A string with one end fixed on a right well , passing over a fixed frictionless pulley at a distance of `2` m from the well , has a point mass `M = 2kg` attached to the it at a distance of I m from the well . A mass `m= 0.5` kg attached at the free end is field at rest as this the
string is horizontal between the wall and the pulley when will be the speed with which the mass `M` with hit the well when the mass `m` is released ?

Answer

B, C

Explanation

Where mass in is released , since `M gt m`, the mass `M` will move on a dotted path with `O` as thecentre . There will be decrease in the potential energy of `M` which will be converted into kinetic energy of `M` and increase in potential energy of `m`
Decrease in P.E. of `M` is `Mgh`
` = 2 xx 9.8 xx1 = 19.61
K.E. of `M = (1)/(2) Mv^(2) `

(Let `V` be the velocity attainned by `M` just before string the wall)
K.E. of `m = (1)/(2) mv^(2)`
from the figure , by velocirty constant
` v = V cos theta`
From `delta OAC`,

`cos theta = (2)/(sqrt(5)`
`:. v = (2V)/(sqrt(5)`
`(OC + CA) - OA = ` beigth attained by `m`
`1 +sqrt(2^(2) + 1^(2) ) - 2 = ` beight attained by `m = sqrt(5) -1 `
increase in P.E. of `m = mgnV = 0.5 xx 9.8 (sqrt(5)-1)`