Two bodies `A` and `B` of each of mass `100 gm` are allowed to move along a frictionless typical path as shown below. In order to have the same kinetic energy for the both the bodies at `M`, the initial velocity that should be given to `B`, if `A` starts from rest is

Options

A. `10m//s`
B. `sqrt(2)m//s`
C. `11m//s`
D. `10sqrt(2)m//s`

Explanation

`(KE_(A))` at `M=mg(10-4)=6J`
`(KE_(a))` at `M=6J` and `(PE_(B))` at `M=(1)/(10)xx10xx4=4J`
`rArr((1)/(2)mv_(B)^(2))_("on ground")=(T_(EB))_(M)=(6+4)J`
`rArrv=sqrt(200)=10sqrt(2)m//s`