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 \(\theta = 60^{0}\).
Solution 1:Lab Frame, use relative motion and momentum conservation
\(m_{1}\) velocity diagram before collision
\(m_{1}\) and \(m_{2}\) velocity diagrams after collision. The line of impact is n- line (normal)

The relative motion equation is
\(e = \frac{v'_{2} - (v'_{1})/2}{u cos30^{0}}\)
that gives, \(u cos30^{0} = v'_{2} - \frac{v'_{1}}{2}\)............(1)
The equation of momentum conservation along the line of motion of m1 is
\(m_{1}u = m_{1}v'_{1}cos30^{0}+m_{2}v'_{2}cos30^{0}\).....(2a)
The equation of momentum conservation along the line  perpendicular to the line of motion of m1 is
\(m_{1}v'_{1}sin30^{0} = m_{2}v'_{2}sin30^{0}\)
that gives \(m_{1}v'_{1} = m_{2}v'_{2}\).........(2b)
Using equation (2a) and (2b) together, we have
\(m_{1}u = 2m_{1}v'_{1}cos30^{0}\)
Divide equation (1) by \(v'_{1}\) get the required result.

Method 2 : Lab Frame : Using momentum conservation and energy conservation
Apart from the equations of momentum conservation write the equation equating kinetic energy before and after collision.
Method 3: Using CM Frame : Well I taught in the last class but if you missed it, tell me to solve again.
There are few more methods based on shortening calculations but who cares these days making calculations short.