Let the heavier mass be $M$ and the lighter mass be $m$. Both move with initial speed $u$ in opposite directions.

Velocity of Center of Mass ($V_{cm}$): taking $M$’s direction as positive.

$$ V_{cm} = \frac{Mu – mu}{M + m} = u \left( \frac{M – m}{M + m} \right) $$

Velocity of M in CM frame ($u’_M$):

$$ u’_M = u – V_{cm} = u – u \left( \frac{M – m}{M + m} \right) = u \left( \frac{2m}{M + m} \right) $$

Since the collision is elastic, the final speed of $M$ in the CM frame is the same: $v’_M = u’_M$.

The relationship between the scattering angle in the Lab frame ($\alpha$) and the CM frame ($\beta$) is given by the vector addition $\vec{v}_{lab} = \vec{V}_{cm} + \vec{v}’_{cm}$.

$$ \tan \alpha = \frac{v’_M \sin \beta}{V_{cm} + v’_M \cos \beta} $$

Dividing numerator and denominator by $v’_M$:

$$ \tan \alpha = \frac{\sin \beta}{\gamma + \cos \beta} \quad \text{where } \gamma = \frac{V_{cm}}{v’_M} $$

Calculating $\gamma$:

$$ \gamma = \frac{u(M-m)/(M+m)}{u(2m)/(M+m)} = \frac{M – m}{2m} $$

Given $\alpha = 30^\circ$ and $\beta = 60^\circ$:

$$ \tan 30^\circ = \frac{\sin 60^\circ}{\gamma + \cos 60^\circ} $$

$$ \frac{1}{\sqrt{3}} = \frac{\sqrt{3}/2}{\gamma + 1/2} $$

$$ \gamma + 0.5 = \sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{3}{2} = 1.5 $$

$$ \gamma = 1.0 $$

Equating to the mass expression:

$$ \frac{M – m}{2m} = 1 $$

$$ M – m = 2m \implies M = 3m $$

$$ \frac{M}{m} = 3 $$