A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to?

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 = 2v/3
Writing momentum equation in Y direction
m(0) + 2m(v) = 3mv2 ⇒ v2 = 2v/3
The velocity of both blocks will be (2v/3)^i + (2v/3)^j
The loss in kinetic energy
1/2m(2v)² + 1/2(2m)(v)² – [1/2(3m)((2v/3)² + (2v/3)²)] = 3mv² - (3mv²/3) = 2mv²
Percentage loss = (2mv²/3mv²) × 100 = 56