A block of mass 'm' is sliding down an inclined plane of mass '10m' (read on)?

Question

A block of mass 'm' is sliding down a wedge of mass '10m' with an acceleration a. Assuming all surfaces to be frictionless, find the acceleration of the wedge in terms of 'a' and 'm'.

Answer 1

F = m*g sin theta. This is the force that causes your block to move down your incline. F = ma

m*g*sin theta = ma; or a = (g*sin theta)

If theta = 90 degrees, then a= g; if theta = 0 then a = 0

Answer 2

the angle-of-track-of-block with horizontal is b°; the acceleration of the block along this angle is a=g*sin(b), hence sin(b) =a/g, cos(b) =√(1-(a/g)^2);

the horizontal component of this acceleration is q1=a*cos(b) (draw a pic!); F=m*q1 is mutual force of the block and the wedge in horizontal direction, that is F=10m*q2, hence

q2=m*q1/(10m) = 0.1*a*cos(b) = 0.1*a*√(1 -(a/g)^2); correct?