Body of mass 1 kg initially at rest explodes and breaks into three fragments of masses in the ratio 1:1: 3. The two pieces of equal masses fly of perpendicular to each other with a speed of 30m/s each. What What is the velocity of heavier fragment

The figure will make clear that movement of COM of the two equally weighed particles is along the dotted axiz (please consult the figure) which is at an angle 45° from both the particles m1 and m2. Now, we consider it as a single particle of mass 400gms (ATQ) [mass of m1+m2] moving with a velocity of [30 cos45°] along the dotted line. 

So, the third particle (m3) which is of mass thrice to the original m1 and m2, and 3/2 of the mass of their centre of mass, must move along the red line with a velocity that balances the net momentum of the rest two particles to zero. Now acc. to Law of Conserv. of Momentum, 

Let, m* and v* be the mass and velocity of centre of mass of m1 and m2 respectively.

m3.v3=m*.v*

.6 is the mass of heavier particle.

.4 is the mass of COM of the two lighter particles.

v3=Unknown

v*= 30cos45°

Solving the above eqn using the given statistics, we get

v3=10√2

which is our required answer

Regards

Arun