Two masses M and m are connected by a light inextensible string which passes over a small pulley as shown in the diagram. If the mass m is moving downward with a velocity v when the string makes an angle of `45^(@)` with the horizontal, find the total K.E. of the two masses. AsSigmae that the mass M moves horizontally.

Explanation

Let y be the depth of the mass m below the top of the pulley. If the total length of the string is `l`, the length of string from M to the pulley is `(l-y)`
The distance of M from the edge of the table `=x=(l-y)cos 45^(@)`.
It is given that, `((dy)/(dt))=v`.
`:.` The velocity of `M=(dx)/(dt)=(d)/(dt)(l-y)cos 45^(@)=-((dy)/(dt))cos45^(@)=-(1)/(sqrt(2))v`.
(The minus sign tell us that x decreases as y increases)
`:.` The net `K.E.=1/2 mv^(2)+1/2 M((dx)/(dt))^(2)=1/2 mv^(2)+1/2M(-(v)/sqrt(2))^(2)=1/2(m+M/2)v^(2)`.