Question 114 of 183

A particle of mass $20g$ is released with an initial velocity $5m{s}^{-1}$ along the curve from the point $A,$ as shown in the figure. The point $A$ is at height $h$ from point $B.$ The particle slides along the frictionless surface. When the particle reaches point $B,$ its angular momentum about $O$ will be: (Take $g=10m{s}^{-2}$)

A$3kg{m}^{2}{s}^{-1}$B$2kg{m}^{2}{s}^{-1}$C$6kg{m}^{2}{s}^{-1}$D$8kg{m}^{2}{s}^{-1}$

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Answer: (C) $6kg{m}^{2}{s}^{-1}$

Applying conservation of energy,

$mgh=\dfrac{1}{2}m{v}_{B}^{2}-\dfrac{1}{2}m{v}_{A}^{2}$
$\Rightarrow {v}_{B}=\sqrt{2gh+{v}_{A}^{2}}$
$\Rightarrow {v}_{B}=\sqrt{2\times 10\times 10+25}$
$\Rightarrow {v}_{B}=15m{s}^{-1}$
Angular momentum about $O$,
${L}_{O}=m{v}_{B}\left(h+a\right)$
$\Rightarrow {L}_{O}=20\times {10}^{-3}\times 15\times 20kg{m}^{2}{s}^{-1}$
$\Rightarrow {L}_{O}=6kg{m}^{2}{s}^{-1}$