Question

A particle of mass $20g$ is released with an initial velocity $5m/s$ 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. $8kg-m^2/s$
• B. $6kg-m^2/s$
• C. $3kg-m^2/s$
• D. $2kg-m^2/s$

Answer 1

The correct option is B $6kg-m^2/s$

Work-Energy Theorem from $A$ to $B$

$mgh=\frac{1}{g}m{v}_B^2-\frac{1}{g}m{v}_A^2$
$2gh=v_B^2-v_A^2$

$2\times 10\times 10=v_B^2-5^2$

$v_B=15m/s$
Angular momentum about $0$

$L_0=mvr$

$=20\times {10}^{-3}\times 20$

$L_0=6kg.m^2/s$