Block A of mass 1 kg is released from rest from a height of 1.8 m on a smooth curved track. At the bottom of the track, it collides perfectly elastically with Block B of mass 2 kg , which is initially at rest. After the collision, Block B slides along a smooth horizontal surface and compresses a horizontal ideal spring of spring constant 8 N m ⁻¹ . The maximum compression in the spring is : (Take g = 10 m s ⁻² )

Step-by-step solution

By conservation of mechanical energy, the velocity of Block A just before the collision is: v₁ = 2gh = 2 10 1.8 = 6 m s ⁻¹ For a perfectly elastic collision, the velocity of Block B just after the collision is given by: v₂' = ( 2m₁ m₁ + m₂ ) v₁ v₂' = ( 2 1 1 + 2 ) 6 = 4 m s ⁻¹ Applying conservation of mechanical energy for Block B and the spring: 1 2 m₂ (v₂')^2 = 1 2 k x^2 1 2 2 (4)^2 = 1 2 8 x^2 16 = 4x^2 x = 2 m Answer: 2 m