Question

A particle of mass $m$ slides on a frictionless surface ABCD, starting from rest as shown in Figure. The part BCD is a circular arc. If it loses contact at point P, the maximum height attained by the particle from point C is :

• A. $R[2+\frac{1}{2\sqrt{2}}]$
• B. $R[2+\frac{1}{2\sqrt{2}}]R$
• C. $3R$
• D. None of these

Answer 1

The correct option is A $R[2+\frac{1}{2\sqrt{2}}]$

$mg\times 2R-mg\frac{R}{\sqrt{2}}=\frac{1}{2}m{v}^2$ __(1)

$V=\sqrt{4gR-\sqrt{2}gR}$

$h_1=\frac{V^2{\sin }^2\theta }{2g}$

$=\frac{V^2{\sin }^{2}45}{2g}$

$H_{max}=R+\frac{R}{\sqrt{2}}+\frac{(4gR-\sqrt{2}gR)}{2g\times \sqrt{2}}$

$H_{max}=R+\frac{R}{\sqrt{2}}+\frac{(4gR-\sqrt{2}gR)}{2g}\times \frac{1}{\sqrt{2}}$

$=R[2+\frac{1}{2\sqrt{2}}]$