Question

A particle of mass $m$ slides on a frictionless surface $ABCD,$ starting from rest as shown in the figure.

Part $BCD$ is a circular arc of radius $R.$ If the particle loses its contact at $D,$ find the maximum height attained by the particle from the ground is.......

• A. $R\left(2+\frac{1}{2\sqrt{2}}\right)$
• B. $R\left(2-\frac{1}{2\sqrt{2}}\right)$
• C. $3R$
• D. $\left(1+\frac{1}{2\sqrt{2}}\right)R$

Answer 1

Answer: (D)

Applying law of conservation of energy
$2mgR-mg\frac{R}{\sqrt{2}}=\frac{1}{2}m{u}^2$
$\Rightarrow u=\sqrt{4gR-\sqrt{2}gR}$
Maximum height attained is
$H_{max}=\frac{R}{\sqrt{2}}+\frac{u^{2}si{n}^{2}45^{{}^{\circ }}}{2g}$
$=\frac{R}{\sqrt{2}}+\frac{4gR-\sqrt{2}gR}{2g\times 2}$
$=\frac{R}{\sqrt{2}}+R-\frac{R}{2\sqrt{2}}$
$=R\left(\frac{2+2\sqrt{2}-1}{2\sqrt{2}}\right)$
$=\left(\frac{1+2\sqrt{2}}{2\sqrt{2}}\right)R=\left(1+\frac{1}{2\sqrt{2}}\right)R$