A body of mass \({M}\) moving at speed \({V}_{0}\) collides elastically with a mass \('m'\) at rest. After the collision, the two masses move at angles \(\theta_{1}\) and \(\theta_{2}\) with respect to the initial direction of motion of the body of mass \({M}\). The largest possible value of the ratio \({M} / {m}\), for which the angles \(\theta_{1}\) and \(\theta_{2}\) will be equal, is :

from momentum conservation

in \(x\)-direction \(M V_{0}=M V_{1} \cos \theta+m V_{2} \cos \theta\)

in \({y}\)-direction \(0={MV}_{1} \sin \theta-{m} {V}_{2} \sin \theta\)

Solving above equations

\({V}_{2}=\frac{{MV}_{1}}{{m}}, {V}_{0}=2 {V}_{1} \cos \theta\)

From energy conservation

\(\frac{1}{2} {MV}_{0}^{2}=\frac{1}{2} {MV}_{1}^{2}+\frac{1}{2} {MV}_{2}^{2}\)

Substituting value of \({V}_{2} \& {V}_{0}\), we will get

\(\frac{{M}}{{m}}+1=4 \cos ^{2} \theta \leq 4\)

\(\frac{{M}}{{m}} \leq 3\)