QNEET-2024-Ph-64: NEET 2024 Physics - Class 11 NEET-2024-Ph-64 Solution: Two bodies A and B of same mass undergo completely inelastic…

In a completely inelastic collision, the two bodies stick together and move with a common final velocity. Here, before the collision, body A is moving with a velocity v 1 and body B is at rest. The conservation of momentum must hold true because no external forces are acting on the system. The momentum before the collision is only due to body A since body B is at rest. Therefore, the total initial momentum p initial of the system is given by: p initial = m A v 1 + m B v 0 = m v 1 + 0 = m v 1 where: m A and m B are the masses of bodies A and B respectively, m is the mass of each body, v 1 is the velocity of body A , v 0 = 0 is the velocity of body B because it is initially at rest. Since they undergo a completely inelastic collision, bodies A and B stick together after the collision and hence move with a common velocity v 2 . The total mass of the combined system post-collision is m A + m B = m + m = 2 m . The momentum after the collision is given by: p final = ( m A + m B ) v 2 = 2 m v 2 Applying the conservation of momentum (since no external force implies momentum is conserved), we equate p initial and p final : m v 1 = 2 m v 2 Dividing through by m yields: v 1 = 2 v 2 Thus, solving for the ratio v 1 v 2 gives: v 1 v 2 = 2 Therefore, the ratio of v 1 to v 2 is 2 : 1 , making the correct answer: Option B: 2 : 1