When they fly apart symmetrically relative to the initial motion direction with the angle of divergence ,
From the conservation of momentum, along the horizontal and vertical direction
and m1v1sin(θ/2)=m2v2sin(θ/2)
or, m1v1=m2v2 ...(2)
Now, from conservation of kinetic energy,
1/2 m1 v12+ 0 =1/2 m1 v12+1/2 m2 v22 ...(3)
From (1) and (2),
m1 u1=cos(θ/2)(m1v1+m1v1m2m2)=2m1v1cos(θ/2)
So, u1=2v1cos(θ/2)...(4)
From (2), (3) and (4)
4m1cos2(θ/2) v12=m1v12+m2m12v12/m22
or, 4cos2(θ/2)=1+m1/m2
or, m1/m2=4(cosθ/2)−1
and putting the value of θ, we get,
m1/m2=2