Circular MotionHard

Question

A small block slides with velocity 0.5 on the horizontal frictionless surface as shown in the Figure. The block leaves the surface at point C. The angle θ in the Figure is :

Options

A.cos-1 (4/9)
B.cos-1 (3/4)
C.cos-1 (1/2)
D.none of the above

Solution

Given vB = 0.5
Assume block leave the contact at C, N = 0
= mg cos θ     .... (1)
from energy conservation mvB2 + mgr (1 − cos θ) = mvC2 ........... (2)
from equation (1) and (2).
m + mg r (1 − cos θ) = mg r cos θ
⇒    cos θ =
⇒    θ = cos-1     Ans. 

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