A `6 kg` block `B` rests as shown on the upper surface of a `15 kg` wedge `A` . Neglecting friction , determine immadiately after the system is released from rest (a) the acceleration of `A` (b) the acceleration of `B` relative to `A` (take `g = 10 m//s^(2))`

Explanation

Block `B` will fall vertically downwards and `A` along the plane.
Writing the equations of motion .
For block `B`
`m_(B) g - N = m_(B) a_(B)`
or ` 60 - N = 6 a_(B)`
`(N +m_(A) g) sin 30^(@) = m_(A) a_(A)`
or `(N + 150) = 30 a_(A)` ...(ii)
Further `a_(B)a_(A) sin 30^(@)`
or `a_(A) = 2a_(B)` ...(iii)
Solving these three equations , we get
(a) `a_(A) = 6.36 m//s^(2)`
(b) `a_(BA) = a_(A) cos 30^(@) = 5.5m//s^(2)`