WAEC Physics 2023 Paper

Concept:The bullet embeds into the block, so they move together after collision as a single body. Momentum is conserved, but kinetic energy is not conserved.Explanation:Mass of bullet, $m_1 = 0.05\ \text{kg}$Initial speed of bullet, $u_1 = 200\ \text{m/s}$Mass of block, $m_2 = 0.95\ \text{kg}$Initial speed of block, $u_2 = 0$Using conservation of momentum:$$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$$$$0.05 \times 200 + 0.95 \times 0 = (0.05 + 0.95)v$$$$10 = 1 \times v$$$$v = 10\ \text{m/s}$$Kinetic energy of the combined system after impact:$$K.E. = \frac{1}{2}(m_1 + m_2) v^2$$$$K.E. = \frac{1}{2} \times 1 \times (10)^2$$$$K.E. = \frac{1}{2} \times 100$$$$K.E. = 50\ \text{J}$$Answer:The kinetic energy of the block after impact is $50\ \text{J}$. Therefore, the correct option is C. 50J.