Question

A particle of mass $20\text{g}$ is released with an initial velocity $5{\text{ms}}^{-1}$ 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:
$(\text{Take}g=10{\text{ms}}^{-2})$

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

Answer 1

The correct option is C $6{\text{kg-m}}^2{\text{s}}^{-1}$

According to work-energy theorem

$mgh=\frac{1}{2}m{v}_B^2-\frac{1}{2}m{v}_A^2$
$\Rightarrow 2gh=v_B^2-v_A^2$
$\Rightarrow 2\times 10\times 10=v_B^2-5^2$
$\Rightarrow v_B=15{\text{ms}}^{-1}$
$\text{Angular momentum about}O,$
$L_O=m{v}_{B}r$
$\text{Here,}r=20\text{m}$

$\Rightarrow L_O=20\times {10}^{-3}\times 15\times 20$

$\Rightarrow L_O=6{\text{kg-m}}^2{\text{s}}^{-1}$

Hence, option $\left(C\right)$ is correct.