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 $\mathrm{M} / \mathrm{m}$, for which the angles $\theta_{1}$ and $\theta_{2}$ will be equal, is :
Correct Option: , 3
given $\theta_{1}=\theta_{2}=\theta$
from momentum conservation
in $\mathrm{x}$-direction $\mathrm{MV}_{0}=\mathrm{MV}_{1} \cos \theta+\mathrm{mV}_{2} \cos \theta$
in $\mathrm{y}$-direction $0=M \mathrm{MV}_{1} \sin \theta-\mathrm{mV}_{2} \sin \theta$
Solving above equations
$\mathrm{V}_{2}=\frac{\mathrm{MV}}{\mathrm{m}}, \mathrm{V}_{0}=2 \mathrm{~V}_{1} \cos \theta$
From energy conservation
$\frac{1}{2} \mathrm{MV}_{0}^{2}=\frac{1}{2} \mathrm{MV}_{1}^{2}+\frac{1}{2} \mathrm{MV}_{2}^{2}$
Substituting value of $V_{2} \& V_{0}$, we will get
$\frac{\mathrm{M}}{\mathrm{m}}+1=4 \cos ^{2} \theta \leq 4$
$\frac{\mathrm{M}}{\mathrm{m}} \leq 3$
Option (3)