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
$2mgR-mg\frac{R}{\sqrt{2}}=\frac{1}{2}mu^{2}$
$\Rightarrow u=\sqrt{4 g R - \sqrt{2} g R}$
Maximum height attained is
$H_{max}=\frac{R}{\sqrt{2}}+\frac{u^{2} sin^{2} 45^{^\circ }}{2 g}$
$=\frac{R}{\sqrt{2}}+\frac{4 g R - \sqrt{2} g R}{2 g \times 2}$
$=\frac{R}{\sqrt{2}}+R-\frac{R}{2 \sqrt{2}}$
$=R\left(\frac{2 + 2 \sqrt{2} - 1}{2 \sqrt{2}}\right)$
$=\left(\frac{1 + 2 \sqrt{2}}{2 \sqrt{2}}\right)R=\left(1 + \frac{1}{2 \sqrt{2}}\right)R$