A small block of mass m is placed at the highest point of a smooth fixed sphere of radius R . It is given a horizontal initial speed v₁ such that it slides down the surface. The block loses contact with the sphere at a point where the radius vector makes an angle with the upward vertical. The ratio of the block's speed v₂ at the point of losing contact to its initial speed v₁ is :

Step-by-step solution

At the point where the block loses contact with the sphere, the normal reaction N becomes zero. The equation of motion along the radial direction is given by: mg - N = m v₂^2 R Setting N = 0 , we get: mg = m v₂^2 R v₂^2 = gR Applying the principle of conservation of mechanical energy between the highest point and the point of losing contact: 1 2 m v₁^2 + mgR = 1 2 m v₂^2 + mgR Multiplying by 2 m throughout: v₁^2 + 2gR = v₂^2 + 2gR v₁^2 = v₂^2 - 2gR(1 - ) From the radial equation, we have gR = v₂^2 . Substituting th