ACCORDING TO LAW OF CONSERVATION OF LINEAR MOMENTUM 
M1V1=mv+mv+3mv
v=30ms-1
30kg*0=30m+30m+3xm
assuming masses m,m and 3m and x=velocity of hevier fragment
o=60m+3xm
-60m/3m=x
-20=x
THEREFORE THE HEVIER FRAGMENT FLIES OFF WITH A SPEED OF 20ms-1 in the perpendicular direction.
plz approve.....!!
Thank you.
						

							
Ah! That is an easy one. Look, 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
CHEGCK
						

							
ration of masses = 1:1:3
m1 = k
m2 = k
m3 = 3k
m1 + m2 + m3 = 1kg
 k = 1/5
m1, m2,m3 are 1/5 , 1/5 , 3/5 kg respectively ...
initial momentam = 0
final momentam = m1v1 + m2v2 + m3v3
since there is no force acting on the system so linear  momentam remains conserved
 m1v1 + m2v2 + m3v3 = 0
after putting value of m1,m2,m3 we get
v3 = -(v1+v2)/3        ....................1
 
now it is given that velocity of particles of same masses are
 perpendicular  with magnitude 30m/s ...so if v1 is 30i then
v2 can be 30j ....                                                                        (i & j are unit vectors along +x & +ve y axis)
v3 = -(30i+30j)/3
v3 = -10i - 10j            ....................3
magnitude of v3 = root(102+102)
                       =10root2 m/s
approve if u like my ans
						

							
mass of the body =1Kg i.e, 1:1:3
(1m,1m,3m)
1m+1m+3m=1Kg
              5m=1Kg
                m=(1/5)Kg -------(1)
Total momentum before explosion,
P=M*V (V=0,AS IT IS IN REST)
Total momentum after explosion,
P1=m*v
    =1m*v (from eq (1 we get)
    =1*(1/5)*30
    =6m/s --------(2)
P2=m*v
    =1*(1/5)*30
    =6m/s --------(3)
P3=m*v
    =3m*v --------(4)
By law of conservation of mass,
total momentum before explotion=total momentum after explosion
i.e, 0=P1+P2+P3
 -P1-P2=P3
-(P1+P2)=P3
P3=-(P1+P2)
(as the questionthe angle between two peices with equal mass is 90 degree
so by using addition of vectors R= root of A^2 +B^2)
thus,
  P3=root of( P1^2+P2^2)
from (2) and (3) we get,
P3=root of (6^2+6^2)
    =root of (2*6^2)
    =6 root of(2)
from eq. (4) we get
   3m*v=P3
  3m*v=6 root of (2)
3*(1/5)*v=6 root of (2)
  v=6 root of (2) *(1/3)*5
   =10 root of (2)=10*1.414
   =14.14m/s
therefore, velocity of the heavier body is 14.14m/s
						

							Let V is the velocity of heavier fragmentm=m*=0.2kgM=0.6kgv=v*=30m/sFrom the principal of conservation of momentum MV=mvcos45*+m*v*cos45*Or 0.6V=0.2*30*1/v2+0.2*30*1/v2V=10v2=14.24m/s