Two identical beads, each of `m=100g`, are connected by an inextensible massless string, which can slide along the two arms AC and BC of a rigid smooth wire frame in a vertical plane. If the system is released from rest, the kinetic energy of the first bead when the second bead has moved down by a distance of `0.1m` is `xxx10^-3J`. Find the value of `x(g=10ms^-2)` (shown situation is after movement of `0.1m`).

Answer

`x=36`

Explanation

Velocity of both beads along the string should be same.
`v_1costheta=v_2sintheta`
`v_14/5=v_23/5impliesv_2=4v_1/3`

Using energy conservation, we get `mgh=1/2mv_1^2+1/2mv_2^2`
`impliesgxx0.1=1/2v_1^2+1/2((4v_1)/(3))^2impliesv_1=3/5sqrt2`
Given `1/2mv_1^2=xxx10^-3`
`implies1/2(100)/(1000)(3/5sqrt2)^2=x xx10^-3=x=36`