A pendulum bob of 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 pendulam bob in mean position,
`v_(1) = sqrt(2gh)= sqrt(2 xx 10xx5xx10^(-2)= 1 ms^(-1)`
When the bob picks up a mass `10^(-3)` kg at the bottom, then by conservation of linear momentum, the velocity of coalesced mass is given by
`m_(1)v_(1) + m_(2)v_(2) = (m_(1) + m_(2))/v`
`10^(-2) + 10^(-3) xx 0 = (10^(-2) + 10^(-3))v`
or `v=(10^(-2)/(1.1 xx 10^(-2))` = 10/11 ms^(-1)`
Now, `h=(v^(2)/(2g)) = (10//11)^(2)/(2 xx 10) = 4.1 xx 10^(-2)m`