A particle of mass 20 g is released with an initial velocity 5m/s alogn 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 (in `kgm^(2)//s)`) about O will be (Take `g=10m//s^(2)`)

Explanation

According to work energy theorem
`mgh=1/2 mv_(B)^(2)-1/2mv_(A)^(2)`
`2gh=v_(B)^(2)=v_(A)^(2)`
`2xx10xx10=v_(B)^(2)-5^(2)`
`impliesv_(B)=15m//s`
Angular momentum about O
`L_(O)=mvr`
`=20xx10^(-3)xx20`
`L_(O)=6kgm^(2)//s`