as the collision is inelastic the masses will move together
Assuming the speed of block A and B becomes v1^i+v2^j
Writing momentum equation in x direction
m(2v)+2m(0)=3mv1⇒v1=2v3
Writing momentum equation in Y direction
m(0)+2m(v)=3mv2⇒v2=2v3
The velocity of both blocks will be 2v3^i+2v3^j
The loss in kinetic energy
12m(2v)2+122m(v)2−123m(√22v3)2
3mv2−43mv2=53mv2
percentage loss53mv23mv2×100=56 %