A pendulum bob os mass `10^(-2)` kg is raised to a height `5 xx 10^(-2)`m and then released. At the bottom of its swing, it picks up a mass `10^(-3)` kg. To what height will the combined mass rise?

Explanation

Velocity of pendulum bob in mean position
`v_(1)=sqrt(2gh)=sqrt(2xx10xx5xx10^(-2))=1m/s`
when the bob picks up a mass `10^(-3)`kg at the bottom, then by conservation of liner momentum, the velocity of the coalesced mass is given by
`m_(1)v_(1)+m_(2)v_(2)=(m_(1)+m_(2))V`
`10^(-2)xx1+10^(-3)xx0=(10^(-2))+10^(-3()V`
or `V=10/11m//s`
if `h'` is the height risen by the combine mass, we have `1/2(m_(1)+m_(2))V^(2)=(m_(1)+m_(2))gh'`
`:. h'=V^(2)/(2g)=((10/11)^(2))/(2xx10)m=5/121m=0.413m`