A small sphere of mass m is attached to a string of length 1 m to form a simple pendulum. It is given a horizontal velocity v at the lowest position. The ratio of its kinetic energy when the string makes an angle of 60^ with the downward vertical to its kinetic energy when the string is horizontal is exactly 3:1 . The initial velocity v at the lowest point is : (Take acceleration due to gravity as 10 m/s ^2 )

Step-by-step solution

Let the lowest point be the reference level for potential energy ( h = 0 ). The initial kinetic energy at the lowest point is: K₀ = 1 2 mv^2 When the string makes an angle of 60^ with the downward vertical, the vertical height of the sphere is: h₁ = L - L (60^ ) = 1 - 1 1 2 = 0.5 m By conservation of mechanical energy, the kinetic energy at this position is: K₁ = K₀ - mgh₁ = 1 2 mv^2 - m(10)(0.5) = 1 2 mv^2 - 5m When the string is horizontal, the angle with the downward vertical is 90^ , and the vertical height is: