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Q.

A small block slides with velocity $0.5 \sqrt{g r}$ on the horizontal frictionless surface as shown in the figure. The block leaves the surface at point C. The value of $\cos θ = \frac{n}{n + 1}$ Find $n$ .

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Detailed Solution

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$\begin{matrix} v_{2} = v_{0}^{2} + 2 g h = {(0.5 \sqrt{g r})}^{2} + 2 g r (1 - \cos θ) \\ = (2.25 g r - 2 g r \cos θ) \end{matrix}$

At the time of leaving contact, $N = 0$

$∴ m g \cos θ = \frac{m v^{2}}{r} = 2.25 m g - 2 m g \cos θ \cos θ = \frac{2.25}{3} = \frac{3}{4} = \frac{n}{n + 1}$

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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.