A string with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of `2m` from the wall, has a point mass M of `2kg` attached to it at a distance of `1m` from the wall. A mass `m` of `0.5kg` is attached to the free end. The system is initially held at rest so that the stirng is horizontal between wall and pulley and vertical beyond the pulley as shown in figure.

What will be the speed with which point mass M will hit the wall when the system is released? `(g=10ms^-2)`

Options

A. `2(sqrt(5+sqrt5)/6)m//s`
B. `2(sqrt(5-sqrt5)/6)m//s`
C. `5(sqrt(5-sqrt5)/6)m//s`
D. `5(sqrt(5+sqrt5)/6)m//s`

Answer

C

Explanation

When `M` strikes the wall, vertically downward component of its displacement from initital position is `1 m` and its distance from pulley `B` is
`C'B=sqrt(1^(2)+2^(2))=sqrt(5)m`
When its initial distance from the pulley was `CB=1m`. It means vertically upward displacement of mass `m` is `(sqrt(5)-1)m`.

Let `M` strike the wall velocity `v`. Since the string between the two blocks always remains taut, therefore at any instant speed of `m` is equal to that component of velocity of `M` which which is along the string `C'B`. Hence , velocity of `m` when `M` strikes the wall is `vcostheta`, where
`costheta=2/sqrt(5)`
`:. (v_(M))/(v_(m))=v/(vcostheta)=(sqrt(5))/2`
According to law of energy, loss of potential energy of `M=`increase in `PE` of `m+KE` of `M+KE` of `m`
`Mgxx1=mg(sqrt(5)-1)+1/2Mv^(2)+1/2m(vcostheta)^(2)`
`v=5sqrt((5-sqrt(5))/6)m//s`