Particle 1 experiences a perfectly elastic collision with a stationary particle 2. Determine their mass ratio if the particles fly apart symmetrically relative to the initial motion direction of particle 1 with the angle of divergence is 60 degree.

When they fly apart symmetrically relative to the initial motion direction with the angle of divergence  $\theta ={60}^{\circ }$,
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