\), m\(_{2}\) = 0.95kg K.E = \(\frac{1}{2}\)m\(_{r}\)v\(^{2}\) m\(_{1}\)u\(_{1}\) = v(m\(_{1}\) + m\(_{2}\)) [law of conversation of momentum] v = \(\frac{0.05 \times 200}{0.05 + 95}\) = 10ms\(^{-1}\) K.E = \(\frac{1}{2}\)(1)10\(^{2}\) = 50J
