A 6kg block B rests as shown on the upper surface of a 15kg wedge A. Neglecting friction, determine immediately after the system is released from rest ,find the acceleration of B relative to A. (Take, $g=10 \mathrm{m}/{\mathrm{s}}^2$ )

Block B will fall vertically downwards and A along the plane.

Writing the equations of motion.

For block B,

$$\begin{gathered} m_{B}g-N=m_B{a}_B \\ 60-N=6a_B \\ \text{ or } \left(N+m_{A}g\right)\sin 30°=m_A{a}_A \\ (N+150)=30a_A \end{gathered}$$

Further

$\text{ or }{a}_A=2a_B$

Solving these three equations, we get

$a_{BA}=a_A\cos 30°=5.5 \mathrm{m}/{\mathrm{s}}^2$