A block of mass m slides down on a wedge of mass M as shown in figure .Let `a_(1)` be the asseleration of the acceleration of and `a_(2)` the the acceleration od block 1`N_(1)` is the normal reaction between block and wedge and `N_(2)` the normal reaction between wedge and gound .frication is absent everwhere .Select the correct alternative(s)

Explanation

As the length of the srting is constant,
`L=sqrt(d^(2)+y^(2))+x`

Since, L is constant, differentiating with respect to time t, we get
`(dL)/(dt)=(1)/(2)(2y)/((d^(2)+y^(2))^((1)/(2)))((dy)/(dt))+(dx)/(dt)=0`
Since `(dy)/(dx)= v_(m)` and `(dx)/(dt)=v_(M)` and
`cos theta = (y)/(sqrt(d^(2)+y^(2)))` so `v_(M)=-v_(m) cos theta`
By differentiating, relation between `a_(m)` and `a_(M)` can be obtained, however, while doing so remember that `cos theta` is not constant, but it is variable. `a_(M)=-a_(m)cos theta`