RRB Technician Grade 1 Model Paper 5

Concept:In a perfectly inelastic collision, linear momentum is conserved, but kinetic energy is partially lost.Explanation:Let $V$ be the common velocity of the block and bullet just after the collision.Using the law of conservation of linear momentum:$$m v_0 = (m + M)V$$Therefore,$$V = \frac{m v_0}{m + M}$$Initial kinetic energy of the bullet:$$KE_i = \frac{1}{2} m v_0^2$$Final kinetic energy of the combined block-bullet system:$$KE_f = \frac{1}{2} (m + M)V^2 = \frac{1}{2}(m + M)\left(\frac{m v_0}{m + M}\right)^2$$$$KE_f = \frac{m^2 v_0^2}{2(m + M)}$$Kinetic energy lost in the collision is the difference between initial and final kinetic energy:$$\Delta KE = KE_i - KE_f$$$$\Delta KE = \frac{1}{2} m v_0^2 - \frac{m^2 v_0^2}{2(m + M)}$$$$\Delta KE = \frac{1}{2} m v_0^2 \left[1 - \frac{m}{m + M}\right]$$$$\Delta KE = \frac{1}{2} m v_0^2 \left(\frac{M}{m + M}\right)$$The maximum height $h$ only confirms the remaining kinetic energy is converted into gravitational potential energy; it is not needed for calculating the loss.Answer:$$\Delta KE = \frac{1}{2} m v_0^2 \left(\frac{M}{m + M}\right)$$Therefore, the correct option is A.