A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (u). The total kinetic energy generated due to explosion is?

Let the initial momentum be pi and the final momentum be pf.
Since the body is at rest initially, pi = 0.
Let the velocities of the three pieces be v1, v2, and v3, and their masses be m, m, and 2m respectively.
Two pieces of mass m move perpendicular to each other with equal speeds u. Let's assume one moves in the x-direction and the other in the y-direction.
Then v1 = ui and v2 = uj
By conservation of momentum:
pi = pf
0 = m(ui) + m(uj) + 2m v3
Solving for v3:
2m v3 = -m(ui) - m(uj)
v3 = -(u/2)i - (u/2)j
|v3| = √[(u/2)² + (u/2)²] = u/√2
The total kinetic energy generated is the sum of the kinetic energies of the three pieces:
KE = (1/2)m u² + (1/2)m u² + (1/2)(2m)(u/√2)²
KE = (1/2)mu² + (1/2)mu² + (1/2)(2m)(u²/2)
KE = mu² + (1/2)mu²
KE = (3/2)mu²
Replacing u with v (as given in the options):
KE = (3/2)mv²
However, there is a discrepancy between the calculated KE and the given options. Let's re-examine the momentum conservation equation. Let's use vector notation.
Initial momentum pi = 0
Final momentum pf = mv1 + mv2 + 2mv3 = 0
Let v1 = vi and v2 = vj
Then 0 = mvi + mvj + 2mv3
=> v3 = -v/2 i - v/2 j
|v3| = √((-v/2)² + (-v/2)²) = v/√2
KE = 1/2 m v² + 1/2 m v² + 1/2 (2m) (v/√2)²
KE = mv² + mv²/2 = 3mv²/2
This still doesn't match the given options. There must be an error in the question or options provided. The closest option is 2mv², but the correct answer based on the given information and momentum conservation is (3/2)mv².