A block of mass M is connected with a particle of mass m by a light inextensible string as shown in fig. Assuming all contaction surfaces as smooth, find the acceleration of the wedge after releasing the system.

Explanation

From the constraint relationship shown in fig.
`\`
`y+4x=l`
`implies (dy)/(dt)+(4dx)/(dt)=0`
`implies (d^(2)y)/(dt)+(4d^(2)x)/(dt)=0`
`implies a_(2)=4a_(1)` (numerically)...(i)
Where `a_(1)` is the acceleration fo M and `a_(2)` is the acceleration of m, with respect to M.
Equation for M
`Sigma F_(x)=Ma_(1)implies4T-N'=Ma_(1)` ...(ii)

For `m, Sigma F_(y)=ma_(2)implies mg - T=ma_(2)` ...(iii)
`Sigma F_(x)=ma_(1)implies N'=ma_(1)` ...(iv)
(ii)+(iv) `implies a_(1)=(4T)/(M+m)` ...(v)

Putting `a_(2)=4a_(1)` from (i) and `T=((M+m)a_(1))/(4)`
From (v) in (iii) we obtain
`mg-((M+m)a_(1))/(4)=m(4a_(1))`

`implies a_(1)=(4mg)/(M+17m)`.