Bibek Thakur

When they fly apart symmetrically relative to the initial motion direction with the angle of divergenceθ=60,
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,...