A small block slides with velocity 0.5 g r on the horizontal frictionless surface as shown in the figure. The block leaves the surface at a point C . What is the value of cosθ ?

Step-by-step solution

m g   cos ⁡ θ - N = m v 2 R ;   N = m g   cos ⁡ θ - m v 2 R N = 0 ⇒ m g   cos ⁡ θ = m v 2 R v = g R   cos ⁡ θ   o r   cos ⁡ θ = v 2 g R   .....(i) From energy conservation 1 2 m v 2 - g R 4 = m g R 1 -   cos ⁡ θ ⇒ v 2 2 -   g R 8 = g R - g R   cos ⁡ θ ...(ii) From (i) and (ii) cosθ = 3 5 = 0 . 6