A small bob of mass 2 kg is released from rest from a certain height h on a smooth curved track. At the bottom of the track, it perfectly elastically strikes a block of mass 6 kg initially at rest on a rough horizontal surface. The 6 kg block slides a distance of 1.6 m on the rough surface before coming to rest. If the coefficient of kinetic friction between the block and the surface is 0.5 , the initial release heig

Step-by-step solution

Let v₂' be the velocity of the 6 kg block just after the collision. Using the work-energy theorem for its motion on the rough surface: 1 2 M (v₂')^2 = M g d (v₂')^2 = 2 g d = 2 0.5 10 1.6 = 16 v₂' = 4 m s ⁻¹ For a perfectly elastic collision, the velocity of the block after collision is related to the bob's initial velocity v₁ by: v₂' = ( 2m m + M ) v₁ 4 = ( 2 2 2 + 6 ) v₁ 4 = 4 8 v₁ v₁ = 8 m s ⁻¹ Applying conservation of mechanical energy for the bob's descent: v₁ = 2gh 8 = 2 10 h 64 = 20h h = 3.2 m Answer: 3.2 m