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Q.

A small block slides with velocity 0.5gr on the horizontal frictionless surface as shown in the figure. The block leaves the surface at point C. The value of cosθ=nn+1  Find n.

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answer is 3.

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Detailed Solution

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v2=v02+2gh=(0.5gr)2+2gr(1−cosθ)=(2.25gr−2grcosθ)

At the time of leaving contact,   N=0

∴mgcosθ=mv2r=2.25mg−2mgcosθ cosθ=2.253=34=nn+1

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