JAMB Physics 2018 Paper

Concept:This is a perfectly inelastic collision because the lead bullet becomes embedded in the lead block.In such collisions, linear momentum is conserved, but kinetic energy is not conserved.The combined system then moves with one common velocity.Explanation:Mass of bullet, $m_1 = 0.05\ \mathrm{kg}$Initial velocity of bullet, $u_1 = 200\ \mathrm{m\,s^{-1}}$Mass of lead block, $m_2 = 0.95\ \mathrm{kg}$The lead block is initially at rest, so $u_2 = 0$.Using the law of conservation of linear momentum:$$m_1 u_1 + m_2 u_2 = (m_1 + m_2)V$$$$(0.05 \times 200) + (0.95 \times 0) = (0.05 + 0.95)V$$$$10 = 1.00 \times V$$$$V = 10\ \mathrm{m\,s^{-1}}$$Now calculate the final kinetic energy of the combined mass:$$K.E. = \frac{1}{2}(m_1 + m_2)V^2$$$$K.E. = \frac{1}{2}(0.05 + 0.95)(10)^2$$$$K.E. = \frac{1}{2} \times 1 \times 100$$$$K.E. = 50\ \mathrm{J}$$Answer:The final kinetic energy after impact is $50\ \mathrm{J}$.