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 ?

Question

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 ?,

MARKS App · Accepted Answer

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

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