A small block A of mass 2 kg is released from rest from the top of a smooth curved track of height 5 m . At the bottom of the track, it enters a rough horizontal surface and immediately undergoes a perfectly inelastic collision with a stationary block B of mass 3 kg . The coefficient of kinetic friction between the combined mass and the horizontal surface is 0.4 . The distance the combined mass slides before coming t

Step-by-step solution

Velocity of block A at the bottom of the smooth track: u = 2gh = 2 10 5 = 10 m s ⁻¹ By conservation of linear momentum during the perfectly inelastic collision: m_A u = (m_A + m_B) v 2 10 = (2 + 3) v v = 4 m s ⁻¹ By the work-energy theorem on the combined mass, the initial kinetic energy is equal to the work done against friction: 1 2 (m_A + m_B)v^2 = (m_A + m_B)gd Solving for the stopping distance d : d = v^2 2 g = 4^2 2 0.4 10 d = 16 8 = 2 m Answer: 2 m