Two blocks `A` and `B` each of equal masses `m` are rleased from the top of a smooth fixed wedge as shown in the figure. Find the magnitude of the accelertion of the centre of mass of the two blocks.

Explanation

Block `A` slides with aceleration `g//2` and block `B` slides wth acceleration `sqrt(3)g//2`. Now the acceleration of centre of mass of the system of blocks `A` and `B` can be given both in `x` and in `y` directions as:
`a_(x)=(3xx(sqrt3g)/2cos60^(@)-2xxg/2cos30^(@))/5=sqrt(3g)/20`
and `a_(y)=(3xx(sqrt(3)g)/2sin60^(@)+2xxg/2sin30^(@))/5=(11g)/20`
Thus acceleration of the centre of mass of the system is given as
`a_(CM)=sqrt(a_(x)^(2)+a_(r)^(2)=sqrt((sqrt(3))/(20)6+((11g)/(20)^(2))=sqrt(31)/1g`