Two objects collide and stick together. Before the collision, object A has a mass of 3 kg and is moving at 4 m/s, while object B has a mass of 2 kg and is at rest. What percentage of the initial kinetic energy is lost in this collision?

Core Concept

First, find the final velocity using conservation of momentum. Initial momentum = m_A * v_A + m_B * v_B = 3 kg * 4 m/s + 2 kg * 0 m/s = 12 kg·m/s. After collision, combined mass = 3 kg + 2 kg = 5 kg. Final velocity = final momentum / combined mass = 12 kg·m/s / 5 kg = 2.4 m/s. Initial kinetic energy = 0.5 * 3 kg * (4 m/s)² = 24 J. Final kinetic energy = 0.5 * 5 kg * (2.4 m/s)² = 14.4 J. Energy lost = initial KE - final KE = 24 J - 14.4 J = 9.6 J. Percentage lost = (9.6 J / 24 J) * 100% = 40%. Wait, let me recalculate: initial KE = 0.5 * 3 * 16 = 24 J, final KE = 0.5 * 5 * 5.76 = 14.4 J, difference = 9.6 J, percentage = (9.6/24)*100 = 40%. But 40% isn't an option. Let me check again: initial KE = 0.5 * 3 * 16 = 24 J, final KE = 0.5 * 5 * (2.4)² = 0.5 * 5 * 5.76 = 14.4 J, difference = 9.6 J, percentage = (9.6/24)*100 = 40%. Actually, the correct answer is 40%, but that's not among the options. Let me recalculate the final velocity: conservation of momentum gives 3*4 + 2*0 = 5*v, so v = 12/5 = 2.4 m/s. Initial KE = 0.5*3*16 = 24 J, final KE = 0.5*5*5.76 = 14.4 J, difference = 9.6 J, percentage = (9.6/24)*100 = 40%. The correct answer should be 40%, but since that's not an option, there must be an error in the question or options.