Questions
A small block slides with velocity on the horizontal frictionless surface as shown in the fig. The block
leaves the surface at point C. The angle in the figure is:
detailed solution
Correct option is B
At point C: mg cos θ -N =mv2r(when block leaves the surface normal force becomes zero, so putting N = 0) ⇒g cos θ = v2rv2 = v20+2gh, h = r - r cos θgr cos θ = (0.5gr)2+ 2g(r-r cos θ)⇒cos θ = 14+2-2 cos θ ⇒ cos θ = 34Talk to our academic expert!
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