In figure, the acceleration of A is a→A=15i^+15j^. Then the acceleration of B is (A remains in contact with B)

From constraint, the acceleration of both block and wedge should be same in a direction perpendicular to the inclined plane as shown in figure.aA⊥=aB⊥,aAX=15,aAY=15aAXcos⁡53∘−aAYcos⁡37∘=aBcos⁡53∘or aB=−5ms−1  or  a→B=−5i^