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)`

Explanation

When M strikes the wall, the vertically downward component of its displacement from initial position is `1m` and its distance from pulley B is, `C^'B=sqrt(1^2+2^2)=sqrt5m`.

while its initial distance from the pulley was `CB=1m`. It means vertically upward displacement of mass m is `(sqrt5-1)m`,
Let M strike the wall with velocity v, Since, string between two blocks always remains taut, therefore at any instant speed of m is equal to that component of velocity of M which is along the string `C^'B`. Hence, the velocity of m when M strikes the wall is `v=costheta`, where
`costheta=2/sqrt5`
According to the law of conservation of energy the loss of potential energy of M
=Increase in PE of m+KE of M=KE of m
`Mgxx1=mg(sqrt5-1)+1/2m(vcostheta)^2`
`v=5sqrt((5-sqrt5)/(6))ms^-1=3.39ms^-1`